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[ "", null, "## International Compressor Engineering Conference\n\n2016\n\n#### Keywords\n\ncompressor, performance, data, representation, uncertainty\n\n#### Abstract\n\nAHRI-540 is the current standard defining the methods for representing compressor performance data. While this standard is widely used across the industry, multiple factors contribute to inaccuracies in data representation including measurement uncertainty, regression uncertainty, compressor to compressor variation, and operation outside of the normal operating envelope (extrapolation). In addition, the number and location of points in the operating envelop also affects the accuracy of the resulting 10-coefficient polynomial. The measurement uncertainty is well known and can be factored into the data reduction. However, the measurement uncertainty is generally not propagated into the regression uncertainty and hence the overall uncertainty in prediction using the polynomial is not known. This uncertainty also changes according to the number of samples used for developing the polynomial. Â As a first step of the evaluation, a regression uncertainty analysis was conducted using a Monte Carlo simulation method. Results showed that the average uncertainty in mass flow rate prediction can be as high as 4% and that in power prediction can be as high as 5%. The worst case maximum absolute error in predicted mass flow rate across all data sets was 17% and that for power was 9%. Error in predicted power and mass flow rate is higher for larger capacity compressors. For most compressors, the high errors occur in the region of the envelope with low suction and low discharge dew point temperatures. Â A study of sampling considering different sample sizes and multiple sampling methods was conducted. Two additional methods of compressor performance representation were also analyzed. This analysis was presented with several challenges, particularly since the compressor operating envelope is a non-rectangular domain. A sampling method using Latin Hypercube Design (LHS) and a proposed alternative sampling method based on polygonal design of experiments (PDOE) were evaluated. The resulting models were validated against a measured data set of more than 600 points encompassing the operating envelope for each compressor. In general, both the LHS and PDOE methods yielded similar errors in mass flow rate for samples sizes of 12, 14 and 16. Thus, for mass flow rate, it is possible to build a model with 12 systematically selected test points. For power prediction, the average error for the LHS and PDOE methods using AHRI540 and two other methods was lower than 2% for all sample sizes.\n\nCOinS" ]

[ null, "https://docs.lib.purdue.edu/assets/md5images/b48b05880dda7ccbf38cc0f9c45421cc.gif", null ]

[ "Courses\n\n# Modern Physics II JEE Advanced - Physics, Solution by DC Pandey NEET Notes | EduRev\n\n## DC Pandey (Questions & Solutions) of Physics: NEET\n\nCreated by: Ciel Knowledge\n\n## NEET : Modern Physics II JEE Advanced - Physics, Solution by DC Pandey NEET Notes | EduRev\n\n``` Page 1\n\nAssertion and Reason\nDirections : Choose the correct option.\n(a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.\n(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) If Assertion is true, but the Reason is false.\n(d) If Assertion is false but the Reason is true.\nQ 1. Assertion: Rate of radioactivity can not be increased or decreased by increasing or decreasing\npressure or temperature.\nReason : Rate depends on number of nuclei present in the radioactive sample.\nQ 2. Assertion: Only those nuclei which are heavier than lead are radioactive.\nReason : Nuclei of elements heavier than lead are unstable.\nQ 3. Assertion: After emission of one particle and two ?-particles, atomic number remains\nunchanged.\nReason : Mass number changes by four.\nQ 4. Assertion : ?-rays are produced by the transition of a nucleus from some higher energy state to\nsome lower energy state.\nReason : Electromagnetic waves are always produced by the transition process.\nQ 5. Assertion : During ?-decay a proton converts into a neutron and an electron. No other particle is\nemitted.\nReason: During ?-decay linear momentum of system should remains constant.\nQ 6. Assertion : If we compare the stability of two nuclei, then that nucleus is more stable whose total\nbinding energy is more.\nReason : More the mass defect during formation of a nucleus more will be the binding energy.\nQ 7. Assertion : In a nuclear process energy is released if total binding energy of daughter nuclei is\nmore than the total binding energy of parent nuclei.\nReason : If energy is released then total mass of daughter nuclei is less than the total mass of\nparent nuclei.\nQ 8. Assertion: Binding energy per nucleon is of the order of MeV.\nReason : 1 MeV = 1.6 X 10\n-13\nJ.\nQ 9. Assertion: 1 amu is equal to 931.48 MeV.\nReason: 1 amu is equal to\n1\nth\n12\nthe mass of C\n12\natom.\nQ 10. Assertion: Between ?, ? and ?\n\nradiations, penetrating power of ?-rays is maximum.\nReason: Ionising power of -rays is least.\nQ 11. Assertion : The nuclear energy can be obtained by the nuclear fission of heavier nuclei as well as\nby fusion of lighter nuclei.\nReason : The binding energy per nucleon with increase in mass number, first increases and then\ndecreases.\nPage 2\n\nAssertion and Reason\nDirections : Choose the correct option.\n(a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.\n(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) If Assertion is true, but the Reason is false.\n(d) If Assertion is false but the Reason is true.\nQ 1. Assertion: Rate of radioactivity can not be increased or decreased by increasing or decreasing\npressure or temperature.\nReason : Rate depends on number of nuclei present in the radioactive sample.\nQ 2. Assertion: Only those nuclei which are heavier than lead are radioactive.\nReason : Nuclei of elements heavier than lead are unstable.\nQ 3. Assertion: After emission of one particle and two ?-particles, atomic number remains\nunchanged.\nReason : Mass number changes by four.\nQ 4. Assertion : ?-rays are produced by the transition of a nucleus from some higher energy state to\nsome lower energy state.\nReason : Electromagnetic waves are always produced by the transition process.\nQ 5. Assertion : During ?-decay a proton converts into a neutron and an electron. No other particle is\nemitted.\nReason: During ?-decay linear momentum of system should remains constant.\nQ 6. Assertion : If we compare the stability of two nuclei, then that nucleus is more stable whose total\nbinding energy is more.\nReason : More the mass defect during formation of a nucleus more will be the binding energy.\nQ 7. Assertion : In a nuclear process energy is released if total binding energy of daughter nuclei is\nmore than the total binding energy of parent nuclei.\nReason : If energy is released then total mass of daughter nuclei is less than the total mass of\nparent nuclei.\nQ 8. Assertion: Binding energy per nucleon is of the order of MeV.\nReason : 1 MeV = 1.6 X 10\n-13\nJ.\nQ 9. Assertion: 1 amu is equal to 931.48 MeV.\nReason: 1 amu is equal to\n1\nth\n12\nthe mass of C\n12\natom.\nQ 10. Assertion: Between ?, ? and ?\n\nradiations, penetrating power of ?-rays is maximum.\nReason: Ionising power of -rays is least.\nQ 11. Assertion : The nuclear energy can be obtained by the nuclear fission of heavier nuclei as well as\nby fusion of lighter nuclei.\nReason : The binding energy per nucleon with increase in mass number, first increases and then\ndecreases.\n1. (b) 2. (d) 3. (b) 4. (c) 5. (d) 6. (d) 7. (a or b) 8. (b) 9. (d) 10. (b) 11. (a or b)\nSolutions\n1. Huge amont of energy is involved in any nuclear process, which cannot be increased or decreased\nby pressure or temperatrue.\n2. Some lighter nuclei are also radioactive.\n3. By emission of one ?-particle atomic number decreases by 2 and mass number by 4. But by the\nemission of one ?-particle, atomic number increases by 1 and mass number reamains uncharged.\n4. In moving from lower energy state to higher energy state electromagnetic waves are absorbed.\n5. Antineutrino is also produced during ?-decay.\n6. Total binding energy per nucleon is more important for stability.\n7. Solution is not required.\n8. Solution is not required.\n9. 19(mu)(c\n2\n) = 931.48 MeV\n10. ?-particles are heaviest. Hence, its ionising power is maximum.\n11. In binding energy per nucleon versus mass number graph binding energy per nucleon of daughter\nnuclei should increase (for release of energy) or, the daughter nuclei should lie towards the peak of\nthe graph.\n\nObjective Questions\nSingle Correct Option\nQ 1. The count rate observed from a radioactive source at t second was N0 and at 4t second it was\n0\nN\n16\n.\n\nThe count rate observed, at\n11\nt\n2\n??\n??\n??\nsecond will be\n(a)\n0\nN\n128\n(b)\n0\nN\n64\n(c)\n0\nN\n32\n(d) None of these\nQ 2. The half lives of a radioactive sample are 30 years and 60 years for two decay processes. If the\nsample decays by both the processes simultaneously. The time after which, only one-fourth of the\nsample will remain is\n(a) 10 years (b) 20 years (c) 40 years (d) 60 years\nQ 3. Consider the nuclear fission reaction W ? Y + Y. What is the Q value (energy released) of the\nreaction?\n\nPage 3\n\nAssertion and Reason\nDirections : Choose the correct option.\n(a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.\n(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) If Assertion is true, but the Reason is false.\n(d) If Assertion is false but the Reason is true.\nQ 1. Assertion: Rate of radioactivity can not be increased or decreased by increasing or decreasing\npressure or temperature.\nReason : Rate depends on number of nuclei present in the radioactive sample.\nQ 2. Assertion: Only those nuclei which are heavier than lead are radioactive.\nReason : Nuclei of elements heavier than lead are unstable.\nQ 3. Assertion: After emission of one particle and two ?-particles, atomic number remains\nunchanged.\nReason : Mass number changes by four.\nQ 4. Assertion : ?-rays are produced by the transition of a nucleus from some higher energy state to\nsome lower energy state.\nReason : Electromagnetic waves are always produced by the transition process.\nQ 5. Assertion : During ?-decay a proton converts into a neutron and an electron. No other particle is\nemitted.\nReason: During ?-decay linear momentum of system should remains constant.\nQ 6. Assertion : If we compare the stability of two nuclei, then that nucleus is more stable whose total\nbinding energy is more.\nReason : More the mass defect during formation of a nucleus more will be the binding energy.\nQ 7. Assertion : In a nuclear process energy is released if total binding energy of daughter nuclei is\nmore than the total binding energy of parent nuclei.\nReason : If energy is released then total mass of daughter nuclei is less than the total mass of\nparent nuclei.\nQ 8. Assertion: Binding energy per nucleon is of the order of MeV.\nReason : 1 MeV = 1.6 X 10\n-13\nJ.\nQ 9. Assertion: 1 amu is equal to 931.48 MeV.\nReason: 1 amu is equal to\n1\nth\n12\nthe mass of C\n12\natom.\nQ 10. Assertion: Between ?, ? and ?\n\nradiations, penetrating power of ?-rays is maximum.\nReason: Ionising power of -rays is least.\nQ 11. Assertion : The nuclear energy can be obtained by the nuclear fission of heavier nuclei as well as\nby fusion of lighter nuclei.\nReason : The binding energy per nucleon with increase in mass number, first increases and then\ndecreases.\n1. (b) 2. (d) 3. (b) 4. (c) 5. (d) 6. (d) 7. (a or b) 8. (b) 9. (d) 10. (b) 11. (a or b)\nSolutions\n1. Huge amont of energy is involved in any nuclear process, which cannot be increased or decreased\nby pressure or temperatrue.\n2. Some lighter nuclei are also radioactive.\n3. By emission of one ?-particle atomic number decreases by 2 and mass number by 4. But by the\nemission of one ?-particle, atomic number increases by 1 and mass number reamains uncharged.\n4. In moving from lower energy state to higher energy state electromagnetic waves are absorbed.\n5. Antineutrino is also produced during ?-decay.\n6. Total binding energy per nucleon is more important for stability.\n7. Solution is not required.\n8. Solution is not required.\n9. 19(mu)(c\n2\n) = 931.48 MeV\n10. ?-particles are heaviest. Hence, its ionising power is maximum.\n11. In binding energy per nucleon versus mass number graph binding energy per nucleon of daughter\nnuclei should increase (for release of energy) or, the daughter nuclei should lie towards the peak of\nthe graph.\n\nObjective Questions\nSingle Correct Option\nQ 1. The count rate observed from a radioactive source at t second was N0 and at 4t second it was\n0\nN\n16\n.\n\nThe count rate observed, at\n11\nt\n2\n??\n??\n??\nsecond will be\n(a)\n0\nN\n128\n(b)\n0\nN\n64\n(c)\n0\nN\n32\n(d) None of these\nQ 2. The half lives of a radioactive sample are 30 years and 60 years for two decay processes. If the\nsample decays by both the processes simultaneously. The time after which, only one-fourth of the\nsample will remain is\n(a) 10 years (b) 20 years (c) 40 years (d) 60 years\nQ 3. Consider the nuclear fission reaction W ? Y + Y. What is the Q value (energy released) of the\nreaction?\n\n(a) E1N1 - (E2N2 + E3N 3) (b) (E2N 2 + E3N3 - E1N1)\n(c) E2N2 +E1N1 -E3N3 (d) E1N1+E3N3 - E2N 2\nQ 4. Consider the following nuclear reaction\nX\n200\n? A\n110\n+ B\n90\n+ Energy\nIf the binding energy per nucleon for X, A and B are 7.4 MeV, 8.2 MeV and 8.2 MeV\nrespectively, the energy released will be\n(a) 90 MeV (b) 110 MeV (c) 200 MeV (d) 160 MeV\nQ 5. The binding energy per nucleon for deuteron\n? ?\n2\n1\nH and helium\n? ?\n4\n2\nHe are 1.1 MeV and 7.0 MeV,\nrespectively. The energy released when two deutrons fuse to form a helium nucleus is\n(a) 47.12 MeV (b) 23.6 MeV (c) 11.8 MeV (d) 34.4 MeV\nQ 6. The energy released by the fission of a single uranium nucleus is 200 MeV. The number of\nfissions of uranium nucleus per second required to produce 16 MW of power is\n(Assume efficiency of the reactor is 50%)\n(a) 2 × 10\n6\n(b) 2.5 × 10\n6\n(c) 5 × 10\n6\n(d) None of these\nQ 7. A radioactive isotope is being produced at a constant rate A. The isotope has a half-life T. Initially\nthere are no nuclei, after a time t>>T, the number of nuclei becomes constant. The value of this\nconstant is\n(a) AT (b)\nA\nln(2)\nT\n(c) AT ln (2) (d)\nAT\nln(2)\n\nQ 8. A moving hydrogen atom makes a head on collision with a stationary hydrogen atom. Before\ncollision both atoms are in the ground state and after collision they move together. What is the\nvelocity of the moving atom if after the collision one of the atom gets minimum excitation energy?\n(Mass of hydrogen atom is 1.673 × 10\n-27\nkg)\n(a) 5.25 × 10\n4\nm/s (b) 4.25 × 10\n4\nm/s (c) 6.25 × 10\n4\nm/s (d) 10.25 × 10\n4\nm/s\nQ 9. A bone containing 200 g carbon-14 has a ?-decay rate of 375 decay/min. Calculate the time that\nhas elapsed since the death of the living one. Given the rate of decay for the living organism is\nequal to 15 decay per min per gram of carbon and half-life of carbon-14 = 5730 years\n(a) 27190 years (b) 1190 years (c) 17190 years (d) None of these\nQ 10. Two identical samples (same material and same amount) P and Q of a radioactive substance\nhaving mean life T are observed to have activities AP and AQ respectively at the time of\nobservation. If P is older than Q, then the difference in their age is\n(a)\nP\nQ\nA\nT ln\nA\n??\n??\n??\n??\n(b)\nQ\nP\nA\nTln\nA\n??\n??\n??\n(c)\nP\nQ\nA\nT\nA\n??\n??\n??\n??\n(d)\nQ\nP\nA\nT\nA\n??\n??\n??\n\nQ 11. A star initially has 10\n40\ndeuterons. It produces energy via the processes\n2 2 2\n1 1 1\nH H H p ? ? ?\n\nand\n2 3 4\n1 1 2\nH H He n ? ? ? .\n\nWhere the masses of the nuclei are :\nm(\n2\nH) = 2.014 amu, m(p)= 1.007 amu, m(n)= 1.008 amu and m(\n4\nHe) = 4.001 amu.\nIf the average power radiated by the star is 10\n16\nW, the deuteron supply of the star is exhausted in\na time of the order of\nPage 4\n\nAssertion and Reason\nDirections : Choose the correct option.\n(a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.\n(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) If Assertion is true, but the Reason is false.\n(d) If Assertion is false but the Reason is true.\nQ 1. Assertion: Rate of radioactivity can not be increased or decreased by increasing or decreasing\npressure or temperature.\nReason : Rate depends on number of nuclei present in the radioactive sample.\nQ 2. Assertion: Only those nuclei which are heavier than lead are radioactive.\nReason : Nuclei of elements heavier than lead are unstable.\nQ 3. Assertion: After emission of one particle and two ?-particles, atomic number remains\nunchanged.\nReason : Mass number changes by four.\nQ 4. Assertion : ?-rays are produced by the transition of a nucleus from some higher energy state to\nsome lower energy state.\nReason : Electromagnetic waves are always produced by the transition process.\nQ 5. Assertion : During ?-decay a proton converts into a neutron and an electron. No other particle is\nemitted.\nReason: During ?-decay linear momentum of system should remains constant.\nQ 6. Assertion : If we compare the stability of two nuclei, then that nucleus is more stable whose total\nbinding energy is more.\nReason : More the mass defect during formation of a nucleus more will be the binding energy.\nQ 7. Assertion : In a nuclear process energy is released if total binding energy of daughter nuclei is\nmore than the total binding energy of parent nuclei.\nReason : If energy is released then total mass of daughter nuclei is less than the total mass of\nparent nuclei.\nQ 8. Assertion: Binding energy per nucleon is of the order of MeV.\nReason : 1 MeV = 1.6 X 10\n-13\nJ.\nQ 9. Assertion: 1 amu is equal to 931.48 MeV.\nReason: 1 amu is equal to\n1\nth\n12\nthe mass of C\n12\natom.\nQ 10. Assertion: Between ?, ? and ?\n\nradiations, penetrating power of ?-rays is maximum.\nReason: Ionising power of -rays is least.\nQ 11. Assertion : The nuclear energy can be obtained by the nuclear fission of heavier nuclei as well as\nby fusion of lighter nuclei.\nReason : The binding energy per nucleon with increase in mass number, first increases and then\ndecreases.\n1. (b) 2. (d) 3. (b) 4. (c) 5. (d) 6. (d) 7. (a or b) 8. (b) 9. (d) 10. (b) 11. (a or b)\nSolutions\n1. Huge amont of energy is involved in any nuclear process, which cannot be increased or decreased\nby pressure or temperatrue.\n2. Some lighter nuclei are also radioactive.\n3. By emission of one ?-particle atomic number decreases by 2 and mass number by 4. But by the\nemission of one ?-particle, atomic number increases by 1 and mass number reamains uncharged.\n4. In moving from lower energy state to higher energy state electromagnetic waves are absorbed.\n5. Antineutrino is also produced during ?-decay.\n6. Total binding energy per nucleon is more important for stability.\n7. Solution is not required.\n8. Solution is not required.\n9. 19(mu)(c\n2\n) = 931.48 MeV\n10. ?-particles are heaviest. Hence, its ionising power is maximum.\n11. In binding energy per nucleon versus mass number graph binding energy per nucleon of daughter\nnuclei should increase (for release of energy) or, the daughter nuclei should lie towards the peak of\nthe graph.\n\nObjective Questions\nSingle Correct Option\nQ 1. The count rate observed from a radioactive source at t second was N0 and at 4t second it was\n0\nN\n16\n.\n\nThe count rate observed, at\n11\nt\n2\n??\n??\n??\nsecond will be\n(a)\n0\nN\n128\n(b)\n0\nN\n64\n(c)\n0\nN\n32\n(d) None of these\nQ 2. The half lives of a radioactive sample are 30 years and 60 years for two decay processes. If the\nsample decays by both the processes simultaneously. The time after which, only one-fourth of the\nsample will remain is\n(a) 10 years (b) 20 years (c) 40 years (d) 60 years\nQ 3. Consider the nuclear fission reaction W ? Y + Y. What is the Q value (energy released) of the\nreaction?\n\n(a) E1N1 - (E2N2 + E3N 3) (b) (E2N 2 + E3N3 - E1N1)\n(c) E2N2 +E1N1 -E3N3 (d) E1N1+E3N3 - E2N 2\nQ 4. Consider the following nuclear reaction\nX\n200\n? A\n110\n+ B\n90\n+ Energy\nIf the binding energy per nucleon for X, A and B are 7.4 MeV, 8.2 MeV and 8.2 MeV\nrespectively, the energy released will be\n(a) 90 MeV (b) 110 MeV (c) 200 MeV (d) 160 MeV\nQ 5. The binding energy per nucleon for deuteron\n? ?\n2\n1\nH and helium\n? ?\n4\n2\nHe are 1.1 MeV and 7.0 MeV,\nrespectively. The energy released when two deutrons fuse to form a helium nucleus is\n(a) 47.12 MeV (b) 23.6 MeV (c) 11.8 MeV (d) 34.4 MeV\nQ 6. The energy released by the fission of a single uranium nucleus is 200 MeV. The number of\nfissions of uranium nucleus per second required to produce 16 MW of power is\n(Assume efficiency of the reactor is 50%)\n(a) 2 × 10\n6\n(b) 2.5 × 10\n6\n(c) 5 × 10\n6\n(d) None of these\nQ 7. A radioactive isotope is being produced at a constant rate A. The isotope has a half-life T. Initially\nthere are no nuclei, after a time t>>T, the number of nuclei becomes constant. The value of this\nconstant is\n(a) AT (b)\nA\nln(2)\nT\n(c) AT ln (2) (d)\nAT\nln(2)\n\nQ 8. A moving hydrogen atom makes a head on collision with a stationary hydrogen atom. Before\ncollision both atoms are in the ground state and after collision they move together. What is the\nvelocity of the moving atom if after the collision one of the atom gets minimum excitation energy?\n(Mass of hydrogen atom is 1.673 × 10\n-27\nkg)\n(a) 5.25 × 10\n4\nm/s (b) 4.25 × 10\n4\nm/s (c) 6.25 × 10\n4\nm/s (d) 10.25 × 10\n4\nm/s\nQ 9. A bone containing 200 g carbon-14 has a ?-decay rate of 375 decay/min. Calculate the time that\nhas elapsed since the death of the living one. Given the rate of decay for the living organism is\nequal to 15 decay per min per gram of carbon and half-life of carbon-14 = 5730 years\n(a) 27190 years (b) 1190 years (c) 17190 years (d) None of these\nQ 10. Two identical samples (same material and same amount) P and Q of a radioactive substance\nhaving mean life T are observed to have activities AP and AQ respectively at the time of\nobservation. If P is older than Q, then the difference in their age is\n(a)\nP\nQ\nA\nT ln\nA\n??\n??\n??\n??\n(b)\nQ\nP\nA\nTln\nA\n??\n??\n??\n(c)\nP\nQ\nA\nT\nA\n??\n??\n??\n??\n(d)\nQ\nP\nA\nT\nA\n??\n??\n??\n\nQ 11. A star initially has 10\n40\ndeuterons. It produces energy via the processes\n2 2 2\n1 1 1\nH H H p ? ? ?\n\nand\n2 3 4\n1 1 2\nH H He n ? ? ? .\n\nWhere the masses of the nuclei are :\nm(\n2\nH) = 2.014 amu, m(p)= 1.007 amu, m(n)= 1.008 amu and m(\n4\nHe) = 4.001 amu.\nIf the average power radiated by the star is 10\n16\nW, the deuteron supply of the star is exhausted in\na time of the order of\n(a) 10\n6\ns (b) 10\n8\ns (c) 10\n12\ns (d) 10\n16\ns\nQ 12. Two radioactive samples of different elements (half lives t 1 and t2 respectively) have same number\nof nuclei at t = 0. The time after which their activities are same is\n(a)\n1 2 2\n2 1 1\nt t t\nln\n0.693(t t ) t ?\n(b)\n1 2 2\n1\nt t t\nln\n0.693 t\n\n(c)\n1 2 2\n1 2 1\nt t t\nln\n0.693(t t ) t ?\n(d) None of these\nQ 13. A nucleus X initially at rest, undergoes alpha decay according to the equation\n\n232 A\nZ 90\nXY ? ? ?\nWhat fraction of the total energy released in the decay will be the kinetic energy of the alpha\nparticle?\n(a)\n90\n92\n(b)\n228\n232\n(c)\n228\n232\n(d)\n1\n2\n\nQ 14. A stationary nucleus of mass 24 amu emits a gamma photon. The energy of the emitted photon is\n7 MeV. The recoil energy of the nucleus is\n(a) 2.2 keV (b) 1.1 keV (c) 3.1 keV (d) 22 keV\nQ 15. A radioactive material of half-life T was kept in a nuclear reactor at two different instants. The\nquantity kept second time was twice of that kept first time. If now their present activities are A 1\nand A2 respectively then their age difference equals\n(a)\n1\n2\n2A T\nln\nln 2 A\n(b)\n1\n2\nA\nT ln\nA\n(c)\n2\n1\nA T\nln\nln 2 2A\n(d)\n2\n1\nA\nT ln\n2A\n\nPassage : (Q. No. 16 to 18)\nThe atomic masses of the hydrogen isotopes are\nHydrogen m 1H\n1\n= 1.007825 amu\nDeuterium m1H\n2\n= 2.014102 amu\nTritium m1H\n3\n= 3.016049 amu\nQ 16. The energy released in the reaction\n1H\n2\n+ 1H\n2\n? 1H\n3\n+ 1H\n1\nis nearly\n(a) 1 MeV (b) 2 MeV (c) 4 MeV (d) 8 MeV\nQ 17. The number of fusion reactions required to generate 1 kWh is nearly\n(a) 10\n8\n(b) 10\n18\n(c) 10\n28\n(d) 10\n38\n\nQ 18. The mass of deuterium, 1 H\n2\nthat would be needed to generate 1 kWh\n(a) 3.7 kg (b) 3.7 g (c) 3.7 × 10\n-5\nkg (d) 3.7 × 10\n-8\nkg\n1.(b) 2.(c) 3.(b) 4.(d) 5.(b) 6.(d) 7.(d) 8.(c) 9.(c) 10.(b) 11.(c) 12.(a) 13.(b) 14.(b) 15.(c) 16.(c)\n17.(b) 18.(d)\nSolutions\n1.\nPage 5\n\nAssertion and Reason\nDirections : Choose the correct option.\n(a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.\n(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) If Assertion is true, but the Reason is false.\n(d) If Assertion is false but the Reason is true.\nQ 1. Assertion: Rate of radioactivity can not be increased or decreased by increasing or decreasing\npressure or temperature.\nReason : Rate depends on number of nuclei present in the radioactive sample.\nQ 2. Assertion: Only those nuclei which are heavier than lead are radioactive.\nReason : Nuclei of elements heavier than lead are unstable.\nQ 3. Assertion: After emission of one particle and two ?-particles, atomic number remains\nunchanged.\nReason : Mass number changes by four.\nQ 4. Assertion : ?-rays are produced by the transition of a nucleus from some higher energy state to\nsome lower energy state.\nReason : Electromagnetic waves are always produced by the transition process.\nQ 5. Assertion : During ?-decay a proton converts into a neutron and an electron. No other particle is\nemitted.\nReason: During ?-decay linear momentum of system should remains constant.\nQ 6. Assertion : If we compare the stability of two nuclei, then that nucleus is more stable whose total\nbinding energy is more.\nReason : More the mass defect during formation of a nucleus more will be the binding energy.\nQ 7. Assertion : In a nuclear process energy is released if total binding energy of daughter nuclei is\nmore than the total binding energy of parent nuclei.\nReason : If energy is released then total mass of daughter nuclei is less than the total mass of\nparent nuclei.\nQ 8. Assertion: Binding energy per nucleon is of the order of MeV.\nReason : 1 MeV = 1.6 X 10\n-13\nJ.\nQ 9. Assertion: 1 amu is equal to 931.48 MeV.\nReason: 1 amu is equal to\n1\nth\n12\nthe mass of C\n12\natom.\nQ 10. Assertion: Between ?, ? and ?\n\nradiations, penetrating power of ?-rays is maximum.\nReason: Ionising power of -rays is least.\nQ 11. Assertion : The nuclear energy can be obtained by the nuclear fission of heavier nuclei as well as\nby fusion of lighter nuclei.\nReason : The binding energy per nucleon with increase in mass number, first increases and then\ndecreases.\n1. (b) 2. (d) 3. (b) 4. (c) 5. (d) 6. (d) 7. (a or b) 8. (b) 9. (d) 10. (b) 11. (a or b)\nSolutions\n1. Huge amont of energy is involved in any nuclear process, which cannot be increased or decreased\nby pressure or temperatrue.\n2. Some lighter nuclei are also radioactive.\n3. By emission of one ?-particle atomic number decreases by 2 and mass number by 4. But by the\nemission of one ?-particle, atomic number increases by 1 and mass number reamains uncharged.\n4. In moving from lower energy state to higher energy state electromagnetic waves are absorbed.\n5. Antineutrino is also produced during ?-decay.\n6. Total binding energy per nucleon is more important for stability.\n7. Solution is not required.\n8. Solution is not required.\n9. 19(mu)(c\n2\n) = 931.48 MeV\n10. ?-particles are heaviest. Hence, its ionising power is maximum.\n11. In binding energy per nucleon versus mass number graph binding energy per nucleon of daughter\nnuclei should increase (for release of energy) or, the daughter nuclei should lie towards the peak of\nthe graph.\n\nObjective Questions\nSingle Correct Option\nQ 1. The count rate observed from a radioactive source at t second was N0 and at 4t second it was\n0\nN\n16\n.\n\nThe count rate observed, at\n11\nt\n2\n??\n??\n??\nsecond will be\n(a)\n0\nN\n128\n(b)\n0\nN\n64\n(c)\n0\nN\n32\n(d) None of these\nQ 2. The half lives of a radioactive sample are 30 years and 60 years for two decay processes. If the\nsample decays by both the processes simultaneously. The time after which, only one-fourth of the\nsample will remain is\n(a) 10 years (b) 20 years (c) 40 years (d) 60 years\nQ 3. Consider the nuclear fission reaction W ? Y + Y. What is the Q value (energy released) of the\nreaction?\n\n(a) E1N1 - (E2N2 + E3N 3) (b) (E2N 2 + E3N3 - E1N1)\n(c) E2N2 +E1N1 -E3N3 (d) E1N1+E3N3 - E2N 2\nQ 4. Consider the following nuclear reaction\nX\n200\n? A\n110\n+ B\n90\n+ Energy\nIf the binding energy per nucleon for X, A and B are 7.4 MeV, 8.2 MeV and 8.2 MeV\nrespectively, the energy released will be\n(a) 90 MeV (b) 110 MeV (c) 200 MeV (d) 160 MeV\nQ 5. The binding energy per nucleon for deuteron\n? ?\n2\n1\nH and helium\n? ?\n4\n2\nHe are 1.1 MeV and 7.0 MeV,\nrespectively. The energy released when two deutrons fuse to form a helium nucleus is\n(a) 47.12 MeV (b) 23.6 MeV (c) 11.8 MeV (d) 34.4 MeV\nQ 6. The energy released by the fission of a single uranium nucleus is 200 MeV. The number of\nfissions of uranium nucleus per second required to produce 16 MW of power is\n(Assume efficiency of the reactor is 50%)\n(a) 2 × 10\n6\n(b) 2.5 × 10\n6\n(c) 5 × 10\n6\n(d) None of these\nQ 7. A radioactive isotope is being produced at a constant rate A. The isotope has a half-life T. Initially\nthere are no nuclei, after a time t>>T, the number of nuclei becomes constant. The value of this\nconstant is\n(a) AT (b)\nA\nln(2)\nT\n(c) AT ln (2) (d)\nAT\nln(2)\n\nQ 8. A moving hydrogen atom makes a head on collision with a stationary hydrogen atom. Before\ncollision both atoms are in the ground state and after collision they move together. What is the\nvelocity of the moving atom if after the collision one of the atom gets minimum excitation energy?\n(Mass of hydrogen atom is 1.673 × 10\n-27\nkg)\n(a) 5.25 × 10\n4\nm/s (b) 4.25 × 10\n4\nm/s (c) 6.25 × 10\n4\nm/s (d) 10.25 × 10\n4\nm/s\nQ 9. A bone containing 200 g carbon-14 has a ?-decay rate of 375 decay/min. Calculate the time that\nhas elapsed since the death of the living one. Given the rate of decay for the living organism is\nequal to 15 decay per min per gram of carbon and half-life of carbon-14 = 5730 years\n(a) 27190 years (b) 1190 years (c) 17190 years (d) None of these\nQ 10. Two identical samples (same material and same amount) P and Q of a radioactive substance\nhaving mean life T are observed to have activities AP and AQ respectively at the time of\nobservation. If P is older than Q, then the difference in their age is\n(a)\nP\nQ\nA\nT ln\nA\n??\n??\n??\n??\n(b)\nQ\nP\nA\nTln\nA\n??\n??\n??\n(c)\nP\nQ\nA\nT\nA\n??\n??\n??\n??\n(d)\nQ\nP\nA\nT\nA\n??\n??\n??\n\nQ 11. A star initially has 10\n40\ndeuterons. It produces energy via the processes\n2 2 2\n1 1 1\nH H H p ? ? ?\n\nand\n2 3 4\n1 1 2\nH H He n ? ? ? .\n\nWhere the masses of the nuclei are :\nm(\n2\nH) = 2.014 amu, m(p)= 1.007 amu, m(n)= 1.008 amu and m(\n4\nHe) = 4.001 amu.\nIf the average power radiated by the star is 10\n16\nW, the deuteron supply of the star is exhausted in\na time of the order of\n(a) 10\n6\ns (b) 10\n8\ns (c) 10\n12\ns (d) 10\n16\ns\nQ 12. Two radioactive samples of different elements (half lives t 1 and t2 respectively) have same number\nof nuclei at t = 0. The time after which their activities are same is\n(a)\n1 2 2\n2 1 1\nt t t\nln\n0.693(t t ) t ?\n(b)\n1 2 2\n1\nt t t\nln\n0.693 t\n\n(c)\n1 2 2\n1 2 1\nt t t\nln\n0.693(t t ) t ?\n(d) None of these\nQ 13. A nucleus X initially at rest, undergoes alpha decay according to the equation\n\n232 A\nZ 90\nXY ? ? ?\nWhat fraction of the total energy released in the decay will be the kinetic energy of the alpha\nparticle?\n(a)\n90\n92\n(b)\n228\n232\n(c)\n228\n232\n(d)\n1\n2\n\nQ 14. A stationary nucleus of mass 24 amu emits a gamma photon. The energy of the emitted photon is\n7 MeV. The recoil energy of the nucleus is\n(a) 2.2 keV (b) 1.1 keV (c) 3.1 keV (d) 22 keV\nQ 15. A radioactive material of half-life T was kept in a nuclear reactor at two different instants. The\nquantity kept second time was twice of that kept first time. If now their present activities are A 1\nand A2 respectively then their age difference equals\n(a)\n1\n2\n2A T\nln\nln 2 A\n(b)\n1\n2\nA\nT ln\nA\n(c)\n2\n1\nA T\nln\nln 2 2A\n(d)\n2\n1\nA\nT ln\n2A\n\nPassage : (Q. No. 16 to 18)\nThe atomic masses of the hydrogen isotopes are\nHydrogen m 1H\n1\n= 1.007825 amu\nDeuterium m1H\n2\n= 2.014102 amu\nTritium m1H\n3\n= 3.016049 amu\nQ 16. The energy released in the reaction\n1H\n2\n+ 1H\n2\n? 1H\n3\n+ 1H\n1\nis nearly\n(a) 1 MeV (b) 2 MeV (c) 4 MeV (d) 8 MeV\nQ 17. The number of fusion reactions required to generate 1 kWh is nearly\n(a) 10\n8\n(b) 10\n18\n(c) 10\n28\n(d) 10\n38\n\nQ 18. The mass of deuterium, 1 H\n2\nthat would be needed to generate 1 kWh\n(a) 3.7 kg (b) 3.7 g (c) 3.7 × 10\n-5\nkg (d) 3.7 × 10\n-8\nkg\n1.(b) 2.(c) 3.(b) 4.(d) 5.(b) 6.(d) 7.(d) 8.(c) 9.(c) 10.(b) 11.(c) 12.(a) 13.(b) 14.(b) 15.(c) 16.(c)\n17.(b) 18.(d)\nSolutions\n1.\nSo, 3t times is equivalent to four half lives. Hence one half life is equal to\n3t\n4\n.\nThe given time\n11 9\nt t t\n22\n?? is equivalent to 6 half lives\n\n2. ? = ?1 + ?2\n\n1\nth\n4\nsample remains after 2 half lives or 40 yr.\n3. Q-value = Final binding energy Initial binding energy\n= E2N2 + E3N3 - E1N1\n4. Energy released = Final binding energy - initial binding energy\n= 110 × 8.2 + 90 × 8.2 - 200 × 7.4\n= 160 MeV\n5. Energy released = final binding energy - initial binding energy\n= (7.0 - 1.1)4\n= 23.6 MeV\n6. It means we are getting only 100 MeV of energy by the fission of one uranium nucleus.\nNumber of nuclei per second\n\n7. When the rate production = rate of disintegration, number of nuclei or maximum.\n?N = A\n\n8. See the hint of Q. No- 24 of section, objective question (single correct option) for JEE Advanced\nof chapter-33. Minimum energy required is 20.4 eV.\n\nSubstituting the value of m, we can get V.\n9. R0 = 15 × 200 = 3000 decay/min from 200g carbon.\nUsing,\n\n? n = number of half lives = 3\n```\nOffer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!\n\n210 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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[ "Index ← Previous Next →\n\n## Translations\n\nΔύο δοθεισῶν εὐθειῶν μέσην ἀνάλογον προσευρεῖν. Ἔστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι αἱ ΑΒ, ΒΓ: δεῖ δὴ τῶν ΑΒ, ΒΓ μέσην ἀνάλογον προσευρεῖν. Κείσθωσαν ἐπ' εὐθείας, καὶ γεγράφθω ἐπὶ τῆς ΑΓ ἡμικύκλιον τὸ ΑΔΓ, καὶ ἤχθω ἀπὸ τοῦ Β σημείου τῇ ΑΓ εὐθείᾳ πρὸς ὀρθὰς ἡ ΒΔ, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΓ. Ἐπεὶ ἐν ἡμικυκλίῳ γωνία ἐστὶν ἡ ὑπὸ ΑΔΓ, ὀρθή ἐστιν. καὶ ἐπεὶ ἐν ὀρθογωνίῳ τριγώνῳ τῷ ΑΔΓ ἀπὸ τῆς ὀρθῆς γωνίας ἐπὶ τὴν βάσιν κάθετος ἦκται ἡ ΔΒ, ἡ ΔΒ ἄρα τῶν τῆς βάσεως τμημάτων τῶν ΑΒ, ΒΓ μέση ἀνάλογόν ἐστιν. Δύο ἄρα δοθεισῶν εὐθειῶν τῶν ΑΒ, ΒΓ μέση ἀνάλογον προσεύρηται ἡ ΔΒ: ὅπερ ἔδει ποιῆσαι.\n\nTo two given straight lines to find a mean proportional. Let AB, BC be the two given straight lines; thus it is required to find a mean proportional to AB, BC. Let them be placed in a straight line, and let the semicircle ADC be described on AC; let BD be drawn from the point B at right angles to the straight line AC, and let AD, DC be joined. Since the angle ADC is an angle in a semicircle, it is right. [III. 31] And, since, in the right-angled triangle ADC, DB has been drawn from the right angle perpendicular to the base, therefore DB is a mean proportional between the segments of the base, AB, BC. [VI. 8, Por.]" ]

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[ "# Article\n\nReferences:\n Lefschetz: Existence of periodic solutions for certain differential equations. Proceedings of the Nat. Acad. of Sc. of the USA, vol. 29, 1943. MR 0007462 | Zbl 0061.19002\n Krylov-Bogoljubov: Introduction to nonlinear mechanics. Princeton 1942.\n Andronov-Chaikin: Theory of oscillations. Princeton 1949.\n Levinson: On the existence of periodic solutions for second order differential equations with forcing term. Journal of Math. and Phys., 1943.\n Levinson: Transformation theory of nonlinear differential equations of the second order. Annals of Math. 1944.\n Cartwright-Littlewood: On nonlinear differential equations of the second order II. Annals of Math. 1947.\n Cartwright: Forced oscillations in nonlinear systems. Contributions to the theory of nonlinear oscillations, Princeton 1950. Zbl 0039.09901\n Massera: The existence of periodic solutions of systems of differential equations. Duke Math. J. 1950. MR 0040512 | Zbl 0038.25002\n Langenhop-Farnell: The existence of forced periodic solutions of second order differential equations near certain equilibrium points of the unforced equation. Contributions to the theory of nonlinear oscillations, Princeton 1950.\n Levinson: Correction. Transformation theory of nonlinear differential equa- tions of the second order. Annals of Math. 1948.\n Massera: The number of subharmonic solutions of nonlinear differential equations of the second order. Annals of math. 1949.\n Levinson: On a nonlinear differential equation of the second order. Journal of Math, and Phys. 1943.\n Cartwright-Littlewood: On nonlinear differential equations of the second order I. Journal of the London Math. Soc. 1945.\n Levinson: A second order differential equation with singular motion. Annals of Math. 1949." ]

