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[ "# youtube to mp3 of “Polynomial approximations of functions (part 5)”\n\nWelcome back. Well, in the last video we took the Maclaurin series for cosine of x, the Maclaurin series representation.\n\nThe following video will tell you about Polynomial approximations of functions (part 5). You can use youtube to mp3 tool to convert the video to mp3.\n\nNotice: download videos and mp3 must be licensed by the version owner and used only for study and research, and not for commercial use and dissemination.\n\n## Polynomial approximations of functions (part 5)\n\nSo I guess we might as well do the same for sine of x. And I’m sounding very nonchalant, but I have a reason behind why I’m doing this, and you’ll see in a few videos from now, when we come up with the grand conclusion. So anyway, let’s just set up f of x. And you might just want to do this yourself instead of watching me do it, because it should be pretty self-explanatory now that you saw cosine of x.\n\nAnd then you could check your work. You can pause right now, and then you can check to see that we got the same answer. You’ll probably be right and I probably made a careless mistake. So f of x is equal to sine of x.\n\nFirst of all, let’s just figure out all of the derivatives of sine of x. We can already guess that it probably cycles similar to cosine of x, with slight variation. So what’s the derivative of sine of x? Well, that’s just cosine of x. What’s the second derivative? I’m just going to stop putting parentheses around these numbers.\n\nThe second derivative, well, that’s just derivative of that. It’s minus sine of x. The third derivative of x. Well, I guess I’ll put the parentheses so you don’t think it’s f to the third times x.\n\nThe third derivative, well, that’s just going to be the minus cosine, rightu0008? Because the derivative of sine is cosine, but then we have that minus sine there. And the fourth derivative. The derivative of cosine is minus sine, but we have a minus there, so we get back to sine. And then the cycle continues.\n\nSo the fifth derivative is just going to be cosine of x, or just the derivative of the sine of x. And then the cycle continues, right? All right. So we know the derivatives, and we can just keep going. So let’s evaluate the derivatives of our function of sine of x at x equals 0.\n\nf of 0. Well, that’s sine of zero. What’s sine of 0? Well, sine of 0 is 0.\n\nf prime of 0 is equal to cosine of 0. That’s equal to 1. The second derivative at 0, that’s minus sine of 0. Well, sine of 0 is still 0.\n\nSo that’s 0. You can see it’s a very similar pattern to what we saw in the Maclaurin series for cosine of x. And then the third derivative evaluated at 0.\n\nThat’s cosine of 0 is 1. But we have a minus sine, so that’s minus 1. This you already know. The fourth derivative is 0.\n\nSine of 0 is 0. And then it starts to cycle again. The fifth derivative, 0 equal to 1. So we start with 0, then positive 1, then 0, then minus 1, then 0, then positive 1.\n\nEvery other number is a 0. Every other, I guess you could say, coefficient in the Maclaurin series is a 0. The coefficient when you don’t include the factorial term. And then the ones in between oscillate between positive 1 and negative 1.\n\nSo what would be the Maclaurin series for sine of x? The Maclaurin series representation? So we could say that sine of x — And remember, I haven’t proven to you that the Maclaurin series representation of sine of x or cosine of x or e to the x really is equal to those functions over the entire domain. I might do that later. Frankly, I’ve been thinking about the proof myself. It hasn’t been completely intuitive how to do that proof.\n\nAlthough if you test them out, it does seem to make a lot of sense. But you shouldn’t just take my word for it. I’m going to look up the proof and I will prove it to you, eventually. But for now, you just have to take it as a bit of a leap of faith that the Maclaurin series representations just don’t approximate those functions around 0, that when you take the infinite series, it actually equals the function.\n\nSo sine of x. The Maclaurin series representation is going to be equal to — well, f of 0, that’s 0 plus f of — so the first derivative, it’s going to be 1 times x to the 1 over 1 factorial, which is just 1. This is just 1. And then we have — this is just plus 0 — and then we have minus — and now this is the third derivative — so, minus 1 times x to the third over 3 factorial.\n\nAnd then you have a 0. Then we have a plus 1. And this is now the fifth derivative, so x to the fifth over five factorial.\n\nAnd we’ll just keep going, but I think you see the pattern as we write and rewrite it. Sine of x is equal to 0, so we get x to the first, so that’s just x minus x to the third over 3 factorial plus x to the fifth over 5 factorial. And then you can imagine the pattern.\n\nWe’d go minus — we’re just taking the odd numbers — x to the seventh over 7 factorial plus x to the ninth over 9 factorial minus x to the eleventh over 11 factorial, and we’ll just keep going. And so we’ll just keep oscillating in sine — that’s a bit of a pun — and we use all of the odd exponents. So if I were to write that in sigma notation, and sigma notation often is the hard part. Well once again, the first term, when the term is 0 — this is the first term, right — we get a positive sine.\n\nBecause we’re going to oscillate in sine, we’re probably going to take negative 1 to some power. It’ll be negative 1 to the n plus 1. So let’s see if that works.\n\nIf this is the first term, this will be a — no, no, no, that won’t work. It’ll be to the 2n plus 1. Actually, I think I should have done that in the previous video, too. I think it should have been negative 1 to the 2n, not negative 1 to the n.\n\nI’m sorry for that mistake. So it’s negative 1 to the 2n plus 1 times x to the 2n. Oh, no, no, sorry, I was right in the previous video.\n\nSee, I’m confusing myself, because I don’t count the 0 terms. It would probably help me to write the sigma down first. So this is equal to — as you can see I do all of this in real time — infinity from n is equal to 0.\n\nAnd so, the first term is positive, so it’ll be negative 1 to the n, right? Because negative 1 to the 0 power is 1, right? So that’s positive and then the second term is negative, then positive, negative, right? And so, the zeroth term is x, so it has to be x to the — let me see — 2n plus 1. Does that work? Right, because the first term would then be 3. Right. x to the 2n plus 1 over 2n plus 1 factorial.\n\nIt’s almost easier when you just write it out like that. Well, that’s pretty interesting. But what’s even more interesting is if you see the similarity between the Maclaurin series representation for sine of x, and then the representation we figured out for cosine of x in the previous video.\n\nWe figured out that cosine of x is equal to 1 minus x squared over 2 factorial plus x to the fourth over 4 factorial minus x to the sixth over six factorial plus x to the eighth over 8 factorial. So they’re almost the opposite, right? They almost complement each other. The cosine, these are all of the even exponents, right? And even factorials. And sine is all of the odd exponents, because this is x to the 0, right? So that’s why you get 1 here.\n\nAnd in sine, it’s all of the odd exponents and all of the odd factorials in the denominator. So that by itself, I think, is pretty neat. What is especially neat, just another fodder for thought, is that we know from trigonometry that sine is just a shifted cosine function or that cosine is just a shifted sine function. But what’s neat is by shifting it by pi over 2 — which is all they are, right, if you were to graph it, they’re just shifted 90 degrees to the left or the right of each other — you can actually represent them differently by essentially picking the odd or even terms of this factorial polynomial series, whatever you want to call it.\n\nBut anyway. Doesn’t matter if you didn’t understand what I said at the end, as long as you appreciate how cool this is. I will see you in the next video.\n\n## youtube to mp3 for video\n\nNow, if you like the video, you can use our best  youtube downloader mp4  to download the video from  youtube or convert it to mp3. If using an android mobile, you can follow the tutorial of youtube video downloader for Android to get it." ]

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[ "# Present Value of $11 in 36 Years When you have a single payment that will be made to you, in this case$11, and you know that it will be paid in a certain number of years, in this case 36 years, you can use the present value formula to calculate what that $11 is worth today. Below is the present value formula we'll use to calculate the present value of$11 in 36 years.\n\n$$Present\\: Value = \\dfrac{FV}{(1 + r)^{n}}$$\n\nWe already have two of the three required variables to calculate this:\n\n• Future Value (FV): This is the $11 • n: This is the number of periods, which is 36 years So what we need to know now is r, which is the discount rate (or rate of return) to apply. It's worth noting that there is no correct discount rate to use here. It's a very personal number than can vary depending on the risk of your investments. For example, if you invest in the market and you earn on average 8% per year, you can use that number for the discount rate. You can also use a lower discount rate, based on the US Treasury ten year rate, or some average of the two. The table below shows the present value (PV) of$11 paid in 36 years for interest rates from 2% to 30%.\n\nAs you will see, the present value of $11 paid in 36 years can range from$0.00 to $5.39. Discount Rate Future Value Present Value 2%$11 $5.39 3%$11 $3.80 4%$11 $2.68 5%$11 $1.90 6%$11 $1.35 7%$11 $0.96 8%$11 $0.69 9%$11 $0.49 10%$11 $0.36 11%$11 $0.26 12%$11 $0.19 13%$11 $0.14 14%$11 $0.10 15%$11 $0.07 16%$11 $0.05 17%$11 $0.04 18%$11 $0.03 19%$11 $0.02 20%$11 $0.02 21%$11 $0.01 22%$11 $0.01 23%$11 $0.01 24%$11 $0.00 25%$11 $0.00 26%$11 $0.00 27%$11 $0.00 28%$11 $0.00 29%$11 $0.00 30%$11 $0.00 As mentioned above, the discount rate is highly subjective and will have a big impact on the actual present value of$11. A 2% discount rate gives a present value of $5.39 while a 30% discount rate would mean a$0.00 present value.\n\nThe rate you choose should be somewhat equivalent to the expected rate of return you'd get if you invested \\$11 over the next 36 years. Since this is hard to calculate, especially over longer periods of time, it is often useful to look at a range of present values (from 5% discount rate to 10% discount rate, for example) when making decisions.\n\nHopefully this article has helped you to understand how to make present value calculations yourself. You can also use our quick present value calculator for specific numbers." ]

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[ "## GUPTA MECHANICAL\n\nIN THIS WEBSITE I CAN TELL ALL ABOUT TECH. TIPS AND TRICKS APP REVIEWS AND UNBOXINGS ALSO TECH. NEWS .............\n\n## [Solution] Sherlock and Parentheses Kick Start Solution\n\nSherlock and Watson have recently enrolled in a computer programming course. Today, the tutor taught them about the balanced parentheses problem. A string $S$ consisting only of characters ( and/or ) is balanced if:\n\n• It is an empty string, or:\n• It has the form ($S$), where $S$ is a balanced string, or:\n• It has the form ${S}_{1}{S}_{2}$, where ${S}_{1}$ is a balanced string and ${S}_{2}$ is a balanced string.\n\nSherlock coded up the solution very quickly and started bragging about how good he is, so Watson gave him a problem to test his knowledge. He asked Sherlock to generate a string $S$ of L + R characters, in which there are a total of $L$ left parentheses ( and a total of $R$ right parentheses ). Moreover, the string must have as many different balanced non-empty substrings as possible. (Two substrings are considered different as long as they start and end at different indexes of the string, even if their content happens to be the same). Note that $S$ itself does not have to be balanced.\nSherlock is sure that once he knows the maximum possible number of balanced non-empty\n\nSolution Click Below:-  👉\n👇👇👇👇👇\n\nsubstrings, he will be able to solve the problem. Can you help him find that maximum number?\n\n### Input\n\nThe first line of the input gives the number of test cases, $T$$T$ test cases follow.\nEach test case consists of one line with two integers: $L$ and $R$.\n\n### Output\n\nFor each test case, output one line containing Case #x$x$: y$y$, where $x$ is the test case number (starting from $1$) and $y$ is the answer, as described above.\n\n### Limits\n\nTime limit: 20 seconds.\nMemory limit: 1 GB.\n$1\\le \\mathbf{T}\\le 100$.\n\n#### Test Set 1\n\n$0\\le \\mathbf{L}\\le 20$.\n$0\\le \\mathbf{R}\\le 20$.\n$1\\le \\mathbf{L}+\\mathbf{R}\\le 20$.\n\n#### Test Set 2\n\n$0\\le \\mathbf{L}\\le {10}^{5}$.\n$0\\le \\mathbf{R}\\le {10}^{5}$.\n$1\\le \\mathbf{L}+\\mathbf{R}\\le {10}^{5}$." ]

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[ "# Predictive Modeling of Leaf Area Removed (LAR) in Wild Mustard in R\n\nThis comprehensive homework with solution uses R functionalities to predict the leaf area removed (LAR) by caterpillars in a species of wild mustard. We explore a range of variables, from leaf glucosinolate and isothiocyanate levels to leaf thickness and trichome count, seeking to build the most accurate predictive model. Follow our journey as we step through data preparation, model refinement, and statistical analysis to identify the key predictors of LAR in this fascinating plant species.\n\n## Problem Description:\n\nIn this R homework, we are tasked with analyzing data on a specific species of wild mustard. The data includes several variables related to the plant, such as leaf glucosinolate levels, isothiocyanate levels, leaf thickness, the number of trichomes, and the average leaf area removed by herbivores (LAR).\n\nThe goal of the homework is to determine the best predictive model for LAR based on the available data. To do this, we will follow a step-by-step approach involving data analysis and regression modeling.\n\nSolution\n\nData.\n\nWe are given the data on a species of wild mustard with the following variables;\n\n• Leaf glucosinolate (ug/mg dry weight) are chemicals that have been implicated in herbivore resistance\n• Isothiocyanates (ug/mg dry weight) are chemicals that have been implicated in herbivore resistance.\n• Leaf thickness (mm) is the average thickness of leaves based on the average of 10 randomly selected leaves.\n• Trichomes (average number per square cm) are small leaf hairs.\n• LAR is a measure of leaf area removed by an herbivore.\n\nFirst we read data into R workspace.\n\n## combining data into one data.frame\n\nmustard.df = data.frame(leaf.glucosinolate = leaf.glucosinolate,\n\nleaf.thickness = leaf.thickness,\n\nisothiocyanates = isothiocyanates,\n\ntrichomes = trichomes,LAR = LAR)\n\n##structure of data\n\nstr(mustard.df)\n\n## 'data.frame': 100 obs. of 5 variables:\n\n## $leaf.glucosinolate: num 129 156 110 105 117 ... ##$ leaf.thickness : num 1.1 0.4 0.8 0.5 0.6 1.2 0.9 0.5 1 0.7 ...\n\n## $isothiocyanates : num 214 169 158 136 150 ... ##$ trichomes : num 22.8 37.8 37.1 29.6 30.9 37.4 32.2 36.1 34.5 43.2 ...\n\n## $LAR : num 31.5 18.5 20.9 17.7 18.7 ... A. Given the data above of average leaf area removed (mm^2) by caterpillars (LAR), what is the best predictive model of leaf area removed based on the available data? First we fit the model with all predictors: LAR_i=β_0+β_1 Leaf.glucosinolate_i+β_2 Leaf.thickness_i+β_3 Isothiocyanates_i+β_4 Trichomes_i+ε_i # A. Given the data above of average leaf area removed ($mm^2$) by caterpillars (LAR), # what is the best predictive model of leaf area removed based on the available data? ## fitting model with all variables lm_all = lm(LAR~., data = mustard.df) ## summary from this model summary(lm_all) ## ## Call: ## lm(formula = LAR ~ ., data = mustard.df) ## ## Residuals: ## Min 1Q Median 3Q Max ## -9.2837 -2.7695 -0.3424 2.5524 15.1614 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) -1.88923 3.87901 -0.487 0.627 ## leaf.glucosinolate 0.01985 0.02529 0.785 0.435 ## leaf.thickness 3.56650 1.68620 2.115 0.037 * ## isothiocyanates 0.10810 0.01134 9.535 1.63e-15 *** ## trichomes 0.02494 0.05835 0.428 0.670 ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 4.5 on 95 degrees of freedom ## Multiple R-squared: 0.6217, Adjusted R-squared: 0.6057 ## F-statistic: 39.03 on 4 and 95 DF, p-value: < 2.2e-16 The summary shows that some variables are not significant. So the full model cannot be the best. We perform stepwise backward regression, where each step we remove one variable and fit the new model. We compare then two models using F test, the variable is excluded if the result of the F test confirms it. The procedure continues until no variable could be excluded using the Anova test. We are call the built-in stepwise regression in R. ## stepwise regression lm_best= step(lm_all, trace = -1) ## Start: AIC=305.71 ## LAR ~ leaf.glucosinolate + leaf.thickness + isothiocyanates + ## trichomes ## ## Df Sum of Sq RSS AIC ## - trichomes 1 3.70 1927.8 303.90 ## - leaf.glucosinolate 1 12.47 1936.6 304.35 ## 1924.1 305.71 ## - leaf.thickness 1 90.61 2014.7 308.31 ## - isothiocyanates 1 1841.41 3765.5 370.85 ## ## Step: AIC=303.9 ## LAR ~ leaf.glucosinolate + leaf.thickness + isothiocyanates ## ## Df Sum of Sq RSS AIC ## - leaf.glucosinolate 1 14.79 1942.6 302.66 ## 1927.8 303.90 ## - leaf.thickness 1 91.93 2019.8 306.56 ## - isothiocyanates 1 1845.92 3773.8 369.07 ## ## Step: AIC=302.66 ## LAR ~ leaf.thickness + isothiocyanates ## ## Df Sum of Sq RSS AIC ## 1942.6 302.66 ## - leaf.thickness 1 81.92 2024.5 304.79 ## - isothiocyanates 1 1851.00 3793.6 367.59 ## summary from this model summary(lm_best) ## ## Call: ## lm(formula = LAR ~ leaf.thickness + isothiocyanates, data = mustard.df) ## ## Residuals: ## Min 1Q Median 3Q Max ## -10.0300 -2.5728 -0.4106 2.4101 15.1797 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 1.41168 1.74043 0.811 0.4193 ## leaf.thickness 3.33593 1.64941 2.022 0.0459 * ## isothiocyanates 0.10834 0.01127 9.614 9.15e-16 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 4.475 on 97 degrees of freedom ## Multiple R-squared: 0.618, Adjusted R-squared: 0.6102 ## F-statistic: 78.47 on 2 and 97 DF, p-value: < 2.2e-16 So, the step wise regression chooses the model with the variables “Leaf.thickness” and “Isothiocyanates”, both are significant. B. Analyze your model using analysis of variance. # B. Analyze your model using analysis of variance. ## anova of the best model anova(lm_best) ## Analysis of Variance Table ## ## Response: LAR ## Df Sum Sq Mean Sq F value Pr(>F) ## leaf.thickness 1 1292.2 1292.22 64.524 2.287e-12 *** ## isothiocyanates 1 1851.0 1851.00 92.425 9.152e-16 *** ## Residuals 97 1942.6 20.03 ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 F test based on ANOVA for the both selected predictors is highly significant. We also obtain coefficient of determination, R^2, for the whole model: ## coefficient of determination: summary(lm_best)$adj.r.squared\n\n## 0.610157\n\nWhich implies that over 61% of variation in LAR could be explained by the model, which is relatively good indicator.\n\nC. Based on your model, what is the most important predictor for leaf area removed (and on what do you base this)?\n\nWe look at Pearson correlation coefficient for the two chosen predictors and the response variable “LAR”:\n\n# C. Based on your model, what is the most important predictor\n\n# for leaf area removed (and on what do you base this)?\n\n# correlation\n\nround(cor(mustard.df[,c(\"LAR\",\"leaf.thickness\",\"isothiocyanates\")]),3)\n\n## LAR leaf.thickness isothiocyanates\n\n## LAR 1.000 0.504 0.776\n\n## leaf.thickness 0.504 1.000 0.509\n\n## isothiocyanates 0.776 0.509 1.000\n\nWe note that variable “isothiocyanates” has higher correlation with the response variable “LAR”. Secondly, looking at the summary of the model (p value and test statistic), here again the variable “isothiocyanates” is highly significant.\n\nWe conclude that the variable “isothiocyanates” is the most important predictor.\n\nD. What would your model predict for leaf area removed (LAR) for a plant with 111.2 ug/mgdwglucosinolate, 149.2 ug/mgdw isothiocyanates, 0.6 mm leaf thickness, and 39.1 trichomes per sq mm?\n\nSince the best model includes only 2 variables, we include use only these two variables for prediction.\n\n# D. What would your model predict for leaf area removed (LAR)\n\n# for a plant with 111.2 ug/mgdwglucosinolate, 149.2 ug/mgdw isothiocyanates,\n\n# 0.6 mm leaf thickness, and 39.1 trichomes per sq mm?\n\nnewdata = data.frame(isothiocyanates = 149.2, leaf.thickness = 0.6)\n\npred.lm = predict(lm_best, newdata = newdata)\n\ncat(\"\\n Predicted LAR for the given plant is \", pred.lm)\n\n##\n\n## Predicted LAR for the given plant is 19.57753" ]

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[ "Solutions by everydaycalculation.com\n\n## Subtract 6/9 from 1/7\n\n1/7 - 6/9 is -11/21.\n\n#### Steps for subtracting fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 7 and 9 is 63\n\nNext, find the equivalent fraction of both fractional numbers with denominator 63\n2. For the 1st fraction, since 7 × 9 = 63,\n1/7 = 1 × 9/7 × 9 = 9/63\n3. Likewise, for the 2nd fraction, since 9 × 7 = 63,\n6/9 = 6 × 7/9 × 7 = 42/63\n4. Subtract the two like fractions:\n9/63 - 42/63 = 9 - 42/63 = -33/63\n5. After reducing the fraction, the answer is -11/21\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

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[ "# Numerical Missing Words Quiz\n\nGive yourself 15 minutes, and take this test. You'll probably know if your answers are correct. This test does not measure intelligence, your fluency with words, creativity or mathematical ability. It will, however, give you some gauge of your mental flexibility. In the three years since the test was developed, few people have been found who could solve more than half of the 26 questions on the first try.\n\nMany, however, reported getting answers long after the testing had been set aside, particularly at unexpected moments when their minds were relaxed. Some reported solving all the questions over a period of several days.\n\nTake the test as your personal challenge. 16 correct answers out of the 26 = Genius.\n\nInstructions: Each equation below contains the initials of words that will make it correct. Furnish the missing words.\n\nFor example: 60 = M in an H. - would be 60 = Minutes in an Hour\n\nGOOD LUCK AND HAVE FUN. REMEMBER: ONLY 15 MINUTES!\n\n• 26 = L of the A\n• 7 = W of the A W\n• 1001 = A N\n• 12 = S of the Z\n• 54 = C in a D (with the J)\n• 9 = P in the S S\n• 88 = P K\n• 13 = S on the A F\n• 18 = H on a G C\n• 32 = D F at which W F\n• 8 = S on a S S\n• 200 = D for P G in M\n• 3 = B M (S H T R)\n• 90 = D in a R A.\n• 4 = Q in a G\n• 24 = H in a D\n• 1 = W on a U\n• 5 = D in a Z C\n• 57 = H V\n• 11 = P on a F T\n• 1000 = W that a P is W\n• 29 = D in F in a L Y\n• 64 = S on a C\n• 40 = D and N of G F\n• 80 = D to G A the W\n• 2 = No. it T to T\n\nUpdated at 15:47 EST on Tue Apr 11, 2006" ]

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[ "The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.", null, "Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A133301 a(n) is the n-th pentagonal number which is the sum of two consecutive pentagonal numbers. 9\n 1, 1926, 850137, 2564464982, 1132138928657, 3415133918621062, 1507685261236261801, 4547981651299964079126, 2007805569980855008712097, 6056610836775865229750164742, 2673822786819976661810784866297, 8065673443881586606920210924732502 (list; graph; refs; listen; history; text; internal format)\n OFFSET 1,2 COMMENTS We solve the equation P(p) = P(r) + P(r+1) with unknowns p and r, equivalent to (6*p-1)^2 = 2*(6*r+2)^2 + 17. The Diophantine equation X^2 = 2*Y^2 + 17 whose solutions give p and r are obtained by (x(n), y(n)) such that: x(1)=5, x(2)=215, x(3)=4517, x(4)=248087 and the same recurrence relation on the odd and even indices x(n+2) = 1154*x(n+1) - x(n); y(1)=2, y(2)=152, y(3)=3194, y(4)=175424 and the same recurrence relation on the odd and even indices y(n+2) = 1154*y(n+1) - y(n). The solutions (p,r) are given by the (u(n),v(n)) such that u(1)=1, u(2)=36, u(3)=753, u(4)=41348 and the same recurrence relation on the odd and even indices u(n+2) = 1154*u(n+1) - u(n) - 192 or u(n+1) = 577*u(n) - 96 + 68*sqrt(72*u(n)^2 - 24*u(n) - 32); v(1)=0, v(2)=25, v(3)=532, v(4)=29237 and the same recurrence relation on the odd and even indices v(n+2) = 1154*v(n+1) - v(n) + 384 or v(n+1) = 577*v(n) + 192 + 68*sqrt(72*u(n)^2 + 48*u(n) + 15). LINKS Colin Barker, Table of n, a(n) for n = 1..300 Index entries for linear recurrences with constant coefficients, signature (1,1331714,-1331714,-1,1). FORMULA For odd and even indices respectively: a(n+2) = 1331714*a(n+1) - a(n) - 416160; on the odd and the even indices respectively we also have a(n+1) = 665857*a(n) - 208080 + 19618*sqrt(1152*a(n)^2 - 720*a(n) - 32). The g.f., h(z), is such that h(z) = a(1)*z + a(2)*z^2 + ... and is given by h(z) = z*(1 + 1925*z - 483503*z^2 + 65395*z^3 + 22*z^4)/((1-z)*(1 - 1331714*z^2 + z^4)). EXAMPLE With P(m) = m*(3*m-1)/2, a(1)=1 because a(1) = P(1) = P(0) + P(1); a(2)=1926 because P(36) = 1926 = P(25) + P(26) = 925 + 1001; a(3)=850137 because P(753) = 850137 = P(532) + P(533) = 424270 + 425867 ... MAPLE a:=proc(m) if type (sqrt(18*m^2-6*m-8)/6-1/3), integer=true then m*(3*m-1)/2 else fi end : seq(a(m), m=1..100000)od; # Emeric Deutsch MATHEMATICA # (3 # - 1)/2 &@ Select[Range[10^6], IntegerQ[Sqrt[18 #^2 - 6 # - 8]/6 - 1/3] &] (* or *) Rest@ CoefficientList[Series[x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/( (1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2)), {x, 0, 12}], x] (* Michael De Vlieger, Jul 14 2016 *) PROG (PARI) Vec(x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/((1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2)) + O(x^20)) \\\\ Colin Barker, Oct 20 2014 (MAGMA) R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/((1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2)) )); // G. C. Greubel, Mar 16 2019 (Sage) a=(x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/((1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2))).series(x, 20).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Mar 16 2019 CROSSREFS Cf. A046173, A144796 (x(n)). Sequence in context: A326018 A202051 A283949 * A258841 A099482 A253337 Adjacent sequences:  A133298 A133299 A133300 * A133302 A133303 A133304 KEYWORD nonn,easy AUTHOR Richard Choulet, Dec 20 2007 EXTENSIONS Fixed typo in g.f. in formula, and more terms from Colin Barker, Oct 20 2014 STATUS approved\n\nLookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam\nContribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent\nThe OEIS Community | Maintained by The OEIS Foundation Inc.\n\nLast modified May 25 02:01 EDT 2020. Contains 334581 sequences. (Running on oeis4.)" ]

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[ "# chemistry\n\nHydrogen gas is produced when zinc reacts with hydrochloric acid. if the actual yeild of this reaction is 85%, how many grams of zinc are needed to produce 112L of H2 at STP?\n\n1. 👍\n2. 👎\n3. 👁\n1. The reaction is:\nZn + 2HCl --> ZnCl2 + H2\n1 mole of any gas at STP is 22.4 L\n112L of H2 / 22.4 L/mol = 5 moles H2\n(5 mol H2)(1 mol Zn / 1 mol H2) = 5 moles of Zn\nThe above relationship assumes that the reaction is 100% efficient. Since the reaction is only 85% efficient, MORE than 5 moles of Zn must be used. I will let you figure that part out on your own. [The multiplier you use has to do with the relationship of 100% to 85%]\n\n1. 👍\n2. 👎\n2. i don't get what to do next, can you please show me??\n\n1. 👍\n2. 👎\n3. multiply the 5 moles of zinc by the factor:\n(1.00/0.85)\n\n1. 👍\n2. 👎\n4. then what do i do??\n\n1. 👍\n2. 👎\n5. 5 mols Zn x (100/85) = ?? mols Zn\nYou want grams, Convert mols Zn to grams by grams = mols x molar mass.\n\n1. 👍\n2. 👎\n6. 380\n\n1. 👍\n2. 👎\n\n## Similar Questions\n\n1. ### Chem\n\nA mixture of chromium and zinc weighing .362 g was reacted with an excess of hydrochloric acid. After all the metals in the mixture reacted 225 mL of dry hydrogen gas was collected at 27 C and 750 torr. Determine the mass percent\n\n2. ### chemistry\n\nIf 20 grams of zinc reacts with hydrochloric acid, how much zinc chloride is produced?\n\n3. ### Chemistry 101\n\nIron reacts with hydrochloric acid to produce iron (2) chloride and hydrogen gas.Fe (s)+ 2 HCL (aq) -FeCl2 (aq) + H2 (g).The hydrogen gas from the reaction 2.2g of iron with excess acid is collected in a 10L flask at 25 degrees\n\n4. ### AP Chemistry\n\nSmall quantities of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc. Zn(s)+2HCl(aq) -> ZnCl2(aq)+ H2(g) Typically, the hydrogen gas is bubbled through water for\n\n1. ### Chemistry - Stoichiometry\n\n16.2g of magnesium reacts exactly with 25.3g of fluorine to produce magnesium fluoride, the only product. How many grams of magnesium fluoride could be produced from the reaction of 10.5g of magnesium with excess fluorine. This is\n\n2. ### AP Chemistry\n\nWhen hydrochloric acid reacts with magnesium metal, hydrogen gas and aqueous magnesium chloride are produced. What volume of 5.0 M HCI is required to react completely with 3.00 g of magnesium?\n\n3. ### chemistry\n\nSuppose that 50.0 cm3 of 2.00 M hydrochloric acid are added to 4.00 grams of zinc metal. What it the limiting reagent? How many grams of zinc are actually consumed? What is the concentration of the hydrochloric acid solution after\n\n4. ### Chemistry\n\nZinc metal reacts with hydrochloric acid according to the following equation: Zn + 2HCl yields ZnCl2 + H2. How many grams of hydrogen gas can be obtained if 50.0 g of zinc react completely?\n\n1. ### Chemistry\n\nZn+ 2HCl -> ZnCL2+h2 what volume of hydrogen at stp is produced when 2.5 of zinc react with an excess of hydrochloric acid\n\n2. ### science\n\nHydrochloric acid is added to a beaker containing a piece if zinc. As a result,zinc chloride is formed and hydrogen gas is released. This is an example of ?\n\n3. ### SCIENCE/CHEM\n\nthe purity of zinc is to be determined by measuring the amount of hydrogen formed when a weighed sample of zinc reacts with an excess of HCL acid.the sample weighs 0.198 grams. what amount of hydrogen gas at STP will be obtained\n\n4. ### Chemistry\n\nHydrogen gas can be produced by the reaction of magnesium metal with hydrochloric acid as shown in the following chemical reaction. Identify the limiting reagent when 6.00 grams of HCl reacts with 5.00 grams of Mg." ]

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[ "# How to draw a triangle as a part of a pentagon using TikZ\n\nI have a regular polygon using TikZ like this:\n\n\\begin{tikzpicture}\n\\node[regular polygon, regular polygon sides=5, shape border rotate=-18, fill=blue!20, minimum size=5cm, thick, draw] (Funfeck) {};\n\\coordinate (P1) at (Funfeck.corner 1);\n\\coordinate (P2) at (Funfeck.corner 2);\n\\coordinate (P3) at (Funfeck.corner 3);\n\\coordinate (P4) at (Funfeck.corner 4);\n\\coordinate (P5) at (Funfeck.corner 5);\n...\n\\end{tikzpicture}\n\n\nIt looks like this:", null, "In another place, I want to have a triangle with the points P5, P1 and P2.\n\nIt should look like this:", null, ", so the part in the document in which these polygons are, should look like this:", null, "The Code would be something like this:\n\n\\draw (P5) -- (P1) -- (P2) -- (P5) -- cycle;\n\n\nBut, as this code is at an entirely different place, this code is in a different tikzpicture and I can't access the coordinates.\n\nNow, I have four solution proposals:\n\nFirst: (My favourite) I get these coordinates without drawing the pentagon. The advantage would be, that I needn't to draw the figure somewhere else if I have the same problem another time.\nSomething like this:\n\n\\coordinate (P1) at (\\node[regular polygon, regular polygon sides=5, shape border rotate=-18, minimum size=5cm].corner 1);\n\\coordinate (P2) at (\\node[regular polygon, regular polygon sides=5, shape border rotate=-18, minimum size=5cm].corner 1);\n\\coordinate (P5) at (\\node[regular polygon, regular polygon sides=5, shape border rotate=-18, minimum size=5cm].corner 1);\n\\draw (P5) -- (P1) -- (P2) -- (P5) -- cycle;\n\n\nSecond: Another possibility would be to get the coordinates out of the other tikzpicture.\nI mean something like this:\n\n\\coordinate (P1) at (tikzpicture.coordinate.P1);\n\\coordinate (P2) at (tikzpicture.coordinate.P2);\n\\coordinate (P5) at (tikzpicture.coordinate.P5);\n\\draw (P5) -- (P1) -- (P2) -- (P5) -- cycle;\n\n\nThird: Another solution, I have fought about was to draw the polygon, get the coordinates and then I remove the polygon. In fact, I can make the node without draw and without fill=blue!20, but then, there is still a whitespace.\nI mean something like this:\n\n\\node[regular polygon, regular polygon sides=5, shape border rotate=-18, minimum size=5cm, draw] (Funfeck) {};\n\\coordinate (P1) at (Funfeck.corner 1);\n\\coordinate (P2) at (Funfeck.corner 2);\n\\coordinate (P3) at (Funfeck.corner 3);\n\\coordinate (P4) at (Funfeck.corner 4);\n\\coordinate (P5) at (Funfeck.corner 5);\n\\remove {Funfeck}\n\\draw (P5) -- (P1) -- (P2) -- (P5) -- cycle;\n\n\nFourth: If nothing of the other methods is possible, the worst solution is to print the Coordinates and insert them3 manually. But I haven't even found a way to do this:\nMy Pseudo-Code would be:\n\n\\begin{tikzpicture}\n\\node[regular polygon, regular polygon sides=5, shape border rotate=-18, fill=blue!20, minimum size=5cm, thick, draw] (Funfeck) {};\n\\coordinate (P1) at (Funfeck.corner 1);\n\\coordinate (P2) at (Funfeck.corner 2);\n\\coordinate (P3) at (Funfeck.corner 3);\n\\print(\"P1, x: \"+P1.X+\" .P1, y: \"+P1.Y);\n\\print(\"P2, x: \"+P2.X+\" .P2, y: \"+P2.Y);\n\\print(\"P3, x: \"+P3.X+\" .P3, y: \"+P3.Y);\n...\n\\end{tikzpicture}\n\n\n...\n\n\\coordinate (P1) at (manually inserted, manually inserted);\n\\coordinate (P2) at (manually inserted, manually inserted);\n\\coordinate (P3) at (manually inserted, manually inserted);\n\n\nCan anybody give me a solution how to solve this problem?\nThank you very much :).\n\n• Welcome to TeX.SX! Please help us (and also you) and add a minimal working example (MWE), that illustrates your problem... compilable code, starting with \\documentclass and ending with \\end{document}. - Does access a coordinate in a nested \\tikzpicture help? – Bobyandbob Oct 30 '17 at 11:48\n• It is very unclear what you are asking. Can you write a document or draw a picture with the expected result or some approximation of it? It is especially unclear what you mean by referring to P1,2,3 from another picture. Do you want to overlay the triangle on the original picture? – Bordaigorl Oct 30 '17 at 11:52\n• @Bordaigorl I have added three pictures to show how the pentagon looks, the triangle should look, and the full document should look :) – Korne127 Oct 30 '17 at 11:59\n• Do you want to draw on the same picture or on a different picture that has the points \"in the same place\"? If the former see section 17.13.1 of the manual -- particular, remember picture and overlay. If the latter, just draw the same points -- for regular polygons just use polar coordinates. – Andrew Oct 30 '17 at 12:01\n\n## 1 Answer\n\nYou can use the overlay option for the node, then it will not be considered for the calculation of the bounding box.\n\nAlternatively, use polar coordinates to calculate the coordinates.\n\n\\documentclass[border=5mm]{standalone}\n\\usepackage{tikz}\n\\usetikzlibrary{shapes.geometric}\n\\begin{document}\n\\begin{tikzpicture}\n\\node[\nregular polygon,\nregular polygon sides=5,\nshape border rotate=-18,\nminimum size=5cm,\noverlay % overlay option means the node isn't included in the calculation of the bounding box\n] (P) {};\n\n\\draw (P.corner 5) -- (P.corner 1) -- (P.corner 2) -- cycle;\n\n% or do as you did, and make new coordinates\n%\\coordinate (P1) at (Funfeck.corner 1);\n%\\coordinate (P2) at (Funfeck.corner 2);\n%\\coordinate (P3) at (Funfeck.corner 3);\n%\\coordinate (P4) at (Funfeck.corner 4);\n%\\coordinate (P5) at (Funfeck.corner 5);\n%\\draw (P5) -- (P1) -- (P2) -- cycle;\n\\end{tikzpicture}\n\n% example using polar coordinates\n\\begin{tikzpicture}\n\\foreach \\N in {1,...,5}\n\\coordinate (P\\N) at ({360/5*(\\N-1)+90-18}:2.5cm);\n\n\\draw (P1) -- (P2) -- (P5) -- cycle;\n\\end{tikzpicture}\n\\end{document}\n\n\nThe second version can be generalized by defining a macro. With inspiration from Ryan Reich's answer to How to create a command with key values?, one possible implementation is shown in the code below.\n\nUsed alone, \\myregpolygon will define coordinates P1 through P5, for a regular five sided polygon with \"radius\" of 1cm, centered on the point (0,0), and the first point at an angle of zero. Each of those parameters can modified with key-value pairs in an optional argument to the macro:\n\n\\myregpolygon[%\nradius=3cm, % radius of circumscribed circle\norigin={(3,2)}, % center point\nstart angle=90, % angle of first coordinate\nname=Q, % prefix for coordinate names\nsides=4 % sides for the polygon\n]\n\n\nOne doesn't have to redefine all at once, of course.\n\n\\documentclass[border=5mm]{standalone}\n\\usepackage{tikz}\n\n\\pgfkeys{\nregpoly/.is family,\n/regpoly,\ndefault/.style={\nradius=1cm,\nstart angle=0,\norigin={(0,0)},\nsides={5},\nname={P}\n},\nradius/.estore in=\\RPradius,\nstart angle/.estore in=\\RPstartangle,\norigin/.estore in=\\RPorigin,\nsides/.estore in=\\RPsides,\nname/.estore in=\\RPname,\n}\n\n\\newcommand\\myregpolygon[]{\n\\pgfkeys{/regpoly,default, #1}\n\\foreach \\N in {1,...,\\RPsides}\n\\coordinate [shift=\\RPorigin] (\\RPname\\N) at ({360/\\RPsides*(\\N-1)+\\RPstartangle}:\\RPradius);\n}\n\\begin{document}\n\\begin{tikzpicture}\n\n\\myregpolygon[start angle=72,radius=2.5cm]\n\n\\draw (P1) -- (P2) -- (P5) -- cycle;\n\\end{tikzpicture}\n\\end{document}\n\n• @Korne127 I added a second option as well. – Torbjørn T. Oct 30 '17 at 12:04\n• I have seen it. Which solution would you prefer? – Korne127 Oct 30 '17 at 12:06\n• I think, I'll use the second solution as it's 'cleaner' and it doesn't contain a hidden node, but calculates the coordinates. – Korne127 Oct 30 '17 at 12:11\n• @TorbjørnT. I was preparing a similar solution with your second (but would give options on a newcommand named \\myregpolygon for radius, starting angle, number of corners etc). Since it is just an extension of yours, I will not upload... I just suggesting to edit and do it this way because it solves more problems and we can keep this question as a guide for future similar requests. Korne127 the second is better for me too... (+1 by the way) – koleygr Oct 30 '17 at 12:19\n• @koleygr There has probably been similar questions before as well, but anyway, I added one possible implementation. – Torbjørn T. Oct 30 '17 at 12:50" ]

[ null, "https://i.stack.imgur.com/sVuW5.png", null, "https://i.stack.imgur.com/wfr41.png", null, "https://i.stack.imgur.com/aDEu4.png", null ]

[ "\"\"\"Data loading components for neighbor sampling\"\"\"\nfrom .. import sampling, subgraph, distributed\nfrom .. import ndarray as nd\nfrom .. import backend as F\n\n[docs]class MultiLayerNeighborSampler(BlockSampler):\n\"\"\"Sampler that builds computational dependency of node representations via\nneighbor sampling for multilayer GNN.\n\nThis sampler will make every node gather messages from a fixed number of neighbors\nper edge type. The neighbors are picked uniformly.\n\nParameters\n----------\nfanouts : list[int] or list[dict[etype, int] or None]\nList of neighbors to sample per edge type for each GNN layer, starting from the\nfirst layer.\n\nIf the graph is homogeneous, only an integer is needed for each layer.\n\nIf None is provided for one layer, all neighbors will be included regardless of\nedge types.\n\nIf -1 is provided for one edge type on one layer, then all inbound edges\nof that edge type will be included.\nreplace : bool, default True\nWhether to sample with replacement\nreturn_eids : bool, default False\nWhether to return the edge IDs involved in message passing in the MFG.\nIf True, the edge IDs will be stored as an edge feature named dgl.EID.\n\nExamples\n--------\nTo train a 3-layer GNN for node classification on a set of nodes train_nid on\na homogeneous graph where each node takes messages from 5, 10, 15 neighbors for\nthe first, second, and third layer respectively (assuming the backend is PyTorch):\n\n... collator.dataset, collate_fn=collator.collate,\n... batch_size=1024, shuffle=True, drop_last=False, num_workers=4)\n... train_on(blocks)\n\nIf training on a heterogeneous graph and you want different number of neighbors for each\nedge type, one should instead provide a list of dicts. Each dict would specify the\nnumber of neighbors to pick per edge type.\n\n... {('user', 'follows', 'user'): 5,\n... ('user', 'plays', 'game'): 4,\n... ('game', 'played-by', 'user'): 3}] * 3)\n\nNotes\n-----\nFor the concept of MFGs, please refer to\n:ref:User Guide Section 6 <guide-minibatch> and\n:doc:Minibatch Training Tutorials <tutorials/large/L0_neighbor_sampling_overview>.\n\"\"\"\ndef __init__(self, fanouts, replace=False, return_eids=False):\nsuper().__init__(len(fanouts), return_eids)\n\nself.fanouts = fanouts\nself.replace = replace\n\n# used to cache computations and memory allocations\n# list[dgl.nd.NDArray]; each array stores the fan-outs of all edge types\nself.fanout_arrays = []\nself.prob_arrays = None\n\n[docs] def sample_frontier(self, block_id, g, seed_nodes):\nfanout = self.fanouts[block_id]\nif isinstance(g, distributed.DistGraph):\nif fanout is None:\n# TODO(zhengda) There is a bug in the distributed version of in_subgraph.\n# let's use sample_neighbors to replace in_subgraph for now.\nfrontier = distributed.sample_neighbors(g, seed_nodes, -1, replace=False)\nelse:\nfrontier = distributed.sample_neighbors(g, seed_nodes, fanout, replace=self.replace)\nelse:\nif fanout is None:\nfrontier = subgraph.in_subgraph(g, seed_nodes)\nelse:\nself._build_fanout(block_id, g)\nself._build_prob_arrays(g)\n\nfrontier = sampling.sample_neighbors(\ng, seed_nodes, self.fanout_arrays[block_id],\nreplace=self.replace, prob=self.prob_arrays)\nreturn frontier\n\ndef _build_prob_arrays(self, g):\n# build prob_arrays only once\nif self.prob_arrays is None:\nself.prob_arrays = [nd.array([], ctx=nd.cpu())] * len(g.etypes)\n\ndef _build_fanout(self, block_id, g):\nassert not self.fanouts is None, \\\n\"_build_fanout() should only be called when fanouts is not None\"\n# build fanout_arrays only once for each layer\nwhile block_id >= len(self.fanout_arrays):\nfor i in range(len(self.fanouts)):\nfanout = self.fanouts[i]\nif not isinstance(fanout, dict):\nfanout_array = [int(fanout)] * len(g.etypes)\nelse:\nif len(fanout) != len(g.etypes):\nraise DGLError('Fan-out must be specified for each edge type '\n'if a dict is provided.')\nfanout_array = [None] * len(g.etypes)\nfor etype, value in fanout.items():\nfanout_array[g.get_etype_id(etype)] = value\nself.fanout_arrays.append(\nF.to_dgl_nd(F.tensor(fanout_array, dtype=F.int64)))\n\n[docs]class MultiLayerFullNeighborSampler(MultiLayerNeighborSampler):\n\"\"\"Sampler that builds computational dependency of node representations by taking messages\nfrom all neighbors for multilayer GNN.\n\nThis sampler will make every node gather messages from every single neighbor per edge type.\n\nParameters\n----------\nn_layers : int\nThe number of GNN layers to sample.\nreturn_eids : bool, default False\nWhether to return the edge IDs involved in message passing in the MFG.\nIf True, the edge IDs will be stored as an edge feature named dgl.EID.\n\nExamples\n--------\nTo train a 3-layer GNN for node classification on a set of nodes train_nid on\na homogeneous graph where each node takes messages from all neighbors for the first,\nsecond, and third layer respectively (assuming the backend is PyTorch):\n\n... collator.dataset, collate_fn=collator.collate,\n... batch_size=1024, shuffle=True, drop_last=False, num_workers=4)\n:ref:User Guide Section 6 <guide-minibatch> and\n:doc:Minibatch Training Tutorials <tutorials/large/L0_neighbor_sampling_overview>." ]

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[ "# Quadrilaterals Posters and Interactive Notebook INB Set", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Subject\nResource Type\nFile Type\n\nPDF\n\n(17 MB|18 pages)\nStandards\nAlso included in:\n1. This is a bundle of all my Geometry and Measurement Poster and Interactive Notebook (INB) Sets. Purchasing this bundle will give you 20% off the original prices.You will get the following sets: Polygons, Triangles, Angle Relationships, Quadrilaterals, Circles, Volume and Surface Area, Types of Angl\n\\$31.25\n\\$25.00\nSave \\$6.25\n• Product Description\n• StandardsNEW\nThis poster and interactive notebook (INB) set covers quadrilaterals, squares, rectangles, rhombus, parallelograms, trapezoids, area of a square, area of a rectangle, area of a rhombus, areas of a parallelogram and area of a trapezoid. This set includes a 2 foldable interactive graphic organizers (INB) and 14 quadrilateral posters (8.5 X 11).\n\nYou will get a poster for the following: 2 quadrilaterals, square, rectangle, rhombus, parallelogram, trapezoid (2 different definitions), area of a square (2 different formulas), area of a rectangle, area of a rhombus, area of a parallelogram, and 2 area of a trapezoid (2 different styles of trapezoids).\n\nIf you like these posters, please check out the other sets I have!\n\n***Save \\$\\$\\$ when you buy this and my other geometry and measurement posters in my Geometry and Measurement Bundle.***\n\nThank you for looking,\nAmy Alvis\nUnderstand that attributes belonging to a category of two-dimensional figures also belong to all subcategories of that category. For example, all rectangles have four right angles and squares are rectangles, so all squares have four right angles.\nClassify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size. Recognize right triangles as a category, and identify right triangles.\nUnderstand that shapes in different categories (e.g., rhombuses, rectangles, and others) may share attributes (e.g., having four sides), and that the shared attributes can define a larger category (e.g., quadrilaterals). Recognize rhombuses, rectangles, and squares as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to any of these subcategories.\nRecognize and draw shapes having specified attributes, such as a given number of angles or a given number of equal faces. Identify triangles, quadrilaterals, pentagons, hexagons, and cubes.\nTotal Pages\n18 pages\nN/A\nTeaching Duration\nN/A\nReport this Resource to TpT\nReported resources will be reviewed by our team. Report this resource to let us know if this resource violates TpT’s content guidelines.", null, "" ]

[ null, "https://ecdn.teacherspayteachers.com/thumbitem/Quadrilaterals-posters-and-graphic-organizer-set-051332400-1381605225-1511537175/original-923010-1.jpg", null, "https://ecdn.teacherspayteachers.com/thumbitem/Quadrilaterals-posters-and-graphic-organizer-set-051332400-1381605225-1511537175/original-923010-1.jpg", null, "https://ecdn.teacherspayteachers.com/thumbitem/Quadrilaterals-posters-and-graphic-organizer-set-051332400-1381605225-1511537175/medium-923010-2.jpg", null, "https://ecdn.teacherspayteachers.com/thumbitem/Quadrilaterals-posters-and-graphic-organizer-set-051332400-1381605225-1511537175/original-923010-2.jpg", null, "https://ecdn.teacherspayteachers.com/thumbitem/Quadrilaterals-posters-and-graphic-organizer-set-051332400-1381605225-1511537175/medium-923010-3.jpg", null, "https://ecdn.teacherspayteachers.com/thumbitem/Quadrilaterals-posters-and-graphic-organizer-set-051332400-1381605225-1511537175/original-923010-3.jpg", null, "https://ecdn.teacherspayteachers.com/thumbitem/Quadrilaterals-posters-and-graphic-organizer-set-051332400-1381605225-1511537175/medium-923010-4.jpg", null, "https://ecdn.teacherspayteachers.com/thumbitem/Quadrilaterals-posters-and-graphic-organizer-set-051332400-1381605225-1511537175/original-923010-4.jpg", null, "https://static1.teacherspayteachers.com/tpt-frontend/releases/production/current/d0b92082f9ec887495db40c424a92829.gif", null ]

[ "## Estimate the pressure exerted on a floor by (a) one pointed heel of area = 0.45 cm2, and (b) one wide heel of area 16 cm2.\n\nQuestion\n\nEstimate the pressure exerted on a floor by (a) one pointed heel of area = 0.45 cm2, and (b) one wide heel of area 16 cm2. The person wearing the shoes has a mass of 56 kg. Answer is both Pa and atm.   I have to solve both in pa and atm thank you.\n\nsolved 0\nGeneral Physics PRemastered 6 years 1 Answer 3932 views 0\n\n1. (a)\n\n(56 * 9.81) / (0.45 * 10^(-4))\n= 12208000 N / m^2.\n\n= 12208000 pa\n\n= 122.08 atm\n\n(b)\n\n(56 * 9.81) / (16 * 10^(-4))\n= 343350 N / m^2.\n\n= 343350 pa\n\n= 3.4335 atm" ]

[ null ]

[ "# [PATCH v6 10/10] ARM: OMAP2+: tusb6010: generic timing calculation\n\nMohammed, Afzal afzal at ti.com\nMon Sep 17 04:39:54 EDT 2012\n\n```Hi Tony,\n\nOn Fri, Sep 14, 2012 at 15:50:02, Mohammed, Afzal wrote:\n> * Mohammed, Afzal: Wednesday, September 12, 2012 3:20 PM\n\n> > But some of the tusb async values is less by one. I need\n> > to get it right.\n>\n> Reason has been identified. It was due to rounding error,\n> no changes are required in the expressions. Moving\n> completely to picoseconds resolves the issue.\n>\n> Can you please try with the attached patch ?\n\nCan you please try with patch attached in previous message\nof this thread and check whether it makes n800 gpmc\nperipherals work properly\n\nChanges as in above mentioned patch has been pasted below again.\n\nRegards\nAfzal\n\n---8<------------------------\n\ndiff --git a/arch/arm/mach-omap2/gpmc.c b/arch/arm/mach-omap2/gpmc.c\nindex d8e5b08..e9d57db 100644\n--- a/arch/arm/mach-omap2/gpmc.c\n+++ b/arch/arm/mach-omap2/gpmc.c\n@@ -289,11 +289,11 @@ static int set_gpmc_timing_reg(int cs, int reg, int st_bit, int end_bit,\nif (time == 0)\nticks = 0;\nelse\n-\t\tticks = gpmc_ns_to_ticks(time);\n+\t\tticks = gpmc_ps_to_ticks(time);\nnr_bits = end_bit - st_bit + 1;\nif (ticks >= 1 << nr_bits) {\n#ifdef DEBUG\n-\t\tprintk(KERN_INFO \"GPMC CS%d: %-10s* %3d ns, %3d ticks >= %d\\n\",\n+\t\tpr_info(\"GPMC CS%d: %-10s* %3d ps, %3d ticks >= %d\\n\",\ncs, name, time, ticks, 1 << nr_bits);\n#endif\nreturn -1;\n@@ -302,10 +302,9 @@ static int set_gpmc_timing_reg(int cs, int reg, int st_bit, int end_bit,\nmask = (1 << nr_bits) - 1;\n#ifdef DEBUG\n-\tprintk(KERN_INFO\n-\t\t\"GPMC CS%d: %-10s: %3d ticks, %3lu ns (was %3i ticks) %3d ns\\n\",\n-\t cs, name, ticks, gpmc_get_fclk_period() * ticks / 1000,\n-\t\t\t(l >> st_bit) & mask, time);\n+\tpr_info(\"GPMC CS%d: %-10s: %3d ticks, %3lu ps (was %3i ticks) %3d ps\\n\",\n+\t\tcs, name, ticks, gpmc_get_fclk_period() * ticks,\n+\t\t(l >> st_bit) & mask, time);\n#endif\nl |= ticks << st_bit;\n@@ -385,8 +384,8 @@ int gpmc_cs_set_timings(int cs, const struct gpmc_timings *t)\nif (l & (GPMC_CONFIG1_READTYPE_SYNC | GPMC_CONFIG1_WRITETYPE_SYNC)) {\n#ifdef DEBUG\n-\t\tprintk(KERN_INFO \"GPMC CS%d CLK period is %lu ns (div %d)\\n\",\n-\t\t\t\tcs, (div * gpmc_get_fclk_period()) / 1000, div);\n+\t\tpr_info(\"GPMC CS%d CLK period is %lu ps (div %d)\\n\",\n+\t\t\t\tcs, div * gpmc_get_fclk_period(), div);\n#endif\nl &= ~0x03;\nl |= (div - 1);\n@@ -922,46 +921,42 @@ static int gpmc_calc_sync_read_timings(struct gpmc_timings *gpmc_t,\n* indirectly necessitates requirement of t_avdp_r & t_avdp_w\n* instead of having a single t_avdp\n*/\n-\t\ttemp = max_t(u32, temp,\tgpmc_t->clk_activation * 1000 +\n-\t\t\t\t\t\t\tdev_t->t_avdh);\n-\t\ttemp = max_t(u32,\n-\t\t\t(gpmc_t->adv_on + gpmc_ticks_to_ns(1)) * 1000, temp);\n+\t\ttemp = max_t(u32, temp,\tgpmc_t->clk_activation + dev_t->t_avdh);\n+\t\ttemp = max_t(u32, gpmc_t->adv_on + gpmc_ticks_to_ps(1), temp);\n}\n-\tgpmc_t->adv_rd_off = gpmc_round_ps_to_ticks(temp) / 1000;\n\n/* oe_on */\ntemp = dev_t->t_oeasu; /* remove this ? */\nif (mux) {\n-\t\ttemp = max_t(u32, temp,\n-\t\t\tgpmc_t->clk_activation * 1000 + dev_t->t_ach);\n-\t\ttemp = max_t(u32, temp, (gpmc_t->adv_rd_off +\n-\t\t\t\tgpmc_ticks_to_ns(dev_t->cyc_aavdh_oe)) * 1000);\n+\t\ttemp = max_t(u32, temp,\tgpmc_t->clk_activation + dev_t->t_ach);\n+\t\ttemp = max_t(u32, temp, gpmc_t->adv_rd_off +\n+\t\t\t\tgpmc_ticks_to_ps(dev_t->cyc_aavdh_oe));\n}\n-\tgpmc_t->oe_on = gpmc_round_ps_to_ticks(temp) / 1000;\n+\tgpmc_t->oe_on = gpmc_round_ps_to_ticks(temp);\n\n/* access */\n/* any scope for improvement ?, by combining oe_on & clk_activation,\n* need to check whether access = clk_activation + round to sync clk ?\n*/\ntemp = max_t(u32, dev_t->t_iaa,\tdev_t->cyc_iaa * gpmc_t->sync_clk);\n-\ttemp += gpmc_t->clk_activation * 1000;\n+\ttemp += gpmc_t->clk_activation;\nif (dev_t->cyc_oe)\n-\t\ttemp = max_t(u32, temp, (gpmc_t->oe_on +\n-\t\t\t\tgpmc_ticks_to_ns(dev_t->cyc_oe)) * 1000);\n-\tgpmc_t->access = gpmc_round_ps_to_ticks(temp) / 1000;\n+\t\ttemp = max_t(u32, temp, gpmc_t->oe_on +\n+\t\t\t\tgpmc_ticks_to_ps(dev_t->cyc_oe));\n+\tgpmc_t->access = gpmc_round_ps_to_ticks(temp);\n\n-\tgpmc_t->oe_off = gpmc_t->access + gpmc_ticks_to_ns(1);\n+\tgpmc_t->oe_off = gpmc_t->access + gpmc_ticks_to_ps(1);\ngpmc_t->cs_rd_off = gpmc_t->oe_off;\n\n/* rd_cycle */\ntemp = max_t(u32, dev_t->t_cez_r, dev_t->t_oez);\ntemp = gpmc_round_ps_to_sync_clk(temp, gpmc_t->sync_clk) +\n-\t\t\t\t\t\t\tgpmc_t->access * 1000;\n+\t\t\t\t\t\t\tgpmc_t->access;\n/* barter t_ce_rdyz with t_cez_r ? */\nif (dev_t->t_ce_rdyz)\n-\t\ttemp = max_t(u32, temp,\n-\t\t\t\tgpmc_t->cs_rd_off * 1000 + dev_t->t_ce_rdyz);\n-\tgpmc_t->rd_cycle = gpmc_round_ps_to_ticks(temp) / 1000;\n+\t\ttemp = max_t(u32, temp,\tgpmc_t->cs_rd_off + dev_t->t_ce_rdyz);\n+\tgpmc_t->rd_cycle = gpmc_round_ps_to_ticks(temp);\n\nreturn 0;\n}\n@@ -976,29 +971,28 @@ static int gpmc_calc_sync_write_timings(struct gpmc_timings *gpmc_t,\ntemp = dev_t->t_avdp_w;\nif (mux) {\ntemp = max_t(u32, temp,\n-\t\t\tgpmc_t->clk_activation * 1000 + dev_t->t_avdh);\n-\t\ttemp = max_t(u32,\n-\t\t\t(gpmc_t->adv_on + gpmc_ticks_to_ns(1)) * 1000, temp);\n+\t\t\tgpmc_t->clk_activation + dev_t->t_avdh);\n+\t\ttemp = max_t(u32, gpmc_t->adv_on + gpmc_ticks_to_ps(1), temp);\n}\n-\tgpmc_t->adv_wr_off = gpmc_round_ps_to_ticks(temp) / 1000;\n\n/* wr_data_mux_bus */\ntemp = max_t(u32, dev_t->t_weasu,\n-\t\t\tgpmc_t->clk_activation * 1000 + dev_t->t_rdyo);\n+\t\t\tgpmc_t->clk_activation + dev_t->t_rdyo);\n/* shouldn't mux be kept as a whole for wr_data_mux_bus ?,\n* and in that case remember to handle we_on properly\n*/\nif (mux) {\ntemp = max_t(u32, temp,\n-\t\t\tgpmc_t->adv_wr_off * 1000 + dev_t->t_aavdh);\n-\t\ttemp = max_t(u32, temp, (gpmc_t->adv_wr_off +\n-\t\t\t\tgpmc_ticks_to_ns(dev_t->cyc_aavdh_we)) * 1000);\n+\t\ttemp = max_t(u32, temp, gpmc_t->adv_wr_off +\n+\t\t\t\tgpmc_ticks_to_ps(dev_t->cyc_aavdh_we));\n}\n-\tgpmc_t->wr_data_mux_bus = gpmc_round_ps_to_ticks(temp) / 1000;\n+\tgpmc_t->wr_data_mux_bus = gpmc_round_ps_to_ticks(temp);\n\n/* we_on */\nif (gpmc_capability & GPMC_HAS_WR_DATA_MUX_BUS)\n-\t\tgpmc_t->we_on = gpmc_round_ps_to_ticks(dev_t->t_weasu) / 1000;\n+\t\tgpmc_t->we_on = gpmc_round_ps_to_ticks(dev_t->t_weasu);\nelse\ngpmc_t->we_on = gpmc_t->wr_data_mux_bus;\n\n@@ -1007,24 +1001,24 @@ static int gpmc_calc_sync_write_timings(struct gpmc_timings *gpmc_t,\ngpmc_t->wr_access = gpmc_t->access;\n\n/* we_off */\n-\ttemp = gpmc_t->we_on * 1000 + dev_t->t_wpl;\n+\ttemp = gpmc_t->we_on + dev_t->t_wpl;\ntemp = max_t(u32, temp,\n-\t\t\t(gpmc_t->wr_access + gpmc_ticks_to_ns(1)) * 1000);\n+\t\t\tgpmc_t->wr_access + gpmc_ticks_to_ps(1));\ntemp = max_t(u32, temp,\n-\t\t(gpmc_t->we_on + gpmc_ticks_to_ns(dev_t->cyc_wpl)) * 1000);\n-\tgpmc_t->we_off = gpmc_round_ps_to_ticks(temp) / 1000;\n+\t\tgpmc_t->we_on + gpmc_ticks_to_ps(dev_t->cyc_wpl));\n+\tgpmc_t->we_off = gpmc_round_ps_to_ticks(temp);\n\n-\tgpmc_t->cs_wr_off = gpmc_round_ps_to_ticks(gpmc_t->we_off * 1000 +\n-\t\t\t\t\t\t\tdev_t->t_wph) / 1000;\n+\tgpmc_t->cs_wr_off = gpmc_round_ps_to_ticks(gpmc_t->we_off +\n+\t\t\t\t\t\t\tdev_t->t_wph);\n\n/* wr_cycle */\ntemp = gpmc_round_ps_to_sync_clk(dev_t->t_cez_w, gpmc_t->sync_clk);\n-\ttemp += gpmc_t->wr_access * 1000;\n+\ttemp += gpmc_t->wr_access;\n/* barter t_ce_rdyz with t_cez_w ? */\nif (dev_t->t_ce_rdyz)\ntemp = max_t(u32, temp,\n-\t\t\t\t gpmc_t->cs_wr_off * 1000 + dev_t->t_ce_rdyz);\n-\tgpmc_t->wr_cycle = gpmc_round_ps_to_ticks(temp) / 1000;\n+\t\t\t\t gpmc_t->cs_wr_off + dev_t->t_ce_rdyz);\n+\tgpmc_t->wr_cycle = gpmc_round_ps_to_ticks(temp);\n\nreturn 0;\n}\n@@ -1038,35 +1032,33 @@ static int gpmc_calc_async_read_timings(struct gpmc_timings *gpmc_t,\ntemp = dev_t->t_avdp_r;\nif (mux)\n-\t\ttemp = max_t(u32,\n-\t\t\t(gpmc_t->adv_on + gpmc_ticks_to_ns(1)) * 1000, temp);\n-\tgpmc_t->adv_rd_off = gpmc_round_ps_to_ticks(temp) / 1000;\n+\t\ttemp = max_t(u32, gpmc_t->adv_on + gpmc_ticks_to_ps(1), temp);\n\n/* oe_on */\ntemp = dev_t->t_oeasu;\nif (mux)\ntemp = max_t(u32, temp,\n-\t\t\tgpmc_t->adv_rd_off * 1000 + dev_t->t_aavdh);\n-\tgpmc_t->oe_on = gpmc_round_ps_to_ticks(temp) / 1000;\n+\tgpmc_t->oe_on = gpmc_round_ps_to_ticks(temp);\n\n/* access */\ntemp = max_t(u32, dev_t->t_iaa, /* remove t_iaa in async ? */\n-\t\t\t\tgpmc_t->oe_on * 1000 + dev_t->t_oe);\n+\t\t\t\tgpmc_t->oe_on + dev_t->t_oe);\ntemp = max_t(u32, temp,\n-\t\t\t\tgpmc_t->cs_on * 1000 + dev_t->t_ce);\n+\t\t\t\tgpmc_t->cs_on + dev_t->t_ce);\ntemp = max_t(u32, temp,\n-\t\t\t\tgpmc_t->adv_on * 1000 + dev_t->t_aa);\n-\tgpmc_t->access = gpmc_round_ps_to_ticks(temp) / 1000;\n+\tgpmc_t->access = gpmc_round_ps_to_ticks(temp);\n\n-\tgpmc_t->oe_off = gpmc_t->access + gpmc_ticks_to_ns(1);\n+\tgpmc_t->oe_off = gpmc_t->access + gpmc_ticks_to_ps(1);\ngpmc_t->cs_rd_off = gpmc_t->oe_off;\n\n/* rd_cycle */\ntemp = max_t(u32, dev_t->t_rd_cycle,\n-\t\t\tgpmc_t->cs_rd_off * 1000 + dev_t->t_cez_r);\n-\ttemp = max_t(u32, temp,\n-\t\t\tgpmc_t->oe_off * 1000 + dev_t->t_oez);\n-\tgpmc_t->rd_cycle = gpmc_round_ps_to_ticks(temp) / 1000;\n+\t\t\tgpmc_t->cs_rd_off + dev_t->t_cez_r);\n+\ttemp = max_t(u32, temp, gpmc_t->oe_off + dev_t->t_oez);\n+\tgpmc_t->rd_cycle = gpmc_round_ps_to_ticks(temp);\n\nreturn 0;\n}\n@@ -1080,37 +1072,35 @@ static int gpmc_calc_async_write_timings(struct gpmc_timings *gpmc_t,\ntemp = dev_t->t_avdp_w;\nif (mux)\n-\t\ttemp = max_t(u32,\n-\t\t\t(gpmc_t->adv_on + gpmc_ticks_to_ns(1)) * 1000, temp);\n-\tgpmc_t->adv_wr_off = gpmc_round_ps_to_ticks(temp) / 1000;\n+\t\ttemp = max_t(u32, gpmc_t->adv_on + gpmc_ticks_to_ps(1), temp);\n\n/* wr_data_mux_bus */\ntemp = dev_t->t_weasu;\nif (mux) {\n-\t\ttemp = max_t(u32, temp,\n-\t\t\tgpmc_t->adv_wr_off * 1000 + dev_t->t_aavdh);\n-\t\ttemp = max_t(u32, temp, (gpmc_t->adv_wr_off +\n-\t\t\t\tgpmc_ticks_to_ns(dev_t->cyc_aavdh_we)) * 1000);\n+\t\ttemp = max_t(u32, temp,\tgpmc_t->adv_wr_off + dev_t->t_aavdh);\n+\t\ttemp = max_t(u32, temp, gpmc_t->adv_wr_off +\n+\t\t\t\tgpmc_ticks_to_ps(dev_t->cyc_aavdh_we));\n}\n-\tgpmc_t->wr_data_mux_bus = gpmc_round_ps_to_ticks(temp) / 1000;\n+\tgpmc_t->wr_data_mux_bus = gpmc_round_ps_to_ticks(temp);\n\n/* we_on */\nif (gpmc_capability & GPMC_HAS_WR_DATA_MUX_BUS)\n-\t\tgpmc_t->we_on = gpmc_round_ps_to_ticks(dev_t->t_weasu) / 1000;\n+\t\tgpmc_t->we_on = gpmc_round_ps_to_ticks(dev_t->t_weasu);\nelse\ngpmc_t->we_on = gpmc_t->wr_data_mux_bus;\n\n/* we_off */\n-\ttemp = gpmc_t->we_on * 1000 + dev_t->t_wpl;\n-\tgpmc_t->we_off = gpmc_round_ps_to_ticks(temp) / 1000;\n+\ttemp = gpmc_t->we_on + dev_t->t_wpl;\n+\tgpmc_t->we_off = gpmc_round_ps_to_ticks(temp);\n\n-\tgpmc_t->cs_wr_off = gpmc_round_ps_to_ticks((gpmc_t->we_off * 1000 +\n-\t\t\t\tdev_t->t_wph)) / 1000;\n+\tgpmc_t->cs_wr_off = gpmc_round_ps_to_ticks(gpmc_t->we_off +\n+\t\t\t\t\t\t\tdev_t->t_wph);\n\n/* wr_cycle */\ntemp = max_t(u32, dev_t->t_wr_cycle,\n-\t\t\t\tgpmc_t->cs_wr_off * 1000 + dev_t->t_cez_w);\n-\tgpmc_t->wr_cycle = gpmc_round_ps_to_ticks(temp) / 1000;\n+\t\t\t\tgpmc_t->cs_wr_off + dev_t->t_cez_w);\n+\tgpmc_t->wr_cycle = gpmc_round_ps_to_ticks(temp);\n\nreturn 0;\n}\n@@ -1125,10 +1115,10 @@ static int gpmc_calc_sync_common_timings(struct gpmc_timings *gpmc_t,\n\ngpmc_t->page_burst_access = gpmc_round_ps_to_sync_clk(\ndev_t->t_bacc,\n-\t\t\t\t\tgpmc_t->sync_clk) / 1000;\n+\t\t\t\t\tgpmc_t->sync_clk);\n\ntemp = max_t(u32, dev_t->t_ces, dev_t->t_avds);\n-\tgpmc_t->clk_activation = gpmc_round_ps_to_ticks(temp) / 1000;\n+\tgpmc_t->clk_activation = gpmc_round_ps_to_ticks(temp);\n\nif (gpmc_calc_divider(gpmc_t->sync_clk) != 1)\nreturn 0;\n@@ -1151,14 +1141,14 @@ static int gpmc_calc_common_timings(struct gpmc_timings *gpmc_t,\nu32 temp;\n\n/* cs_on */\n-\tgpmc_t->cs_on = gpmc_round_ns_to_ticks(dev_t->t_ceasu / 1000);\n+\tgpmc_t->cs_on = gpmc_round_ps_to_ticks(dev_t->t_ceasu);\n\ntemp = dev_t->t_avdasu;\nif (dev_t->t_ce_avd)\ntemp = max_t(u32, temp,\n-\t\t\t\tgpmc_t->cs_on * 1000 + dev_t->t_ce_avd);\n-\tgpmc_t->adv_on = gpmc_round_ns_to_ticks(temp / 1000);\n+\t\t\t\tgpmc_t->cs_on + dev_t->t_ce_avd);\n\ngpmc_calc_sync_common_timings(gpmc_t, dev_t);\ndiff --git a/arch/arm/plat-omap/include/plat/gpmc.h b/arch/arm/plat-omap/include/plat/gpmc.h\nindex e59a932..f1d1d2e 100644\n--- a/arch/arm/plat-omap/include/plat/gpmc.h\n+++ b/arch/arm/plat-omap/include/plat/gpmc.h\n@@ -116,38 +116,38 @@ struct gpmc_timings {\nu32 sync_clk;\n\n/* Chip-select signal timings corresponding to GPMC_CS_CONFIG2 */\n-\tu16 cs_on;\t\t/* Assertion time */\n-\tu16 cs_rd_off;\t\t/* Read deassertion time */\n-\tu16 cs_wr_off;\t\t/* Write deassertion time */\n+\tu32 cs_on;\t\t/* Assertion time */\n+\tu32 cs_rd_off;\t\t/* Read deassertion time */\n+\tu32 cs_wr_off;\t\t/* Write deassertion time */\n\n/* ADV signal timings corresponding to GPMC_CONFIG3 */\n-\tu16 adv_on;\t\t/* Assertion time */\n-\tu16 adv_wr_off;\t\t/* Write deassertion time */\n+\tu32 adv_on;\t\t/* Assertion time */\n+\tu32 adv_wr_off;\t\t/* Write deassertion time */\n\n/* WE signals timings corresponding to GPMC_CONFIG4 */\n-\tu16 we_on;\t\t/* WE assertion time */\n-\tu16 we_off;\t\t/* WE deassertion time */\n+\tu32 we_on;\t\t/* WE assertion time */\n+\tu32 we_off;\t\t/* WE deassertion time */\n\n/* OE signals timings corresponding to GPMC_CONFIG4 */\n-\tu16 oe_on;\t\t/* OE assertion time */\n-\tu16 oe_off;\t\t/* OE deassertion time */\n+\tu32 oe_on;\t\t/* OE assertion time */\n+\tu32 oe_off;\t\t/* OE deassertion time */\n\n/* Access time and cycle time timings corresponding to GPMC_CONFIG5 */\n-\tu16 page_burst_access;\t/* Multiple access word delay */\n-\tu16 access;\t\t/* Start-cycle to first data valid delay */\n-\tu16 rd_cycle;\t\t/* Total read cycle time */\n-\tu16 wr_cycle;\t\t/* Total write cycle time */\n+\tu32 page_burst_access;\t/* Multiple access word delay */\n+\tu32 access;\t\t/* Start-cycle to first data valid delay */\n+\tu32 rd_cycle;\t\t/* Total read cycle time */\n+\tu32 wr_cycle;\t\t/* Total write cycle time */\n\n-\tu16 bus_turnaround;\n-\tu16 cycle2cycle_delay;\n+\tu32 bus_turnaround;\n+\tu32 cycle2cycle_delay;\n\n-\tu16 wait_monitoring;\n-\tu16 clk_activation;\n+\tu32 wait_monitoring;\n+\tu32 clk_activation;\n\n/* The following are only on OMAP3430 */\n-\tu16 wr_access;\t\t/* WRACCESSTIME */\n-\tu16 wr_data_mux_bus;\t/* WRDATAONADMUXBUS */\n+\tu32 wr_access;\t\t/* WRACCESSTIME */\n+\tu32 wr_data_mux_bus;\t/* WRDATAONADMUXBUS */\n\nstruct gpmc_bool_timings bool_timings;\n};\n\n```" ]

[ null ]