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[ "# Chemical Engineering - Stoichiometry\n\n### Exercise :: Stoichiometry - Section 3\n\n1.\n\nWhich of the following is insensitive to changes in pressure ?\n\n A. Heat of vaporisation B. Melting point C. Heat of fusion D. Both (b) and (c)\n\nExplanation:\n\nNo answer description available for this question. Let us discuss.\n\n2.\n\nThe heat capacity of a solid compound is calculated from the atomic heat capacities of its constituent elements with the help of the\n\n A. Trouton's rule B. Kopp's rule C. Antonie equation D. Kistyakowsky equation\n\nExplanation:\n\nNo answer description available for this question. Let us discuss.\n\n3.\n\nHow many phases are present at eutectic point ?\n\n A. 1 B. 2 C. 3 D. unpredictable\n\nExplanation:\n\nNo answer description available for this question. Let us discuss.\n\n4.\n\nThe increase in the temperature of the aqueous solution will result in decrease of its\n\n A. weight % of the solute. B. mole fraction of the solute. C. molarity. D. molality.\n\nExplanation:\n\nNo answer description available for this question. Let us discuss.\n\n5.\n\nWhich of the following gases is the most soluble in water ?\n\n A. NH3 B. CO2 C. H2S D. CH4" ]

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[ "Class Central is learner-supported. When you buy through links on our site, we may earn an affiliate commission.\n\n# Introduction to Formal Concept Analysis\n\nThis course may be unavailable.\n\n### Overview\n\n##### Class Central Tips\nThis online course is an introduction into formal concept analysis (FCA), a mathematical theory oriented at applications in knowledge representation, knowledge acquisition, data analysis and visualization. It provides tools for understanding the data by representing it as a hierarchy of concepts or, more exactly, a concept lattice. FCA can help in processing a wide class of data types providing a framework in which various data analysis and knowledge acquisition techniques can be formulated. In this course, we focus on some of these techniques, as well as cover the theoretical foundations and algorithmic issues of FCA.\n\nUpon completion of the course, the students will be able to use the mathematical techniques and computational tools of formal concept analysis in their own research projects involving data processing. Among other things, the students will learn about FCA-based approaches to clustering and dependency mining.\n\nThe course is self-contained, although basic knowledge of elementary set theory, propositional logic, and probability theory would help. End-of-the-week quizzes include easy questions aimed at checking basic understanding of the topic, as well as more advanced problems that may require some effort to be solved.\n\nDo you have technical problems? Write to us: [email protected]\n\n### Syllabus\n\n• Formal concept analysis in a nutshell\n• This week we will learn the basic notions of formal concept analysis (FCA). We'll talk about some of its typical applications, such as conceptual clustering and search for implicational dependencies in data. We'll see a few examples of concept lattices and learn how to interpret them. The simplest data structure in formal concept analysis is the formal context. It is used to describe objects in terms of attributes they have. Derivation operators in a formal context link together object and attribute subsets; they are used to define formal concepts. They also give rise to closure operators, and we'll talk about what these are, too. We'll have a look at software called Concept Explorer, which is good for basic processing of formal contexts. We'll also talk a little bit about many-valued contexts, where attributes may have many values. Conceptual scaling is used to transform many-valued contexts into \"standard\", one-valued, formal contexts.\n• Concept lattices and their line diagrams\n• This week we'll talk about some mathematical properties of concepts. We'll define a partial order on formal concepts, that of \"being less general\". Ordered in this way, the concepts of a formal concept constitute a special mathematical structure, a complete lattice. We'll learn what these are, and we'll see, through the basic theorem on concept lattices, that any complete lattice can, in a certain sense, be modelled by a formal context. We'll also discuss how a formal context can be simplified without loosing the structure of its concept lattice.\n• Constructing concept lattices\n• We will consider a few algorithms that build the concept lattice of a formal context: a couple of naive approaches, which are easy to use if one wants to build the concept lattice of a small context; a more sophisticated approach, which enumerates concepts in a specific order; and an incremental strategy, which can be used to update the concept lattice when a new object is added to the context. We will also give a formal definition of implications, and we'll see how an implication can logically follow from a set of other implications.\n• Implications\n• This week we'll continue talking about implications. We'll see that implication sets can be redundant, and we'll learn to summarise all valid implications of a formal context by its canonical (Duquenne–Guigues) basis. We'll study one concrete algorithm that computes the canonical basis, which turns out to be a modification of the Next Closure algorithm from the previous week. We'll also talk about what is known in database theory as functional dependencies, and we'll show how they are related to implications.\n• Interactive algorithms for learning implications\n• What if we don't have a direct access to a formal context, but still want to compute its concept lattice and its implicational theory? This can be done if there is a domain expert (or an oracle) willing to answer our queries about the domain. We'll study an approach known as learning with queries that addresses this setting. We'll get to know a few standard types of queries, and we'll see how an implication set can be learnt in time polynomial of its size with so called membership and equivalence queries. We'll then introduce attribute exploration, a method from formal concept analysis, which may require exponential time, but which uses different queries, more suitable for building implicational theories and representative samples of subject domains.\n• Working with real data\n• A concept lattice can be exponentially large in the size of its formal context. Sometimes this can be due to noise in data. We'll study a few heuristics to filter out noisy concepts or select the most interesting concepts in a large lattice built from real data: stability and separation indices, concept probability, iceberg lattices. We will also talk about association rules, which is a name for implications that are supported by strong evidence, but may still have counterexamples in data.\n\nSergei Obiedkov\n\n## Reviews\n\nStart your review of Introduction to Formal Concept Analysis\n\n### Never Stop Learning.\n\nGet personalized course recommendations, track subjects and courses with reminders, and more.", null, "" ]

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[ "# Rust Crash Course\n\n## Fundamentals\n\n### Data Types\n\n• Boolean\n• Integer\n• Double / Float\n• Character\n• String\n\n### Variables\n\n• assign data to a temporary memory location\n• allows programmer to easily work with memory\n• can be set to any value & type (dynamically?)\n• immutable by default, but can be mutable\n``````// notice that no data types are used below\nlet two = 2;\nlet hello = \"hello\";\nlet j = 'j';\nlet my_half = 0.5;\nlet mut my_name = \"Bill\";\nlet quit_program = false;\nlet your_half = my_half;\n``````\n\n### Functions\n\n``````fn add(a: i32, b: i32) -> i32 {\na + b\n}\n``````\n• `fn`: reserved word meaning that we're declaring a function\n• `add`: name of the function\n• `(a: i32, b: i32)`: parameters - `param: type`\n• `-> i32`: return type of the function\n• `a + b`: do the math and return the value (just like in Ruby)\n\n### println macro\n\n• macros expand into additional code\n• `println` prints information to the terminal\n``````let life = 42;\n\n// the '!' symbol means that we're calling a macro\nprintln!(\"hello\");\n\n// in \"{:?}\", the ':?' sequence mean debug print\nprintln!(\"{:?}\", life);\nprintln!(\"{:?} {:?}\", life, life);\nprintln!(\"the meaning is {:?}\", life);\n\n// another notation\nprintln!(\"{life:?}\");\n\n// without the ':?' means it's going to be visible for users\nprintln!(\"{life}\");\n``````\n\n### Control flow using `if`\n\n``````let a = 99;\n\nif a > 200 {\nprintln!(\"Huge number\");\n} else if a > 99 {\nprintln!(\"Big number\");\n} else {\nprintln!(\"Small number\");\n}\n``````\n• parentheses are not mandatory\n\n### Repetition using loops\n\n• types of loops\n• `loop` - infinite loop\n• `while` - conditional loop\n• both types can exit using `break`\n\n`loop` example:\n\n``````let mut a = 0;\n\nloop {\nif a == 5 {\nbreak;\n}\n\nprintln!(\"{:?}\", a);\na = a + 1;\n}\n``````\n\n`while` example:\n\n``````let mut a = 0;\n\nwhile a != 5 {\nprintln!(\"{:?}\", a);\na = a + 1;\n}\n``````\n\n### practice\n\n`cargo` commands:\n\n``````# run a specific file\n# (needs to be at the same directory as Cargo.toml\ncargo run --bin \\${file}\n\n# run without compiling messages\ncarg run --bin \\${file} -q # or --quiet\n``````\n\nThe exercises a1, a2, a3a and a3b are pretty trivial.\n\n### Match expressions\n\nHere at 1:17:00\n\n• similar to `if..else`\n\n• exhaustive\n\n• all options must be accounted for\n• looks like a \"switch case\" statement\n\n• `match` vs `if..else`\n\n• `match` will be checked by the compiler\n• if a new possibility is added, you will be notified when this occurs\n• `if..else` is not checked by the compiler\n• if a new possibility is added, your code may contain a bug\n\nNOTE: prefer `match` over `if..else` when working with a single variable. It's more robust.\n\nexample with boolean:\n\n``````fn main() {\nlet some_bool = true;\nmatch some_bool {\ntrue => println!(\"it's true\"),\nfalse => println!(\"it's false\"),\n}\n}\n``````\n\nexample with int:\n\n``````fn main() {\nlet some_int = 3;\nmatch some_int {\n1 => println!(\"it's 1\"),\n2 => println!(\"it's 2\"),\n3 => println!(\"it's 3\"),\n// underscore _ works like 'default' in C\n_ => println!(\"it's something else\"),\n}\n}\n``````\n\n### Expressions\n\nhttps://youtu.be/lzKeecy4OmQ?t=9364\n\n## Working With Data\n\n### enum\n\n• data that can be one of multiple different possibilities\n• each possibility is called a \"variant\"\n• more robust programs when paired with `match`\n``````enum Direction {\nUp,\nDown,\nLeft,\nRight,\n}\n\nfn which_way(go: Direction) {\n// remember, with match all possibilities\n// must be satisfied. Therefore, if you\n// add a new item to the Direction enum,\n// the compiler will complain if you don't\n// add it to the match block below.\nmatch go {\nDirection::Up => \"up\",\nDirection::Down => \"down\",\nDirection::Left => \"left\",\nDirection::Right => \"right\",\n}\n}\n``````\n\n### struct\n\n• a type containing multiple pieces of data\n• each piece of data is called a \"field\"\n• all fields MUST be populated\n• make working with data easier\n\nExample:\n\n``````struct ShippingBox {\ndepth: i32,\nwidth: i32,\nheight: i32,\n}\n\nfn my_function() {\nlet my_box = ShippingBox {\ndepth: 3,\nwidth: 2,\nheight: 5,\n};\n\n// access individual fields with a '.' dot\nlet tall = my_box.height;\n}\n``````\n\n### tuples\n\n• a type of \"record\"\n• store data anonymously (no need to name fields)\n• useful to return pairs of data from functions\n• can be destructured easily into variables\n• accepts different types\n• to access items of a tuple use the `.` dot followed by the index (kinda unusual notation). example: `my_tuple.0`\n``````\n// declaração com inferência de tipo\nlet numbers = (1, 2, 3);\nlet my_stuff = (1, 2.3, false);\n\n// declaração com tipagem explícita\nlet numbers: (i32, i32, i32) = (1, 2, 3);\nlet my_stuff (i32, f64, bool) = (1, 2.3, false);\n\n// desserialiazação\nlet (a, b, c) = numbers;\n\nfn one_two_three() -> (i32, i32, i32) {\n(1, 2, 3)\n}\n\nlet numbers = one_two_three();\nlet (x, y, z) = one_two_three();\n``````\n\nContinuar em #Expressions" ]

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[ "", null, "LAPACK  3.10.1 LAPACK: Linear Algebra PACKage\n\n## ◆ ssytrf_rk()\n\n subroutine ssytrf_rk ( character UPLO, integer N, real, dimension( lda, * ) A, integer LDA, real, dimension( * ) E, integer, dimension( * ) IPIV, real, dimension( * ) WORK, integer LWORK, integer INFO )\n\nSSYTRF_RK computes the factorization of a real symmetric indefinite matrix using the bounded Bunch-Kaufman (rook) diagonal pivoting method (BLAS3 blocked algorithm).\n\nPurpose:\n``` SSYTRF_RK computes the factorization of a real symmetric matrix A\nusing the bounded Bunch-Kaufman (rook) diagonal pivoting method:\n\nA = P*U*D*(U**T)*(P**T) or A = P*L*D*(L**T)*(P**T),\n\nwhere U (or L) is unit upper (or lower) triangular matrix,\nU**T (or L**T) is the transpose of U (or L), P is a permutation\nmatrix, P**T is the transpose of P, and D is symmetric and block\ndiagonal with 1-by-1 and 2-by-2 diagonal blocks.\n\nThis is the blocked version of the algorithm, calling Level 3 BLAS.\nParameters\n [in] UPLO ``` UPLO is CHARACTER*1 Specifies whether the upper or lower triangular part of the symmetric matrix A is stored: = 'U': Upper triangular = 'L': Lower triangular``` [in] N ``` N is INTEGER The order of the matrix A. N >= 0.``` [in,out] A ``` A is REAL array, dimension (LDA,N) On entry, the symmetric matrix A. If UPLO = 'U': the leading N-by-N upper triangular part of A contains the upper triangular part of the matrix A, and the strictly lower triangular part of A is not referenced. If UPLO = 'L': the leading N-by-N lower triangular part of A contains the lower triangular part of the matrix A, and the strictly upper triangular part of A is not referenced. On exit, contains: a) ONLY diagonal elements of the symmetric block diagonal matrix D on the diagonal of A, i.e. D(k,k) = A(k,k); (superdiagonal (or subdiagonal) elements of D are stored on exit in array E), and b) If UPLO = 'U': factor U in the superdiagonal part of A. If UPLO = 'L': factor L in the subdiagonal part of A.``` [in] LDA ``` LDA is INTEGER The leading dimension of the array A. LDA >= max(1,N).``` [out] E ``` E is REAL array, dimension (N) On exit, contains the superdiagonal (or subdiagonal) elements of the symmetric block diagonal matrix D with 1-by-1 or 2-by-2 diagonal blocks, where If UPLO = 'U': E(i) = D(i-1,i), i=2:N, E(1) is set to 0; If UPLO = 'L': E(i) = D(i+1,i), i=1:N-1, E(N) is set to 0. NOTE: For 1-by-1 diagonal block D(k), where 1 <= k <= N, the element E(k) is set to 0 in both UPLO = 'U' or UPLO = 'L' cases.``` [out] IPIV ``` IPIV is INTEGER array, dimension (N) IPIV describes the permutation matrix P in the factorization of matrix A as follows. The absolute value of IPIV(k) represents the index of row and column that were interchanged with the k-th row and column. The value of UPLO describes the order in which the interchanges were applied. Also, the sign of IPIV represents the block structure of the symmetric block diagonal matrix D with 1-by-1 or 2-by-2 diagonal blocks which correspond to 1 or 2 interchanges at each factorization step. For more info see Further Details section. If UPLO = 'U', ( in factorization order, k decreases from N to 1 ): a) A single positive entry IPIV(k) > 0 means: D(k,k) is a 1-by-1 diagonal block. If IPIV(k) != k, rows and columns k and IPIV(k) were interchanged in the matrix A(1:N,1:N); If IPIV(k) = k, no interchange occurred. b) A pair of consecutive negative entries IPIV(k) < 0 and IPIV(k-1) < 0 means: D(k-1:k,k-1:k) is a 2-by-2 diagonal block. (NOTE: negative entries in IPIV appear ONLY in pairs). 1) If -IPIV(k) != k, rows and columns k and -IPIV(k) were interchanged in the matrix A(1:N,1:N). If -IPIV(k) = k, no interchange occurred. 2) If -IPIV(k-1) != k-1, rows and columns k-1 and -IPIV(k-1) were interchanged in the matrix A(1:N,1:N). If -IPIV(k-1) = k-1, no interchange occurred. c) In both cases a) and b), always ABS( IPIV(k) ) <= k. d) NOTE: Any entry IPIV(k) is always NONZERO on output. If UPLO = 'L', ( in factorization order, k increases from 1 to N ): a) A single positive entry IPIV(k) > 0 means: D(k,k) is a 1-by-1 diagonal block. If IPIV(k) != k, rows and columns k and IPIV(k) were interchanged in the matrix A(1:N,1:N). If IPIV(k) = k, no interchange occurred. b) A pair of consecutive negative entries IPIV(k) < 0 and IPIV(k+1) < 0 means: D(k:k+1,k:k+1) is a 2-by-2 diagonal block. (NOTE: negative entries in IPIV appear ONLY in pairs). 1) If -IPIV(k) != k, rows and columns k and -IPIV(k) were interchanged in the matrix A(1:N,1:N). If -IPIV(k) = k, no interchange occurred. 2) If -IPIV(k+1) != k+1, rows and columns k-1 and -IPIV(k-1) were interchanged in the matrix A(1:N,1:N). If -IPIV(k+1) = k+1, no interchange occurred. c) In both cases a) and b), always ABS( IPIV(k) ) >= k. d) NOTE: Any entry IPIV(k) is always NONZERO on output.``` [out] WORK ``` WORK is REAL array, dimension ( MAX(1,LWORK) ). On exit, if INFO = 0, WORK(1) returns the optimal LWORK.``` [in] LWORK ``` LWORK is INTEGER The length of WORK. LWORK >=1. For best performance LWORK >= N*NB, where NB is the block size returned by ILAENV. If LWORK = -1, then a workspace query is assumed; the routine only calculates the optimal size of the WORK array, returns this value as the first entry of the WORK array, and no error message related to LWORK is issued by XERBLA.``` [out] INFO ``` INFO is INTEGER = 0: successful exit < 0: If INFO = -k, the k-th argument had an illegal value > 0: If INFO = k, the matrix A is singular, because: If UPLO = 'U': column k in the upper triangular part of A contains all zeros. If UPLO = 'L': column k in the lower triangular part of A contains all zeros. Therefore D(k,k) is exactly zero, and superdiagonal elements of column k of U (or subdiagonal elements of column k of L ) are all zeros. The factorization has been completed, but the block diagonal matrix D is exactly singular, and division by zero will occur if it is used to solve a system of equations. NOTE: INFO only stores the first occurrence of a singularity, any subsequent occurrence of singularity is not stored in INFO even though the factorization always completes.```\nFurther Details:\n` TODO: put correct description`\nContributors:\n``` December 2016, Igor Kozachenko,\nComputer Science Division,\nUniversity of California, Berkeley\n\nSeptember 2007, Sven Hammarling, Nicholas J. Higham, Craig Lucas,\nSchool of Mathematics,\nUniversity of Manchester```\n\nDefinition at line 257 of file ssytrf_rk.f.\n\n259 *\n260 * -- LAPACK computational routine --\n261 * -- LAPACK is a software package provided by Univ. of Tennessee, --\n262 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--\n263 *\n264 * .. Scalar Arguments ..\n265  CHARACTER UPLO\n266  INTEGER INFO, LDA, LWORK, N\n267 * ..\n268 * .. Array Arguments ..\n269  INTEGER IPIV( * )\n270  REAL A( LDA, * ), E( * ), WORK( * )\n271 * ..\n272 *\n273 * =====================================================================\n274 *\n275 * .. Local Scalars ..\n276  LOGICAL LQUERY, UPPER\n277  INTEGER I, IINFO, IP, IWS, K, KB, LDWORK, LWKOPT,\n278  \\$ NB, NBMIN\n279 * ..\n280 * .. External Functions ..\n281  LOGICAL LSAME\n282  INTEGER ILAENV\n283  EXTERNAL lsame, ilaenv\n284 * ..\n285 * .. External Subroutines ..\n286  EXTERNAL slasyf_rk, ssytf2_rk, sswap, xerbla\n287 * ..\n288 * .. Intrinsic Functions ..\n289  INTRINSIC abs, max\n290 * ..\n291 * .. Executable Statements ..\n292 *\n293 * Test the input parameters.\n294 *\n295  info = 0\n296  upper = lsame( uplo, 'U' )\n297  lquery = ( lwork.EQ.-1 )\n298  IF( .NOT.upper .AND. .NOT.lsame( uplo, 'L' ) ) THEN\n299  info = -1\n300  ELSE IF( n.LT.0 ) THEN\n301  info = -2\n302  ELSE IF( lda.LT.max( 1, n ) ) THEN\n303  info = -4\n304  ELSE IF( lwork.LT.1 .AND. .NOT.lquery ) THEN\n305  info = -8\n306  END IF\n307 *\n308  IF( info.EQ.0 ) THEN\n309 *\n310 * Determine the block size\n311 *\n312  nb = ilaenv( 1, 'SSYTRF_RK', uplo, n, -1, -1, -1 )\n313  lwkopt = n*nb\n314  work( 1 ) = lwkopt\n315  END IF\n316 *\n317  IF( info.NE.0 ) THEN\n318  CALL xerbla( 'SSYTRF_RK', -info )\n319  RETURN\n320  ELSE IF( lquery ) THEN\n321  RETURN\n322  END IF\n323 *\n324  nbmin = 2\n325  ldwork = n\n326  IF( nb.GT.1 .AND. nb.LT.n ) THEN\n327  iws = ldwork*nb\n328  IF( lwork.LT.iws ) THEN\n329  nb = max( lwork / ldwork, 1 )\n330  nbmin = max( 2, ilaenv( 2, 'SSYTRF_RK',\n331  \\$ uplo, n, -1, -1, -1 ) )\n332  END IF\n333  ELSE\n334  iws = 1\n335  END IF\n336  IF( nb.LT.nbmin )\n337  \\$ nb = n\n338 *\n339  IF( upper ) THEN\n340 *\n341 * Factorize A as U*D*U**T using the upper triangle of A\n342 *\n343 * K is the main loop index, decreasing from N to 1 in steps of\n344 * KB, where KB is the number of columns factorized by SLASYF_RK;\n345 * KB is either NB or NB-1, or K for the last block\n346 *\n347  k = n\n348  10 CONTINUE\n349 *\n350 * If K < 1, exit from loop\n351 *\n352  IF( k.LT.1 )\n353  \\$ GO TO 15\n354 *\n355  IF( k.GT.nb ) THEN\n356 *\n357 * Factorize columns k-kb+1:k of A and use blocked code to\n358 * update columns 1:k-kb\n359 *\n360  CALL slasyf_rk( uplo, k, nb, kb, a, lda, e,\n361  \\$ ipiv, work, ldwork, iinfo )\n362  ELSE\n363 *\n364 * Use unblocked code to factorize columns 1:k of A\n365 *\n366  CALL ssytf2_rk( uplo, k, a, lda, e, ipiv, iinfo )\n367  kb = k\n368  END IF\n369 *\n370 * Set INFO on the first occurrence of a zero pivot\n371 *\n372  IF( info.EQ.0 .AND. iinfo.GT.0 )\n373  \\$ info = iinfo\n374 *\n375 * No need to adjust IPIV\n376 *\n377 *\n378 * Apply permutations to the leading panel 1:k-1\n379 *\n380 * Read IPIV from the last block factored, i.e.\n381 * indices k-kb+1:k and apply row permutations to the\n382 * last k+1 colunms k+1:N after that block\n383 * (We can do the simple loop over IPIV with decrement -1,\n384 * since the ABS value of IPIV( I ) represents the row index\n385 * of the interchange with row i in both 1x1 and 2x2 pivot cases)\n386 *\n387  IF( k.LT.n ) THEN\n388  DO i = k, ( k - kb + 1 ), -1\n389  ip = abs( ipiv( i ) )\n390  IF( ip.NE.i ) THEN\n391  CALL sswap( n-k, a( i, k+1 ), lda,\n392  \\$ a( ip, k+1 ), lda )\n393  END IF\n394  END DO\n395  END IF\n396 *\n397 * Decrease K and return to the start of the main loop\n398 *\n399  k = k - kb\n400  GO TO 10\n401 *\n402 * This label is the exit from main loop over K decreasing\n403 * from N to 1 in steps of KB\n404 *\n405  15 CONTINUE\n406 *\n407  ELSE\n408 *\n409 * Factorize A as L*D*L**T using the lower triangle of A\n410 *\n411 * K is the main loop index, increasing from 1 to N in steps of\n412 * KB, where KB is the number of columns factorized by SLASYF_RK;\n413 * KB is either NB or NB-1, or N-K+1 for the last block\n414 *\n415  k = 1\n416  20 CONTINUE\n417 *\n418 * If K > N, exit from loop\n419 *\n420  IF( k.GT.n )\n421  \\$ GO TO 35\n422 *\n423  IF( k.LE.n-nb ) THEN\n424 *\n425 * Factorize columns k:k+kb-1 of A and use blocked code to\n426 * update columns k+kb:n\n427 *\n428  CALL slasyf_rk( uplo, n-k+1, nb, kb, a( k, k ), lda, e( k ),\n429  \\$ ipiv( k ), work, ldwork, iinfo )\n430\n431\n432  ELSE\n433 *\n434 * Use unblocked code to factorize columns k:n of A\n435 *\n436  CALL ssytf2_rk( uplo, n-k+1, a( k, k ), lda, e( k ),\n437  \\$ ipiv( k ), iinfo )\n438  kb = n - k + 1\n439 *\n440  END IF\n441 *\n442 * Set INFO on the first occurrence of a zero pivot\n443 *\n444  IF( info.EQ.0 .AND. iinfo.GT.0 )\n445  \\$ info = iinfo + k - 1\n446 *\n448 *\n449  DO i = k, k + kb - 1\n450  IF( ipiv( i ).GT.0 ) THEN\n451  ipiv( i ) = ipiv( i ) + k - 1\n452  ELSE\n453  ipiv( i ) = ipiv( i ) - k + 1\n454  END IF\n455  END DO\n456 *\n457 * Apply permutations to the leading panel 1:k-1\n458 *\n459 * Read IPIV from the last block factored, i.e.\n460 * indices k:k+kb-1 and apply row permutations to the\n461 * first k-1 colunms 1:k-1 before that block\n462 * (We can do the simple loop over IPIV with increment 1,\n463 * since the ABS value of IPIV( I ) represents the row index\n464 * of the interchange with row i in both 1x1 and 2x2 pivot cases)\n465 *\n466  IF( k.GT.1 ) THEN\n467  DO i = k, ( k + kb - 1 ), 1\n468  ip = abs( ipiv( i ) )\n469  IF( ip.NE.i ) THEN\n470  CALL sswap( k-1, a( i, 1 ), lda,\n471  \\$ a( ip, 1 ), lda )\n472  END IF\n473  END DO\n474  END IF\n475 *\n476 * Increase K and return to the start of the main loop\n477 *\n478  k = k + kb\n479  GO TO 20\n480 *\n481 * This label is the exit from main loop over K increasing\n482 * from 1 to N in steps of KB\n483 *\n484  35 CONTINUE\n485 *\n486 * End Lower\n487 *\n488  END IF\n489 *\n490  work( 1 ) = lwkopt\n491  RETURN\n492 *\n493 * End of SSYTRF_RK\n494 *\ninteger function ilaenv(ISPEC, NAME, OPTS, N1, N2, N3, N4)\nILAENV\nDefinition: ilaenv.f:162\nsubroutine xerbla(SRNAME, INFO)\nXERBLA\nDefinition: xerbla.f:60\nlogical function lsame(CA, CB)\nLSAME\nDefinition: lsame.f:53\nsubroutine sswap(N, SX, INCX, SY, INCY)\nSSWAP\nDefinition: sswap.f:82\nsubroutine slasyf_rk(UPLO, N, NB, KB, A, LDA, E, IPIV, W, LDW, INFO)\nSLASYF_RK computes a partial factorization of a real symmetric indefinite matrix using bounded Bunch-...\nDefinition: slasyf_rk.f:262\nsubroutine ssytf2_rk(UPLO, N, A, LDA, E, IPIV, INFO)\nSSYTF2_RK computes the factorization of a real symmetric indefinite matrix using the bounded Bunch-Ka...\nDefinition: ssytf2_rk.f:241\nHere is the call graph for this function:\nHere is the caller graph for this function:" ]

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[ "Home / English Posts / Speed of Electromagnetic Signal\n\n# Speed of Electromagnetic Signal", null, "The speed of light in vacuum, commonly denoted by c, is independent of the relative speed of the observer and the source of the radiation and therefore assumes the role of universal constant of physics, such as the gravitational constant G, the electron charge, its mass etc.\n\nAccording to current knowledge c it is the maximum speed at which can travel not only light but also any kind of electromagnetic wave and then in general each type of signal capable of transporting energy and therefore information.\n\nIn the everyday life we use radio frequency signals that propagate through a particular type of cable, known as coaxial cable. Just think about the signal captured by an antenna and sent to a TV or the cables that connect to each other more personal computers on the network and internet. If we think about the everyday normal distances, the signals seem to propagate instantaneously; for this reason, though admitting the existence of a finite limit value, the ways to measure c are not at all trivial, having to estimate the speed of a signal that runs thousands of kilometers per second.\n\nIn the proposed experiment a radio frequency signal will be sent in a coaxial cable having a length of about 10 meters to show that a measure of its speed is feasible even over short distances by means of an oscilloscope that is a device able to display rapid signals.\n\n### Coaxial Cable", null, "A coaxial cable is formed by a pair of cylindrical metallic conductors, isolated and arranged such that one lies within the other. The inner conductor is a solid wire while the outer one is formed from a wire braid. The two conductors are maintained spaced by insulating material. The outer conductor is covered by a sheath of protective material.", null, "Consider a generator that supplies a variable voltage signal and has an internal impedance Zs connected, through a coaxial cable of characteristic impedance Zo, to a load ZL. The propagation of the signal along the cable takes place as a variation of the electric and magnetic fields that are concatenated. In the cable, therefore, an electromagnetic wave travels along the cable and it reaches, after a certain time, the end of the line.\n\nIf ZL is equal to Z0 (characteristic impedance of the cable) then the wave is completely absorbed by the load, if Z0 is different than ZL then the load generates a reflected wave that passes again through the cable. In particular if ZL > Z0 (for example, an open circuit) the load generates a reflected wave of the same polarity, if instead ZL < Z0 (for example a short circuit) then it generates a reflected wave of reversed polarity.\n\n### Experimental Setup\n\nTo make the experiment a function generator has been used, configured to generate 20ns pulses with frequency of 1kHz. The pulses were acquired via a digital oscilloscope with a bandwidth of 100MHz. A 10m long coaxial cable RG-58 with a characteristic impedance of 50Ω has been used as transmission line. The images below show the setup.", null, "Function Generator and Oscilloscope", null, "BNC cable with 50 ohm head\n\n### Results\n\nSome measures have been made, by varying the load impedance value and acquiring the direct wave and reflected wave signals. The results are illustrated in the graphs below. You can see how the reflected wave is absent in the case of adapted impedance (ZL = 50 ohm), whilst for greater and less impedance is present a reflected wave, positive polarity in the first case and negative in the second.", null, "Load impedance = 50 ohm – No reflected pulse", null, "Load Impedance > 50 ohm – Positive reflected pulse", null, "Load Impedance < 50 ohm – Negative reflected pulse\n\n### Determination of pulse speed\n\nFrom the graph that displays both the direct and the reflected wave is possible to easily determine the time interval, and then determine the propagation velocity of the electromagnetic signal in the coaxial cable:\n\ndt = 87 ns = 87 x 10-9\nL = 20 m (10 m x 2)\nV = L / dt = 0,23 x 109m/s = 230.000 km/s\n\nThe result that is obtained is less than the value of c, in fact, the electromagnetic signal does not propagate in a vacuum but in the space between the two conductors filled by the dielectric of the coaxial cable. In this case, the propagation velocity is given by the following relation:\n\nv = 1/√εμ\n\nwhere ε and μ are, respectively, the dielectric constant and the magnetic permeability of the insulating means separating the conductors. Such material is characterized by a permeability μ = μo, where μo is the magnetic permeability of vacuum. By defining the dielectric constant of the medium in relative terms εr is obtained:\n\nv = 1/√μoεoεr = 1/√μoε 1/√εr = c/√εr\n\nFrom which we can derive the dielectric constant of insulating medium:\n\nεr=(c/v)2= 1,69\n\n### Deformation of a pulse on a non-matched load\n\nA load impedance not adapted to the impedance of the coaxial cable causes the deformation of the pulse. This is because the original impulse is added to or subtracts the reflected pulse, this effect is more pronounced the greater is the non-adaptation of the impedances.\nThe results are shown in the graphs below. You can see how the impulse is not deformed in the case of adapted impedance (ZL = 50 ohm), while for higher impedance and lower the pulse is greatly deformed due to reflected wave.", null, "Load Impedence = 50 ohm", null, "Load Impedence > 50 ohm", null, "Load Impedance < 50 ohm\n\nIf you liked this post you can share it on the “social” Facebook, Twitter or LinkedIn with the buttons below. This way you can help us! Thank you !\n\n#### Donation\n\nIf you like this site and if you want to contribute to the development of the activities you can make a donation, thank you !", null, "## Surface Plasmon Resonance\n\nIn the image above the evanescent wave excites the plasmon resonance on the surface of a …" ]

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[ "", null, "# Idon’t understand at all. if you could give answer then explain how or just give answer it would be really", null, "", null, "### Another question on Mathematics", null, "Mathematics, 04.02.2019 01:55\nIn a graph with several intervals o data how does a constant interval appear? what type of scenario produces a constant interval?", null, "Mathematics, 03.02.2019 22:17\nWhat is the value of x in the this question ?", null, "Mathematics, 03.02.2019 02:30\nThe function v(r)=4/3pir^3 can be used to find the volume of air inside a basketball given its radius. what does v(r) represent", null, "Mathematics, 01.02.2019 23:22\nWhat is the correct decimal form of 12%\nIdon’t understand at all. if you could give answer then explain how or just give answer it would be...\nQuestions", null, "", null, "", null, "Mathematics, 26.06.2020 14:39", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Questions on the website: 6628032" ]

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[ "## Saturday, November 17, 2012\n\n### AN OVERVIEW OF BEARING CAPACITY THEORIES.\n\nThe determination of bearing capacity of soil based on the classical earth pressure theory of Rankine (1857) began with Pauker, a Russian military engineer (1889), and was modified by Bell (1915). Pauker's theory was applicable only for sandy soils but the theory of Bell took into account cohesion also. Neither theory took into account the width of the foundation. Subsequent developments led to the modification of Bell's theory to include width of footing also.\n\nThe methods of calculating the ultimate bearing capacity of shallow strip footings by plastic theory developed considerably over the years since Terzaghi (1943) first proposed a method by taking into account the weight of soil by the principle of superposition. Terzaghi extended the theory of Prandtl (1921). Prandtl developed an equation based on his study of the penetration of a long hard metal punch into softer materials for computing the ultimate bearing capacity. He assumed the material was weightless possessing only cohesion and friction. Taylor (1948) extended the equation of Prandtl by taking into account the surcharge effect of the overburden soil at the foundation level.\n\nNo exact analytical solution for computing bearing capacity of footings is available at present because the basic system of equations describing the yield problems is nonlinear. On account of these reasons, Terzaghi (1943) first proposed a semi-empirical equation for computing the ultimate bearing capacity of strip footings by taking into account cohesion, friction and weight of soil, and replacing the overburden pressure with an equivalent surcharge load at the base level of the foundation. This method was for the general shear failure condition and the principle of superposition was adopted. His work was an extension of the work of Prandtl (1921). The final form of the equation proposed by Terzaghi is the same as the one given by Prandtl.\n\nSubsequent to the work by Terzaghi, many investigators became interested in this problem and presented their own solutions. However the form of the equation presented by all these investigators remained the same as that of Terzaghi, but their methods of determining the bearing capacity factors were different.\n\nOf importance in determining the bearing capacity of strip footings is the assumption of plane strain inherent in the solutions of strip footings. The angle of internal friction as determined under an axially symmetric triaxial compression stress state, 0f, is known to be several degrees less than that determined under plane strain conditions under low confining pressures. Thus the bearing capacity of a strip footing calculated by the generally accepted formulas, using 0r, is usually less than the actual bearing capacity as determined by the plane strain footing tests which leads to a conclusion that the bearing capacity formulas are conservative.\n\nThe ultimate bearing capacity, or the allowable soil pressure, can be calculated either from bearing capacity theories or from some of the in situ tests. Each theory has its own good and bad points. Some of the theories are of academic interest only. However, it is the purpose of the author to present here only such theories which are of basic interest to students in particular and professional engineers in general. The application of field tests for determining bearing capacity are also presented which are of particular importance to professional engineers since present practice is to rely more on field tests for determining the bearing capacity or allowable bearing pressure of soil.\n\nSome of the methods that are discussed in this chapter are\n\n1. Terzaghi's bearing capacity theory\n2. The general bearing capacity equation\n3. Field tests" ]

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[ "# [R] Bootstrap CIs for weighted means of paired differences\n\ni.petzev i.petzev at gmail.com\nThu Nov 20 11:23:30 CET 2014\n\n```Hi David,\n\nsorry, I was not clear. The difference comes from defining or not defining “w” in the boot() function. The results with your function and your approach are thus:\n\nset.seed(1111)\nx <- rnorm(50)\ny <- rnorm(50)\nweights <- runif(50)\nweights <- weights / sum(weights)\ndataset <- cbind(x,y,weights)\n\nvw_m_diff <- function(dataset,w) {\ndifferences <- dataset[w,1]-dataset[w,2]\nweights <- dataset[w, \"weights\"]\nreturn(weighted.mean(x=differences, w=weights))\n}\nres_boot <- boot(dataset, statistic=vw_m_diff, R = 1000, w=dataset[,3])\nboot.ci(res_boot)\n\nBOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS\nBased on 1000 bootstrap replicates\n\nCALL :\nboot.ci(boot.out = res_boot)\n\nIntervals :\nLevel Normal Basic\n95% (-0.5657, 0.4962 ) (-0.5713, 0.5062 )\n\nLevel Percentile BCa\n95% (-0.6527, 0.4249 ) (-0.5579, 0.5023 )\nCalculations and Intervals on Original Scale\n\n********************************************************************************************************************\n\nHowever, without defining “w” in the bootstrap function, i.e., running an ordinary and not a weighted bootstrap, the results are:\n\nres_boot <- boot(dataset, statistic=vw_m_diff, R = 1000)\nboot.ci(res_boot)\n\nBOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS\nBased on 1000 bootstrap replicates\n\nCALL :\nboot.ci(boot.out = res_boot)\n\nIntervals :\nLevel Normal Basic\n95% (-0.6265, 0.4966 ) (-0.6125, 0.5249 )\n\nLevel Percentile BCa\n95% (-0.6714, 0.4661 ) (-0.6747, 0.4559 )\nCalculations and Intervals on Original Scale\n\nOn 19 Nov 2014, at 17:49, David Winsemius <dwinsemius at comcast.net> wrote:\n\n>>> vw_m_diff <- function(dataset,w) {\n>>> differences <- dataset[w,1]-dataset[w,2]\n>>> weights <- dataset[w, \"weights\"]\n>>> return(weighted.mean(x=differences, w=weights))\n>>> }\n\n[[alternative HTML version deleted]]\n\n```" ]

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[ "# #0cfa85 Color Information\n\nIn a RGB color space, hex #0cfa85 is composed of 4.7% red, 98% green and 52.2% blue. Whereas in a CMYK color space, it is composed of 95.2% cyan, 0% magenta, 46.8% yellow and 2% black. It has a hue angle of 150.5 degrees, a saturation of 96% and a lightness of 51.4%. #0cfa85 color hex could be obtained by blending #18ffff with #00f50b. Closest websafe color is: #00ff99.\n\n• R 5\n• G 98\n• B 52\nRGB color chart\n• C 95\n• M 0\n• Y 47\n• K 2\nCMYK color chart\n\n#0cfa85 color description : Vivid cyan - lime green.\n\n# #0cfa85 Color Conversion\n\nThe hexadecimal color #0cfa85 has RGB values of R:12, G:250, B:133 and CMYK values of C:0.95, M:0, Y:0.47, K:0.02. Its decimal value is 850565.\n\nHex triplet RGB Decimal 0cfa85 `#0cfa85` 12, 250, 133 `rgb(12,250,133)` 4.7, 98, 52.2 `rgb(4.7%,98%,52.2%)` 95, 0, 47, 2 150.5°, 96, 51.4 `hsl(150.5,96%,51.4%)` 150.5°, 95.2, 98 00ff99 `#00ff99`\nCIE-LAB 87.065, -74.076, 42.419 38.568, 70.138, 33.694 0.271, 0.493, 70.138 87.065, 85.362, 150.203 87.065, -77.404, 69.442 83.749, -64.357, 34.77 00001100, 11111010, 10000101\n\n# Color Schemes with #0cfa85\n\n• #0cfa85\n``#0cfa85` `rgb(12,250,133)``\n• #fa0c81\n``#fa0c81` `rgb(250,12,129)``\nComplementary Color\n• #0cfa0e\n``#0cfa0e` `rgb(12,250,14)``\n• #0cfa85\n``#0cfa85` `rgb(12,250,133)``\n• #0cf8fa\n``#0cf8fa` `rgb(12,248,250)``\nAnalogous Color\n• #fa0e0c\n``#fa0e0c` `rgb(250,14,12)``\n• #0cfa85\n``#0cfa85` `rgb(12,250,133)``\n• #fa0cf8\n``#fa0cf8` `rgb(250,12,248)``\nSplit Complementary Color\n• #fa850c\n``#fa850c` `rgb(250,133,12)``\n• #0cfa85\n``#0cfa85` `rgb(12,250,133)``\n• #850cfa\n``#850cfa` `rgb(133,12,250)``\n• #81fa0c\n``#81fa0c` `rgb(129,250,12)``\n• #0cfa85\n``#0cfa85` `rgb(12,250,133)``\n• #850cfa\n``#850cfa` `rgb(133,12,250)``\n• #fa0c81\n``#fa0c81` `rgb(250,12,129)``\n• #04b65e\n``#04b65e` `rgb(4,182,94)``\n• #04cf6b\n``#04cf6b` `rgb(4,207,107)``\n• #05e878\n``#05e878` `rgb(5,232,120)``\n• #0cfa85\n``#0cfa85` `rgb(12,250,133)``\n• #25fb92\n``#25fb92` `rgb(37,251,146)``\n• #3efb9e\n``#3efb9e` `rgb(62,251,158)``\n• #57fcab\n``#57fcab` `rgb(87,252,171)``\nMonochromatic Color\n\n# Alternatives to #0cfa85\n\nBelow, you can see some colors close to #0cfa85. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0cfa4a\n``#0cfa4a` `rgb(12,250,74)``\n• #0cfa5d\n``#0cfa5d` `rgb(12,250,93)``\n• #0cfa71\n``#0cfa71` `rgb(12,250,113)``\n• #0cfa85\n``#0cfa85` `rgb(12,250,133)``\n• #0cfa99\n``#0cfa99` `rgb(12,250,153)``\n``#0cfaad` `rgb(12,250,173)``\n• #0cfac1\n``#0cfac1` `rgb(12,250,193)``\nSimilar Colors\n\n# #0cfa85 Preview\n\nThis text has a font color of #0cfa85.\n\n``<span style=\"color:#0cfa85;\">Text here</span>``\n#0cfa85 background color\n\nThis paragraph has a background color of #0cfa85.\n\n``<p style=\"background-color:#0cfa85;\">Content here</p>``\n#0cfa85 border color\n\nThis element has a border color of #0cfa85.\n\n``<div style=\"border:1px solid #0cfa85;\">Content here</div>``\nCSS codes\n``.text {color:#0cfa85;}``\n``.background {background-color:#0cfa85;}``\n``.border {border:1px solid #0cfa85;}``\n\n# Shades and Tints of #0cfa85\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, #000704 is the darkest color, while #f3fff9 is the lightest one.\n\n• #000704\n``#000704` `rgb(0,7,4)``\n• #011a0e\n``#011a0e` `rgb(1,26,14)``\n• #012d17\n``#012d17` `rgb(1,45,23)``\n• #014121\n``#014121` `rgb(1,65,33)``\n• #02542b\n``#02542b` `rgb(2,84,43)``\n• #026735\n``#026735` `rgb(2,103,53)``\n• #037a3f\n``#037a3f` `rgb(3,122,63)``\n• #038d49\n``#038d49` `rgb(3,141,73)``\n• #03a153\n``#03a153` `rgb(3,161,83)``\n• #04b45d\n``#04b45d` `rgb(4,180,93)``\n• #04c767\n``#04c767` `rgb(4,199,103)``\n• #04da71\n``#04da71` `rgb(4,218,113)``\n• #05ed7b\n``#05ed7b` `rgb(5,237,123)``\n• #0cfa85\n``#0cfa85` `rgb(12,250,133)``\n• #1ffa8f\n``#1ffa8f` `rgb(31,250,143)``\n• #32fb98\n``#32fb98` `rgb(50,251,152)``\n• #46fba2\n``#46fba2` `rgb(70,251,162)``\n• #59fcac\n``#59fcac` `rgb(89,252,172)``\n• #6cfcb5\n``#6cfcb5` `rgb(108,252,181)``\n• #7ffcbf\n``#7ffcbf` `rgb(127,252,191)``\n• #93fdc9\n``#93fdc9` `rgb(147,253,201)``\n• #a6fdd2\n``#a6fdd2` `rgb(166,253,210)``\n• #b9fedc\n``#b9fedc` `rgb(185,254,220)``\n• #ccfee5\n``#ccfee5` `rgb(204,254,229)``\n• #dffeef\n``#dffeef` `rgb(223,254,239)``\n• #f3fff9\n``#f3fff9` `rgb(243,255,249)``\nTint Color Variation\n\n# Tones of #0cfa85\n\nA tone is produced by adding gray to any pure hue. In this case, #7e8883 is the less saturated color, while #0cfa85 is the most saturated one.\n\n• #7e8883\n``#7e8883` `rgb(126,136,131)``\n• #759183\n``#759183` `rgb(117,145,131)``\n• #6b9b83\n``#6b9b83` `rgb(107,155,131)``\n• #62a484\n``#62a484` `rgb(98,164,132)``\n• #58ae84\n``#58ae84` `rgb(88,174,132)``\n• #4fb784\n``#4fb784` `rgb(79,183,132)``\n• #45c184\n``#45c184` `rgb(69,193,132)``\n• #3cca84\n``#3cca84` `rgb(60,202,132)``\n• #32d484\n``#32d484` `rgb(50,212,132)``\n• #29dd85\n``#29dd85` `rgb(41,221,133)``\n• #1fe785\n``#1fe785` `rgb(31,231,133)``\n• #16f085\n``#16f085` `rgb(22,240,133)``\n• #0cfa85\n``#0cfa85` `rgb(12,250,133)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0cfa85 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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[ "## Monday, August 27, 2012\n\n### AVL tree insertion!!!\n\nLets constrcut the AVL tree with numbers 51, 26, 11, 6, 8, 4, 31, 21, 9, 16.  AVL tree is a balanced Binary search tree. So to construct AVL tree, we need to follow below steps\n• insert the new element in such a way that if the element is less than the current node go left and if the element is greater than go right.\n• Check for the balance at each node after insertion (diff between left subtree and right subtree should be atmost one)\nwe will see step by step after inserting each number into the tree from starting. The numbers are\n51, 26, 11, 6, 8, 4, 31, 21, 9, 16.\n• In step1 the tree is empty , so 51 is the first node, in step2 26 is inserted as a left child to the root node 51 and the tree is balanced.\n• In step3 11 is added as a left child to the node 26, becaus of this root node 51 is un-balanced, as it is shown in the image as red circle. We need to do the left left rotation which is a single rotation.\n• After the rotation, tree is balanced as shown in step4.\n• in step5 we are inserting 6 into the tree and which is as a left child to the node value 11.\n• in step6 we are inserting 8 into the tree and which is as a right child to the node value 6, because of this tree is un-balanced at node value 11 which is marked as red circle in the image.\n• for balancing the node we need to do right-left rotation which is a double rotation.\n• After the rotation, the tree is balanced and it is like as shown in step7.\n\n• In step8 we are inserting 4 into the tree and which is as a left child to node value 6. Becaus of this insertion, the tree is un-balanced at node value 26, and we need to do the left-left rotation as we did it earlier.\n• After the rotation the tree is like shown in step9. After that we need to insert 31 into the tree and which is as a left child to the node value 51 as shown in step10. And the tree is balanced.\n• In step11 we are inserted 21 into the tree whihc is as a right child to the node value 11. In step12 we inserted 9 as a left child to the node value 11. And the tree is balanced.\n• in step13, becaus of inserting 16 as a left child to the node value 21, the tree is un-balanced at node value 8.\n• We need to do the left right rotation  which is a double rotation. And the resultant tree is as shown in step14.\n\n## Friday, August 24, 2012\n\n### AVL tree double rotation!!\n\nDouble rotation: need to do two rotations (left-right or right-left) to balance the tree. In this there are two types of sub rotations\n• Right Left rotation\n• Left Right rotation\n\n", null, "Before Rotation\n Right-Left rotation:  This is double rotation method in which we need to do two rotatios left and right. And this method needs to apply if the tree is as shown in the below image.\n• Here r3 is un balanced and we need to do RL rotation because r2 is right child of r1 and r1 is left child of un balanced r3.\n• \n• Rotation is done in such a way that T1<r1<T2<r2<T3<r3<T4 property should satisfy after rotation.\n•", null, "After Rotation\n• To balance the tree, make r2 as root node and follow the binary search tree property for all other nodes.\n• Sample C code for Right Left rotation is given below. \n```\n//below steps are making the tree as shown in the image before rotation\nr1 = r3-<left;\nr2 = r1-<right;\nt2 = r2-<left;\nt3 = r2-<right;\n\n// actaul rotatiosn happens here\nr1-<right = t2;\nr3-<left = t3;\nr2-<left = r1;\nr2-<right = r3;\n\n//updte the new heihts for r1, r2, r3\nset_height(r1);\nset_height(r2);\nset_height(r3);\n\n*r = r2;\n\n```\nLeft Right rotation: This is double rotation method in which we need to do two rotatios right and left. And this method needs to apply if the tree is as shown in the below image.\n", null, "Before Rotation\n\n• Here r1 is un balanced and we need to do RL rotation because r2 is left child of r3 and r3 is right child of un balanced r1.\n• Rotation is done in such a way that T1<r1<T2<r2<T3<r3<T4 property should satisfy after rotation.\n•", null, "After Rotation\n• To balance the tree, make r2 as root node and follow the binary search tree property for all other nodes.\n• Sample C code for Right Left rotation is given below.\n``` //below steps are making the tree as shown in the image before rotation\nr3 = r1-<right;\nr2 = r3-<left;\nt2 = r2-<left;\nt3 = r2-<right;\n\n// actaul rotatiosn happens here\nr1-<right = t2;\nr3-<left = t3;\nr2-<left = r1;\nr2-<right = r3;\n\n//updte the new heihts for r1, r2, r3\nset_height(r1);\nset_height(r2);\nset_height(r3);\n\n*r = r2;\n\n```\n\n### AVL tree Single rotation!!\n\nSingle rotation: As the name says need to do one rotation (left or right) to balance the tree. In this there are two types of sub rotations\n• Left left rotation\n• Right right rotation\nLeft Left rotation: This is single rotation method which needs only one rotation. Left-Left rotation is required if the un-balanced tree is like as shown in the image.\n", null, "Before Rotation\n• Here r2  is un-balanced, because of adding the new node at T1 sub tree. Newly added node could either left or right. But it changes the balance property of r2.\n• To balance the tree, we need to do left-left rotation. This is because of newly added node is left child of r1 and r1 is a left child of un-balanced node r2.\n• Rotation is done in such a way that T1<r1<T2<r2<T3 property should satisfy after the rotation also.\n•", null, "After Rotation\nAfter the rotation the tree would be as shown in the image.\n• To make the tree balance, make the r2 as the root, and follow the Binary search tree property after that, we will get the resulted tree.\n• Sample code to do left-left rotation\n```r1 = r2->left;\n//we are @r2, below statements are to make the tree like\n// shown in the image. so that rotation becomes easy\nt1 = r1->left;\nt2 = r1->right;\nt3 = r2->right;\n\n//actual rotation happens here\nr1->right = r2;\nr2->left = t2;\n\n// update the r1 , r2 height\nset_height(r1);\nset_height(r2);\n```\n\nRight Right rotation: This is single rotation method which needs only one rotation. Right-Right rotation is required if the un-balanced tree is like as shown in the image.\n\n", null, "Before Rotation\n\n• Here r1 is un-balanced, because of adding new node to the subtree T3 either left or right. But it changes the balanced property of r1.\n• We need to do right-right rotation because newly addded node is right side of r2, and r2 is right side of the unbalanced node r1.\n• Rotation is done in such a way that T1<r1<T2<r2<T3 property should satisfy after the rotation also.\n•", null, "After Rotation\n• After the rotation, resulted tree shouls be as shown in the image.\n• To make the tree balance, make the r2 as the root, and follow the Binary search tree property after that, we will get the resulted tree.\n• See the sample code for right-right rotation below.\n```\n//we are @r1 and below statements are\n// used to arrange the pointers looks like in the image\nr2 = r1-<right;\nt1 = r1-<left;\nt2 = r2-<left;\nt3 = r2-<right;\n\n// actual rotations happens here\nr1-<right = t2;\nr2-<left = r1;\n\nset_height(r1);\nset_height(r2);\n\n*r = r2;\n\n```\n\n## Thursday, August 23, 2012\n\n### AVL tree rotation!!!\n\nAVL tree is balanced binary search tree. When we try to add new node to the AVL tree, it may become un-balanced. Generally the node which is unbalanced is grand parent of the newly inserted node. To balance the AVL tree, we need to do rotations. There are two types of rotations.\n• Single rotation: As the name says need to do one rotation (left or right) to balance the tree. In this there are two types of sub rotations\n• Double rotation: need to do two rotations (left-right or right-left) to balance the tree. In this there are two types of sub rotations\n\n## Tuesday, August 21, 2012\n\nRecently I came across below problem while I was writing the gmock test for our source code. gmock is a unit test framework for C++. gmock is short form of Google Mocking.\n\nAbort\n\nI got the above error two times while I was writing UT test cases. See the two scenarios below.\n1. Extra EXPECT_CALL :  This could be one of the reasons for the above problem. In my test case I have lot of EXPECT_CALL's, because our function is calling lot of other functions, so I need to mock all the functions, in the process of mocking these functions, by mistake I mocked one function which is not been called by this function. That means I added one extra EXPECT_CALL.  By taking stack trace using GDB I found the function and commented that expect call. And it was working fine.\n2. Used Once instead of Repeatedly: This may also causes for the above error message. With the experience of extra expect_call case, I have taken the stack trace or back trace using GDB, and I found that for one expect call I used WillOnce. So changed to WillRepeatedly. It got solved.\n\nAfter my analysis and contacting with our UT team, I came to know that the possible reason for this problem is that calling extra EXPECT_CALL for the unused functions.\n\n## Tuesday, August 14, 2012\n\n### C Program to create threads!!\n\nBelow is the simple C program to create the thread.\n```#include<stdio.h>\n\nvoid *print_data(void *str)\n{\nchar *msg = (char *)str;\nprintf(\"given message is '%s'\\n\",msg);\n}\nmain()\n{\nint ret1,ret2;\nchar msg1[]=\"this is first message\";\nchar msg2[]=\"this is second message\";\n\nprintf(\"th1 return value is %d\\n\",ret1);\nprintf(\"th2 return value is %d\\n\",ret2);\n\n}\n```\n\n## Wednesday, August 1, 2012\n\n### loop in linked list C program!!\n\nBelow is the C  program for finding the loop in a linked list\n```#include<stdio.h>\n#include<stdlib.h>\nstruct node\n{\nint info;\nstruct node *next;\n};\n\ntypedef struct node Node;\n\nNode* append(Node *h, int info)\n{\nNode *newNode = (Node*)malloc(sizeof(Node));\nnewNode->info = info;\nnewNode->next = NULL;\nif(tempHead == NULL) // if the list has zero elements. make new node as a head\n{\nh=newNode;\n}\nelse if(tempHead->next==NULL) // if the list is having only one node\n{\n}\nelse\n{\nNode *tempNode = h;\n// if the list has more than one node, so moving to the last node\nwhile(tempNode->next != NULL)\n{\ntempNode = tempNode->next;\n}\ntempNode->next = newNode; // appending the new node at the end\n}\n// this is for making the list circular/loop\nif(info == 101)\nnewNode->next = h;\nreturn h;\n}\n\n/*****************************************************************************\nfor displaying the nodes in the list.\nDont call this function if the list contains loop.\nif you call you wont get the end, becaue there is no NULL in the list\n*****************************************************************************/\nvoid display(Node *h)\n{\nNode *temp = h;\nwhile (temp->next!=NULL)\n{\nprintf(\"%d->\",temp->info);\ntemp = temp->next;\n}\nprintf(\"%d\\n\",temp->info);\n}\n\n/*************************************************************************************\nlet me explain the concept first. for finding the loop in a single linked list,\nneed to take two temparty nodes tempOne and tempTwo, move temOne node one level\nand move tempTwo two levels(tempTwo->next->next) until tempTwo/tempOne\nreaches NULL, if the list contains loop somewhere, these two nodes must be met/same,\nif not met/same, there is no loop in the list\n***************************************************************************************/\n\n{\nwhile( tempTwo->next!=NULL )\n{\ntempTwo = tempTwo->next; // one move\nif(tempTwo->next!=NULL)\n{\n// if second move is equal to tempOne then there is loop\nif(tempTwo->next==tempOne)\nreturn 1;\nelse\ntempTwo=tempTwo->next; // second move\n}\n\ntempOne = tempOne->next; // one move\n}\nreturn 0;\n}\nmain()\n{\nint i;\nfor (i=1;i<=10;i++)\n{\n}" ]

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[ "# Symmetric strengthening of the Cauchy-Schwarz inequality\n\nIn this great question by Nathaniel Johnston, and in its answers, we can learn the following remarkable inequality: For all $v,w \\in \\mathbb{R}^n$ we have \\begin{align*} \\|v^2\\| \\, \\|w^2\\| - \\langle v^2, w^2 \\rangle \\le \\|v\\|^2 \\|w\\|^2 - \\langle v,w \\rangle^2; \\quad (*) \\end{align*} here, $\\langle\\cdot,\\cdot\\rangle$ denotes the standard inner product on $\\mathbb{R}^n$, $\\|\\cdot\\|$ denotes the Euclidean norm and $v^2,w^2 \\in \\mathbb{R}^n$ denote the elementwise squares of $v$ and $w$. Both sides of $(*)$ are nonnegative by the Cauchy-Schwarz inequality, and the LHS gives a non-zero bound for the right RHS, in general.\n\nWhat strikes me is that the RHS and the LHS of $(*)$ have different (linear) symmetry groups: the RHS does not change if we apply any orthogonal matrix $U \\in \\mathbb{R}^{n \\times n}$ to both $v$ and $w$, while this is not true for the LHS. Hence, we can strengthen $(*)$ to \\begin{align*} \\sup_{U^*U = I}\\Big(\\|(Uv)^2\\| \\, \\|(Uw)^2\\| - \\langle (Uv)^2, (Uw)^2 \\rangle\\Big) \\le \\|v\\|^2 \\|w\\|^2 - \\langle v,w \\rangle^2. \\quad (**) \\end{align*} Unfortunately, I have no idea how to evaluate the LHS of $(**)$.\n\nQuestion. Can we explicitely evaluate the LHS of $(**)$? Or, more generally, is there a version of $(*)$ for which both sides are invariant under multiplying both $v$ and $w$ by (identical) orthogonal matrices?\n\nAdmittedly, this question is a bit vague since it might depend on one's perspective which expressions one considers to be \"explicit\" and which inequalities one considers to be a \"version\" of $(*)$. Nevertheless, I'm wondering whether some people share my intuition that there should be a more symmetric version of $(*)$.\n\nEdit. Maybe it is worthwhile to add the following motivating example: If we choose $n = 2$ and $v = (1,1)/\\sqrt{2}$, $w = (1,-1)/\\sqrt{2}$, then those vectors are orthogonal and the RHS of $(*)$ equals $1$, while the LHS of $(*)$ vanishes since $v^2$ and $w^2$ are linearly dependent. However, the LHS of $(**)$ is also equal to $1$; to see this, choose $U = \\frac{1}{\\sqrt{2}} \\begin{pmatrix} 1 & -1 \\\\ 1 & 1 \\end{pmatrix}$ and observe that $Uv = (0,1)$, $Uw = (1,0)$.\n\n$$(**)$$ is always equality. By homogenuity we may suppose that $$\\|v\\|=\\|w\\|=1$$, then $$Uv,Uw$$ are any two unit vectors with prescribed inner product $$\\langle Uv,Uw\\rangle =\\langle u,v\\rangle$$. I claim that you may find two two-dimensional vectors with prescribed given inner product such that equality in $$(*)$$ holds. Then you may make $$n$$-dimensional examples by adding zero coordinates. Well, just look at GH from MO's proof of (*) and look what we need for equality. If $$n=2$$, it is sufficient that $$F(v,w):=v_1w_1-v_2w_2=0$$. Since $$F((v_1,v_2),(w_1,w_2))=-F((w_2,w_1),(v_2,v_1))$$, we may rotate the vector $$(v_1,v_2)$$ until it becomes equal $$(w_2,w_1)$$, at this moment the vector $$(w_1,w_2)$$ becomes equal to $$(v_2,v_1)$$ (since the inner product and orientation define the pair of unit vectors upto rotation). By continuity $$F$$ attains zero value at some intermediate point.\n\n• Very nice idea to introduce the intermediate value theorem! Do you think we can conclude from your proof that for $v,w$ in general position, there are exactly $\\binom n2$ “essentially different” extremal matrices, one for each pair of coordinates? (By “essentially different” I mean up to permutations and/or sign changes of rows of $U$.) – Wolfgang Nov 19 '18 at 9:37\n\nThis is not an answer, rather a heuristic argument.\n\nI have done some numerical experiments for pairs of randomly chosen $$v,w$$, where I have used products of random Householder matrices for the $$U$$'s. They suggest that the LHS can come arbitrarily close to the RHS. So there should be a set of extremal unitary matrices $$U$$ for which equality is attained. For $$n=2$$ this $$U$$ seems unique up to sign (but the equation for it is already rather messy for the general case of $$v,w$$), and for $$n\\geqslant3$$ there seems to be a whole bunch of them. (a continuum? or rather a finite number of isolated matrices?)\n\nBy homogeneity, we can assume wlog $$\\|v\\|= \\|w\\|=1$$. Then it would be an interesting question to what those extremal unitary matrices, which are defined only by the pair $$\\{v,w\\}\\in\\mathbb R^n\\times\\mathbb R^n$$, correspond geometrically.\n\n• Is the unicity phenomenon you noticed anyhow related to the fact that for $n=2$ one can establish an isomorphism from $\\mathbb{R}^{n}$ to an algebraically closed field? – Sylvain JULIEN Nov 18 '18 at 21:35\n• @SylvainJULIEN I do not see why algebraic closedness should play a role here. But sure if $u,v$ for $n=2$ have, say, algebraic coordinates, the entries of $U$ will be algebraic as well. (Less sure for $n>2$.) – Wolfgang Nov 18 '18 at 21:40\n• My utterly vague idea was to draw a parallel with the fact that in the quaternions division algebra, a given element has \"a whole bunch\" of square roots. – Sylvain JULIEN Nov 18 '18 at 21:49\n• If you know the answer for $n=2$, can't you get the answer in general by a unitary transformation which takes the two-dimensional space generated by $v$ and $w$ to the two-dimensional space generated by $e_1$ and $e_2$? – Will Sawin Nov 18 '18 at 22:39" ]

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[ "to\n\n### Collaborating Authors\n\nGradient-based approaches to direct policy search in reinforcement learning have received much recent attention as a means to solve problems of partial observability and to avoid some of the problems associated with policy degradation in value-function methods. In this paper we introduce GPOMDP, a simulation-based algorithm for generating a biased estimate of the gradient of the average reward in Partially Observable Markov Decision Processes POMDPs controlled by parameterized stochastic policies. A similar algorithm was proposed by (Kimura et al. 1995). The algorithm's chief advantages are that it requires storage of only twice the number of policy parameters, uses one free beta (which has a natural interpretation in terms of bias-variance trade-off), and requires no knowledge of the underlying state. We prove convergence of GPOMDP, and show how the correct choice of the parameter beta is related to the mixing time of the controlled POMDP. We briefly describe extensions of GPOMDP to controlled Markov chains, continuous state, observation and control spaces, multiple-agents, higher-order derivatives, and a version for training stochastic policies with internal states. In a companion paper (Baxter et al., this volume) we show how the gradient estimates generated by GPOMDP can be used in both a traditional stochastic gradient algorithm and a conjugate-gradient procedure to find local optima of the average reward.\n\nThe stochastic meta--descent (SMD) gain adaptation algorithm [3, 4] can considerably accelerate the convergence of stochastic gradient descent.\n\n### Variance Reduction Techniques for Gradient Estimates in Reinforcement Learning\n\nWe consider the use of two additive control variate methods to reduce the variance of performance gradient estimates in reinforcement learning problems.The first approach we consider is the baseline method, in which a function of the current state is added to the discounted value estimate. We relate the performance of these methods, which use sample paths,to the variance of estimates based on iid data. We derive the baseline function that minimizes this variance, and we show that the variance forany baseline is the sum of the optimal variance and a weighted squared distance to the optimal baseline. We show that the widely used average discounted value baseline (where the reward is replaced by the difference between the reward and its expectation) is suboptimal. The second approach we consider is the actor-critic method, which uses an approximate value function. We give bounds on the expected squared error of its estimates. We show that minimizing distance to the true value function is suboptimal in general; we provide an example for which the true value function gives an estimate with positive variance, but the optimal valuefunction gives an unbiased estimate with zero variance. Our bounds suggest algorithms to estimate the gradient of the performance of parameterized baseline or value functions.\n\n### Experiments with Infinite-Horizon, Policy-Gradient Estimation\n\nIn this paper, we present algorithms that perform gradient ascent of the average reward in a partially observable Markov decision process (POMDP). These algorithms are based on GPOMDP, an algorithm introduced in a companion paper (Baxter & Bartlett, this volume), which computes biased estimates of the performance gradient in POMDPs. The algorithm's chief advantages are that it uses only one free parameter beta, which has a natural interpretation in terms of bias-variance trade-off, it requires no knowledge of the underlying state, and it can be applied to infinite state, control and observation spaces. We show how the gradient estimates produced by GPOMDP can be used to perform gradient ascent, both with a traditional stochastic-gradient algorithm, and with an algorithm based on conjugate-gradients that utilizes gradient information to bracket maxima in line searches. Experimental results are presented illustrating both the theoretical results of (Baxter & Bartlett, this volume) on a toy problem, and practical aspects of the algorithms on a number of more realistic problems.\n\n### Experiments with Infinite-Horizon, Policy-Gradient Estimation\n\nIn this paper, we present algorithms that perform gradient ascent of the average reward in a partially observable Markov decision process (POMDP). These algorithms are based on GPOMDP, an algorithm introduced in a companion paper (Baxter and Bartlett, this volume), which computes biased estimates of the performance gradient in POMDPs. The algorithm's chief advantages are that it uses only one free parameter beta, which has a natural interpretation in terms of bias-variance trade-off, it requires no knowledge of the underlying state, and it can be applied to infinite state, control and observation spaces. We show how the gradient estimates produced by GPOMDP can be used to perform gradient ascent, both with a traditional stochastic-gradient algorithm, and with an algorithm based on conjugate-gradients that utilizes gradient information to bracket maxima in line searches. Experimental results are presented illustrating both the theoretical results of (Baxter and Bartlett, this volume) on a toy problem, and practical aspects of the algorithms on a number of more realistic problems." ]

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[ "Explore BrainMass\nShare\n\n# Applications of Exponential and Logarithmic Functions\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\nPart 1: Using the Library, web resources, and/or other materials, find the logarithmic formula that gives the pH of a substance. State what each variable in your equation represents.\n\nFind the hydrogen ion content of a substance of your choice. Using this hydrogen ion content, show how to find the pH of the substance using the formula. Is this substance acidic or basic? Why?\nRound the pH to three decimal places.\n\nPart 2: Suppose that the number of cars, C, on 1st Avenue in a city over a period of time t, in months, is graphed on a rectangular coordinate system where time is on the horizontal axis. Suppose that the number of cars driven on 1st Avenue can be modeled by an exponential function, C= p * at where p is the number of cars on the road on the first day recorded. If you commuted to work each day along 1st Avenue, would you prefer that the value of be between 0 and 1 or larger than 1? Explain your reasoning. Be sure to reference your sources using APA style.\n\nTyping hint: Type formula above as C = p * a^t\n\nhttps://brainmass.com/math/basic-algebra/applications-of-exponential-and-logarithmic-functions-145595\n\n#### Solution Preview\n\nFrom the website\nhttp://www.elmhurst.edu/~chm/vchembook/184ph.html\nwe have the formula for pH:\npH = - log [H+]\nThe variable in this equation is [H+], which is the concentration of hydrogen ion in solution.\n\nSay we have in a 0.01 mol/L solution of HCl it is approximated that there is a concentration of 0.01 mol/L dissolved hydrogen ions. So its pH is:\npH = - log(0.01) ...\n\n#### Solution Summary\n\nApplications of Exponential and Logarithmic Functions are investigated.\n\n\\$2.19" ]

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[ "I’m trying to implement the Curveball optimizer in PyTorch 0.4.0, but I’m running into unexpected behavior from `autograd.grad`.\n\nBelow is a minimal working example that illustrates the problem I’ve encountered.\n\nThe algorithm requires that I compute the product of a vector and the Hessian of the loss function with respect to the network’s output, which means using higher order gradients if I want code that is agnostic about the loss function. To do this, I need to retain the gradient for the network output and then use calls like `HlJfTz = ag.grad(output.grad,output,JfTz,retain_graph=True)` to do the multiplication.\n\nThe first time I do the Hessian-vector product, it all works as expected. However, when I try to compute a second Hessian-vector product, it fails saying that `output.grad.requires_grad` is `False`.\n\nIt appears that `output.grad.requires_grad` changes from `True` to `False` after the statement `Gz = ag.grad(output,self.model.parameters(),HlJfTz,retain_graph=True)`.\n\nI’m wondering if this is intended behavior, and if anyone can suggest a workaround.\n\n``````import torch\nimport torch.nn as nn\n\nclass StochasticRosenbrock(nn.Module):\ndef __init__(self,u,v,lo=0.0,hi=1.0):\nsuper(StochasticRosenbrock, self).__init__()\nself.epsilon = torch.Tensor([1.0])\nself.lo = lo\nself.hi = hi\nreturn\n\ndef forward(self,sample=True):\nout = torch.zeros(2)\nu = self.w\nv = self.w\nif sample:\nself.epsilon.uniform_(self.lo,self.hi)\nout = 1.0 - u\nout = 10.0 * self.epsilon * (v - u**2)\nreturn out\n\nclass CurveBall(object):\ndef __init__(self,model,criterion,lambda_=1.0):\nself.model = model\nself.z = [torch.zeros_like(p) for p in model.parameters()]\nself.criterion = criterion\nself.lambda_ = lambda_\nreturn\n\ndef step(self):\ntarget = torch.zeros(2)\noutput = self.model()\nloss = self.criterion(output,target)\nloss.backward(create_graph=True)\nJ = [p.grad for p in self.model.parameters()]\n# this line mucks up the gradient tracking for output, and breaks the computation of HlJfTdeltaz\ndeltaz = [j.detach() + g + self.lambda_ * z for j, g, z in zip(J, Gz, self.z)]\n# cut here because this is where it breaks\n\ncnn = StochasticRosenbrock(-0.25,-0.25,1.0,1.0)\ncnn.train()\ncriterion = nn.MSELoss(size_average=False)\noptimizer = CurveBall(cnn,criterion)\nloss = optimizer.step()\n``````\n\nThanks!\n\nHi,\n\nI think the option you want to use is `create_graph` not `retain_graph`. Doc is here. I guess `output` is part of the graph you go through and so it messes things up as it’s not creating a higher order graph, just keeping the existing one." ]

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[ "# Calculating Weld Metal Requirements by Hand\n\n### When you don't have your Procedure Handbook Tables", null, "If you need to calculate weld metal requirements and you do not have The Procedure Handbook of Arc Welding", null, "handy you can do this by hand by using simple geometry. What you need to do is find the volume of weld needed in cubic inches. A fillet weld with a flat face forms a right triangle. Using high school math we can see that a right triangle with ¼” legs would have an area of ½ x base x height . Using our 10% allowance for overwelding our legs sizes would be ¼ x 1.1 = 0.275in. So our area = (1/2) x 0.275 x 0.275 = 0.0378 square inches. We have to weld a total of 76.5 inches per part (taken from the example on our previous post: Calculating Weight of Weld Metal Required) so the total weld volume for each part is (0.0378) x (76.5) = 2.89 cubic inches. Since we are welding carbon steel we use its density to calculate pounds of weld metal. The density of steel is 0.283 lb/cubic inch.\n\n2.89 cubic inches x 0.283 pounds/cubic inch = 0.819 pounds per part. We compare this to the values obtained from using Table 12.1 in the Procedure Handbook and we are dead on.\n\nDoing this for fillet welds is easy. Once you throw a more complicated joint like a double-V butt weld it gets a bit harder. Look at the difference on our notebook below:", null, "Calculating the volume of a groove weld gets complicated. We need to break it down into sections to do so.\n\nInstead of a simple triangle, we know take the cross section of the joint and divided into shapes for which we can calculate the area. In the notes above we see that we end up with 2 triangles, 2 circle segments, and 1 rectangle. For this type of joint, table 12-1 of the Procedure Handbook has what you need. However, when you start dealing with J-grooves and other complicated groove joints you may need to do it by hand. There are tables for these joints as well but it seems they are published in obscure books and technical reports.\n\nPlease note: I reserve the right to delete comments that are offensive or off-topic.\n\n## 9 thoughts on “Calculating Weld Metal Requirements by Hand”\n\n1.", null, "Matthew Lobão says:\n\nHi again, I’m with a doubt that is killing me.\nI have to calculate the weld metal in a “V” joint for a job. Only I can not understand how to calculate the weld metal by “The Procedure Handbook of Arc Welding, 14th Edition”. You have step by step calculation for this joint?\n\n•", null, "Matthew Lobão says:\n\nAn application with practical data as the fillet weld. Thank you!\n\n•", null, "The Welding Guys says:\n\nUsing the Procedure Handbook you will need the following information regarding your joint:\n1. Root opening (gap, if any, between the two pieces)\n2. Depth of penetration\n3. Height of top bead reinforcement\n4. Width of top bead reinforcement\n5. Included angle\n6. Radius of the groove (if not square)\n\nThen use Table 12-1 to get and add up all the components of the weld.\nCan you provide the joint details?\n\n•", null, "Matthew Lobao says:\n\nRoot opening = 3/32″\nRoot face = 3/32″\nDepth of penetration = 5/16″\nHeight of top bead reinforcement = 3/32″\nWidth of top bead reinforcement = 13/64″\nGroove Angle = 60°\nBevel Angle = 30°\nPlate thickness = 13/32″\n\nThank you for attention!\n\n•", null, "The Welding Guys says:\n\nMatthew, please see our response via email. Have a few questions regarding the details.\n\n•", null, "Willy Wiranata says:\n\nI really want to learn more about how to calculate the weld metal, can u give me a guide ? im just fresh beginner\nThanks a lot..\n\n•", null, "The Welding Guys says:\n\nWilly, what exactly would you like to calculate? There are tables that can help with calculating weld metal requirements for different types of joints. However, if you don’t have these tables you can always calculate the weld metal needed by calculating the volume that needs to be filled. To do this you need to know the weld joint type and size of weld. You may also need to brush up on geometry. If you have a specific weld in mind please provide the details and we would be happy to walk you through the calculations." ]

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[ "Learning Library\n\n# Multi-Step Word Problems? No Problem!\n\nDo you need extra help for EL students? Try the Writing Questions for Multi-Step Word Problems pre-lesson.\n\nNo standards associated with this content.\n\nWhich set of standards are you looking for?\n\nDo you need extra help for EL students? Try the Writing Questions for Multi-Step Word Problems pre-lesson.\n\nStudents will be able to solve multi-step words problems with mixed operations using a four-step plan.\n\nThe adjustment to the whole group lesson is a modification to differentiate for children who are English learners.\n(5 minutes)\n• Distribute whiteboards to each student.\n• Write the following word problem on the board: \"Nico baked 118 cupcakes. He kept 22 cupcakes at home to share with his family. He took the remaining cupcakes to school to share evenly with his 12 friends. How many cupcakes did each friend get?\"\n• Ask students to solve this word problem on their whiteboards.\n• Call on a student volunteer to share their answer and the steps they took to solve this word problem.\n• Tell students that this is an example of a multi-step word problem, which has more than one step in order to solve it.\n(10 minutes)\n• Explain to students that in order to solve multi-step word problems, they will follow four important steps.\n• Post a piece of chart paper on the board to be used as a reference throughout this lesson. Title this chart paper \"Steps for Solving Multi-Step Problems\" and write the four steps for solving multi-step word problems.\n• Write \"Step 1: Read and understand the word problem.\" Tell students it is important to read and re-read the word problem carefully.\n• Model for students how to read the word problem written on the board about Nico and his cupcakes.\n• Write \"Step 2: Underline the question and circle the important information.\"\n• Model for students how to underline the question being asked (e.g., How many cupcakes did each friend get?) and to circle key numbers (e.g., 118 cupcakes, 22 cupcakes, and 12 friends).\n• Write \"Step 3: Plan a solution and solve.\" Tell students to decide which mathematical operations to use and choose a problem solving strategy (e.g., draw a diagram).\n• Model for students how to first subtract (mathematical operation) 22 from 118 by writing a math equation (problem solving strategy) and then divide that difference (96) by 12 to get the answer (8 cupcakes).\n• Model for students how to check their answers by doing the opposite of whatever operations were used to solve the word problem.\n(15 minutes)\n• Show students a copy of Anna's Birthday: Word Problems on the document camera.\n• Tell students that this worksheet contains multi-step word problems with mixed operations.\n• Model for students how to use the four steps for solving multi-step word problems to complete the first word problem. Point to and reference the chart paper as you move through each step.\n• Distribute a copy of the Anna's Birthday: Word Problems worksheet to each student.\n• Assign students into effective partnerships and tell students to complete the remainder of their worksheet with their partners by following the four important steps.\n• Circulate and offer support.\n• Call students together as a whole class to go over the answers and answer any questions that may have arisen.\n(15 minutes)\n• Tell students that they will now practice this skill independently.\n• Preview and distribute the Word Problems: Flower Power worksheet.\n• Remind students to use the Steps for Solving Multi-Step Problems chart paper as a reference when they are completing this worksheet.\n• Circulate and offer support as students are working independently.\n\nSupport:\n\n• Allow students to use math manipulatives (e.g., counters) during Guided Practice and Independent Work Time.\n• Write out the Steps for Solving Multi-Step Problems four steps on individual pieces of paper to distribute to students to keep at their desks to use throughout the lesson.\n• Provide students with a partially completed copy of the Word Problems: Flower Power worksheet during Independent Work Time.\n• Gather students into a teacher-led small group to complete the Word Problems: Flower Power worksheet with support during Independent Work Time.\n• Allow students to practice with single-step word problems during Independent Work Time (see optional materials).\n\nEnrichment:\n\n• Challenge students by providing them with the Classroom Math: Division Work Problems worksheet, which challenges students to solve multi-step word problems with mixed operations including long division and requires them to use outside knowledge to find the answer (e.g., number of days in two weeks) (see optional materials).\n• Encourage students to create their own word problems worksheet to exchange with a partner. Challenge students to write multi-step word problems with mixed operations.\n(7 minutes)\n• Distribute a blank index card to each student to be used as an exit ticket.\n• Write the following word problem on the board: \"65 bags of marbles are to be divided evenly among 13 students. Each bag contains 15 marbles. How many marbles will each student receive?\"\n• Tell students to rewrite this word problem on their index cards and use the four steps for solving multi-step problems to solve this word problem.\n• Collect students' exit tickets.\n(3 minutes)\n• Tell students that they are going to finish this activity by discussing the three W's.\n• Instruct students to turn to a partner and reflect upon the following questions:\n• What did we learn today?\n• So what? Why was it important and/or useful?\n• Now what? How does this fit into what we are learning and what do you think we will learn next?\n\nCreate new collection\n\n0\n\n### New Collection>\n\n0 items\n\nWhat could we do to improve Education.com?" ]