[ "|\n\n# 中央空调和普通空调哪个更省电?结果让人吃惊\n\n1、效率\n\n多个房间安装中央空调时，仅需要安装一台室外机，在室内仅有少量房间使用空调时负荷较低，整机效率也偏低，使用房间多时效率较高。而普通分体空调必须每个房间安装一台室内机和室外机，如果多房间同时使用的话，负荷加重，效率低。", null, "2、设计安装\n\n分体空调只需要分配好房间内部空间就可以，安装也相对容易。而中央空调则需要专门对安装方式进行设计，并且必须在装修前进行安装，对安装施工要求比较高，安装工作量大，必须由专业施工队完成。\n\n3、清洗维护\n\n普通分体空调清洗相对简单，动手能力强的用户可自行完成。中央空调清洗相对困难，大家一般很难难以自行操作，需要专业的队伍完成，而且中央空调一旦发生故障，所有安装房间都无法正常使用，后续的维护维修也比较麻烦，施工难度大，维护费用高，在这一点上，分体空调要占绝对优势。", null, "4、价格和寿命\n\n中央空调的购买费用要高于分体空调，同样制冷量前提下，中央空调的价格要比普通分体空调高一倍。不过中央空调的设计寿命一般为15-20年，同样也是普通分体空调的两倍，综合来说很难从价格上直接对比出优劣。\n\n简而言之，如果房子面积不大，房间不多，多个房间同时开启空调的机会比较少的话，分体空调优势比较明显。而如果房子比较大，房间较多，层高也较高，同时使用空调的房间多，那么中央空调的优势就相对突出了。\n\n`声明：本文由入驻焦点开放平台的作者撰写，除焦点官方账号外，观点仅代表作者本人，不代表焦点立场错误信息举报电话： 400-099-0099，邮箱：[email protected]，或点此进行意见反馈，或点此进行举报投诉。`", null, "A B C D E F G H J K L M N P Q R S T W X Y Z\nA - B - C - D - E\n• A\n• 鞍山\n• 安庆\n• 安阳\n• 安顺\n• 安康\n• 澳门\n• B\n• 北京\n• 保定\n• 包头\n• 巴彦淖尔\n• 本溪\n• 蚌埠\n• 亳州\n• 滨州\n• 北海\n• 百色\n• 巴中\n• 毕节\n• 保山\n• 宝鸡\n• 白银\n• 巴州\n• C\n• 承德\n• 沧州\n• 长治\n• 赤峰\n• 朝阳\n• 长春\n• 常州\n• 滁州\n• 池州\n• 长沙\n• 常德\n• 郴州\n• 潮州\n• 崇左\n• 重庆\n• 成都\n• 楚雄\n• 昌都\n• 慈溪\n• 常熟\n• D\n• 大同\n• 大连\n• 丹东\n• 大庆\n• 东营\n• 德州\n• 东莞\n• 德阳\n• 达州\n• 大理\n• 德宏\n• 定西\n• 儋州\n• 东平\n• E\n• 鄂尔多斯\n• 鄂州\n• 恩施\nF - G - H - I - J\n• F\n• 抚顺\n• 阜新\n• 阜阳\n• 福州\n• 抚州\n• 佛山\n• 防城港\n• G\n• 赣州\n• 广州\n• 桂林\n• 贵港\n• 广元\n• 广安\n• 贵阳\n• 固原\n• H\n• 邯郸\n• 衡水\n• 呼和浩特\n• 呼伦贝尔\n• 葫芦岛\n• 哈尔滨\n• 黑河\n• 淮安\n• 杭州\n• 湖州\n• 合肥\n• 淮南\n• 淮北\n• 黄山\n• 菏泽\n• 鹤壁\n• 黄石\n• 黄冈\n• 衡阳\n• 怀化\n• 惠州\n• 河源\n• 贺州\n• 河池\n• 海口\n• 红河\n• 汉中\n• 海东\n• 怀来\n• I\n• J\n• 晋中\n• 锦州\n• 吉林\n• 鸡西\n• 佳木斯\n• 嘉兴\n• 金华\n• 景德镇\n• 九江\n• 吉安\n• 济南\n• 济宁\n• 焦作\n• 荆门\n• 荆州\n• 江门\n• 揭阳\n• 金昌\n• 酒泉\n• 嘉峪关\nK - L - M - N - P\n• K\n• 开封\n• 昆明\n• 昆山\n• L\n• 廊坊\n• 临汾\n• 辽阳\n• 连云港\n• 丽水\n• 六安\n• 龙岩\n• 莱芜\n• 临沂\n• 聊城\n• 洛阳\n• 漯河\n• 娄底\n• 柳州\n• 来宾\n• 泸州\n• 乐山\n• 六盘水\n• 丽江\n• 临沧\n• 拉萨\n• 林芝\n• 兰州\n• 陇南\n• M\n• 牡丹江\n• 马鞍山\n• 茂名\n• 梅州\n• 绵阳\n• 眉山\n• N\n• 南京\n• 南通\n• 宁波\n• 南平\n• 宁德\n• 南昌\n• 南阳\n• 南宁\n• 内江\n• 南充\n• P\n• 盘锦\n• 莆田\n• 平顶山\n• 濮阳\n• 攀枝花\n• 普洱\n• 平凉\nQ - R - S - T - W\n• Q\n• 秦皇岛\n• 齐齐哈尔\n• 衢州\n• 泉州\n• 青岛\n• 清远\n• 钦州\n• 黔南\n• 曲靖\n• 庆阳\n• R\n• 日照\n• 日喀则\n• S\n• 石家庄\n• 沈阳\n• 双鸭山\n• 绥化\n• 上海\n• 苏州\n• 宿迁\n• 绍兴\n• 宿州\n• 三明\n• 上饶\n• 三门峡\n• 商丘\n• 十堰\n• 随州\n• 邵阳\n• 韶关\n• 深圳\n• 汕头\n• 汕尾\n• 三亚\n• 三沙\n• 遂宁\n• 山南\n• 商洛\n• 石嘴山\n• T\n• 天津\n• 唐山\n• 太原\n• 通辽\n• 铁岭\n• 泰州\n• 台州\n• 铜陵\n• 泰安\n• 铜仁\n• 铜川\n• 天水\n• 天门\n• W\n• 乌海\n• 乌兰察布\n• 无锡\n• 温州\n• 芜湖\n• 潍坊\n• 威海\n• 武汉\n• 梧州\n• 渭南\n• 武威\n• 吴忠\n• 乌鲁木齐\nX - Y - Z\n• X\n• 邢台\n• 徐州\n• 宣城\n• 厦门\n• 新乡\n• 许昌\n• 信阳\n• 襄阳\n• 孝感\n• 咸宁\n• 湘潭\n• 湘西\n• 西双版纳\n• 西安\n• 咸阳\n• 西宁\n• 仙桃\n• 西昌\n• Y\n• 运城\n• 营口\n• 盐城\n• 扬州\n• 鹰潭\n• 宜春\n• 烟台\n• 宜昌\n• 岳阳\n• 益阳\n• 永州\n• 阳江\n• 云浮\n• 玉林\n• 宜宾\n• 雅安\n• 玉溪\n• 延安\n• 榆林\n• 银川\n• Z\n• 张家口\n• 镇江\n• 舟山\n• 漳州\n• 淄博\n• 枣庄\n• 郑州\n• 周口\n• 驻马店\n• 株洲\n• 张家界\n• 珠海\n• 湛江\n• 肇庆\n• 中山\n• 自贡\n• 资阳\n• 遵义\n• 昭通\n• 张掖\n• 中卫\n\n1室1厅1厨1卫1阳台\n\n1\n2\n3\n4\n5\n\n0\n1\n2\n\n1\n\n1\n\n0\n1\n2\n3", null, "", null, "", null, "报名成功，资料已提交审核", null, "A B C D E F G H J K L M N P Q R S T W X Y Z\nA - B - C - D - E\n• A\n• 鞍山\n• 安庆\n• 安阳\n• 安顺\n• 安康\n• 澳门\n• B\n• 北京\n• 保定\n• 包头\n• 巴彦淖尔\n• 本溪\n• 蚌埠\n• 亳州\n• 滨州\n• 北海\n• 百色\n• 巴中\n• 毕节\n• 保山\n• 宝鸡\n• 白银\n• 巴州\n• C\n• 承德\n• 沧州\n• 长治\n• 赤峰\n• 朝阳\n• 长春\n• 常州\n• 滁州\n• 池州\n• 长沙\n• 常德\n• 郴州\n• 潮州\n• 崇左\n• 重庆\n• 成都\n• 楚雄\n• 昌都\n• 慈溪\n• 常熟\n• D\n• 大同\n• 大连\n• 丹东\n• 大庆\n• 东营\n• 德州\n• 东莞\n• 德阳\n• 达州\n• 大理\n• 德宏\n• 定西\n• 儋州\n• 东平\n• E\n• 鄂尔多斯\n• 鄂州\n• 恩施\nF - G - H - I - J\n• F\n• 抚顺\n• 阜新\n• 阜阳\n• 福州\n• 抚州\n• 佛山\n• 防城港\n• G\n• 赣州\n• 广州\n• 桂林\n• 贵港\n• 广元\n• 广安\n• 贵阳\n• 固原\n• H\n• 邯郸\n• 衡水\n• 呼和浩特\n• 呼伦贝尔\n• 葫芦岛\n• 哈尔滨\n• 黑河\n• 淮安\n• 杭州\n• 湖州\n• 合肥\n• 淮南\n• 淮北\n• 黄山\n• 菏泽\n• 鹤壁\n• 黄石\n• 黄冈\n• 衡阳\n• 怀化\n• 惠州\n• 河源\n• 贺州\n• 河池\n• 海口\n• 红河\n• 汉中\n• 海东\n• 怀来\n• I\n• J\n• 晋中\n• 锦州\n• 吉林\n• 鸡西\n• 佳木斯\n• 嘉兴\n• 金华\n• 景德镇\n• 九江\n• 吉安\n• 济南\n• 济宁\n• 焦作\n• 荆门\n• 荆州\n• 江门\n• 揭阳\n• 金昌\n• 酒泉\n• 嘉峪关\nK - L - M - N - P\n• K\n• 开封\n• 昆明\n• 昆山\n• L\n• 廊坊\n• 临汾\n• 辽阳\n• 连云港\n• 丽水\n• 六安\n• 龙岩\n• 莱芜\n• 临沂\n• 聊城\n• 洛阳\n• 漯河\n• 娄底\n• 柳州\n• 来宾\n• 泸州\n• 乐山\n• 六盘水\n• 丽江\n• 临沧\n• 拉萨\n• 林芝\n• 兰州\n• 陇南\n• M\n• 牡丹江\n• 马鞍山\n• 茂名\n• 梅州\n• 绵阳\n• 眉山\n• N\n• 南京\n• 南通\n• 宁波\n• 南平\n• 宁德\n• 南昌\n• 南阳\n• 南宁\n• 内江\n• 南充\n• P\n• 盘锦\n• 莆田\n• 平顶山\n• 濮阳\n• 攀枝花\n• 普洱\n• 平凉\nQ - R - S - T - W\n• Q\n• 秦皇岛\n• 齐齐哈尔\n• 衢州\n• 泉州\n• 青岛\n• 清远\n• 钦州\n• 黔南\n• 曲靖\n• 庆阳\n• R\n• 日照\n• 日喀则\n• S\n• 石家庄\n• 沈阳\n• 双鸭山\n• 绥化\n• 上海\n• 苏州\n• 宿迁\n• 绍兴\n• 宿州\n• 三明\n• 上饶\n• 三门峡\n• 商丘\n• 十堰\n• 随州\n• 邵阳\n• 韶关\n• 深圳\n• 汕头\n• 汕尾\n• 三亚\n• 三沙\n• 遂宁\n• 山南\n• 商洛\n• 石嘴山\n• T\n• 天津\n• 唐山\n• 太原\n• 通辽\n• 铁岭\n• 泰州\n• 台州\n• 铜陵\n• 泰安\n• 铜仁\n• 铜川\n• 天水\n• 天门\n• W\n• 乌海\n• 乌兰察布\n• 无锡\n• 温州\n• 芜湖\n• 潍坊\n• 威海\n• 武汉\n• 梧州\n• 渭南\n• 武威\n• 吴忠\n• 乌鲁木齐\nX - Y - Z\n• X\n• 邢台\n• 徐州\n• 宣城\n• 厦门\n• 新乡\n• 许昌\n• 信阳\n• 襄阳\n• 孝感\n• 咸宁\n• 湘潭\n• 湘西\n• 西双版纳\n• 西安\n• 咸阳\n• 西宁\n• 仙桃\n• 西昌\n• Y\n• 运城\n• 营口\n• 盐城\n• 扬州\n• 鹰潭\n• 宜春\n• 烟台\n• 宜昌\n• 岳阳\n• 益阳\n• 永州\n• 阳江\n• 云浮\n• 玉林\n• 宜宾\n• 雅安\n• 玉溪\n• 延安\n• 榆林\n• 银川\n• Z\n• 张家口\n• 镇江\n• 舟山\n• 漳州\n• 淄博\n• 枣庄\n• 郑州\n• 周口\n• 驻马店\n• 株洲\n• 张家界\n• 珠海\n• 湛江\n• 肇庆\n• 中山\n• 自贡\n• 资阳\n• 遵义\n• 昭通\n• 张掖\n• 中卫", null, "", null, "• 手机", null, "• 分享\n• 设计\n免费设计\n• 计算器\n装修计算器\n• 入驻\n合作入驻\n• 联系\n联系我们\n• 置顶\n返回顶部" ]

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[ "A Satellite Navigation Signal Scheme Using Zadoff-Chu Sequence for Reducing the Signal Acquisition Space\nA Satellite Navigation Signal Scheme Using Zadoff-Chu Sequence for Reducing the Signal Acquisition Space\nJournal of Positioning, Navigation, and Timing. 2013. Apr, 2(1): 1-8\n• Received : March 29, 2013\n• Accepted : April 28, 2013\n• Published : April 30, 2013", null, "PDF", null, "e-PUB", null, "PubReader", null, "PPT", null, "Export by style\nArticle\nAuthor\nMetrics\nCited by\nTagCloud\nDae-Soon, Park\nDepartment of Information Communication Engineering, Chungnam National University, Daejeon 305-764, Korea\nJeong-been, Kim\nDepartment of Information Communication Engineering, Chungnam National University, Daejeon 305-764, Korea\nJe-won, Lee\nDepartment of Information Communication Engineering, Chungnam National University, Daejeon 305-764, Korea\nKap-Jin, Kim\nAgency for Defense Development, 12, Bugyuseong-daero, 488beon-gil, Yuseong-gu, Daejeon 305-156, Korea\nKiwon, Song\nAgency for Defense Development, 12, Bugyuseong-daero, 488beon-gil, Yuseong-gu, Daejeon 305-156, Korea\nJae Min, Ahn\nDepartment of Information Communication Engineering, Chungnam National University, Daejeon 305-764, Korea\[email protected]\n\nAbstract\nA signal system for improving the code acquisition complexity of Global Navigation Satellite System (GNSS) receiver is proposed and the receiving correlator scheme is presented accordingly. The proposed signal system is a hierarchical code type with a duplexing configuration which consists of the Zadoff-Chu (ZC) code having a good auto-correlation characteristic and the Pseudo Random Noise (PRN) code for distinguishing satellites. The receiving correlator has the scheme that consists of the primary correlator for the ZC code and the secondary correlator which uses the PRN code for the primary correlation results. The simulation results of code acquisition using the receiving correlator of the proposed signal system show that the proposed signal scheme improves the complexity of GNSS receiver and has the code acquisition performance comparable to the existing GNSS signal system using Coarse/Acquisition (C/A) code.\nKeywords\n1. INTRODUCTION\nFor calculating the position of receiver using a Global Navigation Satellite System (GNSS) signal, the distance from satellite to receiver and the coordinate of satellite are necessary, and the positioning in 3-dimensional space can be accomplished when there are at least 4 visible satellites. In order to reduce the error of distance measurement from satellite to receiver, the temporal resolution, which is the basis of distance measurement, needs to be increased, and the signal reaching the ground has weaker intensity than thermal noise. In this regard, every GNSS signal system employs the spread spectrum modulation to achieve the increase of temporal resolution for distance measurement and the processing gain.\nOn the other hand, to distinguish various satellites, the spread spectrum modulation is performed so that spacethe spread spectrum signal generated at each satellite has different sign. When this code for each satellite is received in a mixed form, the code that minimizes the cross-correlation value between the signals is used, and the Coarse/Acquisition (C/A) code of Global Positioning System (GPS) is a representative code for this purpose. Also, for the additional reduction of cross-correlation value, the code period is generally maximized within the limit that minimizes the effect of Doppler shift due to the movements of satellite and receiver. The Doppler shift due to the relative movements of satellite and receiver which is observed at the receiver could be a major factor for degrading the correlation performance of receiving correlator. Therefore, regarding the Doppler shift which is dispersed over a range of more than several kHz, a typical GNSS receiver uses the method that obtains the correlation value by changing the frequency of correlator.\nBased on the above discussion, Fig. 1 shows the search range of receiving correlator for the satellite signal acquisition of typical GNSS receiver. The frequency search interval by the Doppler shift is generally 500Hz, and the number of satellites that a GPS has at present is about 30. For the acquisition range from the code delay axis, the GPS calculates the correlation values of 1023 chips for each cell. However, for the Gailieo from Europe, the calculation for the signal acquisition increases because the correlation values of 10230 chips need to be considered. As the signal acquisition cell formed by the values assigned to each axis increases, the complexity also sharply increases. The increase in the density of hardware technology is trying to cope with the complexity increase of code acquisition correlator, but this could lead to the power and heat problems of portable device. Thus, it is necessary to conduct studies on complexity reduction.", null, "PPT Slide\nLager Image\nAcquisition space of 3-dimension.\nThe purpose of this study was to find a method for reducing the code acquisition complexity of GNSS receiver by improving the signal scheme. In the signal scheme of the existing GNSS, one of the causes for the increase of correlator complexity is that an independent receiving correlator needs to be operated for each satellite. If every satellite generates the spread spectrum signal as the same code but the satellites could still be distinguished, the signal processing procedure for satellite search would be innovatively simplified.\nTo design such signal system, the Zadoff-Chu (ZC) code having a good auto-correlation characteristic was selected, and this code was assigned to the spread spectrum modulation for every satellite. The ZC code is widely used as the preamble for synchronization for the 4th Generation (4G) systems such as Long Term Evolution (LTE) and Ultra Wide Band (UMB) ( Beyme & Leung 2009 ). In designing the system, the satellites are distinguished by making the GNSS signal to be a hierarchical code type via adding different Pseudo Random Noise (PRN) code to the basic period ZC code for each satellite.\nIn Chapter 2, a new GNSS signal system which uses the ZC code and PRN code is introduced, and in Chapter 3, the scheme and operation of the correlator for code acquisition which can be applied to the new signal system are described. In Chapter 4, the correlation characteristic of proposed signal system is presented through the simulation, and in Chapter 5, the complexity effect and the conclusion are included.\n2. DESIGN OF GNSS SCHEME USING ZC\nThe ZC code which has the root of u and the length of prime number length N zc is defined as Eq. (1).", null, "PPT Slide\nLager Image\nThe ZC code has a feature that it does not have a correlation value which excludes the case when the synchronization is matched. An independent code generation is possible depending on the selection of the root u of code which is relatively prime to the code length N zc . Fig. 2 shows the correlation characteristic of ZC code which has the N zc of 1023 that is identical to the C/A code length of GPS. The correlation value is found to be close to 0 for the remaining code delay except for the case that it matches exactly.", null, "PPT Slide\nLager Image\n- 2.2 Modulation scheme using Zadoff-Chu sequence\nThe signal scheme of GNSS should be able to distinguish every satellite and transmit the navigation information that each satellite sends simultaneously. For reducing the complexity, every satellite needs to have the same code, and the code having a good auto-correlation characteristic is more advantageous than the code having a good cross-correlation characteristic. Therefore, using the Zadoff-Chu code having a good auto-correlation characteristic, the signal scheme is proposed as shown in Fig. 3 which is a combined type of the pilot signal for distinguishing satellites and the data signal.", null, "PPT Slide\nLager Image\nSatellite signal scheme.\nEach satellite repeatedly sends the same ZC pilot code, and the satellites are distinguished by hierarchically assigning the PRN code for every ZC pilot code period. When this scheme is applied, the PRN code for distinguishing satellites could be obtained from the correlation value of pilot signal. This requires the calculation for comparing the PRN codes, but as they are relatively short length codes, the calculation will be insignificant when compared to the existing calculation.\nFor the data signal which carries the navigation information, in the pilot signal scheme, the navigation information is assigned to the ZC code which has a length that is the multiple of the PRN code length, and is transmitted simultaneously with the pilot signal. Using formulas, this can be defined as Eqs. (2) and (3).", null, "PPT Slide\nLager Image", null, "PPT Slide\nLager Image\nAs shown in Fig. 3 , the length of ZC code consists of the pilot and the data. The data length is equal to the length which added the code length to the pilot length. Then, the N d , which determines the data length, naturally becomes the multiple of N p , and if the two lengths are designed to be relatively prime to each other, the polyphase code condition could be satisfied ( Chu 1972 ).", null, "PPT Slide\nLager Image", null, "PPT Slide\nLager Image\nIn Eq. (4), for the k-th satellite, the identification number is assigned to the PRN code C k , and as shown in Eq. (5), this is put in the pilot signal and periodically generated for every data length N d .", null, "PPT Slide\nLager Image", null, "PPT Slide\nLager Image\nThe signal sequence D k , that the k-th satellite sends, is also expressed as Eq. (6) depending on the sign of data, and the transmitting signal of k-th satellite which combines these two signals can be expressed as Eq. (7). The pilot and data signals generated in this way are transmitted in a mixed form, and as the code correlation between the signals is insignificant, it will be possible to distinguish the signals.\n3. CORRELATION SCHEME FOR ACQUISITION\nAs for the GPS, for the acquisition of signal, the correlation value is obtained by generating the reference signal that corresponds to each cell shown in Fig. 1 to despread the signal which has been spreaded for about 1ms in the correlator. In the actual receiver environment, the signal arrives while experiencing different delay and attenuation from the satellite that is in operation at each orbit. An ideal receiver without the Doppler effect is assumed. Based on this, the receiving signal can be modeled as Eq. (8). y[n] is the sum of signals where the k-th satellite has the delay of τ k and the attenuation index of g k , among the N s visible satellites.", null, "PPT Slide\nLager Image\n- 3.2 Correlation model\nFor the previously modeled satellite signal, to distinguish satellites from the received signal depending on each code delay, the correlation value which uses the pilot signal can be modeled as Eq. (9).", null, "PPT Slide\nLager Image\nIf the correlation value with the data is relatively prime to the pilot signal, the cross-correlation characteristic of certain numerical value will be observed as shown in Fig. 2 . The diagram for the receiving signal and despreading procedure depending on time delay is shown in Fig. 4 .", null, "PPT Slide\nLager Image", null, "PPT Slide\nLager Image\nIn Eq. (10), the interference component is the interference due to the cross-correlation between the data ZC code and pilot ZC code, and is about 38 dB which is insignificant compared to the auto-correlation peak value as shown in Fig. 5 . When the pilot channel code delay q and receiving delay τ k match exactly, every exponential term becomes 0 and the correlation result is expected to have the peak value. When they do not match exactly, it has a linear phase value with respect to n, and the correlation result is expected to be close to 0 depending on the ZC code characteristic.", null, "PPT Slide\nLager Image\nInterference between data and pilot.\nTherefore, in the receiver, the satellite code could be obtained from the sign of the correlation value of pilot signal. For the data signal, the extraction of data sign would also be possible by obtaining the correlation value, similar to the pilot signal.\n4. SIMULATION AND RESULT\nFor testing the signal scheme proposed in Chapters 2 and 3, by applying the parameters of the environment without noise and Doppler shift assumed in Section 4.1, it is shown, in Sections 4.2 and 4.3, that the satellite could be distinguished and the navigation information could be obtained through the pilot signal and data correlation value. In Section 4.4, the signal power spectrum is analyzed, and in Section 4.5, the applicability of satellite navigation signal scheme is examined by comparing with the signal acquisition performance of GPS system that is currently in operation. In Section 4.6, the complexity of the proposed system is compared with that of GPS.\n- 4.1 Simulation environment and parameter\nFor testing the sign extraction through the signal scheme proposed in Chapter 2 and the correlation value mentioned in Chapter 3, four visible satellites, which are necessary for the navigation of actual receiver, are assumed. Table 1 shows the parameters which apply the receiving signal assumed in Chapter 3 to simulation based on Eq. (8).", null, "PPT Slide\nLager Image\nThe number of satellites (N s ) was 4, and the time delay τ k for each satellite was made to maintain regular interval. Also, for applying the signal attenuation depending on time delay, the signal power was applied so that there is a difference of 3dB relative to S1. For the identical comparison in the future, the pilot signal of 1023 which is the C/A code length of GPS was used, and for the transfer rate of 50 bps, the data length of 1023 x 20 was used.\nFig. 5 shows the analysis of the mutual interference between pilot and data which follow the previously established parameters. When a satellite sends the identification information and data information as shown in Fig. 3 , considerably small interference is observed though they have different lengths.\n- 4.2 Correlation with pilot signal to distinguish satellite\nFig. 6 is a diagram which magnified two periods of pilot signal among the correlation value results. Beginning with the S1 signal which first arrives at the receiver, the correlation value result for each satellite shows gradually decreasing peak value with the delay distance specified in Table 1 . Also, the peak values of this normalized correlation value have the attenuation ratio of about 3 dB. The difference of 3 dB which was set in Table 1 is not precisely observed, which is thought to be the result of self-interference and differential interference. Though there is attenuation, the PRN code of satellite could be extracted for every sample of N p if the threshold value is exceeded.", null, "PPT Slide\nLager Image\nFig. 7 only shows the real number value of the sample which corresponds to peak value from the result of Fig. 6 . The PRN code assigned to each satellite can be identified. Similar to the previous result, the correlation value differs depending on the attenuation ratio which was initially set, and the effect of differential interference is also observed. The PRN code of every satellite, from which the signal is currently received, could be obtained through a single correlation value calculation that corresponds to the length of N p .", null, "PPT Slide\nLager Image\nExtracting PRN code from correlation using pilot signal.\nFig. 8 shows the correlation value between the PRN code extracted from the correlation value of receiving signal for searching the assigned satellite and the PRN code generated inside the receiver. It is found that the result follows the auto-correlation characteristic of PRN code. The PRN code is obtained for every ZC code period based on the peak value shown in Fig. 6 . Thus, if the time delay difference of even a single chip exists between satellite signals, the advantage is that there exists no cross-correlation interference with the PRN code assigned to another satellite.", null, "PPT Slide\nLager Image\nPRN code correlation.\n- 4.3 Correlation with data signal\nThe correlation value was taken as full length for receiving the data transmitted from each satellite, and the result is shown in Fig. 9 . When acquiring the data, the interference from the pilot signal is about -29 dB, which is very small. After the signal acquisition, a stable signal tracking seems to be possible if the frequency synchronization and time synchronization are to some degree stabilized.", null, "PPT Slide\nLager Image\nCorrelation for data acquisition.\n- 4.4 Power spectral density of C/A and Zadoff-Chu code\nThe C/A code has a sinc function shape in the frequency domain because it is the modulation system of Binary-Phase-Shift-Keying, and the ZC code has a constant value in the time and frequency domains as shown in Fig. 10 . The bandwidth of main lobe for the C/A code is 2 MHz, while the ZC code has the main lobe of 1MHz. Thus, when the noise power to bandwidth is identical, the effect of noise for the ZC code is expected to be half of the effect of noise for the C/A code.", null, "PPT Slide\nLager Image\nPSD of CA and Zadoff-Chu code.\n- 4.5 Acquisition performance under noise and Doppler shift environment\nTo consider the actual operation environment, in the environment with noise and Doppler shift, the code acquisition performances for the C/A code of GPS that is currently in operation and the proposed ZC code were analyzed at the same threshold value based on 1023 chips, and the result is shown in Fig. 11 . Also, it was assumed that 10 satellite signals of different phases are received as the same gain. In Fig. 10 , the ZC code was expected to have 3dB less noise effect than the C/A code, but the actual performance difference is insignificant because it is the signal scheme which combines the pilot signal that is 3 dB less and the data signal. However, regarding the interference between the satellite codes, the ZC code has much less interference than the C/A code. Thus, the proposed signal scheme has the gain of about 0.1 dB. Also, regarding the Doppler shift environment, the two codes showed the performance degradation of about 1dB when the Doppler frequency shift is 250 Hz. From the above results, it is shown that the performance of ZC code-based transmitting signal is not inferior to that of C/A code-based signal with respect to noise and Doppler frequency, and that more visible satellites are advantageous in term of code interference.", null, "PPT Slide\nLager Image\nSignal acquisition performance.\n- 4.6 Complexity analysis\nIn the previous sections, the applicability of signal scheme was investigated which was proposed by analyzing the code acquisition for distinguishing satellites from the signal scheme modeled in Chapter 2, the data acquisition for obtaining the satellite navigation information, and the code acquisition performance under noise and Doppler shift environment. In this section, the improvement in the amount of calculation is quantitatively analyzed which can be achieved by the proposed signal scheme when acquiring the data. For this purpose, the number of cells at the code acquisition space is compared between the GPS system that is currently in operation and the proposed system.\nFor the Doppler frequency axis, if it searches the 5 kHz range by the 500 Hz, total 21 intervals are generated. For the code delay axis, if it searches by the half chip for preventing the loss of correlation value, 2,046 intervals are generated, and for the satellite code axis, if it has the intervals as many as the number of satellites (N s ), the GPS has the number of cells of N cell = 21 x 2,046 x N s ( Petovello 2011 ). The proposed system does not have the search for satellite code axis, and only has the number of cells of N cell = 21 x 2046. Each cell has 1023 complex number multiplications and 1022 complex number summations. When every cell is searched for acquiring the signal, the amount of calculation needed is shown in Table 2 and Fig. 12 .\nComparing amount of calculation.", null, "PPT Slide\nLager Image\nComparing amount of calculation.", null, "PPT Slide\nLager Image\nTotal calculation to acquire signal.\nIn the case of the proposed system, the calculation of PRN code correlation value for distinguishing satellites is added, but the amount of calculation is insignificant when compared to the GPS. As the dependence on the number of satellites is markedly different, the proposed system will be more efficient when the number of satellites to be searched is large.\n5. CONCLUSION\nIn this paper, a signal scheme was described which could simplify the cell search of satellite code axis for reducing the complexity when acquiring the satellite signal, and the actual receiving signal was modeled. Using the correlation characteristic of ZC code, by adding the pilot signal which performed the mapping of PRN code for the existing data signal, the PRN code of every satellite, from which the signal is currently received, could be obtained through a single correlation value calculation. The data signal acquisition was also found to be achievable, and the advantage in the code interference was demonstrated by the analysis of code acquisition performance considering the actual environment. Through this, it was verified whether the proposed signal scheme could have the performance of a typical GNSS.\nFor a proper complexity comparison, the parameters which are identical to the GPS system were established, and the amount of calculation was compared and analyzed. As a result, the proposed satellite navigation system could lower the dependence of complexity depending on number of satellites, from the number of cells that the Doppler frequency and code delay axis comprise to the PRN code length, and the improvement of complexity was presented as table and figure.\nHowever, when a signal, without time delay for each satellite signal where the synchronization matches exactly, is received, there could be an interference between the PRN codes. Therefore, the characteristic of PRN code, which is the secondary code for distinguishing satellites, should also be considered in the future.\nAcknowledgements\nThis research was supported by Agency for Defense Development (ADD) of Korea.\nBIO", null, "Dae-Soon Park received the Bachelor of degree in Information Communication from Chung-nam National University in 2012. His research interests include satellite navigation and GNSS signal processing and Next-Generation mobile communication.", null, "Jeong-been Kim received the Master’s degree in Information Communication from Chung-nam National University in 2007. His research interests include satellite navigation and GNSS signal processing.", null, "Je-won Lee received the Master’s degree in Information Communication from Chung-nam National University in 2013. His research interests include satellite navigation and GNSS signal processing.", null, "Kap-Jin Kim received the Masterr’s. degree in Control & Instrumentation from Han-Yang University in 1997. His research interests include satellite navigation and GNSS signal processing.", null, "Kiwon Song received the Doctor’s degree in Electronics from Chung-nam National University in 2002. His research interests include satellite navigation and GNSS signal processing.", null, "Jae Min Ahn received the Doctor’s degree in Electrical and Electronic from KAIST in 1994. His research interests include Physical session of Next-Generation mobile communication and Radio Resource management.\nReferences" ]

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[ "## Xarray set_index example\n\nXarray for multidimensional gridded data¶In last week's lecture, we saw how Pandas provided a way to keep track of additional \"metadata\" surrounding tabular datasets, including \"indexes\" for each row and labels for each column. These features, together with Pandas' many useful routines for all kinds of data munging and analysis, have made Pandas one of the most popular python packages in the pandas.DataFrame.to_xarray¶ DataFrame.to_xarray (self) [source] ¶ Return an xarray object from the pandas object. Returns xarray.DataArray or xarray.Dataset. Data in the pandas structure converted to Dataset if the object is a DataFrame, or a DataArray if the object is a Series.\n\nTo add on to Joe's comment, definitely take a look at the working with pandas section of the xarray docs to see if that helps. If you can set the appropriate pandas.MultiIndex for your data, converting to xarray is usually quite easy. – shoyer Oct 17 '17 at 23:59 xarray with MetPy Tutorial¶. xarray is a powerful Python package that provides N-dimensional labeled arrays and datasets following the Common Data Model. While the process of integrating xarray features into MetPy is ongoing, this tutorial demonstrates how xarray can be used within the current version of MetPy. #Other Examples of Python Set Index Python is an extraordinary language for doing data analysis, primarily because of the fantastic ecosystem of data-centric python packages. Python’s Pandas is one of those packages and makes importing and analyzing data much more comfortable. Keyword arguments with names matching dimensions and values given by lists representing new level orders. Every given dimension must have a multi-index.\n\n## #Other Examples of Python Set Index Python is an extraordinary language for doing data analysis, primarily because of the fantastic ecosystem of data-centric python packages. Python’s Pandas is one of those packages and makes importing and analyzing data much more comfortable.\n\n26 Dec 2019 import xarray as xr import pandas as pd dfcsv=pd.read_csv('sample.csv') dfcsv= dfcsv.set_index(['latitude','longitude','zgodina'])  For example, if your dataset is sorted by time, you can quickly select data for a You can set an index column using the .set_index(column_name) method. Helper object to manage the feature layer, update it's definition, etc For example, if the distance is 100, the query geometry is a point, units is set to meters, and all points within 100 meters df.set_index([\"Person\", \"Single\"]).count (level=\"Person\") Age Person John 2 Myla 1 Return an xarray object from the pandas object. DataFrame.sample([n, frac, replace, …]) 返回随机抽样 axis]) Return data corresponding to axis labels matching criteria DataFrame.set_index(keys[, drop, append, …]) DataFrame.to_xarray() Return an xarray object from the pandas object. In this example, First Name column has been made the index column of Data Frame. filter_none. edit close. play_arrow. link brightness_4 code  Note that set_index() method does not modify the original DataFrame, but returns the DataFrame with the column set as index. Example 1: Set Column as Index in\n\n### xarray.Dataset¶ class xarray.Dataset (data_vars=None, coords=None, attrs=None) ¶ A multi-dimensional, in memory, array database. A dataset resembles an in-memory representation of a NetCDF file, and consists of variables, coordinates and attributes which together form a self describing dataset.\n\nNote that set_index() method does not modify the original DataFrame, but returns the DataFrame with the column set as index. Example 1: Set Column as Index in   xarray.DataArray.set_index¶ DataArray.set_index (self, indexes: Mapping[Hashable, Union[Hashable, Sequence[Hashable]]] = None, append: bool = False, inplace: bool\n\n### xarray supports four kind of indexing. Since we have assigned coordinate labels to the x dimension we can use label-based indexing along that dimension just like pandas. The four examples below all yield the same result (the value at x=10) but at varying levels of convenience and intuitiveness.\n\n16 Nov 2017 I use xarray 0.9.6 for both examples below. With pandas 0.20.3, Dataset. set_index gives me what I expect (i.e., the grid__x data variable  28 Jan 2017 Let's say I want to use xarray to work on some daily weather data, organized in a way that is similar to this example from xarray's documentation  3 Oct 2017 For example, we might imagine that \"Indexes\" are no longer coordinates, but instead their own entry in the repr:   12 Feb 2018 xr.concat(res_da, dim='prop').set_index(prop=['temp', 'pre', 'zon']).unstack('prop'). which results in the 5D array that you desire: >> df = pd\n\n## I use xarray 0.9.6 for both examples below. With pandas 0.20.3, Dataset.set_index gives me what I expect (i.e., the grid__x data variable becomes a coordinate x):\n\nDataFrame.sample([n, frac, replace, …]) 返回随机抽样 axis]) Return data corresponding to axis labels matching criteria DataFrame.set_index(keys[, drop, append, …]) DataFrame.to_xarray() Return an xarray object from the pandas object. In this example, First Name column has been made the index column of Data Frame. filter_none. edit close. play_arrow. link brightness_4 code  Note that set_index() method does not modify the original DataFrame, but returns the DataFrame with the column set as index. Example 1: Set Column as Index in\n\nxarray.DataArray.set_index¶ DataArray.set_index (self, indexes: Mapping[Hashable, Union[Hashable, Sequence[Hashable]]] = None, append: bool = False, inplace: bool xarray.Dataset.set_index¶. Set Dataset (multi-)indexes using one or more existing coordinates or variables. Mapping from names matching dimensions and values given by (lists of) the names of existing coordinates or variables to set as new (multi-)index. If True, append the supplied index(es) to the existing index(es). Unlike pandas, xarray does not guess whether you provide index levels or dimensions when using loc in some ambiguous cases. For example, for mda.loc[{'one': 'a', 'two': 0}] and mda.loc['a', 0] xarray always interprets (‘one Using apply_ufunc¶. Applying unvectorized functions with apply_ufunc. Load data; No errors are raised so our interpolation is working. I use xarray 0.9.6 for both examples below. With pandas 0.20.3, Dataset.set_index gives me what I expect (i.e., the grid__x data variable becomes a coordinate x):" ]

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[ "### How much is 66 milligrams?\n\nConvert 66 milligrams. How much does 66 milligrams weigh? What is 66 milligrams in other units? How big is 66 milligrams? Convert 66 milligrams to lbs, kg, mg, oz, grams, and stone. To calculate, enter your desired inputs, then click calculate. Some units are rounded.\n\n### Summary\n\nConvert 66 milligrams to lbs, kg, mg, oz, grams, and stone.\n\n#### 66 milligrams to Other Units\n\n 66 milligrams equals 0.066 grams 66 milligrams equals 6.6E-5 kg 66 milligrams equals 66 mg\n 66 milligrams equals 0.002328083388 oz 66 milligrams equals 0.0001455052117 lbs 66 milligrams equals 1.039322614E-5 stone" ]

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[ "# Lesson Video: Work Done by a Force Expressed in Vector Notation Mathematics\n\nIn this video, we will learn how to calculate the work done by a constant force vector acting on a body over a displacement vector using the dot product.\n\n13:35\n\n### Video Transcript\n\nIn this lesson, we’ll learn how to calculate the work done by a constant force vector acting on a body over a displacement vector using the dot product.\n\nRemember, work is done when a force causes a body to move over some distance. And in its most simplest form, it can be calculated by finding the product of those two measures, force times displacement. When the force 𝐹 is constant and the angle between the force and the displacement 𝑑 is 𝜃, then work done is alternatively given by 𝐹 times 𝑑 cos 𝜃. And whilst work is a scalar quantity, we can alternatively represent the force and displacement as vectors rather than as the magnitude of vectors.\n\nThen the work done by a force vector 𝐅 with displacement vector 𝐝 is the dot product of the force vector and the displacement vector. This can be alternatively calculated using the magnitude of the force vector times the magnitude of the displacement vector times cos 𝜃. And of course since the dot product or scalar product of two vectors is a scalar quantity, then this definition holds with our understanding of how work is represented. So, given these definitions, we’ll consider a demonstration of how to calculate work done.\n\nA particle moves in a plane in which 𝐢 and 𝐣 are perpendicular unit vectors. A force 𝐅 equals nine 𝐢 plus 𝐣 newtons acts on the particle. The particle moves from the origin to the point with position vector negative nine 𝐢 plus six 𝐣 meters. Find the work done by the force.\n\nRemember, the work done by a force vector 𝐅 with displacement vector 𝐝 is the dot product of 𝐅 and 𝐝. The force acting on our particle is nine 𝐢 plus 𝐣 newtons. But what is its displacement? Well, we’re told it moves from the origin to the point with position vector — let’s call that 𝐩 — negative nine 𝐢 plus six 𝐣. This means that the displacement is simply the vector 𝐨𝐩. That’s negative nine 𝐢 plus six 𝐣 minus zero 𝐢 plus zero 𝐣. And of course, we simply subtract the components. In this case, of course, we’re subtracting the vector zero 𝐢 plus zero 𝐣. So, our original vector remains unchanged. And so the displacement, which we can give in meters, is negative nine 𝐢 plus six 𝐣. So, work done is the dot product or the scalar product of 𝐅 and 𝐝. That’s nine 𝐢 plus 𝐣 dot negative nine 𝐢 plus six 𝐣.\n\nThen, of course, we find the scalar product by finding the sum of the products of the components. So, we multiply the 𝐢-components, nine times negative nine, and then we add the product of the 𝐣-components. Well, that’s one times six. That’s negative 81 plus six, which is, of course, negative 75. Since the force is in newtons and the displacement is in meters, the work done is going to be in joules. Now, we notice that the work done is negative. Well, if the energy of the particle is conserved, then the kinetic energy of the particle will decrease. If the energy of the particle is not conserved, the work done might instead increase the potential energy of the particle. Either way, the work done by the force in this question is negative 75 joules.\n\nIn this question, we considered a single force acting upon a body. But there will be times where multiple forces are acting on the same body. In these cases, we’ll consider the resultant of these forces. And that will help us to calculate the work done. So, in our second example, we’re going to demonstrate what that looks like.\n\nA body moves in a plane in which 𝐢 and 𝐣 are perpendicular unit vectors. Two forces 𝐅 sub one equals nine 𝐢 minus two 𝐣 newtons and 𝐅 sub two equals nine 𝐢 minus seven 𝐣 newtons act on the body. The particle moves from the point with position vector negative six 𝐢 plus two 𝐣 meters to the point two 𝐢 plus three 𝐣 meters. Find the work done by the resultant of the forces.\n\nWe begin by recalling that we can find the work done by a force vector 𝐅 over some displacement vector 𝐝 by finding the dot product of these two vectors. We do need to be a bit careful here though, because we’re working with two forces and we want to find the work done by their resultant. Now, of course, the resultant of two forces is simply their sum. So, the resultant force 𝐅 is the sum of 𝐅 sub one and 𝐅 sub two. In this case, that’s nine 𝐢 minus two 𝐣 plus nine 𝐢 minus seven 𝐣. Now, of course, we simply add the components, so we’ll add the 𝐢-components first. Nine plus nine is 18. Then, we add the 𝐣-components. Negative two minus seven is negative nine. And so, the resultant of the forces in this case is 18𝐢 minus nine 𝐣 newtons.\n\nNow, we need to find the displacement. Now, of course, the particle moves from the point with position vector negative six 𝐢 plus two 𝐣 to the point two 𝐢 plus three 𝐣. Let’s call the first point 𝐴 and the second point 𝐵. Then, the displacement is the vector 𝐀𝐁. Well, that’s the difference between vector 𝐎𝐁 and 𝐎𝐀. So, it’s two 𝐢 plus three 𝐣 minus negative six 𝐢 plus two 𝐣. As before, we simply subtract the individual components, beginning with the 𝐢-components. Two minus negative six is eight, and three minus two is one. So, we find the displacement of the body is eight 𝐢 plus 𝐣. And of course, that’s in meters.\n\nAnd we have everything we need. The work done is simply the dot product or the scalar product of these two vectors. That’s 18 times eight plus negative nine times one, which is 144 minus nine, or 135. The work done by the resultant of the two forces then is 135 joules.\n\nIn our next example, we’ll have a look at how to use our understanding of the features of vectors to find work done by a force.\n\nA particle moved from point 𝐴 seven, negative three to point 𝐵 negative nine, two along a straight line under the action of a force 𝐅 of magnitude eight root 10 newtons acting in the same direction as the vector 𝐜 equals negative three 𝐢 minus 𝐣. Calculate the work done by the force, given that the magnitude of the displacement is measured in meters.\n\nRemember, the work done by a force of vector 𝐅 over some displacement vector 𝐝 is the dot product or scalar product of those two vectors. In this question, we’re told that the particle moves from point 𝐴 to point 𝐵, so we can calculate the displacement of the particle fairly easily. In vector form, we’ll say that the displacement 𝐝 is the vector 𝐀𝐁, and that’s the vector 𝐎𝐁 minus the vector 𝐎𝐀. Now, since we know the position from the origin, that’s the point zero, zero, then the vector 𝐎𝐁 is simply negative nine 𝐢 plus two 𝐣 and the vector 𝐎𝐀 is seven 𝐢 minus three 𝐣. So, the vector 𝐝 is the difference of these, and we can find that by subtracting the individual components. Negative nine minus seven is negative 16, and two minus negative three is five 𝐣. So, we have the displacement vector. It’s negative 16𝐢 plus five 𝐣.\n\nHow do we calculate the force though? We know it has a magnitude of eight root 10, but we don’t know its vector. We do, however, know it acts in the same direction as this vector here, negative three 𝐢 minus 𝐣. So, let’s sketch this out. Vector 𝐜 looks a little something like this. For every three units left, we move one unit down. Vector 𝐅 acts in the same direction as this but has a magnitude of eight root 10. Once again, for every three units left, we still move one unit down. So, we can say that if we define 𝐅 sub 𝑖 as being the 𝐢-component of our force and 𝐅 sub 𝑗 as being the 𝐣-component, then 𝐅 sub 𝑖 is equal to three times 𝐅 sub 𝑗.\n\nAnd, of course, since the 𝐢-components and 𝐣-components of the vectors are perpendicular to one another, we can form a right triangle and apply the Pythagorean theorem to find the value of 𝐅 sub 𝑖 and 𝐅 sub 𝑗. That is, the magnitude of 𝐅 sub 𝑖 squared plus the magnitude of 𝐅 sub 𝑗 squared is equal to eight root 10 all squared. Now, of course, 𝐅 sub 𝑖 is three times 𝐅 sub 𝑗. So, we can replace the magnitude of 𝐅 sub 𝑖 with three times the magnitude of 𝐅 sub 𝑗. Similarly, eight root 10 all squared is 640. Three squared is nine, so we have a total of 10 lots of 𝐅 sub 𝑗 squared. Then, we divide through by 10. And finally, we’ll take the square root of both sides. Now, we only need to take the positive square root of 64 because we’re looking at the magnitude, which is simply a length. And so, the magnitude of 𝐅 sub 𝑗 is equal to eight. Since the magnitude of 𝐅 sub 𝑖 is three times the magnitude of 𝐅 sub 𝑗, it’s three times eight, which is equal to 24. We know the direction in which our force is acting. So we can define the vector force to be negative 24𝐢 minus eight 𝐣 newtons.\n\nLet’s clear some space and we’re ready to take the dot product of our force and our displacement. The work done is the dot or scalar product of negative 24𝐢 minus eight 𝐣 and negative 16𝐢 plus five 𝐣. That’s negative 24 times negative 16 plus negative eight times five, and that’s equal to 344. The work done is in joules. So, we find the work done by our force is 344 joules.\n\nIn our previous examples, we’ve looked at finding the work done when both displacement and force are vectors, where each component is simply a number. In our final example, we’ll demonstrate how work done can also be calculated when the vectors for displacement and/or force are given as expressions in terms of time.\n\nA particle moves in a plane in which 𝐢 and 𝐣 are perpendicular unit vectors. Its displacement from the origin at time 𝑡 seconds is given by 𝐫 equals two 𝑡 squared plus seven 𝐢 plus 𝑡 plus seven 𝐣 meters. And it is acted on by a force 𝐅 equals six 𝐢 plus three 𝐣 newtons. How much work does the force do between 𝑡 equals two seconds and 𝑡 equals three seconds?\n\nSince work done is the dot product of the vector force and the vector for displacement, we’re going to need to calculate the displacement of this particle. We’re given its displacement at some time 𝑡. But we’re also asked how much work the force does between 𝑡 equals two seconds and 𝑡 equals three seconds. So, let’s substitute 𝑡 equals two and 𝑡 equals three into our expressions for the displacement. Beginning with 𝑡 equals three seconds, we get two times three squared plus seven 𝐢 plus three plus seven 𝐣. That’s 25𝐢 plus 10𝐣. And we’re working in meters. So the displacement from the origin and, hence, the position of the particle at three seconds is 25𝐢 plus 10𝐣 meters. In a similar way, we can find the position of the particle at two seconds by substituting 𝑡 equals two into our expression. And we get 15𝐢 plus nine 𝐣 meters.\n\nThe displacement then of the particle between these two times is the difference here. It’s 𝐝 sub three minus 𝐝 sub two. That’s 25𝐢 plus 10𝐣 minus 15𝐢 plus nine 𝐣. And of course, we simply subtract the individual components. 25 minus 15 is 10, and 10 minus nine is one. So, the displacement is simply 10𝐢 plus 𝐣, and that’s in meters. Since the work done is the dot product of the force and displacement, we need to find the dot product of six 𝐢 plus three 𝐣 and the displacement we just calculated. It’s the dot or scalar product of six 𝐢 plus three 𝐣 and 10𝐢 plus 𝐣. That’s six times 10 plus three times one, which is, of course, equal to 63. The work done then is equal to 63 joules.\n\nWe’ll now recap the key points from this lesson. In this lesson, we learned the work done by a constant force vector 𝐅 over a displacement vector 𝐝 is equal to the dot product of 𝐅 and 𝐝. Alternatively, if 𝜃 is the angle between 𝐅 and 𝐝, it’s the magnitude of the force times the magnitude of the displacement times cos 𝜃. And if we express 𝐅 and 𝐝 in component form, we don’t actually need to use this second form. We can simplify the dot product to find the work done." ]

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[ "Hello there! Welcome back to The Mechanical post.\nToday we will discuss the part 2 of what is manometer and its types, along with its advantages and disadvantages. In case if you haven't read the part 1 check it out here.\n\n## Differential Manometer :\n\nDifferential Manometer is more of a \" pressure comparing \" device rather than a \" pressure measuring \" device. Let me explain by defining differential manometer.\n\nDifferential manometer is a type of pressure difference measuring device used for measuring pressure difference between 2 different pipes or 2 points on the same pipe.\n\nDifferential U-tube manometer can be classified into 2 types.\n• U-tube differential manometer.\n• Inverted U-tube differential manometer.\n\n## U-tube differential manometer.\n\nU-tube differential manometer consists of a glass tube bent into U shape. The 2 ends of the U-tube are connected to the points whose pressure is to be measured.\n\nIn the U-tube a manometric liquid is filled ( indicated by the shaded portion in the diagram ). This liquid has specific gravity higher than the liquids present in the pipe. Most of the time mercury is used as a manometric liquid as it has desirable characteristics like high specific gravity, clear visible, it does not stick to glass, can be used at wide range of temperature etc.\n\nWhen the pressurized liquid flows through the pipes, it enters the U-tube manometer. Now depending upon the respective pressure in the pipes, there are 3 possibility.\n\nThe manometer fluid in both right and left limb would be equal. This indicates that the pressure in both the pipes would be the same i.e. a = b.\n\nThe manometric fluid level in the left limb is low whereas the manometric level in the right level is high. This indicates that the pressure in the left hand pipe is greater i.e a  > b.\n\nThe manometric fluid level in the right limb is low whereas the manometric level in the left limb is high. This indicates that the pressure in the right hand pipe is greater i.e a <  b.\n\n#### Pressure Calculations :\n\nConsider the manometer as shown in the figure. Assume that the pressure at point 'a' is greater than point 'b', then the greater pressure at 'a' will force the heavy liquid in U-tube to move downwards, so that heavy liquid will rise in the right limb.\nLet,\nh = Difference of levels of heavy liquid in right limb and left limb.\ns1 = Specific gravity of liquid in the pipes.\ns2 = Specific gravity of heavy liquid.\n\nSince the pressure in the right limb and the left limb above the datum line AB are equal, therefore the difference of the pressures in two points a and b is\n(ha - hb) = h (s2 - s1) m head of water.\n\nSimilarly, when points a and b are at different levels, as shown in the figure, The pressure in the right limb and the left limb above datum are equal.\n\nLet,\nh1 and h3 = Heights of the liquid in the left and right limb.\nh2 = Difference of heavy liquid in left and right limb.\nha and hb = Pressure in pipes a and b respectively.\ns1 and s3 = Specific gravity of liquids in pipes a and b respectively.\ns2 = Specific gravity of heavy liquid.\n\nThen the total pressure in left limb above datum = Total pressure in right limb above datum.\n\n( ha + s1*h1 ) = ( s2*h2 + s3*h3 + hb)\n\nTherefore, ( ha - hb ) = ( s2*h2 + s3* h3 - s1*h1 )\n\n#### You might also like to read:\n\nCentrifugal pump - its construction, working & applications | The Mechanical post.\n\n### Inverted U-tube Differential Manometer:\n\nIt consists of an inverted U-tube containing a light liquid whose specific gravity is less than the specific gravity of the pipe liquid. Manometric liquid such as oil is used mostly. It is used for measuring the difference of low pressure between two points; where accuracy is important.\n\nThe two ends of U-tube are connected to the point whose difference in pressure is to be measured as shown in the figure. Light liquid is used as manometric liquid; because it may be down and flow in low pressure.\n\nLet,\nh1 = Height of liquid in left limb below datum AB.\nh2 = Difference of levels of the light liquid in right limb and left limb.\nh3 = Height of liquid in the right limb below datum AB.\nha = Pressure head in a pipe a.\nhb = Pressure head in pipe b.\ns2 = Specific gravity of light liquid.\ns1, s3 = Specific gravity of liquid in left limb and right limb respectively.\n\nSince the pressure in the left limb and the right limb are equal,\n\nTotal Pressure in left limb below datum = Total Pressure in right limb below datum.\n\n(ha - s1*h1) = hb - s2*h2 - s3*h3\n\n( ha - hb ) = ( s1*h1 - s2*h2 - s3*h3 )\n\n### Applications of Differential manometer:\n\n• It can detect leaks in a pipeline as leakage would cause pressure imbalance, thus imbalancing the manometric fluid.\n• Differential gauges can detect scale formation and blockages in pipeline using pressure pressure difference.", null, "Image credits -  differentialpresssure.com\n• Liquid level in a container can also be measured in a container as shown above.\n• Differential manometer is used in laboratory, for calculating flow rates and velocities of different fluids.\n• It is also used in  industries to measure differential pressures of low velocity fluids.\n\n### Advantages of Differential manometer :\n\n• Simple construction\n• Cost effective\n• Easy maintainance\n• Easy to replace" ]

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[ "", null, "# Vedic Maths for Class 9\n\nStudents in Grade 9 need to be competitive in Maths and therefore it is imperative to develop their math speed and accuracy. Here Vedic Maths for Class 9 has speed-building methods to save their time in math calculations. Vedic Maths learnings have been given by Shri Bharti Krishna Tirathji who is also called as Father of Vedic Maths.\n\nClass 9 students have generally the following syllabus in their schools:-\n\n• Number System\n• Algebra\n• Coordinate Geometry\n• Geometry\n• Trigonometry\n• Mensuration\n• Statistics and Probability\n\nPreparing for exams is requires continuous practice. Preparing well for maths in grade 9 builds a strong foundation for maths in grade 10, grade 11, and grade 12. Which fields you would be selecting in the future – Medical, Non-Medical or Commerce would also depend on your marks scored in 9th,10th, 11th, and 12th.\n\nTherefore, it is important to achieve perfection in the maths syllabus of 9th to understand better the syllabus of grade 10. The above maths concepts in the syllabus of class 9  can be done with great ease with the help of faster calculations techniques used in Vedic Maths.  Knowing the speed-building tricks of Vedic Maths in addition, subtraction, multiplication, division, squares, square roots, cube and cube roots, and higher calculations, one is better prepared for math exams and competitions.\n\nVedic Maths For Class 9 enhances the calculations speed and accuracy in maths. The questions based on the above syllabus can be done faster and time can be saved in solving those questions if we know the Vedic Maths Tricks.\n\nAt Winaum Learning, students of class 9 are well trained to know the Vedic Math Tricks. .The sessions are taken by our trained faculty who are Vedic Maths Experts and Certified Teachers.\n\nAdvantages of Learning Vedic Maths  from Winaum Learning are as follows\n\n• Personalized Attention to every child with classes in small batches.\n• Sessions are taken by our trained teachers.\n• Sessions are online and completely interactive.\n• Students Learn from their home online and hence save time by visiting offline Vedic math centers\n\nThe benefits of Vedic Maths go much beyond the speed and accuracy in maths as it builds a strong math foundation for kids. Winaum Learning provides online classes for students of grade 9 for Vedic Maths, Maths Tutoring, English Grammar, Science Classes, and Handwriting Classes.\n\n## What you Learn in Vedic Maths for Class 9\n\n### Vedic Maths Tricks\n\nDevelop Math Speed and Accuracy\n\n### School Maths Calculation\n\nEnhance your Calculation Methods of Conventional Learning in Schools\n\n### Word Problems\n\nPractice Word Problems\n\n### Maths Games/Activities\n\nMaths Activities , Puzzles and Quizzes build interest even more\n\n## Locations", null, "## Benefits of Vedic Maths for Class 9\n\n1000\n+ Students\n0\n+ Centers and Schools\n\n## FAQs for Class 9 Vedic Maths\n\nVedic Maths is the fastest and the easiest way to learn all the arithmetical calculations. We have four levels of Vedic Maths and each level goes for 2 months. We have designed our program in such a manner that we complete simple maths first and then moving forward towards  Complex maths that makes the foundation of maths stronger for children. It is mainly based on mental calculations which also helps in logical thinking and improves a child’s concentration, side by side while practicing maths in such a creative and easy way it reduces maths phobia.\n\nVedic maths is helpful for all types of students be It senior or junior. Vedic math covers all types of arithmetical calculations and it increase the speed by 12 to 15 times than conventional maths calculations while ensuring to give accurate answers. Too many students suffer from  mathematics anxiety and Vedic maths is also very helpful in reducing maths phobia by practicing maths in the easiest way. It helps students in scoring higher marks in maths exams, achieve top ranks in maths competitions. It save time of students in math calculations.\n\nOur Vedic Maths program is divided into four parts. Accordingly, we have four levels of Vedic Maths and each level goes for 2 months so it becomes an 8 months course.\n\nWinaum Learning makes small batches for students to get the best of individual attention from the teacher. The maximum strength of a batch is 1:6 ie one teacher to six students.\n\nYes, all students of Class 9 who enroll for Vedic Maths go through the assessments.\n\nIt’s very simple just visit our website and request a free trial session. Our education counselor will get in touch with you asap and guide you over a phone call or during online session.\n\n## Vedic Math Syllabus for Class 9\n\nThe syllabus for Vedic Maths for Class 9 has Vedic Maths Tricks/Sutras on addition, subtraction, multiplication, divisions, square roots, squares, cube roots and cubes, and higher calculations. Find out the detailed Vedic Maths Curriculum for Class 9.\n\n## Why Winaum Learning\n\nAt Winaum Learning,  children get complete attention from the teacher for Vedic Maths, Math Tutions, Science, English Grammar, and Handwriting Improvement. The study material is carefully designed in keeping in mind that students are from different boards like CBSE, ICSE, and IB Boards. Students of Grade 10 at Winaum Learning are taught by Well-Trained and Certified Teachers. Children are helped with practice papers of the exams and previous years papers also. Students get personalized attention at a reasonable fee.  Winaum Learning has contributed to 12000+ students at 100+ locations in India and other countries like US, Canada, Dubai, Oman, and Australia.\n\nGrow with Math" ]

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[ "# Multiplication Theorem on Probability and proof\n\n#### alfred\n\n##### New member\nHi Everyone!\n\nI'm with Conditional Probability and I don't understan this theorem.\n\nTheorem:\nIf", null, "then", null, "Proof:\nAll the conditional probabilities are well defined, since", null, "We can rewrite the right site of the equality as follows", null, "Obviously we can simplify the terms through", null, "Can anyone say me how does the simplification work? And why it is so important to be sure that\n\nThank you =)\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nWelcome to MHB, alfred!", null, "The simplification is a consequence of how fractions are multiplied and simplified in general.\nConsider for instance:\n$$\\frac 3 4 \\cdot \\frac 4 5 \\cdot \\frac 5 6 = \\frac 3 {\\cancel 4} \\cdot \\frac {\\cancel 4} {\\cancel 5} \\cdot \\frac {\\cancel 5} 6 = \\frac 3 6$$\n\n#### alfred\n\n##### New member\nThanks! You were right", null, "" ]

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[ "", null, "# Electron in a Nanocrystal Modeled by a Quantum Particle in a Sphere\n\nInitializing live version", null, "Requires a Wolfram Notebook System\n\nInteract on desktop, mobile and cloud with the free Wolfram Player or other Wolfram Language products.\n\nThis Demonstration shows the quantum effects observed on a single electron trapped in a spherical nanoparticle (also called a \"quantum dot\"), modeled as a particle in a sphere. We obtain the relationships among quantum energy levels", null, ", the radius of the nanoparticle", null, ", and the distance of the electron from the center of the nanoparticle", null, "by solving the Schrödinger equation. For spherical symmetry, with", null, "and", null, ":\n\n[more]", null, "The solution is the wavefunction", null, ", shown on the upper left, with the allowed energy levels", null, "for", null, "(For", null, ", the solutions are spherical Bessel functions.)\n\nThe electron's probability density curve is given by the square of the wavefunction, determining the probability of finding the electron at a given radius", null, "from the center of the nanoparticle, as shown on the upper right.\n\nThe lower-left graph shows the probability density in three dimensions.\n\nAt the lower right is an energy level diagram for the electrons, showing the relative spacings of the", null, ".\n\n[less]\n\nContributed by: Yezhi Jin, Kyle Smola, Rahil Ukani (December 2016)\nAdditional contributions by: Eitan Geva (University of Michigan)\nOpen content licensed under CC BY-NC-SA\n\n## Details\n\nReference\n\n T. Kippeny, L. A. Swafford and S. J. Rosenthal, \"Semiconductor Nanocrystals: A Powerful Visual Aid for Introducing the Particle in a Box,\" Journal of Chemical Education, 79(9), 2002 pp. 1094–1100. doi:10.1021/ed079p1094.\n\nSubmission from the Compute-to-Learn course at the University of Michigan.\n\n## Snapshots", null, "", null, "", null, "## Permanent Citation\n\nYezhi Jin, Kyle Smola, Rahil Ukani\n\n Feedback (field required) Email (field required) Name Occupation Organization Note: Your message & contact information may be shared with the author of any specific Demonstration for which you give feedback. Send" ]

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[ "# Advanced Analysis (Math 417) Name: Exam 1 March 14, 2008", null, "```Advanced Analysis (Math 417)\nExam 1\nName:\nMarch 14, 2008\n1. Let f : R2 → R be given by\n 2\nx sin( x1 ) + y 2 sin( y1 )\n\n\n\nx2 sin( 1 )\nx\nf (x, y) =\ny 2 sin( y1 )\n\n\n\n0\nif\nif\nif\nif\nxy 6= 0;\nx 6= 0 and y = 0;\nx = 0 and y =\n6 0;\nx = 0 and y = 0.\n(a) Prove or disprove: f is differentiable at 0 = (0, 0).\n(b) Prove or disprove: f is of class C 1 on an open neighborhood of 0.\n2. Consider the following assertion:\nSuppose f is of class C 2 on an open subset of R2 containing 0, and that ∇f (0) = 0. Then f has either a\nlocal maximum or a local minimum at 0.\n(a) Disprove the assertion with a counterexample.\n(b) Add one more hypothesis to make the assertion true (no proof\nrequired).\n3. Let f : Rn → R be of class C k+1 on Rn , and a ∈ Rn . Suppose\n∂ α f (a) = 0 for all |α| = k + 1. Use the fact that\nX hα Z 1\n(1 − t)k ∂ α f (a + th) dt\nRa,k (h) = (k + 1)\nα! 0\n|α|=k+1\nto prove that\nRa,k (h)\n= 0.\nh→0 khkk+1\nlim\n```" ]

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[ "Cody\n\n# Problem 20. Summing digits\n\nSolution 1312868\n\nSubmitted on 24 Oct 2017 by Bunny Harlan\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\na = 1; b = 2; out = sumDigits(a); assert(isequal(out, b))\n\n2   Pass\na = 10; b = 7; out = sumDigits(a); assert(isequal(out, b))\n\n3   Pass\na = 16; b = 25; out = sumDigits(a); assert(isequal(out, b))" ]

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[ "# Compressing Neural Network Weights\n\n📘\n\nFor Neural Network Format Only\n\nThis page describes the API to compress the weights of a Core ML model that is of type neuralnetwork. For the mlprogram model type, see the API Overview.\n\nThe Core ML Tools package includes a utility to compress the weights of a Core ML neural network model. Weight compression reduces the space occupied by the model. However, the precision of the intermediate tensors and the compute precision of the ops are not altered.\n\nQuantization refers to the process of reducing the number of bits that represent a number. The lower the number of bits, more the chances of degrading the model accuracy. The loss in accuracy varies with the model.\n\nBy default, the Core ML Tools converter produces a model with weights in floating-point 32 bit (float 32) precision. The weights can be quantized to 16 bits, 8 bits, 7 bits, and so on down to 1 bit. The intermediate tensors are kept in float precision (float 32 or float 16 depending on execution unit), while the weights are dequantized at runtime to match the precision of the intermediate tensors. Quantizing from float 32 to float 16 provides up to 2x savings in storage and generally does not affect the model's accuracy.\n\nThe quantize_weights function handles all quantization modes and options:\n\nfrom coremltools.models.neural_network import quantization_utils\n\n# allowed values of nbits = 16, 8, 7, 6, ...., 1\nquantized_model = quantization_utils.quantize_weights(model, nbits)\n\n\nFor a full list of supported arguments, see quantize_weights in the API Reference for models.neural.network. The following examples demonstrate some of these arguments.\n\n## Quantize to Float 16 Weights\n\nQuantizing to float 16, which reduces by half the model's disk size, is the safest quantization option since it generally does not affect the model's accuracy:\n\nimport coremltools as ct\nfrom coremltools.models.neural_network import quantization_utils\n\nmodel_fp32 = ct.models.MLModel('model.mlmodel')\n\nmodel_fp16 = quantization_utils.quantize_weights(model_fp32, nbits=16)\n\n\n## Quantize to 1-8 Bits\n\nQuantizing to 8 bits reduces the disk size to one fourth of the float 32 model. However, it may affect model accuracy, so you should always test the model after quantization, using test data. Depending on the model type, you may be able to quantize to bits lower than 8 without losing accuracy.\n\n# quantize to 8 bit using linear mode\nmodel_8bit = quantize_weights(model_fp32, nbits=8)\n\n# quantize to 8 bit using LUT kmeans mode\nmodel_8bit = quantize_weights(model_fp32, nbits=8,\nquantization_mode=\"kmeans\")\n\n# quantize to 8 bit using linearsymmetric mode\nmodel_8bit = quantize_weights(model_fp32, nbits=8,\nquantization_mode=\"linear_symmetric\")\n\n\nWhen you set nbits to a value between 1 and 8, you can choose one of the following quantization modes:\n\n• linear: The default mode, which uses linear quantization for weights with a scale and bias term.\n• linear_symmetric: Symmetric quantization, with only a scale term.\n• kmeans_lut: Uses a k-means clustering algorithm to construct a lookup table (LUT) quantization of weights.\n\nTry these different algorithms with your model, as some may work better than others depending on the model type.\n\n## Quantization Options\n\nThe following options enable you to experiment with the quantization scheme so that you can find one that works best with your model.\n\n### Custom LUT Function\n\nBy default, the k-means algorithm is used to find the lookup table (LUT). However, you can provide a custom function to compute the LUT by setting quantization_mode = \"custom_lut \".\n\n### Control Which Layers are Quantized\n\nBy default, all the layers that have weight parameters are quantized. However, the model accuracy may be sensitive to certain layers, which shouldn't be quantized. You can choose to skip quantization for certain layers and experiment as follows:\n\n• Use the AdvancedQuantizedLayerSelector class, which lets you set simple properties such as layer types and weight count. For example:\n# Example: 8-bit symmetric linear quantization skipping bias,\n# batchnorm, depthwise-convolution, and convolution layers\n# with less than 4 channels or 4096 elements\n\nskip_layer_types=['batchnorm', 'bias', 'depthwiseConv'],\nminimum_conv_kernel_channels=4,\nminimum_conv_weight_count=4096\n)\n\nquantized_model = quantize_weights(model,\nnbits=8,\nquantization_mode='linear_symmetric',\nselector=selector)\n\n\nFor a list of all the layer types in the Core ML neural network model, see the NeuralNetworkLayer section in the Core ML Format reference for NeuralNetwork.\n\nFor finer control, you can write a custom rule to skip (or not skip) quantizing a layer by extending the QuantizedLayerSelector class:\n\n# Example : 8-bit linear quantization skipping the layer with name 'dense_2'\nfrom coremltools.models.neural_network.quantization_utils import QuantizedLayerSelector\n\nclass MyLayerSelector(QuantizedLayerSelector):\n\ndef __init__(self):\nsuper(MyLayerSelector, self).__init__()\n\ndef do_quantize(self, layer, **kwargs):\nret = super(MyLayerSelector, self).do_quantize(layer)\nif not ret or layer.name == 'dense_2':\nreturn True\n\nselector = MyLayerSelector()\nquantized_model = quantize_weights(\nmlmodel,\nnbits = 8,\nquantization_mode='linear',\nselector=selector\n)" ]

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[ "• Corpus ID: 10569090\n\n# Second-Order Stochastic Optimization for Machine Learning in Linear Time\n\n```@article{Agarwal2017SecondOrderSO,\ntitle={Second-Order Stochastic Optimization for Machine Learning in Linear Time},\nauthor={Naman Agarwal and Brian Bullins and Elad Hazan},\njournal={J. Mach. Learn. Res.},\nyear={2017},\nvolume={18},\npages={116:1-116:40}\n}```\n• Published 12 February 2016\n• Computer Science\n• J. Mach. Learn. Res.\nFirst-order stochastic methods are the state-of-the-art in large-scale machine learning optimization owing to efficient per-iteration complexity. Second-order methods, while able to provide faster convergence, have been much less explored due to the high cost of computing the second-order information. In this paper we develop second-order stochastic methods for optimization problems in machine learning that match the per-iteration cost of gradient based methods, and in certain settings improve…\n127 Citations\n\n### Stochastic second-order optimization for over-parameterized machine learning models\n\nWe consider stochastic second-order methods for minimizing smooth and stronglyconvex functions under an interpolation condition, which can be satisfied by overparameterized machine learning models.\n\n### Stochastic sub-sampled Newton method with variance reduction\n\n• Computer Science, Mathematics\nInt. J. Wavelets Multiresolution Inf. Process.\n• 2019\nStochastic optimization on large-scale machine learning problems has been developed dramatically since stochastic gradient methods with variance reduction technique were introduced. Several\n\n### Progressive Batching for Efficient Non-linear Least Squares\n\n• Computer Science\nACCV\n• 2020\nThis work presents an approach for non-linear least-squares that guarantees convergence while at the same time significantly reduces the required amount of computation, and shows that the proposed method achieves competitive convergence rates compared to traditional second-order approaches on common computer vision problems.\n\n### Oracle Complexity of Second-Order Methods for Finite-Sum Problems\n\n• Computer Science\nICML\n• 2017\nEvidence that the answer to can second-order information indeed be used to solve finite-sum optimization problems more efficiently is provided, at least in terms of worst-case guarantees is provided.\n\n### Approximate Newton Methods and Their Local Convergence\n\n• Computer Science\nICML\n• 2017\nThis paper proposes a unifying framework to analyze local convergence properties of second order methods and proposes a theoretical analysis that matches the performance in real applications.\n\n### Interpolation, growth conditions, and stochastic gradient descent\n\nThe notion of interpolation is extended to stochastic optimization problems with general, first-order oracles, and a simple extension to `2-regularized minimization is provided, which opens the path to proximal-gradient methods and non-smooth optimization under interpolation.\n\n### Finding Local Minima for Nonconvex Optimization in Linear Time\n\n• Computer Science\n• 2016\nA non-convex second-order optimization algorithm that is guaranteed to return an approximate local minimum in time which is linear in the input representation and applies to a very general class of optimization problems including training a neural network and many other non- Convex objectives arising in machine learning.\n\n### SPAN: A Stochastic Projected Approximate Newton Method\n\n• Computer Science\nAAAI\n• 2020\nThis paper proposes SPAN, a novel approximate and fast Newton method that computes the inverse of the Hessian matrix via low-rank approximation and stochastic Hessian-vector products and achieves a better trade-off between the convergence rate and the per-iteration efficiency.\n\n### Approximate Newton Methods\n\n• Computer Science\nJ. Mach. Learn. Res.\n• 2021\nThis paper proposes a unifying framework to analyze both local and global convergence properties of second order methods and presents the theoretical results which match the performance in real applications well.\n\n### Accelerated Stochastic Matrix Inversion: General Theory and Speeding up BFGS Rules for Faster Second-Order Optimization\n\n• Computer Science\nNeurIPS\n• 2018\nThis work develops the first accelerated (deterministic and stochastic) quasi-Newton updates, which lead to provably more aggressive approximations of the inverse Hessian, and lead to speed-ups over classical non-accelerated rules in numerical experiments." ]

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[ "## Stream: general\n\n### Topic: Dealing with heq and Sigma types\n\n####", null, "Patrick Lutz (Nov 11 2020 at 19:49):\n\nWhen proving that some type is equivalent to a Sigma type, how do you deal with heqs that pop up? Here's an example\n\nimport tactic\n\nvariables (A B : Type) (S : set A)\n\nexample : (A → B) ≃ Σ (f : S → B), {φ : A → B // ∀ (s : A) (h : s ∈ S), φ s = f ⟨s, h⟩ } :=\n{ to_fun := λ g, ⟨λ s, g s.1, ⟨g, λ _ _, rfl⟩⟩,\ninv_fun := λ ⟨f, φ⟩, φ.1,\nleft_inv := λ _, rfl,\nright_inv := begin\nrintros ⟨f, φ, h⟩,\next,\n{ cases x, dsimp only, tauto },\n{ -- heq occurs here\nsorry, }\nend }\n\n\nHow should the final sorry be proved? By the way, this example is by Kevin Buzzard. The actual, more complicated, situation that inspired this is here if you're curious.\n\n####", null, "Reid Barton (Nov 11 2020 at 19:58):\n\nTurn back and go another way\n\nexample : (A → B) ≃ Σ (f : S → B), {φ : A → B // ∀ (s : A) (h : s ∈ S), φ s = f ⟨s, h⟩ } :=\n{ to_fun := λ g, ⟨λ s, g s.1, ⟨g, λ _ _, rfl⟩⟩,\ninv_fun := λ p, p.2.1,\nleft_inv := λ _, rfl,\nright_inv := begin\nrintros ⟨f, φ, h⟩,\nhave : f = λ s, φ s.1, sorry,\nsubst this,\nend }\n\n\nohh, nice\n\n####", null, "Patrick Lutz (Nov 11 2020 at 20:03):\n\nwhat exactly does subst do?\n\n####", null, "Kevin Buzzard (Nov 11 2020 at 20:03):\n\n(note that Reid changed inv_fun -- λ ⟨f, φ⟩ is not good style when defining data, as you are here.\n\n####", null, "Kevin Buzzard (Nov 11 2020 at 20:04):\n\nsubst takes a proof of A = B and replaces all the A's everywhere by B's, or possibly the other way around\n\n####", null, "Reid Barton (Nov 11 2020 at 20:04):\n\noh yeah, I did that just in case it would help and then forgot to mention it\n\n####", null, "Bryan Gin-ge Chen (Nov 11 2020 at 20:04):\n\n(Reid's sorry can be filled with simp [h].)\n\n####", null, "Kevin Buzzard (Nov 11 2020 at 20:04):\n\nit's like rw h at * except that it's more powerful because it rewrites under binders. It literally removes A from the picture.\n\noh, I see\n\n####", null, "Patrick Lutz (Nov 11 2020 at 20:05):\n\nKevin Buzzard said:\n\n(note that Reid changed inv_fun -- λ ⟨f, φ⟩ is not good style when defining data, as you are here.\n\n####", null, "Reid Barton (Nov 11 2020 at 20:07):\n\nIf you look at the goal in the original right_inv it's got some _match_1 or whatever junk in it\n\n####", null, "Reid Barton (Nov 11 2020 at 20:08):\n\nmore importantly, the version which matches in the lambda is less likely to reduce when applied to something\n\n####", null, "Reid Barton (Nov 11 2020 at 20:08):\n\nthough maybe in a case as simple as this one it doesn't really matter, not sure\n\n####", null, "Patrick Lutz (Nov 11 2020 at 20:09):\n\nby the way, even though the example was originally by Kevin, I was the one who introduced the angle brackets\n\nLast updated: May 08 2021 at 09:11 UTC" ]

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[ "Solutions by everydaycalculation.com\n\n## Multiply 3/5 with 2/6\n\nThis multiplication involving fractions can also be rephrased as \"What is 3/5 of 2/6?\"\n\n3/5 × 2/6 is 1/5.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 3/5 × 2/6 = 3 × 2/5 × 6 = 6/30\n3. After reducing the fraction, the answer is 1/5\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "The situation with solids is considerably more complicated, with different speeds in different directions, in different kinds of geometries, and differences between transverse and longitudinal waves. Nevertheless, the speed of sound in solids is larger than in liquids and definitely larger than in gases. Young's Modulus for a representative value for the bulk modulus for steel is 160 $$10^9$$ N /$$m^2$$. A list of materials with their typical velocity can be found in the above book. Speed of sound in solid of steel, using a general tabulated value for the bulk modulus, gives a sound speed for structural steel of\n$\\nonumber c = \\sqrt{ E \\over \\rho} = \\sqrt{160 \\times 10^{9} N/m^{2} \\over 7860 Kg /m^3} = 4512 m/s \\tag{19}$" ]

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[ "# How Many Feet In 100 Meters? | Fastest Conversion Tips\n\nIn today’s world, precise measurements and conversions from one unit to another are an essential part of a variety of tasks. Whether you’re measuring the length of a field for agricultural use or converting degrees Celsius to Fahrenheit for scientific applications, understanding and converting various units is key. Among them is the conversion between metric units such as meters and other common systems like feet. In this article, we’ll answer the question: how many feet in 100 meters? We’ll also learn more about how different measuring systems work together in everyday life.\n\n## What Is Meters And How Is It Used?\n\nMeters is a unit of length in the metric system. It is equal to approximately 3 feet and 3.37 inches, or 1.094 yards. Meters are used all over the world as part of everyday life and scientific applications alike. For example, meters are an essential part of manufacturing processes as they allow for precision measurements in both machines and products. In addition, they are used to measure distances in cases such as the length of a road or field.\n\n## What Is Feet And How Is It Used?\n\nFeet is a unit of length in the imperial system. It is equal to 12 inches, or 0.333 yards. Feet are most commonly used in countries that use imperial measurements like the United States and the United Kingdom. For example, feet are often used to measure distance on roads and for construction projects such as building houses or bridges.\n\n## The Importance Of Knowing How Many Feet In 100 Meters\n\nKnowing how many feet in 100 meters is essential for a variety of tasks. For example, if you’re converting the length of a field from metric to imperial units or vice versa, understanding this conversion is key. In addition, knowing how to convert between these two systems can be very useful when traveling. By understanding which measuring system is used in various countries and being able to quickly convert between them, you can make sure that your international travels are easier and more efficient.\n\n## How Many Feet In 100 Meters?\n\nNow, onto the main question: how many feet in 100 meters? To answer this question, we must first convert between the different units. The equation for converting from meters to feet is as follows: 1 meter = 3.281 feet. Applying this equation to our original question of how many feet in 100 meters, we can find the answer by multiplying 100 by 3.281. This gives us 328.1 feet in 100 meters.\n\n## Steps To Convert Feet In 100 Meters\n\nConverting between feet and meters is relatively simple once you understand the equation. Here are a few steps to follow when converting from one unit to the other.\n\n1. Determine which conversion you want to make (e.g., from meters to feet or vice versa).\n2. Find the conversion factor for that particular unit (e.g., 1 meter = 3.281 feet).\n3. Multiply the unit you’re trying to convert by the conversion factor (e.g., 100 meters x 3.281 = 328.1 feet).\n4. The result is your answer (e.g., 100 meters = 328.1 feet).\n\nIn summary, knowing how many feet in 100 meters is important for a variety of tasks. The equation used to convert between these two units is 1 meter = 3.281 feet, and by multiplying the unit you’re converting by this factor, you can quickly get your answer. With this information, you’ll be able to easily convert distances from one measuring system to another.\n\n## Conversion Table For Feet In 100 Meters\n\nFor easy reference, here is a table of some common conversions between feet and meters.\n\n| Meters | Feet |\n\n|:——:|:——–:|\n\n| 100 | 328.1 |\n\n| 200 | 656.2 |\n\n| 300 | 984.3 |\n\n| 400 | 1312.4 |\n\n| 500 | 1640.5 |\n\nAs you can see, knowing how many feet in 100 meters is important for a variety of tasks and projects. With the equation and conversion table provided above, you’ll be able to easily convert distances between metric and imperial units. Understanding and converting between different systems of measurements is an essential part of many professions, so this knowledge can be invaluable for a variety of applications.\n\n## How To Convert Feet In 100 Meters Faster?\n\nIf you need to quickly convert from meters to feet or vice versa, using a conversion calculator is the fastest way to get an accurate result. These calculators are widely available online and can be used for a variety of different units.\n\nIn addition, many apps include conversion calculators that are designed specifically for mobile devices. This makes it easy to use them on the go, without having to find a computer or access the internet.\n\nUsing a conversion calculator is the fastest and easiest way to convert between different units and quickly get an accurate result. Whether you’re converting distances for travel purposes or measuring fields for agricultural use, understanding how to use these tools can make your life easier.\n\n## Tips To Convert Feet In 100 Meters Accurately\n\nWhen converting between different units, it’s important to make sure that you get an accurate answer. Here are a few tips to help you convert feet in 100 meters accurately:\n\n1. Make sure you understand the equation for converting between the two units (e.g., 1 meter = 3.281 feet).\n3. If you’re converting multiple numbers, use a conversion chart to save time and ensure accuracy.\n\nBy following these tips, you can make sure that you get an accurate result when converting between meters and feet. Knowing how many feet are in 100 meters can be useful for travel, construction projects, and other tasks, so making sure that you get an accurate result is essential.\n\n## How To Apply Feet In 100 Meters Conversion In Life?\n\nNow that you know how many feet in 100 meters, it’s time to look at how this information can be applied in everyday life. Here are a few examples of how understanding this conversion can make your life easier:\n\n• If you’re traveling abroad and need to measure distances with the metric system, understanding this conversion can help you quickly convert between units.\n• If you’re measuring a field or other large area, understanding this conversion can help you save time and get an accurate result.\n• Understanding how to convert between the imperial system and metrics can also be helpful when shopping for clothes or furniture in another country.\n\nKnowing how to convert between feet and meters is essential for a variety of tasks and projects. It can make international travel easier, help you quickly measure large areas, and even save time when shopping in another country. With this understanding, you’ll be able to make your life much easier and more efficient.\n\n## Some Objects Use Feet And 100 Meters As Units Of Measurement\n\n1. The International System of Units (SI) is an internationally accepted system that uses meters and liters as its base units of measurement.\n2. In the United States, feet and inches are often used for measuring distances in construction projects, such as building walls or laying out foundations.\n3. In many countries, kilometers and hectometers are used to measure large distances, such as the distance between two cities.\n4. In the United Kingdom, yards and miles are commonly used for measuring distances.\n5. When measuring distances in sports or other activities, meters are often used to measure short distances (e.g., 100 meters).\n6. In many parts of Europe, centimeters and millimeters are often used for measuring small objects or distances.\n\nUnderstanding which unit of measurement is used for a particular task or project is essential for accurate results. In many cases, you’ll need to know how many feet in 100 meters in order to understand the measurements correctly. Knowing this information can save you time and help you get an accurate result when completing tasks that involve converting between different units of measurement.\n\n## Conclusion: How Many Feet In 100 Meters\n\nIn conclusion, learning how many feet in 100 meters is essential for a variety of tasks and projects. The equation used to convert between these two units is 1 meter = 3.281 feet, and multiplying this by the unit you’re converting will give you the answer quickly and easily. With a conversion chart or calculator, it’s even easier to get an accurate result. Knowing how to convert between meters and feet can save you time when traveling abroad, measuring large areas, or even shopping in another country. With a little bit of knowledge about units of measurements and conversions, life is much easier.\n\n## FAQ: Feet In 100 Meters\n\n### Is 100 meters bigger than 5 feet?\n\nDiscover the conversion between meters and feet – a fundamental unit of length. With 100 meters equating to approximately 328 feet, this informative insight reveals the relationship between these two measurements. Learn how a meter forms the basis for various length units within the metric system, equivalent to 100 centimeters, 1/1000th of a kilometer, or approximately 39.37 inches. Meanwhile, a foot stands as a unit of measurement precisely 12 inches or 0.3048 meters in length.\n\n### Is 100 meters equal to 3 feet?\n\nThe length of 100 meters is about 328 feet. To convert meters to feet, simply multiply the number of meters by 3.28.\n\n### Is 100 meters 6 feet?\n\nThe distance of 100 meters cannot be equated to 6 feet. In fact, it is approximately equal to 328 feet or roughly 109 yards. To convert meters to feet, simply multiply the number of meters by 3.28. For instance, if you have 100 meters, you would have 328 feet (100 x 3.28 = 328).\n\n### Is 2 feet bigger than 100 meters?\n\nThe length of 100 meters is about 328 feet. Therefore, two feet are smaller than 100 meters. To convert meters to feet, simply multiply the number of meters by 3.28. For instance, if you have 100 meters, you would have 328 feet (100 x 3.28 = 328).\n\n### Is 10 feet equal to 100 meters?\n\nThe length of 10 feet is not equal to 100 meters. In reality, it is approximately 3.05 meters. To convert feet to meters, simply multiply the number of feet by 0.3048. For example, if you have 10 feet, you would have 3.05 meters (10 x 0.3048 = 3.05). Hence, 10 feet is smaller than 100 meters.\n\n### Is converting feet in 100 meters difficult?\n\nConverting meters to feet doesn’t have to be a challenge. Just multiply the meters by 3.28 and voila! You’ll have the answer in feet. However, to truly master metric to imperial conversions, it’s beneficial to practice regularly, ensuring these conversion skills are readily available when you need them.\n\n### What should be used to measure feet in 100 meters?\n\nHere’s a foolproof method to always get the right answer: Begin with 100 m. Next, multiply 100 by 3.28, which represents the number of feet in 1 meter. The outcome is 328 ft, the exact equivalent in this scenario.\n\n### Is the conversion of feet in 100 meters changed by temperature?\n\nNo, the conversion between feet and meters is not affected by temperature. This is because it’s a mathematical equation, not one that relies on physical factors such as heat or cold. The same equation can be used regardless of the temperature in order to convert between feet and meters accurately.\n\n### Is 100 meters the same as 20 feet?\n\n100 meters is not the same as 20 feet. In fact, 100 meters is equivalent to approximately 328 feet. To calculate this, multiply 100 by 3.28 (the number of feet in 1 meter). The resulting answer will be exactly how many feet there are in 100 meters.\n\n### Is 30 feet equal to 100 meters?\n\nThe distance of 30 feet is not equal to 100 meters. In reality, it is approximately 9.14 meters. To convert feet to meters, simply multiply the number of feet by 0.3048. For instance, if you have 30 feet, you would have 9.14 meters (30 x 0.3048 = 9.14). This shows that 30 feet is smaller than 100 meters. Therefore, this conversion is not accurate." ]

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[ "", null, "# Simpson's Rule Formula\n\nSimpson’s rule is used for approximating the integral using quadratic polynomials where parabolic arcs are present in place of straight line segments used in trapezoidal rule. For approximating the polynomials up to cubic degrees, Simpson’s rule gives the definite result. We do have trapezoidal formula that would take the shape under a curve and find out the area of those area. However to make it more precise and better approximation, Simpson’s rule came to rescue. Through Simpson’s rule parabolas are used to find parts of curve.\n\nThe approximate area under the curve are given by the following formula:\nSimpson’s one-third rule formula is:\n\n$\\large \\int_{a}^{b}f(x)dx=\\frac{h}{3}[(y_{0}+y_{n})+4(y_{1}+y_{3}+….+y_{n-1})+2(y_{2}+y_{4}+…+y_{n-2})]$\n\nSimpson’s three-eighths rule formula is:\n\n$\\large \\int_{a}^{b}f(x)dx=\\frac{3h}{8}[(y_{0}+y_{n})+(y_{1}+y_{2}+y_{4}+….+y_{n-1})+2(y_{3}+y_{6}+…+y_{n-3})]$\n\n### Solved example\n\nQuestion: Calculate the integral of the function f(x) = 2x in the interval (0, 2)?\n\nSolution:\n\nGiven,\nA = 0\nB = 2\nLet n = 6\n\n$h=\\frac{b-a}{n}=\\frac{2-0}{6}=\\frac{1}{3}$\n\n$x_{0}=a=0$\n\n$x_{1}=x_{0}+h=0=\\frac{1}{3}=\\frac{1}{3}$\n\n$x_{1}+h=\\frac{1}{3}+\\frac{1}{3}=\\frac{2}{3}$\n\n$x_{2}+h=\\frac{2}{3}+\\frac{1}{3}=\\frac{3}{3}=1$\n\n$x_{3}+h=\\frac{3}{3}+\\frac{1}{3}=\\frac{4}{3}$\n\n$x_{4}+h=\\frac{4}{3}+\\frac{1}{3}=\\frac{5}{3}$\n\n$x_{5}+h=\\frac{5}{3}+\\frac{1}{3}=\\frac{6}{3}=2$\n\n$x_{6}+h=\\frac{6}{3}+\\frac{1}{3}=1$\n\n$x_{7}=b=1$\n\n$y_{0}=f(0)=2(0)=0$\n\n$y_{1}=2(\\frac{1}{3})=\\frac{2}{3}$\n\n$y_{2}=2(\\frac{2}{3})=\\frac{4}{3}$\n\n$y_{3}=2(\\frac{3}{3})=2$\n\n$y_{4}=2(\\frac{4}{3})=\\frac{8}{3}$\n\n$y_{5}=2(\\frac{5}{3})=\\frac{10}{3}$\n\n$y_{6}=2(\\frac{6}{3})=4$\n\nAccording to the formula\n\n$\\int_{a}^{b}f(x)dx=\\frac{h}{3}[(y_{0}+y_{n})+4(y_{1}+y_{3}+….+y_{n-1})+2(y_{2}+y_{4}+…+y_{n-2})]$\n\n$f(x)dx=\\frac{\\frac{1}{3}}{3}[(0+4)+4(\\frac{2}{3}+2+\\frac{10}{3})+2(\\frac{4}{3}+\\frac{8}{3}+4)]$\n\n=1/9[4 + 24 + 16]\n\n=44/9\n\n= 4.89" ]

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[ "+0\n\n# help\n\n+1\n445\n2\n\nThe number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle? Express your answer in terms of pi and simplest radical form.\n\nAug 13, 2018\n\n#1\n+3\n\nThe number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?\n\nperimeter $$\\triangle$$ = 3a\n\n$$cos(30°)=\\frac{a}{2r}\\\\ r=\\frac{a}{2\\cdot cos(30°)}\\\\ a=2r\\cdot cos(30°)$$\n\n$$\\large\\color{blue}3a=r^2\\pi$$\n\n$$3a=(\\frac{a}{2\\cdot cos(30°)})^2\\cdot \\pi\\\\ 3a=\\frac{\\pi a^2}{4\\cdot cos^2(30°)}\\\\ \\color{blue}a=\\frac{3\\cdot 4\\cdot cos^2(30°) }{\\pi}=\\frac{3\\cdot 4\\cdot (\\frac{1}{2}\\sqrt{3})^2 }{\\pi}=\\frac{3^2}{\\pi}\\\\ \\color{blue}a=2.86479..$$\n\n$$r=\\frac{3\\cdot 4\\cdot cos^2(30°) }{2\\pi\\cdot cos(30°)}\\\\ \\color{blue}r=\\frac{6\\cdot cos(30°)}{\\pi}=\\frac{6\\cdot (\\frac{1}{2}\\sqrt{3})}{\\pi}=\\frac{3\\cdot \\sqrt{3}}{\\pi}\\\\ \\color{blue}r=1.65399$$\n\nproof:\n\n$$perimeter \\triangle=area\\ O\\\\ 3a=r^2\\pi\\\\ 3\\cdot \\frac{3\\cdot 4\\cdot cos^2(30°) }{\\pi}=\\pi\\cdot (\\frac{6\\cdot cos(30°)}{\\pi})^2\\\\ 8.59437=8.59437$$", null, "!\n\nAug 13, 2018\nedited by asinus  Aug 13, 2018\nedited by asinus  Aug 14, 2018\nedited by asinus  Aug 14, 2018\nedited by asinus  Aug 14, 2018\n#2\n+4\n\nThe number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle? Express your answer in terms of pi and simplest radical form.", null, "", null, "Aug 13, 2018" ]

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[ "", null, "MATLAB for Engineers (5th Edition) 5th Edition\n\nPLDZ-3 In Stock\n\\$ 5.00 USD\nDescription\n\nStart at the beginning to introduce your students to MATLAB\n\nMATLAB® For Engineers introduces  students the MATLAB coding language. Developed out of Moore’s experience teaching MATLAB and other languages, the text meets students at their level of mathematical and computer sophistication. Starting with basic algebra, the book shows how MATLAB can be used to solve a wide range of engineering problems. Examples drawn from concepts introduced in early chemistry and physics classes and freshman and sophomore engineering classes stick to a consistent problem-solving methodology.\n\nStudents reading this text should have an understanding of college-level algebra and basic trigonometry. The text includes brief backgrounds when introducing new subjects like statistics and matrix algebra. Sections on calculus and differential equations are introduced near the end and can be used for additional reading material for students with more advanced mathematical backgrounds.\n\nRecent Reviews Write a Review\n0 0 0 0 reviews" ]

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[ "# Network Graphs & Solvers Concepts\n\nKinetica provides a generic and extensible design of networks with the aim of being tailored or used for various real-life applications, such as transportation, utility, social, and geospatial. Networks comprise a graph and a solver. A graph represents topological relationships via nodes that are connected by edges; a solver represents the type of solution appropriate for the issue the graph was made to illustrate.\n\nThe graph server is enabled by default; it can be managed via standard Kinetica Service Management. For more information on using and configuring multiple graph servers, see Distributed Graph Servers. Graph server logs are available in `/opt/gpudb/graph/logs`. See the Configuration Reference for more information on graph server configuration.\n\nTip\n\n• Using SQL or native API, a user can create a graph from several components, solve the graph using one of several solver types, and query the solved graph.\n• Graphs can also be created & previewed using the GAdmin UI or Workbench UI.", null, "## Solvers\n\nKinetica has several different solvers available for presenting solutions to various types of network graph problems. Note that some solvers are only available to /solve/graph and likewise for /match/graph. Consult Solving a Graph or Matching a Graph for more information on the desired operation.\n\n### Solve Graph Solvers\n\nSolver Description CPU Parallel\nALLPATHS Determines all reasonable paths between a source and destination pair. X\nBACKHAUL_ROUTING Determines the optimal routes between remote asset nodes and fixed asset nodes. X\nCENTRALITY Calculates the degree of a node to depict how many pairs of individuals that would have to go through the node to reach one another in the minimum number of hops. Also known as betweenness. X\nCLOSENESS Calculates the centrality closeness score per node as the sum of the inverse shortest path costs to all nodes in the graph. X\nINVERSE_SHORTEST_PATH Determines the shortest path downstream using multiple technician routing. X\nMULTIPLE_ROUTING Calculates the shortest possible route between the nodes and returns to the origin node -- also known as the traveling salesman. X\nPAGE_RANK Calculates how connected the nodes are and determines which nodes are the most important. Weights are not required. X\nPROBABILITY_RANK Calculates the probability of a node being connected to another node using hidden Markov chains. X\nSHORTEST_PATH Determines the shortest path upstream between given source(s) and destination(s). X\nSTATS_ALL Calculates graph statistics such as graph diameter, longest pairs, vertex valences, topology numbers, average and max cluster sizes, etc. X\n\n### Match Graph Solvers\n\nSolver Description CPU Parallel\nMARKOV_CHAIN Matches sample points to the graph using the Hidden Markov Model (HMM)-based method, which conducts a range-tree closest-edge search to find the best combinations of possible road segments for each sample point to create the best route. The route is secured one point at a time, so the prediction is corrected after each point. This solution type is the most accurate, but also the most computationally intensive. X\nMATCH_BATCH_SOLVES Matches each provided sample source and destination pair using the shortest path between the points. X\nMATCH_CHARGING_STATIONS Matches a given sample source and destination pair to the optimal recharging stations along the route (for EVs). X\nMATCH_CLUSTERS\n\nMatches the graph nodes with a cluster index using the Louvain clustering algorithm.\n\nNote\n\nParallel running of this solver is experimental and can be invoked with the parallel_clustering option.\n\nX*\nMATCH_LOOPS Matches closed loops (Eulerian paths) originating and ending at each graph node between min and max hops (levels). X\nMATCH_OD_PAIRS Matches sample points to find the most probable path between origin and destination (OD) pairs with given cost constraints. X\nMATCH_SIMILARITY Computes the Jaccard similarity between vertex pairs and N-level intersections within M hops. X\nMATCH_SUPPLY_DEMAND Matches sample generic supply depots to generic demand points using abstract transportation (referred to as trucks). Each route is determined by a truck's ability (size) to service demand at each demand point. X\n\n## Components and Identifiers\n\nGraph components (e.g., nodes, edges, weights, and/or restrictions) must be defined with varying combinations of the identifiers listed in the tables below.\n\nComponents for create, solve, query, and match operations can be identified in three ways:\n\n• Aliasing existing table columns, e.g., table_name.column_name AS WKTPOINT\n• Using expressions, e.g., ST_LENGTH(wkt) AS VALUESPECIFIED\n• Using constant values, where strings & WKTs are single-quoted, non-decimal numbers will be given an int or long type according to their size (or can be coerced to long by using an L suffix, like 1L), and decimal numbers will be given the double type; e.g.:\n• {0} AS ONOFFCOMPARED\n• {1L, 2, 3L, 4, 5} AS ID\n• {1.1, 12.345689, 123.456789456789} AS NODE1_X\n• {'name1', 'name2'} AS NAME\n• {'POINT(10 15)'} AS WKTPOINT\n\nNote\n\nThere are separate identifiers and combinations for /solve/graph, /query/graph, and /match/graph. See below for /solve/graph identifiers; see the associated Querying a Graph and Matching a Graph sections for their respective endpoint identifiers.\n\nIdentifiers are flagged aliases that the database knows to look for; existing source columns can be used as identifiers. Note that it will often be the case that source columns are reused for different identifiers because the components must naturally be linked together to create a network graph. For example, source table seattle_road_network has columns WKTLINE (a wkt column), TwoWay (an integer column), and time (also an integer column); these columns could be identified via /create/graph like so:\n\nCreate Graph Example\n `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 `````` ``````create_s_graph_response = kinetica.create_graph( graph_name = GRAPH_S, directed_graph = True, nodes = [], edges = [ TABLE_SRN + \".WKTLINE AS WKTLINE\", TABLE_SRN + \".TwoWay AS DIRECTION\" ], weights = [ TABLE_SRN + \".WKTLINE AS EDGE_WKTLINE\", TABLE_SRN + \".TwoWay AS EDGE_DIRECTION\", TABLE_SRN + \".time AS VALUESPECIFIED\" ], restrictions = [], options = { \"recreate\": \"true\" } ) ``````\n\nImportant\n\nIdentifiers with string as a supported type can be any string column type, e.g., char1 - char256 or unrestricted string. Identifiers with wkt as a supported type can include optional z-level values for the provided WKT shape but note that graphs only support z-levels ranging from -4 to +5.\n\n### Nodes\n\nNodes represent fundamental topological units of a graph. Nodes can be defined with an integer ID, a string name, or geospatial information, e.g., WKT point (POINT(X Y [Z])) or XY pair. Nodes are optional, as the start and end points of an edge can implicitly be used as nodes.\n\nIdentifier Supported Types Description\nID int, long A number representing a node identifier\nLABEL string A string value representing a node's label\nNAME string A string value representing a node's name\nPARTITION_BOUNDARY long A number representing the ID of the partition containing this node, when using explicit partitioning to segment the graph; nodes spanning partition boundaries will need to be duplicated in the source table, once for each partition in which the node will reside\nWKTPOINT wkt\n\nA WKT string representing a node's geospatial point; e.g., POINT(X Y [Z])\n\nX float, double, int, long A number representing a node's X or longitude value\nY float, double, int, long A number representing a node's Y or latitude value\n\n### Edges\n\nEdges represent the required fundamental topological units of a graph that typically connect nodes. Edges can be defined with an integer ID, string name, or geospatial information, e.g., WKT point (POINT(X Y [Z])), line (LINESTRING(X1 Y1 [Z1], X2 Y2 [Z2])), or XY pairs. An edge can be implicitly drawn between two nodes. If an edge is defined using WKT linestrings, the graph server is capable of splitting many linestring segments into multiple, separate linestrings, thus creating one edge per linestring segment.\n\nIdentifier Supported Types Description\nID int, long A number representing an edge identifier\nDIRECTION int\n\nA number representing what direction the edge can be traveled:\n\n• 0: forward one-way edge (node 1 ‣ node 2)\n• 1: two-way edge (node 1 ‣ node 2 and node 2 ‣ node 1)\n• 2: backward one-way edge (node 2 ‣ node 1)\nLABEL string A string value representing an edge's label\nNODE1_ID int, long A number representing the edge's first node\nNODE1_NAME string A string value representing the edge's first node's name\nNODE1_WKTPOINT wkt\n\nA WKT string representing the edge's first node's geospatial point; e.g., POINT(X1 Y1 [Z1])\n\nNODE1_X float, double, int, long A number representing the edge's first node's X or longitude value\nNODE1_Y float, double, int, long A number representing the edge's first node's Y or latitude value\nNODE2_ID int, long A number representing the edge's second node\nNODE2_NAME string A string value representing the edge's second node's name\nNODE2_WKTPOINT wkt\n\nA WKT string representing the edge's second node's geospatial point; e.g., POINT(X2 Y2 [Z2])\n\nNODE2_X float, double, int, long A number representing the edge's second node's X or longitude value\nNODE2_Y float, double, int, long A number representing the edge's second node's Y or latitude value\nPARTITION long A number representing the ID of the partition containing this edge, when using explicit partitioning to segment the graph\nWEIGHT_VALUESPECIFIED float, double, int, long A number representing the edge's associated weight value; if this identifier is provided, additional weights do not need to be specified\nWKTLINE wkt\n\nA WKT string representing an edge's geospatial line; e.g., LINESTRING(X1 Y1 [Z1], X2 Y2 [Z2])\n\n### Weights\n\nWeights represent a method of informing the graph solver of the cost of including a given edge in a solution. Weights can be defined using an integer ID, string node names, or spatial information (LINESTRING(X1 Y1 [Z1], X1 Y1 [Z1])) and a static cost value or a cost multiplier. Each edge is associated with one weight, but there can be many edges between two nodes in a graph with directionality (EDGE_DIRECTION), allowing for many different weights along the same edge, which can have useful applications in real-world examples, e.g., different lanes between two junctions may have different speeds of travel.\n\nFor graphs that define edges using complex WKT linestrings (e.g., linestrings with more than two points), weights are applied consistently to each segment of the linestring. For example, if LINESTRING(0 0, 1 3, 4 5) is provided as an edge source and a weight of 5 is assigned to that source, the resulting graph would have two edges, LINESTRING(0 0, 1 3) and LINESTRING(1 3, 4 5), that would both have a weight of 5. See Fitting Road Network Data to a Graph for more information.\n\nNote\n\nIf DIRECTION is specified for an edge in a directed graph, the weight will be the same going in each direction.\n\nIdentifier Supported Types Description\nEDGE_DIRECTION int\n\nA number representing what direction the edge can be traveled:\n\n• 0: forward one-way edge (node 1 ‣ node 2)\n• 1: two-way edge (node 1 ‣ node 2 and node 2 ‣ node 1)\n• 2: backward one-way edge (node 2 ‣ node 1)\nEDGE_ID int, long A number representing a weight's associated edge identifier\nEDGE_NODE1_ID int, long A number representing a weight's associated edge's first node\nEDGE_NODE1_NAME string A string value representing a weight's associated edge's first node's name\nEDGE_NODE2_ID int, long A number representing a weight's associated edge's second node\nEDGE_NODE2_NAME string A string value representing a weight's associated edge's second node's name\nEDGE_PARTITION long A number representing a weight's associated edge's partition ID\nEDGE_WKTLINE wkt\n\nA WKT string representing a weight's associated edge geospatial line; e.g., LINESTRING(X1 Y1 [Z1], X2 Y2 [Z2])\n\nFACTORSPECIFIED float, double, int, long A number representing how much incoming cost values will be multiplied\nFROM_EDGE_ID int, long\n\nA number representing a weight's associated edge identifier that signifies the start of a turn; paired with TO_EDGE_ID to define additionally-weighted traversal from one edge to another through any single node\n\nNote\n\nOnly applicable if the add_turns option was set to true during /create/graph. This identifier can be used to locally override any *_turn_penalty and/or intersection_penalty options previously set. See Using Turn-based Weights & Restrictions for more information.\n\nTO_EDGE_ID int, long\n\nA number representing a weight's associated edge identifier that signifies the end of a turn; paired with FROM_EDGE_ID to define additionally-weighted traversal from one edge to another through any single node\n\nNote\n\nOnly applicable if the add_turns option was set to true during /create/graph. This identifier can be used to locally override any *_turn_penalty and/or intersection_penalty options previously set. See Using Turn-based Weights & Restrictions for more information.\n\nWKTPOINT wkt\n\nA WKT string representing a weight's associated WKT point\n\nNote\n\nUsed exclusively in conjunction with FACTORSPECIFIED to create a combination for graphing weights by the inverse distance weighted averaging of WKT points composing a graph.\n\nVALUESPECIFIED float, double, int, long A number representing the weight's value\n\nImportant\n\nCurrently, FACTORSPECIFIED will only affect the cost if the edge has a VALUESPECIFIED already established. This means that FACTORSPECIFIED should only be used in /solve/graph or in conjunction with a VALUESPECIFIED during /create/graph.\n\n### Restrictions\n\nRestrictions represent a method of informing the graph solver which edges and/or nodes should be ignored for the solution. Restrictions can be defined using an integer ID and a value or as a switch (on or off).\n\nIdentifier Supported Types Description\nEDGE_DIRECTION int\n\nA number representing what direction the edge can be traveled:\n\n• 0: forward one-way edge (node 1 ‣ node 2)\n• 1: two-way edge (node 1 ‣ node 2 and node 2 ‣ node 1)\n• 2: backward one-way edge (node 2 ‣ node 1)\nEDGE_ID int, long A number representing the restriction's associated edge identifier\nEDGE_LABEL string A string value referring to an edge label for restrictive purposes\nEDGE_NODE1_ID int, long A number representing a restriction's associated edge's first node\nEDGE_NODE1_NAME string A string value representing a restriction's associated edge's first node's name\nEDGE_NODE2_ID int, long A number representing a restriction's associated edge's second node\nEDGE_NODE2_NAME string A string value representing a restriction's associated edge's second node's name\nEDGE_PARTITION long A number representing a restriction's associated edge's partition ID\nEDGE_WKTLINE wkt\n\nA WKT string representing a weight's associated edge geospatial line; e.g., LINESTRING(X1 Y1 [Z1], X2 Y2 [Z2])\n\nFROM_EDGE_ID int, long\n\nA number representing the restriction's associated edge identifier that signifies the start of a turn; paired with TO_EDGE_ID to define traversal restriction from one edge to another through any single node\n\nNote\n\nOnly applicable if the add_turns option was set to true during /create/graph. This identifier can be used to locally override any *_turn_penalty and/or intersection_penalty options previously set. See Using Turn-based Weights & Restrictions for more information.\n\nNODE_ID int, long A number representing the restriction's associated node identifier\nNODE_LABEL string A string value referring to a node label for restrictive purposes\nNODE_NAME string A string value representing the restriction's associated node\nNODE_WKTPOINT wkt\n\nA WKT string representing the restriction's associated node's geospatial point, e.g., POINT(X Y [Z])\n\nONOFFCOMPARED int A number representing if the associated node or edge cannot be traversed, with 1 meaning it can be traversed and 0 meaning it cannot\nTO_EDGE_ID int, long\n\nA number representing the restriction's associated edge identifier that signifies the end of a turn; paired with FROM_EDGE_ID to define traversal restriction from one edge to another through any single node\n\nNote\n\nOnly applicable if the add_turns option was set to true during /create/graph. This identifier can be used to locally override any *_turn_penalty and/or intersection_penalty options previously set. See Using Turn-based Weights & Restrictions for more information.\n\nVALUECOMPARED float, double, int, long A number representing the value against which incoming costs will be compared\n\nNote\n\nWhen using VALUECOMPARED, solvers will not use the given node or edge if the current cost is less than the restriction value. When using ONOFFCOMPARED, solvers will not use the given node or edge if the ONOFFCOMPARED value is set to 0 (off).\n\n## Identifier Combinations\n\nFor each component, there's a minimum set of identifiers that must be used to properly create a graph. Each component's identifier combinations must reference columns from the same table, e.g., the node combination of ID and NAME must both use the same table. The columns must also not be nullable. Identifier types across components should match where possible.\n\nImportant\n\nWKT identifiers can be matched to X/Y identifiers (and vice versa) within a user-specified tolerance (merge_tolerance under the /create/graph endpoint's options map). Using ID or NAME identifiers relies on exact matching. The WKTLINE identifiers will use the line's start and end points to map to an XY pair or WKTPOINT.\n\nFor example, if the identifier combination used for nodes is:\n\nNode Identifier Combination Example\n ``````1 2 3 `````` ``````nodes = [ TABLE_TAXI_N + \".id AS ID\" ], ``````\n\nThe edges identifier combination should include a match for the node ID. The following edge combination would match correctly with the node combination; note that matching node point(s) to edge endpoint(s) requires two edge endpoints to make an implicit edge between the points:\n\nEdge Identifier Combination Example\n ``````1 2 3 4 `````` ``````edges = [ TABLE_TAXI_E + \".pickup_id AS NODE1_ID\", TABLE_TAXI_E + \".dropoff_id AS NODE2_ID\" ], ``````\n\nNote\n\nThe above example is not the only edge combinations available for the node ID identifier combination. See the Edges section below for other combinations.\n\nIf using multiple groups of combinations while creating, solving, querying, or matching a graph, the combinations must be separated by empty quotes (\"\") in the respective array, e.g.:\n\nQuery Identifier Combination Example\n ``````1 2 3 4 5 `````` ``````queries = [ \"{'Jane'} AS NODE_NAME\", \"\", \"{'chess'} AS TARGET_NODE_LABEL\" ], ``````\n\nIf specifying identifier combinations as raw values, the number of values within each identifier must match across the combination group, e.g.:\n\n```\"{'Bill', 'Alex'} AS NODE_NAME\",\n\"{0, 0} AS ONOFFCOMPARED\"\n```\n\n### Nodes\n\n• ID\n• ID, NAME\n• ID, WKTPOINT\n• ID, X, Y\n• NAME\n• NAME, WKTPOINT\n• NAME, X, Y\n• WKTPOINT\n• X, Y\n\nThe following identifiers can be added to any valid node combination:\n\n• LABEL\n• PARTITION_BOUNDARY\n\n### Edges\n\n• ID, NODE1_ID, NODE2_ID\n• ID, NODE1_ID, NODE2_ID, DIRECTION\n• ID, NODE1_NAME, NODE2_NAME\n• ID, NODE1_WKTPOINT, NODE2_WKTPOINT\n• ID, NODE1_X, NODE1_Y, NODE2_X, NODE2_Y\n• ID, WKTLINE\n• ID, WKTLINE, DIRECTION\n• NODE1_ID, NODE2_ID\n• NODE1_NAME, NODE2_NAME\n• NODE1_WKTPOINT, NODE2_WKTPOINT\n• NODE1_X, NODE1_Y, NODE2_X, NODE2_Y\n• WKTLINE\n• WKTLINE, DIRECTION\n\nAny or all of the following identifiers can be added to any valid edge combination:\n\n• LABEL\n• PARTITION\n• WEIGHT_VALUESPECIFIED\n\n### Weights\n\n• EDGE_ID, VALUESPECIFIED\n• EDGE_ID, FACTORSPECIFIED\n• EDGE_WKTLINE, EDGE_DIRECTION, VALUESPECIFIED\n• EDGE_WKTLINE, VALUESPECIFIED\n• EDGE_WKTLINE, FACTORSPECIFIED\n• EDGE_NODE1_NAME, EDGE_NODE2_NAME, VALUESPECIFIED\n• EDGE_NODE1_NAME, EDGE_NODE2_NAME, FACTORSPECIFIED\n• EDGE_NODE1_ID, EDGE_NODE2_ID, VALUESPECIFIED\n• EDGE_NODE1_ID, EDGE_NODE2_ID, FACTORSPECIFIED\n• WKTPOINT, FACTORSPECIFIED\n\nIf utilizing turn penalties, the following combination becomes applicable:\n\n• FROM_EDGE_ID, TO_EDGE_ID, VALUESPECIFIED\n\n### Restrictions\n\n• EDGE_ID, ONOFFCOMPARED\n• EDGE_ID, VALUECOMPARED\n• EDGE_LABEL, ONOFFCOMPARED\n• EDGE_NODE1_ID, EDGE_NODE2_ID, ONOFFCOMPARED\n• EDGE_NODE1_NAME, EDGE_NODE2_NAME, ONOFFCOMPARED\n• EDGE_WKTLINE, EDGE_DIRECTION, ONOFFCOMPARED\n• EDGE_WKTLINE, EDGE_DIRECTION, VALUECOMPARED\n• EDGE_WKTLINE, ONOFFCOMPARED\n• EDGE_WKTLINE, VALUECOMPARED\n• NODE_ID, ONOFFCOMPARED\n• NODE_ID, VALUECOMPARED\n• NODE_LABEL, ONOFFCOMPARED\n• NODE_NAME, VALUECOMPARED\n• NODE_NAME, ONOFFCOMPARED\n• NODE_WKTPOINT, ONOFFCOMPARED\n• NODE_WKTPOINT, VALUECOMPARED\n\nIf utilizing turn restrictions, the following combination becomes applicable:\n\n• FROM_EDGE_ID, TO_EDGE_ID, ONOFFCOMPARED\n\n## Using Labels\n\nLabels are a type of identifier that provide additional context to a node or edge and can act as a target for a query. Labels are paired with another identifier in several valid combinations listed above, but each unique label must be part of its own combination; i.e., there cannot be two labels in the same identifier combination. For example, this is a valid multi-label configuration:\n\nValid Multi-Label Configuration\n ``````1 2 3 4 5 6 7 `````` ``````nodes = [ TABLE_P + \".name AS NAME\", TABLE_P + \".interest AS LABEL\", \"\", TABLE_P + \".name AS NAME\", TABLE_P + \".gender AS LABEL\" ], ``````\n\nBut this example is invalid:\n\nInvalid Multi-Label Configuration\n ``````1 2 3 4 5 `````` ``````nodes = [ TABLE_P + \".name AS NAME\", TABLE_P + \".interest AS LABEL\", TABLE_P + \".gender AS LABEL\" ], ``````\n\nThere are several types of labels, some of which can only be referenced in the context of the /query/graph endpoint. See the sections below for more information.\n\n### Node / Edge Labels\n\nThe LABEL node identifiers & edge identifiers are are used to provide additional string-based information about a node or edge respectively, similarly to the node NAME (and other related) identifiers.\n\n### Restriction Labels\n\nThe NODE_LABEL and EDGE_LABEL restriction identifiers are used to restrict the value set defined as node & edge LABEL respectively. These restrictive identifiers must follow an already-defined node or edge LABEL; i.e., they cannot be used on their own at graph creation time. They are used in restriction combinations just like other restriction identifiers.\n\nFor example, if the relation between two people is used as the edge LABEL when creating a graph:\n\nEdge Identifier Combination\n ``````1 2 3 4 5 `````` ``````edges = [ TABLE_K + \".name1 AS NODE1_NAME\", TABLE_K + \".name2 AS NODE2_NAME\", TABLE_K + \".relation AS LABEL\" ], ``````\n\nThen the family relation label can be restricted during the subsequent query using the EDGE_LABEL restriction identifier:\n\nRestriction Identifier Combination Example\n ``````1 2 3 4 `````` ``````restrictions = [ \"{'family'} AS EDGE_LABEL\", \"{0} AS ONOFFCOMPARED\" ], ``````\n\n### Unique Query Labels\n\nThere are several *_LABEL query identifiers that are unique to the /query/graph endpoint.\n\n#### Node / Edge Labels\n\nThe NODE_LABEL and EDGE_LABEL query identifiers are not paired with other query combinations; they are used to retrieve only the nodes and/or edges associated with a given label. For example:\n\nQuery Identifier Combination Example\n ``````1 2 3 `````` ``````queries = [ \"{'male'} AS NODE_LABEL\" ], ``````\n\n#### Target Labels\n\nThe TARGET_NODE_LABEL query identifier is always paired with another query combination to define a source-destination relationship and pathing for a query. This identifier also must follow an already-defined node LABEL. Note that an adjacency output table will have two additional columns if the TARGET_NODE_LABEL identifier is used:\n\n• PATH_ID - an ID that helps identify the different paths taken to arrive at the query target if there is more than one adjacency for a given query.\n• RING_ID - an ID that helps identify the steps taken to arrive at a query target. The RING_ID can also be referred to as the hop ID or number of hops it took to arrive at the query target\n\nFor example, a social graph can be created, representing a set of people, their primary interests, and their relationship to each other. Each node represents a person with their first name as the node NAME and the primary interest as the node LABEL, while each edge represents the relationship between each pair of people. The nodes would be defined as follows, using the name & interest columns from a table containing a given set of people as the NAME & LABEL of each node, respectively:\n\nNode Identifier Combination\n ``````1 2 3 4 `````` ``````nodes = [ TABLE_P + \".name AS NAME\", TABLE_P + \".interest AS LABEL\", ], ``````\n\nThis graph can be used to query for people with an interest in chess who are related either directly, or indirectly via one or more other people, to the person named Jane. This can be accomplished by querying the graph, using \"Jane\" as the query NODE_NAME (which instructs the graph engine to begin the search at the node with a NAME of \"Jane\") and using \"chess\" as the TARGET_NODE_LABEL (which directs the graph engine to only return directly or indirectly connected nodes with a node LABEL of \"chess\"):\n\nQuery Target Identifier Combination Example\n ``````1 2 3 4 5 `````` ``````queries = [ \"{'Jane'} AS NODE_NAME\", \"\", \"{'chess'} AS TARGET_NODE_LABEL\" ], ``````\n\nThe query results in two targets for the output adjacency table (Alex and Tom):\n\nQuery Target Identifier Combination Results\n ``````1 2 3 4 5 6 7 `````` ``````+------------------------+------------------------+--------------------------+--------------------------+-----------+ | QUERY_NODE_ID_SOURCE | QUERY_NODE_ID_TARGET | QUERY_NODE_NAME_SOURCE | QUERY_NODE_NAME_TARGET | RING_ID | +========================+========================+==========================+==========================+===========+ | 4 | 3 | Jane | Alex | 3 | +------------------------+------------------------+--------------------------+--------------------------+-----------+ | 4 | 5 | Jane | Tom | 4 | +------------------------+------------------------+--------------------------+--------------------------+-----------+ ``````\n\n## Using Turn-based Weights & Restrictions\n\nTurn-based weights and restrictions are used to add a cost to solutions utilizing turn types implemented during graph creation or modification. If the add_turns option is set to true during /create/graph or /modify/graph operations, dummy pillowed edges will be added to each intersection of edges in a graph to mimic realistic turns. These dummy edges will not have any weight by default and will have the coordinates of their origin point. The available turn types are as follows:\n\n• Left turns -- turning left from one edge on to another\n• Right turns -- turning right from one edge on to another\n• Intersection -- continuing through an intersection of edges (e.g., a stoplight)\n• Sharp turns -- turning sharply left or right or a u-turn; a sharp turn or u-turn attribution is determined by the angle of the turn (the turn_angle setting)\n\nFor example, say you have a dataset featuring the intersection (designated by the traffic light) below:", null, "Road 1 is a one-way road going north. Road 2 is a one-way road going west. If you were to create a graph from this dataset with add_turns set to false, the edges might look like this:", null, "A solution containing the left turn from road 1 on to road 2 using this graph would incorporate edge 1 ‣ edge 3.\n\nOn the other hand, if you were to create a graph from this dataset with add_turns set to true, the edges might look like this instead (note the two additional dummy edges designated by the dotted lines):", null, "The L and the X associated with the dummy edges designate a left turn and intersection respectively, but these attributions could change depending on the turn angle set. A solution containing the left turn from road 1 on to road 2 using this graph instead would incorporate edge 1 ‣ edge 4 ‣ edge 3.\n\nOnce turns have been enabled and the turn angle has been set, turn penalties can be incurred via the /solve/graph or the /match/graph endpoints using two methods:\n\n• Setting penalties uniformly per turn type across the graph using the left_turn_penalty, right_turn_penalty, intersection_penalty, or sharp_turn_penalty options\n\n• Setting penalties on a per-edge basis using the following weight identifier combinations:\n\nFROM_EDGE_ID, TO_EDGE_ID, VALUESPECIFIED\n\n• Setting restrictions on a per-edge basis using the following restriction identifier combinations:\n\nFROM_EDGE_ID, TO_EDGE_ID, ONOFFCOMPARED\n\nTip\n\nUsing the weight or restriction combinations will locally override any penalty options set.\n\nFor more information on using turn penalties and restrictions, see the turn penalties and restrictions graph example.\n\n## Creating a Graph\n\nCreating a graph is serviced by the /create/graph endpoint; this involves reading from tables with annotated component identifiers and drawing relationships between given nodes and/or edges on a graph, taking into account nodes or edges between nodes that should be favored or ignored.\n\nNote\n\nThough it's recommended edges and weights are kept in the same table, it's not required.\n\nOnce the components are setup, the graph can be created. Requirements for creating a graph include:\n\n• name for the graph\n• if the graph is directed or not\n• edges\n• weights (for most solver types)\n\nImportant\n\nNodes and restrictions are not required to create a graph. If nodes are included, however, they should be kept in a separate table from edges and weights. If restrictions are included, they can exist in either the nodes table and/or edges/weights table(s) or in an entirely separate table.\n\nA graph name must adhere to the standard naming criteria. Each graph exists within a schema and follows the standard name resolution rules for tables.\n\n### Directed Graphs\n\nWhether a graph is directed or not can determine how a graph is solved or queried. Using Components and Identifiers and Identifier Combinations for context, edges connect two nodes together (\"node 1\" and \"node 2\"). When a graph is directed, these nodes that comprise a given edge have an implicit direction: \"node 1\" to \"node 2\". Regardless if these nodes have a spatial context, Kinetica will treat \"node 2\" as if it must be traveled to directly from \"node 1\" (whatever their underlying values may be: Jim and George , POINT(0 0) and POINT(9 10), 32 and 45, etc.).\n\nFor example, given the below non-directed graph:", null, "Attempting to solve for the shortest path from node F to node A would result in a SOLVERS_EDGE_PATH of F, E, D, C, B, A. Querying for other nodes attached to node E would result in two adjacencies: F and D.\n\nOn the other hand, given the below directed graph:", null, "Attempting to solve for the shortest path from node F to node A would be unsuccessful because there are only edges going toward F (and none leading away from F and subsequently toward A). Querying for other nodes attached to node E would result in a single adjacency: F.\n\n## Modifying a Graph\n\nModifying a graph is serviced by the /modify/graph endpoint; this involves using given node(s), edge(s), weight(s), restriction(s), and option(s) to update an existing graph. All parameters and options available to /modify/graph are also available to /create/graph; as such, many of the same principles apply to using /modify/graph.\n\nRequirements for modifying a graph include:\n\n• name for the existing graph\n• components/identifiers and options used to modify the graph\n\n## Solving a Graph\n\nSolving a graph is serviced by the /solve/graph endpoint; this involves using given source node(s), destination node(s), and any weights or restrictions from an existing graph to calculate a given solution type. Off-graph spatial locations are accepted in all solvers, with the results being corrected to snap locations. The calculated solution is then placed in a table in Kinetica; note that many concurrent solves over the same graph are possible. The source node determines from which node the graph solution routing is started; the destination node(s) determines at which node the graph solution will complete routes. Source/destination node(s) can be an ID, name, or WKT point.\n\nRequirements for solving a graph include:\n\n• name of the graph to solve\n• solution type\n\nImportant\n\nAdditional weights and restrictions beyond those defined in the graph creation stage can also be provided. Any provided weights will be added (in the case of VALUESPECIFIED) to or multiplied by (in the case of FACTORSPECIFIED) the existing weight(s). Consult Components and Identifiers for formatting and specifications.\n\nThere are several solution types to choose from:\n\nSolver Description CPU Parallel\nALLPATHS Determines all reasonable paths between a source and destination pair. X\nBACKHAUL_ROUTING Determines the optimal routes between remote asset nodes and fixed asset nodes. X\nCENTRALITY Calculates the degree of a node to depict how many pairs of individuals that would have to go through the node to reach one another in the minimum number of hops. Also known as betweenness. X\nCLOSENESS Calculates the centrality closeness score per node as the sum of the inverse shortest path costs to all nodes in the graph. X\nINVERSE_SHORTEST_PATH Determines the shortest path downstream using multiple technician routing. X\nMULTIPLE_ROUTING Calculates the shortest possible route between the nodes and returns to the origin node -- also known as the traveling salesman. X\nPAGE_RANK Calculates how connected the nodes are and determines which nodes are the most important. Weights are not required. X\nPROBABILITY_RANK Calculates the probability of a node being connected to another node using hidden Markov chains. X\nSHORTEST_PATH Determines the shortest path upstream between given source(s) and destination(s). X\nSTATS_ALL Calculates graph statistics such as graph diameter, longest pairs, vertex valences, topology numbers, average and max cluster sizes, etc. X\n\n### Non-Batch Solving\n\nNon-batch solving is utilized exclusively with every solver except for SHORTEST_PATH, in which case the length of the source and destination node lists will determine if non-batch or Batch Solving is utilized. If the source and destination lists do not match and the SHORTEST_PATH solver is specified, batch solving will be used. If the source and destination node lists match in length (and SHORTEST_PATH is the specified solver), non-batch solving will be used. This means that the first source index would only be matched with the first destination index, the second source index to the second destination index, etc.\n\nFor example, if you attempt to solve three unique sources (node_1, node_2, and node_3) to three unique destinations (dest_1, dest_2, dest_3) by only listing each source and destination once (calling /solve/graph with source_nodes set to [\"node_1\", \"node_2\", \"node_3\"] and destination_nodes set to [\"dest_1\", \"dest_2\", \"dest_3\"]), it would yield a result set like this:\n\n```+--------+--------+\n| SOURCE | TARGET |\n+========+========+\n| node_1 | dest_1 |\n+--------+--------+\n| node_2 | dest_2 |\n+--------+--------+\n| node_3 | dest_3 |\n+--------+--------+\n```\n\n### Batch Solving\n\nIf using the SHORTEST_PATH solver, multiple sources can be routed to multiple destinations using batch solving. A batch will be defined by each unique source node visiting each destination node provided which will yield a Cartesian product.\n\nTip\n\nCalculations from multiple unique sources are faster and more efficient than calculations with one unique source, but results may differ slightly between multiple unique source calculations and single unique source calculations (less than ~1% variance).\n\nFor example, if you wish to batch solve two unique sources (node_1, node_2) each routing to three unique destinations (dest_1, dest_2, dest_3), you'd list each source and each destination, noting that the solution will automatically map node_1 and node_2 to dest_1, dest_2, and dest_3 individually. A call to /solve/graph with source_nodes set to [\"node_1\", \"node_2\"] and destination_nodes set to [\"dest_1\", \"dest_2\", \"dest_3\"] would yield a result set like this:\n\n```+--------+--------+\n| SOURCE | TARGET |\n+========+========+\n| node_1 | dest_1 |\n+--------+--------+\n| node_1 | dest_2 |\n+--------+--------+\n| node_1 | dest_3 |\n+--------+--------+\n| node_2 | dest_1 |\n+--------+--------+\n| node_2 | dest_2 |\n+--------+--------+\n| node_2 | dest_3 |\n+--------+--------+\n```\n\nHowever, if you wish to batch solve three unique sources (node_1, node_2, node_3) to three unique destinations (dest_1, dest_2, dest_3) (or n sources to n destinations for that matter), you'll need to list out all 9 combinations manually, e.g.,:\n\nSource & Destination Node Definitions When the Lists Are the Same Size\n `````` 1 2 3 4 5 6 7 8 9 10 `````` ``````source_nodes=[ \"node1\", \"node1\", \"node1\", \"node2\", \"node2\", \"node2\", \"node3\", \"node3\", \"node3\" ], destination_nodes=[ \"dest_1\", \"dest_2\", \"dest_3\", \"dest_1\", \"dest_2\", \"dest_3\", \"dest_1\", \"dest_2\", \"dest_3\" ], ``````\n\nImportant\n\nRemember, if the source and destination nodes lists match in length, non-batch solving is used.\n\nThe above setup would yield a similar result set as the first example:\n\n```+--------+--------+\n| SOURCE | TARGET |\n+========+========+\n| node_1 | dest_1 |\n+--------+--------+\n| node_1 | dest_2 |\n+--------+--------+\n| node_1 | dest_3 |\n+--------+--------+\n| node_2 | dest_1 |\n+--------+--------+\n| node_2 | dest_2 |\n+--------+--------+\n| node_2 | dest_3 |\n+--------+--------+\n| node_3 | dest_1 |\n+--------+--------+\n| node_3 | dest_2 |\n+--------+--------+\n| node_3 | dest_3 |\n+--------+--------+\n```\n\n## Querying a Graph\n\nQuerying a graph is serviced by the /query/graph endpoint; this involves querying a graph for adjacent nodes (if provided edges) or adjacent edges (if provided nodes) using integer IDs, names, or WKT information. Additional adjacent rings around the specified nodes can also be queried. Results can be exported to a table in Kinetica.\n\nRequirements for querying a graph include:\n\n• name of the graph to query\n• a list of edge or node IDs, names, or WKTs to query\n\nImportant\n\nAdditional restrictions beyond those defined in the graph creation stage can also be provided. Query restrictions can utilize any of the restriction identifiers, including the unique LABEL identifiers.\n\n### Query Identifiers\n\nNodes or edges to be queried can be identified using any of the query-specific identifiers below.\n\nImportant\n\nConsult Components and Identifiers and Identifier Combinations for general information on identifiers and combinations. Note that the same limitations that apply to /create/graph and /solve/graph identifiers also apply to /query/graph identifiers\n\nIdentifier Supported Types Description\nEDGE_ID int, long Defines the edge ID value that will be matched against in the source graph to determine the source for the query\nEDGE_LABEL string\n\nDefines the edge LABEL value that will be matched against in the source graph to determine the source(s) for the query\n\nEDGE_WKTLINE wkt Defines the edge WKTLINE value that will be matched against in the source graph to determine the source for the query\nNODE_ID int, long Defines the node ID value that will be matched against in the source graph to determine the source for the query\nNODE_LABEL string\n\nDefines the node LABEL value that will be matched against in the source graph to determine the source(s) for the query\n\nNODE_NAME string Defines the node NAME value that will be matched against in the the source graph to determine the source for the query\nNODE_WKTPOINT wkt Defines the node WKTPOINT value that will be matched against in the source graph to determine the source for the query\nNODE1_ID int, long Defines the node ID value that will be matched against in the source graph (in conjunction with NODE2_ID) to determine the source for the query\nNODE1_NAME string Defines the node NAME value that will be matched against in the source graph (in conjunction with NODE2_NAME) to determine the source for the query\nNODE1_WKTPOINT wkt Defines the node WKTPOINT value that will be matched against in the source graph (in conjunction with NODE2_WKTPOINT) to determine the source for the query\nNODE2_ID int, long Defines the node ID value that will be matched against in the source graph (in conjunction with NODE1_ID) to determine the source for the query\nNODE2_NAME string Defines the node NAME value that will be matched against in the source graph (in conjunction with NODE1_NAME) to determine the source for the query\nNODE2_WKTPOINT wkt Defines the node WKTPOINT value that will be matched against in the source graph (in conjunction with NODE1_WKTPOINT) to determine the source for the query\nTARGET_NODE_LABEL string Defines the node LABEL value that will be matched against in the source graph to determine the destination for the query as well as the path taken through the graph to arrive at the destination\n\n#### Using Query Identifiers\n\nPair one value with an appropriate identifier to query for features connected to the given value within the provided number of rings:\n\nQuery Identifier Example (Single Key Search)\n ``````1 2 3 `````` ``````queries = [ \"{'male'} AS NODE_LABEL\" ], ``````\n\nPair multiple values with a single appropriate identifier to query for features connected to each value independently (effectively an OR) within the provided number of rings:\n\nQuery Identifier Example (Multiple Key Search, X OR Y)\n ``````1 2 3 `````` ``````queries = [ \"{'female', 'chess'} AS NODE_LABEL\", ], ``````\n\nImportant\n\nA rings value of 0 will return features that match the provided query exactly; in this case, the query would return nodes that have a label of female or chess\n\nProviding any combination with the TARGET_NODE_LABEL identifier (separated by an empty string) will effectively produce a source-destination query where the first combination is the source and the TARGET_NODE_LABEL is the destination (assuming the destination is within the provided number of rings):\n\nQuery Identifier Example (Single Target Search)\n ``````1 2 3 4 5 `````` ``````queries = [ \"{'Jane'} AS NODE_NAME\", \"\", \"{'chess'} AS TARGET_NODE_LABEL\" ], ``````\n\nImportant\n\nImplicitly defined nodes, e.g., from graphs defined with just edges and/or weights, cannot be queried.\n\n### Query Identifier Combinations\n\nTo properly query a graph using identifiers, there's a minimum set of identifiers that must be used. Each identifier combination must reference columns from the same table, e.g., the combination NODE1_ID and NODE2_ID must both use columns from the same table. The columns must also not be nullable.\n\n#### Nodes\n\nThe following combinations will query for edges adjacent to the node associated with the given information:\n\n• NODE_ID\n• NODE_LABEL\n• NODE_NAME\n• NODE_WKTPOINT\n• TARGET_NODE_LABEL\n\n#### Edges\n\nThe following combinations will query for nodes adjacent to the edge associated with the given information:\n\n• EDGE_ID\n• EDGE_LABEL\n• EDGE_WKTLINE\n• NODE1_ID, NODE2_ID\n• NODE1_NAME, NODE2_NAME\n• NODE1_WKTPOINT, NODE2_WKTPOINT\n\n## Matching a Graph\n\nMatching a graph is serviced by the /match/graph endpoint; this involves matching a directed route implied by a given set of latitude/longitude points to an existing underlying road network graph using a given solution type. The solution is then calculated and output to two tables (consult /match/graph for more information):\n\n1. A table containing track information and the mean square error score -- the lower the number, the more accurate a match.\n2. A table containing the coordinate information and how it relates to the track information and segment from the underlying graph.\n\nRequirements for matching a graph include:\n\n• name of the graph to match\n• sample points\n• solution type\n\nSolution types available:\n\nSolver Description CPU Parallel\nMARKOV_CHAIN Matches sample points to the graph using the Hidden Markov Model (HMM)-based method, which conducts a range-tree closest-edge search to find the best combinations of possible road segments for each sample point to create the best route. The route is secured one point at a time, so the prediction is corrected after each point. This solution type is the most accurate, but also the most computationally intensive. X\nMATCH_BATCH_SOLVES Matches each provided sample source and destination pair using the shortest path between the points. X\nMATCH_CHARGING_STATIONS Matches a given sample source and destination pair to the optimal recharging stations along the route (for EVs). X\nMATCH_CLUSTERS\n\nMatches the graph nodes with a cluster index using the Louvain clustering algorithm.\n\nNote\n\nParallel running of this solver is experimental and can be invoked with the parallel_clustering option.\n\nX*\nMATCH_LOOPS Matches closed loops (Eulerian paths) originating and ending at each graph node between min and max hops (levels). X\nMATCH_OD_PAIRS Matches sample points to find the most probable path between origin and destination (OD) pairs with given cost constraints. X\nMATCH_SIMILARITY Computes the Jaccard similarity between vertex pairs and N-level intersections within M hops. X\nMATCH_SUPPLY_DEMAND Matches sample generic supply depots to generic demand points using abstract transportation (referred to as trucks). Each route is determined by a truck's ability (size) to service demand at each demand point. X\n\n### Match Identifiers\n\nMapping a graph to a table requires a set of sample points. Sample points must be provided to the /match/graph endpoint using the unique identifiers below.\n\nImportant\n\nConsult Components and Identifiers and Identifier Combinations for general information on identifiers and combinations. Note that the same limitations that apply to /create/graph and /solve/graph identifiers also apply to /match/graph identifiers\n\nIdentifier Supported Types Description\nDEMAND_ID int, long A number representing a demand's unique identifier\nDEMAND_REGION_ID int, long A number representing a demand source's unique identifier\nDEMAND_SIZE int, long, float A number representing the size of the primary (or only) demand; used in demanding the supply denoted by SUPPLY_SIZE\nDEMAND_SIZE2 int, long, float A number representing the size of the secondary demand; used in demanding the supply denoted by SUPPLY_SIZE2, when there are two different demand type constraints (e.g., weight & volume)\nDEMAND_WKTPOINT wkt A WKT string representing a demand's geospatial source point\nDESTINATION_WKTPOINT wkt\n\nA WKT string representing a sample's geospatial destination point; e.g., POINT(X Y [Z])\n\nID int, long A number representing a sample's unique identifier\nNAME string A string value representing a sample's name\nOD_ID int, long A number representing an OD-pair-related sample's unique identifier\nOD_TIME float, double\n\nA number representing an origin-destination (OD) pair-related sample's cost\n\nNote\n\nOD_TIME may not necessarily depict time, e.g., if the graph's weights were distance-based, OD_TIME could theoretically reflect distance as well, assuming the values are consistent with the values used to create the original weights.\n\nORIGIN_WKTPOINT wkt\n\nA WKT string representing a sample's geospatial origin point; e.g., POINT(X Y [Z])\n\nPRIORITY int A number representing a sample's priority in match processing\nSUPPLY_ID int, long A number representing a supplier's unique identifier\nSUPPLY_ODDEVEN int\n\nA number representing whether the supplier is \"odd\", \"even\", or neither; used in applying odd/even-based restrictions, such as those used in Jakarta, Indonesia:\n\n• 0: supplier is not subject to odd/even-based restrictions\n• 1: supplier is \"odd\" numbered\n• 2: supplier is \"even\" numbered\nSUPPLY_PENALTY int, long, float, double A number representing the cost for unloading this supplier per unit unloaded, regardless of which demand site it is visiting, and in addition to the global unloading cost per unit\nSUPPLY_REGION_ID int, long A number representing a supplier source's unique identifier\nSUPPLY_SIZE int, long, float A number representing a supplier's primary (or only) capacity; used in meeting the demand denoted by DEMAND_SIZE\nSUPPLY_SIZE2 int, long, float A number representing a supplier's secondary capacity; used in meeting the demand denoted by DEMAND_SIZE2, when there are two different capacity type constraints (e.g., weight & volume)\nSUPPLY_WKTPOINT wkt A WKT string representing a supplier's geospatial source point\nTIME long, double\n\nA number representing a sample's time value\n\nNote\n\nTIME could theoretically represent any time unit (seconds, minutes, epoch, etc.) as long as the representations are consistent for the column in the source table.\n\nTRIPID int, long A number representing a unique trip identifier\nWKTPOINT wkt\n\nA WKT string representing a sample's geospatial point; e.g., POINT(X Y [Z])\n\nX float, double A number representing a sample's X or longitude value\nY float, double A number representing a sample's Y or latitude value\n\nNote\n\nRecords with a timestamp of 0 for the TIME column will be ignored when calculating the solution.\n\n### Match Identifier Combinations\n\nTo properly match a graph using identifiers, there's a minimum set of identifiers that must be used. Each identifier combination must reference columns from the same table, e.g., the combination WKTPOINT and TIME must both use columns from the same table. The columns must also not be nullable. The valid match identifier combinations per match graph solver are as follows:\n\n#### Markov Chain\n\n• X, Y, TIME\n• WKTPOINT, TIME\n• X, Y, TIME, TRIPID\n• WKTPOINT, TIME, TRIPID\n\nNote\n\nIf using the TRIPID identifier to match the graph, all trip IDs will be used in the solution.\n\n#### Match Batch Solves\n\n• ORIGIN_WKTPOINT, DESTINATION_WKTPOINT, OD_ID\n\n#### Match Loops\n\n• ID\n• NAME\n• WKTPOINT\n\nNote\n\nUsing no identifier combination will result in loops being searched for across the entire graph\n\n#### Match OD Pairs\n\n• ORIGIN_WKTPOINT, DESTINATION_WKTPOINT, OD_TIME\n• ORIGIN_WKTPOINT, DESTINATION_WKTPOINT, OD_TIME, OD_ID\n\n#### Match Supply Demand\n\nA supply and demand combination are used in conjunction with each other to match suppliers to demand. Reference Identifier Combinations for using multiple combinations syntax.\n\n##### Demand Combinations\n• DEMAND_ID, DEMAND_WKTPOINT, DEMAND_SIZE, DEMAND_REGION_ID\n• DEMAND_ID, DEMAND_WKTPOINT, DEMAND_SIZE, DEMAND_REGION_ID, PRIORITY\n\nAny or all of the following identifiers can be added to a demand combination:\n\n• DEMAND_PENALTY\n• DEMAND_SIZE2 (must be paired with a supply combination using SUPPLY_SIZE2 )\n##### Supply Combinations\n• SUPPLY_REGION_ID, SUPPLY_WKTPOINT, SUPPLY_ID, SUPPLY_SIZE\n\nAny or all of the following identifiers can be added to a supply combination:\n\n• SUPPLY_ODDEVEN\n• SUPPLY_PENALTY\n• SUPPLY_SIZE2 (must be paired with a demand combination using DEMAND_SIZE2 )\n\n## Showing a Graph\n\nUsing /show/graph will provide detailed information about a graph, including number of nodes and edges in the graph, whether the graph is directed or persisted, and more.\n\n## Deleting a Graph\n\nDeleting a graph is serviced by the /delete/graph endpoint; this involves providing a graph name to delete the graph from the graph server (memory) and persist (if applicable).\n\nTip\n\nIf a graph was saved to persist upon creation and then was deleted from the server (but NOT persist), it can be reloaded from persist by recreating the graph using the same graph_name.\n\n## Managing Permissions for a Graph\n\nGraph permissions can be managed either through SQL or through the native API.\n\n## Examples\n\nFor detailed usage of the graph capability, see Guides:\n\n## Limitations and Cautions\n\n• Groups of valid identifier combinations must be from the same table, e.g., node ID & NAME must reference columns from the same table\n• Node, edge, weight, and optional restriction identifiers should be matched to yield a useful graph (node ID ‣ edge NODE1_ID and edge ID ‣ weights EDGE_ID, etc.)\n• Groups of valid numerical identifier combinations must be the same type, e.g., if edge ID is identified from an int column, both edge NODE1_ID & NODE2_ID must also be int\n• Graphs cannot be created using columns with the nullable property\n• If no ID identifier is provided, weights will be matched with edges by table row, e.g., the first record in the weight table will be used for the first record in the edge table (should the weights and edges be separate). If two weights are specified for the same edge, the weights are added (if both are using the VALUESPECIFIED identifier) or multiplied (if one or both are using the FACTORSPECIFIED identifier) together.\n• A node or edge can have up to 64 unique labels." ]

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[ "# AWK get first line\n\nI'm trying to add a column for each line together, and I do it like this:\n\n``````awk '{s+=\\$2} END {print s}'\n``````\n\nNow, when I have the final sum, s, I want to subtract the second column of the first line, like this:\n\n``````awk '{s+=\\$2} {a= COLUMN 2 OF FIRST LINE } END {print s - a}'\n``````\n\nHow would I do that?\n\n``````awk 'FNR>1{s+=\\$2};END{print s}'" ]

[ null ]

[ "", null, "", null, "", null, "Chapter 9, Problem 1QP\n\nChapter\nSection\nTextbook Problem\n\n“The firm’s entire marginal cost curve is its short-run supply curve.” Is the preceding statement true or false? Explain your answer.\n\nTo determine\n\nWhether the given statement is true or false.\n\nExplanation\n\nTechnically, the statement the “firm’s entire marginal cost curve is its short-run supply curve”, which is not true. The marginal cost curve and its short-run supply curve are different. The short-run supply curve is the firm’s entire marginal cost curve only when the price of the goods is above the minimum point of the average cost curve. A firm must cover its variable cost in the short run.\n\nConcept\n\nMarginal cost (MC): Marginal cost refers to the amount of additional costs incurred in the process of increasing one more unit of output. Marginal cost curve shows the additional cost for the different levels of output.\n\nShort-run supply curve: In the short run, it is a curve that shows the relationship between the price level in the economy and the supply in the economy by the firms.\n\nThe Solution to Your Study Problems\n\nBartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!\n\nGet Started\n\nWhat is the relationship between supply and demand?\n\nFoundations of Business (MindTap Course List)\n\nPREFERRED STOCK VALUATION Ezzell Corporation issued perpetual preferred stock with a 10% annual dividend. The s...\n\nFundamentals of Financial Management, Concise Edition (with Thomson ONE - Business School Edition, 1 term (6 months) Printed Access Card) (MindTap Course List)\n\nFind all the answers to your study problems with bartleby.\nTextbook solutions plus Q&A. Get As ASAP arrow_forward", null, "" ]

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[ "# Two locators on two different plots in Manipulate\n\nI am trying to create a Manipulate module that shows two plots, and have locators on each of those plots, as well as two sliders. The locators do do not affect each other's position, but one of the locators affects both sliders, and vice versa. Here is a minimal version of my current code:\n\nManipulate[Row[{\nPlot[(x - a)^2 + b, {x, 0, 1}, ImageSize -> Medium,\nPlotRange -> {0, 1}],\nPlot[{(x - pt[])^2 + pt[], b}, {x, 0, 1}, ImageSize -> Medium,\nPlotRange -> {0, 1}, GridLines -> {{a}, {}},\nEpilog -> {PointSize[Large], Point[Dynamic@{a, b}]}]}],\n{a, 0, 1}, {b, 0, 1}, {{pt, {0, 1}}, Locator}]\n\n\nThis is a screenshot of the scenario:", null, "In the code above only one locator is declared. The other locator I would like to declare should be at {a,b} in the second plot. I've made the point Dynamic, but it doesn't do anything when I try to click-drag with my mouse. I've seen this question about locators and sliders affecting each other, but there only one graph has an interactive locator, and I would like both to have an interactive locator. I also tried to get some intuition from this question about specifying which element of Grid a locator should affect, but I am not an expert Mathematica user, so I couldn't adapt it.\n\nSo, my question is:\n\nHow can I get two independent locators, each in a different plot, one of which depends on sliders?\n\n• What is the second locator supposed to do ? May 16, 2019 at 3:53\n• The second locator should be at the coordinate {a, b} in the second plot, and moving it should change the a and b sliders. This also affects the first plot. May 16, 2019 at 3:55\n\nIs this what you need ? :\n\nDynamicModule[{pt1 = {0.5, 0.5}, a = .6, b = .2}\n, Column[{\nManipulator[Dynamic[a], {0, 1}],\nManipulator[Dynamic[b], {0, 1}],\nRow[{LocatorPane[Dynamic[pt1],\nDynamic @ Plot[(x - a)^2 + b, {x, 0, 1}, ImageSize -> Medium,\nPlotRange -> {0, 1}]]\n, LocatorPane[Dynamic[{a, b}],\nDynamic @\nPlot[{(x - a)^2 + b, b}, {x, 0, 1}, ImageSize -> Medium,\nPlotRange -> {0, 1}, GridLines -> {{a}, {}},\nEpilog -> {PointSize[Large], Point[Dynamic@{a, b}]}]]\n}]\n}]\n]", null, "• Nice! Almost, the second plot should be (x-pt1[])^2+pt1[], and that works as desired. This does answer my question, b̶u̶t̶ ̶-̶ ̶w̶h̶a̶t̶ ̶i̶f̶ ̶I̶ ̶h̶a̶v̶e̶ ̶t̶w̶o̶ ̶l̶o̶c̶a̶t̶o̶r̶s̶ ̶i̶n̶ ̶t̶h̶e̶ ̶f̶i̶r̶s̶t̶ ̶p̶l̶o̶t̶?̶ ̶I̶s̶ ̶t̶h̶e̶r̶e̶ ̶a̶n̶ ̶e̶a̶s̶y̶ ̶w̶a̶y̶ ̶t̶o̶ ̶d̶o̶ ̶t̶h̶a̶t̶ ̶o̶n̶ ̶t̶o̶p̶ ̶o̶f̶ ̶t̶h̶i̶s̶?̶ Figured it out, just put Dynamic[{pt1,pt2}] in LocatorPane. Thanks for your help. May 16, 2019 at 18:57" ]

[ null, "https://i.stack.imgur.com/EFoTD.png", null, "https://i.stack.imgur.com/PLDlj.gif", null ]

[ "Translator Disclaimer\n29 June 2004 Nonmonotone impulse effects in second-order periodic boundary value problems\nIrena Rachůnková, Milan Tvrdý\nAbstr. Appl. Anal. 2004(7): 577-590 (29 June 2004). DOI: 10.1155/S1085337504306299\n\n## Abstract\n\nWe deal with the nonlinear impulsive periodic boundary value problem $u''= f(t,u,u')$, $u(t_i+)=\\mathrm{J}_i(u(t_i))$, $u'(t_i+)=\\mathrm{M}_i(u'(t_i))$, $i=1,2,\\dotsc,m$, $u(0)=u(T)$, $u'(0)= u'(T)$. We establish the existence results which rely on the presence of a well-ordered pair $(\\sigma_1,\\sigma_2)$ of lower/upper functions $(\\sigma_1\\le\\sigma_2 \\text{ on } [0,T])$ associated with the problem. In contrast to previous papers investigating such problems, the monotonicity of the impulse functions $\\mathrm{J}_i$, $\\mathrm{M}_i$ is not required here.\n\n## Citation\n\nIrena Rachůnková. Milan Tvrdý. \"Nonmonotone impulse effects in second-order periodic boundary value problems.\" Abstr. Appl. Anal. 2004 (7) 577 - 590, 29 June 2004. https://doi.org/10.1155/S1085337504306299\n\n## Information\n\nPublished: 29 June 2004\nFirst available in Project Euclid: 7 July 2004\n\nzbMATH: 1109.34024\nMathSciNet: MR2084937\nDigital Object Identifier: 10.1155/S1085337504306299\n\nSubjects:\nPrimary: 34B15 , 34B37\nSecondary: 34C25", null, "", null, "" ]