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[ "# Mathematics of Computation\n\nPublished by the American Mathematical Society since 1960 (published as Mathematical Tables and other Aids to Computation 1943-1959), Mathematics of Computation is devoted to research articles of the highest quality in computational mathematics.\n\nISSN 1088-6842 (online) ISSN 0025-5718 (print)\n\nThe 2020 MCQ for Mathematics of Computation is 1.78.\n\nWhat is MCQ? The Mathematical Citation Quotient (MCQ) measures journal impact by looking at citations over a five-year period. Subscribers to MathSciNet may click through for more detailed information.\n\n## An analysis of a uniformly convergent finite difference/finite element scheme for a model singular-perturbation problemHTML articles powered by AMS MathViewer\n\nby Eugene C. Gartland\nMath. Comp. 51 (1988), 93-106 Request permission\n\n## Abstract:\n\nUniform $\\mathcal {O}({h^2})$ convergence is proved for the El-Mistikawy-Werle discretization of the problem $- \\varepsilon u”+ au’+ bu = f$ on (0,1), $u(0) = A$, $u(1) = B$, subject only to the conditions $a,b,f \\in {\\mathcal {W}^{2,\\infty }}[0,1]$ and $a(x) > 0, 0 \\leq x \\leq 1$. The principal tools used are a certain representation result for the solutions of such problems that is due to the author [Math. Comp., v. 48, 1987, pp. 551-564] and the general stability results of Niederdrenk and Yserentant [Numer. Math., v. 41, 1983, pp. 223-253]. Global uniform $\\mathcal {O}(h)$ convergence is proved under slightly weaker assumptions for an equivalent Petrov-Galerkin formulation.\nSimilar Articles\n• Retrieve articles in Mathematics of Computation with MSC: 65L10, 65L60\n• Retrieve articles in all journals with MSC: 65L10, 65L60" ]

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[ "# Document (#35724)\n\nAuthor\nStöber, T.\nTitle\nServiceangebote der wissenschaftlichen Bibliotheken im Bereich Literaturverwaltung : Ergebnisse einer Umfrage\nSource\nhttp://opus.bibliothek.uni-augsburg.de/volltexte/2010/1611/\nImprint\nAugsburg : Universitätsbilbiothek\nYear\n2010\nPages\n10 S\nAbstract\nDas Thema Literaturverwaltung hat in den letzten Jahren in der Softwareentwicklung, aber auch bei den wissenschaftlichen Bibliotheken einen enormen Aufschwung erfahren: kostenlose Softwarelizenzen sowie Schulungs- und Supportangebote gehören mittlerweile vielerorts zum Dienstleistungsspektrum der Bibliotheken. Die Verbreitung dieser bibliothekarischen Serviceangebote genauer zu benennen und zugleich in der Praxis bewährte Dienstleistungen zu identifizieren, war Ziel einer im Februar 2010 durchgeführten Online-Umfrage, deren Ergebnisse hier vorgestellt werden.\nTheme\nBibliographische Software\nLocation\nD\nArea\nWissenschaftliche Bibliotheken\nRSWK\nLiteraturverwaltung <Programm> , Bibliothek , Dienstleistungsangebot , Umfrage\n\n## Similar documents (content)\n\n1. Stöber, T.; Teichert, A.: Webbasierte Literaturverwaltung : neue Kooperationsformen und Anwendungsszenarien (2008) 0.16\n```0.15618862 = sum of:\n0.15618862 = product of:\n0.6507859 = sum of:\n0.029112307 = weight(abstract_txt:zugleich in 4565) [ClassicSimilarity], result of:\n0.029112307 = score(doc=4565,freq=1.0), product of:\n0.092983395 = queryWeight, product of:\n6.679284 = idf(docFreq=145, maxDocs=42740)\n0.013921162 = queryNorm\n0.31309146 = fieldWeight in 4565, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.679284 = idf(docFreq=145, maxDocs=42740)\n0.046875 = fieldNorm(doc=4565)\n0.038932502 = weight(abstract_txt:genauer in 4565) [ClassicSimilarity], result of:\n0.038932502 = score(doc=4565,freq=1.0), product of:\n0.11286585 = queryWeight, product of:\n1.1017386 = boost\n7.358825 = idf(docFreq=73, maxDocs=42740)\n0.013921162 = queryNorm\n0.34494492 = fieldWeight in 4565, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.358825 = idf(docFreq=73, maxDocs=42740)\n0.046875 = fieldNorm(doc=4565)\n0.011589562 = weight(abstract_txt:einer in 4565) [ClassicSimilarity], result of:\n0.011589562 = score(doc=4565,freq=1.0), product of:\n0.06339802 = queryWeight, product of:\n1.1677507 = boost\n3.8998692 = idf(docFreq=2351, maxDocs=42740)\n0.013921162 = queryNorm\n0.18280637 = fieldWeight in 4565, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.8998692 = idf(docFreq=2351, maxDocs=42740)\n0.046875 = fieldNorm(doc=4565)\n0.033309866 = weight(abstract_txt:wissenschaftlichen in 4565) [ClassicSimilarity], result of:\n0.033309866 = score(doc=4565,freq=1.0), product of:\n0.12815815 = queryWeight, product of:\n1.6602956 = boost\n5.544793 = idf(docFreq=453, maxDocs=42740)\n0.013921162 = queryNorm\n0.2599122 = fieldWeight in 4565, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.544793 = idf(docFreq=453, maxDocs=42740)\n0.046875 = fieldNorm(doc=4565)\n0.031160142 = weight(abstract_txt:bibliotheken in 4565) [ClassicSimilarity], result of:\n0.031160142 = score(doc=4565,freq=1.0), product of:\n0.14032263 = queryWeight, product of:\n2.1277559 = boost\n4.737295 = idf(docFreq=1017, maxDocs=42740)\n0.013921162 = queryNorm\n0.22206071 = fieldWeight in 4565, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.737295 = idf(docFreq=1017, maxDocs=42740)\n0.046875 = fieldNorm(doc=4565)\n0.5066815 = weight(abstract_txt:literaturverwaltung in 4565) [ClassicSimilarity], result of:\n0.5066815 = score(doc=4565,freq=5.0), product of:\n0.52669775 = queryWeight, product of:\n4.122291 = boost\n9.177984 = idf(docFreq=11, maxDocs=42740)\n0.013921162 = queryNorm\n0.9619967 = fieldWeight in 4565, product of:\n2.236068 = tf(freq=5.0), with freq of:\n5.0 = termFreq=5.0\n9.177984 = idf(docFreq=11, maxDocs=42740)\n0.046875 = fieldNorm(doc=4565)\n0.24 = coord(6/25)\n```\n2. Fuchs, C.; Pampel, H.; Vierkant, P.: ORCID in Deutschland : Ergebnisse einer Bestandsaufnahme im Jahr 2016 (2017) 0.14\n```0.13872439 = sum of:\n0.13872439 = product of:\n0.69362193 = sum of:\n0.08416281 = weight(abstract_txt:durchgeführten in 5858) [ClassicSimilarity], result of:\n0.08416281 = score(doc=5858,freq=1.0), product of:\n0.13423589 = queryWeight, product of:\n1.2015218 = boost\n8.025305 = idf(docFreq=37, maxDocs=42740)\n0.013921162 = queryNorm\n0.62697697 = fieldWeight in 5858, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.025305 = idf(docFreq=37, maxDocs=42740)\n0.078125 = fieldNorm(doc=5858)\n0.12950043 = weight(abstract_txt:vielerorts in 5858) [ClassicSimilarity], result of:\n0.12950043 = score(doc=5858,freq=1.0), product of:\n0.17891057 = queryWeight, product of:\n1.3871241 = boost\n9.264996 = idf(docFreq=10, maxDocs=42740)\n0.013921162 = queryNorm\n0.7238278 = fieldWeight in 5858, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.264996 = idf(docFreq=10, maxDocs=42740)\n0.078125 = fieldNorm(doc=5858)\n0.054356717 = weight(abstract_txt:ergebnisse in 5858) [ClassicSimilarity], result of:\n0.054356717 = score(doc=5858,freq=1.0), product of:\n0.12636709 = queryWeight, product of:\n1.6486531 = boost\n5.5059114 = idf(docFreq=471, maxDocs=42740)\n0.013921162 = queryNorm\n0.43014932 = fieldWeight in 5858, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.5059114 = idf(docFreq=471, maxDocs=42740)\n0.078125 = fieldNorm(doc=5858)\n0.0961573 = weight(abstract_txt:wissenschaftlichen in 5858) [ClassicSimilarity], result of:\n0.0961573 = score(doc=5858,freq=3.0), product of:\n0.12815815 = queryWeight, product of:\n1.6602956 = boost\n5.544793 = idf(docFreq=453, maxDocs=42740)\n0.013921162 = queryNorm\n0.75030184 = fieldWeight in 5858, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n5.544793 = idf(docFreq=453, maxDocs=42740)\n0.078125 = fieldNorm(doc=5858)\n0.32944465 = weight(abstract_txt:umfrage in 5858) [ClassicSimilarity], result of:\n0.32944465 = score(doc=5858,freq=2.0), product of:\n0.3816576 = queryWeight, product of:\n3.509093 = boost\n7.8127427 = idf(docFreq=46, maxDocs=42740)\n0.013921162 = queryNorm\n0.8631942 = fieldWeight in 5858, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n7.8127427 = idf(docFreq=46, maxDocs=42740)\n0.078125 = fieldNorm(doc=5858)\n0.2 = coord(5/25)\n```\n3. Franke, F.; Pfister, S.; Schüller-Zwierlein, A.: \"Hätten wir personelle Valenzen, würden wir uns um stärkere Nutzung bemühen.\" : Eine Umfrage zur Vermittlung von lnformationskompetenz an Schüler an den bayerischen wissenschaftlichen Bibliotheken (2007) 0.12\n```0.116299585 = sum of:\n0.116299585 = product of:\n0.5814979 = sum of:\n0.042491697 = weight(abstract_txt:gehören in 2764) [ClassicSimilarity], result of:\n0.042491697 = score(doc=2764,freq=1.0), product of:\n0.09876382 = queryWeight, product of:\n1.0306145 = boost\n6.883767 = idf(docFreq=118, maxDocs=42740)\n0.013921162 = queryNorm\n0.43023545 = fieldWeight in 2764, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.883767 = idf(docFreq=118, maxDocs=42740)\n0.0625 = fieldNorm(doc=2764)\n0.043485377 = weight(abstract_txt:ergebnisse in 2764) [ClassicSimilarity], result of:\n0.043485377 = score(doc=2764,freq=1.0), product of:\n0.12636709 = queryWeight, product of:\n1.6486531 = boost\n5.5059114 = idf(docFreq=471, maxDocs=42740)\n0.013921162 = queryNorm\n0.34411946 = fieldWeight in 2764, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.5059114 = idf(docFreq=471, maxDocs=42740)\n0.0625 = fieldNorm(doc=2764)\n0.06280968 = weight(abstract_txt:wissenschaftlichen in 2764) [ClassicSimilarity], result of:\n0.06280968 = score(doc=2764,freq=2.0), product of:\n0.12815815 = queryWeight, product of:\n1.6602956 = boost\n5.544793 = idf(docFreq=453, maxDocs=42740)\n0.013921162 = queryNorm\n0.4900951 = fieldWeight in 2764, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n5.544793 = idf(docFreq=453, maxDocs=42740)\n0.0625 = fieldNorm(doc=2764)\n0.10992265 = weight(abstract_txt:bibliotheken in 2764) [ClassicSimilarity], result of:\n0.10992265 = score(doc=2764,freq=7.0), product of:\n0.14032263 = queryWeight, product of:\n2.1277559 = boost\n4.737295 = idf(docFreq=1017, maxDocs=42740)\n0.013921162 = queryNorm\n0.78335655 = fieldWeight in 2764, product of:\n2.6457512 = tf(freq=7.0), with freq of:\n7.0 = termFreq=7.0\n4.737295 = idf(docFreq=1017, maxDocs=42740)\n0.0625 = fieldNorm(doc=2764)\n0.3227885 = weight(abstract_txt:umfrage in 2764) [ClassicSimilarity], result of:\n0.3227885 = score(doc=2764,freq=3.0), product of:\n0.3816576 = queryWeight, product of:\n3.509093 = boost\n7.8127427 = idf(docFreq=46, maxDocs=42740)\n0.013921162 = queryNorm\n0.8457542 = fieldWeight in 2764, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n7.8127427 = idf(docFreq=46, maxDocs=42740)\n0.0625 = fieldNorm(doc=2764)\n0.2 = coord(5/25)\n```\n4. Fischer, K.; Wiesenmüller, H.: ¬Der Einsatz der Personalgruppen in der Sacherschließung an wissenschaftlichen Bibliotheken : Ergebnisse einer Umfrage (2016) 0.12\n```0.11594105 = sum of:\n0.11594105 = product of:\n0.57970524 = sum of:\n0.027042309 = weight(abstract_txt:einer in 5129) [ClassicSimilarity], result of:\n0.027042309 = score(doc=5129,freq=1.0), product of:\n0.06339802 = queryWeight, product of:\n1.1677507 = boost\n3.8998692 = idf(docFreq=2351, maxDocs=42740)\n0.013921162 = queryNorm\n0.42654818 = fieldWeight in 5129, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.8998692 = idf(docFreq=2351, maxDocs=42740)\n0.109375 = fieldNorm(doc=5129)\n0.0760994 = weight(abstract_txt:ergebnisse in 5129) [ClassicSimilarity], result of:\n0.0760994 = score(doc=5129,freq=1.0), product of:\n0.12636709 = queryWeight, product of:\n1.6486531 = boost\n5.5059114 = idf(docFreq=471, maxDocs=42740)\n0.013921162 = queryNorm\n0.60220903 = fieldWeight in 5129, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.5059114 = idf(docFreq=471, maxDocs=42740)\n0.109375 = fieldNorm(doc=5129)\n0.07772302 = weight(abstract_txt:wissenschaftlichen in 5129) [ClassicSimilarity], result of:\n0.07772302 = score(doc=5129,freq=1.0), product of:\n0.12815815 = queryWeight, product of:\n1.6602956 = boost\n5.544793 = idf(docFreq=453, maxDocs=42740)\n0.013921162 = queryNorm\n0.60646176 = fieldWeight in 5129, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.544793 = idf(docFreq=453, maxDocs=42740)\n0.109375 = fieldNorm(doc=5129)\n0.07270699 = weight(abstract_txt:bibliotheken in 5129) [ClassicSimilarity], result of:\n0.07270699 = score(doc=5129,freq=1.0), product of:\n0.14032263 = queryWeight, product of:\n2.1277559 = boost\n4.737295 = idf(docFreq=1017, maxDocs=42740)\n0.013921162 = queryNorm\n0.5181416 = fieldWeight in 5129, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.737295 = idf(docFreq=1017, maxDocs=42740)\n0.109375 = fieldNorm(doc=5129)\n0.32613355 = weight(abstract_txt:umfrage in 5129) [ClassicSimilarity], result of:\n0.32613355 = score(doc=5129,freq=1.0), product of:\n0.3816576 = queryWeight, product of:\n3.509093 = boost\n7.8127427 = idf(docFreq=46, maxDocs=42740)\n0.013921162 = queryNorm\n0.8545187 = fieldWeight in 5129, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.8127427 = idf(docFreq=46, maxDocs=42740)\n0.109375 = fieldNorm(doc=5129)\n0.2 = coord(5/25)\n```\n5. Rieder, S.: ¬Das Datenbankprogramm CDS/ISIS und dessen Anwendung in der Bundesrepublik Deutschland : Ergebnisse einer Nutzerstudie (1995) 0.10\n```0.10436089 = sum of:\n0.10436089 = product of:\n0.52180445 = sum of:\n0.06131314 = weight(abstract_txt:verbreitung in 1485) [ClassicSimilarity], result of:\n0.06131314 = score(doc=1485,freq=1.0), product of:\n0.0962432 = queryWeight, product of:\n1.017378 = boost\n6.7953563 = idf(docFreq=129, maxDocs=42740)\n0.013921162 = queryNorm\n0.63706464 = fieldWeight in 1485, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.7953563 = idf(docFreq=129, maxDocs=42740)\n0.09375 = fieldNorm(doc=1485)\n0.09254106 = weight(abstract_txt:programm in 1485) [ClassicSimilarity], result of:\n0.09254106 = score(doc=1485,freq=2.0), product of:\n0.10051109 = queryWeight, product of:\n1.039691 = boost\n6.9443917 = idf(docFreq=111, maxDocs=42740)\n0.013921162 = queryNorm\n0.92070496 = fieldWeight in 1485, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n6.9443917 = idf(docFreq=111, maxDocs=42740)\n0.09375 = fieldNorm(doc=1485)\n0.023179123 = weight(abstract_txt:einer in 1485) [ClassicSimilarity], result of:\n0.023179123 = score(doc=1485,freq=1.0), product of:\n0.06339802 = queryWeight, product of:\n1.1677507 = boost\n3.8998692 = idf(docFreq=2351, maxDocs=42740)\n0.013921162 = queryNorm\n0.36561275 = fieldWeight in 1485, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.8998692 = idf(docFreq=2351, maxDocs=42740)\n0.09375 = fieldNorm(doc=1485)\n0.06522807 = weight(abstract_txt:ergebnisse in 1485) [ClassicSimilarity], result of:\n0.06522807 = score(doc=1485,freq=1.0), product of:\n0.12636709 = queryWeight, product of:\n1.6486531 = boost\n5.5059114 = idf(docFreq=471, maxDocs=42740)\n0.013921162 = queryNorm\n0.5161792 = fieldWeight in 1485, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.5059114 = idf(docFreq=471, maxDocs=42740)\n0.09375 = fieldNorm(doc=1485)\n0.27954307 = weight(abstract_txt:umfrage in 1485) [ClassicSimilarity], result of:\n0.27954307 = score(doc=1485,freq=1.0), product of:\n0.3816576 = queryWeight, product of:\n3.509093 = boost\n7.8127427 = idf(docFreq=46, maxDocs=42740)\n0.013921162 = queryNorm\n0.73244464 = fieldWeight in 1485, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.8127427 = idf(docFreq=46, maxDocs=42740)\n0.09375 = fieldNorm(doc=1485)\n0.2 = coord(5/25)\n```" ]

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[ "# Displaying the time parameter using Timer interrupt\n\n1. ## Displaying the time parameter using Timer interrupt\n\nI have PIC16F886 4mhz crystal 1ms timer interrupt.I have code working as below. Now currently facing updating ms and Sec parameter.\nthe min and sec parameter will update properly but not in msec and sec\n\nCode:\n```void interrupt isr(void) {\n\nasm(\"clrwdt\");\n\nif (TMR1IF) {\nTMR1IF = 0;\n// TMR1H = 0x3C;\n// TMR1L = 0xB0;100ms\n\nTMR1H = 0xFE;\nTMR1L = 0x0C;// timer interrupt for 1ms\nif(Top_Display>Bottom_Value)\n{\nif(LED_SS==0)\n{\nBottom_Value=(Sec*100)+MSec;\n}else\nif(LED_MS==0)\n{\nBottom_Value=(Min*100)+Sec;\n}else\nif(LED_HM==0)\n{\nBottom_Value=(Hour*100)+Min;\n}\n\n}else if(Top_Display==Bottom_Value)\n{\n\n}\n\nMSec++;\nif(MSec>1000)\n{\nMSec=0;\nSec++;\nif(Sec>=59) {\nSec=0;\nMin=Min+1;\nif(Min>=59) {\nMin=0;\nHour=Hour+1;\nif(Hour>99) {\nHour=0;\n}\n}\n}\n}\nDisplay();\n}\n\n}\n\nvoid Process_RUN_MODE()\n{\n//Display_Faults(12,18,27,26,18,11,24,26);\n\n{\n}\nTop_Display=1234;\nLED_MS=0;\nKey_CHK_RESET();\n\nif(LED_MS==0)\n{\nDissect(Top_Display);\nFloat_Value=SplitFloat(Top_Display);\nInG_value=SplitInt(Top_Display);\n\nDissect_2(Bottom_Value);\n}\n\n}\nVoid main()\n{\n\nwhile(1)\n{\nProcess_RUN_MODE();\n}\n\n}\n\nvoid Dissect_2(unsigned int Value) // Spliting of process value in digits form\n{\nunsigned char a,Temp;\nfor(a = 4;a >= 1 ; a--)\n{\n\nTemp = Value%10;\nValue = Value/10;\nLEDBuffer_1[a-1] = DISPTABLE[Temp];\n\n}\n\n}\n\nvoid Dissect(unsigned int Value) // Spliting of process value in digits form\n{\nunsigned char a,Temp;\n\nfor(a = 8;a >= 5 ; a--)\n{\nTemp = Value%10;\nValue = Value/10;\nLEDBuffer_1[a-1] = DISPTABLE[Temp];\n\n}\n\n}```", null, "", null, "Reply With Quote\n\n•\n\n2. ## Re: Displaying the time parameter using Timer interrupt\n\nI think I managed to unscramble it as:\nCode:\n```void interrupt isr(void)\n{\nasm(\"clrwdt\");\n\nif (TMR1IF)\n{\nTMR1IF = 0;\n// TMR1H = 0x3C;\n// TMR1L = 0xB0;100ms\n\nTMR1H = 0xFE;\nTMR1L = 0x0C;// timer interrupt for 1ms\nif(Top_Display>Bottom_Value)\n{\nif(LED_SS==0)\n{\nBottom_Value=(Sec*100)+MSec;\n}\nelse\nif(LED_MS==0)\n{\nBottom_Value=(Min*100)+Sec;\n}\nelse\nif(LED_HM==0)\n{\nBottom_Value=(Hour*100)+Min;\n}\n}\nelse if(Top_Display==Bottom_Value)\n{\n\n}\n\nMSec++;\nif(MSec>1000)\n{\nMSec=0;\nSec++;\nif(Sec>=59)\n{\nSec=0;\nMin=Min+1;\nif(Min>=59)\n{\nMin=0;\nHour=Hour+1;\nif(Hour>99)\n{\nHour=0;\n}\n}\n}\n}\nDisplay();\n}\n}\n\nvoid Process_RUN_MODE()\n{\n//Display_Faults(12,18,27,26,18,11,24,26);\n\n{\n}\nTop_Display=1234;\nLED_MS=0;\nKey_CHK_RESET();\n\nif(LED_MS==0)\n{\nDissect(Top_Display);\nFloat_Value=SplitFloat(Top_Display);\nInG_value=SplitInt(Top_Display);\n\nDissect_2(Bottom_Value);\n}\n\n}\n\nVoid main()\n{\nwhile(1)\n{\nProcess_RUN_MODE();\n}\n\n}\n\nvoid Dissect_2(unsigned int Value) // Spliting of process value in digits form\n{\nunsigned char a,Temp;\nfor(a = 4;a >= 1 ; a--)\n{\nTemp = Value%10;\nValue = Value/10;\nLEDBuffer_1[a-1] = DISPTABLE[Temp];\n}\n}\n\nvoid Dissect(unsigned int Value) // Spliting of process value in digits form\n{\nunsigned char a,Temp;\n\nfor(a = 8;a >= 5 ; a--)\n{\nTemp = Value%10;\nValue = Value/10;\nLEDBuffer_1[a-1] = DISPTABLE[Temp];\n}\n}```\nIf I got it right, your problem is almost certainly because the time taken inside your ISR is longer than the time between intervals between interrupts. As this is a timer of some kind, it would make more sense to ensure it kept time accurately by making the ISR much shorter. i suggest all you do inside the ISR is:\n1. clear the interrupt flag\n3. set a flag to say 100mS has elapsed\n4. exit the ISR\n\nThen in your main loop, poll the flag, do the timing routine and reset the flag.\n\nBrian.", null, "", null, "Reply With Quote\n\n•\n\n3. ## Re: Displaying the time parameter using Timer interrupt\n\nI have actually 2 issue in code.\n\n1) variable declaration. I have declare variable\nunsigned int Top_Display=0;\nif assign Top_Display=0123 display shows 0083\nif assign Top_Display=123 display shows 0123\nSome of the variable it take properly and some are not. if i assign value Top_Display=0129 it gives error digit is out of range.\n\n2)i have created 1ms timer interrupt . every 1ms i am increment msec . i am getting wrong calculation for displaying Msec to sec. other paramter it showing well. somewhere the conversion is affected\n\n1S=1000ms\n60S=1Min\n60M=1H\n\nCode:\n```void Process_RUN_MODE() {\n\nTop_Display=0123;\n\nLED_MS=0;\n\nKey_CHK_RESET();\n\nif(LED_MS==0) {\nLED_SS=1;\nLED_HM=1;\nDissect(Top_Display);\nDissect_2(Bottom_Value);\n\n} else if(LED_HM==0) {\nLED_MS=1;\nLED_SS=1;\nDissect(Top_Display);\nDissect_2(Bottom_Value);\n} else if(LED_SS==0) {\nLED_MS=1;\nLED_HM=1;\nDissect(Top_Display);\nDissect_2(Bottom_Value);\n}\n\n}\n\nvoid interrupt isr(void) {\n\nasm(\"clrwdt\");\n\nif (TMR1IF) {\nTMR1IF = 0;\n//\tTMR1H = 0x3C;\n// TMR1L = 0xB0;100ms\n\nTMR1H = 0xFE;\nTMR1L = 0x0C;// timer interrupt for 1ms\nif(Top_Display>Bottom_Value) {\nif(LED_SS==0) {\nBottom_Value=(Sec*100)+MSec;\n} else if(LED_MS==0) {\nBottom_Value=(Min*100)+Sec;\n} else if(LED_HM==0) {\nBottom_Value=(Hour*100)+Min;\n}\n\n} else if(Top_Display==Bottom_Value) {\n\n}\n\nMSec++;\nif(MSec>1000) {\nMSec=0;\nSec=Sec+1;\n}\nif(Sec>=59) {\nSec=0;\nMin=Min+1;\n}\nif(Min>=59) {\nMin=0;\nHour=Hour+1;\n}\nif(Hour>99) {\nHour=0;\n}\n\nDisplay();\n}\n\n}```", null, "", null, "Reply With Quote\n\n•\n\n4. ## Re: Displaying the time parameter using Timer interrupt\n\nIn C code, a number beginning with \"0\" is an octal number. Octal 0123 = Binary 1010011 = Hex 53 = Decimal 83\n\n0129 is illegal because there is no digit '9' in octal base.\n\nWrong:\nCode:\n```if(MSec>1000)\nif(Sec>=59) {\nif(Min>=59) {```\nCorrect:\nCode:\n```if(MSec >= 1000)\nif(Sec >= 60) {\nif(Min >= 60) {```", null, "", null, "Reply With Quote\n\n5. ## Re: Displaying the time parameter using Timer interrupt\n\nAdditionally, please do what I suggested in the ISR. Reduce it to:\nCode:\n```void interrupt isr(void)\n{\nif (TMR1IF)\n{\nTMR1IF = 0;\nTMR1H = 0xFE;\nTMR1L = 0x0C;// timer interrupt for 1ms\nElapsed100mS = 1;\n}\n}```\nCode:\n```if(Elapsed100mS)\n{\n[clear the WDT]", null, "", null, "Reply With Quote" ]