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[ "Rigid Body Moving Through a fluid\nSuppose we have a solid N0-D body B with surface (N0-1)-curve dB moving with stance bivelocity W through a fluid which we assume to be at rest prior to the passage of B. We expect B to leave a kinematic disturbance (or \"wake\") in the fluid, losing kinetic enregy in the process. Even if the flow is irrotational, for N0=2 there can be nonzero circulations about circuits that enclose B.\nFor concreteness let us consider an aircraft Q flying over a city 4. The aircraft has stance S4¬Q and 2-vector bivelocity W4=-2S4¬QS4¬Q· and we assume for simplicity that the atmospheric flow vp4 at a given event p due to wind, thermals, flak and so forth varies only slightly at the scale of the aircraft so that we can embody the motion of the atmosphere encountered by the plane by a single wind velocity 1-vector w4(c) where c is the instantaneous position of the mass centre of the plane .\nIt is convenenient to cast aircraft bivelocity W and the wind velocity w into the frame of the aircraft and consider the aircraft to be travelling with bivelocity W+ = WQ - wQe¥ through a fluid at rest. We choose Q coordinates such that the aircraft is (approximately) symmetrical in the y=0 plane with e1 coresponding to \"forward\", e2 to \"left\" and e3 to \"up\". We expect the velocity component of W+ = WQ - wQe¥ to approximate a positive multiple of e1e¥ but have some smaller \"sideslip\" and \"drop\" e2e¥ and e3e¥ components.\nThe 2-vector biimpulse T(W+) imparted by the fluid to the aircraft is dependant solely on the surface geometry of the aircraft and is generally not linear in W+; indeed it tends to have quadratic form T(lW+) = l2T(W+) since if we double W+ then in small time interval dt we encounter twice as many fluid particles each of which has twice as much momentum relative to the body. In reality, aerodynamic properties vary with higher mach speeds so we do cannot characterise T purely \"directionally\" via T(W+) = |W+|2 T(W+~) .\nHowever, classical aerodynamic theory usually formulates the force on the body solely due to its velocity as T = åi<jv2 Cij(v~Q) eijW where the ½N0(N0+1) direction-dependant scalar coefficients can be reagrded as coordinate in the wind frame having e1W = v~ ; e2W = ¯v*(e2Q) ; e3W = ¯v*(e3Q) .\nDirection v~Q is typically parameterised by aerodynamic angles a = tan-1(e3Q¿v~ (e1Q¿v~)-1) and b= sin-1(e3Q¿v~) .\n\nBivectors W+ and T(W+) each have ½N(N+1) coordinates since W+ resides in e¥* while T resides in e0*. Thus expressing   each T coordinate as a general quadratic form of the W+ coordinates requires (½N(N+1))3 scalar coefficients. This is 27 for N=2 and 216 for N=3.  Consider approximating T(W+) as åk=1K  (W+¿ak)Ù(W+¿bk) for 2K 3-vectors a1,b1),...,ak,bk). imparted by the fluid to the aircraft is dependant solely on the surface geometry of the", null, "Fluid Mechanics\n\nIntroduction\nTo model the motion of a body through a fluid medium such as air or water we must at least approximate the effects of the fluid on the body which requires simulation at least in part of the motion of the fluid, which is likely to be complicated and frquently chaotic.\n\nClassical Flows\nClassical fluid mechanics typically represents a flow of matter by a ÂN0 nonunit 1-field where N0 is the number of spacial dimensions with MP representing at a given time t the total momentum per unit volume passing through a small test volume at spacial position PÎÂN0; together with scalar 0-field mP representing the density (mass per unit volume) at P (as measured over a small test volume) , constant for an incompressible fluid. We have Mp = mpVp where Vp=mp-1Mp is the (nonunit) average spacial velocity vector of the matter at p= P+te4 is an N=N0+1 dimensional spacetime event. Vp is undefined when mp=0.\nIn thermodynamics we also consider the matter at p to have scalar temperature Qp.\nThe flow satisfies the conservation of matter law (aka. a continuity equation) if ÑP¿Mp  =  -mP/t  =  0 which states that the ammount of matter leaving a small region of volume V at P per unit time as measured over a small test interval equals the fall in the ammount of matter mPV at V at P per unit time We can express the matter conservation law as\nmP/ t + ÑP¿(mpVp)   =   mP/ t + (Vp¿ÑP)mpÑ + mp(ÑP¿Vp)   =   (( / t) + (Vp¿ÑP))mp + mp(ÑP¿Vp)   =   0\nwhere ( / t) + (Vp¿ÑP) is known as the substantial derivative corresponding to the rate of change when \"following\" the flow. Note that this represents the derivative in the direction of the flow scaled by the speed of the flow.\n\nIf we define ÂN 1-vectors p=P+te4 and vp=Vp±e4 where N=N0+1 and e4 is a (non-relativistic) \"temporal\" axis perspendicular to ÂN0 with e42=±1 then the substantial direction becomes (vp¿Ñp) with the differentiating scope of the Ñp extending only rightwards rather than encompassing the vp.\n\nFor a steady flow Mp = MP with vanishing Ðe4 derivatives the matter conservation law becomes (Vp¿ÑP)mpÑ + mp(ÑP¿VpÑ) = 0 and for an incompressible flow with mp=m>0  this yields the incompresability condition ÑP¿Vp = 0, ie. the spacial divergence of the N0-D velocity 1-vector is zero. For vp=Vp±e4 , the incompressability condition is Ñp¿vp=0.\n\nFor an incompressible flow \"momentum\" and \"velocity\" fields become effectively equivalent with Mp=mVp.\n\nClassical Pressure\n\nThe scalar static pressure sp at p is independant of the macroscopic \"net flow\" mp at p , and stems instead from microscopic \"thermal\" agitatation of the particles \"at\" p . The use of the term \"static\" here is traditional, and relates to \"pressure not due to flow\" rather than a restriction on the variability of sp with p. The pressure sp at p the sum of the static pressure and pressure due to the fluid motion,\nConsider a small Â3 3-simplex or 2-sphere V centred at p in a fluid at rest from within which the fluid has been removed. Random microscopic thermal motion of the fluid will tend to cause molecular impacts on the surface of the simplex which we can regard as imparting locally normal \"inward\" momentum uniformly across the boundary surface. Integrating this inward component across the surface and dividing by the surface area content leads us to scalar pressure which we can regard as the magnitude of the thermal e4-forces exterted across a small 2-simplex or 1-sphere centred at p of aribitary 2-blade tangent, divided by the simplex area.\nFor N0||>3, scalar sp represents the inward thermal forces across a (N0-1)-sphre.\nIn an anisotropic ideal gas spV is proportional to nQp where V=|V| is the volume of a test 3-simplex V at p containing n molecules of average temperature Qp. n is proportional to total mass mpV so sp is proportionate to mpQp and we usually write sp = RmpQp where R is the gas constant for the fluid.\nThe entropy proportionate to (spmp-g) = sp - gmp where dimensionless g is the specific heat ratio is conserved by fluid elements in the absence of heat conduction so that (vp¿Ñp)spmp-g = 0 . Entropy has units ( m-4) so can be measured in m .\n\nNow consider some fluid inside V. The total forces acting on this fluid are the external macroscopic forces òV |d3p| fp where fp=mpgp + hp is e4-spacial forces exerted   (typically we might have gp=-ge3 for uniform vertical gravity) , and the static pressure integrated over the boundary òdV|d2p|(-np)sp   =   -òdVd2p e123-1sp   =   -e123-1 òVd3p ÑPsp     where np = I2p-1 e123 = I2p e123-1 is the outward normal at p Î dV  .\nClassicists typically asssume gp=-ÑpGp and hp=-ÑpHp derive from gravitational and nongravitational scalar 0-potentials.\nThe total change in momentum incured by the mass mp|V| of fluid as it follows the flow lines is òV d3p (ÑP¿vp)mp so we have\n(ÑP¿vp)mp = mpgp + hp - Ñpsp     which for constant density mp=m simplifies to the momentum conservation law\n(vp¿ÑP)vp = gp + m-1hp - m-1(ÑPsP)     aka. Euler's equation when combined with ÑP¿VP=0 .\nFor P in Â3 we have (Vp¿ÑP)VpÑ   =   ½ÑP(Vp2) - Vp×(ÑP×VpÑ) so we can express the momentum conservation law as\n½ÑP(Vp2)  -  Vp×(ÑP×VpÑ)  +  m-1ÑPsp - m-1fP   =   0 .\n[ Proof : a×(b×c) = (a¿c)b - (a¿b)c Þ Vp×(ÑP×Vp) = (Vp¿ÑP)VpÑ - (Vp¿VpÑ)ÑP = (Vp¿ÑP)VpÑ - ÑP(VpÑVp) = (Vp¿ÑP)VpÑ - ½ÑP(Vp2)  .]\n\nSmall peturbation sound waves\nLet scalars pressure sp = s0+sp and density mp=m0+mp be small deviations from s0 and _wasmup0\nSince entropy (spmp-g) = (s0+sp)(m0+mp)-g) = _presp0m0-g Þ (1+s0-1sp)(1+m0-1mp)-g = 1 which to first order in s0-1sp and m0-1mp is sp = s2mp where s= (m0-1s0g)½ .\nNow let us suppose that Vp is small so that the matter conservation law mp· = -(Vp¿ÑP)mp - mp(ÑP¿Vp) reduces to mp· = -m0(ÑP¿Vp) giving (ÑP¿Vp) » -m0-1mp· .\nThe momentum conservation law linearises to ÑPsp = (mp+m0)gp + hp - (mp+m0)Vp· » (mp+m0)gp + hp - m0Vp· which gives ÑP2sp = (ÑP¿)((mp+m0)gp + hp - (mp+m0)Vp·) = (ÑPmp)¿gp - m0(ÑP¿)(Vp·) = (ÑPmp)¿gp - m0(ÑP¿Vp)· = (ÑPmp)¿gp + (mp·)· = (ÑPmp)¿gp + (s-2sp·)· Þ sp·· = s2ÑP2sp so the pressure and desnity perturbations satisfy the standard 3D wave equation with radial solution sp = r-1(F(r-st)+G(r+st)) corresponding to an outward wave of amplitude F of radial speed s and an inward wave of amplitude G and radial speed -s.\ns=(m0-1s0g)½ thus corresponds to the propagation speed of density and pressure perturbations from steady state values m0 and s0 and is known as the speed of sound in the fluid.\n\nNavier-Stokes\nSuppose we combine the static pressure force -spnp with a viscous force u((np¿Ñp)mpÑ + Ñp(np¿mpÑ)) ; reducing to um((np¿Ñp)vpÑ + Ñp(np¿vpÑ)) when mp=m.\nIf we regard np as varying negligibly with p compared to the variation of vp (eg. when over a flat surface) then we can view this as the np directed flow derivative plus the gradient of the np-directed flow-speed, equally wieghted by scalar u, leading to the Navier-Stokes equation\n(vp¿Ñp)mp   =   -mp(ÑpGp) - (Ñpsp) - (ÑpHp) + (vp¿Ñp)mpÑvp + uÑp2mp .\nFor incompressible flow mp=m this becomes\n(vp¿Ñp)vp   =   -(ÑpGp) - m-1((Ñpsp) + (ÑpHp)) + uÑp2vp where scalar u is the kinematic viscosity. Water has kinematic viscosity 10-6 m2s-1 at 15 oC while air is fifteen times that, olive oil a hundred. Treacle has viscosity of roughly 1.2×107 m2s-1 at 15 oC, falling rapidly with temeprature.\n\nVorticity\nFor an incompressible flow under gravity, Navier-Stokes becomes\n(Ñp¿vp)vp = -ÑpGp  - m-1((Ñpsp) + uÑp2vp\nand since (vp¿Ñp)vp = vp¿(ÑpÙvp) + ½Ñp(vp2) we have vp¿wp = -ÑpEp + uÑp2vp where Ep = ½vp2 + Gp + m-1sp and wp º ÑpÙvp has Ñp¿wp=0. Hence ÑpÙ(vp¿wp) = uÑpÙ(Ñp2vp) = uÑp3.vp .\n\nBut ÑpÙ(vp¿wp) = (Ñp¿(vpÙwp))i where (N-2)-vector kinematic vorticity wp º wpi-1 = (ÑpÙvp)* so we have Ñp¿(vpÙwp) = 0\n[ Proof : ÑpÙ(vp.wp) = (Ñp¿(vpÙ(wpi-1)))i = (Ñp¿(vpÙwp))i  .]\nFor incompressible flow, Ñp¿(vpÙwp) = (vp¿Ñp)wpÑ - vpÑÙ(wp.Ñp)vpÑ so we have geometric vorticity equation\n(vp¿Ñp)wp = (wp.Ñp)Ùvp + u Ñp¿(Ñp2vpi-1)     with the differentiating scope taken rightwards only.\nFor N=3 inviscid flow (u=0) this reduces to (vp.Ñp)wp = (wp.Ñp)vp with 1-vector vorticity wp = (ÑPÙVP)* = ÑP×VP and is known as the third Helmholtz vortex theorem.\n[ Proof :  (vp¿Ñp)wpÑ + (Ñ¿vpÑ)wp - vpÙ(Ñp¿wpÑ) - vpÑÙ(Ñp¿wp) reduces to result since Ñp¿wp = ÑpÙwp = ÑpÙÑpÙvp vanishes as does Ñp¿vp  . Also u(ÑpÙ(Ñp2vp))i-1 = u(Ñp¿((Ñp2vp)i-1)  .]\n\nThus for N0=3 the flow-directed derivative of the vorticity (vp¿Ñp)wpÑ equals the vorticity-directed derivative of the flow (wp¿Ñp)vpÑ ;   while for N0=2 steady flows we have scalar vorticity preserved along streamlines   (ÑP¿VP)wp=0 since (wp¿Ñp)vp=wÐe3vp=0.\n\nWhen fp=0, integrating along a streamline gives Bernoulli's equation sp + ½mvp2 = s0 , where scalar s0 is known as the stagnation pressure. Bernoulli's equation applies when mp is constant and the flow is static. More generally Ep = sp + mGp + Hp + ½mvp2 is constant along the streamline if mp=m and wp=0 along it.\n[ Proof : sp1   =   sp0 + òt0t1 dt vp¿(Ñpsp)   =   sp0 + òt0t1 dt  vp¿ (-mp(ÑpGp) - (ÑpHp) - (Ñp¿vp)mpÑvp - mp(Ñp¿vp)vp )\n=   sp0 + m(G0-G1) (H0-H1) - m òt0t1 dt  vp¿((Ñp¿vp)vp )   =   sp0 + m(G0-G1) (H0-H1) - m òt0t1 dt  vp¿(vp¿wp + Ñpvp2 )\n=   sp0 + m(G0-G1) (H0-H1) + ½m (v02 - v12) . Hence sp + mGp + Hp + ½mvp2 is constant along the streamline.  .]\n\nCirculation\nThe geometric M-circulation about a closed M-curve CM of a flow mp=mpvp is the <M-1;M+1>-vector GC1 º òCM dMp vp . Circulation is thus usually considered to be an integration of velocity rather than momentum and so has units mM+1 s-1.\nThe 1-circulation about a closed 1-curve (ie. a loop) C1 is thus the <0;2>-vector GC1 º òC1 dp vp and the term circulation traditionally refers to the scalar part GC1 º òC1 dp¿vp .\n\nFor N0=2, C1 bounds a \"solid\" flat surface region C2 and for an irrotational flow with vp=Ñpfp defined over C2 we have\nGC1 º òdC2 dp vp   =   òdC2 dp Ñpfp   =   òd2C2 d0p fp = 0 .\nHence for N0=2 the scalar circulation vanishes provided the region enclosed by C1 contains no regions absent of flow or at which ÑpÙvp is nonzero. In particular the scalar circulation around a solid lamina need not vanish, but is independant of enclosing path C1. Similarly a vortex point with nonzero ÑpÙvp within C2 will contribute a fixed \"residue\" circulation independant of the  enclosing path. |&| òC2fp|d2p| .\n\nThe geometric generalisation of Stoke's law òdCN0-1 dN0-2p.¦(p)   =   (-1)N òCN0-1 (ÑÙ¦(pÑ)).dN0-1p means that the (N-3)-vector component of the (N0-2)-circulation around the (N0-2)-curve boundary of an open hypercurve CN-1 over which vp is defined equals the vorticity integration over CN0-1 (-1)N0 òCN0-1 (ÑÙvp).dN0-1p . Thus if vp is defined and irrotational over CN0-1 the (N0-3) component of GdCN0-1 vanishes.\nThus for N0=3 we have GdC2 º òC1 dp¿vp = -òC2 (ÑÙvp).d2p and so if GC1 is nonzero for a closed 3D 1-curve (ie. a path loop) and C2 is any 3D hypercurve spanning (bounded by) C1 over which vp is everywhere defined, then vp must be irrotational over one or more regions of C2.\n\nIf CN0-1=dCN0 is the hypercurve boundary of an open set CN0 over  which vp is defined   then the <N0-2;N0>-vector hypercirculation GCN-1 º òCN-1 dN-1p vp   =   òCN dNp Ñpvp vanishes for a static incompressible irrotatuonal flow with Ñpvp=0.\nIf CN containis regions over which vp in undefined or zero then GCN-1 may be nonzero but tends to be indendant of CN-1 in that we can deform CN-1 without altering GCN-1 provided we do not move CN-1 across any undefined vp regions .\n\nThe hypercirculation about a (N0-1)-curve CN0-1 is the <N0-2;N0>-vector GCN-1 º òCN-1 dN-1p vp = òCN0-1 ((-1)N0-2 ¯dN0-1p(vp) _dNm1 + (-1)N-1^dN0-1p(vp) dN0-1p) = (-1)N0òCN0-1 (¯dN0-1p(vp) - ^_dN0m1(vp)) dN-1p = (-1)N0 (òCN0-1 ¯dN0-1p(vp)¿dN0-1p) - òCN0-1 ^dN0-1p(vp)ÙdN0-1p) .\nWhen vp=Ñpfp the (N0-2) component    òCN0-1 ¯dN0-1p(vp)¿dN-1p = òCN0-1 ¯dN0-1p(Ñpfp)¿dN0-1p = òCN0-1 (¯dN-1p(Ñp)fp)¿dN0-1p º òCN0-1 (Ñp[CN-1]fp)¿dN0-1p =  òdCN0-1 fp dN0-2p vanishes for closed CN0-1 (or has a constant (N0-2)-vector value comprising residues at enclosed poles).\nThe pseudoscalar hypercicrulation component òCN0-1 ^dN0-1p(vp)Ù_dN0m1 = i òCN0-1 (vp¿np)|_dN0m1| measures the outflow across CN0-1 and vanishes for a steady incompressable flow.  Otherwise for small |CN0-1| it approximates   i|CN0-1|(ÑP¿Vp) = i|CN0-1|ÑP2fp evaluated at any particular p enclosed by CN0-1.\n\nReynolds Number\nIt is frequently the case in Navier-Stokes dynamics that one of (vp¿Ñp)vp and uÑp2vp will dominate to the extent that the other may be neglected and far simpler equations solved. The Reynolds number u-1ua where u is a typical flow speed and a a characteristic length for the problem provides a rough indicator of the likely ratio of the magntiudes of the inertial (vp¿Ñp)vp term to the viscous uÑp2vp term; a high Reynold's number favouring the inviscid approximation forcing u=0.\nThis can fail when the large second derivatives present in thin boundary layers \"seperate\" from the boundary and signifcantly effect the flow far from the boundary.\n\nComplex Potential\nThe complex potential is used to emulate incompressable, irrotational, usually steady, Â2 planar flows.\n\nStream Function\n\nSetting scalar stream function yp º òC dV2 V1 - òC dV1 V2 for any Â2 path C from 0 to p yields\nV1 = Ðe2yp ; V2 = - Ðe1yp    ;     ie. Vp = e12(ÑPyp) = (ÑPyp)e12-1 = Ñp×(ype3) = (ÑPÙ(ype3))e123-1 with Vp2 = (Ñpyp)2 .\nIf yp is any analytic real scalar field over Â2 satisfying this then its analyticity provides incompressability condition Ðe1V1 + Ðe2V2 = 0, ie. ÑPVp=0 and also Ñp2yp = 0. Hence Vp = (ÑP(ype3))e123-1 = -ÑPype12 .\nVp2 = (Ðe2yp)2   + (-Ðe1yp)2 = (Ñpfp)2\nAlso note that ÐVp yp = (Vp¿Ñp)yp = 0 so yp is constant when following the flow. Such 1-curves along which yp is constant are known as streamlines.\n\nVelocity Potential\n\nFor a steady irrotational incompressible flow with ÑpÙVp = 0 we can constuct a 0-potential fp = òC dp¿Vp for any path C from 0 to p so that Vp=Ñpfp , ie. Vi = ei2Ðeifp . If the flow is conserved we also have Ñp¿Vp=0 so ÑpVp = Ñp2fp = 0.\nSetting Fp º fp(ÑPfp) we have ÑPFp = (ÑPfp)2 + fpÑP2fp = (ÑPfp)2 = Vp2 hence for constant density flow the kinetic energy within a volume ½mòV d3p VP2   =      ½mòdV d2p fp(ÑPfp) with pseudoscalar part giving divergence theorem ½mòV |d3p|VP2 = -½m òdV|d2p|fpÐnfp where n is the outward normal to the (N0-1)-curve integration surface dV .\nThus in an incompressibe incompressable flow, we can deduce the kinetic energy insode a volume of fluid from fpÐnfp over the blundary of V\n\nComplex representations of 2D Flows\n\nWhen N0=2 then by taking i=e21=-e12 and setting z = pe1 = x1+x2e21 = x1+x2i ; v(z) = Vpe1 = V1+V2e21 ; we can work in commtuing algebra C @ Â2 +.\nWe define automorphic conjugation x^ º e1xe1 negating e2 and e12 but preserving 1 and e1 so corresponding to complex conjugation over Â2 + and reflection in x2=0 over Â2, giving innerproduct a.b = Real(a^b) and a2 = a^a = |a|+2 .\n\nComplex Potential\n\nThe Complex 0-potential fp º fp + iyp satisfies ÑP2fp = 0 since ÑP2fp and ÑP2yp both vanish. It is defined for x=P where P Î ÂN0=Â2 but by defining z=x1+ix2 = pe1 where p=x1e1+x2e2 and taking i=-e12 we can regard fp = fz,t as a t-dependant regular function mapping C®C whose conjugated derivative provides flow v(z) = Vpe1 = f'(z)^ .\nBy virtue of its construction, the complex potential satisfies the Cauchy-Riemann equations and so any regular complex function provides a complex potential with a direction-independant derivative f'(z,t) º (d/dz)f(z,t) = df/dx1 + idy/dx1 = V1 - iV2 = v^(z,t) .\nHence Vp = f'^ = f^' where ' denotes regular differention with resepect to z and so Vp2 = |f'(z)|+2 .\n\nA regular complex 0-potential fp(t) with Ñp2fp=0 thus fully embodies an incompressible irrotational 2D flow with spacial derivative f' equal to the complex conjugate of the flow, and streamlines indicated by constant Imag(f). If fp is independant of t then the flow is steady.\nApart from at z=0, the reciprocal potential fp(t)-1 is also regular with (fp(t)-1)' = -fp(t)-2 f'p(t) . The fp(t)-2 factor is complex and both redirects and rescales the derivative.\nSimilarly f(lza) has flow (f'(lza)la za-1)^ which is (la za-1)^ times the flow due to f(z). For a=-1 we have f(lz-1) giving flow -(lz-2)^ times the flow due to f(z).\n\nExample: Uniform Flow\nThe uniform flow of speed A and direction subtending a with the horizontal is given by v(z,t) = a = A(ia) and has fp = a^ z = A(-ai) z where p=(x1,x2)=[r,q] and z=x1+ix2.\n\nExample: Line Vortex\nConsider the circular flow up = gr-1eq    , ie. v(z,t) = g|z|+-2iz = gi(z-1)^ = (-giz-1)^ having circulation 2pg |+ e12 ò0a dz ò02p dq gq z = 2pg(1+ e12½pa2). .\nyp=-gr while the velocity potential is fp=gq yielding fp = -gi(z)\nSetting G=2pg we have line vortex potential fp = G(2pi)-1(z) with circulation G(1+e12½pa2)   about any loop enclosing r=a and associated flow up = (2p)-1Gr-1eq   tending to zero at r=¥ . .\nThe r-1 factor on the speed means that the angular momentum mppÙVp = mpgereq = mp(2p)-1Ge12 = mp(2pi)-1G about 0 at p is independant of p for incompressible flow mp=m.\nNote that f(lz-1) = G(2pi)-1(l - z) is the potential for circulation -G plus an irrelevant constant.\n\nFor large D, fp = G(2pi)-1(z-D) approximates uniform flow v(0)^ = (2p)-1GD-1i(-D)~ = (2pi)-1GD-1 so\nfp = a^D(z-D) provides flow close to a for r<<D.\n\nUniform Â2 Flow Past a Circular Boundary\nIf f(z) is the complex potential of a flow having no singularities for |z|<a then f(z) + (f(a2z^-1))^ = f(z) + (f(a2|z|+-2z))^ has the same singularities as f(z) over |z|>a and is purely real when |z|=a (ie. yp=0 when r=a), so the circle r=a is a streamline.\nFrom this we deduce that f(z) = a^z + (a^a2z^-1)^ = a^z + aa2z-1 has flow f'(z)^ approaching a for large r but with zero er component when r=a . By adding a line vortex potential\nf(z,t)   =   a^(z-c) + a2a(z-c)-1 + G(2pi)-1 (z-c)   =   A((-ai)(z-c) + a2(+ai)(z-c)-1) + alai-1(z-c))     where    a=A(ai) and dimensionless scalar l º lr º l(r,G,A) º (2pr)-1 A-1 G = r-1ala we obtain the complex potential for a flow\nv(z) = f'(z)^ = A((ai) - a2(z-c)-2^(-ai)) - G(2pi)-1(z-c)-1^   =   A((ai) - a2(z-c)-2^(-ai) + alai(z-c)-1^ )\nthat approaches uniform flow a=A(ai) as r®¥ but has circle r=a as a y=G(2p)-1a streamline.\n\nTaking a=0, c=0  for a uniform horizontal flow Ae1 at infinity around the origin centered circle we have\nf(z) = A(z+a2z-1) + G(2pi)-1 z   =   A(z+a2z-1 + ala z i-1)   =   A(r(iq) + r-1(-iq)a2) ) + G(2pi)-1(r+qi)\n=   Ar(1+(r-1a)2) cos(q) + (2p)-1qG + Ar(1-(r-1a)2) sin(q)i - (2p)-1rGi\n so our a=0 scalar potential and stream functions are     f(r,q) = Ar(1+(r-1a)2) cos(q) + (2p)-1qG      ;     y(r,q) = Ar(1-(r-1a)2) sin(q) - (2p)-1r↓G .     Vp   =   A( (l - (1+(r-1a)2) sinq)eq + (1-(r-1a)2) cosqer ) .     Vp2     =   A2 ( l(r)2 - 2l(r)(1+(r-1a)2) sin(q) + 1+(r-1a)4-2(r-1a)2 cos(2q) ) .     f(lz-1)   =   A((lz-1)+a2(l-1z) - ala z↓ i-1) + ala l↓ i-1) provides a flow of strength Aa2l about a circular boundary of radius la-1 with circulation -G.     Along the boundary r=a we have constant y = -(2p)-1a↓G and Vp = A(l(a) - 2 sinq)eq parallel to the boundary as required and vanishing at two stagnation points q= sin-1(½l(a)) providing l(a)<2 (ie. G < 4paA) . For l(a)>2 we have just one stagnation point at q=±½p provided l(r) = ±(1+(r-1a)2) Û (r-1a)2 ± l(a)(r-1a) + 1 = 0 Û (r-1a) = ½(-/+ l(a) ± (l(a)2-4)½) Û (r-1a) = ½l(a) + ((½l(a)2-1)½ .", null, "[ Proof :  Vp = -ÑPfpe12 = - (erÐer+eqÐeq)fpe12   =   -(er/r+eqr-1/q) f(r,q,t)e12   =   -er(A(1+(r-1a)2) sin(q) - (2p)-1r-1G)e12 + -eqA(1-(r-1a)2) cos(q)e12  .]\n\nAnalytically we will consider the generalised circular flow f(z) = ab(z-c) + a2a^b-1(z-c)-1 + G(z-c) + Gb with\nf'(z) = ab - a^a2b-1(z-c)-2 + G(z-c)-1 and\nf\"(z) = 2a^a2b-1(z-c)-3 - G(z-c)-2 but in practical cases we require purely imaginary G since the multivalued imaginary component of z = r + qi makes for disconnected streamlines if left within y.\nFurther, we only have a y(z) = Imag(G)(r) circular streamline if b(z-c) = (a2b-1(z-c)-1)^ which requires a2 = |b|+2|z-c|+2 so we invariably take take a2=a2 positive real.\n\nThe total force exerted on the r=a boundary by the fluid is -mAGe2, or more generally mAGae12-1 where a=ae1=A(ae21)e1 is the background flow.\n[ Proof :  Around the r=a streamline we have Vp2 = A2(la - 2 sin(q))2 and in the absence of gravity (Gp=0) and other forces (Hp=0) we have s0 = sp + ½mvp2 Þ m-1sp = m-1s0 - ½Vp2 = m-1s0 -  A2la2 + 2la sin(q) - 2 sin(q)2) .\nThe force exerted on a small element of the boundary is -spa dqer having e2 component -sp sin(q)a dqe2 and integrating this around the boundary gives zero total e1 force component and a total e2 force of -m ò02p dq   a sin(q) A2la2 + 2la sin(q) - 2 sin(q)2)   =   -maA22la ò02p dq sin(q)2)   =   -2pamA2la   =   -mAG  .]\n\nThus there is no \"drag\" on a circular boundary in a steady uniform inviscid incompressible irrotational flow. In reality there is drag, of course. Partly because actual flows are viscid but mostly because they are unsteady and rotational.\n\nCircular Â2 Flow Past a Circular Boundary\nTaking f(z) + f(a2z^-1)^ = f(z) + f(a2|z|+-2z)^ = f(z) + f(a2|z-1)^ for fp = a^D(z-D) provides\nfp = a^D(z-D) + (a^D(a2z-1-D))^ = a^D(z-D) + aD^((a2z-1-D)^)\n\nf(z) = b(z-a)_logncj + G(z-c) has v(0) = -(ba-1 + Gc-1) = a^ provided b = a(Gc-1 - a^) .\nTaking a muich larger than c so that the a(Gc-1 - a^)(z-a) term approximates constant flow a^ near 0 and c  .\n\nÂN flow around hypershpehrical boundary\nAn analagous result for N>2 is that y(p) = ½(f(p-c) + (f(aN-2|p-c|2-N(p-c))») has Ñy(p) = ½(Ñf)(p-c) + ½aN-2|p-c|2-N(Ñf)(aN-2|p-c|2-N(p-c))» since Ñ|x-c|2-N vanishes.\nOver |p-c|=a we have y(p) = ½(f(p-c)+(f(p-c))») and Ñy(p) = ½((Ñf)(p-c)  + ((Ñf)(p-c))») .\n\nBiimpulse exerted on a 2D boundary\nThe D'Alembert paradoxical result of zero drag in an inviscid incompressible steady flow around a circular boundary extends to general 2D boundaries and provides that the force exerted on a 2D body by uniform (at infinity) flow v is rGe12v perpendicular to v.\nThus for steady inviscid irrotational flow the shape of the 2D boundary is irrelevant to the direction of the force, but determines its magnitude by fixing the scalar circulation G. Force perpendicular to fluid velocity is known as lift reagrdless of whether it is \"up\" or \"down\" with regard to e3Q or gravity, eg. a diving plane \"lifts\" horizontally, and F=rGe12v equivalent to F=-rGiv is known as the Kutta-Joukowski Lift Theorem.\nSince the force derives from the pressure which is determined by Vp2, the lift remains the same if we negate the flow, with both v and G changing sign.\n\nSuppse we have a 2D 1-loop C=p(t) where tÎ[0,L] of total length L  that is the y=y0 streamline of a steasdy flow. At a point p(t) = ze1 on the boundary C we have a velocity Vp=ve1=f'(z)^ and force element dF = -spe21dp = (½rpVp2 - s0)e21dp = ½rpVp2e21dtVp~  - s0e21dp = ½rpv^ve21dtv~e1 - s0e21dze1 = -½rpv2v^~e2dt - s0e2dz^ = -½rpv2dz^e2 - s0e2dz^\nComplex force dF = dFe1 = -½rpv2dz^i - s0idz^ with conjuate dF^ = ½irp(v^2)dz + s0dz   =   ½ir(z)f'(z)2dz + s0dz so the net conjugate force exerted on the boundary is òC dF ^ = ½iròC dz f'(z)2 , which is known as Blasius's Theorem.\nIf f'(z)2 is expressed as a Laurent series f'(z) = åk=-n¥ ak(z-c)k for n³1 about some c then traditional complex residue calculus provides òC dz f'(z)2 = 2pia-1 , though of course if there are multiple poles inside C we have multiole contributory residues.\n\nThe moment of dF about c is (p-c)ÙdF = ((p-c)dF)<2> - = ((z-c)e1dFe1)<2> = ((z-c)(dF)^)<2> so the total moment about c is (½iròC dz (z-c)f'(z)2)<2> = ½irReal(òC dz (z-c) f'(z)2 ) = ½irReal(2pia-2) = -pirImag(a-2) in the event of a single pole at c.\nUnlike the lift, the torque does not in general vanish when G=0.\n\nThe deflector or stress tensor vpup-1 deflecting bivelocity up to bivelocity vp is naturaly parameterised over complex time as (t(vpup-1)) up .\nLetting up(t) = (t(t)(vpup-1)) up where t(t) is a path from t(T) to t(T+d).\nThe aerodynamic profile of a body at a particular mach speed S can be characterised by directonal deflection rp_vdp_udin\n\nConformal flow warping\nGiven a conformal map Z = f(z) and a complex potential f(z) we have complex potential F(Z) = f(f-1(Z)) defining a flow in the \"warped\" Z space V(Z) = F'(Z)^ º dF/dZ^ = (df/dz dz/dZ )^ = (df/dz (dZ/dz)-1)^   = v(z) (¦'(z)-1)^   = v(z) (¦-1'(Z))^   .\n\nThe Joukowski transformation Z = ¦(z) = (z-d)+f2(z-d)-1 with dZ/dz = 1 - f2(z-d)-2 ; (d/dz)2Z = 2f2(z-d)-3; and bivalued inverse z = ¦-1(Z) = ½(Z ± (Z2-4f2)½) + d . We frequently take real f2=f2 real and d=0.\n¦ maps z=d±f to Z=±2f and circles to a general \"aerofoil like\" family of 2D 1-curves including ellipses and ridged teardrop shapes.\n\nTo make ¦-1(Z) single valued, we choose the square root q=(Z2-4f2)½ \"nearest\" to Z in that q^*Z ³ 0 and refer to\n¦-1+(Z) = ½(Z+q) + d acting like a d displacement for large r as the principle Joukowski inverse ; and to\n¦-1-(Z)=½(Z-q) + d = f2(¦-1+(Z)-d)-1 + d acting like f2Z-1 + d = f2|Z|+-2 Z^ + d for large Z as the secondary Joukowski inverse.\nThere are discontinuities in ¦-1+ and ¦-1- along the line segment connecting Z = ± 2f acrioss which ¦-1+ and ¦-1- \"switch values\".\n\nA Joukowski aerofoil is the image under ¦ of a circle |z-c|+=a, usually one containing d-f inside it and c+f either inside the circle or on it. Havning d+f on the circle requires c = d+f + a(ig) and |d-f-c|+ £ a whence cos(p+g-z) ³ a-1f where f=f(iz).\nA Joukowski aerofoil has boundary |¦-1(Z)-c|+ = a but there is a difficulty knowing which inverse to use. If c lies on the line connecting d±f then the aerofoil is symmetric and convex and is all \"carried\" back from Z space to z space by ¦-1+. But if c lies sufficiently far from the the d±f connecting centre line, part of the boundary is concave and the region between this and the centre line is carried by ¦-1- .\nThe actual condition for the interior of the aerofoil is |¦-1+(Z)-c|+ £ a  &  |¦-1-(Z)-c|+ £ a and we can move the discontinuity from the centre line portion exterior to the aerofoil to the boundary of the aerofoil by switching to ¦-1+ when |¦-1-(Z)-c|+ £ a .\n\n¦-1±'(Z) = ½(1 ± q-1Z) and ¦-1±\"(Z) = ±½q-1(1 - q-2Z2)    so ¦-1+'(Z)+¦-1-'(Z)=1 and ¦-1+\"(Z)+¦-1-\"(Z)=0 .\n\nÂ2 Flow Past a sharp edged 2D Boundary\n\nJoukowski-Kutta Condition\n\nThe Joukowski warped flow velocity V(Z) = v(z) (1-f2(z-d)-2)-1^ is infinite only when v(z) is or when (z-d)2 = f2 with z lieing outide the boundary and v(z) nonzero. Thus if one of z=d±f lies outside a given closed y(z) streamline conisdered to be a boundary, then only by choosing the circulation G to ensure a stagnation point f'(z)=0 at that z can we eliminate an aphysicsl point of infinite flow.\nFor a generalised circular flow f(z) = a(z-c) + a2a^(z-c)-1 + G(z-c) with f'(z) = a - a2a^(z-c)-2 + G(z-c)-1 finite everywhere but at z=c we have stagnation whenever a(z-c)2 + G(z-c) - a2a^ = 0 which at z=d±f requires G = (d-c±f)-1 (-a(d-c±f)2 + a2a^) .\nIf z=d-f lies within our z space boundary streamline and z=d+f lies on it with c = d+f + a(ig) then for real a=a we have purely imaginary\nG   =   (-a(ig))-1 (-a(-a(ig))2 + a2a^)   =   -a(-a(ig)) + a^(-ig)))   =   2aA sin(g-a)i and hence G = -4paA sin(g-a) .\nFor a symmetric aerofoil with g=p we have G=-4paA sin(a) .\n Joukowski-Kutta condition flow around a Joukowski aerofoil .", null, "Principle flow F+(Z)", null, "Reciprocal flow F-(Z)", null, "Summed flow F+(Z) + F-(Z) yp streamlines in green; fp isopotential lines in red; |Vp| isospeed lines in blue.\n\nIf one or both of d±f lie on a boundary C in z space then the warped boundary ¦(C) in Z space has a gradient discontinuity at ±2f since f'(z) = ½(1 - f2z-2) vanishes at z=±f but f\"(z) =  f2z-3 is ±f-1 there . The reciprocal Joukowski inverse ¦-1-(Z)=½(Z-q) + d  = f2(¦-1+(Z)-c)-1 + d has |¦-1-(Z)-d|+ =   |f2|+|¦-1+(z)-d|+-1\n[ Proof : ¦-1-(Z)-d|+2 = (f2(¦-1+(Z)-d)-1)^ (f2(¦-1+(Z)-d)-1) =   |f2|+2 |¦-1+(z)-d|+-2  .]\n\nConsider a d=0 Joukowski warp Z = ¦(z) = z+f2z-1 acting on a circle of centre c=lf and radius a=(1-l)f chosen so that the circle passes through z=f and z=(2l-1)f and also z=(l±(1-l)i)f while the Joukowski aerofoil passes through Z=2f and Z = (2l-1)+(2l-1)-1)f and Z=(l±(1-l)i + (l±(1-l)i)-1)f   =   (2l±2l(1-l2)(l2+(1-l)2)-1i)f.\nTaking l<0 encloses -f inside the circle and we are left with a boundary gradient discontinuity with stagnation there if G = -4paA sin(a) = -4pf(1-l)A sin(a) .\nThe symmetric aerofoil has length L = (2-(2l-1)-(2l-1)-1)f = (3-2l+(1-2l)-1)f and width W=4l(1-l2)(l2+(1-l)2)-1f.\nSmall negative l yields a thin aerofoil with L=4(1+l2+2l3+...)f » 4f » 4a and W » lf   with critical circulation G=-4paA sin(a) » -pAL sin(a) giving lift F » 4praA2 sin(a) ((ap)i) .   Letting l®0 gives a flat plate of exact length L=4a and critical circulation G=-pLA sin(a).\nLarge negative l yields a \"dimpled circular\" aerofoil with L » W » 2lf » 2a and critical circulation G=-2pLA sin(a).\n\nThese results appear bizarre, with the sin(a) factor suggesting lift is maximised for an aerofoil moving perpendicularly against the flow, and the dimpled circle of a diameter L twice as efficient at generating lift as a flat plate of length L. In practice, the dragless circular flow fails for attack angles exceeding about 20 o or nonthin aerofoils, with seperation of the flow from the aerofoil surface and turbulance in the wake rather than a smooth \"reclosing\" of the flow behind the aerofoil resulting in the lift for increasing a, a pheneomena known as stalling. At the stalling angle, the airflow typically \"seperates\" from the upper surface and forms unsteady turbulence and eddies. Some authors refer to stalling as the point where lift becomes negative, but this is erroneous. Stalling occurs when the still positive lift begins to decrease with increasing attack angle.\nFor small attack angles the sin(a) factor holds resulting in lift being roughly linear in a until the stalling angle is approached.\n\nDeriving the critical Kutta-Joukowski circulation and hence the lift for a general sharp-edged aerofoil is problematic and various iterative methods such as that of Theodorson have been developed.\n\nAerofoil Waffle\n\nLoosely speaking, if a is well inside [-½p,+½p] so that the flow is approximately horizontal left to right, and the circulation is negative (clockwise) then the flow is slower beneath the aerofoil than above it so the pressure s0rpVp2 is greater below the aeorfoil than above it and the lift force exerted on the airofoil is vertically upwards and this is the force that holds up an aircraft. In practice the decrease in pressure above the wing is usually gretaer than the increase in pressure below it with up to 80% of the lift stemming from \"pulling\" of the upper surface wing rather than \"pushing\" of the lower. Both changes are greater near the leading edge of the wing so that the net torque on the wings serves to raise the leading edge and we can think of the centre of pressure - being the point on a designated aerofoil chord line through which we must consider the lift force to act if its couple about the aerofoil centre  of mass is to match the picthing moment - as moving forward.  The picthing moment exerted in the wings is typically balanced by the upward lift on horizontal tail fins.\n\nIf we lower (or raise) a flap at the trailing edge of an airfoil we will increase drag and slow the fluid flow below (or above) the airfoil, resulting in an increase (or decrease) in lift.\nThick airfoils, convex or concave, present the problem of lacking a single unambiguous \"forward\" direction with reference to which one can define an \"attack angle\". The chord line of a (planar section of) an airfoil is the line joining the leading edge to the tailing edge is sometimes used.\n\nLift Proportionate to v2\nSuppose S is a small planar element   of area content s having unit normal n, and centre of mass c travelling with constant velocity v through a fluid of density r initially at rest. Assume n is signed so that n¿v ³ 0 . In small time dt the simplex must displace a small volume s(v~¿n)|v|dt = s(v¿n)dt which we assume to act as a single impactive mass we can consider as having velocity -v and momentum = rs(v¿n)vdt on an unmoving simplex.\nIf the impactive impulse T parallel to n is sufficient to precisely eliminate the n component of the velocity so that the displaced air \"slides off\" the simplex we have T = rs(v¿n)dt (v¿n) = rs(v¿n)2dt     which is percived by the simplex as a \"drag impulse\" - rs(v¿n)2dt (v~¿n) parallel to v and a \"deflective impulse\" - rs(v¿n)2dt (1-(v~¿n)2)½ - rs(v¿n)2dt cos(a) where a = ½p - cos-1()(v~¿n) is known as the angle of attack in the context of a thin planar airfoil.\n\nThis is a gross simplification of airfoil dynamics but substantiates somewhat our assertion that the forces acting on an airfoil are roughly proportionate to the square of the airspeed as well as to the area of the \"leading\" surface.\n\nForce Velocity Coefficients\nBroadly speaking (and for subsonic speeds), the lift and drag are individually proportional to v2 , the density of the medium, and the (XY planar) area of the airfoil and this is usually expressed for N0=3 as\nL   =   ½rSv2 CL,a,b    ;     D   =   ½rSv2 CD,a,b    ;     C   =   ½rSv2 CC,a,b\nwhere r is airdensity, S is wing area and CL,a,b, CD,a,b , CC,a,b are known as lift coefficient , drag coefficent, and crosswind coefficient for attack angle a=qp and yaw angle b=f for the airfoil orientated with e3 upwards amd e1 forward. These are the projections of T into the wind frame , with D parallel to v , L lieing along ¯v*(e3Q) , and (moving to 3D) C  perpendicular to both.\nFor a (Z) symmetric airfoil , CL,0,0 = 0 while CD,0,0 ³ 0 since v is here the velocity of the fluid past the aerofoil rather than the velocity of the aerofoil trhough the fluid. Modern airfoils are asymmetrically cambered to give small CL,0,0 > 0. The crosswind force is frequently neglected, or assumed cancelled by the opposing crosswind force on a symmetricaslly opposing aerofoil.\nFor inviscid irrotational steady flow we have CD,a,b=0 ; CL,a+p,b= CL,a,b ; CC,a+p,b=- CC,a,b but in practice we usually have aerofoils less efficient when travelling backwards with CD,a+p,b >    CD,a,b > 0 and CL,a+p,b £ CL,a,b\n\nAerofoils are frequently characterised by providing CL,a º CL,a,0, CD,a, and CM,a, for a limited range of a typical in normal flight, say -4 to 20 degrees though the values RD L(a)= ( CL,a2+ CD,a2)½ and QD L(a)= tan-1( CL,a/ CD,a)   are arguably more \"natural\" since for N0=2 we have\nF   =   ½rSv2RD L(a)(QD Le21)v~   =   ½rS|v|RD L(a)(QD Le21)v with QD L(a)=±½p for dragless lift.\nSometimes CD,a=(pEA)-1 CL,a2 is assumed where efficiency E»1 and A is a aspect ratio.\nFor aerobatic emulation purposes we typically require values for all a and b. For large a the aerofoil can sometimes be considered a flat plate, although as the dragless circulatory flow model fails for large a for flat plates an alternate high drag model model should be employed.\nCL,a is small or zero for zero a, rising roughly linearly to a maximum at a stalling angle of about 16 degrees, and then falling off in an uncertain manner, possibly to zero value and zero gradient at ½p.\nFor aÎp,p] we might crudely assume CL,a+p=mL CL,a where 0<mL£1 is a reversal inefficiency factor. and characterise or tabulate CL,a only over [-½pp)   . The assumption that the reversal lift profile has the same shape as the forward one is unrealistic but if  we expect an aerofoil to seldom travel backwards it may be acceptable.\nFor aerofoils symmetric in their e1W chordline, the lift for attack angle -a is minus that due to attack angle a.\n\nOne possible strategy is to assume that   for b Î [0,½p] CL,a,±b = cos(b)2 CL,a,0+ sin(b)2 CL,a,±½p and tabulate   forwards and sideways lift coefficients CL,a,0 and CL,a,±½p over aÎ[-½pp) , with CD,a,b, CD,a,b and the moment coefficients implemented similarly. The use of cos(b)2 and sin(b)2 weights ensures that CL,a,b= CL,a,0 for a solid of revelution about e3W disk like aerofoil.\n\nSometimes one or more of the ½, r and S are amalgamated into the coefficients , although r does decrease with altitude and varies across shock waves. However, keeping the density r and area S outside the coffecicients keeps them dimensionless, dependant only on the shape rather than the size of the aerofoil.\n\nTorque Velocity Coefficients\nThe point of a 2D airfoil through which the lift and drag forces can be considered as acting (for a given a), known as the centre of pressure is also relevant, for unless this happens to be the centre of mass of the airfoil, the forces will indice a torque known as a pitching moment tending to rotate the airfoil. This torque is often written as ½rScv2 CM,ap) where CM,ap) is the pitching moment coefficient for attack angle a and point p, about which the torque is computed. c is the chord length. We will again embody the ½rS and now c also into the coefficient and obtain hp   =    v2 CM(a,p) for the picthing moment about a Y-eaxis through point p.\nFor many airfoils there is a point a (more accurately an axis) known as the aerodynamic centre about which the pitching moment for a given airspeed v2 is largely invarient with a (at least over a typical flight range of a of -8 to 16 degrees) so that the moment coefficient CM,aa) is independant of a. This point is usally about ¼ chordlength back from the leading edge of the airfoil, and the constant torque tends to be signed so as to lower the lower nose and raise the tail of an aircraft (or twist the wings from the fuselage!)     The pitching moment tending to rotate the aerofoil is usually taken to be ½rVp2ScCm(a,b) where c is a mean chord length and Cm(a,b) is a dimensionless coefficient. Similarly for a 3D aerofoil we have rolling and yawing torques about the v~ \"wind upward\" axies exerted on the aerleron due to the fluid drift v.\n\nBimpulses due to spin\nEven under our assumption that the fluid velocity is uniform in the vicinity of the body, a nonzero spin w of the body means that the fluid encountered by the body surface will have varying relative velocity. If an aircraft is rolling anticlockwise about e1Q so that its left wing is rising, for example, then the effective angle of attack a of the airstream over the left wing is decreased at a point l out along the wing by roughly lw23|v|-1 while that on the right wing is increased by about the same ammount.\nAssuming that all the lift ½rv2CL derives from the wings, these changes in a result in lift decreasing and increasing by ½r_dlcv2 lw23|v|-1 CL/a on the left and right wings respectively, imparting a clockwise torque r|v_dlcl2w23CL/a acting in conjunction with and probably exceeding any \"fractional\" torque proportionate to (w23)2 opposing the roll.\nWe can regard this as providing an addition ½l|v|-1w23CL/a to the roll coefficient C23(v~) = C23(a,b) providing the characteristic length l used for the effective wing length is the same as the length used to render the roll coefficient dimensionless.\nSimilarly, an anticlockwise yaw about e3Q bringing the right wing forwards will, if the aircraft is travelling horizontally with nose slightly raised, tend to decrease the right wing attack angle but increase its effective speed causing a net increase in lift on the right wing and a decrease on the left, resulting in a torque inducing a roll lowering the wing on the \"inside\" of the yaw. This can be crudely modelled by a further addition ½l|v|-1w12CL/b to the roll coefficient and in general we model the biimpulses due to spin by adding such terms involving a and b derivatives of the velocity based force coeefficients to the ½N0(N0+1) velocity response coefficients.\n\nConsider a long wide aerofoil of stance 1 travelling with velocity v=(-ae31)e1 » e1-ae3 for small a and spinning at w. A segment at le2 has velocity v+le2Ùw.\nLet us first consider yaw rotatiaon with w=we12 with w>0. The segment velocity is then_v-lwe1 so for l>0 the speed decreases but the attack angle increases unless a=0. Matters thus simplify if a=0 since then the attack angle remans constant and the speed becomes v-lw so that the total lift is F = ò-LL dl½rc(v-lw)2 CL,0 e3   =   ½rS CL,0(v2 + 6-1(2L)2w2)e3 where L is the demiwingspan and c=(2L)-1S approximates the chord length for a thin aeorfoil.\nThe total roll torque is\nò-LL dl½rcl(v-lw)2 CL,0 e23   =   -12-1rS CL,0v(2L)2we23    equivalent to adding -6-1 CL,0v-12Lw to the velocity based roll moment coefficient.\n\nFor nonzero a but zero b we have sin(Da)(lw)-1 = sina(v+Dv)-1\n\nF = ò-LL dl ½rc((v cos(a)-lw)2+v2 sin(a)2) CL, tan-1( sin(a)( cos(a)-v-1lw)-1) , typically with  |lw|<<v .\n\nThis is the model used by many flight simulators. The six (½N0(N0+1) for N0=3) velocity induced wind frame force and torque coefficients for the whole aircraft are tabulated, often with thousands of data points perhaps clustered more densely over more common regions , over the encountered ranges of flight (wind) angles a,b and mach speed M.\nEffects due to aircraft spin, acceleration, propellor thrust, flap settings and so forth are the implimented as approximations based on the velocity coefficients such as the addition of ½lw23|v|-1 CL/a to the roll moment velocity coefficient, obtainable via a calculated differential of the tabulated CL,a,bM) surface.\n\nIt is hard to adapt this whole-body model to accomodate damage to particular regions. If a rear tail fin is half shot off, for example, then we lack LUTs for the new assymetric composite aircraft.\n\nFor the emulation of near level flight when velocity is largely forward with speed sufficient for the lift to roughly balance the gravitational weight we can safely extract the Vp2 and parameterise the coefficients solely by airspeed direction but for acrobatic flights involving high spins and accelerations our coeffecients become functions of w as well as v and any control parameters.\nWe can extend our ½N0(N0+1) boy frame coefficients as T = ½r|W|2Såi<jli Cij(W~,m,M)eij¥ where W is the bivelocity of magnitude |W| and M is the mach speed of v.\n\nThe advantage of moving the v2 outside the velocity coefficients where twofold. First, it makes the coefficient dimensionless, and second it makes it a function of only the direction v~. This second advantage is voided if the mach speed M is reintroduced as a parameter so we may as well declare T = ½r|W|2Såi<jli Cij(W,m)eij¥ with dimensionless coefficients that are functions of the unnormalised ½N0(N0+1) coordinate bivector W and a control vector m.\n\nPropellers\nThe principle example of a nonfixed aerofoil subbody is a propellor blade. These will typically be mounted at symmetrically on an aircraft with e3P = e1Q and be spun rapidly about this axis by an engine to provide a forward thrust usually adequate only for subsonic flight, or lift in the case of a helicopter. The path taken by a point on the propellor is a helix, skew if vPQ ¹ e1 and curved if wQ¹0. The instantaneous force exerted on the propellor has a \"thrust\" component along e1_APlane and a \"torque\" conmpoent perpendicur to it which is typically balanced by and opposite torques from the opposite arm of a two-blade propellor, or averages out over a propellor rotation. The net force exerted on the propellor centre is typically parallel to e1_APlane but induces a troque on the aircraft if the propellor centre cPQ lies off the centre line of the aircraft. Propellors are typically symmetrically or centrally mounted but may be above or below the aircraft mass centre and so impart a pitching moment.\nPropellors accelerate the air passing \"through\" them and typically increase the airspeed over the wings and/or tail and hence lift. They also act as gyroscopes resisting pitch and yaw torques.\nConsider a point at -ye2 of a propellor having stance 1 and bivelocity W=(1+½v_e3e)w_e12_frm2(P,Q). Its velocity is ywe1+ve3  which we can regard as a 2D areofoil travelling at speed (y2w2+v2)½ with attack angle g(y)-q(y) where angle g(y) is the pitch of the propellor blade at distance y from its hub and q(y)= tan-1(v(yw)-1). The force along e3 acting on the 2D section is\ndF   =     ½rdS(y)(y2w2+v2)( CL(g(y)-q(y)) cos(q(y) -CD(g(y)-q(y)) sin(q(y))   =   ½rdS(y) RD L(g(y)-q(y)) sin (g(y)-q(y)) cos(q(y) - QD L(g(y)-q(y)) ) where dS(y)=dy L(y)   is the propellor area element. Integrating this over y from 0 to the propellor span gives the instantaneous forward (e3) force exerted by one blade of the propellor.\n\nHowever, and easier way to estiamte the total force T=Te3 exerted by a propellor on its mounting point is to assume that the propollor has a given efficiency EÎ[0,1] so that if the engine exerts a torque Q and so does Qw work then EQw = T_vu3 where _vu3 is the forward speed of the aircraft, so that T=(_vu3)-1Qw.\n\nBody Frame Coefficients\nReturning briefly to 2D, our notion of fluid drift v=A(ai)e1 travelling  for small attack angle a from left to right suggests an aerofoil W aligned with its forward direction e1W=-e1 facing right to left. More generally, we might expect an aircraft Q to be pointing roughly into the enocuntered windpeed with v~ » -e1Q .\nIf we wish to work in the frame of the aerofoil it is sensibkle to take v to be negative the encountered fluide velocity so that directly forward motion of the aireofoil corresponds to a=b=0\n\nMore generally we can define N0N0(N0-1)=½N0(N0+1) dimensionless coefficients defined by expressing the 3-vector bimpulse T = ½rv2Såi<jli Cij(v~)eij¥ in the frame of the body rather than the windframe. Here i,j are from {0,1,..,N0} , li is a characteristic length of the aerofoil in the i dimension with l0=1, and Cij(v~) is a dimensionless scalar coefficient.\nC01(v~) is the forward coefficient with C01(p-a,0) = - cos(a) CD,a,0+ sin(a) CL,a,0 ; C02(v~) is the sideways coefficient ; C03(v~) the upward coefficiecnt with C03(p-a,0) = cos(a) CL,a,0+ sin(a) CD,a,0 ; C13(v~) the body pitching coefficient with C13(p-a,0)= CM,a,0 , and so on.\n\nIn 3D we can approximate these using Cij(a,±b) = cos(b)2Cij(a,0) + sin(b)2Cij(a,±½p) as for the windframe coefficients and because we are now measuring the coefficients in the body frame, for sideways symmetric 3D body symmetric in y=0 we have Cij(a,-b)=±Cij(a,b) with the - occuring when one of i or j is 2.\n\nSpeed of Sound\nCompressable fluids can carry density waves\n\nEven though in theory the velocity based flight coefficients depend only on the direction Vp~ and not on the magnitude Vp2, in practice they do vary with the speed partcularly when distingusihing slow, subsonic, and supersonic flight.\n\nIt is natural to express the airspeed |v| as a fraction of the speed of sound s=(r-1m)½ where fluid stiffness coefficient m has units kg m-1 s-2, ie. we set |v| = M(r-1m)½ for unitless mach number or mach speed M so that dymanic pressure ½rv2 = ½M2m is independant of the density r; or rather its dependance on r is subsumed into M . For M>1 we have supersonic speed and M>5 as hypersonic speed   .\nThe speed of sound varies with density and tempreature and the nature of the fluid but for air at sea level it is roughly 28.4 ms-1\nFurthermore, aircraft typically have control surfaces like aerlerons and flaps whose orientations are varied to effectively alter the flight coefficients of the wings or fins to which they are attached. The settings of such control surfaces are typically parameterised by angles and we have a control vector m fromed from these angles.\n\nBody as rigid amalgamation of simpler subbodies\nAn aletrnative approach is to consider the aircraft as composed of a collection of rigidly connected aerofoils, each with their own aerodynamical coefficients. We might for example consider the left wing to be composed of one or more wingsections.\nAssuming a particular wing section W to be at rest with respect to the aircraft, the instantaneous apparent bivelocity of the fluid as percieved by W is\n-WWW   =   -RW4§§(cQ4dt)e¥ - RWQ§( (cWQ¿wQQ )e¥ + wQQ )RWQ and we can use this to construct the biimpulse about the W centre TW(-WWW) applied to the wing section in the frame of the wing section and then recast this as RWQ§( TW(-WWW) ) into biimpulse about cWQ applied to the aircraft, assuming the imulses and tiorques involved are inadequate to deform or dislodge the wing section.\n\nThe biimpulse about the aircraft centre in the aircraft frame is RWQ§( TW(-WWW) ) + cWQÙ( e¥0¿ RWQ§( TW(-WWW) ) )Ùe¥ and by adding such centred biimpulses for all the aircraft subbodies we obtain a net biimpulse Tc acting on the aircraft about its centre c which we typically take to be be the mass centre.\nThe kinematical responce of the aircraft is then given by DW = H-1Q(Tc).\n\nThis method is again highly artificial.   Even in steady, level flight, the airflow encountered by tail is typically in the downward deflected downwash from the wings, and so has its attack angle lessened. In accelerative motion, vortices or similar flow effects shed from the wings hit the tail planes after a delay roughly proprtionate to |v|-1, a crosswind coming from the left is partially screened from the rightwing by the fuselage, and in general the fluid flow in the vicinity of the body is extremely complicated. However, modelling the fluid itself is typically computationally out of the question and the best we can hope for is broadly plausable behaviour of the body moving through it.\n\nLeftovers\n\nProblems with 1-flows\nImagine a medium consiting of two types of particle. \"Green\" particles of small unit \"mass\" and unit positive \"charge\", and \"red\" particles of unit mass and negative charge which with regard to a partcular observer e4 , are confined to a ring of radius R centred on the origin,. Assume that there are equal large quantities of green and red particles distributed with uniform density r particles per (small) unit volume around the ring and that particles contrive to avoid impacting eachother, or do so only rarely.  Assume that the green partciles are travelling \"anticlockwise\" around the ring with orbital period T while the red particles travel clockwise with the same orbital period.\nThe net spacial flow of mass through a given cross sectional ring element is then Mp=0, although there is kinetic energy 2r½(2pR/T)2 = r(2pR/T)2 . The net spacial current flow is 2 r(2pR/T) anticlockwise . If the red particles were all to reverse direction, and flow counterclockwise, we would have zero current but nonzero momentum.\nThis simple example demonstrates the complications in \"adding up\" flows. Allowing negative and postive charges means that the sum of two charge-weighted timelike vectors may not remain timelike. Generally speaking, when adding four-momentum 1-vectors, relative masses combine additively but relative three-momentum magnitudes incurr a triangle inequality so that proper magnitude is not conseerved and tends to increase .\nWhen physicists speak of the flow \"at\" a particular point they usually mean the averaged flow over a small volume at that point, and by \"small\" they usually actually mean large enough to contain multiple particles. They hope by summing a large number of fundamentally discontinous point-dependant \"path localised\" functions to produce a continuos field. Thus all the matter at a given event p is considered to be effectively flowing with the same direction and speed. We will refer to such a flow as single flow because all the matter at p is effectively doing the same thing.\n\nMore generally we represent a superposition of 1-flows with a 2-tensor vp such that vp(l,d)aÙb provides the number of particles of the matter at p of kind classified by a matter-type indicator  l flowing across spacial 2-plane aÙb in e4-time d. By which we mean that on fixing any timelike e4 perpendicular to a and b the integer number of particles crossing between e4-times t and  t+d divided by d.\nIn the sense of vp representing the documentation of an actual multiparticle flow, we regard vp(l,d)aÙb to be valued in integer multipes of d-1 rather than a full real type.\nAs d®0 the integer multipliers get smaller discontinuously.\n\nIf the flow is steady in the sense that at a suffiicently small d the second derivatives become swamped by the first and we can conceptually replace individual particles with shoals of identical smaller particles having parallel trajectory then we expect vp(l,d)aÙb to approach a limit .\nFlow is thus more naturally regarded as a bivector which we will call 2-flow. For N=3, the 2-flow is dual to the 1-flow .\n\nFlow rond moving sphere\n\nBernouilli's equation concerns a sphere moving through a fluid and states that at anypoint near the surface of a frictionless sphere whose centre moves with low velocity v Î e12 in the x3=0 plane is   sg-1 + x3 + ½v2g-1 = k where s is the fluid pressure , g is gravitational accelaration, and k is a constant. Thus there is a symmetry to the flow and the fluid is left essentially undisturbed, no energy is lost, and drag is zero. If the velocity, or friction, increases, there can instead be eddies or turbulance in the \"wake\" of the sphere. imparting energy to the fluid and so slowing the sphere;s travell. In practice the \"impactive drag\" or form drag on a sphere is about half that on a flat plane of identical leading area travelling \"flat on\" (ap , v~=n) . Optimal streamlining can reduce this drag to about 5%.\nWe can rearrange Bernoulli as sr-1 + ½v2  + x3g = kg and if the sphere is small so that x3g may be neglected we have s + ½rv2  = kgr . If the fluid is considered as incompressible so that r is constant we refer to constant s0 º kgr as that stagnation pressure since it correspends to P when v=0, and ½rv2 = s<0>-s as the dynamic pressure.\nHence the greater |v|, the less s , provided |v| remains small enough (less than half sonic speed) for the Bernouilli equation to approximately hold .", null, "Extremal M-Curves\nA standard problem in classical mechanics is to determine the form of the curve assumed by a rope of length a hanging motionless between two fixed points a distance less than a apart under a uniform gravitational field, which is an extremal 1-curve in N=2 dimensions.\nIf we assume the rope to hold a curve shape p(s)=(x(s),y(x)) (the Y axis being gravitationally vertical) that does not loop \"over\" itself vertically, we have '2 = dx2 + dy2 so that the length constraint is a = ò01 ds = òx0x1 dx(1+y'2)½ where y' º dy/dx.\nThe mechanical constraint is that total gravitational potential energy ò01 ds rgy(s)   =   rg òx0x1 dx y(1+y'2)½ be minimised, where g is uniform vertical gravitational acceleration and r is the mass of unit length of rope.\nMore generally, we might seek to minimise òx0x1 dx¦(x,y,y') subject to a constraint òx0x1 dxg(x,y,y') = a.\nA standard approach is to form h(x,y,y') º ¦(x,y,y') + lg(x,y,y') where scalar l is known as a Lagrange multiplier. One then embodies freedom to vary the path by means of two z0 and z1 such that y(x0,z0,z1)=y(x0) ; y(x1,z0,z1)=y(x1) \" z0,z1 (boundary condition) ; y(x,0,0)=y(x) ; and y(x,z0,z1) is twice continuosuly differentiable in all parameters.\nForming K(z0,z1) = ò+x0x1 dx h(x,y(x,z0,z1),y'(x,z0,z1)) the extremal condition requires K/z0 and K/z1 to vanish at z0=z1=0 and this leads (via integration by parts) to the Euler-Lagrange equation h/y = d/dx(h/y') .\nIn the case of the hanging rope, h(x,y,y') = (rgy-l)(1+y'2)½ and setting l = -rgy0 we have   h(x,y,y') = rg(y-y0)(1+y'2)½ = rg(y-y0)(1+(y-y0)'2)½ for some constant y0 .\nSetting h = y-y0 the Euler-Lagrange equation is rg(1+h'2)½ = d/dx   rghh'(1+h'2) with first order simplification\nh' rghh'(1+(y-y0)'2) - rgh(1+h'2)½ = b Þ rg(1+h'2)h = b having solution h = b cosh(x-c)/b) so the rope has shape y0 + b cosh((x-c)/b) where c,y0 and b are chosen to match the given endpoints.\n\nWe say an M-curve CM is M-extremal for an action functional ¦CM:UN ® UN if some particular magnitude measure (content, scalar part, square, modulus or whatever) of òCM dM-1p ¦CM(p) is maximised (or minimised) by CM in the sense that integrating other any M-curve which deviates only slightly from CM will produce a result no higher (or lower) than the CM integral. CM is a \"locally optimal\" M-curve for a particular integration ¦CM. We specify the dependancy of ¦ on CM to allow action functionals dependant on the geometric properties of the \"sampling curve\" CM , such as tangent or normal vectors , to allow \"direction dependant sampling\" or \"velocity dependance\".\n\nEuler-Lagrange Equations\nFor M=1 we have a path p(s) 1-extremal for FC1 = F(s,p(s),p'(s)).\n\nHamilton's principle provides that the state of a system characterised at time t by k multivector variables x1(t),..,xk(t) varies from time t0 to t0 so as to 1-extremise a (usually real scalar) integral measure\nS(t) º S(t0) + òt0t0 dt L(t,x1(t),..,xk(t),xdt(t),...,xkdt(t))     with the action functional L(t,x1(t),..,x1dt(t)) known as the Lagrangian of the system.\nL is usually assumed to be real-scalar valued and independant of p\"(t) and higher derivatives and is traditionally assumed to seperate into \"kinetic\" and \"potential\" componenets independent of time and velocity respectively as L(t,p(t),p'(t)) = T(p(t),p'(t)) - V(t,p(t)) but more generally we might postulate a multivector-valued Lagrangian of k multivector-valued variables and their first temporal derivatives L(t,x1(t),..,xk(t),x1dt(t),..,xkdt(t)).\nIf the Lagrangian L is itself an integral over some spacial M-curve (typically an M=N-1 hypercurve reprenting a contemporal slice of a Base space) L = òBase(t) dN-1p L(t,p,pdt) that spacially integrated function is known as a Lagrangian density. Note that such requires a \"velocity\" pdt be associated with every point p in Base(t)   .\n\nExtremal Paths\nA standard approach for M=N=1 to 1-extremise ¦(s,x,x') is to embody freedom to vary the path by means of two scalar parameters z0 and z1 such that x(s0,z0,z1)=x(s0) ; x(s1,z0,z1)=x(s1) \" z0,z1 (boundary condition) ; x(s,0,0)=x(s) ; and x(s,z0,z1) is twice continuosuly differentiable in all parameters. Forming K(z0,z1) = òs0s1 ds¦(s,x(s,z0,z1),x'(s,z0,z1)) the extremal condition requires K/z0 and K/z1 to vanish at z0=z1=0 and this leads (via integration by parts) to the Euler-Lagrange equation ¦/x = (d/ds)(¦/x') .\nIn particular cases where ¦(s,x,x') = ¦(x,x') so that ¦/s=0 and there is no explicit s dependence,   we have d/ds(x'¦/x' - ¦)= 0 giving a first order differential equation x'¦/x' - h = constant .\n[ Proof :   y\" h/y' + y'd/dxh/y' - h/x-h/yy' - h/y'y\"   =   y'(d/dxh/y' - h/y) - h/x   =   0  .]\n\nA nongeometric generalisation is to extremise òs0s1 ds¦(s,x1,x2,..xk,x1',x2',..xk') subject to M scalar constraints gi(s,x1,x2,...xk)=0 i=1,2,..,M (Note the absence of any g dependance on the xi'). To do so we form\nh(s,x1,..,_xK,x1',..,xk') = ¦(s,x1,..,_xK,x1',..,xk') + åj=1M lj(s)gj(s,x1,..,xk,x1',..,xk') where lj(s) are M arbitary functions of s and obtain Euler-Lagrange formulae h/xi = (d/ds)(h/xi')     for i=1,2,..k .\n\nGeneralising geometrically, we have F(s,x1,x2,..,xk,x1',x2',..,+xk') with geometric Euler-Lagrange Equations F/xi =  (d/ds)(F/xi')     \" i=1,2,..k and first order equation åi=1k  (xi'*Ñxi')F - F = constant (independant of s) if F/s = 0. Considering the xi as seperate grades of a single multivector argument x we can regard the Euler-Lagrange equations as individual coordinate terms of a single geometric equation\nxF(s,x,x') = (d/ds)(x'F(s,x,x')) where x = åijk.. eijk../xijk.. over all blades comprising x space so that, for example, x = åi=1N ei(/xi).\n\nGeneralised Momentum\n\nWhen F is a scalar-valued Lagrangian we have  (xi'*xi')L = xi'*(xi'L) and multivector m = Ñxi'L(x1,..xk,x1',..,xk) is known as generalised spacial momenta or cononical momenta. Typically xi' is spacial 1-vector valued and so momenutum is a spacial 1-vector.\nIf multivector L is independant of xi so that L/xi=0 then the ith Euler-Lagrange equation provides (d/dt) L/xi' = 0 so Mi = L/xi' is constant, ie. unchanging with t,\n\nThus the constants of a system are consequences of absent dependendencies (aka. symmetries) in the Lagrangian. Energy is conserved when L depends on t only indirectly via x(t) and it derivatives. Momentum is conserved when L depends on x only indirectly via x'.\nFor L(t,x,x') = ½mx'2 - f(x,t) we obtain classical momentum Ñx'½mx'2 = mx' , constant if f(x,t)=f(t).\nAdding an electromagnetic term qc-1ax¿x' to L introduces qc-1ax to the momentum\n\nIf L(t,x1(t),..,xk(t),x1dt(t),..,xkdt(t))   =   L(x1(t),..,xk(t),x1dt(t),..,xkdt(t)) so that there is no explicit t dependance then the Hamiltonian H(x1(t),..,xk(t),x1dt(t),..,xkdt(t)) º åi=1k  (xidt*Ñxidt)L - L     is a constant, the \"energy\" of the system.\nWe can regard the Hamiltonian as generalised temporal momentum. Conservation of energy but varying spacial momentum resulting from only non-relativistic potentials f(x,t)=f(x) .\n\nBut H = - S/t .\n\nEven though, like the Lagrangian, S(t) cannot strictly be regarded as a function of position since it is only defined over our given extremal path, we can nontheless postualte an action field S(t,x) an it can be shown that\nm = (xi'*xi')L = ÑxiS(t,xi) and in particular we have the Hamilton-Jacobi equation H(xi,xi',t) = -S/t\n[ Proof :     For any ¦(x,x')\nd/dt ( åj=1k  xi' ¦/xj' - ¦)   =   åj=1k  xj\" ¦/xj' + åi=1k  xj' d/dt ¦/xj' - ¦/t - åj=1k ¦/xjxj' - åj=1k ¦/xj'xj\"   =   åj=1k  xj' d/dt ¦/xj' - ¦/t - åj=1k ¦/xjxj'   =   åj=1k  xi'( d/dt(¦/xj') - ¦/xj ) - ¦/t\n=   åj=1k  xj'( d/dt(¦/xj') - ¦/xj )     if ¦/t = 0\n=   0 if d/dt(¦/xj') = ¦/xj     j=1,2,..,k which are the Euler-Lagrange equations.\n\nd/dx ( åi=1k  yi' ¦/yi' - ¦)   =   åi=1k  yi\" ¦/yi' + åi=1k  yi' d/dx ¦/yi' - ¦/x - åi=1k ¦/yiyi' - åi=1k ¦/yi'yi\"   =   åi=1k  yi' d/dx ¦/yi' - ¦/x - åi=1k ¦/yiyi'   =   åi=1k  yi'( d/dx(¦/yi') - ¦/yi ) - ¦/x\n=   åi=1k  yi'( d/dx(¦/yi') - ¦/yi )     if ¦/x = 0\n=   0 if d/dx(¦/yi') = ¦/yi     i=1,2,..,k which are the Euler-Lagrange equations.  .]\n\nTheoretical physics then becomes a quest for the One True Action Lagrangian density , usually assumed real scalar and generating kinematic equations involving zeroth, first,  and second differentials only. Geometric (multivector-valued) Lagrangians extermised in the sense that each multivector coordinate is stationary under variation. This is an intrinsically nonrelativistic approach in the case of multiple particulate systems since it requires a favoured temporal parametrisation t with L(t,p,p') = L(p,p')\nAs \"locally shortest routes\", Geodesics are 1-extremals for the scalar path length of ò0t ds|dp¿g(dp)|½ . Integrating a square root can be messy but fortunately such paths also extremise   ò0t ds ds p'(s)¿gp(s))(p'(s)) where p'(s) denotes the (d/ds)p(s) . _Be Thus we set L(p,p',s) = p'¿gp(p')|½ and minimise scalar integral   ò0t ds ds L(p(s),p'(s),s) by solving the N Euler-Lagrange equations d/ds(L/pi') = L/pi) for i=1,2,...N .\nL(p,p',s) will generally depend on p via _gp but if _gp has a symmetry such that L/pi=0 for a particular coordinate i then L/pi' is constant along any geodesic path.\nGeodesic flow can consequently be viewed as arising from a Lagrangian density L(p(t),p'(t)) = |p'(t)¿_gp(p'(t))|½ . with particles following timelike trajectories that extremise (minimally) their proper time (arclength).\nIf t is proper time parameterisation then L(p(t),p'(t)) = |p'(t)2|½ = 1 along a geodesic.\nElectrodynamical forces due to a 1-vector four-potential ap (typically with Ñp¿ap=0) are introduced by adding scalar -q p'(t)¿ ap(t) to the Lagrangian density leading to the Lorentz force law mp\"(t) = qfp.p'(t) where fp = ÑpÙap .\n[ Proof : L = m(-p'2)½ - q p'¿ap Þ ÑpL = Ñp(m(-p'2)½ - q p'¿ ap) = -qÑp(p'¿ap) ; Ñp'L = -m(-p'(t)2) p'(t) - qap     so the Euler-Lagrange equation is qp'¿Ñpap = (d/dt)(m(-p'2) p' + qap ) = mp\" + q(p'¿Ñp)ap Þ mp\" = q(Ñp(ap¿p') - (p'¿Ñp)ap))   =   q(ÑpÙap)¿p'  .]\n\nExtremal surfaces\nA more geometric generalistaion is to consider the problem of finding the M-curve of given content maximising a particular boundary or interior integral. We might, for example, seek the loop starting and ending at a given point a and constrained to lie in a given 2-curve (surface) containing a that maximises enclosed content (area) for a given boundary content (pathlength)\n\nOf fundamental importance is the fact that spacial Lagrangian density (Ñ[e123]yp)2 is extremised for integration over CM (subject to constraint of given boundary values over dCM) if yp satisfies the spacial Laplace equation Ñp [e123]2yp = 0 over CM .\nIn electrodynamics we have L = - 2-4p-1 fp2 + c-1jp¿ap where c is scalar lightspeed and fp = ÑpÙap extremised by solutions to Maxwell equations. ×××××××××××××××××××××××××××××××××××××××", null, "References/Source Material for Multivector Physics\nAnthony Lasenby \"Recent Applications of Conformal Geometric Algebra\"   2005\nhttp://www.mrao.cam.ac.uk/~clifford/publications\n\nGlossary   Contents   Author\nCopyright (c) Ian C G Bell 2006, 2014\nWeb Source: www.iancgbell.clara.net/maths\nLatest Edit: 15 Jun 2014.", null, "" ]

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[ "# Ounce Of Mathematics\n\nprof. Brian Harvey used to say \"an ounce of mathematics is worth pound of computer science\".\n\nThere is how. In famous Peter Norvig's course (on udacity) there is a small sub-task to write a tiny predicate procedure, which returns True if a hand of cards has the same suit.\n\nLike a good student of CS61A I wrote this pile of functional-style code:\n\n```def flush(hand):\nsuits = [s for r,s in hand]\nreturn reduce(lambda x, y: x and y, map(lambda a: a == a,\nzip(suits, suits[1:])))\n```\n\nWhat could possibly go wrong? You could visualize how beautifully a list zips with itself (strictly via references, no data copied!) and how a new list of Booleans is going to be produced, to be folded then into a single value. You could see the flow and all the wonders.\n\nHere is the Norvig's snippet\n\n```def flush(hand):\nsuits = [s for r,s in hand]\nreturn len(set(suits)) == 1\n```\n\nWhat the ...? Well, the set-from-a-list constructor removes all duplicates (due to the nature of a Math set), so if the length of a resulting set is 1, then all elements of a list were the same value.\n\nBut, but we know how to make a set ADT out of conses and write short-circuiting `and` or `every?` for sequences.))" ]

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[ "# Econometrica\n\n## Journal Of The Econometric Society\n\n### An International Society for the Advancement of Economic Theory in its Relation to Statistics and Mathematics\n\nEdited by: Guido W. Imbens • Print ISSN: 0012-9682 • Online ISSN: 1468-0262\n\nEconometrica: Jan, 2009, Volume 77, Issue 1\n\n# Decision Theory Applied to a Linear Panel Data Model\n\nhttps://doi.org/10.3982/ECTA6869\np. 107-133\n\nGary Chamberlain, Marcelo J. Moreira\n\nThis paper applies some general concepts in decision theory to a linear panel data model. A simple version of the model is an autoregression with a separate intercept for each unit in the cross section, with errors that are independent and identically distributed with a normal distribution. There is a parameter of interest γ and a nuisance parameter τ, a × matrix, where is the cross‐section sample size. The focus is on dealing with the incidental parameters problem created by a potentially high‐dimension nuisance parameter. We adopt a “fixed‐effects” approach that seeks to protect against any sequence of incidental parameters. We transform to (δ, ρ, ω), where δ is a × matrix of coefficients from the least‐squares projection of on a × matrix of strictly exogenous variables, ρ is a × symmetric, positive semidefinite matrix obtained from the residual sums of squares and cross‐products in the projection of τ on , and is a (−) × matrix whose columns are orthogonal and have unit length. The model is invariant under the actions of a group on the sample space and the parameter space, and we find a maximal invariant statistic. The distribution of the maximal invariant statistic does not depend upon ω. There is a unique invariant distribution for ω. We use this invariant distribution as a prior distribution to obtain an integrated likelihood function. It depends upon the observation only through the maximal invariant statistic. We use the maximal invariant statistic to construct a marginal likelihood function, so we can eliminate ω by integration with respect to the invariant prior distribution or by working with the marginal likelihood function. The two approaches coincide." ]

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[ "# We use percentages in almost everything.\n\nPercentages are a very important part of our daily lives. They are used in Economics, Cooking, Health, Sports, Mathematics, Science, Jewellery, Geography, Medicine and many other areas.\n\n## Percent of Calculator\n\nCalculate percentage of X, quick & simple.\n\n%\n?\n\n1% of 2656 is:\n26.56\n\n## Percent of - Table For 2656\n\nPercent of Difference\n1% of 2656 is 26.56 2629.44\n2% of 2656 is 53.12 2602.88\n3% of 2656 is 79.68 2576.32\n4% of 2656 is 106.24 2549.76\n5% of 2656 is 132.8 2523.2\n6% of 2656 is 159.36 2496.64\n7% of 2656 is 185.92 2470.08\n8% of 2656 is 212.48 2443.52\n9% of 2656 is 239.04 2416.96\n10% of 2656 is 265.6 2390.4\n11% of 2656 is 292.16 2363.84\n12% of 2656 is 318.72 2337.28\n13% of 2656 is 345.28 2310.72\n14% of 2656 is 371.84 2284.16\n15% of 2656 is 398.4 2257.6\n16% of 2656 is 424.96 2231.04\n17% of 2656 is 451.52 2204.48\n18% of 2656 is 478.08 2177.92\n19% of 2656 is 504.64 2151.36\n20% of 2656 is 531.2 2124.8\n21% of 2656 is 557.76 2098.24\n22% of 2656 is 584.32 2071.68\n23% of 2656 is 610.88 2045.12\n24% of 2656 is 637.44 2018.56\n25% of 2656 is 664 1992\n26% of 2656 is 690.56 1965.44\n27% of 2656 is 717.12 1938.88\n28% of 2656 is 743.68 1912.32\n29% of 2656 is 770.24 1885.76\n30% of 2656 is 796.8 1859.2\n31% of 2656 is 823.36 1832.64\n32% of 2656 is 849.92 1806.08\n33% of 2656 is 876.48 1779.52\n34% of 2656 is 903.04 1752.96\n35% of 2656 is 929.6 1726.4\n36% of 2656 is 956.16 1699.84\n37% of 2656 is 982.72 1673.28\n38% of 2656 is 1009.28 1646.72\n39% of 2656 is 1035.84 1620.16\n40% of 2656 is 1062.4 1593.6\n41% of 2656 is 1088.96 1567.04\n42% of 2656 is 1115.52 1540.48\n43% of 2656 is 1142.08 1513.92\n44% of 2656 is 1168.64 1487.36\n45% of 2656 is 1195.2 1460.8\n46% of 2656 is 1221.76 1434.24\n47% of 2656 is 1248.32 1407.68\n48% of 2656 is 1274.88 1381.12\n49% of 2656 is 1301.44 1354.56\n50% of 2656 is 1328 1328\n51% of 2656 is 1354.56 1301.44\n52% of 2656 is 1381.12 1274.88\n53% of 2656 is 1407.68 1248.32\n54% of 2656 is 1434.24 1221.76\n55% of 2656 is 1460.8 1195.2\n56% of 2656 is 1487.36 1168.64\n57% of 2656 is 1513.92 1142.08\n58% of 2656 is 1540.48 1115.52\n59% of 2656 is 1567.04 1088.96\n60% of 2656 is 1593.6 1062.4\n61% of 2656 is 1620.16 1035.84\n62% of 2656 is 1646.72 1009.28\n63% of 2656 is 1673.28 982.72\n64% of 2656 is 1699.84 956.16\n65% of 2656 is 1726.4 929.6\n66% of 2656 is 1752.96 903.04\n67% of 2656 is 1779.52 876.48\n68% of 2656 is 1806.08 849.92\n69% of 2656 is 1832.64 823.36\n70% of 2656 is 1859.2 796.8\n71% of 2656 is 1885.76 770.24\n72% of 2656 is 1912.32 743.68\n73% of 2656 is 1938.88 717.12\n74% of 2656 is 1965.44 690.56\n75% of 2656 is 1992 664\n76% of 2656 is 2018.56 637.44\n77% of 2656 is 2045.12 610.88\n78% of 2656 is 2071.68 584.32\n79% of 2656 is 2098.24 557.76\n80% of 2656 is 2124.8 531.2\n81% of 2656 is 2151.36 504.64\n82% of 2656 is 2177.92 478.08\n83% of 2656 is 2204.48 451.52\n84% of 2656 is 2231.04 424.96\n85% of 2656 is 2257.6 398.4\n86% of 2656 is 2284.16 371.84\n87% of 2656 is 2310.72 345.28\n88% of 2656 is 2337.28 318.72\n89% of 2656 is 2363.84 292.16\n90% of 2656 is 2390.4 265.6\n91% of 2656 is 2416.96 239.04\n92% of 2656 is 2443.52 212.48\n93% of 2656 is 2470.08 185.92\n94% of 2656 is 2496.64 159.36\n95% of 2656 is 2523.2 132.8\n96% of 2656 is 2549.76 106.24\n97% of 2656 is 2576.32 79.68\n98% of 2656 is 2602.88 53.12\n99% of 2656 is 2629.44 26.56\n100% of 2656 is 2656 0\n\n### Here's How to Calculate 1% of 2656\n\nLet's take a quick example here:\n\nYou have a Target coupon of \\$2656 and you need to know how much will you save on your purchase if the discount is 1 percent.\n\nSolution:\n\nAmount Saved = Original Price x Discount in Percent / 100\n\nAmount Saved = (2656 x 1) / 100\n\nAmount Saved = 2656 / 100\n\nIn other words, a 1% discount for a purchase with an original price of \\$2656 equals \\$26.56 (Amount Saved), so you'll end up paying 2629.44.\n\n### Calculating Percentages\n\nSimply click on the calculate button to get the results of percentage calculations. You will see the result on the next page. If there are errors in the input fields, the result page will be blank. The program allows you to calculate the difference between two numbers in percentages. You can also input a percentage of any number and get the numeric value. Although it is a simple calculator, it can be very useful in many scenarios. Our goal is to give you an easy to use percentage calculator that gives you results you want fast.\n\nPercentage in mathematics refers to fractions based in 100. It is usually represented by “%,” “pct,” or “percentage.” This web app allows a comma or dot as a decimal separator. So you can use both freely.\n\nWe have provided several examples for you to use. You can use the examples to feed in your own data correctly. We hope you will find this site useful for calculang percentages. You can even use it for crosschecking the accuracy of your assignment results.\n\nNB. Americans use “percent,” which the British prefer “per cent.”\n\n#### Examples\n\nExample one\n\nCalculate 20% of 200?\n20% of 200 =____\n(200/100) x 20 = _____\n2 x 20 = 40\n\nIt is quite easy. Just divide 200 by 100 to get one percent. The result is 2. Then multiply it by 20 ( 20% = 20 per hundred) = 20 x 2 = 40\n\nExample two\n\nWhat percentage of 125 is 50?\n\n50 = ---% of 125\n50 x (100/125) = 40%\n\nGet the value of one percent by dividing 100 by 125. After that, multiply the value by 50 to get the percentage value of 50 units, which is 40% That is how to calculate the percentage.\n\nExample three\n\nWhat is the percentage (%) change (increase or decrease) from 120 to 150?\n\n(150-120) x (100/120) = 36\n\nSince 150 represents 100%. One percent will be equal to 100/150. 150-120 is 30. Therefore, 30 units represents 30 x (100/150) = 36 % This is how to calculate the percentage increase.\n\nwe do not use a percentage at all times. There are scenarios where we simply want to show the ratio of numbers. For instance, what is 20% of 50? This can also be interpreted as 20 hundredths of 50. This equates to 20/100 x 50 = 10.\n\nYou can use a calculation trick here. Anyme you want to divide a number by 100, just move the decimal two places to the left. 20/100 x 50 calculated above can also be writen as (20 x 50)/100. Since 20x 50 =1000. You can simply divide 1000 by 100 by moving two decimal places to the left, which gives you 10.\n\nIn another scenario, you want to calculate the percentage increase or decrease. Supposing you have \\$10 and spend \\$2 to buy candy, then you have spent 20% of your money. So how much will be remaining? All the money you have is 100%, if you spend 20%, you will have 80% remaining. You can simply use the percentage reduction tool above to calculate this value.\n\n#### Origin\n\nThe word percent is derived from the Latin word percenter which means per hundred, and it is designated by %" ]