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[ "# Registration\n\nDear SAP Community Member,\nIn order to fully benefit from what the SAP Community has to offer, please register at:\nhttp://scn.sap.com\nThank you,\nThe SAP Community team.\n\n# Code Gallery One more ABAP to JSON Serializer and Deserializer Dynamic Data Accessor Helper Class for ABAP\n\nSkip to end of metadata\nGo to start of metadata\n\nAuthor: Alexey Arseniev\nSubmitted: 12.05.2017\nOther code samples from me:\n\n## Why\n\nSometimes you need to access ABAP data objects dynamically. For example, when:\n\n• you do not know the structure of the ABAP object and to dynamically access value of the field forced to use ASSIGN .. COMPONENT with field symbol\n• you do a cross-release development in ABAP and do not know if some field exists, so need dynamic access to data\n• you access data from optional ABAP component and also need dynamic access to ABAP structure field or internal table row\n• you need a key or index access to a dynamic internal table\n\nNormally, if you need to access ABAP fields/structure component dynamically, you will end up in coding like this:\n\nAccessing Data Objects Dynamically (as suggested in documentation)\n```FIELD-SYMBOLS: <table> TYPE ANY TABLE,\n<line> TYPE ANY,\n<field> TYPE ANY.\n\nASSIGN COMPONENT `FIELD1` OF STRUCTURE <line> TO <field>.\nIF <field> IS ASSIGNED.\nWRITE: <field>.\nENDIF.```\n\nThat may be OK (however still not very convenient), but if you have a dynamic ABAP object with deeper structure, it becomes boring:\n\nAccessing Deeply Nested Data Objects Dynamically\n```FIELD-SYMBOLS:\n<table> TYPE ANY TABLE,\n<line> TYPE ANY,\n<field1> TYPE ANY,\n<field2> TYPE ANY,\n<field3> TYPE ANY,\n<field4> TYPE ANY.\n\nASSIGN COMPONENT `FIELD1` OF STRUCTURE <line> TO <field1>.\nIF <field> IS ASSIGNED.\nASSIGN COMPONENT `CHILD_1` OF STRUCTURE <fielld1> TO <field2>.\nIF <field2> IS ASSIGNED.\nASSIGN COMPONENT `CHILD_11` OF STRUCTURE <fielld2> TO <field3>.\nIF <field3> IS ASSIGNED.\nASSIGN COMPONENT `CHILD_111` OF STRUCTURE <fielld3> TO <field4>.\nIF <field4> IS ASSIGNED.\nWRITE: <field4>.\nENDIF.\nENDIF.\nENDIF.\nENDIF.```\n\nAnd if one thinks about accessing dynamically elements from nested tables it would be at all - hell.\n\nSo, below one can find a helper class, which may help in such cases, in a lean and tasty way.", null, "An original and actual version of the source can be found in class /UI2/CL_DATA_ACCESS delivered with UI2 Add-on (can be applied to SAP_BASIS 700 – 76X). So, you can use this ABAP JSON parser in your standard code mostly on any system. Delivered with note 2526405\n\n## What it can\n\nThe accessor is a single class, with the following features:\n\n• the traversing of any data object (classes are not yet supported) passed as a reference to data or as a data.\n• traversing nested objects without any level limitations\n• automatically resolving reference variables (REF TO fields accessed the same way as standard fields)\n• a method like or XPath-like ways of accessing data\n• read/modification access to elementary types\n• accessing table rows using index or key\n\n## Limitations\n\nThe code of dynamic data accessor class does not pretend to be a complete and fully robust solution, but it may become like this if requests will come", null, "Current limitations are as following:\n\n• only ABAP data structures are supported. Traversing of ABAP objects/classes not yet supported, however possible\n• dynamic modification of tables (not data inside) is not supported. E.g you can not add/remove rows with API, but you can do it via reference to the table\n• the syntax for key access of the rows in dynamic tables does not allow usage of the symbol \",\" as part of the query. No escaping is supported.\n\n## Usage\n\nNecessary declarations:\n\nType and data declarations for example below\n```TYPES:\nBEGIN OF t_properties,\nenabled TYPE abap_bool,\nlength TYPE i,\ndescription TYPE c LENGTH 20,\nEND OF t_properties,\nBEGIN OF t_data,\nid TYPE i,\nproperties TYPE t_properties,\ncontent TYPE string,\nEND OF t_data,\nBEGIN OF t_name_value,\nname TYPE string,\nvalue TYPE string,\nEND OF t_name_value,\nBEGIN OF t_example,\nflag TYPE abap_bool,\nprops TYPE t_data,\nparams TYPE SORTED TABLE OF t_name_value WITH UNIQUE KEY name,\nEND OF t_example.\n\nDATA: ls_data TYPE t_example,\nlr_data TYPE REF TO data,\nlr_ref TYPE REF TO data,\nlv_int TYPE i,\nlo_data TYPE REF TO /ui2/cl_data_access.\n\nFIELD-SYMBOLS: <data> TYPE data.\n\nls_data-props-id = 12345.\nls_data-props-content = `Some Content`.\nls_data-props-properties-enabled = abap_false.\nls_data-props-properties-length = 10.\nls_data-props-properties-description = `My description`.\n\nGET REFERENCE OF ls_data INTO lr_data.```\n\nSimplest usage example:\n\nTraversing ABAP data dynamically\n```CREATE OBJECT lo_data EXPORTING ir_data = lr_data.\n\n\" standard way (does not work on SAP_BASIS 700)\nlr_ref = lo_data->at(`PROPS`)->at(`id`)->ref( ).\nIF lr_ref IS BOUND.\nASSIGN lr_ref->* TO <data>.\nWRITE: <data>.\nENDIF.\n\" XPath like\nlr_ref = lo_data->at(`PROPS-ID`)->ref( ).\nIF lr_ref IS BOUND.\nASSIGN lr_ref->* TO <data>.\nWRITE: <data>.\nENDIF.\n\n\" using helper method for creation\nlr_ref = /ui2/cl_data_access=>create( ir_data = lr_data iv_component = `PROPS-ID`)->ref( ).\nIF lr_ref IS BOUND.\nASSIGN lr_ref->* TO <data>.\nWRITE: <data>.\nENDIF.\n\n\" reading value directly\nlo_data->at(`props-properties-length`)->value( IMPORTING ev_data = lv_int ).\nWRITE: lv_int.\n```\n\nIf you want to use modification operations, you can only operate with references when creating an accessor object.\n\nDynamic data modification in ABAP in nice way\n```\" modifing value\nlr_ref = lo_data->at(`props-properties-length`)->ref( ).\nASSIGN lr_ref->* TO <data>.\n<data> = 25.\n\n\" or even more simple\nlo_data->at(`props-properties-length`)->set( 15 ).```\n\nThe code is robust - accessing of a not existing component of any level will not result in a crash but will return an empty reference or initial value.\n\nAccessing not existing component\n```\" reading not existing value returns initial value\nlo_data->at(`props-properties-not_exist-length-not-exist`)->value( IMPORTING ev_data = lv_int ).\nWRITE: lv_int. \" -> 0```\n\nWorking with tables:\n\nDynamic Access in ABAP in Easy way\n```DATA:\nls_data TYPE t_example,\nls_line LIKE LINE OF ls_data-params,\nlv_value TYPE string,\nlo_data TYPE REF TO /ui2/cl_data_access.\n\nls_line-name = `KEY1`.\nls_line-value = `Value1`.\nINSERT ls_line INTO TABLE ls_data-params.\n\nls_line-name = `KEY2`.\nls_line-value = `Value2`.\nINSERT ls_line INTO TABLE ls_data-params.\n\n/ui2/cl_data_access=>create( iv_data = ls_data iv_component = `params-name`)->value( IMPORTING ev_data = lv_value ).\nWRITE: lv_value.\n\n/ui2/cl_data_access=>create( iv_data = ls_data iv_component = `params[name=KEY1]-value`)->value( IMPORTING ev_data = lv_value ).\nWRITE: lv_value.\n\n/ui2/cl_data_access=>create( iv_data = ls_data iv_component = `params`)->at(`[name=KEY1]-value`)->value( IMPORTING ev_data = lv_value ).\nWRITE: lv_value.\n\n/ui2/cl_data_access=>create( iv_data = ls_data iv_component = `params[name=KEY1, value=Value1]-value`)->value( IMPORTING ev_data = lv_value ).\nWRITE: lv_value.\n```\n\n## CREATE  - Static Method Public Helper method for creating an instance of dynamic accessor\n\n• > IR_DATA (ref to data)Importing Type Ref To DATA Reference to data (allows modification of embedded data)\n• > IV_DATA (data) - any data (modification of embedded data not allowed)\n• > IV_COMPONENT (string) - Sub-component name (XPath-like syntax is supported)\n• < RO_REF (ref to /ui2/cl_data_access) - Reference to accessor object pointing to subcomponent\n\nCONSTRUCTOR   - Instance Method Public Constructor\n\n• > IR_DATA (ref to data)Importing Type Ref To DATA Reference to data (allows modification of embedded data)\n• > IV_DATA (data) - any data (modification of embedded data not allowed)\n\nAT  - Instance Method Public Component accessor\n\n• > IV_COMPONENT (string) - Sub-component name (XPath-like syntax is supported)\n• < RO_REF (ref to /ui2/cl_data_access) - Reference to accessor object pointing to subcomponent\n\nEMPTY  - Instance Method Public Returns TRUE if the embedded object is initial (not bound)\n\n• < RV_VAL (boolean) - ABAP_TRUE if the object is initial and data is not bound\n\nREF  - Instance Method Public Returns a reference to the embedded object\n\n• < RV_DATA (ref to data) - Reference to embedded data\n\nVALUE  - Instance Method Public Returns copy of the value\n\n• < EV_DATA (data) - Copy of the embedded data value, or initial if data is not bound\n\nSET  - Instance Method Public Sets the value of the embedded object, if not initial\n\n• > IV_DATA (data)  - New value for embedded object\n• < RV_SUCCESS (boolean) - ABAP_TRUE, if data was successfully modified\n\n## XPath-like dynamic data access\n\nTo access nested components you can use a nice, object-oriented way, using nested calls of AT method (it is robust, and would not crash accessing not existing components), as\n\n`lo_object->at('subcomp1')->at('subcomp11')->...`\n\nBut if you prefer more compact form, or run code on SAP_BASIS < 702 (nested method calls are not supported), you may use XPath-like syntax for accessing components. Like this:\n\n`lo_object->at('subcomp1-subcomp11-subcomp111')`\n\nor like this\n\n`lo_object->at('subcomp1->subcomp11->subcomp111')`\n\n### The syntax:\n\n• You can use any symbol (or combinations of symbols) as a component separator, except \"[\", \"]\", \"=\", \",\". Recommended separator symbol is \"-\".\n• For dynamic index access of rows in nested tables use \"[index]\" after the component name. E.g. \"table_name\". The indexing starts at 1. If the accessor object references table data object directly, you may skip the component name. E.g. \"\". You may continue accessing components after index access, e.g. : \"table_a-struct-table_b\". If you do out of range access, you got an empty reference back. Index access works only with index tables (STANDARD, SORTED).\n• For dynamic key access of rows in nested tables use \"(key=value)\" for single key lookup, \"(key1=value1, key2=value2)\" for multi-key lookup, \"(value)\" for table line lookup. You can NOT search for values containing \",\". You can use nested lookups: \"table_a(key1=value1)-table_b(key2=value2, key3=value3)\". Values used in a query shall be assignable to field structures.\n\n## The Code\n\nThe actual code for the helper:\n\n/ui2/cl_data_access - Dynamic Data Access For ABAP source code\n```*----------------------------------------------------------------------*\n* CLASS /ui2/cl_data_access DEFINITION\n*----------------------------------------------------------------------*\n* Documentation can be found on SCN:\n\" https://wiki.scn.sap.com/wiki/display/Snippets/Dynamic+Data+Accessor+Helper+Class+for+ABAP\n*----------------------------------------------------------------------*\nCLASS /ui2/cl_data_access DEFINITION.\n\nPUBLIC SECTION.\n\nCLASS-METHODS create\nIMPORTING\nir_data TYPE REF TO data OPTIONAL\niv_data TYPE data OPTIONAL\niv_component TYPE string OPTIONAL\nRETURNING\nVALUE(ro_ref) TYPE REF TO /ui2/cl_data_access .\nMETHODS constructor\nIMPORTING\nir_data TYPE REF TO data OPTIONAL\niv_data TYPE data OPTIONAL .\nMETHODS at\nIMPORTING\niv_component TYPE string OPTIONAL\nRETURNING\nVALUE(ro_ref) TYPE REF TO /ui2/cl_data_access .\nMETHODS empty\nRETURNING\nVALUE(rv_val) TYPE abap_bool .\nMETHODS ref\nRETURNING\nVALUE(rv_data) TYPE REF TO data .\nMETHODS value\nEXPORTING\nev_data TYPE data .\nMETHODS set\nIMPORTING\niv_data TYPE data\nRETURNING\nVALUE(rv_success) TYPE abap_bool .\nPROTECTED SECTION.\n\nDATA mr_data TYPE REF TO data .\n\nCLASS-METHODS deref\nIMPORTING\nir_data TYPE REF TO data\nRETURNING\nVALUE(rr_data) TYPE REF TO data .\nMETHODS at_int\nIMPORTING\niv_component TYPE string OPTIONAL\niv_index TYPE i OPTIONAL\niv_keys TYPE string OPTIONAL\nRETURNING\nVALUE(ro_ref) TYPE REF TO /ui2/cl_data_access .\nENDCLASS.\n\nCLASS /ui2/cl_data_access IMPLEMENTATION.\n\nMETHOD at.\nDATA:\nlv_component TYPE string,\nlv_sindex TYPE string,\nlv_keys TYPE string,\nlv_index TYPE i,\nlt_hier TYPE match_result_tab.\n\nFIELD-SYMBOLS:\n<component> LIKE LINE OF lt_hier,\n<sub_match> TYPE LINE OF submatch_result_tab.\n\nFIND ALL OCCURRENCES OF REGEX `(\\w+)?(?:\\[(?:(\\d+)|([^\\]]+))\\])?` IN iv_component RESULTS lt_hier.\n\nro_ref = me.\n\nLOOP AT lt_hier ASSIGNING <component> WHERE length IS NOT INITIAL.\nCHECK ro_ref->empty( ) EQ abap_false.\nREAD TABLE <component>-submatches INDEX 1 ASSIGNING <sub_match>.\nIF <sub_match>-length IS INITIAL.\nCLEAR lv_component.\nELSE.\nlv_component = iv_component+<sub_match>-offset(<sub_match>-length).\nTRANSLATE lv_component TO UPPER CASE.\nENDIF.\nREAD TABLE <component>-submatches INDEX 2 ASSIGNING <sub_match>.\nIF <sub_match>-length IS INITIAL.\nCLEAR lv_index.\nELSE.\nlv_index = lv_sindex = iv_component+<sub_match>-offset(<sub_match>-length).\nENDIF.\nREAD TABLE <component>-submatches INDEX 3 ASSIGNING <sub_match>.\nIF <sub_match>-length IS INITIAL.\nCLEAR lv_keys.\nELSE.\nlv_keys = iv_component+<sub_match>-offset(<sub_match>-length).\nENDIF.\nro_ref = ro_ref->at_int( iv_component = lv_component iv_index = lv_index iv_keys = lv_keys ).\nENDLOOP.\nENDMETHOD. \"at_int\n\nMETHOD at_int.\n\nDATA: lv_key TYPE string,\nlv_value TYPE string,\nlt_keys TYPE match_result_tab,\nlr_data TYPE REF TO data,\nlo_type TYPE REF TO cl_abap_typedescr.\n\nFIELD-SYMBOLS: <data> TYPE data,\n<comp> TYPE data,\n<key> LIKE LINE OF lt_keys,\n<sub_match> TYPE LINE OF submatch_result_tab,\n<ref> TYPE REF TO data,\n<table> TYPE ANY TABLE,\n<idx_table> TYPE INDEX TABLE.\n\nIF mr_data IS BOUND.\nIF iv_component IS NOT INITIAL.\nASSIGN mr_data->* TO <data>.\nASSIGN COMPONENT iv_component OF STRUCTURE <data> TO <comp>.\nIF <comp> IS ASSIGNED.\nGET REFERENCE OF <comp> INTO lr_data.\nlr_data = deref( lr_data ).\nASSIGN lr_data TO <ref>.\nENDIF.\nELSE.\nASSIGN mr_data TO <ref>.\nENDIF.\nENDIF.\n\nIF <ref> IS ASSIGNED AND ( iv_index IS NOT INITIAL OR iv_keys IS NOT INITIAL ).\nlo_type = cl_abap_typedescr=>describe_by_data_ref( <ref> ).\nIF lo_type->kind EQ cl_abap_typedescr=>kind_table.\n\" check for table index access\nIF iv_index IS NOT INITIAL.\nASSIGN <ref>->* TO <idx_table>.\nIF sy-subrc IS INITIAL.\nREAD TABLE <idx_table> INDEX iv_index REFERENCE INTO <ref>.\nIF sy-subrc IS NOT INITIAL.\nUNASSIGN <ref>.\nENDIF.\nELSE.\nUNASSIGN <ref>.\nENDIF.\nELSEIF iv_keys IS NOT INITIAL.\nASSIGN <ref>->* TO <table>.\nIF sy-subrc IS INITIAL.\nCREATE DATA lr_data LIKE LINE OF <table>.\nASSIGN lr_data->* TO <data>.\nFIND ALL OCCURRENCES OF REGEX `(\\w+)\\s*=\\s*([^,]*),?` IN iv_keys RESULTS lt_keys.\nIF sy-subrc IS INITIAL.\nLOOP AT lt_keys ASSIGNING <key>.\nREAD TABLE <key>-submatches INDEX 1 ASSIGNING <sub_match>.\nlv_key = iv_keys+<sub_match>-offset(<sub_match>-length).\nTRANSLATE lv_key TO UPPER CASE.\nREAD TABLE <key>-submatches INDEX 2 ASSIGNING <sub_match>.\nlv_value = iv_keys+<sub_match>-offset(<sub_match>-length).\nASSIGN COMPONENT lv_key OF STRUCTURE <data> TO <comp>.\nCHECK sy-subrc IS INITIAL.\n<comp> = lv_value.\nENDLOOP.\nELSE.\n<data> = lv_key.\nENDIF.\nREAD TABLE <table> FROM <data> REFERENCE INTO <ref>.\nIF sy-subrc IS NOT INITIAL.\nUNASSIGN <ref>.\nENDIF.\nELSE.\nUNASSIGN <ref>.\nENDIF.\nENDIF.\nENDIF.\nENDIF.\n\nIF <ref> IS ASSIGNED.\nCREATE OBJECT ro_ref\nEXPORTING\nir_data = <ref>.\nELSE.\nCREATE OBJECT ro_ref.\nENDIF.\nENDMETHOD. \"at_int\n\nMETHOD constructor.\nIF ir_data IS NOT INITIAL.\nmr_data = ir_data.\nELSEIF iv_data IS SUPPLIED.\nGET REFERENCE OF iv_data INTO mr_data.\nENDIF.\nmr_data = deref( mr_data ).\nENDMETHOD. \"constructor\n\nMETHOD create.\nIF iv_data IS SUPPLIED.\nCREATE OBJECT ro_ref\nEXPORTING\niv_data = iv_data.\nELSE.\nCREATE OBJECT ro_ref\nEXPORTING\nir_data = ir_data.\nENDIF.\n\nIF iv_component IS NOT INITIAL.\nro_ref = ro_ref->at( iv_component ).\nENDIF.\nENDMETHOD. \"create\n\nMETHOD deref.\n\nDATA: lo_type TYPE REF TO cl_abap_typedescr.\n\nFIELD-SYMBOLS: <data> TYPE data.\n\nrr_data = ir_data.\nIF rr_data IS NOT INITIAL.\nlo_type = cl_abap_typedescr=>describe_by_data_ref( ir_data ).\nIF lo_type->kind EQ cl_abap_typedescr=>kind_ref.\nASSIGN ir_data->* TO <data>.\nrr_data = deref( <data> ).\nENDIF.\nENDIF.\n\nENDMETHOD. \"deref\n\nMETHOD empty.\nIF mr_data IS INITIAL.\nrv_val = abap_true.\nENDIF.\nENDMETHOD. \"empty\n\nMETHOD ref.\nrv_data = mr_data.\nENDMETHOD. \"ref\n\nMETHOD set.\n\nFIELD-SYMBOLS: <data> TYPE data.\n\nIF mr_data IS BOUND.\nASSIGN mr_data->* TO <data>.\n<data> = iv_data.\nrv_success = abap_true.\nENDIF.\n\nENDMETHOD.\n\nMETHOD value.\n\nDATA: lo_type_out TYPE REF TO cl_abap_typedescr,\nlo_type_in TYPE REF TO cl_abap_typedescr.\n\nFIELD-SYMBOLS: <data> TYPE data.\n\nCLEAR ev_data.\n\nIF mr_data IS BOUND.\nASSIGN mr_data->* TO <data>.\nlo_type_out = cl_abap_typedescr=>describe_by_data( ev_data ).\nlo_type_in = cl_abap_typedescr=>describe_by_data( <data> ).\nIF lo_type_out->kind EQ lo_type_in->kind.\nev_data = <data>.\nENDIF.\nENDIF.\n\nENDMETHOD. \"value\n\nENDCLASS.```\n\nAnd unit tests, if you would like to do your modifications and want to validate that it still works:\n\nUnit tests for /ui2/cl_data_access\n```CLASS lc_ut DEFINITION FOR TESTING DURATION SHORT RISK LEVEL HARMLESS.\n\nPUBLIC SECTION.\nTYPES:\nBEGIN OF t_properties,\nenabled TYPE abap_bool,\nlength TYPE i,\ndescription TYPE c LENGTH 20,\nEND OF t_properties,\nBEGIN OF t_data,\nid TYPE i,\nproperties TYPE t_properties,\ncontent TYPE string,\nEND OF t_data,\nBEGIN OF t_name_value,\nname TYPE string,\nvalue TYPE string,\nEND OF t_name_value,\nBEGIN OF t_example,\nflag TYPE abap_bool,\nprops TYPE t_data,\nparams TYPE SORTED TABLE OF t_name_value WITH UNIQUE KEY name,\nEND OF t_example.\n\nMETHODS: test_basic_actions FOR TESTING.\nMETHODS: test_table_access FOR TESTING.\n\nENDCLASS. \"lc_ut\n\nCLASS lc_ut IMPLEMENTATION.\n\nMETHOD test_basic_actions.\n\nDATA: ls_data TYPE t_example,\nlr_data TYPE REF TO data,\nlr_ref TYPE REF TO data,\nlv_int TYPE i,\nlo_data TYPE REF TO /ui2/cl_data_access.\n\nFIELD-SYMBOLS: <data> TYPE data.\n\nls_data-props-id = 12345.\nls_data-props-content = `Some Content`.\nls_data-props-properties-enabled = abap_false.\nls_data-props-properties-length = 10.\nls_data-props-properties-description = `My description`.\n\nGET REFERENCE OF ls_data INTO lr_data.\n\nCREATE OBJECT lo_data EXPORTING ir_data = lr_data.\n\n\" standard way (does not work on SAP_BASIS 700)\nlr_ref = lo_data->at(`PROPS`)->at(`id`)->ref( ).\ncl_aunit_assert=>assert_bound( act = lr_ref msg = `Dynamic access to existing property fails!` ).\n\nASSIGN lr_ref->* TO <data>.\ncl_aunit_assert=>assert_equals( act = <data> exp = ls_data-props-id msg = `Dynamic access to existing property fails!` ).\n\nlr_ref = lo_data->at(`PROPS`)->at(`key`)->ref( ).\ncl_aunit_assert=>assert_not_bound( act = lr_ref msg = `Dynamic access to NOT existing property fails!` ).\n\n\" XPath like\nlr_ref = lo_data->at(`PROPS-ID`)->ref( ).\ncl_aunit_assert=>assert_bound( act = lr_ref msg = `Dynamic access to existing property fails!` ).\n\nASSIGN lr_ref->* TO <data>.\ncl_aunit_assert=>assert_equals( act = <data> exp = ls_data-props-id msg = `Dynamic access to existing property fails!` ).\n\n\" using helper method for creation\nlr_ref = /ui2/cl_data_access=>create( ir_data = lr_data iv_component = `PROPS-ID`)->ref( ).\ncl_aunit_assert=>assert_bound( act = lr_ref msg = `Dynamic access to existing property fails!` ).\n\nASSIGN lr_ref->* TO <data>.\ncl_aunit_assert=>assert_equals( act = <data> exp = ls_data-props-id msg = `Dynamic access to existing property fails!` ).\n\n\" reading value\nlo_data->at(`props-properties-length`)->value( IMPORTING ev_data = lv_int ).\ncl_aunit_assert=>assert_equals( act = lv_int exp = ls_data-props-properties-length msg = `Dynamic read of existing the property fails!` ).\n\nlv_int = 25.\nlo_data->at(`props-properties-len`)->value( IMPORTING ev_data = lv_int ).\ncl_aunit_assert=>assert_initial( act = lv_int msg = `Dynamic read of the NOT existing property fails!` ).\n\n\" modifing value\nlv_int = 25.\nlr_ref = lo_data->at(`props-properties-length`)->ref( ).\nASSIGN lr_ref->* TO <data>.\n<data> = lv_int.\ncl_aunit_assert=>assert_equals( act = lv_int exp = ls_data-props-properties-length msg = `Dynamic modification of the existing property fails!` ).\n\nlv_int = 15.\nlo_data->at(`props-properties-length`)->set( lv_int ).\ncl_aunit_assert=>assert_equals( act = lv_int exp = ls_data-props-properties-length msg = `Dynamic modification of the existing property fails!` ).\n\n\" degenerated example\nlv_int = 15.\nlo_data = /ui2/cl_data_access=>create( iv_data = lv_int ).\nASSIGN lr_ref->* TO <data>.\ncl_aunit_assert=>assert_equals( act = <data> exp = lv_int msg = `Dynamic access of existing the property fails!` ).\n\nENDMETHOD.\n\nMETHOD test_table_access.\n\nDATA:\nls_data TYPE t_example,\nls_line LIKE LINE OF ls_data-params,\nlv_value TYPE string,\nlo_data TYPE REF TO /ui2/cl_data_access.\n\nls_line-name = `KEY1`.\nls_line-value = `Value1`.\nINSERT ls_line INTO TABLE ls_data-params.\n\nls_line-name = `KEY2`.\nls_line-value = `Value2`.\nINSERT ls_line INTO TABLE ls_data-params.\n\nlo_data = /ui2/cl_data_access=>create( iv_data = ls_data iv_component = `params-name`).\nlo_data->value( IMPORTING ev_data = lv_value ).\n\ncl_aunit_assert=>assert_equals( act = lv_value exp = `KEY2` msg = `Dynamic read to table item by index fails!` ).\n\nlo_data = /ui2/cl_data_access=>create( iv_data = ls_data iv_component = `params[name=KEY1]-value`).\nlo_data->value( IMPORTING ev_data = lv_value ).\n\ncl_aunit_assert=>assert_equals( act = lv_value exp = `Value1` msg = `Dynamic read to table item by key index fails!` ).\n\nlo_data = /ui2/cl_data_access=>create( iv_data = ls_data iv_component = `params`)->at(`[name=KEY1]-value`).\nlo_data->value( IMPORTING ev_data = lv_value ).\n\ncl_aunit_assert=>assert_equals( act = lv_value exp = `Value1` msg = `Dynamic read to table item by key index fails!` ).\n\nlo_data = /ui2/cl_data_access=>create( iv_data = ls_data iv_component = `params[name=KEY1, value=Value1]-value`).\nlo_data->value( IMPORTING ev_data = lv_value ).\n\ncl_aunit_assert=>assert_equals( act = lv_value exp = `Value1` msg = `Dynamic read to table item by key index fails!` ).\n\nENDMETHOD.\n\nENDCLASS.```\n• No labels\n\n## 3 Comments\n\n1. Great class but just want to report a bug;\n\nif you are using BW objects (which mainly contains /BIC/ZOBJ/ for field names in custom info object table), it is not working (maybe my parameters are wrong ? ).\n\nSample;\n\nZREFHELPER_CL=>CREATE( IR_DATA = LR_DATA IV_COMPONENT = ``)->AT(`/BIC/Z_YEAR`)->VALUE( IMPORTING EV_DATA = LV_YEAR ).\n\nhere lr_data is a data reference with standard internal table and contains a column named /BIC/Z_YEAR. Above code won't work due to regular expression is causing issue for char `/`. I changed code accordingly to fit my scenario but wanted to ask if there is a way to do it ? (and finally can comment)\n\nwhat i did is; i sent my parameter `BICZ_YEAR` without `/` and inside the AT_INT method; i added below dirty code; (it is working hehe )\n\nreplace first OCCURRENCE OF 'BIC' in lv_Component with '/BIC/'.\n\n2. Hello Bilen,\n\nthanks for feedback and tests!\n\nYes, it is a bug. The code was assuming that ABAP field names can contain only alphabetical characters and numbers. But of course - it is ABAP", null, "I have added support for \"/\" in names now. Hope we do not need some other special characters here", null, "The right regexp would be:\n\n` (?:([\\w\\/]+)|^)(?:\\[(?:(\\d+)|([^\\]]+))\\])?(?:(?:-\\>)|(?:-)|(?:=>)|\\$)`\n\nThe fix would be available in next patch for /UI2/CL_JSON (I am bundling this in same note).\n\nIf you have more feedback or ideas it would be highly appreciated", null, "BR, Alexey.\n\n1. awesome Alexey ! thanks for the quick fix. i changed my code accordingly." ]

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[ "### HOUR(time)\n\nReturns the hour from a given time (Integer).\n\n time The time you want the hour from (Date).\n\n REMARKS\n * The \"time\" can be any string, date or numeric expression.* If time\" is Null, then Null is returned.* The value returned is between 0 and 23.* Returns a Variant (Integer) specifying a whole number between 0 and 23, inclusive, representing the hour of the day.* You can use the MINUTE function to return the minute from a given time.* You can use the SECOND function to return the seconds from a given time.* The equivalent .NET function is Microsoft.VisualBasic.DateAndTime.Hour* For the Microsoft documentation refer to docs.microsoft.com\n\n`Dim tmTime tmTime = #6:25:39 AM# Hour(tmTime)=6 tmTime = #6:25:39 PM# Hour(tmTime)=18 Hour(\"09:34:10\")=9 `" ]

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[ " Convert Yottabyte (YB) (Bytes / Bits)\n\n## Convert Yottabyte (YB)\n\nnumbers in scientific notation\n\nhttps://www.convert-measurement-units.com/convert+Yottabyte.php\n\n## How much is 1 Yottabyte?\n\nMeasurement calculator that can be used to convert Yottabyte, among others.\n\n# Convert Yottabyte (YB):\n\n1. Choose the right category from the selection list, in this case 'Bytes / Bits'.\n2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.\n3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Yottabyte [YB]'.\n4. The value will then be converted into all units of measurement the calculator is familiar with.\n5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.\n\nWith this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '292 Yottabyte'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Yottabyte' or 'YB'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Bytes / Bits'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.\n\nFurthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(90 * 53) YB'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '292 Yottabyte + 876 Yottabyte' or '72mm x 99cm x 36dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.\n\nIf a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 1.608 901 219 926 9×1029. For this form of presentation, the number will be segmented into an exponent, here 29, and the actual number, here 1.608 901 219 926 9. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 1.608 901 219 926 9E+29. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 160 890 121 992 690 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications." ]

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[ "Hit enter after type your search item\n\n# 6 examples of Java array: Create, initialize, and access arrays\n\n### The arrays in Java\n\nThe array is a type of variable in Java that may hold one or more values of the similar type.\n\nFor example, an array of ten elements of type integer can store 10 numbers.\n\nI will explain the elements and important points about Java arrays, first let me show you a few examples along with code for creating, initializing and accessing arrays.\n\n### An example of int array\n\nIn this example, an array of five elements is created. The array is declared with type integer so it may store numeric values. A for loop is used to iterate through the array elements. Inside the for loop, the elements are displayed by using System.out.println.\n\n###### See online demo and code\n\nThe array Java example code:\n\n### An example of creating, initializing and displaying string array\n\nIn this example, a Java string array is created with five elements. The string array contains the names of cities. The array is declared differently than above example. After declaration, the values are assigned to each element as shown in the Java program below:\n\n###### See online demo and code\n\nThe string array code:\n\n### Syntax of creating arrays\n\nAs shown in above two examples, the arrays in Java can be created by different ways.  One way is to use the new keyword as follows:\n\nnumArray = new int;\n\nIn this way, the array is created with five elements. This line of code will allocate the memory of five elements which data type is the integer. Note, the array in Java may contain fixed number of elements. Once created, it cannot be changed.  After creating this array, you may assign values to the array elements as follows:\n\nnumArray = 5;\n\nnumArray = 10;\n\nnumArray = 15;\n\nnumArray = 20;\n\nnumArray = 25;\n\nThere, you may notice the array element index starts at 0. This is important while accessing the array elements as shown in the section below.\n\nCreating array elements with shortcut syntax\n\nYou may also create array elements at the time of declaration. For example:\n\nint[] numArray = { 5, 10, 15,\n\n20, 25, 30,\n\n};\n\nSo, you have to specify the data type which is followed by array name on the left side. In the right side, the array element values are given in the curly braces. Each value is separated by a comma.\n\nIn that way, the number of elements given in the array becomes the length of that array.\n\n### Declaring arrays of other data types\n\nSimilarly, you may create arrays by using any of the above ways for different data types. For example:\n\nbyte[] byteArray; //It will create an array of byte data type\n\nshort[]ShortsArray;\n\nfloat[]FloatsArray;\n\nString[] anArrayOfStrings; //used in one of the above example\n\n### How to access Java array elements\n\nThe array elements can be accessed by using the numeric index that starts at 0. You may use the array index in square brackets to access elements individually. For example:\n\nnumArray;\n\nSee the examples below for accessing single element and using an enhanced for loop for accessing array elements.\n\n### An example of accessing array element by index number\n\nUsing the same array as used in the second example in above section (World cities array), I will access the element number 4 value by using its index as follows:\n\n###### See online demo and code\n\nThe Java code for accessing array element individually:\n\nYou have already seen using the for loop for accessing the same array in above section. To learn more about enhanced for loop go to its chapter here.\n\n### An example of using multi-dimensional array\n\nThe Java also supports multi-dimensional arrays. In this type of array, the array’s components are themselves arrays. The multi-dimensional arrays cantain more than one columns unlike single dimensional arrays used in above section. The rows may vary in length in this type of arrays.\n\nThis is how a multidimensional array can be created:\n\nint[ ][ ] NumArray = new int;\n\nSee this example where an array is created and then its elements are displayed by using the for loop.\n\n###### See online demo and code\n\nThe Java multidimensional array code:\n\n### A program of using array length property\n\nGetting the array length when the number of elements is unknown can be important in different scenarios. The Java provides length property that returns the total number of elements in the specified array. The returned value is an integer.\n\nHave a look at this example where I have used the length property in a for loop to specify the condition for maximum number of times the loop should iterate:\n\n###### See online demo and code\n\nThe array code with length property:\n\nYou see, by using the first System.out.println, the number of elements in the array is displayed by using this:\n\nWorld_cities.length\n\nAfter that, the same is used in the for loop for specifying the maximum counter.\n\n### A demo of arraycopy method\n\nThe arraycopy method of array class can be used for copying the elements from one array to another. The arraycopy method takes a few parameters as described below:\n\nSyntax:\n\nObject copy_from_Array, int sourcePos, Object coptyTo_array, int coptyTo_array_pos, int length\n\nSee an example below to see the arraycopy method in action where an array of numbers is created with nine elements. After that, a new array is created where elements will be copied after using the arraycopy method. Finally, the destination array (after copying) elements are displayed:\n\n###### See online demo and code\n\nThe code of copying array elements:\n\nYou saw, index 1 to 6 elements from the source array are copied to the destination array.\n\n### Summarizing arrays\n\nIn Java, the arrays are useful and powerful concept that can be used for storing multiple values of the similar type. I have shown you a few examples after creating and declaring the array including numeric, string, multidimensional arrays. Let me summarize a few important points about Java arrays below:\n\n• The arrays may contain fixed number of values.\n• The type of the values must be the same or an error will occur.\n• As you declare an array, its length can be specified. Once you have specified the length, it is fixed and cannot be changed.\n• The arrays Java can be created in different ways.\n• You have to specify the type of array e.g. int, byte, short, String etc.\n• The items or elements of an array are accessible by numeric index.\n• The index of the array starts at 0.\n• The array class comes up with a few useful methods and properties to manipulate arrays.\nThis div height required for enabling the sticky sidebar" ]

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[ "# Qr Decomposition Least Squares\n\nThe normal equation approach, which involves forming GTG, can lead to a singular matrix when the underlying G matrix is not singular. A linearly constrained least-squares problem is embedded in the SCF-AA method: min α kDαk s. This method is based on computing the thin QR decomposition of the least squares matrix , where is an -by-matrix with orthogonal columns, and is a -by-upper triangular matrix. This method \"applies the output of DQRDC to compute coordinate transformations, projections, and least squares solutions. Numerical Linear Algebra with Applications is designed for those who want to gain a practical knowledge of modern computational techniques for the numerical solution of linear algebra problems, using MATLAB as the vehicle for computation. •How do we solve least-squares… – without incurring condition-squaring effect of normal equations (A TAx = A b) – when A is singular, “fat”, or otherwise poorly-specified? •QR Factorization – Householder method •Singular Value Decomposition •Total least squares •Practical notes. the QR factorization provided m ˛ n. Householder transformation and the hybrid algorithm were implemented on iPSC/2 and iPSC/860 hypercubes. Learn online and earn valuable credentials from top universities like Yale, Michigan, Stanford, and leading companies like Google and IBM. The cholesky decomposition is a function available in VHLS Linear Algebra Library (the beta version is used. WedefineamatrixQ∈ Rm×m to beJ-orthogonalif QTJQ=J, or, equivalently, QJQT = J, where J is defined in (1. \" We will illustrate the method for the problem of finding a \"least squares. Version: August 12, 2000 79 6 Gram-Schmidt procedure, QR-factorization, Orthog-onal projections, least square Class notes with a few problems and their solutions are found. You may get a few more analytical insights with SVD, but QR is lighter weight machinery (i. The basic idea is the following. Orthogonal Projections and Least Squares. The QRfactorization of Ais a decomposition A= QR, where Qis an m morthogonal matrix and Ris an m nupper triangular matrix.", null, "Parameters: a - Complex matrix to be factored. In signal processing, it is used for adaptive filtering, adaptive beamforming/ interference. The QR decomposition is a popular approach for solving the linear least squares equation. Forwardsubstitution solveAx = b whenA islowertriangularwithnonzerodiagonalelements Algorithm x1 = b1šA11 x2 = „b2 A21x1\"šA22 x3 = „b3 A31x1 A32x2\"šA33 xn. \" I'm a scientist because I believe that discovery and the drive to understand mysteries are among the most important drivers of progress in human civilization. The least squares fitting using non-orthogonal basis We have learned how to nd the least squares approximation of a function fusing an orthogonal basis. the computation per iteration, the QRD-LSL. References. The minimum requires ∂ρ ∂α ˛ ˛ ˛ ˛ β=constant =0 and ∂ρ ∂β ˛ ˛ ˛ ˛ α=constant =0 NMM: Least Squares Curve-Fitting page 8. In other words, [math]A=QR[/math]. solve solves systems of equations via the QR decomposition: if a is a QR decomposition it is the same as solve. Assignment #9: Gram-Schmidt, QR Factorization, Householder Reflections, and Least Squares Due date: Monday, April 22, 2019 (9:15am) For full credit you must show all of your work. However, these al- gorithms have been restricted to problems seeking an estimate. LINEAR ALGEBRA: THEORY. In this lab, you will see how the SVD can be computed, and get the flavor of the standard algorithm used for it. The numerical methods for linear least squares are important because linear regression models are among the most important types of model, both as formal statistical models and for exploration of data-sets. Lab 1 Least squares and Eigenvalues Lab Objective: Use least squares to t curves to data and use QR decomposition to nd eigenvalues.", null, "1) Trefethen (1992) - Definition of Numerical Analysis: Appendix of the Textbook. AU - Chi, Zhipei. Solving the LSP – 2. Matrices involved in least squares problems are SPD, which makes Cholesky an attractive choice for this class of. However, these al- gorithms have been restricted to problems seeking an estimate. SOLVING THE INDEFINITE LEAST SQUARES PROBLEM 919 3. for the QR decomposition; see Question 3. 30-48, 1990. We will have more applications for the QR factorization later in the context of least squares problems. A typical machine learning problem might have several hundred or more variables, while many machine learning algorithms will break down if presented with more than a few dozen. Algorithm: Least Squares via Cholesky factorization 1. The QR decomposition always exists, even if the matrix does not have full rank, so the constructor will never fail. Computing the Moore-Penrose ``pseudo-inverse'' of , and making it possible to solve the system in the least-squares sense. The QR and Cholesky Factorizations §7. Well-suited to large matrices. Usually, using QR. Exercise 1. Use the QR decomposition to solve the least-squares problem Ax ≈ b. Calculating Least-Squares Solutions using the QR Factorization: Theorem. The QR factorization saves us the formation of A T ⁢ A and the solution of the normal equations. Introduction to Python.", null, "You may get a few more analytical insights with SVD, but QR is lighter weight machinery (i. Hyperbolic QR factorization method. decomposition is a good starting point for deriving various theoretical results. Applications Circle fitting, TikhonovRegularization,Image Deblurring. For the hierarchical-tree based QR decomposition, the optimal match between the chosen reduction-tree and the underlying software and hardware layers is, for the most part, system. 1) Trefethen (1992) - Definition of Numerical Analysis: Appendix of the Textbook. least squares (IUS) algorithm and the constrained recursive least squares (CRLS) algorithm based on the numerically stable QR decomposition (QRD) have been of great interest. Least Squares A linear system Ax = b is overdetermined if it has no solutions. Computes the pseudo-inverse of a matrix. 4 High-Performance Cholesky The solutionof overdetermined systems oflinear equations is central to computational science. QR decomposition in R qr package:base R Documentation The QR Decomposition of a Matrix Description: 'qr' computes the QR decomposition of a matrix. However, traditional QR decomposition methods, such as Gram-Schmidt (GS), require high computational complexity and nonlinear operations to achieve high throughput, limiting their usage on resource-limited platforms. The Singular Value Decomposition and Least Squares Problems p. These problems arise in many real-life applications such that curve fitting, statistical modelling and different inverse problems, when some model function should be fitted to the measured data. N2 - In this paper we examine the properties of QR and inverse QR factorizations in the general linear least squares (LS) problem. x = A\\b which produces. Im-plementing the QRD-LSL interpolator merely involves using both forward and backward prediction errors produced at the various stages of a QRD-LSL predictor. qr (a For more information on the qr factorization, Example illustrating a common use of qr: solving of least squares problems. Using QR Decomposition.", null, "A minimizing vector x is called a least squares solution of Ax b. QR decomposition is a matrix factorization technique that decomposes a matrix into a product of an orthogonal matrix Q and an upper triangular matrix R. A matrix library for Java that handles matrices and vectors with complex entries and supports a wide variety of operations, including SVD, LU/QR/Cholesky/Schur decomposition, and efficient determinant computation. Orthogonal decomposition methods of solving the least squares problem are slower than the normal equations method but are more numerically stable because they avoid forming the product XTX. This will fail if isFullRank() returns false. LEAST SQUARE PROBLEMS, QR DECOMPOSITION, AND SVD DECOMPOSITION 3 In general a projector or idempotent is a square matrix Pthat satisfies P2 = P: When v2C(P), then applying the projector results in vitself, i. One of the key benefits of using QR Decomposition over other methods for solving linear least squares is that it is more numerically stable, albeit at the expense of. 1) αi = 1, where D ∈ IRn×k, α = (α 1,,αk) T ∈ IRk and k·k is the Euclidean norm. Least square problem solution is defined in such way that does not require Q matrix, obtained as a result of QR decomposition of the measurement matrix, to be used in calculation and leads to the lower computational complexity. Exercise 1 Find the QR decomposition of A = 2 6 6 4 1 1 1 1 1 0 1 0 1 0 0 1 3 7 7 5; then use that decomposition to solve the least squares problem Ax = 2 6 6 4. NUMERICALLY EFFICIENT METHODS FOR SOLVING LEAST SQUARES PROBLEMS 5 The 2-norm is the most convenient one for our purposes because it is associated with an inner product. BAI, , AND H. Smooth Skinning Decomposition with Rigid Bones [Le and Deng 2012] Rigid bones Highly deformable models Linear solvers. Solve the lower-triangular system R∗y = A∗b 3. Index Terms— Model-order identification, QR factorization, recursive least squares, system identification. Throws: IllegalArgumentException. (Under-constrained problems: see Tygert 2009 and Drineas et al. decomposition. qr, but if a is a rectangular matrix the QR decomposition is computed first. The idea is very simple.", null, "Then Therefore, YXbXX tt ZYQRbYQRRRbRRYQRRbR ttttttttt -- 11 YQRYX RbRQRbQRbQRQRbXX ttt ttttt. QR decomposition, with square, orthogonal Q 2 Rm m, and R 2 Rm n upper triangular (with zeros in its bottom part). Either will handle over- and under-determined systems, providing a least-squares fit if appropriate. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Linear least-squares solution via the normal equations method in and on an equally spaced grid in. qr (a For more information on the qr factorization, Example illustrating a common use of qr: solving of least squares problems. , if is said solution, then is that matrix such that. AU - Parhi, Keshab K. Im-plementing the QRD-LSL interpolator merely involves using both forward and backward prediction errors produced at the various stages of a QRD-LSL predictor. Don't show me this again. The QR and Cholesky Factorizations §7. Recursive least squares notes Recursive least squares notes. Throws: IllegalArgumentException. 6 Least Squares Approximation by QR Factorization 6. In this paper we treat the problem of updating the QR factorization, with applications to the least squares problem. (Dissertation). Given an mxn matrix of rank n, let A=QR be a QR factorization of A.", null, "The first part focuses on various matrix factorizations, such as eigendecomposition, singular value decomposition, Schur decomposition, QZ decomposition and nonnegative factorization. 3 Solution of Rank Deficient Least Squares Problems If rank(A) < n (which is possible even if m < n, i. Any real square matrix A (m x n) may be decomposed as A = QR where Q is an orthogonal matrix (its columns are orthogonal unit vectors meaning QTQ = I) and R is an upper triangular matrix (also called right triangular matrix). Structred Least Squares problems Givens Rotations and Row updating. The rank k of A is determined from the QR decomposition with column pivoting (see Algorithm for details). These problems arise in many real-life applications such that curve fitting, statistical modelling and different inverse problems, when some model function should be fitted to the measured data. From Wikipedia: In linear algebra, a QR decomposition (also called a QR factorization) of a matrix is a decomposition of a matrix A into a product A = QR of an orthogonal matrix Q and an upper triangular matrix R. The QR decomposition always exists, even if the matrix does not have full rank, so the constructor will never fail. We will have more applications for the QR factorization later in the context of least squares problems. 2 Solving large systems computationally Computationally solving systems of equations using traditional methods could introduce huge errors if the matrices used in the calculations are illcondi-tioned. decomposition is a good starting point for deriving various theoretical results. 28/07/2010 : Jon Harrop : London F-Sharp User Group:QR Decomposition Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. Once these factors are calculated, the. solving the linear least squares problem using the singular value decomposition; this collection of routines and sample drivers includes in particular code for the solution of the nonnegative and the bound-constrained LS problems, of the problems arising in spline curve fitting, in least distance programming, as well as a demonstration of singular. A linearly constrained least-squares problem is embedded in the SCF-AA method: min α kDαk s. The computed solution X has at most k nonzero elements per column. • Predict the sparsity structure of the decomposition and allocate storage. The idea is very simple. MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum. Numerical methods for finding the singular value decomposition will also be addressed in this lab. We distinguish our work from previous by focusing on the rigid bones, trying to achieve good approximation even with highly deformable models, and proposing a fast and simple algorithm with all linear solvers. CS 542G: QR, Weighted Least Squares, MLS Robert Bridson October 6, 2008 1 The QR Factorization We established the Gram-Schmidt process last time as a start towards an alternative algorithm for solv-ing least squares problems: while the normal equations approach (using Cholesky factorization) is very. Given the QR orthogonal decomposition X = QR, it is also easy to see. 3, instead one can compute ,. This thesis report aims at introducing the background of QR decomposition and its application.", null, "The least-squares solution to Tx = k is also a solution to TTx = Tk, the normal equations, where the function r(x) = kTx kk2 is minimized. QR decomposition is used in solving linear inverse and least squares problems. Stewart with a date of 8/14/78. less theory is needed to understand it and also it has a smaller coefficient/ faster run time on your computer so its what you'd use in practice for least squares problems). System identi cation 5. Least Squares Using SVD If the matrix A is close to rank-deficient, the QR decomposition method described above has less than ideal stability properties. The least-squares solution to A*X = B is X = E*(R Find the least squares approximate solution to A*x = b with the Q-less QR decomposition and one step of. Least Squares: Numerically, is solving normal equations okay for nice matrices? Generally speaking the QR factorization is a method with a good balancing between. QR decomposition is often used to solve the linear least squares problem and is the basis for a particular eigenvalue algorithm, the. We’ll look at how the QR decomposition can be used for this purpose. We discuss three standard ways to solve teh least square problem: the normal equations, the QR factorization, and the singular value decomposition. New Fast QR Decomposition Least Squares Adaptive Algorithms Athanasios A. In matrix computations, it is used to solve linear equations and least squares problems. In many applications, the system order, in addition to. Approximate Generalized Inverse Preconditioning Methods for Least Squares Problems Abstract A basic problem in science is to fit a model to observations sub ject to errors.", null, "Lecture 5 Least-squares • least-squares (approximate) solution of overdetermined equations Least-squares via QR factorization • A ∈ Rm×n skinny, full rank. While we can use the \\(QR\\) decomposition to solve a system of equations, it is approximately three times more complicated than \\(LU\\) decomposition. The least squares solution is the so-called “basic” solution discussed above and may not be the minimum norm solution. ] [TODO: I need to finish proofs to show that this property is preserved with the right restrictions. • The method of least squares is a standard approach to the approximate solution of overdetermined systems, i. In this paper we treat the problem of updating the QR factorization, with applications to the least squares problem. Since then, the principle of least squares has been the standard procedure for the analysis of scientific data. If any observation has a missing value in any field, that observation is removed before the analysis is carried out. CS 542G: QR, Weighted Least Squares, MLS Robert Bridson October 6, 2008 1 The QR Factorization We established the Gram-Schmidt process last time as a start towards an alternative algorithm for solv-ing least squares problems: while the normal equations approach (using Cholesky factorization) is very. Compute the reduced QR factorization A = QˆRˆ 2. These preliminary QR and LQ factorizations are performed by the drivers xGESVD and xGESDD. In this lab, we introduce linear least squares problems, tools in Python for computing least squares solutions, and two fundamental eigenvalue. Sparse linear systems are an important special case. Linear Least Squares.", null, "Lecture 3 (Aug. Properties of Matrices and Operations on Matrices A very useful factorization is A = QR, where Q is orthogonal and R is upper triangular or trapezoidal. The 'economy' QR decomposition, in which Q ∈ Rm×n (with orthonormal columns) and invertible R, is obtained using [Q,R]=qr(A,0). \"Matrix decomposition refers to the transformation of a given matrix into a given canonical form. Let's compare the solutions of linalg. 1 Formulation of Least Squares Approximation Problems Least-squares problems arise, for instance, when one seeks to determine the relation between an independent variable, say time, and a measured dependent variable, say position or velocity of an object. Let ρ = r 2 2 to simplify the notation. BAI, , AND H. That is where it will use the QR or SVD. Description of multisplitting 2. Skip to content. Our proposed method is mathematically equivalent to an existing method but has several practical advantages over the existing method. solve solves systems of equations via the QR decomposition: if a is a QR decomposition it is the same as solve. Thus, the size of each tells you how much of the total variance is accounted for by each singular vector. QR decomposition is a form of orthogonal iriangularisa-tion which is particularly useful in least squares computations and forms the basis of some very stable numerical algorithms. , if we have an underdetermined problem), then infinitely many solutions exist. Householder transformation and the hybrid algorithm were implemented on iPSC/2 and iPSC/860 hypercubes. less theory is needed to understand it and also it has a smaller coefficient/ faster run time on your computer so its what you'd use in practice for least squares problems). In the present example,. When multiple solutions need to be computed with only minor changes in the underlying data, knowledge of the difference. Open Access in DiVA No full text in DiVA. It relies on a sparse QR factorization coupled with iterative weighted least squares methods. Usage: regressor [-d num] [--qr] -d.", null, "computing the QR decomposition. is the orthogonal matrix from QR decomposition [out] r: is the upper triangular matrix from QR decomposition [out] tau: will contain additional information needed for solving a least squares problem using q and r [in] in: is the input matrix. Stochastic gradient descent to find least square in linear regression Posted on May 23, 2014 by qizele Let’s go back to a very simple yet strong method we use in model: linear regression. For a general matrix, we try to change to the orthogonal case. 4 High-Performance Cholesky The solutionof overdetermined systems oflinear equations is central to computational science. 3 Algebra of least squares The predicted value for y i, using the least squares estimates, is ^y i= Z i ^. If weights are specified then a weighted least squares is performed with the weight given to the jth case specified by the jth entry in wt. QR decomposition using Givens rotations is a efficient method to prevent directly matrix inverse in solving least square minimization problem, which is a typical approach for weight calculation in adaptive beamforming. QR Decomposition is widely used in quantitative finance as the basis for the solution of the linear least squares problem, which itself is used for statistical regression analysis. In general, it is computed using matrix factorization methods such as the QR decomposition , and the least squares approximate solution is given by x^ ls = R 1QTy. QRD is useful for solving least squares' problems and simultaneous equations. Calculating Least-Squares Solutions using the QR Factorization: Theorem. Multisplitting for Ax D b Iterative methods based on a single splitting, A D M−N, are well known . b = R\\C My question is whether I need to be worried about numerical errors here. The residual is computed as a by-product and stored in residual. An algorithm which has better numerical stability for ill-conditioned problems is known as the Tall Skinny QR (TSQR) method. Book Description. , if we have an underdetermined problem), then infinitely many solutions exist. 1 The Singular Value Decomposition. There are three ways to compute this decomposition: 1.\n\nIn this, we determine a factorization A = QR,. It is also the method of choice for solving most linear least-squares problems. less theory is needed to understand it and also it has a smaller coefficient/ faster run time on your computer so its what you'd use in practice for least squares problems). 20 Solution: Observe that A has linearly independent column vectors, so we know in advance that there is a unique least squares. the QR factorization by calling numpy. The QR and Cholesky Factorizations §7. Then we want to find x such that (Ax b)0(Ax b) is minimized. •How do we solve least-squares… – without incurring condition-squaring effect of normal equations (A TAx = A b) – when A is singular, “fat”, or otherwise poorly-specified? •QR Factorization – Householder method •Singular Value Decomposition •Total least squares •Practical notes. DQRDC computes the QR factorization of a real rectangular matrix. The matrix D is initially a single vector (k = 1) and varies from one SCF-AA iteration to the next. NUMERICALLY EFFICIENT METHODS FOR SOLVING LEAST SQUARES PROBLEMS 5 The 2-norm is the most convenient one for our purposes because it is associated with an inner product. \" Use the A = QR factorization to get a \"least squares solution. on QR decomposition based approach for solving least square problems. 1 Least Squares Problems and Pseudo-Inverses The method of least squares is a way of \"solving\" an overdetermined system of linear equations Ax = b, i. QR Factorization for Solving Least Squares Problems I'll briefly review the QR decomposition, which exists for any matrix. Use least square/regression when there are clear explanatary variables and causality Use PCA when there is a set of related variables but no clear causality PCA requires a natural unit for all dimensions. Join Coursera for free and transform your career with degrees, certificates, Specializations, & MOOCs in data science, computer science, business, and dozens of other topics. Once we have an inner product de ned on a vector space, we can de ne both a norm and distance for the inner product space: De nition 3. You may get a few more analytical insights with SVD, but QR is lighter weight machinery (i. Y1 - 2000/12/1. Qr Decomposition Least Squares.\n\nT612019/06/17 16:13: GMT+0530\n\nT622019/06/17 16:13: GMT+0530\n\nT632019/06/17 16:13: GMT+0530\n\nT642019/06/17 16:13: GMT+0530\n\nT12019/06/17 16:13: GMT+0530\n\nT22019/06/17 16:13: GMT+0530\n\nT32019/06/17 16:13: GMT+0530\n\nT42019/06/17 16:13: GMT+0530\n\nT52019/06/17 16:13: GMT+0530\n\nT62019/06/17 16:13: GMT+0530\n\nT72019/06/17 16:13: GMT+0530\n\nT82019/06/17 16:13: GMT+0530\n\nT92019/06/17 16:13: GMT+0530\n\nT102019/06/17 16:13: GMT+0530\n\nT112019/06/17 16:13: GMT+0530\n\nT122019/06/17 16:13: GMT+0530" ]

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[ "# ArgMax : Variable Domain Error [closed]\n\nI'm attempting to utilize the ArgMax function, but am receiving the error \"ArgMax::vdom: Variable domain ... should be either Reals or Integers.\"\n\nArgMax[{objective, cons == 1}, assign, {assign} ∈ {0, 1}]\n\nAs a quick overview, we have edges denoted $$e_{i,j}$$ in $$\\mathbb{R}^3$$, and we want to assign an integer value of 0 or 1 to each such that their sum is at most 1 for each set of edges. So, objective is defined as a set of sets as summations with constant coefficients $$objective = \\{\\{3e_{\\{4, 2, -2\\}, \\{1, 3, 1\\}\\}} + 2e_{\\{4, 2, -2\\}, \\{2, 1, 1\\}\\}}\\}, \\{7e_{\\{5, 3, -1\\}, \\{-2, 6, -1\\}\\}} + 6e_{\\{5, 3, -1\\}, \\{-1, 4, -2\\}\\}}\\}\\}$$\n\nWe also have a set of sets for constraints defined as $$cons = \\{\\{e_{\\{\\{4, 2, -2\\}, \\{1, 3, 1\\}\\}} + e_{\\{\\{4, 2, -2\\}, \\{2, 1, 1\\}\\}}\\}, \\{e_{\\{\\{5, 3, -1\\}, \\{-2, 6, -1\\}\\}} + e_{\\{\\{5, 3, -1\\}, \\{-1, 4, -2\\}\\}}\\}\\}$$.\n\nLastly, we have the variables defined in $$assign = \\{e_{\\{\\{4, 2, -2\\}, \\{1, 3, 1\\}\\}}, e_{\\{\\{4, 2, -2\\}, \\{2, 1, 1\\}\\}}, e_{\\{\\{5, 3, -1\\}, \\{-2, 6, -1\\}\\}}, e_{\\{5, 3, -1\\}, \\{-1, 4, -2\\}\\}}\\}$$.\n\nI'm new to Mathematica, so this may be a trivial fix, but if anyone can point out my error, I would greatly appreciate it.\n\n• Please include expressions for objective in Mathematica format. Is assign a variable? {assign} ∈ {0, 1} is not a legitimate domain for ArgMax Please read the documentation for this function. Jun 9 at 15:27\n• Apologies, I'm not sure how you want me to alter the format. Also, assign is a set containing the variables. Jun 9 at 15:44\n• You'll generally get better answers on this site if you provide all of the code that you are using, rather than describing what the variables are. Can you include the actual code you're using to define objective, cons, and assign? Jun 9 at 15:47\n• There is a large amount of code before this (300-500 lines). I tried to include a snippet of what these sets look like. Jun 9 at 16:10\n\nThe error means what it says: the domain of restriction must be either Reals or Integers. But you can also add additional inequalities in your constraints to ensure that the variables are either 0 or 1. Here's an example of how to do this:\nArgMax[{3 x + 5 y + z,\n\n• @AStrugglingProgrammer - And @@ Thread[0 <= varList <= 1] where varList can have as many variables as you need. Jun 9 at 22:10" ]

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[ "", null, "# How to solve an arithmetic sequence\n\nThis can help the student to understand the problem and How to solve an arithmetic sequence. We can help me with math work.\n\n## How can we solve an arithmetic sequence\n\nWe can do your math homework for you, and we'll make sure that you understand How to solve an arithmetic sequence. Another way to get algebra help online is to use a resource like Khan Academy. This website provides a wealth of helpful videos and articles on algebra, so you can learn at your own pace. This is a great\n\nSolving for x in a quadratic equation can be done using the quadratic formula, which is: x = (-b +/- sqrt(b^2 - 4ac)) / 2a. In this equation, a, b, and c are the coefficients of the quadratic equation (x^2 + bx + c = 0), and +/- represents taking the positive or negative root.\n\nOnce you have the roots, you can use them to determine which values of x satisfy the inequality. If the roots are real, you will need to use the sign of the quadratic equation to determine which values of x satisfy the inequality. If the roots are complex, you will need to use the conjugate roots to determine which values of x satisfy the inequality.\n\nIf the batteries seem fine, try resetting the camera. This can be done by taking out the batteries and then putting them back in. If neither of these fixes work, you may need to take your camera to a professional to have it repaired or replaced.\n\nIf you're struggling to solve a math problem, there's help available online. There are websites and apps that can walk you through the steps to get the answer. You can also watch videos that show how to solve similar problems. And if you're still stuck, you can ask for help from a tutor or a friend who's good at math. With a little effort, you can usually find a way to solve even the most challenging math problem.\n\nThere are various homework cheat websites available online that students can use to get help with their homework. Some of these websites provide answers to specific questions, while others provide a step-by-step guide on how to solving a particular problem. However, using these websites may not be the best idea as it may lead to students not understanding the material and getting lower grades.\n\n## Help with math\n\nOne of the best learning tools for almost anyone, even calculus doesn’t slow down the app. The step-by-step function makes it very easy to legitimately learn with. Helpful good job. But it has a VIP too that you need to pay so it's not completely free but five stars!", null, "Tatum Perry\nVery useful and nice app. It solves questions of algebra and irrational number so quickly than another usual calculator. It also shows the steps how to solve the problem which is very helpful also for question like algebra it draws the graph automatically. If there were few more updates like for fractional exponent and many other. The app will be the best in all calculating app", null, "Talia Rogers\nHow to solve commission math problems Solve piecewise functions Online photo math How to solve sigma notation Precalculus equation solver Math scaner" ]

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[ "The University of Southampton\nCourses\n\n# MATH2010 Statistical Modelling I\n\n## Module Overview\n\nSimple linear regression is developed for one explanatory variable using the principle of least squares. The extension to two explanatory variables raises the issue of whether both variables are needed for a well-fitting model, or whether one is sufficient and, if so, which one. These ideas are generalised to many explanatory variables (multiple regression), for which the necessary theory of linear models is developed in terms of vectors and matrices. Checking model adequacy is introduced, e.g. by examining plots of the residuals. Widening the class of models that can be considered by the use of dummy variables for qualitative explanatory variables to assess treatment effects. The methods are implemented using a suitable software and students gain experience and advice through weekly worksheets. One of the pre-requisites for MATH3012, MATH3013, MATH3014, MATH6021, MATH6025, MATH6027 and MATH6135\n\n### Aims and Objectives\n\n#### Learning Outcomes\n\n##### Knowledge and Understanding\n\nHaving successfully completed this module, you will be able to demonstrate knowledge and understanding of:\n\n• Interpret the output from such an analysis including the meaning of interactions and terms based on qualitative factors\n• Understand how to make a critical appraisal of a fitted model .\n• Use the theory of linear models and matrix algebra to investigate standard and non-standard problems\n##### Subject Specific Practical Skills\n\nHaving successfully completed this module you will be able to:\n\n• Carry out t-tests and calculate confidence intervals by hand and by computer\n##### Learning Outcomes\n\nHaving successfully completed this module you will be able to:\n\n• Carry out simple linear regression by computer\n• Fit multiple regression models using the adopted software package\n• Using a variety of procedures for variable selection\n\n### Syllabus\n\n• Analysis of simple data sets, confidence intervals and t-tests; • Regression analysis, simple linear regression with one independent variable in detail, including analysis of variance, confidence intervals, plotting ideas for diagnostic assessment; • Method of least squares using matrix notation, deriving the general results from first principles; • Comparison with simple linear regression. • Multiple linear regression. Examples with two, three and many independent variables; • Model selection; • Indicator variables for qualitative factors - Interpretation of interactions\n\n### Learning and Teaching\n\n#### Teaching and learning methods\n\nLectures, coursework, problem classes, workshops, computer labs, private study.\n\nTypeHours\nTeaching48\nIndependent Study102\nTotal study time150\n\nS. Weisberg. Applied Linear Regression.\n\n### Assessment\n\n#### Summative\n\nMethodPercentage contribution\nCoursework 40%\nWritten exam 60%\n\n#### Referral\n\nMethodPercentage contribution\nWritten exam 100%\n\n#### Repeat Information\n\nRepeat type: Internal & External\n\nPre-requisites: MATH1024 AND MATH2011\n\n### Costs\n\n#### Costs associated with this module\n\nStudents are responsible for meeting the cost of essential textbooks, and of producing such essays, assignments, laboratory reports and dissertations as are required to fulfil the academic requirements for each programme of study.\n\nIn addition to this, students registered for this module typically also have to pay for:\n\n##### Books and Stationery equipment\n\nRecommended texts for this module may be available in limited supply in the University Library and students may wish to purchase reading texts as appropriate.\n\nCourse texts are provided by the library and there are no additional compulsory costs associated with the module.\n\nPlease also ensure you read the section on additional costs in the University’s Fees, Charges and Expenses Regulations in the University Calendar available at www.calendar.soton.ac.uk." ]

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[ "# A new fractional analytical approach for treatment of system of physical models by using Laplace Transform\n\nAuthor\n\nDepartment of Mathematics, National Institute of Technology, Jamshedpur, 831014, Jharkhand India\n\nAbstract\n\nIn this study, the homotopy perturbation transform method (HPTM) is performed to give approximate and analytical solutions of the first order linear and nonlinear system of time fractional partial differential equation. The HPTM is a combined form of the Laplace transform, the homotopy perturbation method, and He’s polynomials. The nonlinear terms can be easily handled by the use of He’s polynomials. The proposed scheme finds the solutions without any discretization or restrictive assumptions and is free from round-off errors and therefore, reduces the numerical computations to a great extent. The speed of convergence of the method is based on a rapidly convergent series with easily computable components. The fractional derivatives are described here in the Caputo sense. Numerical results show that the HPTM is easy to implement and accurate when applied to time- fractional system of partial differential equations..\n\nKeywords" ]

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[ "# Number 85204\n\n\n\n## Description of the number 85204\n\n85204 is a natural number (hence integer, rational and real) of 5 digits that follows 85203 and precedes 85205.\n\n85204 is an even number, since it is divisible by 2.\n\nThe number 85204 is a unique number, with its own characteristics that, for some reason, has caught your attention. It is logical, we use numbers every day, in multiple ways and almost without realizing it, but knowing more about the number 85204 can help you benefit from that knowledge, and be of great use. If you keep reading, we will give you all the facts you need to know about the number 85204, you will see how many of them you already knew, but we are sure you will also discover some new ones.\n\n## how to write 85204 in letters?\n\nNumber 85204 in English is written as eighty-five thousand two hundred four\nThe number 85204 is pronounced digit by digit as (8) eight (5) five (2) two (0) zero (4) four.\n\n## What are the divisors of 85204?\n\nThe number 85204 has 24 divisors, they are as follows:\n\nThe sum of its divisors, excluding the number itself is 96236, so it is an abundant number and its abundance is 11032\n\n## Is 85204 a prime number?\n\nNo, 85204 is not a prime number since it has more divisors than 1 and the number itself\n\n## What are the prime factors of 85204?\n\nThe factorization into prime factors of 85204 is:\n\n22*71*171*1791\n\n## What is the square root of 85204?\n\nThe square root of 85204 is. 291.89724219321\n\n## What is the square of 85204?\n\nThe square of 85204, the result of multiplying 85204*85204 is. 7259721616\n\n## How to convert 85204 to binary numbers?\n\nThe decimal number 85204 into binary numbers is.10100110011010100\n\n## How to convert 85204 to octal?\n\nThe decimal number 85204 in octal numbers is246324\n\n## How to convert 85204 to hexadecimal?\n\nThe decimal number 85204 in hexadecimal numbers is14cd4\n\n## What is the natural or neperian logarithm of 85204?\n\nThe neperian or natural logarithm of 85204 is.11.352803660072\n\n## What is the base 10 logarithm of 85204?\n\nThe base 10 logarithm of 85204 is4.9304599837004\n\n## What are the trigonometric properties of 85204?\n\n### What is the sine of 85204?\n\nThe sine of 85204 radians is.-0.76151152801666\n\n### What is the cosine of 85204?\n\nThe cosine of 85204 radians is. -0.64815136557576\n\n### What is the tangent of 85204?\n\nThe tangent of 85204 radians is.1.1748976681399" ]

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[ "", null, "# Java Program to Find Frequency of Characters in a String\n\nFinding characters frequency of a String means calculating the occurrence of  each alphabet present in the string. Since there's no pre-build function present in Java for calculating the occurrence of the characters or alphabets so, we have to write a program which will give the frequency of characters present in the string. Let's get started.", null, "## Logic: Characters Frequency in a String\n\nFirst we have take an integer array of size 26 having all its elements equal to zero.\nThan we have to traverse the string 26 times in search of all the 26 alphabets one by one.\nIf the string character matches to the searching alphabet than we have to increment the value stored at that particular index of int array by 1.\n\nconfused? have a look at the source code given below.\n\n## Source Code\n\n```class CharFrequency {\nstatic int[] charFreq(String s){\ns=s.toLowerCase();  //convert to lowercase alphabets\nint freq[]=new int; //array for counting freq of all alphabets\n\nfor(int i=0,c=97; i<26; i++,c++){\nfor(int j=0; j<s.length(); j++){\nchar ch=s.charAt(j); //traversing the string\nif(ch==c) //checking string for 'a' i.e 97 then 'b' i.e 98 and so on\nfreq[i]++; //increasing count of those aplhabets which are present in the string\n}\n}\nreturn freq;    //returning array having freq of alphabests present in String s\n}\npublic static void main(String args[]){\n\nString s = \"Java Programming\";\n\n//Getting the frequency\nint freq[]=charFreq(s);\n\n//printing the output\nSystem.out.println(\"Alphabet\\tFrequency\");\nfor(int i=0,c=97; i<26; i++,c++){\nif(freq[i]!=0){ //alphabet having freq > 0\nchar ch = (char) c;\nSystem.out.println(ch+\"\\t\\t\"+freq[i]);\n}\n}\n}\n}```\n\n### Output", null, "### Explanation\n\nThis is the most basic approach for finding the characters frequency.\nHere we made a function named charFreq which takes the string as an argument and returns an array of integers which have the calculated frequency of all the alphabets present in the input string.\nFirstly we converted the entire string to lower-case characters for the further operations and defined an integer array freq[] of  size 26 (default value of all the elements in an integer array is zero).\nThen as mentioned in the Logic above we have to traverse the string 26 times for each and every character on by one. Here we are taking alphabets as 97, 98, 99 and so on which is the ascii value of lower-case alphabets. when the string character matches with our searching alphabet then we increase the value at that index in an array freq[i]++. Finally at the end the function returns an array of integers having the frequency of characters.\n\nUsing the Function: int array[] = charFreq( \"you string here\" );\n\nExtracting value from array: array[ index_between_1_to_26 ]\ne.g  array will give the occurrence of 'c'\n\n### Conclusion\n\nThere are many other logics by which we can perform the same task, above logic is the most simple approach for finding the character frequency of a string. You may try your own logic and do share with us!\nIf you like this post please like and share with your friends and feel free to coment if you have any queries related to this program." ]

[ null, "https://2.bp.blogspot.com/-NbB1oHj7PWg/VCfcEM1T9TI/AAAAAAAAB40/ARG33MNnmbc/s1600/java-char-freq.png", null, "https://2.bp.blogspot.com/-NbB1oHj7PWg/VCfcEM1T9TI/AAAAAAAAB40/ARG33MNnmbc/s1600/java-char-freq.png", null, "https://2.bp.blogspot.com/-sikTsF5tXbM/VCfglMo8qzI/AAAAAAAAB5A/TwCjayd5QCg/s1600/output-java-char-freq.png", null ]

[ "# For every degree of longitude, the time difference is four minutes. Explain.\n\nEarth is divided into 360⁰ of longitude lines. They are imaginary lines that run from north to south of the globe. And the time for one rotation of Earth is 24 hours. We find it as day and night completed in 24 hours.\n\nIf 360⁰ = 24 hours,\n\nThen, 1 hour =", null, "= 15⁰, each hour is further divided into minutes, and minutes into seconds.\n\nSo, 1 hour = 60 minutes,\n\nSo, earth moves 15 degrees in 60 minutes.\n\nThen, in moving 1⁰ the earth takes", null, "", null, "So for every degree of longitude, the time difference is four minutes.\n\nRate this question :\n\nHow useful is this solution?\nWe strive to provide quality solutions. Please rate us to serve you better." ]

[ null, "https://gradeup-question-images.grdp.co/liveData/PROJ22193/1539589237984461.png", null, "https://gradeup-question-images.grdp.co/liveData/PROJ22193/1539589238848876.png", null, "https://gradeup-question-images.grdp.co/liveData/PROJ22193/1539589239614857.png", null ]

[ "# Is it possible to stop time? [duplicate]\n\nThis question already has an answer here:\n\nAssuming the spacetime principle, if the space is modified the time does too. So if the velocity in the space is increase, does the time slow down? What happens if the speed is the speed of light, does the time stops?\n\n## marked as duplicate by Kyle Kanos, John Rennie, JamalS, ACuriousMind♦, Chris MuellerMar 5 '15 at 15:07\n\nThis question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.\n\n## 3 Answers\n\nNo, for several reasons.\n\nFirst, the idea of time \"slowing down\" is a little bit of a misnomer. If you were traveling at relativistic speeds, you would not perceive the passage of time any differently than you do right now. It's only when you compare your clocks to an observer in another reference frame (let's me, sitting in my living room, at rest with respect to the ground) that you would notice that your clock shows a smaller passage of time than mine does. So within your own frame of reference, it makes no sense to talk about time ticking slow or fast, it's only when you compare your clock to a clock in another reference frame that you can compare passages in time.\n\nAlso, and this is probably the larger hang-up to your question, is that a particle with mass can never reach the speed of light. With truly astounding amounts of energy, you could get arbitrarily close to the speed of light, but never reach it. The reason is because at relativistic speeds, the equation for kinetic energy is different than the classical equation for kinetic energy that you're used to.\n\nClassically, the kinetic energy of a moving particle is given by $$KE=\\frac{1}{2}mv^2$$ where we can reach any finite velocity that we want, so long as we do enough work on the object to increase its kinetic energy to the proper amount. However, it turns out that this is equation is only an approximation for small velocities. Relativistically, the equation for kinetic energy is $$KE=\\frac{mc^2}{\\sqrt{1-\\frac{v}{c}^2}}-mc^2$$. What this equation shows you is that for any finite amount of kinetic energy on an object $v<c$. If you have trouble seeing this, attack the equation from a different angle. Let's say I'm crazy, and it is possible to travel at the speed of light. So let's take $v=c$ and substitute that into our relativistic kinetic energy equation. Oh no! You'll see that we get a divide by zero error, because our entire denominator goes to zero. This means we would need an infinite amount of kinetic energy for our object and that's just not possible.\n\nSo I hate to burst your bubble, but no light-speed travel for you anytime soon. (Or ever).\n\nIts not possible to stop time but using relativity it can be thought of to be slowed down .\nNothing can be faster than the speed of light so its not possible .\nEven when we near it , energy tends to become infinity .\n\nThings will be bigger/smaller/slower with speed as $\\frac{1}{\\sqrt{1-\\frac{v^2}{c^2}}}$.\n\nThus, as $v \\rightarrow c$,\n\n1. Mass (or energy) goes to infinite,\n2. Time goes to zero.\n\nBut it is only a limit, which is unreachable (at least in the special relativity), because of (1).\n\n• This is reachable for photons however, since those are already massless. – ahemmetter Mar 5 '15 at 11:03\n• This is a very incomplete answer. For example: What is $1/\\sqrt{1-v^2/c^2}$? Where did it come from? What \"things\" do you mean? How does mass/energy go to $\\infty$? How does time go to 0? – Kyle Kanos Mar 5 '15 at 15:21\n• @KyleKanos $v$, $c$, $\\infty$ and such have their common meanings. \"Things\" means here mass points moving slower than light. Thank you. – peterh Mar 5 '15 at 23:46\n• @KyleKanos It was a simple question, which deserved a simple answer. I couldn't include a 200 page long book with the axiomatical description of the special relativity. I don't really understand, what is not clear to you. You know also SR, probably better as me. But this was exactly (the Lorentz-transformation of the spacetime coordinates) which couldn't been written there, because the question required clearly the possible simplest answer. – peterh Mar 6 '15 at 4:04\n• I'm also not suggesting that I don't understand your post. I know what you mean to say, but you are doing a terrible job at actually saying it. What I am saying is that John Q Public who doesn't have any physics knowledge would look at your post would have no clue what you were talking about. – Kyle Kanos Mar 6 '15 at 4:12" ]

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[ "# #20dbe8 Color Information\n\nIn a RGB color space, hex #20dbe8 is composed of 12.5% red, 85.9% green and 91% blue. Whereas in a CMYK color space, it is composed of 86.2% cyan, 5.6% magenta, 0% yellow and 9% black. It has a hue angle of 183.9 degrees, a saturation of 81.3% and a lightness of 51.8%. #20dbe8 color hex could be obtained by blending #40ffff with #00b7d1. Closest websafe color is: #33ccff.\n\n• R 13\n• G 86\n• B 91\nRGB color chart\n• C 86\n• M 6\n• Y 0\n• K 9\nCMYK color chart\n\n#20dbe8 color description : Vivid cyan.\n\n# #20dbe8 Color Conversion\n\nThe hexadecimal color #20dbe8 has RGB values of R:32, G:219, B:232 and CMYK values of C:0.86, M:0.06, Y:0, K:0.09. Its decimal value is 2153448.\n\nHex triplet RGB Decimal 20dbe8 `#20dbe8` 32, 219, 232 `rgb(32,219,232)` 12.5, 85.9, 91 `rgb(12.5%,85.9%,91%)` 86, 6, 0, 9 183.9°, 81.3, 51.8 `hsl(183.9,81.3%,51.8%)` 183.9°, 86.2, 91 33ccff `#33ccff`\nCIE-LAB 80.063, -37.85, -18.651 40.488, 56.792, 85.168 0.222, 0.311, 56.792 80.063, 42.196, 206.232 80.063, -59.066, -23.993 75.361, -35.98, -14.253 00100000, 11011011, 11101000\n\n# Color Schemes with #20dbe8\n\n• #20dbe8\n``#20dbe8` `rgb(32,219,232)``\n• #e82d20\n``#e82d20` `rgb(232,45,32)``\nComplementary Color\n• #20e891\n``#20e891` `rgb(32,232,145)``\n• #20dbe8\n``#20dbe8` `rgb(32,219,232)``\n• #2077e8\n``#2077e8` `rgb(32,119,232)``\nAnalogous Color\n• #e89120\n``#e89120` `rgb(232,145,32)``\n• #20dbe8\n``#20dbe8` `rgb(32,219,232)``\n• #e82077\n``#e82077` `rgb(232,32,119)``\nSplit Complementary Color\n• #dbe820\n``#dbe820` `rgb(219,232,32)``\n• #20dbe8\n``#20dbe8` `rgb(32,219,232)``\n• #e820db\n``#e820db` `rgb(232,32,219)``\n• #20e82d\n``#20e82d` `rgb(32,232,45)``\n• #20dbe8\n``#20dbe8` `rgb(32,219,232)``\n• #e820db\n``#e820db` `rgb(232,32,219)``\n• #e82d20\n``#e82d20` `rgb(232,45,32)``\n• #12a0aa\n``#12a0aa` `rgb(18,160,170)``\n• #14b6c1\n``#14b6c1` `rgb(20,182,193)``\n• #16ccd8\n``#16ccd8` `rgb(22,204,216)``\n• #20dbe8\n``#20dbe8` `rgb(32,219,232)``\n• #37dfea\n``#37dfea` `rgb(55,223,234)``\n• #4ee2ed\n``#4ee2ed` `rgb(78,226,237)``\n• #65e6ef\n``#65e6ef` `rgb(101,230,239)``\nMonochromatic Color\n\n# Alternatives to #20dbe8\n\nBelow, you can see some colors close to #20dbe8. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #20e8c3\n``#20e8c3` `rgb(32,232,195)``\n• #20e8d4\n``#20e8d4` `rgb(32,232,212)``\n• #20e8e4\n``#20e8e4` `rgb(32,232,228)``\n• #20dbe8\n``#20dbe8` `rgb(32,219,232)``\n• #20cae8\n``#20cae8` `rgb(32,202,232)``\n• #20bae8\n``#20bae8` `rgb(32,186,232)``\n• #20a9e8\n``#20a9e8` `rgb(32,169,232)``\nSimilar Colors\n\n# #20dbe8 Preview\n\nThis text has a font color of #20dbe8.\n\n``<span style=\"color:#20dbe8;\">Text here</span>``\n#20dbe8 background color\n\nThis paragraph has a background color of #20dbe8.\n\n``<p style=\"background-color:#20dbe8;\">Content here</p>``\n#20dbe8 border color\n\nThis element has a border color of #20dbe8.\n\n``<div style=\"border:1px solid #20dbe8;\">Content here</div>``\nCSS codes\n``.text {color:#20dbe8;}``\n``.background {background-color:#20dbe8;}``\n``.border {border:1px solid #20dbe8;}``\n\n# Shades and Tints of #20dbe8\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, #010808 is the darkest color, while #f5fdfe is the lightest one.\n\n• #010808\n``#010808` `rgb(1,8,8)``\n• #03181a\n``#03181a` `rgb(3,24,26)``\n• #05292c\n``#05292c` `rgb(5,41,44)``\n• #063a3e\n``#063a3e` `rgb(6,58,62)``\n• #084b4f\n``#084b4f` `rgb(8,75,79)``\n• #0a5b61\n``#0a5b61` `rgb(10,91,97)``\n• #0c6c73\n``#0c6c73` `rgb(12,108,115)``\n• #0e7d85\n``#0e7d85` `rgb(14,125,133)``\n• #108e96\n``#108e96` `rgb(16,142,150)``\n• #119ea8\n``#119ea8` `rgb(17,158,168)``\n• #13afba\n``#13afba` `rgb(19,175,186)``\n• #15c0cc\n``#15c0cc` `rgb(21,192,204)``\n• #17d1de\n``#17d1de` `rgb(23,209,222)``\n• #20dbe8\n``#20dbe8` `rgb(32,219,232)``\n• #32deea\n``#32deea` `rgb(50,222,234)``\n• #44e1ec\n``#44e1ec` `rgb(68,225,236)``\n• #55e4ee\n``#55e4ee` `rgb(85,228,238)``\n• #67e6ef\n``#67e6ef` `rgb(103,230,239)``\n• #79e9f1\n``#79e9f1` `rgb(121,233,241)``\n• #8becf3\n``#8becf3` `rgb(139,236,243)``\n• #9ceff5\n``#9ceff5` `rgb(156,239,245)``\n• #aef2f7\n``#aef2f7` `rgb(174,242,247)``\n• #c0f5f9\n``#c0f5f9` `rgb(192,245,249)``\n• #d2f8fa\n``#d2f8fa` `rgb(210,248,250)``\n• #e4fbfc\n``#e4fbfc` `rgb(228,251,252)``\n• #f5fdfe\n``#f5fdfe` `rgb(245,253,254)``\nTint Color Variation\n\n# Tones of #20dbe8\n\nA tone is produced by adding gray to any pure hue. In this case, #7f8989 is the less saturated color, while #0debfb is the most saturated one.\n\n• #7f8989\n``#7f8989` `rgb(127,137,137)``\n• #759193\n``#759193` `rgb(117,145,147)``\n• #6c999c\n``#6c999c` `rgb(108,153,156)``\n• #62a1a6\n``#62a1a6` `rgb(98,161,166)``\n• #59aaaf\n``#59aaaf` `rgb(89,170,175)``\n• #4fb2b9\n``#4fb2b9` `rgb(79,178,185)``\n• #46bac2\n``#46bac2` `rgb(70,186,194)``\n• #3cc2cc\n``#3cc2cc` `rgb(60,194,204)``\n• #33cbd5\n``#33cbd5` `rgb(51,203,213)``\n• #29d3df\n``#29d3df` `rgb(41,211,223)``\n• #20dbe8\n``#20dbe8` `rgb(32,219,232)``\n• #17e3f1\n``#17e3f1` `rgb(23,227,241)``\n• #0debfb\n``#0debfb` `rgb(13,235,251)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #20dbe8 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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[ "# Article\n\nKeywords:\nfunction space; pointwise convergence; linearly ordered topological space; Lindelöf space; Cantor tree\nSummary:\nA.V. Arkhangel'skii asked that, is it true that every space \\$Y\\$ of countable tightness is homeomorphic to a subspace (to a closed subspace) of \\$C_p(X)\\$ where \\$X\\$ is Lin\\-de\\-löf? \\$C_p(X)\\$ denotes the space of all continuous real-valued functions on a space \\$X\\$ with the topology of pointwise convergence. In this note we show that the two arrows space is a counterexample for the problem by showing that every separable compact linearly ordered topological space is second countable if it is homeomorphic to a subspace of \\$C_p(X)\\$ where \\$X\\$ is Lindelöf. Other counterexamples for the problem are also given by making use of the Cantor tree. In addition, we remark that every separable supercompact space is first countable if it is homeomorphic to a subspace of \\$C_p(X)\\$ where \\$X\\$ is Lindelöf.\nReferences:\n Arkhangel'skii A.V.: Problems in \\$C_p\\$-theory. in: J. van Mill and G.M. Reed, Eds., {Open Problems in Topology}, North-Holland, 1990, 601-615. MR 1078667 | Zbl 0994.54020\n Engelking R.: General Topology. Sigma Series in Pure Math. 6, Helderman Verlag, Berlin, 1989. MR 1039321 | Zbl 0684.54001\n Lutzer D.J.: On generalized ordered spaces. Dissertationes Math. 89 (1971). MR 0324668 | Zbl 0228.54026\n Mill J. van: Supercompactness and Wallman spaces. Mathematical Centre Tracts 85 (1977). MR 0464160\n Mill J. van, Mills C.F.: On the character of supercompact spaces. Top. Proceed. 3 (1978), 227-236. MR 0540493" ]

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[ "# Find rows of a matrix corresponds to accumarray results\n\n3 ビュー (過去 30 日間)\nMaii 2020 年 2 月 24 日\n\nHi\nI have a matrix with 4 columns.\nfor rows that the 1st column is the same , check the 4th column and keep the row with maximum 4th column value.\nfor example\nA=[ 4 5 6 7; 3 6 7 0 ; 4 1 9 3;1 2 3 4 ; 7 1 5 6; 3 1 4 3]\nand I like to have output as\nout=[ 4 5 6 7 ;1 2 3 4 ; 7 1 5 6;3 1 4 3]\nI used accumarray and it gave me [7 4 6 3], but I dont know how to call the rows that their 4th column values are the same.\njust note that, In my real code the number of rows of matrix A will be different in each iteration and I should run code later for 10000 times.\n\n#### 0 件のコメント\n\nサインイン to comment.\n\n### 回答 (1 件)\n\nJon 2020 年 2 月 24 日\n\nI was having a little difficulty understanding the description of your problem but I think this accomplishes what you are trying to do.\n% generate test data\nA = randi(5,100,4); % random matrix with max value of 5\n% find the unique column one values\nuniqueVals = unique(A(:,1));\n% loop through unique values\nnumUnique = length(uniqueVals); % number of unique values\nout = zeros(numUnique,4); % preallocate array to hold results\nfor k = 1:numUnique\n% find rows where this unique value occurs\nidx = find(A(:,1)== uniqueVals(k));\n% for the rows with this unique value in first column find the one\n% with the maximum value in the last column and just keep that row\n[~,imax] = max(A(idx,4));\n% just keep that row\nout(k,:) = A(idx(imax),:);\nend\nNote that if there is more than one row that has the same element in the first column and the same maximum value in the fourth column then the answer returned in the out array will not be unique.\n\n#### 0 件のコメント\n\nサインイン to comment.\n\nサインイン してこの質問に回答します。" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Re: Hypergeometric2F1\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg93197] Re: [mg93183] Re: Hypergeometric2F1\n• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>\n• Date: Fri, 31 Oct 2008 03:03:51 -0500 (EST)\n• References: <[email protected]>\n\n```On 30 Oct 2008, at 16:02, Bill Rowe wrote:\n\n> On 10/29/08 at 5:50 AM, DWCantrell at sigmaxi.net (David W. Cantrell)\n> wrote:\n>\n>> Artur <grafix at csl.pl> wrote:\n>>> Dear Mathematica Gurus! Who know which Mathematica procedure to use\n>>> to find such a,b,c that\n>>> ArcCosh/ArcCosh[2-x]==Hypergeometric2F1[a,b,c,x] for\n>>> {x,-Infinity,1}\n>\n>> It seems that you are wanting to determine a,b,c such that\n>\n>> ArcCosh/ArcCosh[2-x]==Hypergeometric2F1[a,b,c,x]\n>\n>> would be an identity for x < 1. But that is not possible. What made\n>> you think it would be possible?\n>\n> Given\n>\n> In:= Hypergeometric2F1[a, b, c, 0]\n>\n> Out= 1\n>\n> It appears there are infinitely many solutions. Is the result\n> returned by Mathematica for Hypergeometric2F1[a,b,c,x] incorrect\n> when x = 0? What am I missing here?\n>\n\nIt's a matter of mind reading ;-)\n\nThe questioner \"obviously meant\" \"for all x<1\".\n\nAndrzej Kozlowski\n\n```\n\n• Prev by Date: Re: Re: Is there a way to make Mathematica commands and functions\n• Next by Date: Re: I can't evaluate cells in Mathematica\n• Previous by thread: Re: Hypergeometric2F1\n• Next by thread: Re: Re: Hypergeometric2F1" ]

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[ "# 闭包的起源\n\n### 为什么要有闭包？\n\n#### 闭包的定义\n\n``````class A {\nint sum = 0;\n\nint sum(int a) {\nreturn a = a + sum;\n}\n}\n\n#### 闭包的由来\n\n##### Lambda演算\n\nλ演算（英语：lambda calculus，λ-calculus）是一套从数学逻辑中发展，以变量绑定和替换的规则，来研究函数如何抽象化定义、函数如何被应用以及递归的形式系统。\n\n1. Lambda演算的基本定义：\n\nλx. E，x是参数，并且有且仅有一个参数，E是函数体。\n\n2. 函数的应用：\n\nE1 E2，E1是个函数，E2也是个函数，并且每个函数都有返回值，E2的返回值当成λx. E的x，带人到E1的函数中，在返回E1的结果。\n\n1. Lambda演算的基本定义：\n\nλx. x + 2，x是参数，x + 2函数体。\n\n2. 函数的应用：\n\n当有个参数λa.a 3的时候就是这个样子(λx. x + 2) (λa.a 3)，可以写成也就符合上面的E1 E2格式。结果就是(λx. x + 2)3 = 3 + 2 = 5\n\n``````returnValue funcitonName(parameter){\nmethodBody\n}\n\n##### Lambda演算的加法问题\n\nLambda演算只支持一个参数，我想计算f(x, y) = x + y怎么算呢，λx y. x + y，这样是违反规则的，不要着急我们可以采用Currying的方式，λx.λy.x + y，调用的时候是这样的（λx.λy.x + y) (7 2) = （λy.7 + y)(2) = (7 + 2) = 9。想一下程序怎样写。\n\n``````function add(x){\nreturn (function(y) {\nreturn x+y;\n});\n}" ]