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[ "User Real IP - 3.237.94.109\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => Array\n(\n => 45.116.232.56\n)\n\n => Array\n(\n => 45.118.166.91\n)\n\n => Array\n(\n => 180.92.230.234\n)\n\n => Array\n(\n => 157.40.57.160\n)\n\n => Array\n(\n => 110.38.38.130\n)\n\n => Array\n(\n => 72.255.57.183\n)\n\n => Array\n(\n => 182.68.81.85\n)\n\n => Array\n(\n => 39.57.202.122\n)\n\n => Array\n(\n => 119.152.154.36\n)\n\n => Array\n(\n => 5.62.62.141\n)\n\n => Array\n(\n => 119.155.54.232\n)\n\n => Array\n(\n => 39.37.141.22\n)\n\n => Array\n(\n => 183.87.12.225\n)\n\n => Array\n(\n => 107.170.127.117\n)\n\n => Array\n(\n => 125.63.124.49\n)\n\n => Array\n(\n => 39.42.191.3\n)\n\n => Array\n(\n => 116.74.24.72\n)\n\n => Array\n(\n => 46.101.89.227\n)\n\n => Array\n(\n => 202.173.125.247\n)\n\n => Array\n(\n => 39.42.184.254\n)\n\n => Array\n(\n => 115.186.165.132\n)\n\n => Array\n(\n => 39.57.206.126\n)\n\n => Array\n(\n => 103.245.13.145\n)\n\n => Array\n(\n => 202.175.246.43\n)\n\n => Array\n(\n => 192.140.152.150\n)\n\n => Array\n(\n => 202.88.250.103\n)\n\n => Array\n(\n => 103.248.94.207\n)\n\n => Array\n(\n => 77.73.66.101\n)\n\n => Array\n(\n => 104.131.66.8\n)\n\n => Array\n(\n => 113.186.161.97\n)\n\n => Array\n(\n => 222.254.5.7\n)\n\n => Array\n(\n => 223.233.67.247\n)\n\n => Array\n(\n => 171.249.116.146\n)\n\n => Array\n(\n => 47.30.209.71\n)\n\n => Array\n(\n => 202.134.13.130\n)\n\n => Array\n(\n => 27.6.135.7\n)\n\n => Array\n(\n => 107.170.186.79\n)\n\n => Array\n(\n => 103.212.89.171\n)\n\n => Array\n(\n => 117.197.9.77\n)\n\n => Array\n(\n => 122.176.206.233\n)\n\n => Array\n(\n => 192.227.253.222\n)\n\n => Array\n(\n => 182.188.224.119\n)\n\n => Array\n(\n => 14.248.70.74\n)\n\n => Array\n(\n => 42.118.219.169\n)\n\n => Array\n(\n => 110.39.146.170\n)\n\n => Array\n(\n => 119.160.66.143\n)\n\n => Array\n(\n => 103.248.95.130\n)\n\n => Array\n(\n => 27.63.152.208\n)\n\n => Array\n(\n => 49.207.114.96\n)\n\n => Array\n(\n => 102.166.23.214\n)\n\n => Array\n(\n => 175.107.254.73\n)\n\n => Array\n(\n => 103.10.227.214\n)\n\n => Array\n(\n => 202.143.115.89\n)\n\n => Array\n(\n => 110.93.227.187\n)\n\n => Array\n(\n => 103.140.31.60\n)\n\n => Array\n(\n => 110.37.231.46\n)\n\n => Array\n(\n => 39.36.99.238\n)\n\n => Array\n(\n => 157.37.140.26\n)\n\n => Array\n(\n => 43.246.202.226\n)\n\n => Array\n(\n => 137.97.8.143\n)\n\n => Array\n(\n => 182.65.52.242\n)\n\n => Array\n(\n => 115.42.69.62\n)\n\n => Array\n(\n => 14.143.254.58\n)\n\n => Array\n(\n => 223.179.143.236\n)\n\n => Array\n(\n => 223.179.143.249\n)\n\n => Array\n(\n => 103.143.7.54\n)\n\n => Array\n(\n => 223.179.139.106\n)\n\n => Array\n(\n => 39.40.219.90\n)\n\n => Array\n(\n => 45.115.141.231\n)\n\n => Array\n(\n => 120.29.100.33\n)\n\n => Array\n(\n => 112.196.132.5\n)\n\n => Array\n(\n => 202.163.123.153\n)\n\n => Array\n(\n => 5.62.58.146\n)\n\n => Array\n(\n => 39.53.216.113\n)\n\n => Array\n(\n => 42.111.160.73\n)\n\n => Array\n(\n => 107.182.231.213\n)\n\n => Array\n(\n => 119.82.94.120\n)\n\n => Array\n(\n => 178.62.34.82\n)\n\n => Array\n(\n => 203.122.6.18\n)\n\n => Array\n(\n => 157.42.38.251\n)\n\n => Array\n(\n => 45.112.68.222\n)\n\n => Array\n(\n => 49.206.212.122\n)\n\n => Array\n(\n => 104.236.70.228\n)\n\n => Array\n(\n => 42.111.34.243\n)\n\n => Array\n(\n => 84.241.19.186\n)\n\n => Array\n(\n => 89.187.180.207\n)\n\n => Array\n(\n => 104.243.212.118\n)\n\n => Array\n(\n => 104.236.55.136\n)\n\n => Array\n(\n => 106.201.16.163\n)\n\n => Array\n(\n => 46.101.40.25\n)\n\n => Array\n(\n => 45.118.166.94\n)\n\n => Array\n(\n => 49.36.128.102\n)\n\n => Array\n(\n => 14.142.193.58\n)\n\n => Array\n(\n => 212.79.124.176\n)\n\n => Array\n(\n => 45.32.191.194\n)\n\n => Array\n(\n => 105.112.107.46\n)\n\n => Array\n(\n => 106.201.14.8\n)\n\n => Array\n(\n => 110.93.240.65\n)\n\n => Array\n(\n => 27.96.95.177\n)\n\n => Array\n(\n => 45.41.134.35\n)\n\n => Array\n(\n => 180.151.13.110\n)\n\n => Array\n(\n => 101.53.242.89\n)\n\n => Array\n(\n => 115.186.3.110\n)\n\n => Array\n(\n => 171.49.185.242\n)\n\n => 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Array\n(\n => 103.224.54.190\n)\n\n => Array\n(\n => 49.36.141.210\n)\n\n => Array\n(\n => 5.62.63.218\n)\n\n => Array\n(\n => 39.59.9.18\n)\n\n => Array\n(\n => 111.88.86.45\n)\n\n => Array\n(\n => 178.54.139.5\n)\n\n => Array\n(\n => 116.68.105.241\n)\n\n => Array\n(\n => 119.160.96.187\n)\n\n => Array\n(\n => 182.189.192.103\n)\n\n => Array\n(\n => 119.160.96.143\n)\n\n => Array\n(\n => 110.225.89.98\n)\n\n => Array\n(\n => 169.149.195.134\n)\n\n => Array\n(\n => 103.238.104.54\n)\n\n => Array\n(\n => 47.30.208.142\n)\n\n => Array\n(\n => 157.46.179.209\n)\n\n => Array\n(\n => 223.235.38.119\n)\n\n => Array\n(\n => 42.106.180.165\n)\n\n => Array\n(\n => 154.122.240.239\n)\n\n => Array\n(\n => 106.223.104.191\n)\n\n => Array\n(\n => 111.93.110.218\n)\n\n => Array\n(\n => 182.183.161.171\n)\n\n => Array\n(\n => 157.44.184.211\n)\n\n => Array\n(\n => 157.50.185.193\n)\n\n => Array\n(\n => 117.230.19.194\n)\n\n => Array\n(\n => 162.243.246.160\n)\n\n => Array\n(\n => 106.223.143.53\n)\n\n => Array\n(\n => 39.59.41.15\n)\n\n => Array\n(\n => 106.210.65.42\n)\n\n => Array\n(\n => 180.243.144.208\n)\n\n => Array\n(\n => 116.68.105.22\n)\n\n => Array\n(\n => 115.42.70.46\n)\n\n => Array\n(\n => 99.72.192.148\n)\n\n => Array\n(\n => 182.183.182.48\n)\n\n => Array\n(\n => 171.48.58.97\n)\n\n => Array\n(\n => 37.120.131.188\n)\n\n => Array\n(\n => 117.99.167.177\n)\n\n => Array\n(\n => 111.92.76.210\n)\n\n => Array\n(\n => 14.192.144.245\n)\n\n => Array\n(\n => 169.149.242.87\n)\n\n => Array\n(\n => 47.30.198.149\n)\n\n => Array\n(\n => 59.103.57.140\n)\n\n => Array\n(\n => 117.230.161.168\n)\n\n => Array\n(\n => 110.225.88.173\n)\n\n => Array\n(\n => 169.149.246.95\n)\n\n => Array\n(\n => 42.106.180.52\n)\n\n => Array\n(\n => 14.231.160.157\n)\n\n => Array\n(\n => 123.27.109.47\n)\n\n => Array\n(\n => 157.46.130.54\n)\n\n => Array\n(\n => 39.42.73.194\n)\n\n => Array\n(\n => 117.230.18.147\n)\n\n => Array\n(\n => 27.59.231.98\n)\n\n => Array\n(\n => 125.209.78.227\n)\n\n => Array\n(\n => 157.34.80.145\n)\n\n => Array\n(\n => 42.201.251.86\n)\n\n => Array\n(\n => 117.230.129.158\n)\n\n => Array\n(\n => 103.82.80.103\n)\n\n => Array\n(\n => 47.9.171.228\n)\n\n => Array\n(\n => 117.230.24.92\n)\n\n => Array\n(\n => 103.129.143.119\n)\n\n => Array\n(\n => 39.40.213.45\n)\n\n => Array\n(\n => 178.92.188.214\n)\n\n => Array\n(\n => 110.235.232.191\n)\n\n => Array\n(\n => 5.62.34.18\n)\n\n => Array\n(\n => 47.30.212.134\n)\n\n => Array\n(\n => 157.42.34.196\n)\n\n => Array\n(\n => 157.32.169.9\n)\n\n => Array\n(\n => 103.255.4.11\n)\n\n => Array\n(\n => 117.230.13.69\n)\n\n => Array\n(\n => 117.230.58.97\n)\n\n => Array\n(\n => 92.52.138.39\n)\n\n => Array\n(\n => 221.132.119.63\n)\n\n => Array\n(\n => 117.97.167.188\n)\n\n => Array\n(\n => 119.153.56.58\n)\n\n => Array\n(\n => 105.50.22.150\n)\n\n => Array\n(\n => 115.42.68.126\n)\n\n => Array\n(\n => 182.189.223.159\n)\n\n => Array\n(\n => 39.59.36.90\n)\n\n => Array\n(\n => 111.92.76.114\n)\n\n => Array\n(\n => 157.47.226.163\n)\n\n => Array\n(\n => 202.47.44.37\n)\n\n => Array\n(\n => 106.51.234.172\n)\n\n => Array\n(\n => 103.101.88.166\n)\n\n => Array\n(\n => 27.6.246.146\n)\n\n => Array\n(\n => 103.255.5.83\n)\n\n => Array\n(\n => 103.98.210.185\n)\n\n => Array\n(\n => 122.173.114.134\n)\n\n => Array\n(\n => 122.173.77.248\n)\n\n => Array\n(\n => 5.62.41.172\n)\n\n => Array\n(\n => 180.178.181.17\n)\n\n => Array\n(\n => 37.120.133.224\n)\n\n => Array\n(\n => 45.131.5.156\n)\n\n => Array\n(\n => 110.39.100.110\n)\n\n => Array\n(\n => 176.110.38.185\n)\n\n => Array\n(\n => 36.255.41.64\n)\n\n => Array\n(\n => 103.104.192.15\n)\n\n => Array\n(\n => 43.245.131.195\n)\n\n => Array\n(\n => 14.248.111.185\n)\n\n => Array\n(\n => 122.173.217.133\n)\n\n => Array\n(\n => 106.223.90.245\n)\n\n => Array\n(\n => 119.153.56.80\n)\n\n => Array\n(\n => 103.7.60.172\n)\n\n => Array\n(\n => 157.46.184.233\n)\n\n => Array\n(\n => 182.190.31.95\n)\n\n => Array\n(\n => 109.87.189.122\n)\n\n => Array\n(\n => 91.74.25.100\n)\n\n => Array\n(\n => 182.185.224.144\n)\n\n => Array\n(\n => 106.223.91.221\n)\n\n => Array\n(\n => 182.190.223.40\n)\n\n => Array\n(\n => 2.58.194.134\n)\n\n => Array\n(\n => 196.246.225.236\n)\n\n => Array\n(\n => 106.223.90.173\n)\n\n => Array\n(\n => 23.239.16.54\n)\n\n => Array\n(\n => 157.46.65.225\n)\n\n => Array\n(\n => 115.186.130.14\n)\n\n => Array\n(\n => 103.85.125.157\n)\n\n => Array\n(\n => 14.248.103.6\n)\n\n => Array\n(\n => 123.24.169.247\n)\n\n => Array\n(\n => 103.130.108.153\n)\n\n => Array\n(\n => 115.42.67.21\n)\n\n => Array\n(\n => 202.166.171.190\n)\n\n => Array\n(\n => 39.37.169.104\n)\n\n => Array\n(\n => 103.82.80.59\n)\n\n => Array\n(\n => 175.107.208.58\n)\n\n => Array\n(\n => 203.192.238.247\n)\n\n => Array\n(\n => 103.217.178.150\n)\n\n => Array\n(\n => 103.66.214.173\n)\n\n => Array\n(\n => 110.93.236.174\n)\n\n => Array\n(\n => 143.189.242.64\n)\n\n => Array\n(\n => 77.111.245.12\n)\n\n => Array\n(\n => 145.239.2.231\n)\n\n => Array\n(\n => 115.186.190.38\n)\n\n => Array\n(\n => 109.169.23.67\n)\n\n => Array\n(\n => 198.16.70.29\n)\n\n => Array\n(\n => 111.92.76.186\n)\n\n => Array\n(\n => 115.42.69.34\n)\n\n => Array\n(\n => 73.61.100.95\n)\n\n => Array\n(\n => 103.129.142.31\n)\n\n => Array\n(\n => 103.255.5.53\n)\n\n => Array\n(\n => 103.76.55.2\n)\n\n => Array\n(\n => 47.9.141.138\n)\n\n => Array\n(\n => 103.55.89.234\n)\n\n => Array\n(\n => 103.223.13.53\n)\n\n => Array\n(\n => 175.158.50.203\n)\n\n => Array\n(\n => 103.255.5.90\n)\n\n => Array\n(\n => 106.223.100.138\n)\n\n => Array\n(\n => 39.37.143.193\n)\n\n => Array\n(\n => 206.189.133.131\n)\n\n => Array\n(\n => 43.224.0.233\n)\n\n => Array\n(\n => 115.186.132.106\n)\n\n => Array\n(\n => 31.43.21.159\n)\n\n => Array\n(\n => 119.155.56.131\n)\n\n => Array\n(\n => 103.82.80.138\n)\n\n => Array\n(\n => 24.87.128.119\n)\n\n => Array\n(\n => 106.210.103.163\n)\n\n => Array\n(\n => 103.82.80.90\n)\n\n => Array\n(\n => 157.46.186.45\n)\n\n => Array\n(\n => 157.44.155.238\n)\n\n => Array\n(\n => 103.119.199.2\n)\n\n => Array\n(\n => 27.97.169.205\n)\n\n => Array\n(\n => 157.46.174.89\n)\n\n => Array\n(\n => 43.250.58.220\n)\n\n => Array\n(\n => 76.189.186.64\n)\n\n => Array\n(\n => 103.255.5.57\n)\n\n => Array\n(\n => 171.61.196.136\n)\n\n => Array\n(\n => 202.47.40.88\n)\n\n => Array\n(\n => 97.118.94.116\n)\n\n => Array\n(\n => 157.44.124.157\n)\n\n => Array\n(\n => 95.142.120.13\n)\n\n => Array\n(\n => 42.201.229.151\n)\n\n => Array\n(\n => 157.46.178.95\n)\n\n => Array\n(\n => 169.149.215.192\n)\n\n => Array\n(\n => 42.111.19.48\n)\n\n => Array\n(\n => 1.38.52.18\n)\n\n => Array\n(\n => 145.239.91.241\n)\n\n => Array\n(\n => 47.31.78.191\n)\n\n => Array\n(\n => 103.77.42.60\n)\n\n => Array\n(\n => 157.46.107.144\n)\n\n => Array\n(\n => 157.46.125.124\n)\n\n => Array\n(\n => 110.225.218.108\n)\n\n => Array\n(\n => 106.51.77.185\n)\n\n => Array\n(\n => 123.24.161.207\n)\n\n => Array\n(\n => 106.210.108.22\n)\n\n => Array\n(\n => 42.111.10.14\n)\n\n => Array\n(\n => 223.29.231.175\n)\n\n => Array\n(\n => 27.56.152.132\n)\n\n => Array\n(\n => 119.155.31.100\n)\n\n => Array\n(\n => 122.173.172.127\n)\n\n => Array\n(\n => 103.77.42.64\n)\n\n => Array\n(\n => 157.44.164.106\n)\n\n => Array\n(\n => 14.181.53.38\n)\n\n => Array\n(\n => 115.42.67.64\n)\n\n => Array\n(\n => 47.31.33.140\n)\n\n => Array\n(\n => 103.15.60.234\n)\n\n => Array\n(\n => 182.64.219.181\n)\n\n => Array\n(\n => 103.44.51.6\n)\n\n => Array\n(\n => 116.74.25.157\n)\n\n => Array\n(\n => 116.71.2.128\n)\n\n => Array\n(\n => 157.32.185.239\n)\n\n => Array\n(\n => 47.31.25.79\n)\n\n => Array\n(\n => 178.62.85.75\n)\n\n => Array\n(\n => 180.178.190.39\n)\n\n => Array\n(\n => 39.48.52.179\n)\n\n => Array\n(\n => 106.193.11.240\n)\n\n => Array\n(\n => 103.82.80.226\n)\n\n => Array\n(\n => 49.206.126.30\n)\n\n => Array\n(\n => 157.245.191.173\n)\n\n => Array\n(\n => 49.205.84.237\n)\n\n => Array\n(\n => 47.8.181.232\n)\n\n => Array\n(\n => 182.66.2.92\n)\n\n => Array\n(\n => 49.34.137.220\n)\n\n => Array\n(\n => 209.205.217.125\n)\n\n => Array\n(\n => 192.64.5.73\n)\n\n => Array\n(\n => 27.63.166.108\n)\n\n => Array\n(\n => 120.29.96.211\n)\n\n => Array\n(\n => 182.186.112.135\n)\n\n => Array\n(\n => 45.118.165.151\n)\n\n => Array\n(\n => 47.8.228.12\n)\n\n => Array\n(\n => 106.215.3.162\n)\n\n => Array\n(\n => 111.92.72.66\n)\n\n => Array\n(\n => 169.145.2.9\n)\n\n => Array\n(\n => 106.207.205.100\n)\n\n => Array\n(\n => 223.181.8.12\n)\n\n => Array\n(\n => 157.48.149.78\n)\n\n => Array\n(\n => 103.206.138.116\n)\n\n => Array\n(\n => 39.53.119.22\n)\n\n => Array\n(\n => 157.33.232.106\n)\n\n => Array\n(\n => 49.37.205.139\n)\n\n => Array\n(\n => 115.42.68.3\n)\n\n => Array\n(\n => 93.72.182.251\n)\n\n => Array\n(\n => 202.142.166.22\n)\n\n => Array\n(\n => 157.119.81.111\n)\n\n => Array\n(\n => 182.186.116.155\n)\n\n => Array\n(\n => 157.37.171.37\n)\n\n => Array\n(\n => 117.206.164.48\n)\n\n => Array\n(\n => 49.36.52.63\n)\n\n => Array\n(\n => 203.175.72.112\n)\n\n => Array\n(\n => 171.61.132.193\n)\n\n => Array\n(\n => 111.119.187.44\n)\n\n => Array\n(\n => 39.37.165.216\n)\n\n => Array\n(\n => 103.86.109.58\n)\n\n => Array\n(\n => 39.59.2.86\n)\n\n => Array\n(\n => 111.119.187.28\n)\n\n => Array\n(\n => 106.201.9.10\n)\n\n => Array\n(\n => 49.35.25.106\n)\n\n => Array\n(\n => 157.49.239.103\n)\n\n => Array\n(\n => 157.49.237.198\n)\n\n => Array\n(\n => 14.248.64.121\n)\n\n => Array\n(\n => 117.102.7.214\n)\n\n => Array\n(\n => 120.29.91.246\n)\n\n => Array\n(\n => 103.7.79.41\n)\n\n => Array\n(\n => 132.154.99.209\n)\n\n => Array\n(\n => 212.36.27.245\n)\n\n => Array\n(\n => 157.44.154.9\n)\n\n => Array\n(\n => 47.31.56.44\n)\n\n => Array\n(\n => 192.142.199.136\n)\n\n => Array\n(\n => 171.61.159.49\n)\n\n => Array\n(\n => 119.160.116.151\n)\n\n => Array\n(\n => 103.98.63.39\n)\n\n => Array\n(\n => 41.60.233.216\n)\n\n => Array\n(\n => 49.36.75.212\n)\n\n => Array\n(\n => 223.188.60.20\n)\n\n => Array\n(\n => 103.98.63.50\n)\n\n => Array\n(\n => 178.162.198.21\n)\n\n => Array\n(\n => 157.46.209.35\n)\n\n => Array\n(\n => 119.155.32.151\n)\n\n => Array\n(\n => 102.185.58.161\n)\n\n => Array\n(\n => 59.96.89.231\n)\n\n => Array\n(\n => 119.155.255.198\n)\n\n => Array\n(\n => 42.107.204.57\n)\n\n => Array\n(\n => 42.106.181.74\n)\n\n => Array\n(\n => 157.46.219.186\n)\n\n => Array\n(\n => 115.42.71.49\n)\n\n => Array\n(\n => 157.46.209.131\n)\n\n => Array\n(\n => 220.81.15.94\n)\n\n => Array\n(\n => 111.119.187.24\n)\n\n => Array\n(\n => 49.37.195.185\n)\n\n => Array\n(\n => 42.106.181.85\n)\n\n => Array\n(\n => 43.249.225.134\n)\n\n => Array\n(\n => 117.206.165.151\n)\n\n => Array\n(\n => 119.153.48.250\n)\n\n => Array\n(\n => 27.4.172.162\n)\n\n => Array\n(\n => 117.20.29.51\n)\n\n => Array\n(\n => 103.98.63.135\n)\n\n => Array\n(\n => 117.7.218.229\n)\n\n => Array\n(\n => 157.49.233.105\n)\n\n => Array\n(\n => 39.53.151.199\n)\n\n => Array\n(\n => 101.255.118.33\n)\n\n => Array\n(\n => 41.141.246.9\n)\n\n => Array\n(\n => 221.132.113.78\n)\n\n => Array\n(\n => 119.160.116.202\n)\n\n => Array\n(\n => 117.237.193.244\n)\n\n => Array\n(\n => 157.41.110.145\n)\n\n => Array\n(\n => 103.98.63.5\n)\n\n => Array\n(\n => 103.125.129.58\n)\n\n => Array\n(\n => 183.83.254.66\n)\n\n => Array\n(\n => 45.135.236.160\n)\n\n => Array\n(\n => 198.199.87.124\n)\n\n => Array\n(\n => 193.176.86.41\n)\n\n => Array\n(\n => 115.97.142.98\n)\n\n => Array\n(\n => 222.252.38.198\n)\n\n => Array\n(\n => 110.93.237.49\n)\n\n => Array\n(\n => 103.224.48.122\n)\n\n => Array\n(\n => 110.38.28.130\n)\n\n => Array\n(\n => 106.211.238.154\n)\n\n => Array\n(\n => 111.88.41.73\n)\n\n => Array\n(\n => 119.155.13.143\n)\n\n => Array\n(\n => 103.213.111.60\n)\n\n => Array\n(\n => 202.0.103.42\n)\n\n => Array\n(\n => 157.48.144.33\n)\n\n => Array\n(\n => 111.119.187.62\n)\n\n => Array\n(\n => 103.87.212.71\n)\n\n => Array\n(\n => 157.37.177.20\n)\n\n => Array\n(\n => 223.233.71.92\n)\n\n => Array\n(\n => 116.213.32.107\n)\n\n => Array\n(\n => 104.248.173.151\n)\n\n => Array\n(\n => 14.181.102.222\n)\n\n => Array\n(\n => 103.10.224.252\n)\n\n => Array\n(\n => 175.158.50.57\n)\n\n => Array\n(\n => 165.22.122.199\n)\n\n => Array\n(\n => 23.106.56.12\n)\n\n => Array\n(\n => 203.122.10.146\n)\n\n => Array\n(\n => 37.111.136.138\n)\n\n => Array\n(\n => 103.87.193.66\n)\n\n)\n```\n3rd place? Not bad.: Keeping It Frugal\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > 3rd place? Not bad.", null, "", null, "", null, "April 5th, 2008 at 10:04 pm\n\nI found out today how I did in our church's 20 week fitness challenge. Out of the 39 remaining contenders, I placed third. The top 5 individuals were all women, which really surprised me.\n\nI'm not sure why they did this, but they gave cash prizes to the top 3 teams, instead of just the top team. But then they gave a cash prize just to the top individual, and gave *bath sets* to the 2nd and 3rd place people. Um . . not to complain or anything, but I think I'd have preferred whatever cash they spent on the bath set . . .\n\nI was a bit embarrassed though - two reasons. One is that as I was walking up front to accept my \"prize\" one of my shoes fell off! The other reason is that they had a graph/slide listing the top ten individual weight loss amounts with the names next to the amounts - showing to the whole church!!\n\nI am dissapointed that I didn't win the \\$200 first place money. But I do know that I had more than one occassion of slacking off during the 20 weeks, so coming in third is still doing pretty well.\n\nNow I'm about to jet off to my babysitting job where I'll be coming home with cash in my pocket - not \\$200, but still more than I'll be arriving there with.", null, "### 10 Responses to “3rd place? Not bad.”\n\n1. compulsive debtor Says:\n\nGood for you, Laura! Now sell that bath set on eBay and you'll at least have a little bit more cash than you started out with.\n\n2. Ima saver Says:\n\nGood going!!\n\n3. frugaltexan75 Says:\n\nCD,\n\nThanks! Never thought about selling bath sets on ebay - this one is probably one of the nicer ones ... may have to do some research and see if it'd be worth it.", null, "Thanks Ima!\n\n4. debtfreeme Says:\n\ncongratulations! just think how far you came in 20 weeks!\n\n5. reginaastralis Says:\n\nCongrats!\n\nMy mom gets bath sets for Christmas all the time as an elementary teacher.\n\nNeedless to say, they are regifted quite often. I finally got around to asking her to request my favorite sent (the kids ask her for a list of things she likes) so now I get LOTS to bathsets.\n\n6. frugaltexan75 Says:\n\nThanks dfm!\n\nRegina - I didn't get very many bath sets as a teacher - what I got a lot of were coffee mugs and candles! That's nice that you get to benefit from your mom's students gifts.", null, "7. SicilyYoder Says:\n\nThe Shelbyville church is doing a challenge too- a vegan weight loss challenge. I heard about it at Women's Minisitries meeting, but I didn't make it to the retreat- i was out of town. I noticed you mentioned Sabbaths- are you an Adventist? I am.\n\n8. frugaltexan75 Says:\n\nSicily,\n\nYes, I am an SDA.", null, "My dad is a retired SDA minister, and both my mom and I taught in Adventist schools.\n\n9. SicilyYoder Says:\n\nCool- I just read your post about Southern- I always dreamed of going there, but ended up at a public university. Happy Sabbath!! I did a booksigning at the Welcome center- handed out alot of literature too. They had a Bluegrass band playing next too me.\n\n10. frugaltexan75 Says:\n\nI knew I was going to be going to Southern at the same time I knew I was going to be a teacher -- in the first grade.", null, "My mom attended Southern, and like me, took elementary ed as her major, and like me, ended up not graduating from Southern.\n\nShe ended up graduating (with her BA) from Columbia Union (near Washington D.C.) then getting her Master's from Southwestern (TX). I got my BS from Southwestern. My dad got his BA from Pacific Union (as did my brother and his wife) and went to Andrews University for his seminary degree. I actually took a few master's level classes at Pacific Union as well (when I was teaching out in CA), but never went any further than that.\n\nSo . . . my family is well versed in Adventist education.", null, "(Note: If you were logged in, we could automatically fill in these fields for you.)\n Name: * Email: Will not be published. Subscribe: Notify me of additional comments to this entry. URL: Verification: * Please spell out the number 4.  [ Why? ]\n\nvB Code: You can use these tags: [b] [i] [u] [url] [email]" ]

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[ "# joule\n\nsuomi-englanti sanakirja\n\n1. joule\n\n## joule englanniksi\n\n1. Joule\n\n1. In the System of Units, the unit of energy, work and heat; the work required to exert a force of one newton for a distance of one metre. Also equal to the energy of one watt of power for a duration of one second. Symbol: J\n\n2. joule (gloss)\n\n3. joule (unit of measure)\n\n4. (l)\n\n5. (l) (gloss)\n\n6. (syn)" ]

[ null ]

[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "RE: Random Trouble\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg39207] RE: [mg39178] Random Trouble\n• From: Rob Pratt <rpratt at email.unc.edu>\n• Date: Mon, 3 Feb 2003 01:10:55 -0500 (EST)\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```On Sun, 2 Feb 2003, Donald A. Darling wrote:\n\n> > I have a program which uses a random number in several places - the same\n> > number in a given run of the program. When I implement the program however\n> > the number changes in every new call to it. I've tried to overcome this\n> > using \"Which\", \"Hold\", \"Verbatim\", and others, all to no avail. Any help\n> > greatly appreciated.\n>\n> Try using = instead of :=.\n>\n> r = Random[];\n>\n> Then use r throughout the program. Using r := Random[] (or just Random[])\n> instead would cause a fresh random number to be sampled each time r appears.\n>\n> Rob Pratt\n> Department of Operations Research\n> The University of North Carolina at Chapel Hill\n>\n> rpratt at email.unc.edu\n>\n> http://www.unc.edu/~rpratt/\n>\n> Rob - This doesn't work for me. If I try\n>\n> F[n_]=Table[Random[Real,{-1,1}],{n}]\n>\n> I get a different sequence for F each time I use it.\n>\n> Don Darling\n\nThe code I suggested does work for a single random number.\n\nYour command to obtain and remember a random n-vector yields an error\nmessage. Try the code below. The first call to F assigns a random\n5-vector to F. Subsequent calls to F just look up the stored value.\n\nIn:= F[n_] := F[n] = Table[Random[Real, {-1, 1}], {n}]\n\nIn:= F\n\nOut= {0.601045, -0.776375, -0.269912, -0.172019, -0.589347}\n\nIn:= F\n\nOut= {0.601045, -0.776375, -0.269912, -0.172019, -0.589347}\n\nRob Pratt\nDepartment of Operations Research\nThe University of North Carolina at Chapel Hill\n\nrpratt at email.unc.edu\n\nhttp://www.unc.edu/~rpratt/\n\n```\n\n• Prev by Date: Re: Re: Writing a program to hunt for a prime between n^2 and (n+1)^2\n• Next by Date: Re: Re: Writing a program to hunt for a prime between n^2 and (n+1)^2\n• Previous by thread: RE: Random Trouble\n• Next by thread: Re: Random Trouble" ]

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[ "wavedecomp\nCompute the wavelet transform of the series.\nSyntax\nSeries View: series_name.wavelet(options)\nOptions\n\n transform=arg (default=“dwt”) Wavelet transform type: “dwt” (discrete wavelet transform – DWT), “modwt” (maximum overlap DWT – MODWT), “mra” (DWT multiresolution analysis – DWT MRA), or “momra” (MODWT MRA).Note that when performing DWT or MRA, if the series length is not dyadic, a dyadic fix may be set with the “fixlen=” option fixlen=arg (default=“mean”) Fix dyadic lengths in DWT and MRA transforms: “zeros” (pad remainder with zeros), “mean” (pad remainder with mean of series), “median” (pad remainder with median of series), “shorten” (cut series length to dyadic length preceding series length). maxscale=integer (default = max possible) Maximum scale for wavelet transform. The max possible is obtained as follows. Let", null, "denote the series length and decompose", null, "into its dyadic component and a remainder:", null, ",", null, ". The default maxscale", null, "is then set with the following rules:DWT: (1) if", null, "then", null, ", otherwise (2) if expanding the series,", null, "and (3) if contracting the series", null, ".MODWT:", null, ". filter=arg (default=“h”) Wavelet filter class: “h” (Haar), “d” (Daubechies), “la” (least asymmetric).If “filter=h” or “filter=la”, the filter length may be specified using “flen=”.Wavelet filter boundary conditions are specified using the “bound=” option flen=integer Wavelet filter excess length as an even number between 2 and 20.For use when “filter=d” (default= 4) or “filter=la” (default=8). bound=arg (default = “p”) Filter boundary handling: “p” (periodic), “r” (reflective). hidebound Wavelet filter coefficients affected by the boundary will not be highlighted in the output graphs. prompt Force the dialog to appear from within a program. p Print results.\nExamples\ndgp.wavelet(maxscale=7)\nThe line above will perform the discrete wavelet transform of the series DGP using a Haar wavelet filter and up to the seventh wavelet scale.\ndgp.wavelet(transform=modwt, maxscale=3, lter=D)\nThe line above will perform the maximum overlap discrete wavelet transform using a Daubechies wavelet filter of length 4 and up to the third wavelet scale.\ndgp.wavelet(transform=mra, maxscale=4, lter=la, en=10, xlen=zeros)\nThe line above will perform a DWT multi-resolution analysis of the series DGP using a least asymmetric wavelet filter of length 10 and up to the fourth wavelet scale. It will also fix the non-dyadic length of the series by padding with zeros.\ndgp.wavelet(transform=momra, maxscale=4, filter=d, flen=12, hidebound)\nThe line above will perform a MODWT multi-resolution analysis using a Daubechies wavelet filter of length 12 and up to the fourth wavelet scale. It will also turn off highlighting of wavelet coefficients on the boundary.\nCross-references\nSee “Wavelet Analysis” and “Wavelet Transforms” for discussion. See also “Wavelet Objects”." ]

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[ "# Solve System of Algebraic Equations\n\nThis topic shows you how to solve a system of equations symbolically using Symbolic Math Toolbox™. This toolbox offers both numeric and symbolic equation solvers. For a comparison of numeric and symbolic solvers, see Select Numeric or Symbolic Solver.\n\n### Handle the Output of `solve`\n\nSuppose you have the system\n\n`$\\begin{array}{l}{\\mathit{x}}^{2}{\\mathit{y}}^{2}=0\\\\ \\mathit{x}-\\frac{\\mathit{y}}{2}=\\alpha \\text{\\hspace{0.17em}},\\end{array}$`\n\nand you want to solve for $x$ and $y$. First, create the necessary symbolic objects.\n\n`syms x y a`\n\nThere are several ways to address the output of `solve`. One way is to use a two-output call. The call returns the following.\n\n`[solx,soly] = solve(x^2*y^2 == 0, x-y/2 == a)`\n```solx =  $\\left(\\begin{array}{c}a\\\\ 0\\end{array}\\right)$```\n```soly =  $\\left(\\begin{array}{c}0\\\\ -2 a\\end{array}\\right)$```\n\nModify the first equation to ${x}^{2}{y}^{2}=1$. The new system has more solutions. Four distinct solutions are produced.\n\n`[solx,soly] = solve(x^2*y^2 == 1, x-y/2 == a)`\n```solx =  $\\left(\\begin{array}{c}\\frac{a}{2}-\\frac{\\sqrt{{a}^{2}-2}}{2}\\\\ \\frac{a}{2}-\\frac{\\sqrt{{a}^{2}+2}}{2}\\\\ \\frac{a}{2}+\\frac{\\sqrt{{a}^{2}-2}}{2}\\\\ \\frac{a}{2}+\\frac{\\sqrt{{a}^{2}+2}}{2}\\end{array}\\right)$```\n```soly =  $\\left(\\begin{array}{c}-a-\\sqrt{{a}^{2}-2}\\\\ -a-\\sqrt{{a}^{2}+2}\\\\ \\sqrt{{a}^{2}-2}-a\\\\ \\sqrt{{a}^{2}+2}-a\\end{array}\\right)$```\n\nSince you did not specify the dependent variables, `solve` uses `symvar` to determine the variables.\n\nThis way of assigning output from `solve` is quite successful for “small” systems. For instance, if you have a 10-by-10 system of equations, typing the following is both awkward and time consuming.\n\n```[x1,x2,x3,x4,x5,x6,x7,x8,x9,x10] = solve(...) ```\n\nTo circumvent this difficulty, `solve` can return a structure whose fields are the solutions. For example, solve the system of equations `u^2 - v^2 = a^2`, `u + v = 1`, `a^2 - 2*a = 3`. The solver returns its results enclosed in a structure.\n\n```syms u v a S = solve(u^2 - v^2 == a^2, u + v == 1, a^2 - 2*a == 3)```\n```S = struct with fields: a: [2x1 sym] u: [2x1 sym] v: [2x1 sym] ```\n\nThe solutions for `a` reside in the “`a`-field” of `S`.\n\n`S.a`\n```ans =  $\\left(\\begin{array}{c}-1\\\\ 3\\end{array}\\right)$```\n\nSimilar comments apply to the solutions for `u` and `v`. The structure `S` can now be manipulated by the field and index to access a particular portion of the solution. For example, to examine the second solution, you can use the following statement to extract the second component of each field.\n\n`s2 = [S.a(2),S.u(2),S.v(2)]`\n`s2 = $\\left(\\begin{array}{ccc}3& 5& -4\\end{array}\\right)$`\n\nThe following statement creates the solution matrix `M` whose rows comprise the distinct solutions of the system.\n\n`M = [S.a,S.u,S.v]`\n```M =  $\\left(\\begin{array}{ccc}-1& 1& 0\\\\ 3& 5& -4\\end{array}\\right)$```\n\nClear `solx` and `soly` for further use.\n\n`clear solx soly`\n\n### Solve a Linear System of Equations\n\nLinear systems of equations can also be solved using matrix division. For example, solve this system.\n\n```clear u v x y syms u v x y eqns = [x + 2*y == u, 4*x + 5*y == v]; S = solve(eqns); sol = [S.x;S.y]```\n```sol =  $\\left(\\begin{array}{c}\\frac{2 v}{3}-\\frac{5 u}{3}\\\\ \\frac{4 u}{3}-\\frac{v}{3}\\end{array}\\right)$```\n```[A,b] = equationsToMatrix(eqns,x,y); z = A\\b```\n```z =  $\\left(\\begin{array}{c}\\frac{2 v}{3}-\\frac{5 u}{3}\\\\ \\frac{4 u}{3}-\\frac{v}{3}\\end{array}\\right)$```\n\nThus, `sol` and `z` produce the same solution, although the results are assigned to different variables.\n\n### Return the Full Solution of a System of Equations\n\n`solve` does not automatically return all solutions of an equation. To return all solutions along with the parameters in the solution and the conditions on the solution, set the `ReturnConditions` option to `true`.\n\nConsider the following system of equations:\n\n`$\\begin{array}{l}\\mathrm{sin}\\left(\\mathit{x}\\right)+\\mathrm{cos}\\left(\\mathit{y}\\right)=\\frac{4}{5}\\\\ \\mathrm{sin}\\left(\\mathit{x}\\right)\\text{\\hspace{0.17em}}\\mathrm{cos}\\left(\\mathit{y}\\right)=\\frac{1}{10}\\end{array}$`\n\nVisualize the system of equations using `fimplicit`. To set the x-axis and y-axis values in terms of `pi`, get the axes handles using `axes` in `a`. Create the symbolic array `S` of the values `-2*pi` to `2*pi` at intervals of `pi/2`. To set the ticks to `S`, use the `XTick` and `YTick` properties of `a`. To set the labels for the x-and y-axes, convert `S` to character vectors. Use `arrayfun` to apply `char` to every element of `S` to return `T`. Set the `XTickLabel` and `YTickLabel` properties of `a` to `T`.\n\n```syms x y eqn1 = sin(x)+cos(y) == 4/5; eqn2 = sin(x)*cos(y) == 1/10; a = axes; fimplicit(eqn1,[-2*pi 2*pi],'b'); hold on grid on fimplicit(eqn2,[-2*pi 2*pi],'m'); L = sym(-2*pi:pi/2:2*pi); a.XTick = double(L); a.YTick = double(L); M = arrayfun(@char, L, 'UniformOutput', false); a.XTickLabel = M; a.YTickLabel = M; title('Plot of System of Equations') legend('sin(x)+cos(y) == 4/5','sin(x)*cos(y) == 1/10',... 'Location','best','AutoUpdate','off')```", null, "The solutions lie at the intersection of the two plots. This shows the system has repeated, periodic solutions. To solve this system of equations for the full solution set, use `solve` and set the `ReturnConditions` option to `true`.\n\n`S = solve(eqn1,eqn2,'ReturnConditions',true)`\n```S = struct with fields: x: [2x1 sym] y: [2x1 sym] parameters: [z z1] conditions: [2x1 sym] ```\n\n`solve` returns a structure `S` with the fields `S.x` for the solution to `x`, `S.y` for the solution to `y`, `S.parameters` for the parameters in the solution, and `S.conditions` for the conditions on the solution. Elements of the same index in `S.x`, `S.y`, and `S.conditions` form a solution. Thus, `S.x(1)`, `S.y(1)`, and `S.conditions(1)` form one solution to the system of equations. The parameters in `S.parameters` can appear in all solutions.\n\nIndex into `S` to return the solutions, parameters, and conditions.\n\n`S.x`\n```ans =  $\\left(\\begin{array}{c}{z}_{1}\\\\ {z}_{1}\\end{array}\\right)$```\n`S.y`\n```ans =  $\\left(\\begin{array}{c}z\\\\ z\\end{array}\\right)$```\n`S.parameters`\n`ans = $\\left(\\begin{array}{cc}z& {z}_{1}\\end{array}\\right)$`\n`S.conditions`\n```ans =  ```\n\n### Solve a System of Equations Under Conditions\n\nTo solve the system of equations under conditions, specify the conditions in the input to `solve`.\n\nSolve the system of equations considered above for `x` and `y` in the interval `-2*pi` to `2*pi`. Overlay the solutions on the plot using `scatter`.\n\n```Srange = solve(eqn1, eqn2, -2*pi < x, x < 2*pi, -2*pi < y, y < 2*pi, 'ReturnConditions', true); scatter(Srange.x,Srange.y,'k')```", null, "### Work with Solutions, Parameters, and Conditions Returned by `solve`\n\nYou can use the solutions, parameters, and conditions returned by `solve` to find solutions within an interval or under additional conditions. This section has the same goal as the previous section, to solve the system of equations within a search range, but with a different approach. Instead of placing conditions directly, it shows how to work with the parameters and conditions returned by `solve`.\n\nFor the full solution `S` of the system of equations, find values of `x` and `y` in the interval `-2*pi` to `2*pi` by solving the solutions `S.x` and `S.y` for the parameters `S.parameters` within that interval under the condition `S.conditions`.\n\nBefore solving for `x` and `y` in the interval, assume the conditions in `S.conditions` using `assume` so that the solutions returned satisfy the condition. Assume the conditions for the first solution.\n\n`assume(S.conditions(1))`\n\nFind the parameters in `S.x` and `S.y`.\n\n`paramx = intersect(symvar(S.x),S.parameters)`\n`paramx = ${z}_{1}$`\n`paramy = intersect(symvar(S.y),S.parameters)`\n`paramy = $z$`\n\nSolve the first solution of `x` for the parameter `paramx`.\n\n`solparamx(1,:) = solve(S.x(1) > -2*pi, S.x(1) < 2*pi, paramx)`\n```solparamx =  $\\left(\\begin{array}{cccc}\\pi +\\mathrm{asin}\\left(\\frac{\\sqrt{6}}{10}-\\frac{2}{5}\\right)& \\mathrm{asin}\\left(\\frac{\\sqrt{6}}{10}-\\frac{2}{5}\\right)-\\pi & -\\mathrm{asin}\\left(\\frac{\\sqrt{6}}{10}-\\frac{2}{5}\\right)& -2 \\pi -\\mathrm{asin}\\left(\\frac{\\sqrt{6}}{10}-\\frac{2}{5}\\right)\\end{array}\\right)$```\n\nSimilarly, solve the first solution of `y` for `paramy`.\n\n`solparamy(1,:) = solve(S.y(1) > -2*pi, S.y(1) < 2*pi, paramy)`\n```solparamy =  $\\left(\\begin{array}{cccc}\\mathrm{acos}\\left(\\frac{\\sqrt{6}}{10}+\\frac{2}{5}\\right)& \\mathrm{acos}\\left(\\frac{\\sqrt{6}}{10}+\\frac{2}{5}\\right)-2 \\pi & -\\mathrm{acos}\\left(\\frac{\\sqrt{6}}{10}+\\frac{2}{5}\\right)& 2 \\pi -\\mathrm{acos}\\left(\\frac{\\sqrt{6}}{10}+\\frac{2}{5}\\right)\\end{array}\\right)$```\n\nClear the assumptions set by `S.conditions(1)` using `assume`. Call `asumptions` to check that the assumptions are cleared.\n\n```assume(S.parameters,'clear') assumptions```\n``` ans = Empty sym: 1-by-0 ```\n\nAssume the conditions for the second solution.\n\n`assume(S.conditions(2))`\n\nSolve the second solution to `x` and `y` for the parameters `paramx` and `paramy`.\n\n`solparamx(2,:) = solve(S.x(2) > -2*pi, S.x(2) < 2*pi, paramx)`\n```solparamx =  ```\n`solparamy(2,:) = solve(S.y(2) > -2*pi, S.y(2) < 2*pi, paramy)`\n```solparamy =  ```\n\nThe first rows of `paramx` and `paramy` form the first solution to the system of equations, and the second rows form the second solution.\n\nTo find the values of `x` and `y` for these values of `paramx` and `paramy`, use `subs` to substitute for `paramx` and `paramy` in `S.x` and `S.y`.\n\n```solx(1,:) = subs(S.x(1), paramx, solparamx(1,:)); solx(2,:) = subs(S.x(2), paramx, solparamx(2,:))```\n```solx =  ```\n```soly(1,:) = subs(S.y(1), paramy, solparamy(1,:)); soly(2,:) = subs(S.y(2), paramy, solparamy(2,:))```\n```soly =  ```\n\nNote that `solx` and `soly` are the two sets of solutions to `x` and to `y`. The full sets of solutions to the system of equations are the two sets of points formed by all possible combinations of the values in `solx` and `soly`.\n\nPlot these two sets of points using `scatter`. Overlay them on the plot of the equations. As expected, the solutions appear at the intersection of the plots of the two equations.\n\n```for i = 1:length(solx(1,:)) for j = 1:length(soly(1,:)) scatter(solx(1,i), soly(1,j), 'k') scatter(solx(2,i), soly(2,j), 'k') end end```", null, "### Convert Symbolic Results to Numeric Values\n\nSymbolic calculations provide exact accuracy, while numeric calculations are approximations. Despite this loss of accuracy, you might need to convert symbolic results to numeric approximations for use in numeric calculations. For a high-accuracy conversion, use variable-precision arithmetic provided by the `vpa` function. For standard accuracy and better performance, convert to double precision using `double`.\n\nUse `vpa` to convert the symbolic solutions `solx` and `soly` to numeric form.\n\n`vpa(solx)`\n```ans =  $\\left(\\begin{array}{cccc}2.9859135500977407388300518406219& -3.2972717570818457380952349259371& 0.15567910349205249963259154265761& -6.1275062036875339772926952239014\\\\ 0.70095651347102524787213653614929& 2.4406361401187679905905068471302& -5.5822287937085612290531502304097& -3.8425491670608184863347799194288\\end{array}\\right)$```\n`vpa(soly)`\n```ans =  $\\left(\\begin{array}{cccc}0.86983981332387137135918515549046& -5.4133454938557151055661016110685& -0.86983981332387137135918515549046& 5.4133454938557151055661016110685\\\\ 1.4151172233028441195987301489821& -4.8680680838767423573265566175769& -1.4151172233028441195987301489821& 4.8680680838767423573265566175769\\end{array}\\right)$```\n\n### Simplify Complicated Results and Improve Performance\n\nIf results look complicated, `solve` is stuck, or if you want to improve performance, see, Troubleshoot Equation Solutions from solve Function." ]

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[ "# How can I plot y-z plane slices in a 3D volume?\n\n조회 수: 125(최근 30일)\nNathan Lauer 2016년 5월 3일\n댓글: Josh 2020년 12월 11일\nI have a series of plots representing the vorticity field behind an aircraft wing, at various downstream distances. Each is a 2D plot, and I would like to display them in 3D, one behind the other, in order to get a full 3D sense of the vorticity field behind the wing. In my first attempt, I made a 3D meshgrid, set it to zero, and filled in 4 z-slices with 4 vorticity plots, and that worked. However, I could not get it to the viewpoint that I wanted. Here is the plot that resulted:", null, "This is rather confusing because in this plot, the wing is beneath the plot, and therefore is more likely to confuse the intended audience than do any good. Further, in attempting to change the viewing angle with Matlab's view command, I was not able to get to the desired viewing angle.\nSo then I made a second attempt, where instead of plotting z-slices I plot x-slices. This does indeed give me the desired viewing angle, as seen here (where the slices are just zeros):", null, "Here, the wing is to the left of the plot, and it's much easier for the audience to get a sense of the full 3D vorticity field. However, I am unable to get the vorticity plots to display. Here is my code:\nFullVectorField = zeros(200,14,13);\n[x,y,z] = meshgrid(1:1:14, 1:1:200, 1:1:13);\nFullVectorField(50,:,:) = vorticity70;\nFullVectorField(100,:,:) = vorticity80;\nFullVectorField(150,:,:) = vorticity90;\nFullVectorField(200,:,:) = vorticity100;\nzslice = [];\nxslice = [50,100,150,200];\nyslice = [];\nA = FullVectorField(50,:,:);\nfigure\nslice(x,y,z,FullVectorField,xslice,yslice,zslice);\npbaspect([3 1 1])\naxis([0, 200, 0, 14, 0, 13])\ncolormap jet\ncolorbar\nxlabel('x axis (cm)')\nylabel('y axis (cm)')\nzlabel('z axis (cm)')\nvorticity70, vorticity80, etc are the matrices containing the data points I would like to plot. Any help in getting the vorticity plots to display along the xslices? Thanks!\n\n댓글을 달려면 로그인하십시오.\n\n### 채택된 답변\n\nMike Garrity 2016년 5월 3일\n편집: Mike Garrity 2016년 5월 3일\nYou don't really need to build a full 3D array and then slice it. You can just place individual 2D slices in a 3D axes.\n[y,z] = meshgrid(linspace(0,10,40));\nfor off=50:50:200\nx = off + zeros(size(z));\n% My standin for your vorticity data\nc = cos((x+y)/5) .* cos((x+z)/5);\nsurf(x,y,z,c)\nhold on\nend\nhold off\nxlim([0 200])", null, "##### 댓글 수: 4표시숨기기 이전 댓글 수: 3\nJosh 2020년 12월 11일\n\n댓글을 달려면 로그인하십시오.\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]

[ null, "https://www.mathworks.com/matlabcentral/answers/uploaded_files/163339/image.jpeg", null, "https://www.mathworks.com/matlabcentral/answers/uploaded_files/163340/image.jpeg", null, "https://www.mathworks.com/matlabcentral/answers/uploaded_files/174237/image.png", null ]

[ "Journal of Operational Risk", null, "Various approximations of the total aggregate loss quantile function with application to operational risk\n\nRoss Griffiths and Walid Mnif\n\nNeed to know\n\n• We explain the mechanics of the empirical aggregate loss bootstrap distribution and develop an analytic approximation to its quantile function.\n• We present various other approximations of the empirical bootstrap quantile function based on techniques in the literature.\n• We study the impact of empirical moments and large losses in the context of the empirical bootstrap measure of operational risk capital.\n\nAbstract\n\nA compound Poisson distribution is the sum of independent and identically distributed random variables over a count variable that follows a Poisson distribution. Generally, this distribution is not tractable. However, it has many practical applications that require the estimation of the quantile function at a high percentile, eg, the 99.9th percentile. Without loss of generality, this paper focuses on the application to operational risk. We assume that the support of random variables is nonnegative, discrete and finite. We investigate the mechanics of the empirical aggregate loss bootstrap distribution and suggest different approximations of its quantile function. Furthermore, we study the impact of empirical moments and large losses on the empirical bootstrap capital at the 99.9% confidence level." ]

[ null, "https://www.risk.net/sites/risk/files/styles/print_logo/public/2018-09/print-logo.png", null ]

[ "# Python Language Loops While Loop\n\n## Example\n\nA `while` loop will cause the loop statements to be executed until the loop condition is falsey. The following code will execute the loop statements a total of 4 times.\n\n``````i = 0\nwhile i < 4:\n#loop statements\ni = i + 1\n``````\n\nWhile the above loop can easily be translated into a more elegant `for` loop, `while` loops are useful for checking if some condition has been met. The following loop will continue to execute until `myObject` is ready.\n\n``````myObject = anObject()\n``````\n\n`while` loops can also run without a condition by using numbers (complex or real) or `True`:\n\n``````import cmath\n\ncomplex_num = cmath.sqrt(-1)\nwhile complex_num: # You can also replace complex_num with any number, True or a value of any type\nprint(complex_num) # Prints 1j forever\n``````\n\nIf the condition is always true the while loop will run forever (infinite loop) if it is not terminated by a break or return statement or an exception.\n\n``````while True:\nprint \"Infinite loop\"\n# Infinite loop\n# Infinite loop\n# Infinite loop\n# ...\n``````", null, "PDF - Download Python Language for free" ]

[ null, "https://riptutorial.com/Images/icon-pdf-2.png", null ]

[ "# Preprint 50/2011\n\n## Stability of solutions to abstract evolution equations with delay\n\n### Alexander Ramm\n\nContact the author: Please use for correspondence this email.\nSubmission date: 03. Aug. 2011\nPages: 9\npublished in: Journal of mathematical analysis and applications, 396 (2012) 2, p. 523-527", null, "DOI number (of the published article): 10.1016/j.jmaa.2012.06.033\nBibtex\nMSC-Numbers: 34E05, 35R30, 74J25, 34G20, 34K20, 37L05\nKeywords and phrases: stability, evolution problems, abstract evolution equations, equations with delay\nAn equation", null, "= A(t)u + B(t)F(t,u(t - τ)), u(t) = v(t),-τ t 0 is considered, A(t) and B(t) are linear operators in a Hilbert space H,", null, "=", null, ", F : H H is a non-linear operator, τ > 0 is a constant. Under some assumption on A(t),B(t) and F(t,u) sufficient condittions are given for the solution u(t) to exist globally, i.e, for all t 0, to be globally bounded, and to tend to zero as t →∞." ]

[ null, "https://sfx.mpg.de/sfx_local/sfx.gif", null, "https://www.mis.mpg.de/fileadmin/preprint_img/2011/tex_1676a0x.png", null, "https://www.mis.mpg.de/fileadmin/preprint_img/2011/tex_1676a1x.png", null, "https://www.mis.mpg.de/fileadmin/preprint_img/2011/tex_1676a2x.png", null ]

[ "", null, "", null, "### Home > CC3 > Chapter Ch4 > Lesson 4.1.5 > Problem4-44\n\n4-44.\n\nComplete a table for the rule $y=3x−2$. 4-44 HW eTool (Desmos). Homework Help ✎\n\n$\\left. \\begin{array} { | c | c | c | c | c | c | } \\hline x & { - 2 } & { - 1 } & { 0 } & { 1 } & { 2 } \\\\ \\hline y & { - 8 } & { - 5 } & { - 2 } & { 1 } & { 4 } \\\\ \\hline \\end{array} \\right.$\n\n1. Draw a complete graph for this rule.\n\n2. Is $(−50,−152)$ a point on the graph? Explain how you know.\n\nTo check if the point is on the graph, substitute the given $x$- and $y$- coordinates into the rule $y=3x−2$.\n\nIf the values on each side are equal, then the point is on the graph.\n\nComplete the table in the eTool below to view the graph.\nClick the link at right for the full version of the eTool: CC3 4-44 HW eTool" ]

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", null ]

[ "# Properties\n\n Label 58800gj Number of curves $2$ Conductor $58800$ CM no Rank $1$ Graph", null, "# Related objects\n\nShow commands for: SageMath\nsage: E = EllipticCurve(\"gj1\")\n\nsage: E.isogeny_class()\n\n## Elliptic curves in class 58800gj\n\nsage: E.isogeny_class().curves\n\nLMFDB label Cremona label Weierstrass coefficients j-invariant Discriminant Torsion structure Modular degree Faltings height Optimality\n58800.e2 58800gj1 $$[0, -1, 0, -653, -9183]$$ $$-40960/27$$ $$-20329747200$$ $$[]$$ $$54432$$ $$0.67887$$ $$\\Gamma_0(N)$$-optimal\n58800.e1 58800gj2 $$[0, -1, 0, -59453, -5559903]$$ $$-30866268160/3$$ $$-2258860800$$ $$[]$$ $$163296$$ $$1.2282$$\n\n## Rank\n\nsage: E.rank()\n\nThe elliptic curves in class 58800gj have rank $$1$$.\n\n## Complex multiplication\n\nThe elliptic curves in class 58800gj do not have complex multiplication.\n\n## Modular form 58800.2.a.gj\n\nsage: E.q_eigenform(10)\n\n$$q - q^{3} + q^{9} - 6q^{11} + 5q^{13} - 6q^{17} + 5q^{19} + O(q^{20})$$", null, "## Isogeny matrix\n\nsage: E.isogeny_class().matrix()\n\nThe $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the Cremona numbering.\n\n$$\\left(\\begin{array}{rr} 1 & 3 \\\\ 3 & 1 \\end{array}\\right)$$\n\n## Isogeny graph\n\nsage: E.isogeny_graph().plot(edge_labels=True)\n\nThe vertices are labelled with Cremona labels.", null, "" ]

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vW8m7ZszfjW8b/0TGMFy5c3e4L%2Bpy/2%2BI1jXp9D332Z06xdZI/h5bkNDPSyCRMnZuOW7r1wbt6yNX/7jXvzw8e2Z9OWRzJz/g1Jkvf95tnZ%2BPAP8uCPftKtW3J01cbvb82ECRN7/HEZ%2BOwxDC4CEHpZa2trHnviyexo7953mn5m1W0580Oz8%2B7L5mb33mfzD1/483z3ew/nXb//qVz4m2fnv0z6jz065472jmx7cmdaW1t79HEZHOwxDC4tDbdKh17V3t6esWPH5k8/dkmunPn%2BZo/zS9244n/m6v/%2B5Wzbti2jRo1q9jj0M/YYBhdnAKGXjRo1Kh/4wAeydPWdOXToULPHeVEHDx7MstV3Zvr06V40eVH2GAYXAQh9YPbs2Xl0247cvb6t2aO8qLvXt%2BXRbTsye/bsl/9lyrLHMHi4BAx9oNFopHX8%2BIw8qiV333Rdv/p0YqPRyHmXz8tPDyQb29r61Wz0L/YYBg9nAKEPtLS0ZP6CBblnw%2BasvHtts8f5OSvu%2Bmbu2bA518yf70WTl2SPYfBwBhD60IwZM/L1O76WB1YuzZhRI5s9Trbv7MjbZszKey%2B6KCtWrGj2OAwQ9hgGPgEIfaijoyNnnH56znrzKVl94zVNPVPRaDQy5cr52fDIo3nwoYcycmTzX8gZGOwxDHwuAUMfGjlyZJYuW5avfXtdrv3SyqbOsmD5itzxnfVZdvPNXjTpFnsMA58AhD42derULFy4MAu%2BuCKfWXVbU2b4zKrbcu3ylbnuuusyZcqUpszAwGaPYWDzXcDQBHPnzs0zzzyTKxctytN79uSaj8zsk8tojUYjC5avyLXLV2bu3Ln51Kc%2B1evPyeBlj2HgEoDQBC0tLVm4cGFGjBiRefPmZfOWH2bpVZf16hvqt%2B/syKwbFueO76zPokWLvGjyqtljGLh8CASabM2aNbn04x/Pvr17c9MVl2bm5Hf16FmURqORFXd9M5d/dlmGDhuWZTff7HIZPc4ew8AiAKEf6OjoyJw5c7Jq1aqcO3FcPvnb0zL5303IEUe88rfpHjx4MHevb8tffuWruWfD5syYMSOLFy/OCSec0IOTwwvsMQwcAhD6kTVr1mTB/PnZfN99OXnsmFw69YJcMnlSty6p7WjvyJfvWptlq%2B/Mo9t2ZNyZZ2b%2BggW56KKLenFyeIE9hv5PAEI/02g0sn79%2BixZsiS33HJLOjs7c9LrTkzraW/K%2BNNOzdtOeWOOGT4sQ448Mp0HDuSZPXvzwI9%2BnE1btmbj97dm25M7M3To0EyfPj2zZ8/OWWed5ZsR6HP2GPo3AQj9WHt7e%2B69995s3LgxbW0b09bWll27nvqF3zv%2B%2BOPS2tqaCRMmprW1Neecc05GjRrVhInhF9lj6H8EIAwgjUYjO3fuzN69e7Nv374MHTo0w4YNy%2BjRo50dYcCwx9B8AhAAoBjfBAIAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADF/H%2BAORPNfpuyWgAAAABJRU5ErkJggg%3D%3D", null, "https://www.lmfdb.org/static/images/t2r.png", null, 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vW8m7ZszfjW8b/0TGMFy5c3e4L%2Bpy/2%2BI1jXp9D332Z06xdZI/h5bkNDPSyCRMnZuOW7r1wbt6yNX/7jXvzw8e2Z9OWRzJz/g1Jkvf95tnZ%2BPAP8uCPftKtW3J01cbvb82ECRN7/HEZ%2BOwxDC4CEHpZa2trHnviyexo7953mn5m1W0580Oz8%2B7L5mb33mfzD1/483z3ew/nXb//qVz4m2fnv0z6jz065472jmx7cmdaW1t79HEZHOwxDC4tDbdKh17V3t6esWPH5k8/dkmunPn%2BZo/zS9244n/m6v/%2B5Wzbti2jRo1q9jj0M/YYBhdnAKGXjRo1Kh/4wAeydPWdOXToULPHeVEHDx7MstV3Zvr06V40eVH2GAYXAQh9YPbs2Xl0247cvb6t2aO8qLvXt%2BXRbTsye/bsl/9lyrLHMHi4BAx9oNFopHX8%2BIw8qiV333Rdv/p0YqPRyHmXz8tPDyQb29r61Wz0L/YYBg9nAKEPtLS0ZP6CBblnw%2BasvHtts8f5OSvu%2Bmbu2bA518yf70WTl2SPYfBwBhD60IwZM/L1O76WB1YuzZhRI5s9Trbv7MjbZszKey%2B6KCtWrGj2OAwQ9hgGPgEIfaijoyNnnH56znrzKVl94zVNPVPRaDQy5cr52fDIo3nwoYcycmTzX8gZGOwxDHwuAUMfGjlyZJYuW5avfXtdrv3SyqbOsmD5itzxnfVZdvPNXjTpFnsMA58AhD42derULFy4MAu%2BuCKfWXVbU2b4zKrbcu3ylbnuuusyZcqUpszAwGaPYWDzXcDQBHPnzs0zzzyTKxctytN79uSaj8zsk8tojUYjC5avyLXLV2bu3Ln51Kc%2B1evPyeBlj2HgEoDQBC0tLVm4cGFGjBiRefPmZfOWH2bpVZf16hvqt%2B/syKwbFueO76zPokWLvGjyqtljGLh8CASabM2aNbn04x/Pvr17c9MVl2bm5Hf16FmURqORFXd9M5d/dlmGDhuWZTff7HIZPc4ew8AiAKEf6OjoyJw5c7Jq1aqcO3FcPvnb0zL5303IEUe88rfpHjx4MHevb8tffuWruWfD5syYMSOLFy/OCSec0IOTwwvsMQwcAhD6kTVr1mTB/PnZfN99OXnsmFw69YJcMnlSty6p7WjvyJfvWptlq%2B/Mo9t2ZNyZZ2b%2BggW56KKLenFyeIE9hv5PAEI/02g0sn79%2BixZsiS33HJLOjs7c9LrTkzraW/K%2BNNOzdtOeWOOGT4sQ448Mp0HDuSZPXvzwI9%2BnE1btmbj97dm25M7M3To0EyfPj2zZ8/OWWed5ZsR6HP2GPo3AQj9WHt7e%2B69995s3LgxbW0b09bWll27nvqF3zv%2B%2BOPS2tqaCRMmprW1Neecc05GjRrVhInhF9lj6H8EIAwgjUYjO3fuzN69e7Nv374MHTo0w4YNy%2BjRo50dYcCwx9B8AhAAoBjfBAIAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADFCEAAgGIEIABAMQIQAKAYAQgAUIwABAAoRgACABQjAAEAihGAAADF/H%2BAORPNfpuyWgAAAABJRU5ErkJggg%3D%3D", null ]

[ "# ERP visualization: Three conditions\n\nIn an earlier post I took a look at visualizing ERPs from two conditions at a single electrode. This time I’m going to look at three conditions. As in the previous post, I’ll assume a basic familiarity with ERPs.\n\nFirst I’ll load in the full dataset, which contains ERPs for all conditions for all subjects, and whip it into shape.\n\n``````library(ggplot2)\nlibrary(tidyverse)\nlibrary(afex)\nlibrary(Rmisc)\nlibrary(magrittr)\n\ncol_names = c(\"Object.BB\",\"Object.HSF\",\"Object.LSF\",\"Non-Object.LSF\",\"Non-Object.HSF\",\"Non-Object.BB\",\"Time\",\"Subject\")\n)\nlevCatGAall <- levCatGAall[c(1,2,3,6,5,4,7,8)]\n\nlevCatGAall <- levCatGAall %>%\nfilter(Time >= -100 & Time <= 400) %>%\ngather(condition,amplitude,-Subject,-Time) %>%\nseparate(condition,c(\"Object\",\"Frequency\"),sep= \"[.]\",extra=\"merge\")``````\n\n## The Plots\n\nLet’s start off with a simple ERP plot with both within- and between-subject 95% confidence intervals (see previous post). Although the dataset is for a 2 x 3 design (Object X Spatial Frequency - for further details, check our article Early and late effects of objecthood and spatial frequency on event-related potentials and gamma band activity) - I’ll be focussing entirely on the Spatial Frequency factor here. Frequency has three levels: high (HSF), broadband (BB), and low (LSF).\n\n``````# basic plot setup\nERP.plot <- ggplot(levCatGAall,aes(Time, amplitude))+\nscale_color_brewer(palette = \"Dark2\") +\nscale_fill_brewer(palette = \"Dark2\") +\ntheme_minimal()\n\n## Calculate running within-subject CIs\nrunningCIs <- levCatGAall %>%\ngroup_by(Frequency, Time, Subject) %>%\ndplyr::summarise(amplitude = mean(amplitude)) %>%\nsplit(.\\$Time) %>%\nmap(~summarySEwithin(data = .,\nmeasurevar = \"amplitude\",\nwithinvars = c(\"Frequency\"),\nidvar = \"Subject\"))\n\nWSCI <- map_df(runningCIs,magrittr::extract) %>%\nmutate(\nTime = rep(unique(levCatGAall\\$Time),each =3) #Note, you'll have to change 3 to match the number of conditions\n)\n\nERP.plot+\ngeom_ribbon(data = WSCI,\naes(ymin = amplitude-ci,\nymax = amplitude+ci,\nfill = Frequency,\ncolour = Frequency),\nlinetype=\"dashed\",\nalpha = 0.3) +\nguides(fill = \"none\")+\nstat_summary(fun.data = mean_cl_normal,\ngeom = \"ribbon\",\nsize = 1,\naes(colour = Frequency),\nlinetype = \"dotted\",\nfill = NA,\nalpha = 0.8)+\nstat_summary(fun.y = mean,\ngeom = \"line\",\nsize = 1,\naes(colour = Frequency),\nalpha = 0.8)+\nlabs(x = \"Time (ms)\",\ny = expression(paste(\"Amplitude (\",mu,\"V)\")),\ncolour = \"\")+\ngeom_vline(xintercept = 0,linetype = \"dashed\" )+\ngeom_hline(yintercept = 0,linetype = \"dashed\")``````", null, "Looks like there are consistently higher amplitudes for the BB images from the P1 onwards until after the P2. The within-subject CIs definitely help here - between-subject CIs would have a lot of overlap, as you can probably see. Having both types of CIs on this plot is making it a little messy - I wouldn’t normally do this. We’ll plow on adding individual subject data.\n\n``````ERP.plot +\nstat_summary(fun.y = mean,geom = \"line\",alpha = 0.4,aes(group =interaction(Subject,Frequency),colour = Frequency),size = 0.7) +\nguides(alpha= \"none\") +\ngeom_ribbon(data = WSCI,\naes(ymin = amplitude-ci,\nymax = amplitude+ci,\nfill = Frequency,\ncolour = Frequency),\nlinetype=\"dashed\",\nalpha = 0.3) +\nstat_summary(fun.y = mean,geom = \"line\",size = 1,aes(colour = Frequency),alpha = 0.8) +\nlabs(x = \"Time (ms)\",y = expression(paste(\"Amplitude (\",mu,\"V)\")),colour = \"\") +\nguides(colour= \"none\") +\ngeom_vline(xintercept = 0,linetype = \"dashed\") +\ngeom_hline(yintercept = 0,linetype = \"dashed\")``````", null, "Now it’s becoming a real mess, which can can only get worse when you add more conditions. Let’s split the conditions up.\n\n``````ERP.plot+\nfacet_wrap(~Frequency)+\nstat_summary(fun.y = mean,geom = \"line\",aes(group = Subject),alpha = 0.3)+\ngeom_ribbon(data = WSCI,\naes(ymin = amplitude-ci,\nymax = amplitude+ci,\nfill = Frequency,\ncolour = Frequency),\nlinetype=\"dashed\",\nalpha = 0.3) +\nguides(fill= \"none\")+\nstat_summary(fun.y = mean,geom = \"line\",size = 1)+\nlabs(x = \"Time (ms)\",y = expression(paste(\"Amplitude (\",mu,\"V)\")),colour = \"\")+\ngeom_vline(xintercept = 0,linetype = \"dashed\" )+\ngeom_hline(yintercept = 0,linetype = \"dashed\")``````", null, "Now we have three subplots, each informative about a particular condition. The mental gymnastics to work out where the differences are are becoming harder, and I find myself relying on the condition means. Couple of points of note - you have a couple of people who look a bit odd in the 160-200 ms time window - where most people are showing negative-going deflections, they aren’t. After 200 ms or so, the data in general seems much more variable. I can tell that from the confidence intervals anyway, and it doesn’t seem to differ much across conditions.\n\nSince there’s no way to tell which line belongs to which participant, it’s impossible to know if it’s the same participants showing the same patterns across conditions (although I’d say it’s very likely in the early parts of the ERP at least). As usual, what I’m really interested in here are the within-subject differences across conditions.\n\n## Difference waves\n\nIn the two condition post, life was simple. Two conditions only need a single difference wave. But as the number of conditions increases, the number of pairwise differences increases. For three conditions, you’d need three difference waves (here: BB-HSF, BB-LSF, HSF-LSF). For four conditions, you’d need six difference waves. But of course, you’d start off with an F-test to test whether any of the means differ from each other, and, if they do, we’d then run post-hoc t-tests to check which pairs of means differ. A simple starting point is thus the difference between each condition mean and the grand mean, the mean across all conditions.\n\nAs described in a previous post on running t-tests, we can also add the results of a running ANOVA to our plot. We’ll use afex to run each ANOVA and purrr to run them across each timepoint and collect the results. We’ll stick with uncorrected p values for now, but of course they can be corrected as required. Note that you have to be careful to make sure that you choose the p-values from the right term in the ANOVA. With a one-way ANOVA this is not so tough, but here I’m actually running a 2 by 3 ANOVA, so there are actually three terms (Object, Frequency, and Object by Frequency). For the purposes of this post I’m focussing solely on Frequency. I’ve also included 95% within-subject confidence intervals.\n\n``````# Run the ANOVA on each timepoint\nFtests <- levCatGAall %>%\nsplit(.\\$Time) %>%\nmap(~ aov_ez(\"Subject\",\"amplitude\",.,within = c(\"Object\",\"Frequency\")))\n\n#Extract the p-values and correct them as desired.\nlevCatGAall\\$crit <- 0+(levCatGAall\\$pval <= .05)\nlevCatGAall\\$crit[levCatGAall\\$crit == 0] <- NA\n\n#calculate the grand average ERP across all subjects and conditions\n\n#re-doing the CIs. Note it's not necessary to recalculate them, as they're the same - this is just an easy way for me to get them in the right format. It's not very efficient so feel free to improve it ;)\nrunningCIs <- levCatGAall %>%\ngroup_by(Frequency,Time,Subject)%>%\nsplit(.\\$Time) %>%\nmap(~summarySEwithin(data = .,\nmeasurevar = \"amplitude\",\nwithinvars = c(\"Frequency\"),\nidvar = \"Subject\"))\n\nWSCI <- map_df(runningCIs,magrittr::extract) %>%\nmutate(\nTime = rep(unique(levCatGAall\\$Time),each =3) #Note, you'll have to change 3 to match the number of conditions\n)\n\nERPdiff.plot <- ggplot(levCatGAall,aes(Time,amplitude))+\nscale_color_brewer(palette = \"Dark2\")+\nscale_fill_brewer(palette = \"Dark2\")+\ntheme_minimal()\n\nERPdiff.plot+\nguides(fill = \"none\")+\nlabs(x = \"Time (ms)\",\ny = expression(paste(\"Amplitude (\",mu,\"V)\")),\ncolour = \"\")+\ngeom_ribbon(data = WSCI,\naes(ymin = amplitude-ci,\nymax = amplitude+ci,\nfill = Frequency,\ncolour = Frequency),\nlinetype=\"dashed\",\nalpha = 0.3) +\nstat_summary(fun.y = mean,\ngeom = \"line\",\nsize = 1,\ncolour = Frequency)\n)+\ngeom_line(aes(x = Time,\ny = crit-3),\nna.rm = TRUE,\nsize = 2)+\ngeom_vline(xintercept = 0,linetype = \"dashed\" )+\ngeom_hline(yintercept = 0,linetype = \"dashed\")``````", null, "This shows that there is a significant effect of frequency around 40 ms, from around 80 to 360 ms, and from around 370 - 375 ms. Of course, we can’t tell from an F-test alone which means are significantly different from each other at any given time point. But it’s pretty clear that BB images elicit more positive ERPs than HSF and LSF images from around 80 - 300 ms. The CIs also help here - by around 300ms the HSF and BB image ERPs are converging, but LSF is staying more negative. Really we need to do post-hoc tests - I’ll try that out in a follow-up post.\n\nIn theory, we could also add lines for each person’s difference from the grand mean…\n\n``````ERPdiff.plot+\nguides(fill = \"none\")+\nlabs(x = \"Time (ms)\", y = expression(paste(\"Amplitude (\",mu,\"V)\")),colour = \"\")+\nstat_summary(fun.y = mean,geom = \"line\",size = 1,aes(y = GAdiff,group = interaction(Subject,Frequency),colour = Frequency),alpha = 0.4)+\ngeom_ribbon(data = WSCI,\naes(ymin = amplitude-ci,\nymax = amplitude+ci,\nfill = Frequency,\ncolour = Frequency),\nlinetype=\"dashed\",\nalpha = 0.3)+\nstat_summary(fun.y = mean,geom = \"line\",size = 2,aes(y = GAdiff,colour = Frequency))+\ngeom_vline(xintercept = 0,linetype = \"dashed\" )+\ngeom_hline(yintercept = 0,linetype = \"dashed\")``````", null, "…but if you can make sense of that, you have better eyesight than I have. Splitting them up by condition helps, but still leaves you with having to perform mental gymnastics to compare across conditions. But coupled with one of the first difference plot, that in itself might not be a huge issue.\n\n``````ERPdiff.plot+\nguides(fill = \"none\")+\nlabs(x = \"Time (ms)\", y = expression(paste(\"Amplitude (\",mu,\"V)\")),colour = \"\")+\nstat_summary(fun.y = mean,geom = \"line\",size = 1,aes(y = GAdiff,group = Subject),alpha = 0.3)+\nstat_summary(fun.y = mean,geom = \"line\",size = 2,aes(y = GAdiff))+\ngeom_vline(xintercept = 0,linetype = \"dashed\" )+\ngeom_hline(yintercept = 0,linetype = \"dashed\")+\nfacet_wrap(~Frequency)``````", null, "So at a first pass, subtracting the grand average ERP from the condition ERPs seems like a plausible way to show overall differences across three conditions. I’ll post soon on doing pairwise comparisons between those three conditions.", null, "" ]

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[ "The square of a number is the number multiplied by itself. 52 = 5 * 5 = 25 In computer applications, squaring and other raising to a power is usually indicated by a ^ symbol. 5^2 = 5 * 5 = 25 The square root of a number is a number such that if it were squared it would return the original number. Technically, all positive real numbers have two square roots, a positive one and a negative one. sqrt(25) = +5 and -5 However, in most applications, the negative root is ignored. In statistics, the square root of 25 is 5. Also remember that negative numbers do not have square roots (in the real number system). sqrt(-25) is undefined or nonexistent or NaN (not a number) The square root of 25 can also be written 251/2. Other roots generalize from that notation. The cube root of 25 is 251/3, and so on. Return To Main Page" ]

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[ "## μ-Dots and Totient Sandwiches\n\n(Eingestellt am 16. Juni 2022, 20:22 Uhr von Hrafnsvaengr)\n\nWhat started out as a very silly idea turned into this, a very niche but fun little puzzle based on two functions from number theory. It shouldn't be too difficult, although the rules may seem more complicated at first than they actually are.", null, "A link for the puzzle on f-puzzles can be found here and on CtC here.\n\n• Normal sudoku rules apply.\n\n• μ-Dots:\n• If the product of two orthogonally adjacent digits is the product of an even number of distinct primes, there is a blue dot between them.\n• If the product of two orthogonally adjacent digits is the product of an odd number of distinct primes, there is an orange dot between them.\n• If the product of two orthogonally adjacent digits is the product of a perfect square (other than 1) and any other number, there is no dot between them.\n\n• Totient Sandwiches:\n• The clues around the edge of the grid are totient sandwiches.\n• Let N be the sum of the digits that appear between the 1 and 9 in a given row or column.\n• The clue adjacent to that row or column is equal to the number of positive integers, up to N, which share no factors with N.\n\nAlternatively, the rules can be simplified as so, if you are familiar with the Möbius function, μ(n), and the Euler totient function, φ(n).\n\n• Normal sudoku rules apply.\n• Let a,b be two orthogonally adjacent cells. If there is a blue dot between them, μ(ab)=1; if there is an orange dot between them, μ(ab)=-1; if there is no dot, μ(ab)=0.\n• Let si be the digits appearing between the 1 and 9 in a row or column. A clue outside the grid is equal to φ(Σ si) in the corresponding row or column.\n\nLösungscode: Row 4 and then column 3, without commas or spaces.\n\nZuletzt geändert am 16. Juni 2022, 20:23 Uhr\n\nGelöst von Arashdeep Singh, kublai, RJBlarmo, baris, SKORP17, Xean\nKomplette Liste\n\n### Kommentare\n\nam 17. Juni 2022, 00:54 Uhr von RJBlarmo\nFun puzzle, nothing too difficult once you really consider how the Mobius function works. Needed to have a totient function table up though :)\n\nam 16. Juni 2022, 22:13 Uhr von kublai\nWOW! Once I got the mu and totient constraints sorted it wasn't so bad, but that part was tricky enough to bump it up a little on difficulty.\n\nam 16. Juni 2022, 21:23 Uhr von Arashdeep Singh\nNice puzzle. Learned the functions through it. Thanks :)\n\nam 16. Juni 2022, 20:23 Uhr von Hrafnsvaengr\nTypo correction\n\n Schwierigkeit:", null, "Bewertung: N/A Gelöst: 6 mal Beobachtet: 3 mal ID: 000A97\n\nLösungscode:\n\n## Anmelden", null, "" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Boundary problem with parameter using NDSolve\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg93355] Re: Boundary problem with parameter using NDSolve\n• From: dh <dh at metrohm.ch>\n• Date: Wed, 5 Nov 2008 04:51:25 -0500 (EST)\n• References: <gemjh2\\$4s0\\[email protected]>\n\n```\nHi Mikhail,\n\nif you expand the wavefunction into base-functions, your problem becomes\n\nan Eigenvalue problem and can be solved directly.\n\nIf you integrate the DE numerically you will have to search the enery\n\n(shooting method).\n\nhope this helps, Daniel\n\nMikhail Lemeshko wrote:\n\n> Dear friends, I hope you can help me with the following problem.\n\n>\n\n> I try to solve the Schr=F6dinger equation for some arbitrary potential\n\n> V(x):\n\n>\n\n> -f''[x] + V[x]*f[x] == ee*f[x]\n\n>\n\n> Considering discrete energy levels (ee<0), there are also boundary\n\n> conditions f==f[some_very_big_x]==0, which are satisfied only at\n\n> some values of the parameter ee. I wish to find the values of energy\n\n> ee and the corresponding wavefunctions.\n\n>\n\n> Is there an easy method to do it with Mathematica? I mean, to write a\n\n> code \"here are the equations with boundary conditions, please find all\n\n> values of ee in the interval from E1 to E2, where the boundary\n\n> conditions are satisfied, and return me the list of them\"?\n\n>\n\n> Or, I have to implement a cycle, which will increase the energy step-\n\n> by-step and check whether the boundary conditions are satisfied?\n\n>\n\n>\n\n> Thank you a lot in advance,\n\n> Mikhail.\n\n>\n\n--\n\nDaniel Huber\n\nMetrohm Ltd.\n\nOberdorfstr. 68\n\nCH-9100 Herisau\n\nTel. +41 71 353 8585, Fax +41 71 353 8907\n\nE-Mail:<mailto:dh at metrohm.com>\n\nInternet:<http://www.metrohm.com>\n\n```\n\n• Prev by Date: Re: Constructing a Label\n• Next by Date: Re: Mathematica Help Urgent (Very New Beginner)\n• Previous by thread: Boundary problem with parameter using NDSolve\n• Next by thread: Re: Mathematica 6.0: How to collect data with Manipulate?" ]

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[ "# Ex.5.1 Q3 Lines and Angles - NCERT Maths Class 7\n\nGo back to  'Ex.5.1'\n\n## Question\n\nIdentify which of the following pairs of angles are complementary and which are supplementary:\n\n(i) $$65^\\circ, 115^\\circ$$\n\n(ii) $$63^\\circ, 27^\\circ$$\n\n(iii) $$112^\\circ, 68 ^\\circ$$\n\n(iv) $$130^\\circ, 50^\\circ$$\n\n(v) $$45^\\circ, 45^\\circ$$\n\n(vi) $$80^\\circ, 10^\\circ$$\n\nNOTE: The sum of the measure of complementary angle is $$90^\\circ$$ and that of supplementary angle is $$180^\\circ$$\n\nVideo Solution\nLines & Angles\nEx 5.1 | Question 3\n\n## Text Solution\n\nReasoning:\n\nFind out the sum of two given angles,and then check whether it is $$180^\\circ$$ or $$90^\\circ.$$ If  the sum of two angles is either equal to $$90^\\circ$$, the angles are complementary and if the sum of the two angles is $$180^\\circ$$, the angles are complementary.\n\nSteps:\n\nSolve for supplementary angle or complementary angle:\n\n(i) $$65^\\circ, 115^\\circ$$\n\nSum of measure of these two angles $$= 65^\\circ + 115^\\circ = 180^\\circ$$\n\nTherefore, these two angles are supplementary.\n\n(ii) $$63^\\circ, 27^\\circ$$\n\nSum of measure of these two angles $$= 63^\\circ + 27^\\circ = 90^\\circ$$\n\nTherefore, these two angles are complementary.\n\n(iii) $$112^\\circ, 68^\\circ$$\n\nSum of measure of these two angles $$= 112^\\circ + 68^\\circ = 180^\\circ$$\n\nTherefore, these two angles are supplementary.\n\n(iv) $$130^\\circ, 50^\\circ$$\n\nSum of measure of these two angles $$= 130^\\circ + 50^\\circ = 180^\\circ$$\n\nTherefore, these two angles are supplementary.\n\n(v) $$45^\\circ, 45^\\circ$$\n\nSum of measure of these two angles $$= 45^\\circ + 45^\\circ = 90^\\circ$$\n\nTherefore, these two angles are complementary.\n\n(vi) $$80^\\circ, 10^\\circ$$\n\nSum of measure of these two angles $$= 80^\\circ + 10^\\circ = 90^\\circ$$\n\nTherefore, these two angles are complementary.\n\nLearn from the best math teachers and top your exams\n\n• Live one on one classroom and doubt clearing\n• Practice worksheets in and after class for conceptual clarity\n• Personalized curriculum to keep up with school" ]

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[ "# VBScript array_map function\n\nA VBScript equivalent of PHP’s array_map\n\n``````\nFunction array_map(callback, arr)\n\nDim key\nDim tmp_ar\n\nIf isArray( arr ) Then\n\nIf Len( callback ) = 0 Then\narray_map = arr\nExit Function\nEnd If\n\nReDim tmp_ar( uBound(arr) )\nFor key = 0 to uBound( arr )\nIf isObject( arr(key) ) Then\nexecute(\"set tmp_ar(key) = \" & callback & \"(arr(key))\")\nElse\nexecute(\"tmp_ar(key) = \" & callback & \"(arr(key))\")\nEnd If\nNext\n\narray_map = tmp_ar\n\nElseIf isObject( arr ) Then\n\nIf Len( callback ) = 0 Then\nset array_map = arr\nExit Function\nEnd If\n\nDim return_val\n\nset tmp_ar = Server.CreateObject(\"Scripting.Dictionary\")\nFor Each key In arr\nreturn_val = \"\nIf isObject( arr.Item(key) ) Then\nexecute(\"set return_val = \" & callback & \"(arr.Item(key))\")\nElse\nexecute(\"return_val = \" & callback & \"(arr.Item(key))\")\nEnd If" ]

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[ "# fixed point in regular expressions\n\nI've posted this question first on StackOverflow but this section seems more suited for this kind of questions. Also I'm not trying to simply solve this exercise (it is a \"parsing\" exercise, once I'll figure out the regular expression I'll derive the equivalent Finite Automaton that accepts the language generated by the grammar), but instead I'm trying to understand the methodology to derive regular expressions (if they exists, from context-free grammars)\n\nI have this productions of a context free grammar (axiom S, e is the empty word)\n\nS->AS|b|A A->abA|Ab|e I have to figure out a regular expression (if exists) that generates a language equivalent to the one generated by that grammar.\n\nSo far i wrote that\n\nL(S)=L(A)L(S) + b + L(A) = L(A)*(L(A) + b) (by Arden Rule)\nL(A)=(ab)*(L(A)b+e)\n\n\nI've tried the method of finding the fixed point (I'd like more info about that since seems I can't get the hang of it\n\nfor A: 0 -> e -> (ab)*(b+e) -> (ab)*[(ab)*(b+e)](b+e)\n\n\nshould i check if (ab)(b+e) = (ab)(ab)*(b+e) (if that's is the fixed point? Or should i go ahead?) It's not the first exercise I'm having trouble with so any help would be appreaciated, Thanks.\n\n• Also posted on Stack Overflow. Please do not post the same question on multiple sites. Each community should have an honest shot at answering without anybody's time being wasted. If you don't get a satisfying answer after a week or so, feel free to flag for migration. – D.W. Jul 6 '15 at 3:01\n• I've deleded the post in StackOverflow, this section seems more appropriate – Crysis85 Jul 6 '15 at 9:35\n\nAs all productions in $A\\rightarrow \\mathrm{ab}A\\text{ }|\\text{ }A\\mathrm{b}\\text{ }|\\text{ }ε$ are linear, and (crucially) there is only one non-terminal symbol, all derivation sequences can be reordered into equivalent ones where left-side derivations always precede (or always succeed) right-side ones.\n\nFor instance:\n\n$wAv \\Rightarrow w\\mathrm{ab}Av \\Rightarrow w\\mathrm{ab}A\\mathrm{b}v \\Rightarrow w\\mathrm{abab}A\\mathrm{b}v \\Rightarrow w\\mathrm{abab}A\\mathrm{bb}v\\equiv \\\\wAv \\Rightarrow wA\\mathrm{b}v \\Rightarrow wA\\mathrm{bb}v \\Rightarrow w\\mathrm{ab}A\\mathrm{bb}v \\Rightarrow w\\mathrm{abab}A\\mathrm{bb}v$\n\n(This is a hint of what an inductive proof using fixed-points would look like in this particular case. Check this if you are determined to go down that road).\n\nThis means that this grammar can be converted into a regular grammar in two steps, by grouping left-linear and right-linear productions:\n\n1.\n\n\n$A \\rightarrow A'A''\\\\A' \\rightarrow \\mathrm{ab}A'\\text{ }|\\text{ }ε$\n$A'' \\rightarrow A''\\mathrm{b}\\text{ }|\\text{ }ε$\n\n2.\n\n\n$A \\rightarrow A'A''\\\\A' \\rightarrow \\mathrm{ab}A'\\text{ }|\\text{ }ε$\n$A'' \\rightarrow \\mathrm{b}A''\\text{ }|\\text{ }ε\\\\$\n\nSo $\\mathcal{L}(A) = (\\mathrm{ab})^*\\mathrm{b}^*$" ]

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[ "# Arduino Tutorial 13: Understanding Arduino If Statements\n\nIn many real world applications, you want your programs to branch into different directions depending on values that you are reading from sensors. For such cases, you need to us IF statements. IF statements allow you to do different things depending on conditions. This lesson shows you how to use IF statements, and the different conditions that can be used with these commands. If you want to follow along at home, you can order the Arduino Kit we are using HERE.\n\n# Arduino Tutorial 4: Understanding Arduino Variables\n\nIn our earlier lessons we would program using “constants” when we needed numbers. For example, if we wanted to set pin 13 to an output, we would use the command:\n\npinMode(13, OUTPUT);\n\nThe problem with using constants like the number 13, if you decided later to use pin 8 instead, you would have to edit every line of code that used that number. It is much better practice to use variables when coding. In this video we show you have to program using variables. We start by doing things the wrong way, using constants, then show you the advantages of using variables.\n\nHopefully you can see from this video how much better it is, and more efficient to use variables instead of constants. In all the future lessons, we expect you to use variables." ]

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[ "Search a number\nBaseRepresentation\nbin100101111011\n310022220\n4211323\n534202\n615123\n710035\noct4573\n93286\n102427\n111907\n1214a3\n131149\n14c55\n15abc\nhex97b\n\n2427 has 4 divisors (see below), whose sum is σ = 3240. Its totient is φ = 1616.\n\nThe previous prime is 2423. The next prime is 2437. The reversal of 2427 is 7242.\n\n2427 is an esthetic number in base 15, because in such base its adjacent digits differ by 1.\n\nIt is a semiprime because it is the product of two primes.\n\nIt is a cyclic number.\n\nIt is not a de Polignac number, because 2427 - 22 = 2423 is a prime.\n\nIt is a d-powerful number, because it can be written as 25 + 211 + 4 + 73 .\n\nIt is a D-number.\n\n2427 is a modest number, since divided by 27 gives 24 as remainder.\n\n2427 is a lucky number.\n\nIt is a plaindrome in base 13 and base 15.\n\nIt is a nialpdrome in base 14.\n\nIt is not an unprimeable number, because it can be changed into a prime (2423) by changing a digit.\n\nIt is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 402 + ... + 407.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (810).\n\n2427 is a deficient number, since it is larger than the sum of its proper divisors (813).\n\n2427 is an equidigital number, since it uses as much as digits as its factorization.\n\n2427 is an evil number, because the sum of its binary digits is even.\n\nThe sum of its prime factors is 812.\n\nThe product of its digits is 112, while the sum is 15.\n\nThe square root of 2427 is about 49.2645917470. The cubic root of 2427 is about 13.4386793629.\n\nAdding to 2427 its sum of digits (15), we get a palindrome (2442).\n\nAdding to 2427 its reverse (7242), we get a palindrome (9669).\n\nThe spelling of 2427 in words is \"two thousand, four hundred twenty-seven\", and thus it is an iban number.\n\nDivisors: 1 3 809 2427" ]

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[ "# Zusammenfassung für den Vortrag am 26.11.2007 (17:00 Uhr)\n\nOberseminar Statistical Mechanics\n\nMichael Stolz (Ruhr University Bochum)\n\nLimit theorems for random matrix ensembles associated to symmetric spaces\n\nThis talk surveys recent work which develops aspects of classical random matrix theory in the broader framework of matrix ensembles associated to classical symmetric spaces. It is a classical result of Wigner that for an hermitian matrix with independent entries on and above the diagonal, the mean empirical eigenvalue distribution converges weakly to the semicircle law as matrix size tends to infinity. In joint work with Katrin Hofmann-Credner (Bochum), this has been generalized to random matrices taken from all infinitesimal versions of classical symmetric spaces. Like Wigner's, this result is universal in that it only depends on certain assumptions about the moments of the matrix entries, but not on the specifics of their distributions. Joint work with Benoît Collins (Lyon/Ottawa) points into a different direction. Here random vectors of the form (Tr(A^(1) V),..., Tr(A^(r) V)) are studied, where V is a uniformly distributed element of a matrix version of a classical compact symmetric space, and the A^() are deterministic parameter matrices. It is proven that for increasing matrix sizes these random vectors converge to a joint Gaussian limit. This generalizes work of Diaconis et al. on the compact classical groups.\n\n06.06.2018, 07:19" ]

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[ "", null, "Download your favorite Linux distribution at LQ ISO.", null, "LinuxQuestions.org", null, "Getting started using the GDB debugger to debug C/C++ source.\n\nNotices\n\n By edbarx at 2014-05-18 04:38 This howto is intended for anyone who never succeeded to use gdb successfully. gdb, being a CLI application, may not appeal to the eye, but my experience in using it, has proven that it is a very good debugger. Step 1: Compile your source code using the g++/gcc -g option. This tells the compiler to include debugging information in the compiled executable. Step2: While at the directory containing the freshly compiled executable, invoke gdb as follows: Code: `gdb executable-name` Step 3: Set breakpoints as follows: For the main source file, ie where main() is implemented, use: Code: `b 51` The 51 is the line number where the breakpoint is placed. In your case, it will be a different line number. For other files, say parser.cpp, use: Code: `b parser.cpp:125` Your project will most likely have differently named files. parser.cpp should be accompanied by a parser.h file: I don't know whether opting to only use .cpp files works. Step 4: Run the program by typing 'r' and by pressing Enter. The actual debugging.To step over a line of code use: 'n'. Stepping a line a code forces the debugger to fully execute that line of code. That means, if there is a function call, say: Code: `length = getMinimalLength(a, b, c);` the getMinimalLength function will be executed until it returns and the return value will be saved in length. To step into a line of code, for instance, when a function is called, use 's'. If you are unsure what is happening in your code, this is a best tool. Stepping into a line of code forces the debugger to step into any invoked functions. In our example, that means, the debugger will start debugging the code comprising the getMinimalLength function. To list a few lines of code use: 'list'. If you are unware which line of code the debugger is at, use list. This will list about ten lines of code showing the current line of code in the middle. To watch variable values, use 'p variable-name'. Debugging without being able to watch variable values is next to useless. This feature is central to any serious debugging. Let us assume, variable length is in the current scope and that we are debugging the line just after where getMinimalLength(..) function is called. To do this is as easy as: Code: `p length` An Example: For this example we will consider the following c++ program: Code: ```#include using namespace std; long double fact (int n) { int i; long double r = 1.0; for (i = 1; i <= n; i++) r = r * i; return r; } int main() { long double arr ; int i, j, z; cout << \"Start getting factorials from: \"; cin >> z; cout << \"Enter the number of iterations: \"; cin >> j; if (j < 0 || j > 100) { cout << \"Iteration count must lie within 0 and 100 including both ends\\n\"; return 0; } for (i = 0; i < j; i++) { arr[i] = fact(z); z++; } for (i = 0; i < j; i++) cout << arr[i] << '\\n'; return 0; }``` We will now compile the code. Assume the code is saved in a file named test07.cpp. Open a terminal, cd to the directory where you save test07.cpp, and invoke the g++ compiler in this way: Code: `g++ -g test07.cpp -o test07` Take special notice of the -g directive. This tells the compiler that we want debugging information in our executable file. Now, invoke the gdb debugger. The latter needs to know what executable you want to debug - pass its name as a parameter. Code: `gdb test07` gdb will present its interface. This should look like the following: Code: ```\\$ gdb test07 GNU gdb (GDB) 7.4.1-debian Copyright (C) 2012 Free Software Foundation, Inc. License GPLv3+: GNU GPL version 3 or later This is free software: you are free to change and redistribute it. There is NO WARRANTY, to the extent permitted by law. Type \"show copying\" and \"show warranty\" for details. This GDB was configured as \"x86_64-linux-gnu\". For bug reporting instructions, please see: ... Reading symbols from /home/edbarx/projects/test05/test07...done. (gdb)``` Suppose we want to debug the fact function. Line 7 in our code is a good place for a breakpoint because this will precede the portion of the function where the actual calculation takes place. To insert a breakpoint we use the b command as follows: Code: `(gdb) b 7` In the case our line of code we are interested in is in another file in the project, we would use the command as follows: Code: `(gdb) b test07.cpp:7` The debugger will tell us that the breakpoint has been accepted. Code: ```Breakpoint 1 at 0x4009f3: file test07.cpp, line 7. (gdb)``` Now, we run the code: Code: ```(gdb) r Starting program: /path-to-our-program/test07 Start getting factorials from:``` Follow, the program's prompts until the debugger halts the program at the breakpoint. Our program will display information similar to this. Code: ```(gdb) r Starting program: /path-to-our-program/test07 Start getting factorials from: 1 Enter the number of iterations: 2 Breakpoint 1, fact (n=1) at test07.cpp:7 7 long double r = 1.0; (gdb)``` As you can see, the debugger duly stopped at the breakpoint. Typing n for next, moves the execution by one line as follows. Code: ```(gdb) n 9 for (i = 1; i <= n; i++)``` Another n will move the point of execution by another line. Code: ```(gdb) n 10 r = r * i;``` To see what values are in our variables we use the p command as follows: Code: ```(gdb) p i \\$1 = 1 (gdb) p n \\$2 = 1 (gdb) p r \\$3 = 1 (gdb)``` This is the first iteration where we are calculating the factorial of 1 which is 1. The for loop should exit immediately after this step. Let us see: Code: ```(gdb) n 9 for (i = 1; i <= n; i++) (gdb) n 12 return r; (gdb)``` As you can confirm, the function is about to return the value r which should be 1. Let us check, if this is the case: Code: ```(gdb) p r \\$4 = 1``` ... and we have r = 1 as expected. To list the code where return r is situated, we use the list command: Code: ```(gdb) list 7 long double r = 1.0; 8 9 for (i = 1; i <= n; i++) 10 r = r * i; 11 12 return r; 13 } 14 15 int main() 16 { (gdb)``` Stepping further, ie using n some more times, the debugger will exit from the fact function returning to the main function as follows: Code: ```(gdb) n 13 } (gdb) n main () at test07.cpp:35 35 z++; (gdb) n 32 for (i = 0; i < j; i++) (gdb)``` Now, we will examine the second fact call which should return the factorial of 2. Note that 2! = 2. Please, note that the use of s, forces the debugger to move more slowly showing all steps even those behing the scenes. In this case, it is telling us that the returned value from fact is saved in the arr array. As you can see, sometimes, s, behaves like n. This depens on the complexity of the line at the point of execution. Function calls force the debugger to branch into them when s is used. Code: ```(gdb) s 34 arr[i] = fact(z); (gdb) s Breakpoint 1, fact (n=2) at test07.cpp:7 7 long double r = 1.0; (gdb) s 9 for (i = 1; i <= n; i++) (gdb) s 10 r = r * i; (gdb) s 9 for (i = 1; i <= n; i++) (gdb) s 10 r = r * i; (gdb) p i \\$5 = 2 (gdb) s 9 for (i = 1; i <= n; i++) (gdb) s 12 return r; (gdb) p r \\$6 = 2 (gdb)``` The result from the function is 2 as expected. To run the program until it exits or it meets another breakpoint use c (continue). Code: ```(gdb) c Continuing. 1 2 [Inferior 1 (process 5331) exited normally]``` Needless to state the obvious, the process number need not agree with mine. To quit the debugger, use the q command as follows: Code: `(gdb) q` Enjoy. :)\n\n by rtmistler on Fri, 2015-02-27 08:43 I like what you've written and I'm a proponent of using GDB as well. You should make that a blog entry. Another thing you should note in there if you agree, is that GDB can be used through emacs. I don't use it much that way but it acts more like a real debugger and one can use the mouse for things directly within the screen, plus there are menus also accessible via the mouse. You also can pull up child windows for things like the stack and watch variables, memory views, etc. Edit a simple C source in emacs Issue compile command at the prompt, default is \"make -k\", just change that to be: gcc -ggdb -o , this part is not necessary, but you can compile within emacs and a benefit is that you can browse for next error using the editor where it jumps right to each error and corrects line alignment if you edit/fix problems Issue debug command \"gdb\" at the prompt, it allows you to enter in options, but usually will pick good defaults and detect the executable file matching your source, especially if you compiled with it. From within the debugger you can do things like mouse left click near the left border of a source line to make a breakpoint, there are menu options to go, step, step-over, step-in, look at data, open sub-windows for things like watches, stack and so forth. For those who can type and remember stuff, it's not always the best. Plus on a target system which may or may not have a display manager, knowing command line GDB is best.", null, "All times are GMT -5. The time now is 09:08 AM.\n\n Contact Us - Advertising Info - Rules - Privacy - LQ Merchandise - Donations - Contributing Member - LQ Sitemap -\n My LQ", null, "" ]

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[ "", null, "", null, "### Home > INT1 > Chapter 2 > Lesson 2.3.2 > Problem2-102\n\n2-102.\n\nWrite the equation for the line containing the points listed in the table below.\n\n $x$ $y$ $–1$ $1$ $3$ $5$ $2$ $16$ $30$ $44$\n\nFind the rate of growth per unit. This is the slope.\nDo this by finding the pattern in the graph, or by plotting two of the points and making a slope triangle.\n\nIf $−1$ is $2$ and $1$ is $16$, what is $0$?\nThis is the $y$-intercept. Use it along with the slope to write a $y = mx + b$ equation.\n\nUse the eTool below to write the equation for the function in the table.\nClick the link at the right to view full version of the eTool: 2-102 HW eTool" ]

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", null ]

[ "# The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago Vitiated These Books, are Corrected and Some of Euclid's Demonstrations are Restored. Also, The Book of Euclid's Data, in Like Manner Corrected. viz. The first six books, together with the eleventh and twelfth\n\nA. Foulis, 1781 - 466 páginas\n0 Opiniones\nLas opiniones no están verificadas, pero Google revisa que no haya contenido falso y lo quita si lo identifica\n\n### Comentarios de la gente -Escribir un comentario\n\nNo encontramos ningún comentario en los lugares habituales.\n\n### Pasajes populares\n\nPágina 156 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.\nPágina 3 - A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another.\nPágina 323 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.\nPágina 92 - If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it ; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square- of the line which meets it, the line which meets shall touch the circle.\nPágina 80 - EA : and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each ; and the angle ADE is equal to the angle CDE, for each of them is a right angle ; therefore the base AE is equal (4.\nPágina 52 - If a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced.\nPágina 36 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.\nPágina 2 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle; and the straight line which stands on the other is called a perpendicular to it.\nPágina 54 - AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.\nPágina 74 - The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle..." ]

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[ "# Shared Flashcard Set\n\n## Details\n\nPenny's Final Review\nFinal Review for NMHS Pre-AICE Chem.\n18\nChemistry\n06/02/2009\n\n## Additional Chemistry Flashcards\n\nTerm\n standard tempurature and pressure\nDefinition\n 0°C and 1 atm\nTerm\n percent compostion\nDefinition\n the percent by mass of each element in a compound\nTerm\nDefinition\n number of representative particles in 1 mole of that substance\nTerm\n representative particle\nDefinition\n an atom, an ion, or a molecule, depending upon the way a substance commonly exists\nTerm\n empirical formula\nDefinition\n the smallest whole number ratio of the atoms in a compound\nTerm\n mole\nDefinition\n SI unit used to measure amount of substance\nTerm\n atomic mass\nDefinition\n number of grams of an element that is numerically equal to the atomic mass of the element in amu\nTerm\n molar volume\nDefinition\n the volume occupied by a mole of any gas at STP\nTerm\n molar mass\nDefinition\n the mass of a  mole of any element or compund\nTerm\n gas at STP\nDefinition\n The molar mass of a substance can be calculated from its density alone, if that substance is a __________.\nTerm\n empirical formula\nDefinition\n the lowest whole number ratio of the elements in a compound is called the _________.\nTerm\n mole\nDefinition\n Avogadro's number of representative particles is equal to one __________.\nTerm\n skeleton equasion\nDefinition\n an equasion that does not indicate relative amounts of reactants and products\nTerm\n product\nDefinition\n a new substance formed in a chemical reaction\nTerm\n reactant\nDefinition\n a starting substance in a chemical reaction\nTerm\n chemical equasion\nDefinition\n a concise representation of a chemical reaction\nTerm\n balanced equasion\nDefinition\n an equasion in which each side has the same number of atoms of each element\nSupporting users have an ad free experience!" ]

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[ "×\nGet Full Access to Statistics Through Applications - 2 Edition - Chapter 2 - Problem 2.24\nGet Full Access to Statistics Through Applications - 2 Edition - Chapter 2 - Problem 2.24\n\n×\n\n# Travel time Th e dotplot on the facing page displays data", null, "ISBN: 9781429219747 133\n\n## Solution for problem 2.24 Chapter 2\n\nStatistics Through Applications | 2nd Edition\n\n• Textbook Solutions\n• 2901 Step-by-step solutions solved by professors and subject experts\n• Get 24/7 help from StudySoup virtual teaching assistants", null, "Statistics Through Applications | 2nd Edition\n\n4 5 1 347 Reviews\n19\n5\nProblem 2.24\n\nTravel time Th e dotplot on the facing page displays data on students responses to the question How long does it usually take you to travel to school? (a) Make a well-labeled histogram of the data. (b) Describe the shape, center, and spread of the distribution. Are there any outliers?\n\nStep-by-Step Solution:\nStep 1 of 3\n\nLab #1 The Penny Problem: Review The Scientific Method Scientists use the scientific method to answer questions and provide explanations about natural phenomena. It is a logical process based on careful observation and experimentation. 1. Observation: leads to a question and discussion about the observed phenomena. 2. Hypothesis: based on the observation scientists generate a hypothesis or tentative explanation for the observed phenomena. Hypothesis are written as two statements; one called the null hypothesis, and the other called alternate hypotheses. - The Null Hypothesis – is written in a way that shows no difference between the groups under study. Ex: There is statistically significant relationship between\n\nStep 2 of 3\n\nStep 3 of 3\n\n##### ISBN: 9781429219747\n\nThe answer to “Travel time Th e dotplot on the facing page displays data on students responses to the question How long does it usually take you to travel to school? (a) Make a well-labeled histogram of the data. (b) Describe the shape, center, and spread of the distribution. Are there any outliers?” is broken down into a number of easy to follow steps, and 50 words. This textbook survival guide was created for the textbook: Statistics Through Applications, edition: 2. Since the solution to 2.24 from 2 chapter was answered, more than 369 students have viewed the full step-by-step answer. This full solution covers the following key subjects: data, Travel, make, describe, displays. This expansive textbook survival guide covers 10 chapters, and 710 solutions. Statistics Through Applications was written by and is associated to the ISBN: 9781429219747. The full step-by-step solution to problem: 2.24 from chapter: 2 was answered by , our top Statistics solution expert on 11/10/17, 06:04PM.\n\nUnlock Textbook Solution" ]

[ null, "https://studysoup.com/cdn/2cover_2603882", null, "https://studysoup.com/cdn/2cover_2603882", null ]

[ "Community\ncancel\nShowing results for\nDid you mean:", null, "Beginner\n82 Views\n\n## What kind of speed increase should I be seeing?\n\nHi everyone,\n\nI've got a routine here that I want to make faster. The size of the input data is roughly 4000x4000 (similar to my previous problem) and I want to see some kind of speedup using intel primitives, but I'm just not seeing it.\n\nHere's the code:\n//#define INTELPRIMITIVES\n\n#ifdef INTELPRIMITIVES\n#include \"ipp.h\"\n#endif\nstatic void Linearize(unsigned short* inData, float* outData,\nconst int inXSize, const int inYSize, float inLinearizeConst = -9.9546e-5){\n\n//float a = -9.9546e-5 * (log(2.0)/log(1.778));\nfloat a = inLinearizeConst * (log(2.0)/log(1.778));\n\nint i;\nconst int theSize = inXSize*inYSize;\n#ifndef INTELPRIMITIVES\nfor (i = 0; i < theSize; i++){\noutData = (65535.0f - 65535.0f * exp(a * (float)inData));\n}\n#else\nippStaticInit();//just to make sure\n//first, copy the given data vector into a float vector\n//then, multiply each element in the vector with a\n//then, raise the vector by exp\n//then, multiply by 65535\n//then, subtract the vector from 65535\n//then, place into the output vector\n//and this should be faster?\n\nIpp32f* pStart = ippsMalloc_32f(inXSize*inYSize);\nIpp32f* pTmp = ippsMalloc_32f(inXSize*inYSize);\n\nippsConvert_16u32f(inData, pStart, theSize);\n\nippsMulC_32f(pStart, a, pTmp, theSize);\n//now, do the raising, back into start\nippsExp_32f(pTmp, pStart, theSize);\n//now, the multiplication, going back the other way\nippsMulC_32f(pStart, 65535.0f, pTmp, theSize);\n//now, the element-by-element subtraction\n//first, set the pStart array to be 65535\n//then place into a third array, which will be copied out and returned.\nippsSet_32f(65535.0f, pStart, theSize);\n\n//Ipp32f* pFinal = ippsMalloc_32f(theSize);\nippsSub_32f(pTmp, pStart, outData, theSize);\n\nippsFree(pStart);\nippsFree(pTmp);\n\n#endif\n};\n\nIf I add or remove that #define, no appreciable difference in speed is seen. Is it that that loop is already vectorized by the intel compiler, and so the explicit unrolling of the math isn't doing anything? Is usin g ::GetTickCount() accurate enough for this kind of timing? On images of that size, I'm seeing timing of around half a second for this routine; should that be faster, on a 2.4 Ghz core 2 duo?\n\nThanks!", null, "Employee\n82 Views\n\nHello,\n\nin your case there is a huge (4Kx4K!) intermediate buffers, which cause your algorithm trash the processor cache at each processing stage. You'd better your slicing to process images by relatevely small parts, which fit into cache and you may gat up to 3X speedup just because of that.\n\nIf you have IPP 6.0 beta I would recommend you to take a look on Intel Deferred Mode Image Processing Layer which was specifically developed to simplify coding of calculation pipelines with slicing andutilizing threading capability of modern processors to parallelize processing on slice level.\n\nRegards,", null, "" ]

[ null, "https://community.intel.com/t5/image/serverpage/avatar-name/intelcharcoal/avatar-theme/candy/avatar-collection/Intel_Customer/avatar-display-size/message/version/2", null, "https://community.intel.com/t5/image/serverpage/avatar-name/intelorangeinverse/avatar-theme/candy/avatar-collection/Intel_Customer/avatar-display-size/message/version/2", null, "https://community.intel.com/skins/images/195C61C8999AB3A290E8180F0D972123/responsive_peak/images/icon_anonymous_message.png", null ]

[ "", null, "WORKSPACE\nACCOUNT\nGALLERY\nVersion: 3.4.0\nMix & Match #61757E\nSearch colors...\n#61757E\n\n## Mix\n\n#61757E can be mixed using 38% of RED, 7% of BLUE, 45% of CYAN, 3% of GREEN, 7% of BLACK.\n\n## Match\n\nWe found 50 paints that are close to #61757E from brands like Ameritone Devoe, Premier Paints, Dutch Standard, Huls, Ace, California Paints.\n\nMix guide\n\n#61757E\n#61757E match: 100%\nRED\nBLUE\nCYAN\nGREEN\nBLACK\nMixing process step by step\n\n1\n=\n1\n+\n1\n=\n1\n+\n2\n=\n1\n+\n3\n=\n1\n+\n4\n=\n1\n+\n5\n=\n1\n+\n6\n=\n1\n+\n7\n=\n1\n+\n8\n=\n1\n+\n9\n=\n1\n+\n10\n=\n1\n+\n11\n=\n1\n+\n12\n=\n1\n+\n13\n=\n1\n+\n14\n=\n1\n+\n15\n=\n1\n+\n16\n=\n1\n+\n17\n=\n1\n+\n18\n=\n1\n+\n19\n=\n1\n+\n20\n=\n1\n+\n21\n=\n1\n+\n22\n=\n1\n+\n23\n=\n1\n+\n24\n=\n1\n+\n25\n=\n1\n+\n26\n=\n1\n+\n27\n=\n1\n+\n28\n=\nConversion table\n\nHEX\n#61757E\n\nHSV\n199°, 23, 49\n\nHSL\n199°, 13, 44\n\nCIE Lab\n47.97, -5.19, -7.38\n\nRGB decimal\n97, 117, 126\n\nRGB percent\n38%, 45.9%, 49.4%\n\nCMYK\n23, 7, 0, 51\n\nColor name" ]

[ null, "https://trycolors.com/_next/image", null ]

[ "# Linear Block Code using MATLAB\n\n• Last Updated : 05 Sep, 2020\n\nAny linear blend of codewords is likewise a code word only. So in coding, a linear code is a mistake correcting code. Linear codes are generally partitioned into block codes and convolutional codes, despite the fact that turbo codes can be viewed as a half breed of these two sorts. Linear codes take into account more productive encoding and deciphering calculations than different codes.\n\nLinear codes are utilized in forward mistake adjustment and are applied in methods for techniques for sending symbols (e.g., bits) on a communications channel so that, if mistakes occur in the communication, some mistakes can be amended or recognized by the beneficiary of a message block. The code words in a linear block code are blocks of symbols that are encoded utilizing a greater number of symbols than the first incentive to be sent.\n\nLet us see the MATLAB code for Linear Block Code.\n\n `% Given H Matrix``H = [1 0 1 1 1 0 0;``     ``1 1 0 1 0 1 0;``     ``0 1 1 1 0 0 1]`` ` `k = 4;``n = 7;`` ` `% Generating G Matrix`` ` `% Taking the H Matrix Transpose``P = H';  `` ` `% Making a copy of H Transpose Matrix``L = P;  `` ` `% Taking the last 4 rows of L and storing``L((5:7), : ) = [];  `` ` `% Creating a Identity matrix of size K x K``I = eye(k);  `` ` `% Making a 4 x 7 Matrix``G = [I L]  `` ` `% Generate U data vector, denoting all information sequences``no = 2 ^ k`` ` `% Iterate through an Unit-Spaced Vector``for` `i = 1 : 2^k  `` ` `  ``% Iterate through Vector with Specified Increment ``  ``% or in simple words here we are decrementing 4 till we get 1    ``  ``for` `j = k : -1 : 1 ``    ``if` `rem(i - 1, 2 ^ (-j + k + 1)) >= 2 ^ (-j + k)``      ``u(i, j) = 1;``    ``else``      ``u(i, j) = 0;``    ``end``     ` `    ``% To avoid displaying each iteration/loop value``    ``echo off; ``  ``end``end`` ` `echo on;``u`` ` `% Generate CodeWords``c = rem(u * G, 2)`` ` `% Find the min distance``w_min = min(sum((c(2 : 2^k, :))'))`` ` `% Given Received codeword``r = [0 0 0 1 0 0 0];``r`` ` `p = [G(:, n - k + 2 : n)];`` ` `%Find Syndrome``ht = transpose(H)`` ` `s = rem(r * ht, 2)`` ` `for` `i = 1 : 1 : size(ht)``  ``if``(ht(i,1:3)==s)``    ``r(i) = 1-r(i);``    ``break``;``  ``end``end`` ` `disp(``'The Error is in bit:'``)``disp(i)`` ` `disp(``'The Corrected Codeword is :'``)``disp(r)`\n\nOutput:\n\n```H =\n\n1 0 1 1 1 0 0\n1 1 0 1 0 1 0\n0 1 1 1 0 0 1\n\nG =\n\n1 0 0 0 1 1 0\n0 1 0 0 0 1 1\n0 0 1 0 1 0 1\n0 0 0 1 1 1 1\n\nno = 16\nu =\n\n0 0 0 0\n0 0 0 1\n0 0 1 0\n0 0 1 1\n0 1 0 0\n0 1 0 1\n0 1 1 0\n0 1 1 1\n1 0 0 0\n1 0 0 1\n1 0 1 0\n1 0 1 1\n1 1 0 0\n1 1 0 1\n1 1 1 0\n1 1 1 1\n\nc =\n\n0 0 0 0 0 0 0\n0 0 0 1 1 1 1\n0 0 1 0 1 0 1\n0 0 1 1 0 1 0\n0 1 0 0 0 1 1\n0 1 0 1 1 0 0\n0 1 1 0 1 1 0\n0 1 1 1 0 0 1\n1 0 0 0 1 1 0\n1 0 0 1 0 0 1\n1 0 1 0 0 1 1\n1 0 1 1 1 0 0\n1 1 0 0 1 0 1\n1 1 0 1 0 1 0\n1 1 1 0 0 0 0\n1 1 1 1 1 1 1\n\nw_min = 3\n\nr\nr =\n\n0 0 0 1 0 0 0\n\nht =\n\n1 1 0\n0 1 1\n1 0 1\n1 1 1\n1 0 0\n0 1 0\n0 0 1\n\ns =\n\n1 1 1\n\nThe Error is in bit:\n4\n\nThe Corrected Codeword is :\n0 0 0 0 0 0 0\n```\n\nMy Personal Notes arrow_drop_up" ]

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[ "# amplify.IsingMatrix¶\n\nclass IsingMatrix\n\nUpper triangular matrix representation of the Ising model with real coefficients.\n\nThe Ising model will be the form of $$H = s^{\\mathrm{T}} J s - \\mathrm{Tr}J + s^{\\mathrm{T}} \\cdot \\operatorname{diag} J$$, where $$s$$ is a vector of variables and $$J$$ is a square matrix.\n\nThis class denotes the upper triangular matrix representation of $$J$$.\n\nIn the descriptions of class methods, $$J$$ and $$s$$ are the matrix and the vector this class represents, respectively.\n\nThe following operators are defined for the class.\n• Indexing: a[slices] (__getitem__(), __setitem__())\n\n• Equality: a == b (__eq__())\n\n• Inequality: a != b (__ne__())\n\n• Addition: a + b (__add__(), __radd__(), __iadd__())\n\n• Subtraction: a - b (__sub__(), __rsub__(), __isub__())\n\n• Multiplication: a * b (__mul__(), __rmul__(), __imul__())\n\n• Division: a / b (__truediv__(), __rtruediv__(), __itruediv__())\n\n• Floor Division: a // b (__floordiv__(), __rfloordiv__(), __ifloordiv__())\n\n__init__(size)\n\nReturns a zero-matirx with specified size.\n\nパラメータ:\n\nsize (int) -- Size of the matrix.\n\nExample\n\n>>> from amplify import IsingMatrix\n>>> m = IsingMatrix(3)\n>>> m[0, 1] = 1\n>>> m[0, 2] = 2\n>>> m[1, 2] = 3\n>>> m\n[[0, 1, 2],\n[0, 0, 3],\n[0, 0, 0]]\n\n\nMethods\n\n __init__(size) Returns a zero-matirx with specified size. evaluate(self, object) Evaluates the matrix with array s. resize(self, size) Resizes the matrix. size(self) Returns the matrix size. to_BinaryMatrix(self[, ascending]) Converts the matrix to BinaryMatrix. to_IsingMatrix(self[, ascending]) Converts the matrix to IsingMatrix. to_Poly(self) Converts the matrix to IsingPoly. to_numpy(self) no docstring\nevaluate(self, object)\n\nEvaluates the matrix with array s.\n\nパラメータ:\n\nobject -- array-like. The size of arary object should be equal to the matrix size.\n\n$$s^{\\mathrm{T}} J s - \\mathrm{Tr}J + s^{\\mathrm{T}} \\cdot \\operatorname{diag} J$$\n\nExample\n\n>>> from amplify import IsingMatrix\n>>> m = IsingMatrix(3)\n>>> m[0, 1] = 1\n>>> m[0, 2] = 2\n>>> m[1, 2] = 3\n>>> m\n[[0, 1, 2],\n[0, 0, 3],\n[0, 0, 0]]\n>>> m.evaluate([0, 1, 1])\n3.0\n\nresize(self: amplify.IsingMatrix, size: int) None\n\nResizes the matrix.\n\nExample\n\n>>> from amplify import IsingMatrix\n>>> m = IsingMatrix(2)\n>>> m[0, 1] = 1\n>>> m\n[[0, 1],\n[0, 0]]\n>>> m.resize(3)\n>>> m\n[[0, 1, 0],\n[0, 0, 0],\n[0, 0, 0]]\n\nsize(self: amplify.IsingMatrix) int\n\nReturns the matrix size.\n\nsize of the matrix\n\nint\n\nExample\n\n>>> from amplify import IsingMatrix\n>>> m = IsingMatrix(3)\n>>> m.size()\n3\n\nto_BinaryMatrix(self: amplify.IsingMatrix, ascending: bool = True) Tuple[amplify.BinaryMatrix, float]\n\nConverts the matrix to BinaryMatrix.\n\nBy this function, the Ising model formulation $$s^{\\mathrm{T}} J s - \\mathrm{Tr}J + s^{\\mathrm{T}} \\cdot \\operatorname{diag} J$$ would be transformed to the QUBO formulation $$q^T Q q$$, where $$q$$ is a vector of variables and $$Q$$ is an upper triangular square matrix. The conversion is along $$s = 2q - \\left\\{1 \\right\\}$$.\n\nThis function returns the pair of the converted BinaryMatrix $$Q$$ and the constant term $$c$$ in the converted QUBO formulation.\n\n$$(Q, c)$$\n\nExample\n\n>>> from amplify import IsingMatrix\n>>> m = IsingMatrix(3)\n>>> m[0, 1] = 1\n>>> m[0, 2] = 2\n>>> m[1, 2] = 3\n>>> m\n[[0, 1, 2],\n[0, 0, 3],\n[0, 0, 0]]\n>>> m.to_BinaryMatrix()\n([[-6, 4, 8],\n[0, -8, 12],\n[0, 0, -10]], 6.0)\n\nto_IsingMatrix(self: amplify.IsingMatrix, ascending: bool = True) Tuple[amplify.IsingMatrix, float]\n\nConverts the matrix to IsingMatrix.\n\nThis function returns the pair of the converted IsingMatrix $$J$$ and the constant term $$c$$ in the converted Ising model formulation.\n\n$$(J, c)$$\n\nExample\n\n>>> from amplify import IsingMatrix\n>>> m = IsingMatrix(3)\n>>> m[0, 1] = 1\n>>> m[0, 2] = 2\n>>> m[1, 2] = 3\n>>> m\n[[0, 1, 2],\n[0, 0, 3],\n[0, 0, 0]]\n>>> m.to_IsingMatrix()\n([[0, 1, 2],\n[0, 0, 3],\n[0, 0, 0]], 0.0)\n\nto_Poly(self: amplify.IsingMatrix)\n\nConverts the matrix to IsingPoly.\n\npolynomial expression of the matrix\n\nIsingPoly\n\nExample\n\n>>> from amplify import IsingMatrix\n>>> m = IsingMatrix(3)\n>>> m[0, 1] = 1\n>>> m[0, 2] = 2\n>>> m[1, 2] = 3\n>>> m\n[[0, 1, 2],\n[0, 0, 3],\n[0, 0, 0]]\n>>> m.to_Poly()\ns_0 s_1 + 2.000000 s_0 s_2 + 3.000000 s_1 s_2\n\nto_numpy(self: amplify.IsingMatrix) numpy.ndarray[numpy.float64]\n\nno docstring" ]

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[ " Function Macros | Enterprise Architect User Guide\n Prev Next\n\n# Function Macros\n\nFunction macros are a convenient way of manipulating and formatting various element data items. Each function macro returns a result string. There are two primary ways to use the results of function macros:\n\n• Direct substitution of the returned string into the output, such as: %TO_LOWER(attName)%\n• Storing the returned string as part of a variable definition such as: \\$name = %TO_LOWER(attName)%\n\nFunction macros can take parameters, which can be passed to the macros as:\n\n• String literals, enclosed within double quotation marks\n• Direct substitution macros without the enclosing percent signs\n• Variable references\n• Numeric literals\n\nMultiple parameters are passed using a comma-separated list.\n\nFunction macros are named according to the All-Caps style, as in:\n\n%CONVERT_SCOPE(opScope)%\n\nThe available function macros are described here. Parameters are denoted by square brackets, as in:\n\nFUNCTION_NAME([param]).\n\n## CONVERT_SCOPE([umlScope])\n\nFor use with supported languages, to convert [umlScope] to the appropriate scope keyword for the language being generated. This table shows the conversion of [umlScope] with respect to the given language.\n\nLanguage\n\nConversions\n\nC++\n\nPackage ==> public\n\nPublic ==> public\n\nPrivate ==> private\n\nProtected ==> protected\n\nC#\n\nPackage ==> internal\n\nPublic ==> public\n\nPrivate ==> private\n\nProtected ==> protected\n\nDelphi\n\nPackage ==> protected\n\nPublic ==> public\n\nPrivate ==> private\n\nProtected ==> protected\n\nJava\n\nPackage ==> {blank}\n\nPublic ==> public\n\nPrivate ==> private\n\nProtected ==> protected\n\nPHP\n\nPackage ==> public\n\nPublic ==> public\n\nPrivate ==> private\n\nProtected ==> protected\n\nVB\n\nPackage ==> Protected\n\nPublic ==> Public\n\nPrivate ==> Private\n\nProtected ==> Protected\n\nVB .Net\n\nPackage ==> Friend\n\nPublic ==> Public\n\nPrivate ==> Private\n\nProtected ==> Protected\n\n## COLLECTION_CLASS([language])\n\nGives the appropriate collection Class for the language specified for the current linked attribute.\n\n## CSTYLE_COMMENT([wrap_length])\n\nConverts the notes for the element currently in scope to plain C-style comments, using /* and */.\n\n## DELPHI_PROPERTIES([scope], [separator], [indent])\n\nGenerates a Delphi property.\n\n## DELPHI_COMMENT([wrap_length])\n\nConverts the notes for the element currently in scope to Delphi comments.\n\nInvokes an Enterprise Architect Add-In function, which can return a result string.\n\n[addin_name] and [function_name] specify the names of the Add-In and function to be invoked.\n\nParameters to the Add-In function can be specified via parameters [prm_1] to [prm_n].\n\nAny function that is to be called by the EXEC_ADD_IN macro must have two parameters: an EA.Repository object, and a Variant array that contains any additional parameters from the EXEC_ADD_IN call. Return type should be Variant.\n\nPublic Function ProcessOperation(Repository As EA.Repository, args As Variant) As Variant\n\n## FIND([src], [subString])\n\nPosition of the first instance of [subString] in [src]; -1 if none.\n\n## GET_ALIGNMENT()\n\nReturns a string where all of the text on the current line of output is converted into spaces and tabs.\n\nConverts the notes for the element currently in scope to javadoc -style comments.\n\n## LEFT([src], [count])\n\nThe first [count] characters of [src].\n\n## LENGTH([src])\n\nLength of [src]. Returns a string.\n\nIn a code template or DDL template, these three macros perform, respectively, the mathematical functions of:\n\n• Multiplication (x*y) and\n• Subtraction (x-y)\n\nThe arguments x and y can be integers or variables, or a combination of the two. Consider these examples, as used in a 'Class' template for C++ code generation:\n\n• \\$b = %MATH_SUB(10,3)%\n• \\$c = %MATH_MULT(2,3)%\n• \\$e = %MATH_SUB(\\$b,\\$c)%\n• \\$f = %MATH_MULT(\\$a,\\$b)%\n• \\$g = %MATH_MULT(\\$a,10)%\n• \\$h = %MATH_MULT(10,\\$b)%\n\nThese compute, in the same sequence, to:\n\n• a = 3 + 4 = \\$a\n• b = 10 - 3 = \\$b\n• c = 2 * 3 = \\$c\n• d = a + b = \\$d\n• e = b - c = \\$e\n• f = a * b = \\$f\n• g = a * 10 = \\$g\n• h = 10 * b = \\$h\n\nWhen the code is generated, the .h file (for C++) contains these corresponding strings:\n\n• a = 3 + 4 = 7\n• b = 10 - 3 = 7\n• c = 2 * 3 = 6\n• d = a + b = 14\n• e = b - c = 1\n• f = a * b = 49\n• g = a * 10 = 70\n• h = 10 * b = 70\n\n## MID([src], [start]) MID([src], [start], [count])\n\nSubstring of [src] starting at [start] and including [count] characters. Where [count] is omitted the rest of the string is included.\n\n## PI([option], [value], {[option], [value]})\n\nSets the PI for the current template to [value]. Valid values for [value] are:\n\n• \"\\n\"\n• \"\\t \"\n• “ “\n• “”\n\n<option> controls when the new PI takes effect. Valid values for <option> are:\n\n• I, Immediate: the new PI is generated before the next non-empty template line\n• N, Next: the new PI is generated after the next non-empty template line\n\nMultiple pairs of options are allowed in one call. An example of the situation where this would used is where one keyword is always on a new line, as illustrated here:\n\n%PI=\" \"%\n\n%classAbstract ? \"abstract\"%\n\n%if classTag:\"macro\" != \"\"%\n\n%PI(\"I\", \"\\n\", \"N\", \" \")%\n\n%classTag:\"macro\"%\n\n%endIf%\n\nclass\n\n%className%\n\nFor more details, see The Processing Instruction (PI) Macro.\n\n## PROCESS_END_OBJECT([template_name])\n\nEnables the Classes that are one Class further away from the base Class, to be transformed into objects (such as attributes, operations, Packages, parameters and columns) of the base Class. [template_name] refers to the working template that temporarily stores the data.\n\n## REMOVE_DUPLICATES([source], [separator])\n\nWhere [source] is a [separator] separated list; this removes any duplicate or empty strings.\n\n## REPLACE([string], [old], [new])\n\nReplaces all occurrences of [old] with [new] in the given string <string>.\n\n## RESOLVE_OP_NAME()\n\nResolves clashes in interface names where two method-from interfaces have the same name.\n\n## RESOLVE_QUALIFIED_TYPE() RESOLVE_QUALIFIED_TYPE([separator]) RESOLVE_QUALIFIED_TYPE([separator], [default])\n\nGenerates a qualified type for the current attribute, linked attribute, linked parent, operation, or parameter. Enables the specification of a separator other than. and a default value for when some value is required.\n\n## RIGHT([src], [count])\n\nThe last [count] characters of [src].\n\n## TO_LOWER([string])\n\nConverts [string] to lower case.\n\n## TO_UPPER([string])\n\nConverts [string] to upper case.\n\n## TRIM([string]) TRIM([string], [trimChars])\n\nRemoves trailing and leading white spaces from [string]. If [trimChars] is specified, all leading and trailing characters in the set of <trimChars> are removed.\n\n## TRIM_LEFT([string]) TRIM_LEFT([string], [trimChars])\n\nRemoves the specified leading characters from <string>.\n\n## TRIM_RIGHT([string]) TRIM_RIGHT([string], [trimChars])\n\nRemoves the specified trailing characters from <string>.\n\n## VB_COMMENT([wrap_length])\n\nConverts the notes for the element currently in scope to Visual Basic style comments.\n\n## WRAP_COMMENT([comment], [wrap_length], [indent], [start_string])\n\nWraps the text [comment] at width [wrap_length] putting [indent] and [start_string] at the beginning of each line.\n\n\\$behavior = %WRAP_COMMENT(opBehavior, \"40\", \" \", \"//\")%\n\n<wrap_length> must still be passed as a string, even though WRAP_COMMENT treats this parameter as an integer.\n\n## WRAP_LINES([text], [wrap_length], [start_string] {, [end_string] })\n\nWraps [text] as designated to be [wrap_length], adding [start_string] to the beginning of every line and [end_string] to the end of the line if it is specified.\n\n## XML_COMMENT([wrap_length])\n\nConverts the notes for the element currently in scope to XML-style comments." ]

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[ "# Resizing and rotating a QGraphicsItem results in odd shape\n\nI can't understand how scaling and rotation are applied to `QGraphicsItem`.\n\nI need to be able to apply rotation and scaling (not necessarily keeping aspect ratio) and I get fully unexpected results.\n\nRotation must be around the item center. I seem to have no problem doing that - yet if I try to debug the bounding rectangle, I get seemingly wrong values.\n\nIf I don't keep aspect ratio, instead of rotation I get a very weird skew, and I have been struggling for quite a while to find the cause and correct it. I hope anybody can find a solution.\n\nFor many items - like rectangles - my solution was to give up on resize - and just replace the item with a new one of given size. I even did that for pixmap (though it probably will affect performance a lot).\nBut I don't know how to do that for text, or a few other types (svg...).\nI am trying to understand how scaling is applied, on rotated items, and how to get it applied correctly.\n\nThe code below is an experiment where I scale and rotate a text item, and the result... see attached image\n\n``````#include <QApplication>\n#include <QGraphicsView>\n#include <QGraphicsTextItem>\n\nvoid experimentScaling(QGraphicsScene* s)\n{\nQGraphicsTextItem* ref = new QGraphicsTextItem(); // a reference, not resized\nref->setPlainText(\"hello world\");\nref->setDefaultTextColor(Qt::red);\nref->setRotation(45);\n\nQGraphicsTextItem* t = new QGraphicsTextItem(); // text item to be experimented on\nt->setPlainText(\"hello world\");\n\nQTransform transform; // scale\ntransform.scale(10, 1);\nt->setTransform(transform);\nt->update();\n\nQPointF _center = t->boundingRect().center();\nqDebug(\"%f %f %f %f\", t->boundingRect().left(), t->boundingRect().top(), t->boundingRect().right(), t->boundingRect().bottom()); // seems to be unscaled...\n\nt->setTransformOriginPoint(_center); // rotation must be around item center - and seems to work even though the bounding rect gives wrong values above\nt->setRotation(45); // skewed\nt->update();\n}\n\nint main(int argc, char *argv[])\n{\nQApplication app(argc, argv);\nQGraphicsScene s;\nQGraphicsView view(&s);\ns.setSceneRect(-20, -20, 800, 600);\nview.show();\nexperimentScaling(&s);\nreturn app.exec();\n}\n``````\n\nReference (red) text rotated 45 degrees, text rotated 45 degrees and resized 10,1:", null, "The resized (black) text should have the same height as the reference (red) - yet is much taller;\nThe bounding rectangle is no longer a rectangle - it is skewed;\nThe angle looks much smaller than 45;\n\nAdded a resized but not rotated reference as well:", null, "I have tried looking into `QGraphicsRotation` but I can't figure out how to apply it... All I get is a move instead of rotation.\n\n• Is that transform.scale(10, 1) correct? Shouldn't it be transform.scale(10, 10) ? – juzzlin Aug 24 '15 at 16:56\n• @juzzlin: I don't want to keep aspect ratio. I am trying to stretch the text in one direction or both – Thalia Aug 24 '15 at 17:01\n\nAs documented, the item's transformations are mathematically applied in a certain order - this is the order you'd be multiplying the transform matrices in and is, conceptually, the reverse of the order you'd normally think of.\n\n1. The `transform` is applied. The origin point must be included in the transform itself, by applying translations during the transform.\n2. The `transformations` are applied - each of them can specify its own center.\n3. `rotation` then `scale` are applied, both relative to `transformOriginPoint`.\n\nWhen you set `transform` to scaling, and set `rotation`, the rotation is performed before scaling. The scaling applies to the rotated result - it simply stretches the rotated version horizontally in your case.\n\nYou need to somehow enforce the reverse order of operations. The only two ways to do that are:\n\n1. Stack the transforms in correct order and pass them to `transform`, or.\n\n2. Pass a list of correct transformations to `transformations`.\n\nI'll demonstrate how to do it either way, in an interactive fashion where you can adjust the transform parameters using sliders.\n\nTo obtain the correct result using `transform`:\n\n``````QGraphicsItem * item = ....;\nQTransform t;\nQPointF xlate = item->boundingRect().center();\nt.translate(xlate.x(), xlate.y());\nt.rotate(angle);\nt.scale(xScale, yScale);\nt.translate(-xlate.x(), -xlate.y());\nitem->setTransform(t);\n``````\n\nTo obtain the correct result using `transformations`:\n\n``````QGraphicsItem * item = ....;\nQGraphicsRotation rot;\nQGraphicsScale scale;\nauto center = item->boundingRect().center();\nrot.setOrigin(QVector3D(center));\nscale.setOrigin(QVector3D(center()));\nitem->setTransformations(QList<QGraphicsTransform*>() << &rot << &scale);\n``````\n\nFinally, the example:", null, "``````// https://github.com/KubaO/stackoverflown/tree/master/questions/graphics-transform-32186798\n#include <QtWidgets>\n\nstruct Controller {\npublic:\nQSlider angle, xScale, yScale;\nController(QGridLayout & grid, int col) {\nangle.setRange(-180, 180);\nxScale.setRange(1, 10);\nyScale.setRange(1, 10);\n}\ntemplate <typename F> void connect(F && f) { connect(f, f, std::forward<F>(f)); }\ntemplate <typename Fa, typename Fx, typename Fy> void connect(Fa && a, Fx && x, Fy && y) {\nQObject::connect(&angle, &QSlider::valueChanged, std::forward<Fa>(a));\nQObject::connect(&xScale, &QSlider::valueChanged, std::forward<Fx>(x));\nQObject::connect(&yScale, &QSlider::valueChanged, std::forward<Fy>(y));\n}\nQTransform xform(QPointF xlate) {\nQTransform t;\nt.translate(xlate.x(), xlate.y());\nt.rotate(angle.value());\nt.scale(xScale.value(), yScale.value());\nt.translate(-xlate.x(), -xlate.y());\nreturn t;\n}\n};\n\nint main(int argc, char **argv)\n{\nauto text = QStringLiteral(\"Hello, World!\");\nQApplication app(argc, argv);\nQGraphicsScene scene;\nQWidget w;\nQGridLayout layout(&w);\nQGraphicsView view(&scene);\nController left(layout, 0), right(layout, 4);\n\nauto ref = new QGraphicsTextItem(text); // a reference, not resized\nref->setDefaultTextColor(Qt::red);\nref->setTransformOriginPoint(ref->boundingRect().center());\nref->setRotation(45);\n\nauto leftItem = new QGraphicsTextItem(text); // controlled from the left\nleftItem->setDefaultTextColor(Qt::green);\n\nauto rightItem = new QGraphicsTextItem(text); // controlled from the right\nrightItem->setDefaultTextColor(Qt::blue);\n\nQGraphicsRotation rot;\nQGraphicsScale scale;\nrightItem->setTransformations(QList<QGraphicsTransform*>() << &rot << &scale);\nrot.setOrigin(QVector3D(rightItem->boundingRect().center()));\nscale.setOrigin(QVector3D(rightItem->boundingRect().center()));\n\nleft.connect([leftItem, &left]{ leftItem->setTransform(left.xform(leftItem->boundingRect().center()));});\nright.connect([&rot](int a){ rot.setAngle(a); },\n[&scale](int s){ scale.setXScale(s); }, [&scale](int s){ scale.setYScale(s); });\nright.angle.setValue(45);\nright.xScale.setValue(3);\nright.yScale.setValue(1);\n\nview.ensureVisible(scene.sceneRect());\nw.show();\nreturn app.exec();\n}\n``````\n• This is great, thank you for the very clear explanation - and for showing me how to use the transformations (`QGraphicsRotation` and `QGraphicsScale `) – Thalia Aug 24 '15 at 19:55\n• I also have a similar problem with this (your example) as I do with using transformations: if I try to \"pin\" scaling to top let, the object also moves - I am experimenting with translating it back (using m31, m32) but getting inconsistent results... Is there a way to compensate for the translation without messing up the center-based rotation ? – Thalia Sep 2 '15 at 14:28\n• @Thalia Pin the scaling in which coordinate system, at what point? The systems can be the parent or the text item's. – Unslander Monica Sep 2 '15 at 14:58\n• The item's coordinate system, boundingRect().topLeft()... I added a sample in this question: stackoverflow.com/questions/32356453/… – Thalia Sep 2 '15 at 15:12\n\nI was able to make that work by using two separate QTransforms and multiplying them together. Note that the order of transformations matter:\n\n``````QTransform transform1;\ntransform1.scale(10, 1);\n\nQTransform transform2;\ntransform2.rotate(45);\n\nt->setTransform(transform1 * transform2);\n``````" ]

[ null, "https://i.stack.imgur.com/K3zt6.png", null, "https://i.stack.imgur.com/MD0KJ.png", null, "https://i.stack.imgur.com/wyppC.png", null ]

[ "An Analytical Comparison of the Signal to Noise Ratios Between CMY and RGB Imaging Techniques\n\nBy Ed Grafton\n\nOne of the most significant obstacles, which must be overcome in imaging astronomical objects, is the inherent faintness of these objects. From the early days of photographic film to present day CCDs, great strides have been made in capturing and rendering these faint objects. With the advent of CCD technology an astro-imager was no longer plagued by photographic film reciprocity failure and finally had a linear detector to capture these objects. The most efficient way (in terms of detector speed) at present to capture these faint objects in color is to take three filtered images and combine them into a single color image. The traditional methodology of producing color astronomical images is to take an image through a red, green, and blue filter and combine the three into a color image. This process was first developed in the 19th century and applied to film emulsions to produce the first color photographs.\n\nIn 1998 Al kelly, Richard Berry, Chuck Shaw and Ed Grafton began to experiment with using cyan, magenta and yellow filters to produce color astronomical images in stead of using the traditional red, green, and blue methodology. Since the cyan magenta and yellow are each comprised of two colors each (and thereby passing twice as much data per unit exposure time), it seemed as if this methodology would be more efficient that the traditional red, green, blue (RGB) methodology. The charts below show the typical band pass of characteristics of RGB and CMY filter sets.\n\nThe Bandpass Characteristics of RGB and CMY Filters", null, "In order to construct an image from the CMY data set, the images must first be converted in to an RGB data set. The relationship between RGB and CMY is:\n\nR = M + Y - C\n\nG = C + Y - M\n\nB = C + M - Y\n\nThe derivation of the RGBs from CMY data has the effect of degrading the S/N ratio of the RGBs derived from CMY filters. The reason is that noise adds as the root sum of squares so the RGB derivatives adds sqrt(3) to the noise. Since the signal is twice as large the S/N is degraded by a factor of sqrt(2/3). The S/N of an RGB derived from a CMY data set has 81% the S/N ratio of RGBs from RGB filters. So the CMY RGBs are noisier than the RGBs from RGB filters.\n\nIn practice many if not most imagers are producing images that are LRGB or WCMY. So what does this mean to the S/N of a LRGB compared to a WCMY image?\n\nSuppose we wanted to bring the CMY RGBs up to the S/N level of the RGBs from RGB filters. Let the R, G and B filter Signal=S, then the C, M and Y filter signal=2S and an IR blocked \"L\" or \"W\" image would have a signal of 3S. Let the noise for each image = N.\n\n1) The RGB S/N is = S/sqrt(S)\n\n2) The RGB S/N for CMY derived RGBs is = 2S/(Sqrt(sqrt(2S)^2 + sqrt(2S)^2 + sqrt(2S)^2)) = 2S/Sqrt(6S). If the exposure of the CMYs were increased by 50% then:\n\n3) Then equation 2 becomes 3S/(sqrt(sqrt(3S)^2 + sqrt(3S)^2 + sqrt(3S)^2)) = 3S/sqrt(9S) = S/sqrt(s) = to equation 1\n\nThe math shows that you would need 50% longer exposures for the CMYs to get to the same S/N level as the RGBs from RGB filters.\n\nNow lets construct the luminance (W) from these 50% longer CMY images (as I typically do) and compare them to the luminance constructed from the RGB filters and also do the same for the RGBs\n\n4) The W for the 50% longer CMY exposures = 3s+3s+3S/(sqrt(sqrt(3S)^2 + sqrt(3S)^2 + sqrt(3S)^2))= 9S/sqrt(9S)\n\n5) The L for the RGBs is = 3S/sqrt(3S)\n\n6) Dividing eq.4 by equation 5 = 1.73 = 173%\n\nThe math shows that the SNR of the constructed W from the CMYs is 173% better than the L constructed from the RGB filters. So with a 50% longer exposure for the CMY you will get 173% better luminance S/N and equal S/N on the chrominance.\n\nNow instead of equalizing the chrominance, lets equalize the luminance by taking longer RGB exposures.\n\nSince the RGB signal = S and CMY signal = 2S:\n\nThe math shows that you would need to take 100% longer exposures through the RGBs to construct an L that has the same S/N of the CMY constructed W image, but the RGBs (do some more math) would have a chrominance S/N that is 173% better than the CMY chrominance derived RGBs.\n\nNow lets go back to where we had equal chrominance S/N (this is where we had to have 50% longer CMYs) and a W S/N that was 173% better than the RGBs. In order to bring the LRGB L image up to the S/N of the W (173% better S/N) from the 50% longer CMYs, we will take a no filtered L image and add it to the RGBs to construct a L that has the same S/N as the W. This means that the chrominance and the luminance will then have the same S/N for each method.\n\n7) From equation 4 the W image has a S/N of 9S/sqrt(9S).\n\n8) The L (R+G+B+L) image would have a S/N = S+S+S+3XS/(sqrt(sqrt(S)^2 + sqrt(S)^2 + sqrt(S)^2 + sqrt(3XS)^2)). Where X is the multiplication factor applied to the no filter image so the R+G+B+L constructed luminance has the same S/N as eq.7 (i.e. the S/N of LRGB Luminance = to the S/N of the WCMY Luminance). Now if X=2 then:\n\n9) The S/N of the LRGB luminance is S+S+S+6S/(sqrt(9S) = 9S/sqrt(9S) which is equal to the W luminance in equation 7.\n\nSo now we have a WCMY from 3 images and a LRGB from 4 images with equal Luminance and chrominance. The exposure time units for the LRGB was 1+1+1 and 2(the X component in equation 8) for the no filtered image and is equal to 5 time units.\n\nThe WCMY exposure time was (1+1+1) times 1.5 (the 50% increase in exposure time that was necessary to bring the S/N of the CMY derived RGBs up to the chrominance of S/N of the RGBs from RGB filters) = 4.5 time units.\n\nThe Bottom-Line (all things being equal which they rarely are!):\n\nA WCMY with an equal S/N in both the chrominance and luminance to the S/N of a LRGB will require 3 images that are a total duration of 90% (4.5/5) as much as the total duration of the 4 images needed by the LRGB.\n\nHow does this work in real life? Below are two images, with equal exposure times, of the Globular Cluster M3.", null, "M3 was imaged with equal exposures using CMY filters and RGB filters. The CMY image shows fainter stars and a greater expanse of the globular M3. Since each of the CMY color filters pass two colors of R, G, and B; the CMY image reaches to a fainter magnitude than the RGB image with equal exposures. Taken with a C14 @ f/7. The RGB is 5min. R, 10min. G, 15 min. B. The CMY image is 10min each C, M and Y.\n\n............Ed Grafton\n\nYou are visitor number", null, "since a while back." ]

[ null, "http://www.egrafton.com/bp.gif", null, "http://www.egrafton.com/m3comptn.jpg", null, "http://www.ghgcorp.com/cgi-bin/Count.cgi", null ]

[ "# Number 36884\n\nNumber 36,884 spell 🔊, write in words: thirty-six thousand, eight hundred and eighty-four . Ordinal number 36884th is said 🔊 and write: thirty-six thousand, eight hundred and eighty-fourth. The meaning of number 36884 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 36884. What is 36884 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 36884.\n\n## What is 36,884 in other units\n\nThe decimal (Arabic) number 36884 converted to a Roman number is (X)(X)(X)(V)MDCCCLXXXIV. Roman and decimal number conversions.\n\n#### Weight conversion\n\n36884 kilograms (kg) = 81314.5 pounds (lbs)\n36884 pounds (lbs) = 16730.5 kilograms (kg)\n\n#### Length conversion\n\n36884 kilometers (km) equals to 22919 miles (mi).\n36884 miles (mi) equals to 59360 kilometers (km).\n36884 meters (m) equals to 121010 feet (ft).\n36884 feet (ft) equals 11243 meters (m).\n36884 centimeters (cm) equals to 14521.3 inches (in).\n36884 inches (in) equals to 93685.4 centimeters (cm).\n\n#### Temperature conversion\n\n36884° Fahrenheit (°F) equals to 20473.3° Celsius (°C)\n36884° Celsius (°C) equals to 66423.2° Fahrenheit (°F)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n36884 seconds equals to 10 hours, 14 minutes, 44 seconds\n36884 minutes equals to 3 weeks, 4 days, 14 hours, 44 minutes\n\n### Zip codes 36884\n\n• Zip code 36884 Petan (San Xian), Galicia, Pontevedra, Spain a map\n• Zip code 36884 Aguilares, Guanajuato, Salamanca, Mexico a map\n• Zip code 36884 El Socorro, Guanajuato, Salamanca, Mexico a map\nZip code areas 36884\n\n### Codes and images of the number 36884\n\nNumber 36884 morse code: ...-- -.... ---.. ---.. ....-\nSign language for number 36884:", null, "", null, "", null, "", null, "", null, "Number 36884 in braille:", null, "Images of the number\nImage (1) of the numberImage (2) of the number", null, "", null, "More images, other sizes, codes and colors ...\n\n#### Number 36884 infographic", null, "## Share in social networks", null, "## Mathematics of no. 36884\n\n### Multiplications\n\n#### Multiplication table of 36884\n\n36884 multiplied by two equals 73768 (36884 x 2 = 73768).\n36884 multiplied by three equals 110652 (36884 x 3 = 110652).\n36884 multiplied by four equals 147536 (36884 x 4 = 147536).\n36884 multiplied by five equals 184420 (36884 x 5 = 184420).\n36884 multiplied by six equals 221304 (36884 x 6 = 221304).\n36884 multiplied by seven equals 258188 (36884 x 7 = 258188).\n36884 multiplied by eight equals 295072 (36884 x 8 = 295072).\n36884 multiplied by nine equals 331956 (36884 x 9 = 331956).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 36884\n\nHalf of 36884 is 18442 (36884 / 2 = 18442).\nOne third of 36884 is 12294,6667 (36884 / 3 = 12294,6667 = 12294 2/3).\nOne quarter of 36884 is 9221 (36884 / 4 = 9221).\nOne fifth of 36884 is 7376,8 (36884 / 5 = 7376,8 = 7376 4/5).\nOne sixth of 36884 is 6147,3333 (36884 / 6 = 6147,3333 = 6147 1/3).\nOne seventh of 36884 is 5269,1429 (36884 / 7 = 5269,1429 = 5269 1/7).\nOne eighth of 36884 is 4610,5 (36884 / 8 = 4610,5 = 4610 1/2).\nOne ninth of 36884 is 4098,2222 (36884 / 9 = 4098,2222 = 4098 2/9).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 36884\n\n#### Is Prime?\n\nThe number 36884 is not a prime number. The closest prime numbers are 36877, 36887.\n\n#### Factorization and factors (dividers)\n\nThe prime factors of 36884 are 2 * 2 * 9221\nThe factors of 36884 are 1 , 2 , 4 , 9221 , 18442 , 36884\nTotal factors 6.\nSum of factors 64554 (27670).\n\n#### Powers\n\nThe second power of 368842 is 1.360.429.456.\nThe third power of 368843 is 50.178.080.055.104.\n\n#### Roots\n\nThe square root √36884 is 192,052076.\nThe cube root of 336884 is 33,287359.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 36884 = loge 36884 = 10,515533.\nThe logarithm to base 10 of No. log10 36884 = 4,566838.\nThe Napierian logarithm of No. log1/e 36884 = -10,515533.\n\n### Trigonometric functions\n\nThe cosine of 36884 is -0,131072.\nThe sine of 36884 is 0,991373.\nThe tangent of 36884 is -7,563557.\n\n### Properties of the number 36884\n\nIs a Friedman number: No\nIs a Fibonacci number: No\nIs a Bell number: No\nIs a palindromic number: No\nIs a pentagonal number: No\nIs a perfect number: No\n\n## Number 36884 in Computer Science\n\nCode typeCode value\n36884 Number of bytes36.0KB\nUnix timeUnix time 36884 is equal to Thursday Jan. 1, 1970, 10:14:44 a.m. GMT\nIPv4, IPv6Number 36884 internet address in dotted format v4 0.0.144.20, v6 ::9014\n36884 Decimal = 1001000000010100 Binary\n36884 Decimal = 1212121002 Ternary\n36884 Decimal = 110024 Octal\n36884 Decimal = 9014 Hexadecimal (0x9014 hex)\n36884 BASE64MzY4ODQ=\n36884 MD59fc0baa4d24c7eb0a4c171d49e174a3d\n36884 SHA224d88f31385dba97f010a0c182120b787bc2cc919093fb52e1648458d4\nMore SHA codes related to the number 36884 ...\n\nIf you know something interesting about the 36884 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 36884\n\n### Character frequency in number 36884\n\nCharacter (importance) frequency for numerology.\n Character: Frequency: 3 1 6 1 8 2 4 1\n\n### Classical numerology\n\nAccording to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 36884, the numbers 3+6+8+8+4 = 2+9 = 1+1 = 2 are added and the meaning of the number 2 is sought.\n\n## Interesting facts about the number 36884\n\n### Asteroids\n\n• (36884) 2000 SN158 is asteroid number 36884. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 9/27/2000.\n\n## Number 36,884 in other languages\n\nHow to say or write the number thirty-six thousand, eight hundred and eighty-four in Spanish, German, French and other languages. The character used as the thousands separator.\n Spanish: 🔊 (número 36.884) treinta y seis mil ochocientos ochenta y cuatro German: 🔊 (Anzahl 36.884) sechsunddreißigtausendachthundertvierundachtzig French: 🔊 (nombre 36 884) trente-six mille huit cent quatre-vingt-quatre Portuguese: 🔊 (número 36 884) trinta e seis mil, oitocentos e oitenta e quatro Chinese: 🔊 (数 36 884) 三万六千八百八十四 Arabian: 🔊 (عدد 36,884) ستة و ثلاثون ألفاً و ثمانمائةأربعة و ثمانون Czech: 🔊 (číslo 36 884) třicet šest tisíc osmset osmdesát čtyři Korean: 🔊 (번호 36,884) 삼만 육천팔백팔십사 Danish: 🔊 (nummer 36 884) seksogtredivetusinde og ottehundrede og fireogfirs Dutch: 🔊 (nummer 36 884) zesendertigduizendachthonderdvierentachtig Japanese: 🔊 (数 36,884) 三万六千八百八十四 Indonesian: 🔊 (jumlah 36.884) tiga puluh enam ribu delapan ratus delapan puluh empat Italian: 🔊 (numero 36 884) trentaseimilaottocentottantaquattro Norwegian: 🔊 (nummer 36 884) tretti-seks tusen, åtte hundre og åtti-fire Polish: 🔊 (liczba 36 884) trzydzieści sześć tysięcy osiemset osiemdziesiąt cztery Russian: 🔊 (номер 36 884) тридцать шесть тысяч восемьсот восемьдесят четыре Turkish: 🔊 (numara 36,884) otuzaltıbinsekizyüzseksendört Thai: 🔊 (จำนวน 36 884) สามหมื่นหกพันแปดร้อยแปดสิบสี่ Ukrainian: 🔊 (номер 36 884) тридцять шiсть тисяч вiсiмсот вiсiмдесят чотири Vietnamese: 🔊 (con số 36.884) ba mươi sáu nghìn tám trăm tám mươi bốn Other languages ...\n\n## News to email\n\nPrivacy Policy.\n\n## Comment\n\nIf you know something interesting about the number 36884 or any natural number (positive integer) please write us here or on facebook." ]

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[ "", null, "", null, "Question\n\n# Given that a:b::c:d, then which of the following follows Componendo and Dividendo property? d(a−b)=b(c−d) (a−b):b=(c−d):d a−ba+b=c−dc+d Both a and b\n\nSolution\n\n## The correct options are C a−ba+b=c−dc+d When a, b, c, and d are in proportion i.e. when a : b :: c : d, then ab=cd Applying componendo and dividendo, a−ba+b=c−dc+d", null, "", null, "Suggest corrections", null, "", null, "", null, "", null, "", null, "" ]

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[ "# Rewrite as a Set of Linear Factors (three x plus three) divide(x minus three) equals nine\n\nRewrite as a Set of Linear Factors (3x+3)/(x-3)=9\nMove to the left side of the equation by subtracting it from both sides.\nClarify .\nFactor out of .\nFactor out of .\nFactor out of .\nFactor out of .\nTo write as a fraction with a common denominator, First, multiply by .\nClarify terms.\nAdd numerators over the common denominator.\nClarify the numerator.\nFactor out of .\nFactor out of .\nFactor out of .\nUse the distributive law.\nMultiply by .\nSubtract from ." ]

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[ "# What are Logarithms?\n\nThe concept of logarithms was published in 1614 by John Napier (but were probably independently invented by swiss clockmaker Joost Bürgi around 1588). The idea behind logarithms is to make multiplication much faster by using the simpler techniques of addition and subtraction.\n\nIf your are given the equation a^b^=c, your can re-write the equation as log,,a,,(c)=b, read as, \"the log (logarithm) in base a of c equals b,\" to solve for b. For example, 10^4^=10,000, so mathematically speaking, log,,10,,(10,000)=4. In other words, the base 10 log of 10,000 is 4.\n\n# Base 10 Logarithms\n\nWe'll start with logarithms in base 10, because that is the number base familiar to most people.\n\n## Powers of 10\n\nThe logarithm of any power of 10 in base 10 is simple to figure out, as they are always whole numbers equal to the number of trailing zeros.\n\nlog,,10,,1=0\nlog,,10,,10=1\nlog,,10,,100=2\nlog,,10,,1,000=3\nlog,,10,,10,000=4\nlog,,10,,100,000=5\n\n...and so on.\n\n## Logs of Other Whole Numbers\n\nThe base 10 log of other numbers are expressed with trailing decimal places. Since log,,10,,1=0 and log,,10,,10=1, all the logarithms of the numbers from 1 through 10 will be between 0 and 1.\n\n### Logs of Primes Between 1 and 10\n\nAs you already know, the base 10 log of 1 is 0. Here are the logarithms to four decimal places of all the primes between 1 and 10:\n\nlog,,10,,2=.3010 (because 10^.3010^=2)\nlog,,10,,3=.4771\nlog,,10,,5=.6990\nlog,,10,,7=.8451\n\nYou can use MemoryTechnique to remember these numbers, but using logs will be simpler and more direct if you learn these numbers through repetition.\n\n(Some people prefer to memorize things the \"other way around\". Instead of memorizing the logs of the integers 1, 2, 3, 4, ... 10, they memorize the values that, after you take the log of them and multiply them by 10, give the integers 0, 1, 2, 3, ... 10. See http://en.wikipedia.org/wiki/Decibel#Reckoning ).\n\n### Logs of Other Non-Primes Between 1 and 10\n\nHere are the logs of the remaining numbers between 1 and 10 to four decimal places:\n\nlog,,10,,4=.6020\nlog,,10,,6=.7781\nlog,,10,,8=.9031\nlog,,10,,9=.9542\n\nThese logs, you don't have to remember. Why not? Because you can use the prime base 10 logs you learned in the previous section to determine these!\n\nIf you want to know the logarithms for any number from 1 to 10, simply break it down into it's prime factors, and add the logs for those prime factors together. Mathematically speaking, the formula looks like this:\n\nlog,,10,,(x*y)=log,,10,,(x) + log,,10,,(y)\n\nTo figure out these log for 4, the process would look something like this:\n\nlog,,10,,4=log,,10,,(2*2)=log,,10,,2 + log,,10,,2=.3010 + .3010=.6020\n\nNotice that you solved a multiplication problem by addition? You've just had you first taste of the power of logarithms!\n\nHere is how the other non-prime logarithms break down:\n\nlog,,10,,6=log,,10,,(2*3)=log,,10,,2 + log,,10,,3=.3010 + .4771=.7781\nlog,,10,,8=log,,10,,(2*2*2)=log,,10,,2 + log,,10,,2 + log,,10,,2=.3010 + .3010 + .3010=.9030\nlog,,10,,9=log,,10,,(3*3)=log,,10,,3 + log,,10,,3=.4771 + .4771=.9542\n\nNote that log,,10,,8 isn't exact, but it is only off by 1/10,000th!\n\nPractice quizzing yourself on the base 10 logs of all the numbers from 1 to 10 before moving on to the next sections.\n\nAlso, note that the logs don't progess in a simple linear scale. As they get closer to 10, the amount the logs increase by gets smaller and smaller. This, not surprisingly, is referred to as a \"logarithmic\" scale.\n\n## Logs of Other Numbers Greater Than 10 With Trailing Zeros\n\nJust as all the base 10 logs of the numbers 1 through 10 fall between 0 and 1, the logs of numbers between 10 and 100 would all fall between 1 and 2. The base 10 logs of numbers between 100 and 1000 would fall between 2 and 3, and so on.\n\nThe process of finding the logs for any number that ends in trailing zeros adds just one step to the process you've already learned. Start by writing the number in exponential form (a * 10^b^). Write down \"b\" as the first number. Next, take the leftmost digit (the \"a\" in a * 10^b^), and write down its base 10 log. These two numbers together will give a good approximation of the base 10 log for that number.\n\nAs an example, let's find the log of 20. 20=2 * 10^1^, so we start by writing \"1\". The leftmost digit is a 2, and the base 10 log of 2, as we learned earlier, is .3010, so we write \".3010\" next to the \"1\" we wrote down previously, giving us a result of 1.3010.\n\nHow about a larger number like 300,000? 300,000=3 * 10^5^, so we write \"5\". The leftmost digit is a 3, so we add \".4771\" to the 5, so we know that log,,10,,300,000=5.4771!\n\nWhat does log,,10,,8,000 equal? It's \"3\" (8 * 10^3^) plus \".9030\" for the leftmost 8 (log,,10,,2 + log,,10,,2 + log,,10,,2), giving us log,,10,,8,000=3.9030!\n\n## Logs of Other Numbers Greater Than 10 Without Trailing Zeros\n\nFor most numbers, you will only be able to approximate the log. With what you've learned already, however, you'll be able to create an amazing approximation of the log, however.\n\nAs a starting example, we'll try an approximate the log of 2,087. We start by converting the number into its exponential form, which gives us 2.087 * 10^3^.\n\nNext, we need to round the number to the nearest 10th, so we have an easier equation to work with. Rounding our example up, we get 2.1 * 10^3^. We already known from this equation that the first number we need is 3.\n\nIn this example, we know that the log,,10,,2=.3010 and log,,10,,3=.4771, so how do figure out log,,10,,2.1? We can get a good estimate by assuming that the log,,10,,2.1 is 1/10 of the distance between .3010 and .4771. The distance between these numbers is about .17, so 1/10 of the distance is .017. Adding .017 to .3010 gives us .3180.\n\nAdding this to our previous result, gives 3.3180 as an approximate base 10 log for 2,087. Checking with a calculator, we find that the actual answer for log,,10,,2,087 is 3.3195. Our mental approximation was only off by little more than 1/1000th!\n\nTo help reinforce the process, let's try a trickier challenge, such as log,,10,,85,412. 85,412 translates roughly to 8.5 * 10^4^.\n\nApproximating 5/10 of the distance between .9030 (log,,10,,8) and .9542 (log,,10,,9), we get .929, so we can approximate log,,10,,85,412 as 4.9286. A calculator will show that log,,10,,85,412 is 4.9315, giving us a difference of little more than 5/1000! That's not a bad approximation for some brief mental work.\n\nThe more you practice with these logs, the better a feel you'll get for better approximation, too. Notice too, that while the logs run on a logarithmic scale, approximating with a linear scale still helps get quite close." ]