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[ "# Recursion: Thinking Recursively\n\nThis lesson has one primary goal: to show that the thought process followed in writing the `sum` function follows a common recursive programming “pattern.” Indeed, when you write recursive functions you’ll generally follow the three-step process shown in this lesson.\n\nI don’t want to make this too formulaic, but the reality is that if you follow these three steps in your thinking, it will make it easier to write recursive functions, especially when you first start.\n\nAs I mentioned in the previous lessons, when I sit down to write a recursive function, I think of three things:\n\n• What is the function signature?\n\n• What is the end condition for this algorithm?\n\n• What is the actual algorithm? For example, if I’m processing all of the elements in a `List`, what does my algorithm do when the function receives a non-empty `List`?\n\nLet’s take a deep dive into each step in the process to make more sense of these descriptions.\n\nOnce I know that I’m going to write a recursive function, the first thing I ask myself is, “What is the signature of this function?”\n\nIf you can describe the function verbally, you should find that you know (a) the parameters that will be passed into the function and (b) what the function will return. In fact, if you don’t know these things, you’re probably not ready to write the function yet.\n\nIn the `sum` function the algorithm is to add all of the integers in a given list together to return a single integer result. Therefore, because I know the function takes a list of integers as its input, I can start sketching the function signature like this:\n\n``````def sum(list: List[Int]) ...\n``````\n\nBecause the description also tells me that the function returns an `Int` result, I add the function’s return type:\n\n``````def sum(list: List[Int]): Int = ???\n``````\n\nThis is the Scala way to say that “the `sum` function takes a list of integers and returns an integer result,” which is what I want. In FP, sketching the function signature is often half of the battle, so this is actually a big step.\n\nThe next thing I usually think about is, “How will this algorithm end? What is its end condition?”\n\nBecause a recursive function like `sum` keeps calling itself over and over, it’s of the utmost importance that there is an end case. If a recursive algorithm doesn’t have an end condition, it will keep calling itself as fast as possible until either (a) your program crashes with a `StackOverflowError`, or (b) your computer’s CPU gets extraordinarily hot. Therefore, I offer this tip:\n\nAlways have an end condition, and write it as soon as possible.\n\nIn the `sum` algorithm you know that you have a `List`, and you want to march through the entire `List` to add up the values of all of its elements. You may not know it at this point in your recursive programming career, but right away this statement is a big hint about the end condition. Because (a) you know that you’re working with a `List`, (b) you want to operate on the entire `List`, and (c) a `List` ends with the `Nil` element, (d) you can begin to write the end condition `case` expression like this:\n\n``````case Nil => ???\n``````\n\nTo be clear, this end condition is correct because you’re working with a `List`, and you know that the algorithm will operate on the entire `List`. Because the `Nil` element is to a `List` as a caboose is to a train, you’re guaranteed that it’s always the last element of the `List`.\n\nNote: If your algorithm will not work on the entire `List`, the end condition will be different than this.\n\nNow the next question is, “What should this end condition return?”\n\nA key here is that the function signature states that it returns an `Int`. Therefore, you know that this end condition must return an `Int` of some sort. But what `Int`? Because this is a “sum” algorithm, you also know that you don’t want to return anything that will affect the sum. Hmmm … what `Int` can you return when the `Nil` element is reached that won’t affect the sum?\n\nThe answer is `0`.\n\n(More on this shortly.)\n\nGiven that answer, I can update the first `case` condition:\n\n``````def sum(list: List[Int]): Int = list match {\ncase Nil => 0\ncase ???\n}\n``````\n\nThat condition states that if the function receives an empty `List` — denoted by `Nil` — the function will return `0`.\n\nNow we’re ready for the third step.\n\nI’ll expand more on the point of returning `0` in this algorithm in the coming lessons, but for now it may help to know that there’s a mathematical theory involved in this decision. What’s happening here is that you’re returning something known as an “identity” element for the current data set and algorithm. As a quick demonstration of what I’m talking about, here are a few other identity elements for different data sets and algorithms:\n\n1) Imagine that you want to write a “product” algorithm for a list of integers. What would you return for the end condition in this case? The correct answer is `1`. This is because the product involves multiplying all elements of the list, and multiplying any number by `1` gives you the original number, so this doesn’t affect the final result in any way.\n\n2) Imagine that you’re writing a concatenation algorithm for a `List[String]`. What would you return for the end condition in this case? The correct answer is ``, an empty`String` (because once again, it does not affect the final result).\n\nNow that you’ve defined the function signature and the end condition, the final question is, “What is the algorithm at hand?”\n\nWhen your algorithm will operate on all of the elements in a `List` and the first `case` condition handles the “empty list” case, this question becomes, “What should my function do when it receives a non-empty `List`?”\n\nThe answer for a “sum” function is that it should add all of the elements in the list. (Similarly, the answer for a “product” algorithm is that it should multiply all of the list elements.)\n\nAt this point I go back to the original statement of the `sum` algorithm:\n\n“The sum of a list of integers is the sum of the head element, plus the sum of the tail elements.”\n\nBecause the first `case` expression handles the “empty list” case, you know that the second `case` condition should handle the case of the non-empty list. A common way to write the pattern for this `case` expression is this:\n\n``````case head :: tail => ???\n``````\n\nThis pattern says, “`head` will be bound to the value of the first element in the `List`, and `tail` will contain all of the remaining elements in the `List`.”\n\nBecause my description of the algorithm states that the sum is “the sum of the head element, plus the sum of the tail elements,” I start to write a `case` expression, starting by adding the head element:\n\n``````case head :: tail => head + ???\n``````\n\nand then I write this code to represent “the sum of the tail elements”:\n\n``````case head :: tail => head + sum(tail)\n``````\n\nThat is a Scala/FP recursive way of expressing the thought, “The sum of a list of integers is the sum of the head element, plus the sum of the tail elements.”\n\n(I described that thought process in detail in the previous lessons, so I won’t repeat all of that thought process here.)\n\nNow that we have the function signature, the end condition, and the main algorithm, we have the completed function:\n\n``````def sum(list: List[Int]): Int = list match {\ncase Nil => 0\n}\n``````\n\nAs I noted in the previous lessons, when FP developers work with lists, they often prefer to use the variable name `x` to refer to a single element and `xs` to refer to multiple elements, so this function is more commonly written with these variable names:\n\n``````def sum(list: List[Int]): Int = list match {\ncase Nil => 0\ncase x :: xs => x + sum(xs)\n}\n``````\n\n(But you don’t have to use those names; use whatever is easiest for you to read.)\n\nIn practice, the first step — sketching the function signature — is almost always the first step in the process. As I mentioned, it’s hard to write a function if you don’t know what the inputs and output will be.\n\nBut the last two steps — defining the end condition, and writing the algorithm — are interchangeable, and even iterative. For instance, if you’re working on a `List` and you want to do something for every element in the list, you know the end condition will occur when you reach the `Nil` element. But if you’re not going to operate on the entire list, or if you’re working with something other than a `List`, it can help to bounce back and forth between the end case and the main algorithm until you come to the solution.\n\nNote that the `sum` algorithm I’ve shown specifically works on a Scala `List`, which ends with a `Nil` element. It will not work with other sequences like `Vector`, `ArrayBuffer`, `ListBuffer`, or other sequences that do not have a `Nil` value as the last element in the sequence. I discuss the handling of those other sequences later in the book.\n\nWhen I sit down to write a recursive function, I generally think of three things:\n\n• What is the function signature?\n\n• What is the end condition for this algorithm?\n\n• What is the main algorithm?\n\nTo solve the problem I almost always write the function signature first, and after that I usually write the end condition next, though the last two steps can also be an iterative process.\n\nNow that you’ve seen this “general pattern” of writing recursive functions, the next two lessons are exercises that give you a taste of how to use the patterns to write your own recursive functions.\n\nFirst, I’ll have you write another recursive function to operate on all of the elements in a `List`, and then you’ll work on a recursive algorithm that operates on only a subset of a `List`." ]

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[ "", null, "CMSIS-DSP  Version 1.9.0 CMSIS DSP Software Library\nBayesian estimators\n\n## Functions\n\nuint32_t arm_gaussian_naive_bayes_predict_f16 (const arm_gaussian_naive_bayes_instance_f16 *S, const float16_t *in, float16_t *pOutputProbabilities, float16_t *pBufferB)\nNaive Gaussian Bayesian Estimator. More...\n\nuint32_t arm_gaussian_naive_bayes_predict_f32 (const arm_gaussian_naive_bayes_instance_f32 *S, const float32_t *in, float32_t *pOutputProbabilities, float32_t *pBufferB)\nNaive Gaussian Bayesian Estimator. More...\n\n## Description\n\nImplement the naive gaussian Bayes estimator. The training must be done from scikit-learn.\n\nThe parameters can be easily generated from the scikit-learn object. Some examples are given in DSP/Testing/PatternGeneration/Bayes.py\n\n## Function Documentation\n\n uint32_t arm_gaussian_naive_bayes_predict_f16 ( const arm_gaussian_naive_bayes_instance_f16 * S, const float16_t * in, float16_t * pOutputProbabilities, float16_t * pBufferB )\nParameters\n [in] *S points to a naive bayes instance structure [in] *in points to the elements of the input vector. [out] *pOutputProbabilities points to a buffer of length numberOfClasses containing estimated probabilities [out] *pBufferB points to a temporary buffer of length numberOfClasses\nReturns\nThe predicted class\n uint32_t arm_gaussian_naive_bayes_predict_f32 ( const arm_gaussian_naive_bayes_instance_f32 * S, const float32_t * in, float32_t * pOutputProbabilities, float32_t * pBufferB )\nParameters\n [in] *S points to a naive bayes instance structure [in] *in points to the elements of the input vector. [out] *pOutputProbabilities points to a buffer of length numberOfClasses containing estimated probabilities [out] *pBufferB points to a temporary buffer of length numberOfClasses\nReturns\nThe predicted class" ]

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[ "ORANGIAUSA\n\nVector Collection Storage", null, "", null, "", null, "", null, "", null, "Write Vector Polar Cartesian Form Find Magnitude Direction Resultant Vector Q\n\nThis post categorized under Vector and posted on January 10th, 2019.", null, "This Write Vector Polar Cartesian Form Find Magnitude Direction Resultant Vector Q has 2046 x 1016 pixel resolution with jpeg format. How To Find The Resultant Vector Of 3 Vectors, How To Find Direction Of A Vector, Resultant Of Two Vectors Formula, How To Find The Magnitude Of The Resultant Of Three Vectors, Vector Addition Formula, Magnitude And Direction Of Resultant Vector, How To Find The Magnitude Of A Vector With 3 Components, Magnitude Of Two Vectors, Resultant Of Two Vectors Formula, Vector Addition Formula, How To Find The Magnitude Of A Vector With 3 Components, Finding The Resultant Vector, Calculating Angles Of Resultant Vectors, Resultant Of Three Vectors, Angle Of Resultant Vector, Magnitude Of The Resultant, Resultant Angle Formula was related topic with this Write Vector Polar Cartesian Form Find Magnitude Direction Resultant Vector Q. You can download the Write Vector Polar Cartesian Form Find Magnitude Direction Resultant Vector Q picture by right click your mouse and save from your browser.\n\nMechanics of solids - Basic principles In addressing any problem in continuum or solid mechanics three factors must be considered (1) the Newtonian equations of motion in the more general form recognized by Euler expressing conservation of linear and angular momentum for finite bodies (rather than just for point particles) and the related concept of stress as formalized by Cauchy (2 Gives control of the alphamatte channel of an image. Used to set a flag on an image indicating whether or not to use existing alpha channel data to create an alpha channel or to perform other operations on the alpha channel.Some notes and solutions to Russell and Norvigs Artificial Intelligence A Modern Approach (AIMA 3rd edition)\n\nQUESTION I am confused if I put a book from ground to a table at height h note that the book is at rest before and after putting total work done on book is 0 as change in k.e.With the recent publication of PHYSICS IS there are now three Ask the Physicist books Click on the book images below for information on the content of the books and for information on ordering.Four-vector Derivations of relativity graphicetime diagrams Differential geometry Curved graphicetime Mathematics of general relativity graphicetime topology\n\nAbelian In abstract algebra a commutative group or ring. Absolute In the study of logic something observed similarly by most observers or something agreed upon or which has the same value each time measured.Something not in dispute unarguable and independent of other state.As opposed to contextual. AC Alternating Current Electrical power which repeatedly reverses direction of flow.is and in to a was not you i of it the be he his but for are this that by on at they with which she or from had we will have an what been one if would who has her" ]

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[ "43oh\n\nPutting a While Loop within a While Loops\n\nRecommended Posts\n\nI'm having an issue with Energia where it isnt allowing me to put a while loops within another while loops or any loop for that matter. I need my program to run a loop for as long as the source is correct but run another loop inside of it a specific number of times. It always gives me an error asking for a \";\" before a numeric value. Here is my code.\n\nconst int voltPin = P1_5;\nconst int voltoutPin = P1_7;\nconst int voltoutPin2 = P1_6;\nfloat voltage1;\nfloat voltage2;\nfloat voltage3;\n\nvoid setup() {\n\npinMode(voltoutPin, OUTPUT);\npinMode(voltoutPin2, OUTPUT);\npinMode(voltPin, INPUT);\nanalogFrequency(10000);\n\n}\n\nvoid loop() {\nwhile(voltage3 >= 12){\n}\n}\n}\nwhile(voltage3 < 12){\n}\n}\n}\n\nfloat voltage;\n\nvoltage2 = (voltage1 / 1024) * 3.3;\n\nvoltage3 = voltage2 * 5.65;\n\n} // void loop close\n\nLet me know how that I can imbed a loop within another loop or just a better way to do this function. Essentially I'm reading a voltage then when the voltage reaches a certain limit I start to cutoff the voltage source by cutting off the PWM. When I cut it off instantly it shut down the whole system, probably from a voltage surge. So I want it now to ramp it down to avoid that.\n\nShare on other sites\n\nwhat is \"++10\" or \"--10\"?  never seen that notation\n\nusually it's += 10 or -= 10\n\nalso can you please indent the code correctly?  It's hard to ascertain what right-braces \"}\" corresponds to which loop the way it's formatted here.\n\nShare on other sites\n\nHere's my updated code, it should make more sense.  Instead of trying to do a while within a while, I just changed it so that the values are constrained. It isnt working the way it should though. The output will go to 100% duty cycle but when It is suppose to fade out it just starts going haywire. I don't think its fading to 0 for some reason. Look at this updated code and let me know what can be done. Thanks.\n\nconst int voltPin = P1_5;\nconst int voltoutPin = P1_7;\nconst int voltoutPin2 = P1_6;\nfloat voltage1;\nfloat voltage2;\nfloat voltage3;\n\nvoid setup() {\n\npinMode(voltoutPin, OUTPUT);\npinMode(voltoutPin2, OUTPUT);\npinMode(voltPin, INPUT);\nanalogFrequency(10000);\n\n}\n\nvoid loop() {\n\n//Obtain RAW voltage data\nvoltage2 = (voltage1 / 1024) * 3.3;\nvoltage3 = voltage2 * 5.65;\n\n//If the priority is high\nwhile(voltage3 >= 12) {\nvoltage2 = (voltage1 / 1024) * 3.3;\nvoltage3 = voltage2 * 5.65;\n}\n\n// If priority is low\nwhile(voltage3 < 12) {\nvoltage2 = (voltage1 / 1024) * 3.3;\nvoltage3 = voltage2 * 5.65;\n}\n\n}\n\nShare on other sites\n\nWell I don't see anything glaring wrong with the code (edit: other than using float's, although if it doesn't need blistering performance that might still be OK), but I should add if this is MSP430 there's a glitch in the way analogWrite() works where it will do short-cut updates to the PWM registers (without waiting for the period to reset) when you repeatedly analogWrite a port.  The result is glitchy garbage.  What chip (or launchpad) are you using?\n\nShare on other sites\n\n@@Derekspeegle - maybe...:\n\nconst int voltPin = P1_5;\nconst int voltoutPin = P1_7;\nconst int voltoutPin2 = P1_6;\nfloat voltage1;\nfloat voltage2;\nfloat voltage3;\n\n// get voltages function\nvoid getVoltages() {\n//Obtain RAW voltage data\nvoltage2 = (voltage1 / 1024) * 3.3;\nvoltage3 = voltage2 * 5.65;\n}\n\nvoid setup() {\n\n// configure input and output pins as required:\npinMode(voltPin, INPUT);\npinMode(voltoutPin, OUTPUT);\npinMode(voltoutPin2, OUTPUT);\n\n// configure analog/PWM output frequency:\nanalogFrequency(10000);\n\n// get our voltages\ngetVoltages();\n\n// set initial fade values based on voltage3 priority\n// may need some work + additional math so initial values scale properly\nif (voltage3 >=12) { // high priority\nfadeValue = 0; // will start at 0 and count up\nfadeValue2 = 255; // will start at 255 and count down\n}\nelse { // low priority\nfadeValue = 255; // will start at 255 and count down\nfadeValue2 = 0; // will start at 0 and count up\n}\n}\n\nvoid loop() {\n// initialized in setup() so we don't need to get them again, at the moment.\n// will have been re-calculated within loop(), so all we need to do is get\n// voltage3 before we loop() again, so fadeValue and fadeValue2 can be\n// re-calculated before the outputs are changed.\n\n// Decide what to do based on voltage3 priority:\nif (voltage3 >= 12) { // priority is high\n\n}\nelse { // priority is low\n\n}\n\n// update voltout pins:\n\n// get our voltages for the next pass through loop()\ngetVoltages();\n}\nShare on other sites\n\nwhat is \"++10\" or \"--10\"?  never seen that notation\n\nusually it's += 10 or -= 10\n\nalso can you please indent the code correctly?  It's hard to ascertain what right-braces \"}\" corresponds to which loop the way it's formatted here.\n\nshould be interpreted as\n\nAlas, the result of this equation is not assigned to anything and thus discarded. A compiler might even optimize it away. So, it is valid syntax, but totally not what it's meant to do.\n\nfloat voltage1, voltage2, voltage3;\n\nvoltage2 = (voltage1 / 1024) * 3.3;\nvoltage3 = voltage2 * 5.65;\n\n// could better be\n\nvoltage = analogRead(voltPin) / 1024.0 * 3.3 * 5.65;\n\n// since now the compiler can pre-calculate 1024.0^-1 * 3.3 * 5.65,\n// resulting in something like voltage = analogRead(voltPin) * 0.0182080078125;\n\n// To make your code much more speedy, consider that\n// 1/55 = 0.01818... which is very close to 0.0182080078125 or only 0.144% off\n// So you could write voltage = analogRead(voltPin) / 55;\n// This would suffer from integer rounding, but it might be acceptable.\n\nint voltage;\nShare on other sites\nvoid loop()\n{\n//Obtain RAW voltage data\n\n//If the priority is high\nwhile(voltage >= 12)\n{\n}\n\n// If priority is low\nwhile(voltage < 12)\n{\n}\n}\n\nI cleaned up the spacing in your code, as you can see, the only difference between the two blocks is the sign of the changes to fadeValue and fadeValue2. You could consider moving the rest of the code to a common place. Or, to move the constrain() call into the update of fadeValue(2).\n\nvoid loop()\n{\n//Obtain RAW voltage data\n\n//If the priority is high\nwhile(voltage >= 12)\n{\n}\n\n// If priority is low\nwhile(voltage < 12)\n{\n}\n}\n\nAlso, I guess you're trying make some kind of fade in/out function, which is not what you're really doing. The while loops will be so fast that there is no real fading. The gross result will be that the outputs will drift to some mean point, or maybe oscillate a little around that value at a very high frequency. I assumed in this that voltoutPin and voltPin are tied together, an analog lagging filter would of course result in some kind of slower sinoid signal.\n\nIt triggers me that your two cases seem to be exactly opposite in behavior. You might consider removing those while loops alltogether, since loop() is a loop itself.\n\nvoid loop()\n{\n//Obtain RAW voltage data\n\nif(voltage >= 12) // high priority\n{\n}\nelse // low priority\n{\n}\n}\n\nThen, in setup() you set either fadeValue or fadeValue2 to 255, the other to 0. Why not do that calculation on the spot, since you'd always have the sum of fadeValue and fadeValue2 be 255.\n\nvoid loop()\n{\n//Obtain RAW voltage data\n\nif(voltage >= 12) // high priority\nelse // low priority\n}\n\nAnd lastly, I'd like to introduce you to the ternary operator, use it with care, also it might make your code a little more obscure.\n\nThe ternary operator is written like this\n\n(condition) ? (true-value) : (false-value)\n\nIn contrast to the if() ... else ... construction, the ternary operator returns a value, it is not intended to run statements. Example:\n\nfood = (animal == monkey) ? banana : grass;\n\nso if the animal variable is monkey, food becomes banana, otherwise it becomes grass.\n\nAnother example, now with real values\n\nabs_val = (val > 0) ? val : -val;\n\nThis will check is val is positive, then abs_val becomes that val, otherwise abs_val becomes the negative of val, and the negative of a non-positive number is a positive number itself. So now we have abs_val containing the absolute value of val.\n\nIn your code, you either add 10 or subtract 10 depending on a condition (voltage >= 12), adding -10 is the same as subtracting 10, so we could either add 10 or -10 depending on your condition:\n\nfadeValue = constrain(fadeValue + ((voltage >= 12) ? 10 : -10), 0, 255);\n\nNote that for readabilities sake I put the ternary operator between braces, though that's not necessary.\n\nvoid loop()\n{\n//Obtain RAW voltage data\n\nfadeValue = constrain(fadeValue + (voltage >= 12 ? 10 : -10), 0, 255);\n}\n\nYou see that the code might become a little harder to read, also it does not speed your code up; the compiler will do something very similar to what it already did.\n\nNote that voltage is now only used once, so you could slide that in too, but your code would become even more obscure and it would not speed anything up either.\n\nvoid loop()\n{\n//Obtain RAW voltage data\n\nfadeValue += (voltage >= 12 ? 10 : -10);\n}\nShare on other sites\n\nLike the OP code, your code's fadeValue and fadeValue2 do not have any value assigned before entering loop() and being accessed in constrain() functions so are undefined.\n\nvoid loop()\n{\n//Obtain RAW voltage data\n\n//If the priority is high\nwhile(voltage >= 12)\n{\n}\n\n// If priority is low\nwhile(voltage < 12)\n{\n}\n}\n\nShare on other sites\n\nLike the OP code, your code's fadeValue and fadeValue2 do not have any value assigned before entering loop() and being accessed in constrain() functions so are undefined.\n\nThat is not my code, that's OP's code, but with nice indentation", null, "As I said directly below that code block:\n\nI cleaned up the spacing in your code\n\nEventually the values of OP's code would settle to 0 and 255 or mean out to anything that's a sum of those. Since I later optimize fadeValue2 away, the only thing undefined is fadeValue, but that's limited to the 0 to 255 range because of the constrain() call. This means the only thing undefined is where in the settling/oscillation loop the sketch begins, which is okay-ish.\n\nA little more speedy, based on my latest code example.\n\nAs you know you can haul values to the other side of the equation\n\na / b = c\n\na = c * b\n\nYou can do this too to the voltage variable, allowing for you to get rid of the float variable alltogether:\n\nvoid loop()\n{\n//Obtain RAW voltage data\n\nfadeValue += (voltage >= 12 * 55.0 ? 10 : -10);\n}\n\nSee how I moved the / 55.0 to the other side of the equation in the ternary operator.\n\nNow since voltage is always an integer value (analogRead gives us an integer value), we can know for sure that the float behaviour is not required:\n\nint voltage;\n\nvoid loop()\n{\n//Obtain RAW voltage data\n\nfadeValue += (voltage >= 12 * 55 ? 10 : -10);\n}\n\nsee how I changes 55.0 to 55, no float constants, so no float calculations.\n\nNow consider that 55 is a bit magic, (as is 12), consider moving those to constants.\n\n#define Threshold (12)\n#define VoltScale (55) /* 5.65 times the voltage at the analog pin ( 3.3 * 5.65 / 1024) */\nint voltage;\n\nvoid loop()\n{\n//Obtain RAW voltage data\n\nfadeValue += (voltage >= Threshold * VoltScale ? 10 : -10);\n}\n\nWell, now there is really no good use for the voltage temporary variable:\n\n#define Threshold (12)\n#define VoltScale (55) /* 5.65 times the voltage at the analog pin; 1/(3.3 * 5.65 / 1024) */\n\nvoid loop()\n{\n}\nShare on other sites\n\nWow, holy cow. Thank you guys for your amazing input. I have a lot to work through now, I will return and let you guys know how it all went. Thanks a ton. Do you guys think you could look over my other post, ADC with the MSP430? That one is still unanswered. Also, to answer spirilis's questions, I am using the mps430fr4133, can you elaborate more on the problem with analogwrite? I'm using it for another project which involves using analogwrite for rgb values and ive noticed the glitching. Is there a work around?\n\nShare on other sites\n\nWow, holy cow. Thank you guys for your amazing input. I have a lot to work through now, I will return and let you guys know how it all went. Thanks a ton. Do you guys think you could look over my other post, ADC with the MSP430? That one is still unanswered. Also, to answer spirilis's questions, I am using the mps430fr4133, can you elaborate more on the problem with analogwrite? I'm using it for another project which involves using analogwrite for rgb values and ive noticed the glitching. Is there a work around?\n\n@@spirilis or @@energia are the ones most able to address the glitches.\nShare on other sites\n• 2 weeks later...\n\nJoin the conversation\n\nYou can post now and register later. If you have an account, sign in now to post with your account.", null, "×   Pasted as rich text.   Paste as plain text instead\n\nOnly 75 emoji are allowed.\n\n×   Your previous content has been restored.   Clear editor\n\n×   You cannot paste images directly. Upload or insert images from URL.\n\n×\n×\n• Blog\n\n• Activity\n\n×\n• Create New..." ]

[ null, "https://forum.43oh.com/uploads/emoticons/default_wink.png", null, "https://forum.43oh.com/uploads/set_resources_2/84c1e40ea0e759e3f1505eb1788ddf3c_default_photo.png", null ]

[ "# Corresponding Angles formed by Parallel Lines and their transversal", null, "In the case of intersection of two parallel lines by their transversal, four pairs of angles have same relative position at each intersection of every two straight lines and they are known as corresponding angles of the transversal of parallel lines.\n\nLook at the picture, an angle has similarity in position with another angle. Each pair of such angles are corresponding angles and total four pairs of corresponding angles are possibly formed geometrically when a pair of parallel lines are intersected by their transversal line.\n\n## Property\n\nThe corresponding angles which have relative position are equal when two parallel lines are cut by their transversal and the property can be proved geometrically.", null, "$\\angle XPB$ and $\\angle XQD$ are two angles and they are relatively similar. Therefore, they both are called as corresponding angles. The two corresponding angles are equal geometrically when two parallel lines $\\overleftrightarrow{AB}$ and $\\overleftrightarrow{CD}$ are intersected by their transversal $\\overleftrightarrow{XY}$.\n\n$\\angle XPB = \\angle XQD$\n\nSimilarly, the corresponding angles $\\angle YPB$ and $\\angle YQD$ are also equal geometrically.\n\n$\\angle YPB = \\angle YQD$", null, "The corresponding angles $\\angle YQC$ and $\\angle YPA$ are equal because of similarity in position.\n\n$\\angle YQC = \\angle YPA$\n\nFinally, the pair of corresponding angles $\\angle XQC$ and $\\angle XPA$ are also equal.\n\n$\\angle XQC = \\angle XPA$\n\nTherefore, the animation tutorial has proved that each pair of corresponding angles are equal geometrically when two parallel lines are cut by their transversal line.\n\nLatest Math Topics\nJun 26, 2023\nJun 23, 2023\n\nLatest Math Problems\nJul 01, 2023\nJun 25, 2023\n###### Math Questions\n\nThe math problems with solutions to learn how to solve a problem.\n\nLearn solutions\n\nPractice now\n\n###### Math Videos\n\nThe math videos tutorials with visual graphics to learn every concept.\n\nWatch now\n\n###### Subscribe us\n\nGet the latest math updates from the Math Doubts by subscribing us." ]

[ null, "https://www.mathdoubts.com/pics/transversal/corresponding-angles-parallel-lines.gif", null, "https://www.mathdoubts.com/pics/transversal/corresponding-angles-parallel-right.gif", null, "https://www.mathdoubts.com/pics/transversal/corresponding-angles-parallel-left.gif", null ]

[ "Heart of Algebra\nPassport to Advanced Mathematics\nProblem Solving & Data Analysis\nAdditional Topics in Math\nNo Calculator Allowed!\n100\nIf a/b = 2, what is the value of 4b/a?\nWhat is 2?\n100\nIf 3x − y = 12, what is the value of 8^x/2^y? a. 2^12 b. 4^4 c. 8^2 d. It cannot be determined from information given.\nWhat is a?\n100\nIf y = kx, where k is a constant, and y = 24 when x = 6, what is the value of y when x = 5 ?\nWhat is 20?\n100\nwhat is the sum (7 + 3i) + (−8 + 9i) ?\nWhat is -1+12i?\n100\nIf (x− 1)/3=k and k = 3, what is the value of x ?\nWhat is 10?\n200\nA line in the xy-plane passes through the origin and has a slope of 1/7. Which of the following points lies on the line? A) (0, 7) B) (1, 7) C) (7, 7) D) (14, 2)\nWhat is D?\n200\nIf (ax + 2)(bx + 7) = 15x^2 + cx + 14 for all values of x, and a + b = 8, what are the two possible values for c ? A) 3 and 5 B) 6 and 35 C) 10 and 21 D) 31 and 41\nWhat is D?\n200\nAlma bought a laptop computer at a store that gave a 20 percent discount off its original price. The total amount she paid to the cashier was p dollars, including an 8 percent sales tax on the discounted price. Which of the following represents the original price of the computer in terms of p ?\nWhat is p/( .8 * 1.08)?\n200\nIn a right triangle, one angle measures x°, where sin x° =4/5. What is cos(90° − x°) ?\nWhat is 4/5?\n200\nOn Saturday afternoon, Armand sent m text messages each hour for 5 hours, and Tyrone sent p text messages each hour for 4 hours. Which of the following represents the total number of messages sent by Armand and Tyrone on Saturday afternoon?\nWhat is 5m+4p?\n300\nThe posted weight limit for a covered wooden bridge in Pennsylvania is 6000 pounds. A delivery truck that is carrying x identical boxes each weighing 14 pounds will pass over the bridge. If the combined weight of the empty delivery truck and its driver is 4500 pounds, what is the maximum possible value for x that will keep the combined weight of the truck, driver, and boxes below the bridge’s posted weight limit?\nWhat is 107?\n300\nh = − 4.9t^2 + 25t The equation above expresses the approximate height h, in meters, of a ball t seconds after it is launched vertically upward from the ground with an initial velocity of 25 meters per second. After approximately how many seconds will the ball hit the ground?\nWhat is 5.1?\n300\nKatarina is a botanist studying the production of pears by two types of pear trees. She noticed that Type A trees produced 20 percent more pears than Type B trees did. Based on Katarina’s observation, if the Type A trees produced 144 pears, how many pears did the Type B trees produce?\nWhat is 120?\n300\nWhich of the following is an equation of a circle in the xy-plane with center (0, 4) and a radius with endpoint (4/3, 5)?\nWhat is x^2+(y-4)^2=25/9?\n300\nKathy is a repair technician for a phone company. Each week, she receives a batch of phones that need repairs. The number of phones that she has left to fix at the end of each day can be estimated with the equation P = 108 − 23d, where P is the number of phones left and d is the number of days she has worked that week. What is the meaning of the value 108 in this equation?\nWhat is Kathy will start each week with 108 phones to fix?\n400\nWyatt can husk at least 12 dozen ears of corn per hour and at most 18 dozen ears of corn per hour. Based on this information, what is a possible amount of time, in hours, that it could take Wyatt to husk 72 dozen ears of corn?\nWhat is any number between 4-6 inclusive?\n400\nFor a polynomial p(x), the value of p(3) is −2. Which of the following must be true about p(x) ? A) x − 5 is a factor of p(x). B) x − 2 is a factor of p(x). C) x + 2 is a factor of p(x). D) The remainder when p(x) is divided by x − 3 is −2.\nWhat is D?\n400\nA local television station sells time slots for programs in 30-minute intervals. If the station operates 24 hours per day, every day of the week, what is the total number of 30-minute time slots the station can sell for Tuesday and Wednesday?\nWhat is 96?\n400\nA dairy farmer uses a storage silo that is in the shape of the right circular cylinder with height 8. If the volume of the silo is 72π cubic yards, what is the diameter of the base of the cylinder, in yards?\nWhat is 6?\n400\nh = 3a + 28.6 A pediatrician uses the model above to estimate the height h of a boy, in inches, in terms of the boy’s age a, in years, between the ages of 2 and 5. Based on the model, what is the estimated increase, in inches, of a boy’s height each year?\nWhat is 3?\n500\ny<-x+a, y>x+b In the xy-plane, if (0, 0) is a solution to the system of inequalities above, which of the following relationships between a and b must be true?\nWhat is a>b\n500\nJessica opened a bank account that earns 2 percent interest compounded annually. Her initial deposit was \\$100, and she uses the expression \\$100(x)^t to find the value of the account after t years. What is the value of x in the expression?\nWhat is 1.02?\n500\nThe graph above displays the total cost C, in dollars, of renting a boat for h hours. What does the C-intercept represent in the graph? A) The initial cost of renting the boat B) The total number of boats rented C) The total number of hours the boat is rented D) The increase in cost to rent the boat for each additional hour\nWhat is A?\n500\nA x feet D B A summer camp counselor wants to find a length, x, in feet, across a lake as represented in the sketch I draw. The lengths represented by AB, EB, BD, and CD on the sketch were determined to be 1800 feet, 1400 feet, 700 feet, and 800 feet, respectively. Segments AC and DE intersect at B, and ∠AEB and ∠CDB have the same measure. What is the value of x ?\nWhat is 1600?\n500\ng(x) = ax^2 + 24 For the function g defined above, a is a constant and g(4) = 8. What is the value of g(−4) ?\nWhat is 8?\nClick to zoom" ]

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[ "src/Pure/General/pretty.ML\n author wenzelm Wed Jan 13 15:18:02 1999 +0100 (1999-01-13) changeset 6118 caa439435666 parent 6116 8ba2f25610f7 child 6271 957d8aa4a06b permissions -rw-r--r--\nfixed titles;\n``` 1 (* Title: Pure/General/pretty.ML\n```\n``` 2 ID: \\$Id\\$\n```\n``` 3 Author: Lawrence C Paulson\n```\n``` 4 Copyright 1991 University of Cambridge\n```\n``` 5\n```\n``` 6 Generic pretty printing module.\n```\n``` 7\n```\n``` 8 Loosely based on\n```\n``` 9 D. C. Oppen, \"Pretty Printing\",\n```\n``` 10 ACM Transactions on Programming Languages and Systems (1980), 465-483.\n```\n``` 11\n```\n``` 12 The object to be printed is given as a tree with indentation and line\n```\n``` 13 breaking information. A \"break\" inserts a newline if the text until\n```\n``` 14 the next break is too long to fit on the current line. After the newline,\n```\n``` 15 text is indented to the level of the enclosing block. Normally, if a block\n```\n``` 16 is broken then all enclosing blocks will also be broken. Only \"inconsistent\n```\n``` 17 breaks\" are provided.\n```\n``` 18\n```\n``` 19 The stored length of a block is used in breakdist (to treat each inner block as\n```\n``` 20 a unit for breaking).\n```\n``` 21 *)\n```\n``` 22\n```\n``` 23 type pprint_args = (string -> unit) * (int -> unit) * (int -> unit) *\n```\n``` 24 (unit -> unit) * (unit -> unit);\n```\n``` 25\n```\n``` 26 signature PRETTY =\n```\n``` 27 sig\n```\n``` 28 type T\n```\n``` 29 val str: string -> T\n```\n``` 30 val strlen: string -> int -> T\n```\n``` 31 val sym: string -> T\n```\n``` 32 val spc: int -> T\n```\n``` 33 val brk: int -> T\n```\n``` 34 val fbrk: T\n```\n``` 35 val blk: int * T list -> T\n```\n``` 36 val lst: string * string -> T list -> T\n```\n``` 37 val quote: T -> T\n```\n``` 38 val commas: T list -> T list\n```\n``` 39 val breaks: T list -> T list\n```\n``` 40 val fbreaks: T list -> T list\n```\n``` 41 val block: T list -> T\n```\n``` 42 val strs: string list -> T\n```\n``` 43 val enclose: string -> string -> T list -> T\n```\n``` 44 val list: string -> string -> T list -> T\n```\n``` 45 val str_list: string -> string -> string list -> T\n```\n``` 46 val big_list: string -> T list -> T\n```\n``` 47 val string_of: T -> string\n```\n``` 48 val writeln: T -> unit\n```\n``` 49 val str_of: T -> string\n```\n``` 50 val pprint: T -> pprint_args -> unit\n```\n``` 51 val setdepth: int -> unit\n```\n``` 52 val setmargin: int -> unit\n```\n``` 53 end;\n```\n``` 54\n```\n``` 55 structure Pretty : PRETTY =\n```\n``` 56 struct\n```\n``` 57\n```\n``` 58 (*printing items: compound phrases, strings, and breaks*)\n```\n``` 59 datatype T =\n```\n``` 60 Block of T list * int * int | (*body, indentation, length*)\n```\n``` 61 String of string * int | (*text, length*)\n```\n``` 62 Break of bool * int; (*mandatory flag, width if not taken*);\n```\n``` 63\n```\n``` 64 (*Add the lengths of the expressions until the next Break; if no Break then\n```\n``` 65 include \"after\", to account for text following this block. *)\n```\n``` 66 fun breakdist (Block(_,_,len)::es, after) = len + breakdist(es, after)\n```\n``` 67 | breakdist (String (s, len) :: es, after) = len + breakdist (es, after)\n```\n``` 68 | breakdist (Break _ :: es, after) = 0\n```\n``` 69 | breakdist ([], after) = after;\n```\n``` 70\n```\n``` 71 fun repstring a 0 = \"\"\n```\n``` 72 | repstring a 1 = a\n```\n``` 73 | repstring a k =\n```\n``` 74 if k mod 2 = 0 then repstring(a^a) (k div 2)\n```\n``` 75 else repstring(a^a) (k div 2) ^ a;\n```\n``` 76\n```\n``` 77 (*** Type for lines of text: string, # of lines, position on line ***)\n```\n``` 78\n```\n``` 79 type text = {tx: string, nl: int, pos: int};\n```\n``` 80\n```\n``` 81 val emptytext = {tx=\"\", nl=0, pos=0};\n```\n``` 82\n```\n``` 83 fun blanks wd {tx,nl,pos} =\n```\n``` 84 {tx = tx ^ repstring \" \" wd,\n```\n``` 85 nl = nl,\n```\n``` 86 pos = pos+wd};\n```\n``` 87\n```\n``` 88 fun newline {tx,nl,pos} =\n```\n``` 89 {tx = tx ^ \"\\n\",\n```\n``` 90 nl = nl+1,\n```\n``` 91 pos = 0};\n```\n``` 92\n```\n``` 93 fun string s len {tx,nl,pos:int} =\n```\n``` 94 {tx = tx ^ s,\n```\n``` 95 nl = nl,\n```\n``` 96 pos = pos + len};\n```\n``` 97\n```\n``` 98\n```\n``` 99 (*** Formatting ***)\n```\n``` 100\n```\n``` 101 (* margin *)\n```\n``` 102\n```\n``` 103 (*example values*)\n```\n``` 104 val margin = ref 80 (*right margin, or page width*)\n```\n``` 105 and breakgain = ref 4 (*minimum added space required of a break*)\n```\n``` 106 and emergencypos = ref 40; (*position too far to right*)\n```\n``` 107\n```\n``` 108 fun setmargin m =\n```\n``` 109 (margin := m;\n```\n``` 110 breakgain := !margin div 20;\n```\n``` 111 emergencypos := !margin div 2);\n```\n``` 112\n```\n``` 113 val () = setmargin 76;\n```\n``` 114\n```\n``` 115\n```\n``` 116 (*Search for the next break (at this or higher levels) and force it to occur*)\n```\n``` 117 fun forcenext [] = []\n```\n``` 118 | forcenext (Break(_,wd) :: es) = Break(true,0) :: es\n```\n``` 119 | forcenext (e :: es) = e :: forcenext es;\n```\n``` 120\n```\n``` 121 (*es is list of expressions to print;\n```\n``` 122 blockin is the indentation of the current block;\n```\n``` 123 after is the width of the following context until next break. *)\n```\n``` 124 fun format ([], _, _) text = text\n```\n``` 125 | format (e::es, blockin, after) (text as {pos,nl,...}) =\n```\n``` 126 (case e of\n```\n``` 127 Block(bes,indent,wd) =>\n```\n``` 128 let val blockin' = (pos + indent) mod !emergencypos\n```\n``` 129 val btext = format(bes, blockin', breakdist(es,after)) text\n```\n``` 130 (*If this block was broken then force the next break.*)\n```\n``` 131 val es2 = if nl < #nl(btext) then forcenext es else es\n```\n``` 132 in format (es2,blockin,after) btext end\n```\n``` 133 | String (s, len) => format (es,blockin,after) (string s len text)\n```\n``` 134 | Break(force,wd) => (*no break if text to next break fits on this line\n```\n``` 135 or if breaking would add only breakgain to space *)\n```\n``` 136 format (es,blockin,after)\n```\n``` 137 (if not force andalso\n```\n``` 138 pos+wd <= Int.max(!margin - breakdist(es,after),\n```\n``` 139 blockin + !breakgain)\n```\n``` 140 then blanks wd text (*just insert wd blanks*)\n```\n``` 141 else blanks blockin (newline text)));\n```\n``` 142\n```\n``` 143\n```\n``` 144 (*** Exported functions to create formatting expressions ***)\n```\n``` 145\n```\n``` 146 fun length (Block (_, _, len)) = len\n```\n``` 147 | length (String (_, len)) = len\n```\n``` 148 | length (Break(_, wd)) = wd;\n```\n``` 149\n```\n``` 150 fun str s = String (s, size s);\n```\n``` 151 fun strlen s len = String (s, len);\n```\n``` 152 fun sym s = String (s, Symbol.size s);\n```\n``` 153\n```\n``` 154 fun spc n = str (repstring \" \" n);\n```\n``` 155\n```\n``` 156 fun brk wd = Break (false, wd);\n```\n``` 157 val fbrk = Break (true, 0);\n```\n``` 158\n```\n``` 159 fun blk (indent, es) =\n```\n``` 160 let\n```\n``` 161 fun sum([], k) = k\n```\n``` 162 | sum(e :: es, k) = sum (es, length e + k);\n```\n``` 163 in Block (es, indent, sum (es, 0)) end;\n```\n``` 164\n```\n``` 165 (*Join the elements of es as a comma-separated list, bracketted by lp and rp*)\n```\n``` 166 fun lst(lp,rp) es =\n```\n``` 167 let fun add(e,es) = str\",\" :: brk 1 :: e :: es;\n```\n``` 168 fun list(e :: (es as _::_)) = str lp :: e :: foldr add (es,[str rp])\n```\n``` 169 | list(e::[]) = [str lp, e, str rp]\n```\n``` 170 | list([]) = []\n```\n``` 171 in blk(size lp, list es) end;\n```\n``` 172\n```\n``` 173\n```\n``` 174 (* utils *)\n```\n``` 175\n```\n``` 176 fun quote prt =\n```\n``` 177 blk (1, [str \"\\\"\", prt, str \"\\\"\"]);\n```\n``` 178\n```\n``` 179 fun commas prts =\n```\n``` 180 flat (separate [str \",\", brk 1] (map (fn x => [x]) prts));\n```\n``` 181\n```\n``` 182 fun breaks prts = separate (brk 1) prts;\n```\n``` 183\n```\n``` 184 fun fbreaks prts = separate fbrk prts;\n```\n``` 185\n```\n``` 186 fun block prts = blk (2, prts);\n```\n``` 187\n```\n``` 188 val strs = block o breaks o (map str);\n```\n``` 189\n```\n``` 190 fun enclose lpar rpar prts =\n```\n``` 191 block (str lpar :: (prts @ [str rpar]));\n```\n``` 192\n```\n``` 193 fun list lpar rpar prts =\n```\n``` 194 enclose lpar rpar (commas prts);\n```\n``` 195\n```\n``` 196 fun str_list lpar rpar strs =\n```\n``` 197 list lpar rpar (map str strs);\n```\n``` 198\n```\n``` 199 fun big_list name prts =\n```\n``` 200 block (fbreaks (str name :: prts));\n```\n``` 201\n```\n``` 202\n```\n``` 203\n```\n``` 204 (*** Pretty printing with depth limitation ***)\n```\n``` 205\n```\n``` 206 val depth = ref 0; (*maximum depth; 0 means no limit*)\n```\n``` 207\n```\n``` 208 fun setdepth dp = (depth := dp);\n```\n``` 209\n```\n``` 210 (*Recursively prune blocks, discarding all text exceeding depth dp*)\n```\n``` 211 fun pruning dp (Block(bes,indent,wd)) =\n```\n``` 212 if dp>0 then blk(indent, map (pruning(dp-1)) bes)\n```\n``` 213 else str \"...\"\n```\n``` 214 | pruning dp e = e;\n```\n``` 215\n```\n``` 216 fun prune dp e = if dp>0 then pruning dp e else e;\n```\n``` 217\n```\n``` 218\n```\n``` 219 fun string_of e = #tx (format ([prune (!depth) e], 0, 0) emptytext);\n```\n``` 220\n```\n``` 221 val writeln = writeln o string_of;\n```\n``` 222\n```\n``` 223\n```\n``` 224 (*Create a single flat string: no line breaking*)\n```\n``` 225 fun str_of prt =\n```\n``` 226 let\n```\n``` 227 fun s_of (Block (prts, _, _)) = implode (map s_of prts)\n```\n``` 228 | s_of (String (s, _)) = s\n```\n``` 229 | s_of (Break (false, wd)) = repstring \" \" wd\n```\n``` 230 | s_of (Break (true, _)) = \" \";\n```\n``` 231 in\n```\n``` 232 s_of (prune (! depth) prt)\n```\n``` 233 end;\n```\n``` 234\n```\n``` 235 (*Part of the interface to the Poly/ML and New Jersey ML pretty printers*)\n```\n``` 236 fun pprint prt (put_str, begin_blk, put_brk, put_fbrk, end_blk) =\n```\n``` 237 let\n```\n``` 238 fun pp (Block (prts, ind, _)) = (begin_blk ind; pp_lst prts; end_blk ())\n```\n``` 239 | pp (String (s, _)) = put_str s\n```\n``` 240 | pp (Break (false, wd)) = put_brk wd\n```\n``` 241 | pp (Break (true, _)) = put_fbrk ()\n```\n``` 242 and pp_lst [] = ()\n```\n``` 243 | pp_lst (prt :: prts) = (pp prt; pp_lst prts);\n```\n``` 244 in\n```\n``` 245 pp (prune (! depth) prt)\n```\n``` 246 end;\n```\n``` 247\n```\n``` 248\n```\n``` 249 end;\n```" ]