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[ "# Class 11 NCERT Solutions- Chapter 10 Straight Lines – Exercise 10.2 | Set 1\n\n• Last Updated : 05 Apr, 2021\n\n### Question 1. Write the equations for the x-and y-axes.\n\nSolution:\n\nThe y-coordinate of every point on x-axis is 0 and the x-coordinate of every point on y-axis is 0\n\nAttention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12.\n\nThe students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.\n\nSo, the equation of x-axis is y = 0 and the equation of y-axis is y = 0.\n\n### Question 2. Passing through the point (– 4, 3) with slope 1/2.\n\nSolution:\n\nGiven that point p(x1, y1) is (-4, 3) and slope m = 1/2\n\nEquation of Line can be derived by formula y – y1 = m (x – x1),\n\nWhere m is slope of line and (x1, y1) is co-ordinate of p from which line passes.\n\ny – 3 = 1/2(x – (-4))\n\ny – 3 = 1/2 (x + 4)\n\n2(y – 3) = x + 4\n\n2y – 6 = x + 4\n\nx + 4 – (2y – 6) = 0\n\nx + 4 – 2y + 6 = 0\n\nx – 2y + 10 = 0\n\nSo, the equation of the line is x – 2y + 10 = 0.\n\n### Question 3. Passing through (0, 0) with slope m.\n\nSolution:\n\nGiven that Point p(x1, y1) is (0, 0) and slope is m.\n\nSo, the equation of Line can be derived by formula y – y1 = m (x – x1),\n\nWhere m is slope of line and (x1, y1) is co-ordinate from which line passes.\n\nSo, y – 0 = m (x – 0)\n\ny = mx\n\ny – mx = 0\n\nSo, the equation of the line is y – mx = 0.\n\n### Question 4. Passing through (2, 2√3) and inclined with the x-axis at an angle of 75o.\n\nSolution:\n\nGiven that point p(x1, y1) is (2, 2√3) and θ = 75°\n\nSo, the equation of line is (y – y1) = m (x – x1)\n\nwhere, m = slope of line = tan θ and (x1, y1) are the points through which line passes\n\nSo, m = tan 75°\n\nNow, finding tan 75° using below formula:\n\ntan(A + B) = (tanA + tanB)/(1 – tanA.tanB)\n\ntan 75° = tan(45° + 30°)\n\ntan 75° = (tan 45° + tan 30°)/(1 – tan 30°.tan 45°)\n\ntan 75° = (1 + 1/√3)/(1 – 1/√3)\n\ntan 75° = (√3 + 1)/(√3 – 1)\n\nBy rationalizing, we get\n\ntan 75° = 2 + √3\n\nEquation of Line will be,\n\n(y – 2√3) = (2 + √3)(x – 2)\n\ny – 2√3 = 2 x – 4 + √3x – 2 √3\n\ny = 2 x – 4 + √3x\n\n(2 + √3)x – y – 4 = 0\n\nSo, the equation of the line is (2 + √3)x – y – 4 = 0.\n\n### Question 5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.\n\nSolution:\n\nGiven that Slope m = –2, x-intercept = 3,\n\nthat means Line is passing from point p(x1, y1) that is (3, 0)\n\nEquation of line will be\n\ny – 0 = (–2) × (x − 3).\n\ny = (–2) × (x + 3)\n\ny = –2x – 6\n\n2x + y + 6 = 0\n\nSo, the equation of the line is 2x + y + 6 = 0.\n\n### Question 6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with the positive direction of the x-axis.\n\nSolution:\n\nGiven that θ = 30° then slope(m) will be = tan θ\n\nm = tan30° = (1/√3)\n\nAs Y intercept is 2 that means line is passing from (0, 2) point p(x1, y1) = (0, 2)\n\nEquation of Line will be\n\ny – 2 = (1/√3)x\n\ny = (1/√3)x + 2\n\ny = (x + 2√3) / √3\n\n√3 y = x + 2√3\n\nx – √3 y + 2√3 = 0\n\nSo, the equation of the line is x – √3 y + 2√3 = 0.\n\n### Question 7. Passing through the points (–1, 1) and (2, – 4).\n\nSolution:\n\nGiven that point p1(x1, y1) is (–1, 1) and point p2(x2, y2) is (2, –4),\n\nIt is mention in question that line passes from p1 and p2.\n\nThat means, the Slope(m) of line will be (y2 – y1)/(x2 – x1)\n\nm = (–4 – 1)/(2 – (–1))\n\nm = –5/3\n\nEquation of Line will be\n\n(y – y1) = m(x – x1)\n\ny – 1 = –5/3 (x + 1)\n\n3 (y – 1) = (–5)(x + 1)\n\n3y – 3 = –5x – 5\n\n3y – 3 + 5x + 5 = 0\n\n5x + 3y + 2 = 0\n\nSo, the equation of the line is 5x + 3y + 2 = 0.\n\n### Question 8. Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30°.\n\nSolution:\n\nGiven that p = 5 and ω = 30°\n\nWe know that the equation of the line having normal distance p from the origin and\n\nangle ω which the normal makes with the positive direction of x-axis is given by x cos ω + y sin ω = p.\n\nOn substituting the values in the equation, we get\n\nx cos30° + y sin30° = 5\n\nx(√3 / 2) + y(1/2) = 5\n\n√3 x + y = 5(2) = 10\n\n√3 x + y – 10 = 0\n\nSo, the equation of the line is √3 x + y – 10 = 0.\n\n### Question 9. The vertices of ΔPQR are P (2, 1), Q (–2, 3), and R (4, 5). Find the equation of the median through the vertex R.\n\nSolution:\n\nGiven Vertices of ΔPQR that are P (2, 1), Q (–2, 3) and R (4, 5)\n\nLet RS be the median of vertex R => S is a midpoint of PQ.\n\nAs S is midpoint of PQ => S = (P + Q)/2\n\nS = (2 – 2, 1 + 3)/2\n\nS = (0, 2)\n\nEquation of the line passing through the points (x1, y1) and (x2, y2) is given by", null, "y – 5 = –3/ –4(x – 4)\n\n(–4)(y – 5) = (–3)(x – 4)\n\n–4y + 20 = –3x + 12\n\n–4y + 20 + 3x – 12 = 0\n\n3x – 4y + 8 = 0\n\nSo, the equation of median through the vertex R is 3x – 4y + 8 = 0.\n\n### Question 10. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).\n\nSolution:\n\nGiven that Points are (2, 5) and (-3, 6).\n\nSo, the slope, m1 = (y2 – y1)/(x2 – x1)\n\n= (6 – 5)/(–3 – 2)\n\n= 1/–5 = –1/5\n\nAs we know that two non-vertical lines are perpendicular to each other\n\nif their slopes are negative reciprocals of each other.\n\nThen, m = (–1/m1)\n\n= –1/(–1/5)\n\n= 5\n\nAs we know that the point p(x, y) lies on the line with\n\nslope m through the fixed point (x1, y1),\n\nIf its coordinates satisfy the equation y – y1 = m (x – x1)\n\nThen, y – 5 = 5(x – (–3))\n\ny – 5 = 5x + 15\n\n5x + 15 – y + 5 = 0\n\n5x – y + 20 = 0\n\nSo, the equation of the line is 5x – y + 20 = 0\n\n### Chapter 10 Straight Lines – Exercise 10.2 | Set 2\n\nMy Personal Notes arrow_drop_up" ]

[ null, "https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-9b6290f52b029688a1a79e051bc3141d_l3.png", null ]

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\"+(e.bdUrl||n.getPageUrl()))})},c=function(){window.print()},h=function(){window._bd_share_main.F.use(\"trans/trans_bdxc\",function(e){e&&e.run()})},p=function(e){window._bd_share_main.F.use(\"trans/trans_bdysc\",function(t){t&&t.run(e)})},d=function(e){window._bd_share_main.F.use(\"trans/trans_weixin\",function(t){t&&t.run(e)})},v=function(e){o(e)};t.run=v,i()});\n\n【字体： 】  【打印文章 上一篇：已经没有了 下一篇：山东省政府旧址:一座红色庄园的百年沉浮(组图)\n" ]

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[ "Medical Imaging Interaction Toolkit  2018.4.99-cfdaacfc Medical Imaging Interaction Toolkit\nImage Arithmetrics\n\nAllows to perform basic arithmetic operations with all images. By default, the resulting image will be of the same type as the first input image. This also means that all operations will be perfomed using the same datatype as the input image. This can lead to rounding errors if an integer data type is used. It is possible to avoid this problem by creating a double-based image as result image by selecting the corresponding option.\n\nThere are three different types of operations:\n\n• Single Image Operations: These are operations that are performed on each voxel value and do not need any additional parameter. Typical operations are calculating the absolute value of an image, or calculating an trigometric value.\n• Single Image and Double Value Operation: These are the basic mathematical operations (add, subtract, multiply, and dividive).\n• Two Images Operations: Allows to perfom basic operations between two images.\n\nAllows to perfom a arithmetic operation to each voxel of an image independently. Be aware that some operations expect a specific input range. This range is usually not tested for, and it might cause an error if the range if values outside of the expceted range occure. Also, some operations are rather expensive with respect to the calculation time and take some time to finish.\n\nPerforms a basic mathematical operation (add, subtract, multiply, and divide) between an image and a floating point value. The operation is execuded for each voxel independently. There are two options if the order of the input is relevant (i.e. for subtraction and division) to enable both ways.\n\nThis allows to perfom mathematical operations between two images. The second image needs to be specified within this panel. Both images must match in size, it is also assumed (but not checked) that both images share the same geometry. This means that the corresponding voxels of both images share the same geometrical space.\n\nThe first image, that would also be used for other operations, is used as \"Image\", while the image specified in the panel is used as \"Second Image\". In general, the first image is also used as first parameter in the mathematical operation." ]

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[ "", null, "A common question when viewing a microscopy image on a large screen is: What is the total magnification I am viewing on my monitor?\" In order to correctly calculate the answer to this question, we must know the answer to the following four questions.\n\n1. What is the microscopes objective's magnification?\n2. What is the c-mount adapter's magnification?\n3. What is the diagonal measurement of the monitor in mm?\n4. What is the size of the camera sensor in mm?", null, "The microscope objective lens magnification is printed on the side of the lens as shown in the image at left (1). If you are using a stereo microscope the objective lens value would be printed on the zoom knob or on the objective turret that is turned to change the microscope magnification.", null, "The microscope c-mount adapter (2) is the piece used to connect the microscope and the microscope camera. The c-mount adapter will have a number printed on the side of it. In the image at right, the c-mount adapter has a 0.5x lens built into it.", null, "The next number we need to determine is the diagonal measurement of the monitor in mm (3). Typically a monitor's measurement is provided in inches. We will just simply convert the inches measurement to mm. For example, if you are using a 19\" monitor, multiply this number by 25.4 to convert it to mm. 19\" x 25.4 = 482.6mm diagonal monitor measurement.", null, "The final number we will need in order to determine the on-screen magnification is the size of the camera sensor in mm (4). The size of the sensor differs slightly from the size of the chip, so the best way to find this number is to use the chart located below. The microscope camera sensor sizes are listed at the top of the chart, and the corresponding diagonal sensor size in mm is shown below.\n\n Sensor 1/4\" 1/3.2\" 1/3\" 1/2.5\" 1/2\" 1/1.8\" 2/3\" Diagonal (mm) 4.00 5.68 6.00 7.18 8.00 8.93 11.0\n\nThe formula for calculating microscope on-screen magnification is:\n\nTotal Magnification = Optical Magnification x Digital Magnification\n\nHow is optical magnification calculated? This one is fairly simple.\n\nOptical Magnification = Objective Magnification x C-Mount Adapter Magnification.\n\nIf we use our examples shown above from (1) and (2), we would calculate 4 x 0.5x = 2.\n\nWhat about digital magnification, how do we calculate this?\n\nDigital Magnification = Screen Size / Sensor Size.\n\nIf we are using the 19\" monitor we mentioned earlier, to convert this to mm we multiple 19 x 25.4 = 482.6mm screen size.  For this example, let's say we are using a microscope camera with a 1/2\" camera sensor in it. Based on the chart above we would calculate digital magnification by using 482.6mm / 8.00 = 60.325.\n\nNow we can find the total on-screen magnification by multiplying optical magnification x digital magnification. In our example 2 x 60.325 = 120.65x on-screen magnification.\n\nIf you have questions regarding calculating on-screen microscope magnification contact Microscope World and we will be happy to help." ]

[ null, "https://www.microscopeworld.com/images/product/medium/3375.jpg", null, "https://www.microscopeworld.com/Images/objective-magnification.jpg", null, "https://www.microscopeworld.com/Images/cmount-magnification.jpg", null, "https://www.microscopeworld.com/Images/monitor-diagonal.jpg", null, "https://www.microscopeworld.com/Images/camera-chip-diagonal.jpg", null ]

[ "# Controlling Program Flow in Java\n\n1. Using Java operators\n2. Execution control\n3. Summary\n4. Exercises\n• Print\nThis excerpt from Thinking In Java shows you how to use operators and execution control statements. It covers precedence, assignment, regular expressions, relational and logical operators, and short circuiting. You'll learn to use Java's execution control statements, including loops and switches.\nThis chapter is from the book\n\n### This chapter is from the book \n\nLike a sentient creature, a program must manipulate its world and make choices during execution.\n\nIn Java you manipulate data using operators, and you make choices with execution control statements. Java was inherited from C++, so most of these statements and operators will be familiar to C and C++ programmers. Java has also added some improvements and simplifications.\n\nIf you find yourself floundering a bit in this chapter, make sure you go through the multimedia CD ROM bound into this book: Foundations for Java. It contains audio lectures, slides, exercises, and solutions specifically designed to bring you up to speed with the fundamentals necessary to learn Java.\n\n## Using Java operators\n\nAn operator takes one or more arguments and produces a new value. The arguments are in a different form than ordinary method calls, but the effect is the same. Addition (+), subtraction and unary minus (-), multiplication (*), division (/), and assignment (=) all work much the same in any programming language.\n\nAll operators produce a value from their operands. In addition, an operator can change the value of an operand. This is called a side effect. The most common use for operators that modify their operands is to generate the side effect, but you should keep in mind that the value produced is available for your use, just as in operators without side effects.\n\nAlmost all operators work only with primitives. The exceptions are ‘=’, ‘==’ and ‘!=’, which work with all objects (and are a point of confusion for objects). In addition, the String class supports ‘+’ and ‘+=’.\n\n### Precedence\n\nOperator precedence defines how an expression evaluates when several operators are present. Java has specific rules that determine the order of evaluation. The easiest one to remember is that multiplication and division happen before addition and subtraction. Programmers often forget the other precedence rules, so you should use parentheses to make the order of evaluation explicit. For example:\n\n`a = x + y - 2/2 + z; `\n\nhas a very different meaning from the same statement with a particular grouping of parentheses:\n\n`a = x + (y - 2)/(2 + z); `\n\n### Assignment\n\nAssignment is performed with the operator =. It means “take the value of the right-hand side (often called the rvalue) and copy it into the left-hand side (often called the lvalue).” An rvalue is any constant, variable, or expression that can produce a value, but an lvalue must be a distinct, named variable. (That is, there must be a physical space to store the value.) For instance, you can assign a constant value to a variable:\n\n`a = 4; `\n\nbut you cannot assign anything to constant value—it cannot be an lvalue. (You can’t say 4 = a;.)\n\nAssignment of primitives is quite straightforward. Since the primitive holds the actual value and not a reference to an object, when you assign primitives, you copy the contents from one place to another. For example, if you say a = b for primitives, then the contents of b are copied into a. If you then go on to modify a, b is naturally unaffected by this modification. As a programmer, this is what you’ve come to expect for most situations.\n\nWhen you assign objects, however, things change. Whenever you manipulate an object, what you’re manipulating is the reference, so when you assign “from one object to another,” you’re actually copying a reference from one place to another. This means that if you say c = d for objects, you end up with both c and d pointing to the object that, originally, only d pointed to. Here’s an example that demonstrates this behavior:\n\n```//: c03:Assignment.java\n// Assignment with objects is a bit tricky.\nimport com.bruceeckel.simpletest.*;\n\nclass Number {\nint i;\n}\n\npublic class Assignment {\nstatic Test monitor = new Test();\npublic static void main(String[] args) {\nNumber n1 = new Number();\nNumber n2 = new Number();\nn1.i = 9;\nn2.i = 47;\nSystem.out.println(\"1: n1.i: \" + n1.i +\n\", n2.i: \" + n2.i);\nn1 = n2;\nSystem.out.println(\"2: n1.i: \" + n1.i +\n\", n2.i: \" + n2.i);\nn1.i = 27;\nSystem.out.println(\"3: n1.i: \" + n1.i +\n\", n2.i: \" + n2.i);\nmonitor.expect(new String[] {\n\"1: n1.i: 9, n2.i: 47\",\n\"2: n1.i: 47, n2.i: 47\",\n\"3: n1.i: 27, n2.i: 27\"\n});\n}\n} ///:~ ```\n\nFirst, notice that something new has been added. The line:\n\n`import com.bruceeckel.simpletest.*; `\n\nimports the “simpletest” library that has been created to test the code in this book, and is explained in Chapter 15. At the beginning of the Assignment class, you see the line:\n\n`static Test monitor = new Test(); `\n\nThis creates an instance of the simpletest class Test, called monitor. Finally, at the end of main( ), you see the statement:\n\n```monitor.expect(new String[] {\n\"1: n1.i: 9, n2.i: 47\",\n\"2: n1.i: 47, n2.i: 47\",\n\"3: n1.i: 27, n2.i: 27\"\n}); ```\n\nThis is the expected output of the program, expressed as an array of String objects. When the program is run, it not only prints out the output, but it compares it to this array to verify that the array is correct. Thus, when you see a program in this book that uses simpletest, you will also see an expect( ) call that will show you what the output of the program is. This way, you see validated output from the program.\n\nThe Number class is simple, and two instances of it (n1 and n2) are created within main( ). The i value within each Number is given a different value, and then n2 is assigned to n1, and n1 is changed. In many programming languages you would expect n1 and n2 to be independent at all times, but because you’ve assigned a reference, you’ll see the output in the expect( ) statement. Changing the n1 object appears to change the n2 object as well! This is because both n1 and n2 contain the same reference, which is pointing to the same object. (The original reference that was in n1, that pointed to the object holding a value of 9, was overwritten during the assignment and effectively lost; its object will be cleaned up by the garbage collector.)\n\nThis phenomenon is often called aliasing, and it’s a fundamental way that Java works with objects. But what if you don’t want aliasing to occur in this case? You could forego the assignment and say:\n\n`n1.i = n2.i; `\n\nThis retains the two separate objects instead of tossing one and tying n1 and n2 to the same object, but you’ll soon realize that manipulating the fields within objects is messy and goes against good object-oriented design principles. This is a nontrivial topic, so it is left for Appendix A, which is devoted to aliasing. In the meantime, you should keep in mind that assignment for objects can add surprises.\n\n#### Aliasing during method calls\n\nAliasing will also occur when you pass an object into a method:\n\n```//: c03:PassObject.java\n// Passing objects to methods may not be what\n// you're used to.\nimport com.bruceeckel.simpletest.*;\n\nclass Letter {\nchar c;\n}\n\npublic class PassObject {\nstatic Test monitor = new Test();\nstatic void f(Letter y) {\ny.c = 'z';\n}\npublic static void main(String[] args) {\nLetter x = new Letter();\nx.c = 'a';\nSystem.out.println(\"1: x.c: \" + x.c);\nf(x);\nSystem.out.println(\"2: x.c: \" + x.c);\nmonitor.expect(new String[] {\n\"1: x.c: a\",\n\"2: x.c: z\"\n});\n}\n} ///:~ ```\n\nIn many programming languages, the method f( ) would appear to be making a copy of its argument Letter y inside the scope of the method. But once again a reference is being passed, so the line\n\n`y.c = 'z'; `\n\nis actually changing the object outside of f( ). The output in the expect( ) statement shows this.\n\nAliasing and its solution is a complex issue and, although you must wait until Appendix A for all the answers, you should be aware of it at this point so you can watch for pitfalls.\n\n### Mathematical operators\n\nThe basic mathematical operators are the same as the ones available in most programming languages: addition (+), subtraction (-), division (/), multiplication (*) and modulus (%, which produces the remainder from integer division). Integer division truncates, rather than rounds, the result.\n\nJava also uses a shorthand notation to perform an operation and an assignment at the same time. This is denoted by an operator followed by an equal sign, and is consistent with all the operators in the language (whenever it makes sense). For example, to add 4 to the variable x and assign the result to x, use: x += 4.\n\nThis example shows the use of the mathematical operators:\n\n```//: c03:MathOps.java\n// Demonstrates the mathematical operators.\nimport com.bruceeckel.simpletest.*;\nimport java.util.*;\n\npublic class MathOps {\nstatic Test monitor = new Test();\n// Shorthand to print a string and an int:\nstatic void printInt(String s, int i) {\nSystem.out.println(s + \" = \" + i);\n}\n// Shorthand to print a string and a float:\nstatic void printFloat(String s, float f) {\nSystem.out.println(s + \" = \" + f);\n}\npublic static void main(String[] args) {\n// Create a random number generator,\n// seeds with current time by default:\nRandom rand = new Random();\nint i, j, k;\n// Choose value from 1 to 100:\nj = rand.nextInt(100) + 1;\nk = rand.nextInt(100) + 1;\nprintInt(\"j\", j); printInt(\"k\", k);\ni = j + k; printInt(\"j + k\", i);\ni = j - k; printInt(\"j - k\", i);\ni = k / j; printInt(\"k / j\", i);\ni = k * j; printInt(\"k * j\", i);\ni = k % j; printInt(\"k % j\", i);\nj %= k; printInt(\"j %= k\", j);\n// Floating-point number tests:\nfloat u,v,w; // applies to doubles, too\nv = rand.nextFloat();\nw = rand.nextFloat();\nprintFloat(\"v\", v); printFloat(\"w\", w);\nu = v + w; printFloat(\"v + w\", u);\nu = v - w; printFloat(\"v - w\", u);\nu = v * w; printFloat(\"v * w\", u);\nu = v / w; printFloat(\"v / w\", u);\n// the following also works for\n// char, byte, short, int, long,\n// and double:\nu += v; printFloat(\"u += v\", u);\nu -= v; printFloat(\"u -= v\", u);\nu *= v; printFloat(\"u *= v\", u);\nu /= v; printFloat(\"u /= v\", u);\nmonitor.expect(new String[] {\n\"%% j = -?\\\\d+\",\n\"%% k = -?\\\\d+\",\n\"%% j \\\\+ k = -?\\\\d+\",\n\"%% j - k = -?\\\\d+\",\n\"%% k / j = -?\\\\d+\",\n\"%% k \\\\* j = -?\\\\d+\",\n\"%% k % j = -?\\\\d+\",\n\"%% j %= k = -?\\\\d+\",\n\"%% v = -?\\\\d+\\\\.\\\\d+(E-?\\\\d)?\",\n\"%% w = -?\\\\d+\\\\.\\\\d+(E-?\\\\d)?\",\n\"%% v \\\\+ w = -?\\\\d+\\\\.\\\\d+(E-?\\\\d)??\",\n\"%% v - w = -?\\\\d+\\\\.\\\\d+(E-?\\\\d)??\",\n\"%% v \\\\* w = -?\\\\d+\\\\.\\\\d+(E-?\\\\d)??\",\n\"%% v / w = -?\\\\d+\\\\.\\\\d+(E-?\\\\d)??\",\n\"%% u \\\\+= v = -?\\\\d+\\\\.\\\\d+(E-?\\\\d)??\",\n\"%% u -= v = -?\\\\d+\\\\.\\\\d+(E-?\\\\d)??\",\n\"%% u \\\\*= v = -?\\\\d+\\\\.\\\\d+(E-?\\\\d)??\",\n\"%% u /= v = -?\\\\d+\\\\.\\\\d+(E-?\\\\d)??\"\n});\n}\n} ///:~ ```\n\nThe first thing you will see are some shorthand methods for printing: the printInt( ) prints a String followed by an int and the printFloat( ) prints a String followed by a float.\n\nTo generate numbers, the program first creates a Random object. Because no arguments are passed during creation, Java uses the current time as a seed for the random number generator. The program generates a number of different types of random numbers with the Random object simply by calling the methods: nextInt( ) and nextFloat( ) (you can also call nextLong( ) or nextDouble( )).\n\nThe modulus operator, when used with the result of the random number generator, limits the result to an upper bound of the operand minus 1 (99 in this case).\n\n#### Regular expressions\n\nSince random numbers are used to generate the output for this program, the expect( ) statement can’t just show literal output as it did before, since the output will vary from one run to the next. To solve this problem, regular expressions, a new feature introduced in Java JDK 1.4 (but an old feature in languages like Perl and Python) will be used inside the expect( ) statement. Although coverage of this intensely powerful tool doesn’t occur until Chapter 12, to understand these statements you’ll need an introduction to regular expressions. Here, you’ll learn just enough to read the expect( ) statements, but if you want a full description, look up java.util.regex.Pattern in the downloadable JDK documentation.\n\nA regular expression is a way to describe strings in general terms, so that you can say: “If a string has these things in it, then it matches what I’m looking for.” For example, to say that a number might or might not be preceded by a minus sign, you put in the minus sign followed by a question mark, like this:\n\n`-? `\n\nTo describe an integer, you say that it’s one or more digits. In regular expressions, a digit is ‘\\d’, but in a Java String you have to “escape” the backslash by putting in a second backslash: ‘\\\\d’. To indicate “one or more of the preceding expression” in regular expressions, you use the ‘+’. So to say “possibly a minus sign, followed by one or more digits,” you write:\n\n`-?\\\\d+ `\n\nWhich you can see in the first lines of the expect( ) statement in the preceding code.\n\nOne thing that is not part of the regular expression syntax is the ‘%%’ (note the space included for readability) at the beginning of the lines in the expect( ) statement. This is a flag used by simpletest to indicate that the rest of the line is a regular expression. So you won’t see it in normal regular expressions, only in simpletest expect( ) statements.\n\nAny other characters that are not special characters to regular expression searches are treated as exact matches. So in the first line:\n\n`%% j = -?\\\\d+ `\n\nThe ‘j = ’ is matched exactly. However, in the third line, the ‘+’ in ‘j + k’ must be escaped because it is a special regular expression character, as is ‘*’. The rest of the lines should be understandable from this introduction. Later in the book, when additional features of regular expressions are used inside expect( ) statements, they will be explained.\n\n#### Unary minus and plus operators\n\nThe unary minus (-)and unary plus (+) are the same operators as binary minus and plus. The compiler figures out which use is intended by the way you write the expression. For instance, the statement\n\n`x = -a; `\n\nhas an obvious meaning. The compiler is able to figure out:\n\n`x = a * -b; `\n\nbut the reader might get confused, so it is clearer to say:\n\n`x = a * (-b); `\n\nUnary minus inverts the sign on the data. Unary plus provides symmetry with unary minus, although it doesn’t have any effect.\n\n### Auto increment and decrement\n\nJava, like C, is full of shortcuts. Shortcuts can make code much easier to type, and either easier or harder to read.\n\nTwo of the nicer shortcuts are the increment and decrement operators (often referred to as the auto-increment and auto-decrement operators). The decrement operator is -- and means “decrease by one unit.” The increment operator is ++ and means “increase by one unit.” If a is an int, for example, the expression ++a is equivalent to (a = a + 1). Increment and decrement operators not only modify the variable, but also produce the value of the variable as a result.\n\nThere are two versions of each type of operator, often called the prefix and postfix versions. Pre-increment means the ++ operator appears before the variable or expression, and post-increment means the ++ operator appears after the variable or expression. Similarly, pre-decrement means the -- operator appears before the variable or expression, and post-decrement means the -- operator appears after the variable or expression. For pre-increment and pre-decrement, (i.e., ++a or --a), the operation is performed and the value is produced. For post-increment and post-decrement (i.e. a++ or a--), the value is produced, then the operation is performed. As an example:\n\n```//: c03:AutoInc.java\n// Demonstrates the ++ and -- operators.\nimport com.bruceeckel.simpletest.*;\n\npublic class AutoInc {\nstatic Test monitor = new Test();\npublic static void main(String[] args) {\nint i = 1;\nSystem.out.println(\"i : \" + i);\nSystem.out.println(\"++i : \" + ++i); // Pre-increment\nSystem.out.println(\"i++ : \" + i++); // Post-increment\nSystem.out.println(\"i : \" + i);\nSystem.out.println(\"--i : \" + --i); // Pre-decrement\nSystem.out.println(\"i-- : \" + i--); // Post-decrement\nSystem.out.println(\"i : \" + i);\nmonitor.expect(new String[] {\n\"i : 1\",\n\"++i : 2\",\n\"i++ : 2\",\n\"i : 3\",\n\"--i : 2\",\n\"i-- : 2\",\n\"i : 1\"\n});\n}\n} ///:~ ```\n\nYou can see that for the prefix form, you get the value after the operation has been performed, but with the postfix form, you get the value before the operation is performed. These are the only operators (other than those involving assignment) that have side effects. (That is, they change the operand rather than using just its value.)\n\nThe increment operator is one explanation for the name C++, implying “one step beyond C.” In an early Java speech, Bill Joy (one of the Java creators), said that “Java=C++--” (C plus plus minus minus), suggesting that Java is C++ with the unnecessary hard parts removed, and therefore a much simpler language. As you progress in this book, you’ll see that many parts are simpler, and yet Java isn’t that much easier than C++.\n\n### Relational operators\n\nRelational operators generate a boolean result. They evaluate the relationship between the values of the operands. A relational expression produces true if the relationship is true, and false if the relationship is untrue. The relational operators are less than (<), greater than (>), less than or equal to (<=), greater than or equal to (>=), equivalent (==) and not equivalent (!=). Equivalence and nonequivalence work with all built-in data types, but the other comparisons won’t work with type boolean.\n\n#### Testing object equivalence\n\nThe relational operators == and != also work with all objects, but their meaning often confuses the first-time Java programmer. Here’s an example:\n\n```//: c03:Equivalence.java\nimport com.bruceeckel.simpletest.*;\n\npublic class Equivalence {\nstatic Test monitor = new Test();\npublic static void main(String[] args) {\nInteger n1 = new Integer(47);\nInteger n2 = new Integer(47);\nSystem.out.println(n1 == n2);\nSystem.out.println(n1 != n2);\nmonitor.expect(new String[] {\n\"false\",\n\"true\"\n});\n}\n} ///:~ ```\n\nThe expression System.out.println(n1 == n2) will print the result of the boolean comparison within it. Surely the output should be true and then false, since both Integer objects are the same. But while the contents of the objects are the same, the references are not the same and the operators == and != compare object references. So the output is actually false and then true. Naturally, this surprises people at first.\n\nWhat if you want to compare the actual contents of an object for equivalence? You must use the special method equals( ) that exists for all objects (not primitives, which work fine with == and !=). Here’s how it’s used:\n\n```//: c03:EqualsMethod.java\nimport com.bruceeckel.simpletest.*;\n\npublic class EqualsMethod {\nstatic Test monitor = new Test();\npublic static void main(String[] args) {\nInteger n1 = new Integer(47);\nInteger n2 = new Integer(47);\nSystem.out.println(n1.equals(n2));\nmonitor.expect(new String[] {\n\"true\"\n});\n}\n} ///:~ ```\n\nThe result will be true, as you would expect. Ah, but it’s not as simple as that. If you create your own class, like this:\n\n```//: c03:EqualsMethod2.java\nimport com.bruceeckel.simpletest.*;\n\nclass Value {\nint i;\n}\n\npublic class EqualsMethod2 {\nstatic Test monitor = new Test();\npublic static void main(String[] args) {\nValue v1 = new Value();\nValue v2 = new Value();\nv1.i = v2.i = 100;\nSystem.out.println(v1.equals(v2));\nmonitor.expect(new String[] {\n\"false\"\n});\n}\n} ///:~ ```\n\nyou’re back to square one: the result is false. This is because the default behavior of equals( ) is to compare references. So unless you override equals( ) in your new class you won’t get the desired behavior.\n\nUnfortunately, you won’t learn about overriding until Chapter 7 and about the proper way to define equals( ) until Chapter 11, but being aware of the way equals( ) behaves might save you some grief in the meantime.\n\nMost of the Java library classes implement equals( ) so that it compares the contents of objects instead of their references.\n\n### Logical operators\n\nEach of the logical operators AND (&&), OR (||) and NOT (!) produces a boolean value of true or false based on the logical relationship of its arguments. This example uses the relational and logical operators:\n\n```//: c03:Bool.java\n// Relational and logical operators.\nimport com.bruceeckel.simpletest.*;\nimport java.util.*;\n\npublic class Bool {\nstatic Test monitor = new Test();\npublic static void main(String[] args) {\nRandom rand = new Random();\nint i = rand.nextInt(100);\nint j = rand.nextInt(100);\nSystem.out.println(\"i = \" + i);\nSystem.out.println(\"j = \" + j);\nSystem.out.println(\"i > j is \" + (i > j));\nSystem.out.println(\"i < j is \" + (i < j));\nSystem.out.println(\"i >= j is \" + (i >= j));\nSystem.out.println(\"i <= j is \" + (i <= j));\nSystem.out.println(\"i == j is \" + (i == j));\nSystem.out.println(\"i != j is \" + (i != j));\n// Treating an int as a boolean is not legal Java:\n//! System.out.println(\"i && j is \" + (i && j));\n//! System.out.println(\"i || j is \" + (i || j));\n//! System.out.println(\"!i is \" + !i);\nSystem.out.println(\"(i < 10) && (j < 10) is \"\n+ ((i < 10) && (j < 10)) );\nSystem.out.println(\"(i < 10) || (j < 10) is \"\n+ ((i < 10) || (j < 10)) );\nmonitor.expect(new String[] {\n\"%% i = -?\\\\d+\",\n\"%% j = -?\\\\d+\",\n\"%% i > j is (true|false)\",\n\"%% i < j is (true|false)\",\n\"%% i >= j is (true|false)\",\n\"%% i <= j is (true|false)\",\n\"%% i == j is (true|false)\",\n\"%% i != j is (true|false)\",\n\"%% \\\\(i < 10\\\\) && \\\\(j < 10\\\\) is (true|false)\",\n\"%% \\\\(i < 10\\\\) \\\\|\\\\| \\\\(j < 10\\\\) is (true|false)\"\n});\n}\n} ///:~ ```\n\nIn the regular expressions in the expect( ) statement, parentheses have the effect of grouping an expression, and the vertical bar ‘|’ means OR. So:\n\n`(true|false) `\n\nMeans that this part of the string may be either ‘true’ or ‘false’. Because these characters are special in regular expressions, they must be escaped with a ‘\\\\’ if you want them to appear as ordinary characters in the expression.\n\nYou can apply AND, OR, or NOT to boolean values only. You can’t use a non-boolean as if it were a boolean in a logical expression as you can in C and C++. You can see the failed attempts at doing this commented out with a //! comment marker. The subsequent expressions, however, produce boolean values using relational comparisons, then use logical operations on the results.\n\nNote that a boolean value is automatically converted to an appropriate text form if it’s used where a String is expected.\n\nYou can replace the definition for int in the preceding program with any other primitive data type except boolean. Be aware, however, that the comparison of floating-point numbers is very strict. A number that is the tiniest fraction different from another number is still “not equal.” A number that is the tiniest bit above zero is still nonzero.\n\n#### Short-circuiting\n\nWhen dealing with logical operators, you run into a phenomenon called “short circuiting.” This means that the expression will be evaluated only until the truth or falsehood of the entire expression can be unambiguously determined. As a result, the latter parts of a logical expression might not be evaluated. Here’s an example that demonstrates short-circuiting:\n\n```//: c03:ShortCircuit.java\n// Demonstrates short-circuiting behavior.\n// with logical operators.\nimport com.bruceeckel.simpletest.*;\n\npublic class ShortCircuit {\nstatic Test monitor = new Test();\nstatic boolean test1(int val) {\nSystem.out.println(\"test1(\" + val + \")\");\nSystem.out.println(\"result: \" + (val < 1));\nreturn val < 1;\n}\nstatic boolean test2(int val) {\nSystem.out.println(\"test2(\" + val + \")\");\nSystem.out.println(\"result: \" + (val < 2));\nreturn val < 2;\n}\nstatic boolean test3(int val) {\nSystem.out.println(\"test3(\" + val + \")\");\nSystem.out.println(\"result: \" + (val < 3));\nreturn val < 3;\n}\npublic static void main(String[] args) {\nif(test1(0) && test2(2) && test3(2))\nSystem.out.println(\"expression is true\");\nelse\nSystem.out.println(\"expression is false\");\nmonitor.expect(new String[] {\n\"test1(0)\",\n\"result: true\",\n\"test2(2)\",\n\"result: false\",\n\"expression is false\"\n});\n}\n} ///:~ ```\n\nEach test performs a comparison against the argument and returns true or false. It also prints information to show you that it’s being called. The tests are used in the expression:\n\n`if(test1(0) && test2(2) && test3(2)) `\n\nYou might naturally think that all three tests would be executed, but the output shows otherwise. The first test produced a true result, so the expression evaluation continues. However, the second test produced a false result. Since this means that the whole expression must be false, why continue evaluating the rest of the expression? It could be expensive. The reason for short-circuiting, in fact, is that you can get a potential performance increase if all the parts of a logical expression do not need to be evaluated.\n\n### Bitwise operators\n\nThe bitwise operators allow you to manipulate individual bits in an integral primitive data type. Bitwise operators perform Boolean algebra on the corresponding bits in the two arguments to produce the result.\n\nThe bitwise operators come from C’s low-level orientation, where you often manipulate hardware directly and must set the bits in hardware registers. Java was originally designed to be embedded in TV set-top boxes, so this low-level orientation still made sense. However, you probably won’t use the bitwise operators much.\n\nThe bitwise AND operator (&) produces a one in the output bit if both input bits are one, otherwise it produces a zero. The bitwise OR operator (|) produces a one in the output bit if either input bit is a one and produces a zero only if both input bits are zero. The bitwise EXCLUSIVE OR, or XOR (^), produces a one in the output bit if one or the other input bit is a one, but not both. The bitwise NOT (~, also called the ones complement operator) is a unary operator; it takes only one argument. (All other bitwise operators are binary operators.) Bitwise NOT produces the opposite of the input bit—a one if the input bit is zero, a zero if the input bit is one.\n\nThe bitwise operators and logical operators use the same characters, so it is helpful to have a mnemonic device to help you remember the meanings: because bits are “small,” there is only one character in the bitwise operators.\n\nBitwise operators can be combined with the = sign to unite the operation and assignment: &=, |= and ^= are all legitimate. (Since ~ is a unary operator, it cannot be combined with the = sign.)\n\nThe boolean type is treated as a one-bit value, so it is somewhat different. You can perform a bitwise AND, OR, and XOR, but you can’t perform a bitwise NOT (presumably to prevent confusion with the logical NOT). For booleans, the bitwise operators have the same effect as the logical operators except that they do not short circuit. Also, bitwise operations on booleans include an XOR logical operator that is not included under the list of “logical” operators. You’re prevented from using booleans in shift expressions, which are described next.\n\n### Shift operators\n\nThe shift operators also manipulate bits. They can be used solely with primitive, integral types. The left-shift operator (<<) produces the operand to the left of the operator shifted to the left by the number of bits specified after the operator (inserting zeroes at the lower-order bits). The signed right-shift operator (>>) produces the operand to the left of the operator shifted to the right by the number of bits specified after the operator. The signed right shift >> uses sign extension: if the value is positive, zeroes are inserted at the higher-order bits; if the value is negative, ones are inserted at the higher-order bits. Java has also added the unsigned right shift >>>, which uses zero extension: regardless of the sign, zeroes are inserted at the higher-order bits. This operator does not exist in C or C++.\n\nIf you shift a char, byte, or short, it will be promoted to int before the shift takes place, and the result will be an int. Only the five low-order bits of the right-hand side will be used. This prevents you from shifting more than the number of bits in an int. If you’re operating on a long, you’ll get a long result. Only the six low-order bits of the right-hand side will be used, so you can’t shift more than the number of bits in a long.\n\nShifts can be combined with the equal sign (<<= or >>= or >>>=). The lvalue is replaced by the lvalue shifted by the rvalue. There is a problem, however, with the unsigned right shift combined with assignment. If you use it with byte or short, you don’t get the correct results. Instead, these are promoted to int and right shifted, but then truncated as they are assigned back into their variables, so you get -1 in those cases. The following example demonstrates this:\n\n```//: c03:URShift.java\n// Test of unsigned right shift.\nimport com.bruceeckel.simpletest.*;\n\npublic class URShift {\nstatic Test monitor = new Test();\npublic static void main(String[] args) {\nint i = -1;\nSystem.out.println(i >>>= 10);\nlong l = -1;\nSystem.out.println(l >>>= 10);\nshort s = -1;\nSystem.out.println(s >>>= 10);\nbyte b = -1;\nSystem.out.println(b >>>= 10);\nb = -1;\nSystem.out.println(b>>>10);\nmonitor.expect(new String[] {\n\"4194303\",\n\"18014398509481983\",\n\"-1\",\n\"-1\",\n\"4194303\"\n});\n}\n} ///:~ ```\n\nIn the last shift, the resulting value is not assigned back into b, but is printed directly, so the correct behavior occurs.\n\nHere’s an example that demonstrates the use of all the operators involving bits:\n\n```//: c03:BitManipulation.java\n// Using the bitwise operators.\nimport com.bruceeckel.simpletest.*;\nimport java.util.*;\n\npublic class BitManipulation {\nstatic Test monitor = new Test();\npublic static void main(String[] args) {\nRandom rand = new Random();\nint i = rand.nextInt();\nint j = rand.nextInt();\nprintBinaryInt(\"-1\", -1);\nprintBinaryInt(\"+1\", +1);\nint maxpos = 2147483647;\nprintBinaryInt(\"maxpos\", maxpos);\nint maxneg = -2147483648;\nprintBinaryInt(\"maxneg\", maxneg);\nprintBinaryInt(\"i\", i);\nprintBinaryInt(\"~i\", ~i);\nprintBinaryInt(\"-i\", -i);\nprintBinaryInt(\"j\", j);\nprintBinaryInt(\"i & j\", i & j);\nprintBinaryInt(\"i | j\", i | j);\nprintBinaryInt(\"i ^ j\", i ^ j);\nprintBinaryInt(\"i << 5\", i << 5);\nprintBinaryInt(\"i >> 5\", i >> 5);\nprintBinaryInt(\"(~i) >> 5\", (~i) >> 5);\nprintBinaryInt(\"i >>> 5\", i >>> 5);\nprintBinaryInt(\"(~i) >>> 5\", (~i) >>> 5);\n\nlong l = rand.nextLong();\nlong m = rand.nextLong();\nprintBinaryLong(\"-1L\", -1L);\nprintBinaryLong(\"+1L\", +1L);\nlong ll = 9223372036854775807L;\nprintBinaryLong(\"maxpos\", ll);\nlong lln = -9223372036854775808L;\nprintBinaryLong(\"maxneg\", lln);\nprintBinaryLong(\"l\", l);\nprintBinaryLong(\"~l\", ~l);\nprintBinaryLong(\"-l\", -l);\nprintBinaryLong(\"m\", m);\nprintBinaryLong(\"l & m\", l & m);\nprintBinaryLong(\"l | m\", l | m);\nprintBinaryLong(\"l ^ m\", l ^ m);\nprintBinaryLong(\"l << 5\", l << 5);\nprintBinaryLong(\"l >> 5\", l >> 5);\nprintBinaryLong(\"(~l) >> 5\", (~l) >> 5);\nprintBinaryLong(\"l >>> 5\", l >>> 5);\nprintBinaryLong(\"(~l) >>> 5\", (~l) >>> 5);\nmonitor.expect(\"BitManipulation.out\");\n}\nstatic void printBinaryInt(String s, int i) {\nSystem.out.println(\ns + \", int: \" + i + \", binary: \");\nSystem.out.print(\" \");\nfor(int j = 31; j >= 0; j--)\nif(((1 << j) & i) != 0)\nSystem.out.print(\"1\");\nelse\nSystem.out.print(\"0\");\nSystem.out.println();\n}\nstatic void printBinaryLong(String s, long l) {\nSystem.out.println(\ns + \", long: \" + l + \", binary: \");\nSystem.out.print(\" \");\nfor(int i = 63; i >= 0; i--)\nif(((1L << i) & l) != 0)\nSystem.out.print(\"1\");\nelse\nSystem.out.print(\"0\");\nSystem.out.println();\n}\n} ///:~ ```\n\nThe two methods at the end, printBinaryInt( ) and printBinaryLong( ), take an int or a long, respectively, and print it out in binary format along with a descriptive string. You can ignore the implementation of these for now.\n\nYou’ll note the use of System.out.print( ) instead of System.out.println( ). The print( ) method does not emit a newline, so it allows you to output a line in pieces.\n\nIn this case, the expect( ) statement takes a file name, from which it reads the expected lines (which may or may not include regular expressions). This is useful in situations where the output is too long or inappropriate to include in the book. The files ending with “.out” are part of the code distribution, available for download from www.BruceEckel.com, so you can open the file and look at it to see what the output should be (or simply run the program yourself).\n\nAs well as demonstrating the effect of all the bitwise operators for int and long, this example also shows the minimum, maximum, +1, and -1 values for int and long so you can see what they look like. Note that the high bit represents the sign: 0 means positive and 1 means negative. The output for the int portion looks like this:\n\n```-1, int: -1, binary:\n11111111111111111111111111111111\n+1, int: 1, binary:\n00000000000000000000000000000001\nmaxpos, int: 2147483647, binary:\n01111111111111111111111111111111\nmaxneg, int: -2147483648, binary:\n10000000000000000000000000000000\ni, int: 59081716, binary:\n00000011100001011000001111110100\n~i, int: -59081717, binary:\n11111100011110100111110000001011\n-i, int: -59081716, binary:\n11111100011110100111110000001100\nj, int: 198850956, binary:\n00001011110110100011100110001100\ni & j, int: 58720644, binary:\n00000011100000000000000110000100\ni | j, int: 199212028, binary:\n00001011110111111011101111111100\ni ^ j, int: 140491384, binary:\n00001000010111111011101001111000\ni << 5, int: 1890614912, binary:\n01110000101100000111111010000000\ni >> 5, int: 1846303, binary:\n00000000000111000010110000011111\n(~i) >> 5, int: -1846304, binary:\n11111111111000111101001111100000\ni >>> 5, int: 1846303, binary:\n00000000000111000010110000011111\n(~i) >>> 5, int: 132371424, binary:\n00000111111000111101001111100000 ```\n\nThe binary representation of the numbers is referred to as signed two’s complement.\n\n### Ternary if-else operator\n\nThis operator is unusual because it has three operands. It is truly an operator because it produces a value, unlike the ordinary if-else statement that you’ll see in the next section of this chapter. The expression is of the form:\n\n`boolean-exp ? value0 : value1 `\n\nIf boolean-exp evaluates to true, value0 is evaluated, and its result becomes the value produced by the operator. If boolean-exp is false, value1 is evaluated and its result becomes the value produced by the operator.\n\nOf course, you could use an ordinary if-else statement (described later), but the ternary operator is much terser. Although C (where this operator originated) prides itself on being a terse language, and the ternary operator might have been introduced partly for efficiency, you should be somewhat wary of using it on an everyday basis—it’s easy to produce unreadable code. The conditional operator can be used for its side effects or for the value it produces, but in general you want the value, since that’s what makes the operator distinct from the if-else. Here’s an example:\n\n```static int ternary(int i) {\nreturn i < 10 ? i * 100 : i * 10;\n} ```\n\nYou can see that this code is more compact than what you’d need to write without the ternary operator:\n\n```static int alternative(int i) {\nif (i < 10)\nreturn i * 100;\nelse\nreturn i * 10;\n} ```\n\nThe second form is easier to understand, and doesn’t require a lot more typing. So be sure to ponder your reasons when choosing the ternary operator—it’s generally warranted when you’re setting a variable to one of two values.\n\n### The comma operator\n\nThe comma is used in C and C++ not only as a separator in function argument lists, but also as an operator for sequential evaluation. The sole place that the comma operator is used in Java is in for loops, which will be described later in this chapter.\n\n### String operator +\n\n```int x = 0, y = 1, z = 2;\nString sString = \"x, y, z \";\nSystem.out.println(sString + x + y + z); ```\n\nHere, the Java compiler will convert x, y, and z into their String representations instead of adding them together first. And if you say:\n\n`System.out.println(x + sString); `\n\nJava will turn x into a String.\n\n### Common pitfalls when using operators\n\nOne of the pitfalls when using operators is attempting to leave out the parentheses when you are even the least bit uncertain about how an expression will evaluate. This is still true in Java.\n\nAn extremely common error in C and C++ looks like this:\n\n```while(x = y) {\n// ....\n} ```\n\nThe programmer was clearly trying to test for equivalence (==) rather than do an assignment. In C and C++ the result of this assignment will always be true if y is nonzero, and you’ll probably get an infinite loop. In Java, the result of this expression is not a boolean, but the compiler expects a boolean and won’t convert from an int, so it will conveniently give you a compile-time error and catch the problem before you ever try to run the program. So the pitfall never happens in Java. (The only time you won’t get a compile-time error is when x and y are boolean, in which case x = y is a legal expression, and in the preceding example, probably an error.)\n\nA similar problem in C and C++ is using bitwise AND and OR instead of the logical versions. Bitwise AND and OR use one of the characters (& or |) while logical AND and OR use two (&& and ||). Just as with = and ==, it’s easy to type just one character instead of two. In Java, the compiler again prevents this, because it won’t let you cavalierly use one type where it doesn’t belong.\n\n### Casting operators\n\nThe word cast is used in the sense of “casting into a mold.” Java will automatically change one type of data into another when appropriate. For instance, if you assign an integral value to a floating-point variable, the compiler will automatically convert the int to a float. Casting allows you to make this type conversion explicit, or to force it when it wouldn’t normally happen.\n\nTo perform a cast, put the desired data type (including all modifiers) inside parentheses to the left of any value. Here’s an example:\n\n```void casts() {\nint i = 200;\nlong l = (long)i;\nlong l2 = (long)200;\n} ```\n\nAs you can see, it’s possible to perform a cast on a numeric value as well as on a variable. In both casts shown here, however, the cast is superfluous, since the compiler will automatically promote an int value to a long when necessary. However, you are allowed to use superfluous casts to make a point or to make your code more clear. In other situations, a cast may be essential just to get the code to compile.\n\nIn C and C++, casting can cause some headaches. In Java, casting is safe, with the exception that when you perform a so-called narrowing conversion (that is, when you go from a data type that can hold more information to one that doesn’t hold as much), you run the risk of losing information. Here the compiler forces you to do a cast, in effect saying “this can be a dangerous thing to do—if you want me to do it anyway you must make the cast explicit.” With a widening conversion an explicit cast is not needed, because the new type will more than hold the information from the old type so that no information is ever lost.\n\nJava allows you to cast any primitive type to any other primitive type, except for boolean, which doesn’t allow any casting at all. Class types do not allow casting. To convert one to the other, there must be special methods. (String is a special case, and you’ll find out later in this book that objects can be cast within a family of types; an Oak can be cast to a Tree and vice-versa, but not to a foreign type such as a Rock.)\n\n#### Literals\n\nOrdinarily, when you insert a literal value into a program, the compiler knows exactly what type to make it. Sometimes, however, the type is ambiguous. When this happens, you must guide the compiler by adding some extra information in the form of characters associated with the literal value. The following code shows these characters:\n\n```//: c03:Literals.java\n\npublic class Literals {\nchar c = 0xffff; // max char hex value\nbyte b = 0x7f; // max byte hex value\nshort s = 0x7fff; // max short hex value\nint i1 = 0x2f; // Hexadecimal (lowercase)\nint i2 = 0X2F; // Hexadecimal (uppercase)\nint i3 = 0177; // Octal (leading zero)\n// Hex and Oct also work with long.\nlong n1 = 200L; // long suffix\nlong n2 = 200l; // long suffix (but can be confusing)\nlong n3 = 200;\n//! long l6(200); // not allowed\nfloat f1 = 1;\nfloat f2 = 1F; // float suffix\nfloat f3 = 1f; // float suffix\nfloat f4 = 1e-45f; // 10 to the power\nfloat f5 = 1e+9f; // float suffix\ndouble d1 = 1d; // double suffix\ndouble d2 = 1D; // double suffix\ndouble d3 = 47e47d; // 10 to the power\n} ///:~ ```\n\nHexadecimal (base 16), which works with all the integral data types, is denoted by a leading 0x or 0X followed by 0-9 or a-f either in uppercase or lowercase. If you try to initialize a variable with a value bigger than it can hold (regardless of the numerical form of the value), the compiler will give you an error message. Notice in the preceding code the maximum possible hexadecimal values for char, byte, and short. If you exceed these, the compiler will automatically make the value an int and tell you that you need a narrowing cast for the assignment. You’ll know you’ve stepped over the line.\n\nOctal (base 8) is denoted by a leading zero in the number and digits from 0-7. There is no literal representation for binary numbers in C, C++, or Java.\n\nA trailing character after a literal value establishes its type. Uppercase or lowercase L means long, upper or lowercase F means float and uppercase or lowercase D means double.\n\nExponents use a notation that I’ve always found rather dismaying: 1.39 e-47f. In science and engineering, ‘e’ refers to the base of natural logarithms, approximately 2.718. (A more precise double value is available in Java as Math.E.) This is used in exponentiation expressions such as 1.39 x e-47, which means 1.39 x 2.718-47. However, when FORTRAN was invented, they decided that e would naturally mean “ten to the power,” which is an odd decision because FORTRAN was designed for science and engineering, and one would think its designers would be sensitive about introducing such an ambiguity.1 At any rate, this custom was followed in C, C++ and now Java. So if you’re used to thinking in terms of e as the base of natural logarithms, you must do a mental translation when you see an expression such as 1.39 e-47f in Java; it means 1.39 x 10-47.\n\nNote that you don’t need to use the trailing character when the compiler can figure out the appropriate type. With\n\n`long n3 = 200; `\n\nthere’s no ambiguity, so an L after the 200 would be superfluous. However, with\n\n`float f4 = 1e-47f; // 10 to the power `\n\nthe compiler normally takes exponential numbers as doubles, so without the trailing f, it will give you an error telling you that you must use a cast to convert double to float.\n\n#### Promotion\n\nYou’ll discover that if you perform any mathematical or bitwise operations on primitive data types that are smaller than an int (that is, char, byte, or short), those values will be promoted to int before performing the operations, and the resulting value will be of type int. So if you want to assign back into the smaller type, you must use a cast. (And, since you’re assigning back into a smaller type, you might be losing information.) In general, the largest data type in an expression is the one that determines the size of the result of that expression; if you multiply a float and a double, the result will be double; if you add an int and a long, the result will be long.\n\n### Java has no “sizeof”\n\nIn C and C++, the sizeof( ) operator satisfies a specific need: it tells you the number of bytes allocated for data items. The most compelling need for sizeof( ) in C and C++ is portability. Different data types might be different sizes on different machines, so the programmer must find out how big those types are when performing operations that are sensitive to size. For example, one computer might store integers in 32 bits, whereas another might store integers as 16 bits. Programs could store larger values in integers on the first machine. As you might imagine, portability is a huge headache for C and C++ programmers.\n\nJava does not need a sizeof( ) operator for this purpose, because all the data types are the same size on all machines. You do not need to think about portability on this level—it is designed into the language.\n\n### Precedence revisited\n\nUpon hearing me complain about the complexity of remembering operator precedence during one of my seminars, a student suggested a mnemonic that is simultaneously a commentary: “Ulcer Addicts Really Like C A lot.”\n\n Mnemonic Operator type Operators Ulcer Unary + - ++-- Addicts Arithmetic (and shift) * / % + - << >> Really Relational > < >= <= == != Like Logical (and bitwise) && || & | ^ C Conditional (ternary) A > B ? X : Y A Lot Assignment = (and compound assignment like *=)\n\nOf course, with the shift and bitwise operators distributed around the table it is not a perfect mnemonic, but for non-bit operations it works.\n\n### A compendium of operators\n\nThe following example shows which primitive data types can be used with particular operators. Basically, it is the same example repeated over and over, but using different primitive data types. The file will compile without error because the lines that would cause errors are commented out with a //!.\n\n```//: c03:AllOps.java\n// Tests all the operators on all the primitive data types\n// to show which ones are accepted by the Java compiler.\n\npublic class AllOps {\n// To accept the results of a boolean test:\nvoid f(boolean b) {}\nvoid boolTest(boolean x, boolean y) {\n// Arithmetic operators:\n//! x = x * y;\n//! x = x / y;\n//! x = x % y;\n//! x = x + y;\n//! x = x - y;\n//! x++;\n//! x--;\n//! x = +y;\n//! x = -y;\n// Relational and logical:\n//! f(x > y);\n//! f(x >= y);\n//! f(x < y);\n//! f(x <= y);\nf(x == y);\nf(x != y);\nf(!y);\nx = x && y;\nx = x || y;\n// Bitwise operators:\n//! x = ~y;\nx = x & y;\nx = x | y;\nx = x ^ y;\n//! x = x << 1;\n//! x = x >> 1;\n//! x = x >>> 1;\n// Compound assignment:\n//! x += y;\n//! x -= y;\n//! x *= y;\n//! x /= y;\n//! x %= y;\n//! x <<= 1;\n//! x >>= 1;\n//! x >>>= 1;\nx &= y;\nx ^= y;\nx |= y;\n// Casting:\n//! char c = (char)x;\n//! byte B = (byte)x;\n//! short s = (short)x;\n//! int i = (int)x;\n//! long l = (long)x;\n//! float f = (float)x;\n//! double d = (double)x;\n}\nvoid charTest(char x, char y) {\n// Arithmetic operators:\nx = (char)(x * y);\nx = (char)(x / y);\nx = (char)(x % y);\nx = (char)(x + y);\nx = (char)(x - y);\nx++;\nx--;\nx = (char)+y;\nx = (char)-y;\n// Relational and logical:\nf(x > y);\nf(x >= y);\nf(x < y);\nf(x <= y);\nf(x == y);\nf(x != y);\n//! f(!x);\n//! f(x && y);\n//! f(x || y);\n// Bitwise operators:\nx= (char)~y;\nx = (char)(x & y);\nx = (char)(x | y);\nx = (char)(x ^ y);\nx = (char)(x << 1);\nx = (char)(x >> 1);\nx = (char)(x >>> 1);\n// Compound assignment:\nx += y;\nx -= y;\nx *= y;\nx /= y;\nx %= y;\nx <<= 1;\nx >>= 1;\nx >>>= 1;\nx &= y;\nx ^= y;\nx |= y;\n// Casting:\n//! boolean b = (boolean)x;\nbyte B = (byte)x;\nshort s = (short)x;\nint i = (int)x;\nlong l = (long)x;\nfloat f = (float)x;\ndouble d = (double)x;\n}\nvoid byteTest(byte x, byte y) {\n// Arithmetic operators:\nx = (byte)(x* y);\nx = (byte)(x / y);\nx = (byte)(x % y);\nx = (byte)(x + y);\nx = (byte)(x - y);\nx++;\nx--;\nx = (byte)+ y;\nx = (byte)- y;\n// Relational and logical:\nf(x > y);\nf(x >= y);\nf(x < y);\nf(x <= y);\nf(x == y);\nf(x != y);\n//! f(!x);\n//! f(x && y);\n//! f(x || y);\n// Bitwise operators:\nx = (byte)~y;\nx = (byte)(x & y);\nx = (byte)(x | y);\nx = (byte)(x ^ y);\nx = (byte)(x << 1);\nx = (byte)(x >> 1);\nx = (byte)(x >>> 1);\n// Compound assignment:\nx += y;\nx -= y;\nx *= y;\nx /= y;\nx %= y;\nx <<= 1;\nx >>= 1;\nx >>>= 1;\nx &= y;\nx ^= y;\nx |= y;\n// Casting:\n//! boolean b = (boolean)x;\nchar c = (char)x;\nshort s = (short)x;\nint i = (int)x;\nlong l = (long)x;\nfloat f = (float)x;\ndouble d = (double)x;\n}\nvoid shortTest(short x, short y) {\n// Arithmetic operators:\nx = (short)(x * y);\nx = (short)(x / y);\nx = (short)(x % y);\nx = (short)(x + y);\nx = (short)(x - y);\nx++;\nx--;\nx = (short)+y;\nx = (short)-y;\n// Relational and logical:\nf(x > y);\nf(x >= y);\nf(x < y);\nf(x <= y);\nf(x == y);\nf(x != y);\n//! f(!x);\n//! f(x && y);\n//! f(x || y);\n// Bitwise operators:\nx = (short)~y;\nx = (short)(x & y);\nx = (short)(x | y);\nx = (short)(x ^ y);\nx = (short)(x << 1);\nx = (short)(x >> 1);\nx = (short)(x >>> 1);\n// Compound assignment:\nx += y;\nx -= y;\nx *= y;\nx /= y;\nx %= y;\nx <<= 1;\nx >>= 1;\nx >>>= 1;\nx &= y;\nx ^= y;\nx |= y;\n// Casting:\n//! boolean b = (boolean)x;\nchar c = (char)x;\nbyte B = (byte)x;\nint i = (int)x;\nlong l = (long)x;\nfloat f = (float)x;\ndouble d = (double)x;\n}\nvoid intTest(int x, int y) {\n// Arithmetic operators:\nx = x * y;\nx = x / y;\nx = x % y;\nx = x + y;\nx = x - y;\nx++;\nx--;\nx = +y;\nx = -y;\n// Relational and logical:\nf(x > y);\nf(x >= y);\nf(x < y);\nf(x <= y);\nf(x == y);\nf(x != y);\n//! f(!x);\n//! f(x && y);\n//! f(x || y);\n// Bitwise operators:\nx = ~y;\nx = x & y;\nx = x | y;\nx = x ^ y;\nx = x << 1;\nx = x >> 1;\nx = x >>> 1;\n// Compound assignment:\nx += y;\nx -= y;\nx *= y;\nx /= y;\nx %= y;\nx <<= 1;\nx >>= 1;\nx >>>= 1;\nx &= y;\nx ^= y;\nx |= y;\n// Casting:\n//! boolean b = (boolean)x;\nchar c = (char)x;\nbyte B = (byte)x;\nshort s = (short)x;\nlong l = (long)x;\nfloat f = (float)x;\ndouble d = (double)x;\n}\nvoid longTest(long x, long y) {\n// Arithmetic operators:\nx = x * y;\nx = x / y;\nx = x % y;\nx = x + y;\nx = x - y;\nx++;\nx--;\nx = +y;\nx = -y;\n// Relational and logical:\nf(x > y);\nf(x >= y);\nf(x < y);\nf(x <= y);\nf(x == y);\nf(x != y);\n//! f(!x);\n//! f(x && y);\n//! f(x || y);\n// Bitwise operators:\nx = ~y;\nx = x & y;\nx = x | y;\nx = x ^ y;\nx = x << 1;\nx = x >> 1;\nx = x >>> 1;\n// Compound assignment:\nx += y;\nx -= y;\nx *= y;\nx /= y;\nx %= y;\nx <<= 1;\nx >>= 1;\nx >>>= 1;\nx &= y;\nx ^= y;\nx |= y;\n// Casting:\n//! boolean b = (boolean)x;\nchar c = (char)x;\nbyte B = (byte)x;\nshort s = (short)x;\nint i = (int)x;\nfloat f = (float)x;\ndouble d = (double)x;\n}\nvoid floatTest(float x, float y) {\n// Arithmetic operators:\nx = x * y;\nx = x / y;\nx = x % y;\nx = x + y;\nx = x - y;\nx++;\nx--;\nx = +y;\nx = -y;\n// Relational and logical:\nf(x > y);\nf(x >= y);\nf(x < y);\nf(x <= y);\nf(x == y);\nf(x != y);\n//! f(!x);\n//! f(x && y);\n//! f(x || y);\n// Bitwise operators:\n//! x = ~y;\n//! x = x & y;\n//! x = x | y;\n//! x = x ^ y;\n//! x = x << 1;\n//! x = x >> 1;\n//! x = x >>> 1;\n// Compound assignment:\nx += y;\nx -= y;\nx *= y;\nx /= y;\nx %= y;\n//! x <<= 1;\n//! x >>= 1;\n//! x >>>= 1;\n//! x &= y;\n//! x ^= y;\n//! x |= y;\n// Casting:\n//! boolean b = (boolean)x;\nchar c = (char)x;\nbyte B = (byte)x;\nshort s = (short)x;\nint i = (int)x;\nlong l = (long)x;\ndouble d = (double)x;\n}\nvoid doubleTest(double x, double y) {\n// Arithmetic operators:\nx = x * y;\nx = x / y;\nx = x % y;\nx = x + y;\nx = x - y;\nx++;\nx--;\nx = +y;\nx = -y;\n// Relational and logical:\nf(x > y);\nf(x >= y);\nf(x < y);\nf(x <= y);\nf(x == y);\nf(x != y);\n//! f(!x);\n//! f(x && y);\n//! f(x || y);\n// Bitwise operators:\n//! x = ~y;\n//! x = x & y;\n//! x = x | y;\n//! x = x ^ y;\n//! x = x << 1;\n//! x = x >> 1;\n//! x = x >>> 1;\n// Compound assignment:\nx += y;\nx -= y;\nx *= y;\nx /= y;\nx %= y;\n//! x <<= 1;\n//! x >>= 1;\n//! x >>>= 1;\n//! x &= y;\n//! x ^= y;\n//! x |= y;\n// Casting:\n//! boolean b = (boolean)x;\nchar c = (char)x;\nbyte B = (byte)x;\nshort s = (short)x;\nint i = (int)x;\nlong l = (long)x;\nfloat f = (float)x;\n}\n} ///:~ ```\n\nNote that boolean is quite limited. You can assign to it the values true and false, and you can test it for truth or falsehood, but you cannot add booleans or perform any other type of operation on them.\n\nIn char, byte, and short, you can see the effect of promotion with the arithmetic operators. Each arithmetic operation on any of those types produces an int result, which must be explicitly cast back to the original type (a narrowing conversion that might lose information) to assign back to that type. With int values, however, you do not need to cast, because everything is already an int. Don’t be lulled into thinking everything is safe, though. If you multiply two ints that are big enough, you’ll overflow the result. The following example demonstrates this:\n\n```//: c03:Overflow.java\n// Surprise! Java lets you overflow.\nimport com.bruceeckel.simpletest.*;\n\npublic class Overflow {\nstatic Test monitor = new Test();\npublic static void main(String[] args) {\nint big = 0x7fffffff; // max int value\nSystem.out.println(\"big = \" + big);\nint bigger = big * 4;\nSystem.out.println(\"bigger = \" + bigger);\nmonitor.expect(new String[] {\n\"big = 2147483647\",\n\"bigger = -4\"\n});\n}\n} ///:~ ```\n\nYou get no errors or warnings from the compiler, and no exceptions at run time. Java is good, but it’s not that good.\n\nCompound assignments do not require casts for char, byte, or short, even though they are performing promotions that have the same results as the direct arithmetic operations. On the other hand, the lack of the cast certainly simplifies the code.\n\nYou can see that, with the exception of boolean, any primitive type can be cast to any other primitive type. Again, you must be aware of the effect of a narrowing conversion when casting to a smaller type, otherwise you might unknowingly lose information during the cast.\n\n• 🔖 Save To Your Account\n\n### InformIT Promotional Mailings & Special Offers\n\nI would like to receive exclusive offers and hear about products from InformIT and its family of brands. I can unsubscribe at any time.\n\n## Overview\n\nPearson Education, Inc., 221 River Street, Hoboken, New Jersey 07030, (Pearson) presents this site to provide information about products and services that can be purchased through this site.\n\nThis privacy notice provides an overview of our commitment to privacy and describes how we collect, protect, use and share personal information collected through this site. 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[ "# Difficulties while understanding Convolution [duplicate]\n\nI have just started convolution sum. I am able to do Mathematical convolution sum. But I am unable to make a figure of it in my mind. So I tried to go through the graphical convolution. Below is the picture", null, "I understand decomposed part but I don't understand that how each decomposed part is generating x[ 0 ]h[ n ],x[ 1 ]h[ n-1 ]....?\n\nI guess a little help makes me to understand the Convolution sum. Thanks\n\n• It's on ee.se, but maybe this answer helps? – a concerned citizen Jun 1 '18 at 6:27\n• Read this answer on this site and then edit your question especially the figure labels based on your understanding of what that answer says. – Dilip Sarwate Jun 1 '18 at 13:20\n• – MBaz Jun 1 '18 at 15:16\n\nThe convolution sum is \\begin{equation} y[n] = \\sum_k x[k]h[n-k] \\end{equation} If you take the term corresponding to $k=0$, the contribution is $xh[n]$. Similarly the contribution of $k=1$ is $xh[n-1]$. Here, $h[n]$ is delayed by $n=1$. Similarly you can see how each term of the sum contributes to overall sum. The sum is taken over values $k$ where $x[k] \\neq 0$ because only those terms contribute to the sum.\n\n• I know mathematically how to convolved two signal. Please can you tell me what is Xh[n]...? – Anurag Rag Jun 1 '18 at 13:12\n• To understand that, we need to delve deep into the basics of LTI systems. For an LTI system, a linearly scaled time-shifted input results in the same linearly scaled time-shifted output. If we represent x[n] by multiple shifted-scaled versions of delta function (the basic signal - d[n]) , the k=0 term is xd[n]. When we pass this component through LTI system represented by h[n], we get the output xh[n]. Similarly for other terms of k. – jithin Jun 2 '18 at 14:52\n\nFor convolution to apply the system must be Linear Time Invariant. This means that when the input is the sum of 2 signals then the output is equal to the sum of the individual outputs (when each signal is applied separately). It also means that you can scale the input and the output will be scaled by the same factor. The input signal can also be shifted in time and the output signal will also be shifted in time by the same amount. The decomposition you described is using these properties.\n\nGiven that an impulse $\\delta[n]$ applied to a system the output is given by $h[n]$. If the input is $A\\delta[n]$ , where $A$ is a scalar then the output is given by $Ah[n]$ - it simply scales the impulse response. If the input is $A\\delta[n-k]$ , where $k$ is a constant, then the output is $Ah[n-k]$.\n\nIn your picture, they decompose the input into a series of scaled delta functions i.e. $x$ , $x$ and so on. The output from $x$ is $xh[n]$ i.e. it scales the impulse response by the scalar value $x$. Similarly the output corresponding to $x$ is $xh[n-1]$. Since $x$ is a weighted impulse delayed by one sample the impulse response must be weighted and delay by one sample as well. Note that $x = x\\cdot \\delta[n-1]$ This continues for all the samples in $x[n]$. The total output is given by adding these weighted and delayed impulse responses together.\n\nThere are a few ways of approaching convolution. Linear algebra is one, let $x[n]$ where $x[n]=0$ for $n <0$ be input and $y[n]$ output. A causal filter can be expressed as a matrix vector product.\n\n$$\\left| \\begin{array}{c} y \\\\ y \\\\ y \\\\ y \\\\ \\vdots \\end{array} \\right| = \\left| \\begin{array}{ccccc }h & 0 & 0 & 0 & \\dots\\\\ h & h & 0 & 0 & \\dots\\\\ h & h & h & 0 & \\dots\\\\ h & h & h & h & \\ddots \\\\ \\vdots & \\vdots & \\vdots & \\vdots & \\ddots \\end{array}\\right| \\left| \\begin{array}{c} x \\\\ x \\\\ x \\\\ x \\\\ \\vdots \\end{array} \\right|$$ For an IIR Filter, the matrix is infinite, for a FIR filter, the matrix dimension depends on the length of $x[n]$.\n\nYou can write out each term of $y[n]$ and see it is discrete convolution.\n\nWe don't do infinite graphical convolution so you can use this construction to draw your pictures.\n\nThis form of matrix is called Toeplitz." ]

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[ "# XPY (X To The Power Of Y) Ladder Logic Instruction\n\nThe XPY instruction is a ladder logic rung output instruction that takes the Source X to the power of Source Y and places the result in the destination. The instruction is enabled when the preceding logic is true and disabled otherwise. The values can be constants, tags or any combination.\n\nExample:\n\nX to the power of  Y Where X = 2 and Y = 4\n\n= X * X * X * X\n\n= 2 * 2 * 2 * 2\n\n= 16\n\nList of possible arguments\n\nSOURCE X,Y and the DESTINATION can be any combination of the following data types.\n\n• CONSTANT             Can’t be the DESTINATION obviously\n• SINT\n• INT\n• DINT\n• REAL\n\nNote: Arithmetic status flags are affected (S:V, S:Z, S:N, S:C). Overflow, Zero, Negative, and Carry. If Source X is negative and Source Y is not an integer value a minor fault will occur." ]

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[ "# Logic Design - Multiplexer, Encoder and Decoder Circuits", null, "Hello I'm back and this time we will talk about more Advanced Circuits! I will start off with Multiplexers, continue with Encoders and finish with Decoders! So, without further do let's get started!\n\n# Multiplexers:\n\nA Multiplexer is a circuit that selects one of 2^n inputs from n selection lines and gives 1 specific Output! A MUX is also called a Data Selector. So, we have 2^n-to-1 Multiplexers, where n is the number of selection lines. For example, MUX 2-to-1, 4-to-1, 8-to-1 etc.", null, "4-to-1 Mux and Truth Table!\n\nAn 4-to-1 mux for example will have 2 Selection Lines and will work for 3 Variable Input!\n\nTo define those Inputs we have to know what kind of Outputs we have in every Input combination. So, we will use N-1 of the Inputs Variables as Selection Inputs to define minterm groups and afterwards compare the value of the other Variable (01) with the one of the Outputs of the minterm groups (you will understand it better in a minute). This Comparation will give us our \"I\" Inputs for the Multiplexer!\n\nExample:\n\nLet's take F(X, Y, Z) for example that has minterms at the binary values of 1, 3, 4, 5.\n\nThe Truth Table looks like this:", null, "We will take Y, Z as selections and form groups of 01 for X, where YZ need to be the same (like minterms). So, we will create an array that contains X, the Output F and I that will be the comparition result. Taking the first and fifth row for example we see that only X changes so we have X and F be 01 and so I is X! We continue like this and end up with this:", null, "I0 = X, I1 = 1, I2 = 0 and I3 = X'. That will be the Inputs for our Multiplexer having Y and Z be the Selection Line Inputs! The Output of this 4-to-1 Mux will follow the Truth Table of F(X, Y, Z).\n\nWe can do the same using XZ or XY as Selections, again forming groups where the other value changes (01) but the Selections stay the same! We can also use this concept in bigger MUX's with ease.\n\nThe Circuit in Multisim can looks like this for example:", null, "You can see that we put X, 1, 0,X' as \"I\" Inputs and Y, Z as Selections. Having XYZ be 001 (1) we see that the Output becomes 1, because it follows the Truth Table of F!\n\n# Encoders:\n\nEncoders convert information from one format to another to compress data. In Circuits they compress multiple lines of inputs into a smaller number of outputs! In Encoders the number of Outputs depends on the number of Input \"Rows\" and they are complicated and need an implementation every time! That's why we mostly use Priority Encoders and those are the ones that I will talk about! In them when having more then 1 Inputs be active at once, the one in the most left has the highest priority and gets selected! The number of Output lines depends on the different Inputs we can have, so we are able of having 4-to-2 , 8-to-3 encoders and so on...that follow the rule of 2^n-to-n Encoder", null, "4-to-2 Encoder\n\nWe will have to write a Circuit for an Encoder using the Inputs and Outputs of an Truth Table that represents the functionality of such an Encoder!\n\nExample:\n\nLet's take this Truth Table for example that defines an 4-to-2 Encoder:", null, "The X's represent don't cares. We put them cause the most left input has the highest priority on so the value of the others is ignored! You can also see that we have one more Output called Z, whose value tells us if we have an valid Output!\n\nSo, to implement this Encoder we will have to write a function that will be translated into a Circuit for each of the Outputs! It's not too difficult in this one!\n\nUsing Karnaugh Map for each of those I ended up with this Circuit:", null, "# Decoders:\n\nDecoders are the opposite of an Encoder and they actually can also be called Demultiplexers or DEMUX, cause they also work as the opposite of an Multiplexer! The only difference between DEMUX's and Decoders is actually that the first uses only 1 Input and gives us 2^n Outputs, and the second uses n Inputs and gives us 2^n Outputs! I will talk about Decoders to keep things simple, cause DEMUX are more complicated!", null, "As I already said the Decoder is the opposite of an Encoder, so using the Outputs of an Encoder as Inputs in it's corresponding Decoder we get the Inputs that gave us this Outputs in the Encoder! So, we get the Inputs back! But, this is not the way we use them, cause it can be made much simpler!\n\nAn Decoder can't have more then 1 Outputs lines have a value of 1! Cause each Output line represents an Combination (binary number) of Inputs. So, each Output line represents an minterm or maxterm of our function! A Decoder mostly has an inverted Output so we will have to use maxterms or use minterms and invert the Outputs using an NAND Gate!\n\nUsing that logic we can implement any Circuit, knowing only where it has minterms or maxterms!\n\nExample:\n\nLast time we talked about Full-Adders. Let's implement one using an Decoder! We have to use an 3-to-8 Decoder, cause we have 3 Inputs.\n\nThe Truth Table looks like this (if you remember):", null, "To use an Decoder for that Circuit we simply have to put a Multi-Input NAND Gate (if Outputs are inverted, else OR) for each of the Outputs and the Inputs will be the corresponding Outputs the Decoder gives us for each Input Combination (Minterms)!\n\nSo, we end up with this Circuit:", null, "This is the end of today's post! Hope you enjoyed it!\n\nNext time we will talk about Clocks and Waveforms, that will help us get into Latches and Flip-Flops that are used in Sequentials Circuits that we will talk about also some posts later on!\n\nUntil tomorrow...Bye!\n\nH2\nH3\nH4\n3 columns\n2 columns\n1 column\nJoin the conversation now" ]

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[ "# Finding two languages satisfying conditions\n\nLet $$E_{TM} = \\left \\{ \\langle M\\rangle \\mid L(M) = \\emptyset \\right\\}$$\n\nProve that there are two languages $$L_1, L_2$$ such that\n\n1. $$L_1, L_2$$ are infinite.\n2. $$L_1 \\cup L_2 = E_{TM}$$\n3. $$L_1 \\cap L_2 = \\emptyset$$\n4. $$L_1$$ is decidable, $$L_2$$ is not recognizable.\n\nI'm finding it really hard to find two languages that satisfy these conditions.\n\nEspecially the second condition, which two language unify to $$E_{TM}$$?\n\n• Hint 1, can you construct one $M$ such that $L(M)=\\emptyset$? Hint 2, can you construct infinitely many such $M$? Hint 3, let $L_2=E_{TM}\\setminus L_1$. Mar 27, 2019 at 20:04\n• Thank you, I'll try again!\n– Alan\nMar 28, 2019 at 8:30\n• If you has figured out this problem, please write an answer (yes, you can answer your own question). Mar 28, 2019 at 8:53\n• Certainly, I will\n– Alan\nMar 28, 2019 at 8:55\n• The padding lemma states that there is a total computable function that given any (encoding of) TM $M$, it returns a TM $N$ having a \"longer\" encoding than the one of $M$, where $L(M)=L(N)$. Essentially, the function adds a few redundant states to $M$ to make its encoding \"longer\". You could try to apply this function many times, iterating it.\n– chi\nMar 28, 2019 at 12:30\n\nA slight improvement:\n\n$$L_2 = E_{TM} \\backslash L_1$$ (since we require that $$L_1 \\cap L_2 = \\emptyset$$).\n\n$$L_1$$ must be a language of turing machines, i.e. $$L_1 = \\{\\langle M\\rangle \\mid \\dots\\}$$, so that $$L_1 \\cup L_2 = E_{TM}$$.\n\nSo let's set $$L_1 =$$ $$\\{\\langle M\\rangle\\mid C_0\\text{ is a reject state}\\}$$.\n\nThe following holds:\n\n• $$L_1$$ is infinite and decidable (easy to decide given $$M$$, what the configuration of the first state is).\n\n• $$L_2$$ is infinite and unrecognizable." ]

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[ "×\nAiming for the stars? Great, but did you have to build a rocket first.\n--Your friends at LectureNotes\n\n# Note for Strength Of Materials - SOM by Muthuraman Mohan\n\n• Strength Of Materials - SOM\n• Note\n• Anna university - ACEW\n• Civil Engineering\n• B.Tech\n• 1 Topics\n• 63 Views\n• Uploaded 10 months ago\nTouch here to read\nPage-1\n\n#### Note for Strength Of Materials - SOM by Muthuraman Mohan\n\nTopic:", null, "/ 259\n\n0 User(s)\n\n### Leave your Comments\n\n#### Text from page-1\n\nVisit : Civildatas.blogspot.in CE6306 STREGNTH OF MATERIALS Question Bank Unit-I STRESS, STRAIN, DEFORMATION OF SOLIDS PART-A sp ot. in 1. Define Poison’s Ratio May/June 2009 2. What is thermal stress? May/June 2009 3. Estimate the load carried by a bar if the axial stress is 10 N/mm2 and diameter of the bar is 10 mm. Nov/Dec 2006 4. Give the relation between the modulus of elasticity and modulus of rigidity. May/June 2007. May/June 2007. 6. Define the terms poisson’s ratio and bulk modulus. Nov/Dec 2007. 7. Explain the effect of change of temperature in a composite bar. Nov/Dec 2007. 8. Define the terms: Resilience and Modulus of Resilience. May/June 2012. 9. State Hooke,s law. May/June 2009 s.b 10.List out the various elastic constant. log 5. Write the concepts used for finding stresses in compound bars. 11. Define shear strain and volumetric strain. May/June 2009 Apl/May 2009. 12. Mention the relationship between the modulus of elasticity, modulus of rigidity, modulus of elasticity and bulk modulus. Apl/Mayj 2009. Nov/Dec 2010. 14. Define shear stress Nov/Dec 2010. 15. Define volumetric strain Nov/Dec 2011. 16.What is proof resilience Nov/Dec 2011. 17. Define Hook’s law May/June 2013. 18.Define the term modulus of resilience. May/June 2013 vil d ata 13. State the principle of Superposition 19. Define Principal Stress. Ci 20. Define Hydrostatic Pressure. 1 Visit : Civildatas.blogspot.in\n\n#### Text from page-2\n\nVisit : Civildatas.blogspot.in PART-B 1. An alloy circular bar ABCD is 3m long is subjected to a tensile force of 50kN as shown in fig. If the stress in the middle portion BC is not to exceed 150Mpa, then what would be its diameter? Also find the length of the middle portion, if the total extension of the bar should not exceed by (Nov/Dec 2010) sp ot. in 3mm. Take E=100GPa. (12 Marks) 2. Find the stresses in each section of the bar shown in fig and also find the total extension of the bar. E = 2.1x105N/mm2 (16 Marks) s.b log (Nov/Dec 2006) ata 3. Find the value of P and the change in length of each component and the total change in length Ci vil d of the bar shown in fig. 2 Visit : Civildatas.blogspot.in\n\n#### Text from page-3\n\nVisit : Civildatas.blogspot.in 4. A steel rod of 25mm diameter is placed inside a copper tube of 30mm internal diameter and 5mm thickness and the ends are rigidly connected. The assembly is subjected to a compressive load of 250kN. Determine the stresses induced in the steel rod and copper tube. Take the modulus of elasticity of steel and copper as 200GPa and 80GPa respectively. (10 Marks). sp ot. in (NOV/DEC 2006). 5. An aluminum cylinder of diameter 60mm located inside a steel cylinder of internal diameter 60mm and wall thickness 15mm. The assembly is subjected to a compressive force of 200kN. What are the forces carried and stresses developed in steel and aluminum? Take modulus of elasticity for steel as 200GPa and aluminum as 50GPa. (8 Marks) (Nov/Dec 2008) 6. A reinforced concrete column 500 mm x 500 mm in section is reinforced with 4 steel bars of 25 mm diameter; one in each corner, the column is carrying a load of 1000 kN. Find the stresses in the concrete and steel bars. Take Esteel = 210x103N/mm2 and ECon = 14x103N/mm2. (Nov/Dec log 2007) 7. A reinforced concrete circular column of 400 mm diameter has 4 steel bars of 20mm diameter embedded in it. Find the maximum load which the column can carry, if the stresses in steel and 18 times that of concrete. s.b concrete are not to exceed 120MPa and 5MPa respectively. Take modulus of elasticity of steel as 8. A steel tube of 20mm internal diameter and 30mm external diameter encases a copper rod of 15mm diameter to which it is rigidly joined at each end. If the temperature of the assembly is ata raised by 80 C. Calculate the stresses produced in the tube. ES=2x105N/mm2, EC=1x105N/mm2, Co-efficient of linear expansion of steel and copper are 11x10-6 per C and 18x10-6 per C. (MAY/JUNE 2009). vil d 9. A steel tube 30mm external diameter and 25mm internal diameter encloses a gun metal rod 20mm diameter to which it is rigidly joined at each end. The temperature of the whole assembly is raised to 150 C. Find the intensity of stress in the rod when the common temperature has fallen to 20 C. The value of the young’s modulus for steel and gun metal are 2.1x105N/mm2 and Ci 1x105N/mm2 respectively. The co-efficient of linear expansion for steel is 12x10-6/ C and for gun metal 20x10-6/ C. (MAY/JUNE 2009). 10. The composite bar shown in fig is rigidly fixed at the ends. An axial pull of P=15kN is applied at B at 20 C. Find the stresses in each material at 80 C. Take αs=11x10-6/ C, αa=24x106 / C, Es= 210kN/mm2, Ea= 70kN/mm2. 3 Visit : Civildatas.blogspot.in\n\n#### Text from page-4\n\nVisit : Civildatas.blogspot.in 11. A bar of 30mm in diameter was subjected to a tensile load of 54kN and measured extension on 300mm gauge length was 0.112 mm and change in diameter was 0.00366 mm. Calculate the Poisson’s ratio and the values of three elastic modulii. (MAY/JUNE 2009). 12. A steal rod 5mm long and 25 mm in diameter is subjected to an axial tensile load of 50kN. sp ot. in Determine the change in length, diameter and volume of the rod. Take E=2x105N/mm2 and 1/m=0.30 (MAY/JUNE2007) 13. A 25mm diameter bar is subjected to an axial tensile load of 100 kN. Under the action of this load a 200mm gauge length is found to extend 0.19 mm.Determine the modulus of elasticity of the material. If in order to reduce the weight whilst keeping the external diameter constant, the bar is bored axially to produce a hollow cylinder of uniform thickness, what is the maximum diameter of bore possible given that the maximum allowable stress is 240MPa? The load can be assumed to remain constant at 100kN. What will be the change in the outer diameter of the bar log under the above load? Taking 1/m=0.3, also calculate the bulk modulus and the shear modulus of the material. (APL/MAY 2010) 2 14. The modulus of rigidity of a material is 38 kN/mm . A 10mm diameter rod of the material is subjected to an axial tensile force of 5kN and the change in its diameter is observed to be s.b 0.002mm. Calculate the Poisson’s ratio, modulus of elasticity and bulk modulus of the material. 15. In an experiment, a bar of 30mm diameter is subjected to a pull of 60kN. The measured extension on gauge length of 200mm is 0.09mm and the change in diameter is 0.0039 mm. ata Calculate the Poisson’s ratio and the values of the three moduli. 16. For a given material, Young’s modulus is 120 GPa and modulus of rigidity is 40 GPa. Find the bulk modulus and lateral contraction of a round bar of 50mm diameter and 2.5m long, when vil d stretched 2.5 mm, Take 1/m as 0.25. 17. A steel plate 300mm long, 60mm wide and 30mm deep is acted upon by the forces shown in fig. Determine the change in volume. Take E=200 kN/mm2 and Poisson’s ratio=0.3. 18. A metallic bar 250mm x 50mm is loaded as shown in fig. Find the change in volume. Take Ci E=2x105 N/mm2 and 1/m =0.25. Also find the change that would be made in the 4MN load, in order that there should be no change in the volume of the bar. (NON/DEC 2008) 19. The maximum extension produced by an unknown falling weight through a height of 40mm in a vertical bar of length 3m and cross-sectional area of 500mm2 is 2.1mm. Determine the 4 Visit : Civildatas.blogspot.in" ]

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[ "### [Zurück]\n\nZeitschriftenartikel:\n\nC. Carstensen, D. Praetorius:\n\"Convergence of adaptive boundary element methods\";\nJournal of Integral Equations and Applications, 24 (2012), 1; S. 1 - 23.\n\nKurzfassung englisch:\nIn many applications, adaptive mesh-refinement is observed to be an\nefficient tool for the numerical solution of partial differential equations\nand integral equations. Convergence of adaptive schemes to the correct\nsolution, however, is so far only understood for certain kind of differential\nequations. In general, it cannot be excluded that the adaptive algorithm\ncomputes a convergent sequence of discrete approximations with a limit\nwhich is not the correct solution. This work proposes a feedback loop\nwhich guarantees the convergence of the computed discrete approximations\nto the correct solution. Although stated for Symm's integral equation of\nthe first kind, the main part of this work is written for a general audience\nin the context of weak forms as Riesz representations in Hilbert spaces.\nNumerical examples illustrate the adaptive strategies.\n\n\"Offizielle\" elektronische Version der Publikation (entsprechend ihrem Digital Object Identifier - DOI)\n\nErstellt aus der Publikationsdatenbank der Technischen Universität Wien." ]

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[ "Cantera  2.3.0\nIdealMolalSoln.h\nGo to the documentation of this file.\n1 /**\n2  * @file IdealMolalSoln.h\n3  * ThermoPhase object for the ideal molal equation of\n4  * state (see \\ref thermoprops\n5  * and class \\link Cantera::IdealMolalSoln IdealMolalSoln\\endlink).\n6  *\n7  * Header file for a derived class of ThermoPhase that handles variable pressure\n8  * standard state methods for calculating thermodynamic properties that are\n9  * further based upon activities on the molality scale. The Ideal molal solution\n10  * assumes that all molality-based activity coefficients are equal to one. This\n11  * turns out to be highly nonlinear in the limit of the solvent mole fraction\n12  * going to zero.\n13  */\n14\n15 // This file is part of Cantera. See License.txt in the top-level directory or\n16 // at http://www.cantera.org/license.txt for license and copyright information.\n17\n18 #ifndef CT_IDEALMOLALSOLN_H\n19 #define CT_IDEALMOLALSOLN_H\n20\n21 #include \"MolalityVPSSTP.h\"\n22\n23 namespace Cantera\n24 {\n25\n26 /**\n27  * This phase is based upon the mixing-rule assumption that all molality-based\n28  * activity coefficients are equal to one.\n29  *\n30  * This is a full instantiation of a ThermoPhase object. The assumption is that\n31  * the molality-based activity coefficient is equal to one. This also implies\n32  * that the osmotic coefficient is equal to one.\n33  *\n34  * Note, this does not mean that the solution is an ideal solution. In fact,\n35  * there is a singularity in the formulation as the solvent concentration goes\n36  * to zero.\n37  *\n38  * The mechanical equation of state is currently assumed to be that of an\n39  * incompressible solution. This may change in the future. Each species has its\n40  * own molar volume. The molar volume is a constant.\n41  *\n42  * Class IdealMolalSoln represents a condensed phase. The phase and the pure\n43  * species phases which comprise the standard states of the species are assumed\n44  * to have zero volume expansivity and zero isothermal compressibility. Each\n45  * species does, however, have constant but distinct partial molar volumes equal\n46  * to their pure species molar volumes. The class derives from class\n47  * ThermoPhase, and overloads the virtual methods defined there with ones that\n48  * use expressions appropriate for incompressible mixtures.\n49  *\n50  * The standard concentrations can have three different forms depending on the\n51  * value of the member attribute m_formGC, which is supplied in the XML file.\n52  *\n53  * | m_formGC | ActivityConc | StandardConc |\n54  * | -------- | -------------------------------- | ------------------ |\n55  * | 0 | \\f${m_k}/ { m^{\\Delta}}\\f$ | \\f$1.0 \\f$ |\n56  * | 1 | \\f$m_k / (m^{\\Delta} V_k)\\f$ | \\f$1.0 / V_k \\f$ |\n57  * | 2 | \\f$m_k / (m^{\\Delta} V^0_0)\\f$ | \\f$1.0 / V^0_0\\f$ |\n58  *\n59  * \\f$V^0_0 \\f$ is the solvent standard molar volume. \\f$m^{\\Delta} \\f$ is a\n60  * constant equal to a molality of \\f$1.0 \\quad\\mbox{gm kmol}^{-1} \\f$.\n61  *\n62  * The current default is to have mformGC = 2.\n63  *\n64  * The value and form of the activity concentration will affect reaction rate\n65  * constants involving species in this phase.\n66  *\n67  * <thermo model=\"IdealMolalSoln\">\n68  * <standardConc model=\"solvent_volume\" />\n69  * <solvent> H2O(l) </solvent>\n70  * <activityCoefficients model=\"IdealMolalSoln\" >\n71  * <idealMolalSolnCutoff model=\"polyExp\">\n72  * <gamma_O_limit> 1.0E-5 </gamma_O_limit>\n73  * <gamma_k_limit> 1.0E-5 <gamma_k_limit>\n74  * <X_o_cutoff> 0.20 </X_o_cutoff>\n75  * <C_0_param> 0.05 </C_0_param>\n76  * <slope_f_limit> 0.6 </slope_f_limit>\n77  * <slope_g_limit> 0.0 </slope_g_limit>\n78  * </idealMolalSolnCutoff>\n79  * </activityCoefficients>\n80  * </thermo>\n81  *\n82  * @ingroup thermoprops\n83  */\n85 {\n86 public:\n87  /// Constructor\n89\n91  IdealMolalSoln& operator=(const IdealMolalSoln&);\n92  virtual ThermoPhase* duplMyselfAsThermoPhase() const;\n93\n94  //! Constructor for phase initialization\n95  /*!\n96  * This constructor will initialize a phase, by reading the required\n97  * information from an input file.\n98  *\n99  * @param inputFile Name of the Input file that contains information\n100  * about the phase\n101  * @param id id of the phase within the input file\n102  */\n103  IdealMolalSoln(const std::string& inputFile, const std::string& id = \"\");\n104\n105  //! Constructor for phase initialization\n106  /*!\n107  * This constructor will initialize a phase, by reading the required\n108  * information from XML_Node tree.\n109  *\n110  * @param phaseRef reference for an XML_Node tree that contains\n111  * the information necessary to initialize the phase.\n112  * @param id id of the phase within the input file\n113  */\n114  IdealMolalSoln(XML_Node& phaseRef, const std::string& id = \"\");\n115\n116  virtual std::string type() const {\n117  return \"IdealMolalSoln\";\n118  }\n119\n120  //! @}\n121  //! @name Molar Thermodynamic Properties of the Solution\n122  //! @{\n123\n124  //! Molar enthalpy of the solution. Units: J/kmol.\n125  /*!\n126  * Returns the amount of enthalpy per mole of solution. For an ideal molal\n127  * solution,\n128  * \\f[\n129  * \\bar{h}(T, P, X_k) = \\sum_k X_k \\bar{h}_k(T)\n130  * \\f]\n131  * The formula is written in terms of the partial molar enthalpies.\n132  * \\f$\\bar{h}_k(T, p, m_k) \\f$.\n133  * See the partial molar enthalpy function, getPartialMolarEnthalpies(),\n134  * for details.\n135  *\n136  * Units: J/kmol\n137  */\n138  virtual doublereal enthalpy_mole() const;\n139\n140  //! Molar internal energy of the solution: Units: J/kmol.\n141  /*!\n142  * Returns the amount of internal energy per mole of solution. For an ideal\n143  * molal solution,\n144  * \\f[\n145  * \\bar{u}(T, P, X_k) = \\sum_k X_k \\bar{u}_k(T)\n146  * \\f]\n147  * The formula is written in terms of the partial molar internal energy.\n148  * \\f$\\bar{u}_k(T, p, m_k) \\f$.\n149  */\n150  virtual doublereal intEnergy_mole() const;\n151\n152  //! Molar entropy of the solution. Units: J/kmol/K.\n153  /*!\n154  * Returns the amount of entropy per mole of solution. For an ideal molal\n155  * solution,\n156  * \\f[\n157  * \\bar{s}(T, P, X_k) = \\sum_k X_k \\bar{s}_k(T)\n158  * \\f]\n159  * The formula is written in terms of the partial molar entropies.\n160  * \\f$\\bar{s}_k(T, p, m_k) \\f$.\n161  * See the partial molar entropies function, getPartialMolarEntropies(),\n162  * for details.\n163  *\n164  * Units: J/kmol/K.\n165  */\n166  virtual doublereal entropy_mole() const;\n167\n168  //! Molar Gibbs function for the solution: Units J/kmol.\n169  /*!\n170  * Returns the Gibbs free energy of the solution per mole of the solution.\n171  *\n172  * \\f[\n173  * \\bar{g}(T, P, X_k) = \\sum_k X_k \\mu_k(T)\n174  * \\f]\n175  *\n176  * Units: J/kmol\n177  */\n178  virtual doublereal gibbs_mole() const;\n179\n180  //! Molar heat capacity of the solution at constant pressure. Units: J/kmol/K.\n181  /*!\n182  * \\f[\n183  * \\bar{c}_p(T, P, X_k) = \\sum_k X_k \\bar{c}_{p,k}(T)\n184  * \\f]\n185  *\n186  * Units: J/kmol/K\n187  */\n188  virtual doublereal cp_mole() const;\n189\n190  //@}\n191  /** @name Mechanical Equation of State Properties\n192  *\n193  * In this equation of state implementation, the density is a function only\n194  * of the mole fractions. Therefore, it can't be an independent variable.\n195  * Instead, the pressure is used as the independent variable. Functions\n196  * which try to set the thermodynamic state by calling setDensity() may\n197  * cause an exception to be thrown.\n198  */\n199  //@{\n200\n201 protected:\n202  /**\n203  * Calculate the density of the mixture using the partial molar volumes and\n204  * mole fractions as input\n205  *\n206  * The formula for this is\n207  *\n208  * \\f[\n209  * \\rho = \\frac{\\sum_k{X_k W_k}}{\\sum_k{X_k V_k}}\n210  * \\f]\n211  *\n212  * where \\f$X_k\\f$ are the mole fractions, \\f$W_k\\f$ are the molecular\n213  * weights, and \\f$V_k\\f$ are the pure species molar volumes.\n214  *\n215  * Note, the basis behind this formula is that in an ideal solution the\n216  * partial molar volumes are equal to the pure species molar volumes. We\n217  * have additionally specified in this class that the pure species molar\n218  * volumes are independent of temperature and pressure.\n219  */\n220  void calcDensity();\n221\n222 public:\n223  /**\n224  * Overridden setDensity() function is necessary because the density is not\n225  * an independent variable.\n226  *\n227  * This function will now throw an error condition\n228  *\n229  * @internal May have to adjust the strategy here to make the eos for these\n230  * materials slightly compressible, in order to create a condition where\n231  * the density is a function of the pressure.\n232  *\n233  * @param rho Input Density\n234  */\n235  virtual void setDensity(const doublereal rho);\n236\n237  /**\n238  * Overridden setMolarDensity() function is necessary because the density\n239  * is not an independent variable.\n240  *\n241  * This function will now throw an error condition.\n242  *\n243  * @param rho Input Density\n244  */\n245  virtual void setMolarDensity(const doublereal rho);\n246\n247  //! The isothermal compressibility. Units: 1/Pa.\n248  /*!\n249  * The isothermal compressibility is defined as\n250  * \\f[\n251  * \\kappa_T = -\\frac{1}{v}\\left(\\frac{\\partial v}{\\partial P}\\right)_T\n252  * \\f]\n253  *\n254  * It's equal to zero for this model, since the molar volume doesn't change\n255  * with pressure or temperature.\n256  */\n257  virtual doublereal isothermalCompressibility() const;\n258\n259  //! The thermal expansion coefficient. Units: 1/K.\n260  /*!\n261  * The thermal expansion coefficient is defined as\n262  *\n263  * \\f[\n264  * \\beta = \\frac{1}{v}\\left(\\frac{\\partial v}{\\partial T}\\right)_P\n265  * \\f]\n266  *\n267  * It's equal to zero for this model, since the molar volume doesn't change\n268  * with pressure or temperature.\n269  */\n270  virtual doublereal thermalExpansionCoeff() const;\n271\n272  /**\n273  * @}\n274  * @name Activities and Activity Concentrations\n275  *\n276  * The activity \\f$a_k\\f$ of a species in solution is related to the\n277  * chemical potential by \\f[ \\mu_k = \\mu_k^0(T) + \\hat R T \\log a_k. \\f] The\n278  * quantity \\f$\\mu_k^0(T)\\f$ is the chemical potential at unit activity,\n279  * which depends only on temperature and the pressure.\n280  * @{\n281  */\n282\n283  virtual void getActivityConcentrations(doublereal* c) const;\n284  virtual doublereal standardConcentration(size_t k=0) const;\n285\n286  /*!\n287  * Get the array of non-dimensional activities at the current solution\n288  * temperature, pressure, and solution concentration.\n289  *\n290  * (note solvent is on molar scale)\n291  *\n292  * @param ac Output activity coefficients. Length: m_kk.\n293  */\n294  virtual void getActivities(doublereal* ac) const;\n295\n296  /*!\n297  * Get the array of non-dimensional molality-based activity coefficients at\n298  * the current solution temperature, pressure, and solution concentration.\n299  *\n300  * (note solvent is on molar scale. The solvent molar\n301  * based activity coefficient is returned).\n302  *\n303  * @param acMolality Output Molality-based activity coefficients.\n304  * Length: m_kk.\n305  */\n306  virtual void getMolalityActivityCoefficients(doublereal* acMolality) const;\n307\n308  //@}\n309  /// @name Partial Molar Properties of the Solution\n310  //@{\n311\n312  //!Get the species chemical potentials: Units: J/kmol.\n313  /*!\n314  * This function returns a vector of chemical potentials of the species in\n315  * solution.\n316  *\n317  * \\f[\n318  * \\mu_k = \\mu^{o}_k(T,P) + R T \\ln(\\frac{m_k}{m^\\Delta})\n319  * \\f]\n320  * \\f[\n321  * \\mu_w = \\mu^{o}_w(T,P) +\n322  * R T ((X_w - 1.0) / X_w)\n323  * \\f]\n324  *\n325  * \\f$w \\f$ refers to the solvent species.\n326  * \\f$X_w \\f$ is the mole fraction of the solvent.\n327  * \\f$m_k \\f$ is the molality of the kth solute.\n328  * \\f$m^\\Delta \\f$ is 1 gmol solute per kg solvent.\n329  *\n330  * Units: J/kmol.\n331  *\n332  * @param mu Output vector of species chemical potentials. Length: m_kk.\n333  */\n334  virtual void getChemPotentials(doublereal* mu) const;\n335\n336  //! Returns an array of partial molar enthalpies for the species in the\n337  //! mixture.\n338  /*!\n339  * Units (J/kmol). For this phase, the partial molar enthalpies are equal to\n340  * the species standard state enthalpies.\n341  * \\f[\n342  * \\bar h_k(T,P) = \\hat h^{ref}_k(T) + (P - P_{ref}) \\hat V^0_k\n343  * \\f]\n344  * The reference-state pure-species enthalpies, \\f$\\hat h^{ref}_k(T) \\f$,\n345  * at the reference pressure,\\f$P_{ref} \\f$, are computed by the species\n346  * thermodynamic property manager. They are polynomial functions of\n347  * temperature.\n348  * @see MultiSpeciesThermo\n349  *\n350  * @param hbar Output vector of partial molar enthalpies.\n351  * Length: m_kk.\n352  */\n353  virtual void getPartialMolarEnthalpies(doublereal* hbar) const;\n354\n355  //! Returns an array of partial molar entropies of the species in the\n356  //! solution. Units: J/kmol.\n357  /*!\n358  * Maxwell's equations provide an insight in how to calculate this\n359  * (p.215 Smith and Van Ness)\n360  * \\f[\n361  * \\frac{d(\\mu_k)}{dT} = -\\bar{s}_i\n362  * \\f]\n363  * For this phase, the partial molar entropies are equal to the standard\n364  * state species entropies plus the ideal molal solution contribution.\n365  *\n366  * \\f[\n367  * \\bar{s}_k(T,P) = s^0_k(T) - R \\ln( \\frac{m_k}{m^{\\triangle}} )\n368  * \\f]\n369  * \\f[\n370  * \\bar{s}_w(T,P) = s^0_w(T) - R ((X_w - 1.0) / X_w)\n371  * \\f]\n372  *\n373  * The subscript, w, refers to the solvent species. \\f$X_w \\f$ is the mole\n374  * fraction of solvent. The reference-state pure-species entropies,\\f$375 * s^0_k(T) \\f$, at the reference pressure, \\f$P_{ref} \\f$, are computed by\n376  * the species thermodynamic property manager. They are polynomial functions\n377  * of temperature.\n378  * @see MultiSpeciesThermo\n379  *\n380  * @param sbar Output vector of partial molar entropies.\n381  * Length: m_kk.\n382  */\n383  virtual void getPartialMolarEntropies(doublereal* sbar) const;\n384\n385  // partial molar volumes of the species Units: m^3 kmol-1.\n386  /*!\n387  * For this solution, the partial molar volumes are equal to the constant\n388  * species molar volumes.\n389  *\n390  * Units: m^3 kmol-1.\n391  * @param vbar Output vector of partial molar volumes.\n392  */\n393  virtual void getPartialMolarVolumes(doublereal* vbar) const;\n394\n395  //! Partial molar heat capacity of the solution:. UnitsL J/kmol/K\n396  /*!\n397  * The kth partial molar heat capacity is equal to the temperature\n398  * derivative of the partial molar enthalpy of the kth species in the\n399  * solution at constant P and composition (p. 220 Smith and Van Ness).\n400  * \\f[\n401  * \\bar{Cp}_k(T,P) = {Cp}^0_k(T)\n402  * \\f]\n403  *\n404  * For this solution, this is equal to the reference state heat capacities.\n405  *\n406  * Units: J/kmol/K\n407  *\n408  * @param cpbar Output vector of partial molar heat capacities.\n409  * Length: m_kk.\n410  */\n411  virtual void getPartialMolarCp(doublereal* cpbar) const;\n412\n413  //@}\n414\n415  // -------------- Utilities -------------------------------\n416\n417  virtual bool addSpecies(shared_ptr<Species> spec);\n418\n419  virtual void initThermoXML(XML_Node& phaseNode, const std::string& id=\"\");\n420\n421  //! Report the molar volume of species k\n422  /*!\n423  * units - \\f$m^3 kmol^{-1} \\f$\n424  *\n425  * @param k Species index.\n426  */\n427  double speciesMolarVolume(int k) const;\n428\n429  /*!\n430  * Fill in a return vector containing the species molar volumes\n431  * units - \\f$m^3 kmol^{-1} \\f$\n432  *\n433  * @param smv Output vector of species molar volumes.\n434  */\n435  void getSpeciesMolarVolumes(double* smv) const;\n436  //@}\n437\n438 protected:\n439  //! Species molar volume \\f$m^3 kmol^{-1} \\f$\n441\n442  /**\n443  * The standard concentrations can have three different forms depending on\n444  * the value of the member attribute m_formGC, which is supplied in the XML\n445  * file.\n446  *\n447  * | m_formGC | ActivityConc | StandardConc |\n448  * | -------- | -------------------------------- | ------------------ |\n449  * | 0 | \\f${m_k}/ { m^{\\Delta}}\\f$ | \\f$1.0 \\f$ |\n450  * | 1 | \\f$m_k / (m^{\\Delta} V_k)\\f$ | \\f$1.0 / V_k \\f$ |\n451  * | 2 | \\f$m_k / (m^{\\Delta} V^0_0)\\f$ | \\f$1.0 / V^0_0\\f$ |\n452  */\n453  int m_formGC;\n454\n455 public:\n456  //! Cutoff type\n458\n459 private:\n460  //! vector of size m_kk, used as a temporary holding area.\n461  mutable vector_fp m_tmpV;\n462\n463  //! Logarithm of the molal activity coefficients\n464  /*!\n465  * Normally these are all one. However, stability schemes will change that\n466  */\n468 public:\n469  //! value of the solute mole fraction that centers the cutoff polynomials\n470  //! for the cutoff =1 process;\n471  doublereal IMS_X_o_cutoff_;\n472\n473  //! gamma_o value for the cutoff process at the zero solvent point\n474  doublereal IMS_gamma_o_min_;\n475\n476  //! gamma_k minimum for the cutoff process at the zero solvent point\n477  doublereal IMS_gamma_k_min_;\n478\n479  //! Parameter in the polyExp cutoff treatment. This is the slope of the f\n480  //! function at the zero solvent point. Default value is 0.6\n481  doublereal IMS_slopefCut_;\n482\n483  //! Parameter in the polyExp cutoff treatment. This is the slope of the g\n484  //! function at the zero solvent point. Default value is 0.0\n485  doublereal IMS_slopegCut_;\n486\n487  //! @name Parameters in the polyExp cutoff treatment having to do with rate\n488  //! of exp decay\n489  //! @{\n490  doublereal IMS_cCut_;\n491  doublereal IMS_dfCut_;\n492  doublereal IMS_efCut_;\n493  doublereal IMS_afCut_;\n494  doublereal IMS_bfCut_;\n495  doublereal IMS_dgCut_;\n496  doublereal IMS_egCut_;\n497  doublereal IMS_agCut_;\n498  doublereal IMS_bgCut_;\n499  //! @}\n500\n501 private:\n502  //! This function will be called to update the internally stored\n503  //! natural logarithm of the molality activity coefficients\n504  /*!\n505  * Normally the solutes are all zero. However, sometimes they are not,\n506  * due to stability schemes.\n507  *\n508  * gamma_k_molar = gamma_k_molal / Xmol_solvent\n509  *\n510  * gamma_o_molar = gamma_o_molal\n511  */\n512  void s_updateIMS_lnMolalityActCoeff() const;\n513\n514  //! Calculate parameters for cutoff treatments of activity coefficients\n515  /*!\n516  * Some cutoff treatments for the activity coefficients actually require\n517  * some calculations to create a consistent treatment.\n518  *\n519  * This routine is called during the setup to calculate these parameters\n520  */\n521  void calcIMSCutoffParams_();\n522 };\n523\n524 }\n525\n526 #endif\nvirtual ThermoPhase * duplMyselfAsThermoPhase() const\nDuplication routine for objects which inherit from ThermoPhase.\nvirtual doublereal intEnergy_mole() const\nMolar internal energy of the solution: Units: J/kmol.\nvirtual std::string type() const\nString indicating the thermodynamic model implemented.\nvirtual doublereal standardConcentration(size_t k=0) const\nReturn the standard concentration for the kth species.\nvirtual void getPartialMolarVolumes(doublereal *vbar) const\nvoid getSpeciesMolarVolumes(double *smv) const\ndoublereal IMS_slopefCut_\nParameter in the polyExp cutoff treatment.\ndoublereal IMS_slopegCut_\nParameter in the polyExp cutoff treatment.\nvirtual void getPartialMolarEntropies(doublereal *sbar) const\nReturns an array of partial molar entropies of the species in the solution.\nIdealMolalSoln()\nConstructor.\nvirtual void getActivities(doublereal *ac) const\nvector_fp m_tmpV\nvector of size m_kk, used as a temporary holding area.\nClass XML_Node is a tree-based representation of the contents of an XML file.\nDefinition: xml.h:97\nvirtual void getMolalityActivityCoefficients(doublereal *acMolality) const\nvirtual void getPartialMolarCp(doublereal *cpbar) const\nPartial molar heat capacity of the solution:. UnitsL J/kmol/K.\ndoublereal IMS_gamma_k_min_\ngamma_k minimum for the cutoff process at the zero solvent point\nvirtual void setDensity(const doublereal rho)\nOverridden setDensity() function is necessary because the density is not an independent variable...\nvirtual void setMolarDensity(const doublereal rho)\nOverridden setMolarDensity() function is necessary because the density is not an independent variable...\nvirtual doublereal cp_mole() const\nMolar heat capacity of the solution at constant pressure. Units: J/kmol/K.\ndoublereal IMS_gamma_o_min_\ngamma_o value for the cutoff process at the zero solvent point\nvector_fp m_speciesMolarVolume\nSpecies molar volume .\nHeader for intermediate ThermoPhase object for phases which employ molality based activity coefficien...\nBase class for a phase with thermodynamic properties.\nDefinition: ThermoPhase.h:93\nvirtual void getActivityConcentrations(doublereal *c) const\nThis method returns an array of generalized concentrations.\nvirtual doublereal entropy_mole() const\nMolar entropy of the solution. Units: J/kmol/K.\nvirtual void initThermoXML(XML_Node &phaseNode, const std::string &id=\"\")\nImport and initialize a ThermoPhase object using an XML tree.\nvirtual doublereal thermalExpansionCoeff() const\nThe thermal expansion coefficient. Units: 1/K.\nvirtual doublereal enthalpy_mole() const\nMolar enthalpy of the solution. Units: J/kmol.\nvirtual bool addSpecies(shared_ptr< Species > spec)\nint IMS_typeCutoff_\nCutoff type.\nvoid s_updateIMS_lnMolalityActCoeff() const\nThis function will be called to update the internally stored natural logarithm of the molality activi...\nint m_formGC\nThe standard concentrations can have three different forms depending on the value of the member attri...\nstd::vector< double > vector_fp\nTurn on the use of stl vectors for the basic array type within cantera Vector of doubles.\nDefinition: ct_defs.h:157\nvirtual doublereal isothermalCompressibility() const\nThe isothermal compressibility. Units: 1/Pa.\nvirtual void getChemPotentials(doublereal *mu) const\nGet the species chemical potentials: Units: J/kmol.\nvirtual void getPartialMolarEnthalpies(doublereal *hbar) const\nReturns an array of partial molar enthalpies for the species in the mixture.\nvoid calcDensity()\nCalculate the density of the mixture using the partial molar volumes and mole fractions as input...\nThis phase is based upon the mixing-rule assumption that all molality-based activity coefficients are...\nNamespace for the Cantera kernel.\nDefinition: application.cpp:29\nvirtual doublereal gibbs_mole() const\nMolar Gibbs function for the solution: Units J/kmol.\nvoid calcIMSCutoffParams_()\nCalculate parameters for cutoff treatments of activity coefficients.\ndouble speciesMolarVolume(int k) const\nReport the molar volume of species k.\ndoublereal IMS_X_o_cutoff_\nvalue of the solute mole fraction that centers the cutoff polynomials for the cutoff =1 process; ...\nvector_fp IMS_lnActCoeffMolal_\nLogarithm of the molal activity coefficients." ]

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[ "# A question in set-theoretic geology: if $M[G][K]=M[H][K],$ then can we conlude $M[G]=M[H]$?\n\nI was recently asked this interesting question on set-theoretic geology by Iian Smythe, a set-theory post-doc at Rutgers University; the problem arose in the context of one of this current research projects.\n\nQuestion. Assume that two product forcing extensions are the same $$M[G][K]=M[H][K],$$ where $M[G]$ and $M[H]$ are forcing extensions of $M$ by the same forcing notion $\\mathbb{P}$, and $K\\subset\\mathbb{Q}\\in M$ is both $M[G]$ and $M[H]$-generic with respect to this further forcing $\\mathbb{Q}$.  Can we conclude that $$M[G]=M[H]\\ ?$$ Can we make this conclusion at least in the special case that $\\mathbb{P}$ is adding a Cohen real and $\\mathbb{Q}$ is collapsing the continuum?", null, "It seems natural to hope for a positive answer, because we are aware of many such situations that arise with forcing, where indeed $M[G]=M[H]$. Nevertheless, the answer is negative. Indeed, we cannot legitimately make this conclusion even when both steps of forcing are adding merely a Cohen real. And such a counterexample implies that there is a counterexample of the type mentioned in the question, simply by performing further collapse forcing.\n\nTheorem. For any countable model $M$ of set theory, there are $M$-generic Cohen reals $c$, $d$ and $e$, such that\n\n1. The Cohen reals $c$ and $e$ are mutually generic over $M$.\n2. The Cohen reals $d$ and $e$ are mutually generic over $M$.\n3. These two pairs produce the same forcing extension $M[c][e]=M[d][e]$.\n4. But  the intermediate models are different $M[c]\\neq M[d]$.\n\nProof. Fix $M$, and let $c$ and $e$ be any two mutually generic Cohen reals over $M$. Let us view them as infinite binary sequences, that is, as elements of Cantor space. In the extension $M[c][e]$, let $d=c+e \\mod 2$, in each coordinate. That is, we get $d$ from $c$ by flipping bits, but only on coordinates that are $1$ in $e$. This is the same as applying a bit-flipping automorphism of the forcing, which is available in $M[e]$, but not in $M$. Since $c$ is $M[e]$-generic by reversing the order of forcing, it follows that $d$ also is $M[e]$-generic, since the automorphism is in $M[e]$. Thus, $d$ and $e$ are mutually generic over $M$. Further, $M[c][e]=M[d][e]$, because $M[e][c]=M[e][d]$, as $c$ and $d$ were isomorphic generic filters by an isomorphism in $M[e]$. But finally, $M[c]$ and $M[d]$ are not the same, because from $c$ and $d$ together we can construct $e$, because we can tell exactly which bits were flipped. $\\Box$\n\nIf one now follows the $e$ forcing with collapse forcing, one achieves a counterexample model of the type mentioned in the question, namely, with $M[c][e*K]=M[d][e*K]$, but $M[c]\\neq M[d]$.\n\nI have a feeling that my co-authors on a current paper in progress, Set-theoretic blockchains, on the topic of non-amalgamation in the generic multiverse, will tell me that the argument above is an instance of some of the theorems we prove in the latter part of that paper. (Miha, please tell me in the comments, if you see this, or tell me where I have seen this argument before; I think I made this argument or perhaps seen it before.) The paper is\n\n•", null, "M. E. Habič, J. D. Hamkins, L. D. Klausner, J. Verner, and K. J. Williams, “Set-theoretic blockchains,” Archive for Mathematical Logic, 2019.\n[Bibtex]\n@ARTICLE{HabicHamkinsKlausnerVernerWilliams2018:Set-theoretic-blockchains,\nauthor = {Miha E. Habič and Joel David Hamkins and Lukas Daniel Klausner and Jonathan Verner and Kameryn J. Williams},\ntitle = {Set-theoretic blockchains},\njournal=\"Archive for Mathematical Logic\",\nyear=\"2019\",\nmonth=\"Mar\",\nday=\"26\",\nabstract=\"Given a countable model of set theory, we study the structure of its generic multiverse, the collection of its forcing extensions and ground models, ordered by inclusion. Mostowski showed that any finite poset embeds into the generic multiverse while preserving the nonexistence of upper bounds. We obtain several improvements of his result, using what we call the blockchain construction to build generic objects with varying degrees of mutual genericity. The method accommodates certain infinite posets, and we can realize these embeddings via a wide variety of forcing notions, while providing control over lower bounds as well. We also give a generalization to class forcing in the context of second-order set theory, and exhibit some further structure in the generic multiverse, such as the existence of exact pairs.\",\nissn=\"1432-0665\",\ndoi=\"10.1007/s00153-019-00672-z\",\nnote = {},\nabstract = {},\neprint = {1808.01509},\narchivePrefix = {arXiv},\nprimaryClass = {math.LO},\nkeywords = {},\nsource = {},\nurl = {http://wp.me/p5M0LV-1M8},\n}\n\n.\n\n## 4 thoughts on “A question in set-theoretic geology: if $M[G][K]=M[H][K],$ then can we conlude $M[G]=M[H]$?”\n\n1.", null, "Iian Smythe on said:\n\nHi Joel, and thanks again for your answer to this question. The solution is very clear.\n\nOut of curiosity, could you say a bit about why various known results in set-theoretic geology, e.g., the definability of the ground model in a forcing extension, do not work here?\n\n• Thanks, it was a great question.\n\nFor the definability of the ground model, indeed, the model $M$ is definable in $M[c]$ and $M[c][e]$ and so on, using the parameter $P(\\omega)^M$. And $M[c]$ and $M[d]$ are each definable in the common extension $M[c][e]=M[d][e]$, using the parameters $P(\\omega)^{M[c]}$ and $P(\\omega)^{M[d]}$, respectively, but those parameters are not the same, and so this is why the ground-model definability theorem does not give you $M[c]=M[d]$.\n\n2. Joel, it is not clear to me that your argument follows from something in the paper (or at least I cannot see a straight path from one to the other). I think you advocated using bit-flipping automorphisms like this when we were discussing surgery some time ago, but I am not sure anything exactly like this argument came up.\n\n• OK, thanks for the comment. I guess there is a family resemblance in all the bit-flipping arguments." ]

[ null, "http://jdh.hamkins.org/wp-content/uploads/2018/07/Augustus_Leopold_Egg_-_The_Travelling_Companions_-_Google_Art_Project-2-1024x851.jpg", null, "http://jdh.hamkins.org/wp-content/plugins/papercite/img/external.png", null, "http://0.gravatar.com/avatar/6116fdf0d6dfdedc74d63c88e49e7121", null ]

[ "## Friday, April 10, 2009\n\nI have a word problem: The price of a pen was \\$5, The price of a pencil was \\$2. Miss Lee bought a number of pens and pencils for \\$26. How many pens and pencils did she buy?\nThe day before yesterday I taught my pupils this: I made a list for the pens : 1 x \\$5 = \\$5 , 2 x \\$5 = \\$10 and so on. And also a list for the pencils : 1 x \\$2 = \\$2 and so on. Miss Lee bought 4 pens and 3 pencils because (4 x \\$5) + (3 x \\$2) = \\$26. However, I realize that this method cannot be used in big numbers.\nMerry\nThe method you used is called make a list. It is a common problem-solving heuristic. Please continue to use it with your younger students. I wonder if Miss Lee could also buy 2 pens and some pencils. I know 5 pens is not possible because the money left is an odd number \\$1 and the price of a pencil is \\$2. Similarly, 1 pen or 3 pens are not possible. Students learn reasoning.\nIf students know algebra, they can set up equation 5x + 2y =26 where x is the number of pens and y is the number of pencils. As there is only one condition, you still need to use guess-and-check to solve this equation. Unless the problem says something about x + y.\nIf the value is not 26 but larger then the equation is 5x + 2y = k where k is the large number. A graph can be plotted and possible solutions seen on the graph. (With larger k the number of solutions increases).\nFor the young students, the method you use is probably the best. When they are older, they will learn to solve the same problem with larger k values.\n\n#### 1 comment:\n\n1.", null, "In my opinion, it is much easier for kids to visualise the mathematical patterns and relationships when the numbers are in the table form such as the guess-and-check method, which I use when coaching kids esp. when they deal with large numbers, when they are in the pre-algebra stage. (:" ]

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[ "Contents: CCRL 40/40 Downloads and Statistics June 15, 2019 Testing summary: Total: 1'059'406 games played by 2'473 programs 144801 CPU days (X2 4600+) White wins: 368'317 (34.8%) Black wins: 274'199 (25.9%) Draws: 416'890 (39.4%) White score: 54.4%\n\nEngine Details\n\n Options Show each game results\nLc0 0.20.1 w36089 64-bit#44  (3048+22\n−21\n)Quote\n Author: Gary Linscott and others Links: Homepage, Networks\nThis is one of the 6 Leela Chess versions we tested: Compare them!\n Opponent Elo Diff Results Score LOS Perf –   Laser 1.7 64-bit 3212 +15−15 (+164) 10.5 − 21.5(+2−13=17) 32.8%10.5 / 32 0.0% +62 = 0 0 0 = 0 = = = = 0 0 1 0 0 0 = 0 0 = = 1 = = = = 0 = = 0 = = –   Booot 6.3.1 64-bit 3203 +11−11 (+155) 9.5 − 22.5(+2−15=15) 29.7%9.5 / 32 0.0% +29 = 0 0 0 = 1 0 = 0 = = = = 0 1 0 = = 0 0 0 0 0 0 = 0 = = = = = 0 –   Fizbo 2 64-bit 3200 +9−9 (+152) 9.5 − 22.5(+3−16=13) 29.7%9.5 / 32 0.0% +23 = 0 = = = = 0 1 0 1 0 = 0 0 0 0 0 0 = = 0 0 0 0 = 0 0 = = = = 1 –   Andscacs 0.95 64-bit 3186 +12−12 (+138) 10.5 − 21.5(+4−15=13) 32.8%10.5 / 32 0.0% +27 0 0 0 0 = = 0 0 = 1 = 0 = = 0 0 = = 0 0 0 0 0 1 0 = = = 1 = = 1 –   Xiphos 0.4 64-bit 3152 +13−13 (+104) 11 − 21(+4−14=14) 34.4%11.0 / 32 0.0% +3 1 = = 0 = = 1 0 = 0 = 0 0 0 1 0 = 0 = 0 = 0 = 0 = 0 0 0 = = = 1 –   Arasan 21.3 64-bit 3115 +19−19 (+67) 6 − 8(+2−4=8) 42.9%6.0 / 14 0.0% +27 = = = 1 0 1 0 = = 0 0 = = = –   Chiron 4 64-bit 3114 +9−9 (+66) 12 − 20(+5−13=14) 37.5%12.0 / 32 0.0% −12 = 1 0 = 0 = 1 = 0 = 0 = 0 1 0 = 0 0 0 = 0 1 = 1 = = = 0 = 0 = 0 –   Schooner 2.0.34 64-bit 3110 +12−12 (+62) 18 − 13(+12−7=12) 58.1%18.0 / 31 0.0% +118 1 1 = 1 0 0 = 1 1 1 1 0 = 0 = 0 = = = = 1 1 1 0 1 = = = 0 = 1 –   Pedone 1.9 64-bit 3086 +19−19 (+38) 11.5 − 8.5(+6−3=11) 57.5%11.5 / 20 0.4% +85 = 1 = 1 0 = = 1 = = = 1 0 0 = 1 = = = 1 –   Wasp 3.60 64-bit 3043 +22−22 (−5) 8 − 8(+4−4=8) 50.0%8.0 / 16 63.1% −7 1 = 0 0 = = = = = = 1 0 1 = 1 0 –   RubiChess 1.4 64-bit 3040 +18−18 (−8) 7 − 9(+5−7=4) 43.8%7.0 / 16 69.9% −54 0 = 0 1 1 0 = 1 0 1 1 0 0 = 0 = –   RubiChess 1.3 64-bit 2979 +21−21 (−69) 8.5 − 5.5(+7−4=3) 60.7%8.5 / 14 100.0% +9 = 1 = 1 1 1 0 = 0 1 0 1 0 1 –   Pirarucu 2.9.5 64-bit 2967 +20−20 (−81) 9.5 − 4.5(+6−1=7) 67.9%9.5 / 14 100.0% +29 = 1 = = = 1 = 1 1 0 = = 1 1 –   Amoeba 2.8 64-bit 2926 +13−13 (−122) 20.5 − 11.5(+17−8=7) 64.1%20.5 / 32 100.0% −20 = = 1 1 1 0 1 = 0 0 1 0 1 1 = 1 = 1 1 1 0 0 1 = 0 1 = 0 1 1 1 1 –   Bobcat 8.0 64-bit 2924 +13−13 (−124) 19 − 13(+12−6=14) 59.4%19.0 / 32 100.0% −63 = 1 1 = 1 1 1 = 0 1 0 1 = = 1 = = = 0 0 1 = = = = = 1 = 1 0 1 0 –   Cheng 4.39 64-bit 2922 +11−11 (−126) 17.5 − 14.5(+13−10=9) 54.7%17.5 / 32 100.0% −91 1 1 0 1 = 1 = 0 0 1 = = = 1 0 0 = = = = 0 1 1 1 1 0 0 1 0 0 1 1 –   Spike 1.4 Leiden 2916 +8−8 (−132) 24.5 − 7.5(+20−3=9) 76.6%24.5 / 32 100.0% +56 0 1 1 = 1 1 0 1 1 1 1 1 1 = 1 1 = 1 1 1 = = = 1 = 1 = = 1 1 0 1 –   Deuterium 2018.1.35.514 64-bit 2913 +13−13 (−135) 20.5 − 11.5(+15−6=11) 64.1%20.5 / 32 100.0% −44 = = 0 1 1 = = 1 1 1 1 = 1 = 1 0 = 1 1 1 1 0 = 1 = 0 1 0 = 1 = 0 –   Quazar 0.4 64-bit 2906 +10−11 (−142) 19.5 − 12.5(+15−8=9) 60.9%19.5 / 32 100.0% −66 1 1 1 1 = 1 0 1 0 1 0 1 = = = 0 1 = = = 1 1 = 1 1 0 0 1 0 1 0 = –   Toga II 4.0 64-bit 2904 +14−14 (−144) 21.5 − 10.5(+17−6=9) 67.2%21.5 / 32 100.0% −24 1 1 1 = 1 1 = 1 0 = 1 1 1 0 1 0 1 = 1 1 = = = 0 1 1 1 = = 0 1 0 –   Atlas 3.91 64-bit 2900 +14−14 (−148) 23.5 − 8.5(+19−4=9) 73.4%23.5 / 32 100.0% +18 1 1 = 1 1 = 1 1 = 1 = = 1 1 1 1 1 0 = 1 1 1 = 0 0 1 0 1 1 1 = = –   Spark 1.0 64-bit 2894 +12−12 (−154) 22.5 − 9.5(+19−6=7) 70.3%22.5 / 32 100.0% −2 1 1 0 1 = = 1 = 1 1 1 1 = 1 1 1 0 0 1 0 1 = 1 0 1 = 1 = 1 1 1 0 –   Gaviota 1.0 64-bit 2860 +11−11 (−188) 23 − 9(+18−4=10) 71.9%23.0 / 32 100.0% −36 1 0 = = 1 = 1 1 1 1 1 0 1 1 1 1 = = 1 = 0 1 1 = 1 0 1 = = = 1 1 –   MinkoChess 1.3 64-bit 2854 +14−14 (−194) 25.5 − 6.5(+21−2=9) 79.7%25.5 / 32 100.0% +20 1 1 1 1 0 1 = 1 1 1 1 1 = = = = 1 1 1 = 1 1 = 1 1 1 = 1 = 0 1 1 –   Nemo 1.0.1 64-bit 2848 +12−12 (−200) 23 − 9(+17−3=12) 71.9%23.0 / 32 100.0% −58 1 1 1 0 1 0 1 1 1 = = = = 0 1 = 1 1 1 1 = 1 = = = = = 1 1 1 = 1 –   DisasterArea 1.65 64-bit 2846 +14−14 (−202) 26 − 6(+21−1=10) 81.3%26.0 / 32 100.0% +17 1 1 = = 0 1 1 1 = = 1 1 1 1 1 = 1 1 = = = 1 1 = 1 1 = 1 1 1 1 1 –   Tornado 8.0 64-bit 2844 +14−14 (−204) 23.5 − 8.5(+18−3=11) 73.4%23.5 / 32 100.0% −44 1 1 = 1 = = 0 = 1 0 = 1 = = 1 1 1 0 1 1 = 1 1 1 = 1 1 1 = 1 1 =\n\nRating changes by day", null, "Rating changes with played games", null, "Created in 2005-2013 by CCRL team Last games added on June 15, 2019" ]

[ null, "http://ccrl.chessdom.com/ccrl/4040/rating-history-by-day-graphs/Lc0_0_20_1_w36089_64-bit.png", null, "http://ccrl.chessdom.com/ccrl/4040/rating-history-by-day-graphs-2/Lc0_0_20_1_w36089_64-bit.png", null ]

[ "Search a number\nBaseRepresentation\nbin1011111010111…\n…10000010100011\n320222011112002021\n411331132002203\n5201044444112\n613531202311\n72322661630\noct575360243\n9228145067\n1099999907\n11514a1457\n12295a6397\n1317943757\n14d3d1187\n158ba4907\nhex5f5e0a3\n\n99999907 has 8 divisors (see below), whose sum is σ = 120300800. Its totient is φ = 81202824.\n\nThe previous prime is 99999847. The next prime is 99999931. The reversal of 99999907 is 70999999.\n\nIt is a sphenic number, since it is the product of 3 distinct primes.\n\nIt is a cyclic number.\n\nIt is not a de Polignac number, because 99999907 - 215 = 99967139 is a prime.\n\nIt is a Duffinian number.\n\nIt is not an unprimeable number, because it can be changed into a prime (99999307) by changing a digit.\n\nIt is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 375807 + ... + 376072.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (15037600).\n\nAlmost surely, 299999907 is an apocalyptic number.\n\n99999907 is the 5774-th hex number.\n\n99999907 is a deficient number, since it is larger than the sum of its proper divisors (20300893).\n\n99999907 is a wasteful number, since it uses less digits than its factorization.\n\n99999907 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 751905.\n\nThe product of its (nonzero) digits is 3720087, while the sum is 61.\n\nThe square root of 99999907 is about 9999.9953499989. The cubic root of 99999907 is about 464.1587394720.\n\nThe spelling of 99999907 in words is \"ninety-nine million, nine hundred ninety-nine thousand, nine hundred seven\".\n\nDivisors: 1 7 19 133 751879 5263153 14285701 99999907" ]

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[ "Cody\n\n# Problem 627. Compute a dot product of two vectors x and y\n\nSolution 2088119\n\nSubmitted on 10 Jan 2020 by Paul Johnson\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\nx = [1 2]; y= [1 3]; d_correct=7; assert(isequal(vector_dot(x,y),d_correct))\n\n2   Pass\nx = [1 -1]; y= [1 1]; d_correct=0; assert(isequal(vector_dot(x,y),d_correct))" ]

[ null ]

[ "Courses\n\n# Solutions of Reflection of Light (Page No- 209) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev\n\n## Class 10 : Solutions of Reflection of Light (Page No- 209) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev\n\nThe document Solutions of Reflection of Light (Page No- 209) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Physics Solutions By Lakhmir Singh & Manjit Kaur.\nAll you need of Class 10 at this link: Class 10\n\nLakhmir Singh Physics Class 10 Solutions Page No:209\n\nQuestion 1: An object is kept at a distance of 5 cm in front of a convex mirror of focal length 10 cm. Calculate the position and magnification of the image and state its nature.\n\nSolution :\n\nu = -5cm f -10cm\n\nWe know that", null, "The position of the image is 3.33 cm behind the convex mirror. Magnification. m= -v/u = -3.33/-5 = 0.66 The image is virtual and erect.\n\nQuestion 2: An object is placed at a distance of 10 cm from a convex mirror of focal length 5 cm.\nDraw a ray-diagram showing the formation of image.\nState two characteristics of the image formed.\nCalculate the distance of the image from mirror.\n\nSolution :", null, "(ii) The image formed is diminished and erect.\n(iii) u=-10cm. f = 5cm", null, "Question 3: An object is placed at a distance of 6 cm from a convex mirror of focal length 12 cm. Find the position and nature of the image.\n\nSolution : u= -6cm f = 12cm. v = ?\n\nWe know that", null, "Image will be formed 4 cm behind the mirror.\n\nSince the image is formed behind the convex mirror, it is virtual and erect.\n\nQuestion 4: An object placed 20 cm in front of a mirror is found to have an image 15 cm (a) in front of it, (b) behind the mirror. Find the focal length of the mirror and the kind of mirror in each case.\n\nSolution : (a) u =-20cm, v = -15cm\n\nWe know that", null, "The mirror is a concave mirror.\n(b) u = -20cm v = 15cm\n\nWe know that", null, "The mirror is a convex mirror.\n\nQuestion 5: An arrow 2.5 cm high is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the nature, position and size of the image formed.\n\nSolution : h= 2.5cm.u = -25cm. f = 20cm\n\nWe know that", null, "The image is formed 11.1cm behind the convex mirror.\n\nNow.", null, "The image is virtual. erect and 1.11cm tall.\n\nQuestion 6: A convex mirror used as a rear-view mirror in a car has a radius of curvature of 3 m. If a bus is located at a distance of 5 m from this mirror, find the position of image. What is the nature of the image ?\n\nSolution : Ra = 3 m  u = -5m", null, "We knoe that", null, "The image is formed 1.15 m behind the mirror.\n\nThe image is virtual and erect.\n\nQuestion 7: A diverging mirror of radius of curvature 40 cm forms an image which is half the height of the object. Find the object and image positions.\n\nSolution : R = 40cm\n\nf = R/2 = 10/2 = 20tm\n\nImage is half the height of the object", null, "So. the object is placed 20 cm in front of the mirror and the image is formed 10cm behind the mirror.\n\nQuestion 8: The radius of curvature of a convex mirror used as a rear view mirror in a moving car is 2.0 m. A truck is coming from behind it at a distance of 3.5 m. Calculate (a) position, and (b) size, of the image relative to the size of the truck. What will be the nature of the image ?\n\nSolution :", null, "", null, "So, the image is formed 0.77m behind the mirror.", null, "As m is positive, so image formed is virtual and erect\n\nQuestion 9: (a) Draw a diagram to represent a convex mirror. On this diagram mark principal axis, principal focus F\nand the centre of curvature C if the focal length of convex mirror is 3 cm.\n(b) An object 1 cm tall is placed 30 cm in front of a convex mirror of focal length 20 cm. Find the size and position of the image formed by the convex mirror.\n\nSolution :", null, "(b) h1 = 1cm. u = -30cm f = 20cm. h2= ?, v = ?", null, "The image is formed 12 cm behind the mirror.", null, "Question 10: A shop security mirror 5.0 m from certain items displayed in the shop produces one-tenth magnification.\n(a) What is the type of mirror ?\n(b) What is the radius of curvature of the mirror ?\n\nSolution :\n\n(a) The mirror is of convex type.\n(b) u = -5cm. m = 1/10\n\nWe have", null, "", null, "Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!\n\n94 docs\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;" ]

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[ "# #181424 Color Information\n\nIn a RGB color space, hex #181424 is composed of 9.4% red, 7.8% green and 14.1% blue. Whereas in a CMYK color space, it is composed of 33.3% cyan, 44.4% magenta, 0% yellow and 85.9% black. It has a hue angle of 255 degrees, a saturation of 28.6% and a lightness of 11%. #181424 color hex could be obtained by blending #302848 with #000000. Closest websafe color is: #000033.\n\n• R 9\n• G 8\n• B 14\nRGB color chart\n• C 33\n• M 44\n• Y 0\n• K 86\nCMYK color chart\n\n#181424 color description : Very dark (mostly black) blue.\n\n# #181424 Color Conversion\n\nThe hexadecimal color #181424 has RGB values of R:24, G:20, B:36 and CMYK values of C:0.33, M:0.44, Y:0, K:0.86. Its decimal value is 1578020.\n\nHex triplet RGB Decimal 181424 `#181424` 24, 20, 36 `rgb(24,20,36)` 9.4, 7.8, 14.1 `rgb(9.4%,7.8%,14.1%)` 33, 44, 0, 86 255°, 28.6, 11 `hsl(255,28.6%,11%)` 255°, 44.4, 14.1 000033 `#000033`\nCIE-LAB 7.424, 6.558, -10.353 0.945, 0.822, 1.778 0.267, 0.232, 0.822 7.424, 12.255, 302.353 7.424, 0.517, -6.833 9.066, 2.746, -5.281 00011000, 00010100, 00100100\n\n# Color Schemes with #181424\n\n• #181424\n``#181424` `rgb(24,20,36)``\n• #202414\n``#202414` `rgb(32,36,20)``\nComplementary Color\n• #141824\n``#141824` `rgb(20,24,36)``\n• #181424\n``#181424` `rgb(24,20,36)``\n• #201424\n``#201424` `rgb(32,20,36)``\nAnalogous Color\n• #182414\n``#182414` `rgb(24,36,20)``\n• #181424\n``#181424` `rgb(24,20,36)``\n• #242014\n``#242014` `rgb(36,32,20)``\nSplit Complementary Color\n• #142418\n``#142418` `rgb(20,36,24)``\n• #181424\n``#181424` `rgb(24,20,36)``\n• #241814\n``#241814` `rgb(36,24,20)``\n• #142024\n``#142024` `rgb(20,32,36)``\n• #181424\n``#181424` `rgb(24,20,36)``\n• #241814\n``#241814` `rgb(36,24,20)``\n• #202414\n``#202414` `rgb(32,36,20)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #020203\n``#020203` `rgb(2,2,3)``\n• #0d0b14\n``#0d0b14` `rgb(13,11,20)``\n• #181424\n``#181424` `rgb(24,20,36)``\n• #231d34\n``#231d34` `rgb(35,29,52)``\n• #2e2645\n``#2e2645` `rgb(46,38,69)``\n• #392f55\n``#392f55` `rgb(57,47,85)``\nMonochromatic Color\n\n# Alternatives to #181424\n\nBelow, you can see some colors close to #181424. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #141424\n``#141424` `rgb(20,20,36)``\n• #151424\n``#151424` `rgb(21,20,36)``\n• #171424\n``#171424` `rgb(23,20,36)``\n• #181424\n``#181424` `rgb(24,20,36)``\n• #191424\n``#191424` `rgb(25,20,36)``\n• #1b1424\n``#1b1424` `rgb(27,20,36)``\n• #1c1424\n``#1c1424` `rgb(28,20,36)``\nSimilar Colors\n\n# #181424 Preview\n\nThis text has a font color of #181424.\n\n``<span style=\"color:#181424;\">Text here</span>``\n#181424 background color\n\nThis paragraph has a background color of #181424.\n\n``<p style=\"background-color:#181424;\">Content here</p>``\n#181424 border color\n\nThis element has a border color of #181424.\n\n``<div style=\"border:1px solid #181424;\">Content here</div>``\nCSS codes\n``.text {color:#181424;}``\n``.background {background-color:#181424;}``\n``.border {border:1px solid #181424;}``\n\n# Shades and Tints of #181424\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, #07060b is the darkest color, while #fdfdfe is the lightest one.\n\n• #07060b\n``#07060b` `rgb(7,6,11)``\n• #100d17\n``#100d17` `rgb(16,13,23)``\n• #181424\n``#181424` `rgb(24,20,36)``\n• #201b31\n``#201b31` `rgb(32,27,49)``\n• #29223d\n``#29223d` `rgb(41,34,61)``\n• #31294a\n``#31294a` `rgb(49,41,74)``\n• #3a3056\n``#3a3056` `rgb(58,48,86)``\n• #423763\n``#423763` `rgb(66,55,99)``\n• #4a3e70\n``#4a3e70` `rgb(74,62,112)``\n• #53457c\n``#53457c` `rgb(83,69,124)``\n• #5b4c89\n``#5b4c89` `rgb(91,76,137)``\n• #645395\n``#645395` `rgb(100,83,149)``\n• #6c5aa2\n``#6c5aa2` `rgb(108,90,162)``\n• #7766aa\n``#7766aa` `rgb(119,102,170)``\n• #8272b1\n``#8272b1` `rgb(130,114,177)``\n• #8d7fb8\n``#8d7fb8` `rgb(141,127,184)``\n• #988cbf\n``#988cbf` `rgb(152,140,191)``\n• #a498c6\n``#a498c6` `rgb(164,152,198)``\n• #afa5cd\n``#afa5cd` `rgb(175,165,205)``\n• #bab2d4\n``#bab2d4` `rgb(186,178,212)``\n• #c5bedb\n``#c5bedb` `rgb(197,190,219)``\n• #d1cbe2\n``#d1cbe2` `rgb(209,203,226)``\n• #dcd7e9\n``#dcd7e9` `rgb(220,215,233)``\n• #e7e4f0\n``#e7e4f0` `rgb(231,228,240)``\n• #f2f1f7\n``#f2f1f7` `rgb(242,241,247)``\n• #fdfdfe\n``#fdfdfe` `rgb(253,253,254)``\nTint Color Variation\n\n# Tones of #181424\n\nA tone is produced by adding gray to any pure hue. In this case, #1b1a1e is the less saturated color, while #0e0137 is the most saturated one.\n\n• #1b1a1e\n``#1b1a1e` `rgb(27,26,30)``\n• #1a1820\n``#1a1820` `rgb(26,24,32)``\n• #191622\n``#191622` `rgb(25,22,34)``\n• #181424\n``#181424` `rgb(24,20,36)``\n• #171226\n``#171226` `rgb(23,18,38)``\n• #161028\n``#161028` `rgb(22,16,40)``\n• #150e2a\n``#150e2a` `rgb(21,14,42)``\n• #140b2d\n``#140b2d` `rgb(20,11,45)``\n• #13092f\n``#13092f` `rgb(19,9,47)``\n• #120731\n``#120731` `rgb(18,7,49)``\n• #100533\n``#100533` `rgb(16,5,51)``\n• #0f0335\n``#0f0335` `rgb(15,3,53)``\n• #0e0137\n``#0e0137` `rgb(14,1,55)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #181424 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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[ "# Derivative homework help\n\nDerivative homework help Rated 5 stars, based on 26 customer reviews From \\$8.35 per page Available! Order now!\n\nOur tutors are available round-the- clock to meet your tight assignment deadline. We shall verify a special case of this theorem at the end of this section. The value of the derivatives fluctuates with the fluctuations in the assets on whose performance the derivate is dependent. Nevertheless, they have to be ready on time. Derivative at a value slope at a value tangent lines normal lines points of horizontal tangents rolle's theorem mean value theorem intervals of increase and decrease intervals of concavity relative extrema absolute extrema optimization curve sketching comparing a function and its derivatives motion along a line related rates differentials. This said entity is commonly referred to as the. First and second derivatives math homework help my class. Challenge homework: differentiation rules 30m. You will soon see those trig derivatives are instrumental in modeling situations in the real-world like motion, vibrations, waves and more. I would appreciate the help, thank you. Understand your high school math homework by watching free math videos online from your own free math help tutor.\n\nBe certain to bookmark this page so that you. One way derivative homework help to do that is through some trigonometric identities. Studying for tomorrow's ap calculus ab derivative homework help exam at midnight. A derivative is a security or a contract whose value is derived from the performance of the underlying asset or assets. Indeed, we have so we will use the product formula to get which implies using the trigonometric formula, we get once this is done, you may ask about the derivative of. Live homework help and tutoring. Ap calculus absolute extrema 5. Our tutors have many years of industry experience and have had years of experience providing first principle of derivatives homework help. From computation of derivatives to u-substitution, our tutors know it all. A student solutions manual (available in the bookstore) has complete solutions for odd-numbered problems in the text. This is where we step in, the.\n\nIn so doing, we'll see. Based on that glorious table, calculate the following: (a) (b. I can help you to solve your math problems asap in the following subjects: calculus, probability, statistics, algebra, trigonometry and all math topics you are needing i help students around the world since 2005 and you will be satisfied with my work for sure. About us careers blog contact us. Have no idea of what a binomial is and what its role. Symbolab: equation search and math solver - cv writing service oxfordshire solves algebra, trigonometry and calculus problems step by step. Begin studying for the ap(r) calculus ab or bc test by examining limits and continuity. Instant help solution: we offer instant derivative market help solution to students who often need last-minute help.\n\n:\n• Let c + h be any other point of this interval which lies to the right or left of c according as h is positive or n-gative\n• Homework help homework help online homework helper\n• F(x) = (x2 (ovo o (abs o max 2) 4) absolute max of 8 at x absolute min of -19 at x = 2 absolute max of 27 at x = 2 absolute min of -l at x -o\n• As a consequence, derivatives have dramatically increased in popularity buy my essay for a procedure of generating income\n• Derivatives have no direct worth in and of themselves- their worth is based upon the anticipated future cost motions of their hidden possession\n• Differentiation is a method to calculate the rate of change (or the slope at a point on the graph)\n• Financial derivatives assignment homework help\n\nGeomatry, math december 11, 2012 leave a comment. The derivative is simply an agreement between a couple of parties. Typically, the seller receives money in exchange for an agreement. Find the derivative of f(x). Best derivative homework help assignment help online. Most of the students without wasting time in asking 'do my derivative market assignment.\n\n:\n1. Do my math homework for me - online hw help -\n2. Free 24*7 customer support is available for any questions related to assignment or homework help\n3. Failure to adhere to all the crucial steps can result in a wrong answer\n4. Close o posted by 40 minutes ago\n5. Core homework: differentiation rules 30m\n\nBy keeping y as constant, z = f(x,y) naturally becomes a function of x alone and so, the derivative of the function primary homework help river thames facts can be calculated with respect to x. Tutorpace provides online tutoring, homework help, microsoft excel homework help test prep for k-12 and college students. Procrastination can have bad consequences, as the number of assignments one hasn't completed can become a real problem. Ap calculus help, notes, equations and answers.\n\n:\n2. Need help figuring out the table of derivatives in your calculus primary homework help victorian christmas homework\n3. Morris has suggested that past research and what pompeii looked like, covered with volcanic ash\n4. These assignments tend to have short deadlines making them hard to complete in timely manner\n5. Still need help after using our trigonometry resources\n6. Because the slopes of perpendicular lines (neither of which is vertical) are negative reciprocals of one another, the slope of the normal line to the graph of f(x) is -1/ f\n\nHelp with government homework o best website for writing. Solve any calculus problem with step-by-step math problem. When a student is not able to complete their project given by schools and colleges marks from exams get deducted and this is a very bad thing for that student's career. Math introduction to derivatives free derivative homework help homework help.\n\nDerivatives homework help & derivatives tutors provide 24 * 7 services. Let us the black death homework help find the derivative of. We always guarantee high quality papers within your given timeline 24/7 customer support +1(805) 568 7317.\n\n:\n1. The derivative of would just derivative homework help be\n2. We have a team of over 150 writers from the nursing field\n3. Derivatives assignment & homework help service\n4. Its worth is set by changes in the said asset\n5. That is, the derivative of the integral of/ with respect to the upper limit of integration x is equal to f evaluated at x\n\nLearn more about subplot graph matlab, derivative subplot graph matlab, homework, no attempt. The number of tasks may vary greatly from subject to subject. For the student who needs help with calculus, this is a reference to help provide easy access to derivative rules.\n\n:\n• Homework problems, but each student should submit derivative homework help their own work\n• Connect to our tutors now for all subjects help, available 24*7\n• The most typical kinds of derivatives are futures, options, forwards and swaps\n• Creating a properly-researched derivative market assignment in such limited time becomes impossible\n• Symbolab math solver - step by step calculator\n• I need help with this: compute the derivative of the given\n• This is a 160 point assignment\n• F(x)=9x7/5-5x2+104 using the rules that the derivative of a sum of functions is the sum of the derivatives, the derivative of xn = n*x(n-1), the derivative of\n\nGiven that fact, don't be afraid to ask friends, tas, or instructors for help when you need it.\n\n:\n1. Derivatives are a financial concept, a derivative can be defined as a contract which derives its worth on the basis of the performances of an underlying entity in a business, an entity which can be an index, asset or an interest rate\n2. The derivative is a concept that is at the root of are two ways of introducing this concept, the geometrical way (as the slope of a curve), and the physical way (as a rate of change)\n3. This derivative is partial derivative\n4. We have videos explaining derivative notion, instantaneous rate of change, and more\n5. Math homework help, get assistance with your math home work\n6. The derivative - calculus - math - homework resources\n7. How to evaluate the limit of a function as x goes to infinity (or minus infinity), and how to determine the horizontal asymptote of its graph\n\nX* +6x2 +9 8'(x) =_-122 using the second derivative test what are the x-values where local extrema exist in. The following variables and constants are reserved: e = euler's number, the base of the exponential function (); i = imaginary number (i. Derivatives assignment help the most typical underlying possessions include stocks, bonds, products, currencies, interest rates, and market indexes. See screenshots and samples to view zgrapher in action. Home math homework help partial derivatives. Get albert's free 2020 ap(r) calculus ab-bc review guide to help with your exam prep here. My homework help is always there to help the students with the derivatives homework help whoever needs derivative homework help this help. Derivatives market homework help what is a derivative. Calculus applications of the derivative. There are instances when students remember about their assignment at the last minute. Want to learn algebra from the basics. At statistics assignment experts, you are guaranteed the best possible support at an economical rate. How to compute the average velocity of an object moving in one dimension from a graph of its position vs. Results of your work http://beta.archivitamins.com/prize.php?c=proposal-examples&contentID=205&essay-editing-cheap may be exported to word or saved as bmp, gif and pcx pictures. Hints help you try the next step on your own. The second derivative calculator is a tool which helps to find the derivative of the derivative of function f. Our tutors who provide first principle of derivatives help are highly qualified. They offer practical ways of mitigating risks, locking in prices, and hedging against unfavorable movements at a limited. The derivative market has three types of traders or we can say three types of participants in. Again we will see how. By continuing to use this site you consent to the use of cookies on your device as described in our cookie policy unless you have disabled them. Cheap dissertation help that based homework help for tens. Note: the online texts for math 1300, 1310, 1313, 1314, 1330, 2303, 2311 and 3321 have moved to courseware. Derivatives assignment help and homework help service. Optimization using the second derivative test calculus applications of the derivative.\n\nAnalyze various representations of essay writers online cheap functions and form the conceptual foundation of all calculus: limits.\n\n:\n1. Some values of the functions and their derivatives are given in the following table\n2. Start the homework at least a few days before it is due to be derivative homework help sure you can give each problem your best effort\n3. Our calculators will derivative homework help give you the answer and take you through the whole process, step-by-step\n4. Our experts are available 24x7 globally in order to help the students of different academic levels\n5. Turn in the homework to the assignments tab in your courseware file\n6. Did the california cellphone ban minimize accidents\n7. Are you aware that all homework assignments is among the best derivatives online tutors\n8. To send me them now you can top content writing companies in bangalore contact me in our\n9. Example i illustrates theorem l" ]

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[ "# #d0d1dc Color Information\n\nIn a RGB color space, hex #d0d1dc is composed of 81.6% red, 82% green and 86.3% blue. Whereas in a CMYK color space, it is composed of 5.5% cyan, 5% magenta, 0% yellow and 13.7% black. It has a hue angle of 235 degrees, a saturation of 14.6% and a lightness of 83.9%. #d0d1dc color hex could be obtained by blending #ffffff with #a1a3b9. Closest websafe color is: #cccccc.\n\n• R 82\n• G 82\n• B 86\nRGB color chart\n• C 5\n• M 5\n• Y 0\n• K 14\nCMYK color chart\n\n#d0d1dc color description : Light grayish blue.\n\n# #d0d1dc Color Conversion\n\nThe hexadecimal color #d0d1dc has RGB values of R:208, G:209, B:220 and CMYK values of C:0.05, M:0.05, Y:0, K:0.14. Its decimal value is 13685212.\n\nHex triplet RGB Decimal d0d1dc `#d0d1dc` 208, 209, 220 `rgb(208,209,220)` 81.6, 82, 86.3 `rgb(81.6%,82%,86.3%)` 5, 5, 0, 14 235°, 14.6, 83.9 `hsl(235,14.6%,83.9%)` 235°, 5.5, 86.3 cccccc `#cccccc`\nCIE-LAB 84.058, 1.713, -5.549 61.729, 64.178, 76.842 0.304, 0.317, 64.178 84.058, 5.808, 287.15 84.058, -1.183, -8.816 80.111, -2.653, -0.793 11010000, 11010001, 11011100\n\n# Color Schemes with #d0d1dc\n\n• #d0d1dc\n``#d0d1dc` `rgb(208,209,220)``\n• #dcdbd0\n``#dcdbd0` `rgb(220,219,208)``\nComplementary Color\n• #d0d7dc\n``#d0d7dc` `rgb(208,215,220)``\n• #d0d1dc\n``#d0d1dc` `rgb(208,209,220)``\n• #d5d0dc\n``#d5d0dc` `rgb(213,208,220)``\nAnalogous Color\n• #d7dcd0\n``#d7dcd0` `rgb(215,220,208)``\n• #d0d1dc\n``#d0d1dc` `rgb(208,209,220)``\n• #dcd5d0\n``#dcd5d0` `rgb(220,213,208)``\nSplit Complementary Color\n• #d1dcd0\n``#d1dcd0` `rgb(209,220,208)``\n• #d0d1dc\n``#d0d1dc` `rgb(208,209,220)``\n• #dcd0d1\n``#dcd0d1` `rgb(220,208,209)``\n• #d0dcdb\n``#d0dcdb` `rgb(208,220,219)``\n• #d0d1dc\n``#d0d1dc` `rgb(208,209,220)``\n• #dcd0d1\n``#dcd0d1` `rgb(220,208,209)``\n• #dcdbd0\n``#dcdbd0` `rgb(220,219,208)``\n• #a4a6bb\n``#a4a6bb` `rgb(164,166,187)``\n• #b3b4c6\n``#b3b4c6` `rgb(179,180,198)``\n• #c1c3d1\n``#c1c3d1` `rgb(193,195,209)``\n• #d0d1dc\n``#d0d1dc` `rgb(208,209,220)``\n• #dfdfe7\n``#dfdfe7` `rgb(223,223,231)``\n• #edeef2\n``#edeef2` `rgb(237,238,242)``\n• #fcfcfd\n``#fcfcfd` `rgb(252,252,253)``\nMonochromatic Color\n\n# Alternatives to #d0d1dc\n\nBelow, you can see some colors close to #d0d1dc. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #d0d4dc\n``#d0d4dc` `rgb(208,212,220)``\n• #d0d3dc\n``#d0d3dc` `rgb(208,211,220)``\n• #d0d2dc\n``#d0d2dc` `rgb(208,210,220)``\n• #d0d1dc\n``#d0d1dc` `rgb(208,209,220)``\n• #d0d0dc\n``#d0d0dc` `rgb(208,208,220)``\n• #d1d0dc\n``#d1d0dc` `rgb(209,208,220)``\n• #d2d0dc\n``#d2d0dc` `rgb(210,208,220)``\nSimilar Colors\n\n# #d0d1dc Preview\n\nThis text has a font color of #d0d1dc.\n\n``<span style=\"color:#d0d1dc;\">Text here</span>``\n#d0d1dc background color\n\nThis paragraph has a background color of #d0d1dc.\n\n``<p style=\"background-color:#d0d1dc;\">Content here</p>``\n#d0d1dc border color\n\nThis element has a border color of #d0d1dc.\n\n``<div style=\"border:1px solid #d0d1dc;\">Content here</div>``\nCSS codes\n``.text {color:#d0d1dc;}``\n``.background {background-color:#d0d1dc;}``\n``.border {border:1px solid #d0d1dc;}``\n\n# Shades and Tints of #d0d1dc\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, #070709 is the darkest color, while #fdfdfd is the lightest one.\n\n• #070709\n``#070709` `rgb(7,7,9)``\n• #0f1014\n``#0f1014` `rgb(15,16,20)``\n• #181820\n``#181820` `rgb(24,24,32)``\n• #20212b\n``#20212b` `rgb(32,33,43)``\n• #282a36\n``#282a36` `rgb(40,42,54)``\n• #313241\n``#313241` `rgb(49,50,65)``\n• #393b4d\n``#393b4d` `rgb(57,59,77)``\n• #414358\n``#414358` `rgb(65,67,88)``\n• #4a4c63\n``#4a4c63` `rgb(74,76,99)``\n• #52556e\n``#52556e` `rgb(82,85,110)``\n• #5b5d7a\n``#5b5d7a` `rgb(91,93,122)``\n• #636685\n``#636685` `rgb(99,102,133)``\n• #6b6e90\n``#6b6e90` `rgb(107,110,144)``\n• #767999\n``#767999` `rgb(118,121,153)``\n• #8184a1\n``#8184a1` `rgb(129,132,161)``\n• #8d8faa\n``#8d8faa` `rgb(141,143,170)``\n• #989ab2\n``#989ab2` `rgb(152,154,178)``\n• #a3a5bb\n``#a3a5bb` `rgb(163,165,187)``\n• #aeb0c3\n``#aeb0c3` `rgb(174,176,195)``\n• #babbcb\n``#babbcb` `rgb(186,187,203)``\n• #c5c6d4\n``#c5c6d4` `rgb(197,198,212)``\n• #d0d1dc\n``#d0d1dc` `rgb(208,209,220)``\n• #dbdce4\n``#dbdce4` `rgb(219,220,228)``\n• #e6e7ed\n``#e6e7ed` `rgb(230,231,237)``\n• #f2f2f5\n``#f2f2f5` `rgb(242,242,245)``\n• #fdfdfd\n``#fdfdfd` `rgb(253,253,253)``\nTint Color Variation\n\n# Tones of #d0d1dc\n\nA tone is produced by adding gray to any pure hue. In this case, #d3d4d9 is the less saturated color, while #adb4ff is the most saturated one.\n\n• #d3d4d9\n``#d3d4d9` `rgb(211,212,217)``\n• #d0d1dc\n``#d0d1dc` `rgb(208,209,220)``\n• #cdcedf\n``#cdcedf` `rgb(205,206,223)``\n• #cacce2\n``#cacce2` `rgb(202,204,226)``\n• #c7c9e5\n``#c7c9e5` `rgb(199,201,229)``\n• #c3c6e9\n``#c3c6e9` `rgb(195,198,233)``\n• #c0c4ec\n``#c0c4ec` `rgb(192,196,236)``\n• #bdc1ef\n``#bdc1ef` `rgb(189,193,239)``\n• #babff2\n``#babff2` `rgb(186,191,242)``\n• #b7bcf5\n``#b7bcf5` `rgb(183,188,245)``\n• #b4b9f8\n``#b4b9f8` `rgb(180,185,248)``\n• #b0b7fc\n``#b0b7fc` `rgb(176,183,252)``\n``#adb4ff` `rgb(173,180,255)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #d0d1dc 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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