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[ "# A short interval result for the extension of the exponential divisor function.\n\n[section]1. Introduction\n\nIn 1972, M. V. Subbarao established the definition of exponential divisor: Let n > 1 be an integer of canonical from [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. The integer [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is called an exponential divisor of n if [b.sub.i]|[a.sub.i] for every i [member of] {1, 2, ..., s}, notation: d[|.sub.e]n. By convention 1[|.sub.e]1. Besides, he also studied the mean value problem of exponential divisor function [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and obtained:\n\n[summation over (n [less than or equal to] x)][[tau].sup.(e)](n) = Ax + E(x),\n\nwhere E(x) = O([x.sup.1/2]).\n\nJ. Wu improved the result of M. V. Subbarao and obtained\n\n[summation over (n [less than or equal to] x)][[tau].sup.(e)](n) = Ax + B[x.sup.1/2] + O([x.sup.2/9]log x),\n\nwhere\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].\n\nM. V. Subbarao also proved that for any integers r, we have the estimate\n\n[summation over (n [less than or equal to] x)][([[tau].sup.(e)](n)).sup.r] ~ [A.sub.r]x,\n\nwhere\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].\n\nL. TOth [1,2] proved\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],\n\nhere [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is a polynomial of degree [2.sup.r] - 2 of t, [u.sub.r] = [2.sup.r + 1]-1/[2.sup.r + 2] + 1.\n\nSimilar to the generalization from d(n) to [d.sup.k](n), we extended [[tau].sup.(e)](n) and established a definition as follows:\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].\n\nObviously [[tau].sup.2](e)(n) = [[tau].sup.(e)](n). In this paper we studied the case of k = 3 which means to study the properties of [([[tau].sup.(e).sub.3](n)).sup.2] and obviously [([[tau].sup.(e).sub.3](n)).sup.2] is a multiplicative function. The aim of this paper is to study the short interval case of [([[tau].sup.(e).sub.3](n)).sup.2] and prove the following theorem.\n\nTheorem 1.1. If [x.sup.1/5 + 2[epsilon]] < y [less than or equal to] x, then\n\n[summation over (x < n [less than or equal to] x + y)][([[tau].sup.(e).sub.3](n)).sup.2] = [c.sub.1]y + O(yx - ii + 3/2[epsilon] + x). (1)\n\nwhere [c.sub.1] = [Res.sub.s = 1]F(s) and F(s);= [[summation].sup.[infinity].sub.n = 1] [([[tau].sup.(e).sub.3](n)).sup.2]/[n.sup.s].\n\nNotations. Throughout this paper, [epsilon] always denotes a fixed but sufficiently small positive constant. We assume that 1 [less than or equal to] a [less than or equal to] b are fixed integers, and we denote by d(a, b; k) the number of representations of k as k = [n.sub.1]a[n.sup.2]b, where [n.sub.1], [n.sup.2] are natural numbers, that is,\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],\n\nand [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] will be used freely.\n\n[section]2. Proof of the theorem\n\nIn order to prove our theorem, we need the following lemmas.\n\nLemma 2.1. Suppose s = [sigma] + it is a complex number (Rs > 1), then\n\nF(s): = [[summation].sup.[infinity].sub.n = 1][([[tau].sup.(e).sub.3](n)).sup.2]/[n.sup.s] = [zeta](s)[[zeta].sup.2](2s)/[[zeta].sup.3](5s)G(s),\n\nwhere the Dirichlet series G(s); = [[summation].sup.[infinity].sub.n = 1] g(n)/[n.sup.s][infinity] is absolutely convergent for Rs = [sigma] > 1/6.\n\nProof. Here [[tau].sup.(e).sub.3](n) is multiplicative and by Euler product formula we have for [sigma] > 1 that\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (2)\n\nNow we write G(s): = [[summation].sup.[infinity].sub.n = 1]g(n)/[n.sup.s][infinity]. It is easily seen the Dirichlet series is absolutely convergent for Rs = [sigma] > 1/6.\n\nLemma 2.2. Let k [greater than or equal to] 2 be a fixed integer, 1 < y [less than or equal to] x be large real numbers and\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].\n\nThen we have\n\nB(x, y; k, [epsilon]) [much less than] y[x.sup.-[epsilon]] + [x.sup.1/2k + 1]logx. (3)\n\nProof. This lemma is very important when studying the short interval distribution of 1-free number; see for example .\n\nLet [a.sub.1](n), [a.sup.2](n), [a.sup.3](n) and [a.sub.4](n) be arithmetic functions defined by the following Dirichlet series (for Rs > 1):\n\n[[infinity].summation over (n = 1)][a.sub.1](n)/[n.sup.s] = [zeta](s)G(s). (4)\n\n[[infinity].summation over (n = 1)][a.sub.2](n)/[n.sup.2s] = [[zeta].sup.8](2s). (5)\n\n[[infinity].summation over (n = 1)][a.sub.3](n)/[n.sup.4s] = [[zeta].sup.-9](4s). (6)\n\n[[infinity].summation over (n = 1)][a.sub.4](n)/[n.sup.5s] = [[zeta].sup.-27](5s). (7)\n\nLemma 2.3. Let [a.sub.1](n) be an arithmetic function defined by (4), then we have\n\n[[summation].sub.n [less than or equal to] x][a.sub.1](n) = Cx + O([x.sup.1/6 + [epsilon]]), (8)\n\nwhere C = [Res.sub.s = 1][zeta](s)G(s).\n\nProof. Using lemma 1.1, it is easy to see that\n\n[summation over (n [less than or equal to] x)][absolute value of g(n)] [much less than] [x.sup.1/6 + [epsilon]]\n\nTherefore from the definition of g(n) and (4), it follows that\n\n[summation over (n [less than or equal to] x)][a.sub.1](n) = [[summation].sub.mn [less than or equal to] x]g(n) (9)\n\n= [summation over (n [less than or equal to] x)]g(n)[summation over (m [less than or equal to] x/n1)] (10)\n\n= [summation over (n [less than or equal to] x)]g(n)(x/n + O(1)) (11)\n\n= Cx + O([x.sub.1/6 + [epsilon]]), (12)\n\nand C = [Res.sub.s = 1][zeta](s)G(s).\n\nNext we prove our Theorem. From lemma 2.3 and the definition of [a.sub.1](n), [a.sup.2](n), [a.sup.3](n) and [a.sub.4](n), we get\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],\n\nand\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (13)\n\nSo we have\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (14)\n\nwhere\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (15)\n\nIn view of lemma 3,\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (16)\n\nwhere [c.sub.1] = [Res.sub.s = 1]F(s).\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (17)\n\nSimilarly we have\n\n[summation over 3] [much less than] [yx.sup.- [member of] /2] + [x.sup.1/5 + 3/2[epsilon]], [summation over 4] [much less than] [yx.sup.- [member of] /2] + [x.sup.1/5 + 3/2[epsilon]]. (18)\n\nNow our theorem follows from (9)-(14).\n\nAcknowledgements. The author deeply thank the referee for this careful reading of the manuscript and many valuable suggestions.\n\nThis work is supported by Natural Science Foundation of Jiangxi province of China (Grant Nos. 2012ZBAB211001) and Natural Science Foundation of Jiangxi province of China (Grant Nos. 20132BAB2010031).\n\nReferences Laszlo Toth, An order result for the exponential divisor function, Publ. Math. Debrean. 71 (2007), No. 1-2, 165-171.\n\n Laszlo Toth, On certain arithmetic function involing exponential divisors, II. Annales Univ. Sci. Budapest. Sect. Comp., 27 (2007), 155-156.\n\n M. V. Subbarao, The theory of arithmetic functions (Proc. Conf., Western Michigan Univ., Kalamazoo, Mich., 1971), Lecture Notes in Math., Springer, Berlin, (1972), 251.\n\n W. G. Zhai, Square-free numbers as sum of two squares, Number Theory: Tradition and modernization, Spring, New York, (2006), 219-227.\n\n J. WU, Problem of exponential divisors and exponentially square-free integers, J. Theor. Nombres Bordeaux, 7(1995), No. 1, 133-141.\n\nLi Yang\n\nDepartment of Mathematics, Nanchang University\n\nNanchang, Jiangxi 330031, P. R. China\n\nE-mail: [email protected]\nAuthor:", null, "Printer friendly", null, "Cite/link", null, "Email", null, "Feedback Yang, Li Scientia Magna Report Sep 1, 2013 1319 The adjacent vertex distinguishing I-total chromatic number of ladder graph. Terminal hosoya polynomial of thorn graphs. Exponential functions Functions, Exponential Mathematical analysis" ]

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[ "# Solve Time Complexity problem using Time Hierarchy\n\nI am trying to understand Time Hierarchy. I have an example that is solvable using the rules of Time Hierarchy. I would like an explanation on how to solve so that I may understand better how to use Time Hierarchy.\n\nI need to show that the following are either equal, on inclusive of one another...\n\nTIME($2^{n^2}$) and TIME($n^3 \\times 2^n$)\n\nMy intuition tells me that...\n\nTIME($n^3 \\times 2^n$) $\\subset$ TIME($2^{n^2}$)\n\nCan someone explain how I can prove this using the rules of Time Hierarchy?\n\nYour intuition that $\\text{TIME}(n^3 \\times 2^n) \\subsetneq \\text{TIME}(2^{n^2})$ is correct.\nIt is easy to see that $2^{n^2}$ is time-constructible. Given $n$, you can compute $n^2$ in $\\log^2 n$ time, then write 100...00 with $n^2$ zeros, in time $O(n^2)$.\nThe Time-Hierarchy Theorem states that, for such time-constructible $f(n)$,\n$$\\text{TIME}(o(\\frac{f(n)}{\\log f(n)})) \\subsetneq \\text{TIME}(f(n))$$\nIt is clear that $n^3 \\times 2^n \\in o(\\frac{2^{n^2}}{\\log 2^{n^2}}) = o(\\frac{2^{n^2}}{n^2})$.\nTherefore, $\\text{TIME}(n^3 \\times 2^n) \\subseteq \\text{TIME}(o(\\frac{2^{n^2}}{\\log 2^{n^2}})) \\subsetneq \\text{TIME}(2^{n^2})$." ]

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[ "Home - 6th Grade Math Word Problem Worksheets\n\n# 6th Grade Math Word Problem Worksheets\n\nHere is the 6th Grade Math Word Problem Worksheets section. Here you will find all we have for 6th Grade Math Word Problem Worksheets. For instance there are many worksheet that you can print here, and if you want to preview the 6th Grade Math Word Problem Worksheets simply click the link or image and you will take to save page section." ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "#### *trace-indent-width*\n\n##### Summary\n\nThe amount of extra indentation in the trace output for each level of nesting.\n\nhcl\n\n2\n\n##### Description\n\n`*trace-indent-width*` is the extra amount by which the traced output for function calls is indented upon entering a deeper level of nesting (that is, a traced call from a function that is itself traced). If it is 0 then no indentation occurs.\n\n##### Example\n`CL-USER 1 > (setq *trace-indent-width* 4`\n` *max-trace-indent* 50)`\n`50`\n` `\n`CL-USER 2 > (defun quad (a b c) (- (* b b) (* 4 a c)))`\n`QUAD`\n` `\n`CL-USER 3 > (trace quad *)`\n`(QUAD *)`\n` `\n`CL-USER 4 > (quad 4 3 14)`\n`0 QUAD > ...`\n` >> A : 4`\n` >> B : 3`\n` >> C : 14`\n` 1 * > ...`\n` >> SYSTEM::ARGS : (3 3)`\n` 1 * < ...`\n` << VALUE-0 : 9`\n` 1 * > ...`\n` >> SYSTEM::ARGS : (4 4 14)`\n` 1 * < ...`\n` << VALUE-0 : 224`\n`0 QUAD < ...`\n` << VALUE-0 : -215`\n`-215`\n##### Notes\n\n`*trace-indent-width*` is an extension to Common Lisp." ]

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[ "## 高分必备！48条数学秒杀型公式和方法汇总\n\n2022-09-18 16:56 高考直通车综合 评论\n\n1.适用条件：[直线过焦点]，必有ecosA=(x-1)/(x+1)，其中A为直线与焦点所在轴夹角，是锐角。x为分离比，必须大于1。\n\n2.函数的周期性问题(记忆三个)：\n(1)若f(x)=-f(x+k)，则T=2k;\n(2)若f(x)=m/(x+k)(m不为0)，则T=2k;\n(3)若f(x)=f(x+k)+f(x-k)，则T=6k。\n\n3.关于对称问题(无数人搞不懂的问题)总结如下：\n(1)若在R上(下同)满足：f(a+x)=f(b-x)恒成立，对称轴为x=(a+b)/2;\n(2)函数y=f(a+x)与y=f(b-x)的图像关于x=(b-a)/2对称;\n(3)若f(a+x)+f(a-x)=2b，则f(x)图像关于(a，b)中心对称\n\n4.函数奇偶性：\n(1)对于属于R上的奇函数有f(0)=0;\n(2)对于含参函数，奇函数没有偶次方项，偶函数没有奇次方项\n(3)奇偶性作用不大，一般用于选择填空\n\n5.数列爆强定律：\n(1)等差数列中：S奇=na中，例如S13=13a7(13和7为下角标);\n(2)等差数列中：S(n)、S(2n)-S(n)、S(3n)-S(2n)成等差(3)等比数列中，上述2中各项在公比不为负一时成等比，在q=-1时，未必成立4，等比数列爆强公式：S(n+m)=S(m)+q²mS(n)可以迅速求q\n\n6.数列的终极利器，特征根方程。(如果看不懂就算了)。首先介绍公式：对于an+1=pan+q(n+1为下角标，n为下角标)，a1已知，那么特征根x=q/(1-p)，则数列通项公式为an=(a1-x)p²(n-1)+x，这是一阶特征根方程的运用。二阶有点麻烦，且不常用。所以不赘述。希望同学们牢记上述公式。当然这种类型的数列可以构造(两边同时加数)\n\n7.函数详解补充：\n(1)复合函数奇偶性：内偶则偶，内奇同外\n(2)复合函数单调性：同增异减\n(3)重点知识关于三次函数：恐怕没有多少人知道三次函数曲线其实是中心对称图形。它有一个对称中心，求法为二阶导后导数为0，根x即为中心横坐标，纵坐标可以用x带入原函数界定。另外，必有唯一一条过该中心的直线与两旁相切。\n\n8.常用数列bn=n×(2²n)求和Sn=(n-1)×(2²(n+1))+2记忆方法：前面减去一个1，后面加一个，再整体加一个2\n\n9.适用于标准方程(焦点在x轴)爆强公式\nk椭=-{(b²)xo｝/{(a²)yo｝k双={(b²)xo｝/{(a²)yo｝\nk抛=p/yo\n\n10.强烈推荐一个两直线垂直或平行的必杀技：已知直线L1：a1x+b1y+c1=0直线L2：a2x+b2y+c2=0若它们垂直：(充要条件)a1a2+b1b2=0；\n\n11.经典中的经典：相信邻项相消大家都知道。下面看隔项相消：对于Sn=1/(1×3)+1/(2×4)+1/(3×5)+…+1/[n(n+2)] =1/2[1+1/2-1/(n+1)-1/(n+2)]\n\n12.爆强△面积公式：S=1/2∣mq-np∣其中向量AB=(m，n)，向量BC=(p，q)注：这个公式可以解决已知三角形三点坐标求面积的问题！\n\n13.你知道吗?空间立体几何中，以下命题均错：\n(1)空间中不同三点确定一个平面；\n(2)垂直同一直线的两直线平行；\n(3)两组对边分别相等的四边形是平行四边形；\n(4)如果一条直线与平面内无数条直线垂直，则直线垂直平面；\n(5)有两个面互相平行，其余各面都是平行四边形的几何体是棱柱；\n(6)有一个面是多边形，其余各面都是三角形的几何体都是棱锥注：对初中生不适用。\n\n14.一个小知识点：所有棱长均相等的棱锥可以是三、四、五棱锥。\n\n15.求f(x)=∣x-1∣+∣x-2∣+∣x-3∣+…+∣x-n∣(n为正整数)的最小值。\n\n16.√〔(a²+b²)〕/2≥(a+b)/2≥√ab≥2ab/(a+b)(a、b为正数，是统一定义域)\n\n17.椭圆中焦点三角形面积公式：S=b²tan(A/2)\n\n18.爆强定理：空间向量三公式解决所有题目：\ncosA=|{向量a.向量b｝/[向量a的模×向量b的模]|\nA为线线夹角；A为线面夹角(但是公式中cos换成sin)；A为面面夹角注：以上角范围均为[0，派/2]。\n\n19.爆强公式1²+2²+3²+…+n²=1/6(n)(n+1)(2n+1)；\n1²3+2²3+3²3+…+n²3=1/4(n²)(n+1)²\n\n20.爆强切线方程记忆方法：写成对称形式，换一个x，换一个y。\n\n21.爆强定理：(a+b+c)²n的展开式[合并之后]的项数为：Cn+22，n+2在下，2在上\n\n22.[转化思想]切线长l=√(d²-r²)d表示圆外一点到圆心得距离，r为圆半径，而d最小为圆心到直线的距离。\n\n23.对于y²=2px，过焦点的互相垂直的两弦AB、CD，它们的和最小为8p。\n\n24.关于一个重要绝对值不等式的介绍爆强：∣|a|-|b|∣≤∣a±b∣≤∣a∣+∣b∣\n\n25.关于解决证明含ln的不等式的一种思路：\n\n26.爆强简洁公式：向量a在向量b上的射影是：〔向量a×向量b的数量积〕/[向量b的模]。记忆方法：在哪投影除以哪个的模\n\n27.说明一个易错点：若f(x+a)[a任意]为奇函数，那么得到的结论是f(x+a)=-f(-x+a)〔等式右边不是-f(-x-a)〕，同理如果f(x+a)为偶函数，可得f(x+a)=f(-x+a)牢记!\n\n28.离心率爆强公式：e=sinA/(sinM+sinN)注：P为椭圆上一点，其中A为角F1PF2，两腰角为M，N\n\n29.椭圆的参数方程也是一个很好的东西，它可以解决一些最值问题。比如x²/4+y²=1求z=x+y的最值。解：令x=2cosay=sina再利用三角有界即可。比你去=0不知道快多少倍!\n\n30.[仅供有能力的童鞋参考]]爆强公式：\n\n31.爆强定理：直观图的面积是原图的√2/4倍。\n\n32.三角形垂心爆强定理：\n(1)向量OH=向量OA+向量OB+向量OC(O为三角形外心，H为垂心)\n(2)若三角形的三个顶点都在函数y=1/x的图象上，则它的垂心也在这个函数图象上。\n\n33.维维安尼定理(不是很重要(仅供娱乐))，--正三角形内(或边界上)任一点到三边的距离之和为定值，这定值等于该三角形的高。\n\n34.爆强思路：如果出现两根之积x1x2=m，两根之和x1+x2=n，我们应当形成一种思路，那就是返回去构造一个二次函数，再利用△大于等于0，可以得到m、n范围。\n\n35.常用结论：过(2p，0)的直线交抛物线y²=2px于A、B两点。O为原点，连接AO.BO。必有角AOB=90度\n\n36.爆强公式：ln(x+1)≤x(x>-1)该式能有效解决不等式的证明问题。\n\n37.函数y=(sinx)/x是偶函数。在(0，派)上它单调递减，(-派，0)上单调递增。利用上述性质可以比较大小。\n\n38.函数y=(lnx)/x在(0，e)上单调递增，在(e，+无穷)上单调递减。另外y=x²(1/x)与该函数的单调性一致。\n\n39.几个数学易错点：\n(1)f`(x)<0是函数在定义域内单调递减的充分不必要条件；\n(2)在研究函数奇偶性时，忽略最开始的也是最重要的一步：考虑定义域是否关于原点对称！\n(3)不等式的运用过程中，千万要考虑\"=\"号是否取到！\n(4)研究数列问题不考虑分项，就是说有时第一项并不符合通项公式，所以应当极度注意：数列问题一定要考虑是否需要分项！\n\n40.提高计算能力五步曲：\n(1)扔掉计算器；\n(2)仔细审题(提倡看题慢，解题快)，要知道没有看清楚题目，你算多少都没用；\n(3)熟记常用数据，掌握一些速算技巧；\n(4)加强心算，估算能力；\n(5)[检验]！\n41.一个美妙的公式：爆强!已知三角形中AB=a，AC=b，O为三角形的外心，则向量AO×向量BC(即数量积)=(1/2)[b²-a²]强烈推荐！证明：过O作BC垂线，转化到已知边上\n\n42.(1)函数单调性的含义：大多数同学都知道若函数在区间D上单调，则函数值随着自变量的增大(减小)而增大(减小)，但有些意思可能有些人还不是很清楚，若函数在D上单调，则函数必连续(分段函数另当别论)这也说明了为什么不能说y=tanx在定义域内单调递增，因为它的图像被无穷多条渐近线挡住，换而言之，不连续。\n\n(2)函数周期性：这里主要总结一些函数方程式所要表达的周期设f(x)为R上的函数，对任意x∈R：\n①f(a±x)=f(b±x)T=(b-a)(加绝对值，下同)\n②f(a±x)=-f(b±x)T=2(b-a)\n③f(x-a)+f(x+a)=f(x)T=6a\n④设T≠0，有f(x+T)=M[f(x)]其中M(x)满足M[M(x)]=x,且M(x)≠x则函数的周期为2\n\n43.奇偶函数概念的推广：\n(1)对于函数f(x)，若存在常数a，使得f(a-x)=f(a+x)，则称f(x)为广义(Ⅰ)型偶函数，且当有两个相异实数a，b满足时，f(x)为周期函数T=2(b-a)\n\n(2)若f(a-x)=-f(a+x)，则f(x)是广义(Ⅰ)型奇函数，当有两个相异实数a，b满足时，f(x)为周期函数T=2(b-a)\n\n(3)有两个实数a，b满足广义奇偶函数的方程式时，就称f(x)是广义(Ⅱ)型的奇，偶函数。\n\n44.函数对称性：\n(1)若f(x)满足f(a+x)+f(b-x)=c则函数关于(a+b/2，c/2)成中心对称\n(2)若f(x)满足f(a+x)=f(b-x)则函数关于直线x=a+b/2成轴对称\n\n(1)若f(xy)=f(x)+f(y)(x>0,y>0),则f(x)=㏒ax\n(2)若f(xy)=f(x)f(y)(x>0,y>0),则f(x)=x²u(u由初值给出)\n(3)f(x+y)=f(x)f(y)则f(x)=a²x\n(4)若f(x+y)=f(x)+f(y)+kxy,则f(x)=ax2+bx\n(5)若f(x+y)+f(x-y)=2f(x),则f(x)=ax+b\n\n45.与三角形有关的定理或结论中学数学平面几何最基本的图形就是三角形\n(1)正切定理(我自己取的，因为不知道名字)：在非Rt△中，有tanA+tanB+tanC=tanAtanBtanC\n(2)任意三角形射影定理(又称第一余弦定理)：在△ABC中a=bcosC+ccosB;b=ccosA+acosC;c=acosB+bcosA\n(3)任意三角形内切圆半径r=2S/a+b+c(S为面积)，外接圆半径应该都知道了吧\n(4)梅涅劳斯定理：设A1,B1，C1分别是△ABC三边BC，CA，AB所在直线的上的点，则A1，B1，C1共线的充要条件是CB1/B1A·BA1/A1C·AC1/C1B=1\n\n46.易错点：\n(1)函数的各类性质综合运用不灵活，比如奇偶性与单调性常用来配合解决抽象函数不等式问题。\n(2)三角函数恒等变换不清楚，诱导公式不迅捷。\n(3)忽略三角函数中的有界性，三角形中角度的限定，比如一个三角形中，不可能同时出现两个角的正切值为负。\n(4)三角的平移变换不清晰，说明：由y=sinx变成y=sinwx的步骤是将横坐标变成原来的1/∣w∣倍。\n(5)数列求和中，常常使用的错位相减总是粗心算错，规避方法：在写第二步时，提出公差，括号内等比数列求和，最后除掉系数。\n\n(6)数列中常用变形公式不清楚，如：an=1/[n(n+2)]的求和保留四项。\n(7)数列未考虑a1是否符合根据sn-sn-1求得的通项公式。\n(8)数列并不是简单的全体实数函数，即注意求导研究数列的最值问题过程中是否取到问题。\n(9)向量的运算不完全等价于代数运算。\n(10)在求向量的模运算过程中平方之后，忘记开方。比如这种选择题中常常出现2，√2的答案…，基本就是选√2，选2的就是因为没有开方。\n(11)复数的几何意义不清晰。\n\n47.关于辅助角公式：asint+bcost=[√(a²+b²)]sin(t+m)其中tanm=b/a[条件：a>0]\n\n48.A、B为椭圆x²/a²+y²/b²=1上任意两点。若OA垂直OB，则有1/∣OA∣²+1/∣OB∣²=1/a²+1/b\n\nGKZTC\n\n1. 本站注明稿件来源为其他媒体的文/图等稿件均为转载稿，本站转载出于非商业性的教育和科研之目的，并不意味着赞同其观点或证实其内容的真实性。如转载稿涉及版权等问题，请作者在两周内速来电或来函联系，本站根据实际情况会进行下架处理。\n2. 任何媒体、网站或个人未经本网协议授权不得转载、链接、转贴或以其他方式复制发表，违者本站将依法追究责任。", null, "" ]

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[ "# Regression In Python\n\n## What Is Regression?\n\nRegression is statistical processes for find relationship between depends variable and independent variables. depended variable also called as predict or outcome variable. independent variable also call as predictors, covariates, or features variable. independent may be one or more variables.\n\nRegression analysis use for prediction, forecasting and analyse relationship between dependent and independent variable.\n\n## Simple Linear Regression\n\nA model that Predict a linear relationship between the independent variable (x) and the depend (output) variable (y) called as Linear regression or linear model.\n\n```#import library\nimport numpy as np\nimport pandas as pd\nimport scipy.stats as stats\nfrom sklearn.linear_model import LinearRegression\nimport matplotlib.pyplot as plt\n#import data set\n#spilt data set for independent and depend future\nx = Emp_data.iloc[:, :-1].values\ny = Emp_data.iloc[:,-1:].values\n#qq plot\nstats.probplot(Emp_data.Churn_out_rate, dist=\"norm\", plot=plt)\nplt.title(\"Normal Q-Q plot\")\nplt.show()\nstats.probplot(Emp_data.Salary_hike, dist=\"norm\", plot=plt)\nplt.title(\"Normal Q-Q plot\")\nplt.show()\nplt.plot(Emp_data.Churn_out_rate,Emp_data.Salary_hike)\nplt.show()\n#Multicollinearity check\ncorr = Emp_data.corr()\n#create model\nreg = LinearRegression()\n#fit model\nreg.fit(x,y)\nprint(reg.score(x, y))\n#transform future for better accuracy\nreg.fit(np.log(x),y)\nprint(reg.score(np.log(x), y))\nreg.fit(np.log(x),np.log(y))\nprint(reg.score(np.log(x),np.log(y)))\n```\n\n### Linear Regression Assumptions :\n\nRelationship : A must be linear relationship between independent and predict variables.\n\nNo Collinearity : Remove multicollinearity between predictors variables. because model difficult to predict which predictor variable are affect depend variable which not. independent variables depend from each other call multicollinearity\n\nAuto correlations: No Residual Errors Dependent On Each Other. Most Of It is Occur in time series models because where the next instant is dependent on previous instant.\n\nHeteroskedasticity : No Heteroskedasticity, in the scatter plot Should be clear pattern distribution of data called homoscedasticity.\n\nNormal distribution: random variables should be normally distributed. This Is Check using Q-Q Plot.\n\n## Multiple Linear Regression\n\nMultiple linear regression is predict relationship between one continuous predict variable and two or more predictors variables. The predictors variables can be continuous or categorical. if categorical then need to convert them dummy variables.\n\n```#import libarary\nimport pandas as pd\nimport numpy as np\nimport matplotlib.pyplot as pltfrom\nfrom sklearn.linear_model import LinearRegression\n#Find Correlaton\ncorr = ComputerData.corr()\n#split data using columan name\nx = pd.DataFrame(ComputerData, columns = ['speed', 'hd', 'ram', 'screen', 'ads', 'trend'])\ny = pd.DataFrame(ComputerData, columns = ['price'])\n# Scatter plot between the variables along with histograms\nimport seaborn as sns\nsns.pairplot(ComputerData)\n# Preparing model\nreg = LinearRegression()\nreg.fit(x,y)\n#check score\nreg.score(x,y)```\n\n## Polynomial Regression\n\nPolynomial Regression: If (Y)Depened And (X)Indepened variable is correlated but relationship is not liner.\n\nBroad range of function will be fit under it. but too sensitive to the outliers. The presence of one or two outliers within the data can seriously affect the results of the nonlinear analysis.Polynomial basically fits wide selection of curvature.\n\n```# Import libraries\nimport numpy as np\nimport matplotlib.pyplot as plt\nimport pandas as pd\nfrom sklearn.preprocessing import PolynomialFeatures\nfrom sklearn.linear_model import LinearRegression\n# Import the dataset\ndatas\nX = pd.DataFrame(datas, columns = ['Temperature'])\ny = pd.DataFrame(datas, columns = ['Pressure'])\n# Fitting Linear Regression\nlin = LinearRegression()\nlin.fit(X, y)\n# Fitting Polynomial Regression\npoly = PolynomialFeatures(degree = 4)\nX_poly = poly.fit_transform(X)\npoly.fit(X_poly, y)\nlin2 = LinearRegression()\nlin2.fit(X_poly, y)\n# Visualise Linear Regression results\nplt.scatter(X, y, color = 'blue')\nplt.plot(X, lin.predict(X), color = 'red')\nplt.title('Linear Regression')\nplt.xlabel('Temperature')\nplt.ylabel('Pressure')\nplt.show()\n# Visualise Polynomial Regression results\nplt.scatter(X, y, color = 'blue')\nplt.plot(X, lin2.predict(poly.fit_transform(X)), color = 'red')\nplt.title('Polynomial Regression')\nplt.xlabel('Temperature')\nplt.ylabel('Pressure')\nplt.show() ```\n\n## Support Vector Regression (SVR)\n\nSupport Vector regression is a part of Support vector machine that supports both linear and non-linear regression.\n\n```from sklearn.svm import SVR\nregressor = SVR(kernel = 'rbf')\nregressor.fit(X, y)\n#predicte new value\ny_pred = regressor.predict(6.5)\ny_pred = sc_y.inverse_transform(y_pred)\nview raw```\n\n## Decision Tree Regression\n\nWhen Out Predicted Variable is continuous (real numbers) then applies Decision Tree Regression\n\n```# create a decisiontreeregressor model\nregressor = DecisionTreeRegressor(random_state = 0)\n# fit the regressor with X and Y data\nregressor.fit(X, y)\n```\n\n## Random Forest Regression\n\nA Random Forest is an ensemble technique. opposite to build a single decision tree. random forest build many decision trees. Then combine every decision tree output and give stable output. this technique called Bootstrap Aggregation also known as bagging.\n\n```# import the regressor\nfrom sklearn.ensemble import RandomForestRegressor\n# create regressor object\nregressor = RandomForestRegressor(n_estimators = 100, random_state = 0)\n# fit the regressor with x and y data\nregressor.fit(X, y) ```\n\nGithub Code : Click Hear\n\n## Conclusion\n\nWhen Predicted Variable is Should Be Continuous. if not then create dummy variable. in python most of NumPy, scikit-learn, and statsmodels library used.\n\n### 1 thought on “Regression In Python”\n\n1.", null, "Keep up the great work, I read few blog posts on this site and I believe that your website is really interesting and has loads of good info. Lovely blog ..! I really enjoyed reading this article. keep it up!!" ]

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[ "# How do you translate graphs of square root functions?\n\nDec 17, 2014\n\nIn order to translate any function to the right or left, place an addition or subtraction \"inside\" of the Parent function. In the case of the square root function, it would look like y = $\\sqrt{x - 2}$ or y = $\\sqrt{x + 5}$.\n\nLet's look at the effect of the addition or subtraction. First, the domain will be altered. You must set x - 2 $\\ge$ 0 , or say that you understand that the square root function has a domain of $x \\ge 2$.\nThis implies a horizontal shift/translation of 2 units to the right. (see graph)\n\nNow repeat for x + 5 $\\ge$ 0, or $x \\ge - 5$. This graph will be translated 5 units to the left. (see graph)", null, "Now, let's explore how to translate a square root function vertically. y = $\\sqrt{x} + 3$ or y = $\\sqrt{x} - 4$. The addition or subtraction on the OUTSIDE of the square root function will cause the graph to translate up or down. Adding 3 will raise the graph up, and subtracting 4 will lower the graph by 4 units. (see graph)", null, "If you are ready for a challenge, we can try to translate in more than one direction at a time!\n\ny = $\\sqrt{x + 2} - 7$ First, make a prediction...will the graph be translated right 2 or left 2, up 7 or down 7?\n\nFind the domain by setting x + 2 $\\ge$ 0 for starters.\n$x \\ge - 2$ (that means left 2, of course!)\nAnd, subtraction of 7, must mean down 7. (see graph)", null, "" ]

[ null, "https://useruploads.socratic.org/W7kWMDkgQnOJTB8YgkfQ_12-16-2014%20Image007.jpg", null, "https://useruploads.socratic.org/wIbeh3KSmz6IIqemBzYg_12-16-2014%20Image008.jpg", null, "https://useruploads.socratic.org/4GwBXmbQ1iTxRh68U6tc_12-16-2014%20Image009.jpg", null ]

[ "", null, "# Range\n\nIt is the difference between the largest and smallest values. Can you find the range of these numbers?\nWhat is the range of the following numbers?\nExample: Find the range of the following: 1, 2, 4, 5, 5\nStep 1: Sort the list → 1, 2, 4, 5, 5\nStep 2: Highest number → 5; Lowest number → 1\nStep 3: Difference between the highest and the lowest numbers in the list → 5 - 1 = 4. Therefore, the range is 4." ]

[ null, "https://www.ezschool.com/gif/games/choosemeicon.png", null ]

[ "", null, "", null, "", null, "", null, "Question\n\n# A hemispherical earth plate is buried into the earth. Find the potential difference between the feet of a man standing on ground as shown in figure. Current through earth plate is I. Conductivity of earth is [δ=10−3π,I=1A](up to two decimal places)", null, "Open in App\nSolution\n\n## j=σE Assuming that the current density is uniform in all directions , ⇒i2πr2=σE where, 2πr2 is the surface area of the hemisphere ⇒E=i2πr2σ⇒ΔV=∫E.dr=∫32i dr2πr2δ ΔV=i2πδ(−1r)32⇒ΔV=12×π×π×10−3×16=10312π2", null, "", null, "Suggest Corrections", null, "", null, "0", null, "", null, "", null, "", null, "", null, "", null, "Similar questions\nJoin BYJU'S Learning Program\nSelect...", null, "", null, "Related Videos", null, "", null, "", null, "Dielectrics\nPHYSICS\nWatch in App", null, "", null, "Explore more\nJoin BYJU'S Learning Program\nSelect..." ]

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[ "Community Profile", null, "# IDN\n\nLast seen: 20 dagen ago Active since 2021\n\n#### Statistics\n\n•", null, "•", null, "•", null, "#### Content Feed\n\nView by\n\nRunning Rate of Change given two different columns if a criteria is met\nGot it combining 2 for loops. Thanks!\n\n3 maanden ago | 0\n\n| accepted\n\nQuestion\n\nRunning Rate of Change given two different columns if a criteria is met\nGood Morning all, I am looking to accomplish the following. I have 6 Columns i wan to calculate a running rate of change that s...\n\n3 maanden ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nFor Loop \"Array indices must be positive integers or logical values\"\nGood Morning all, I have the below statement and its not working when i swith signs and i dont understand why. Appreciate the h...\n\n5 maanden ago | 2 answers | 0\n\n### 2\n\nQuestion\n\nLoop input variables into function to get the various output in table\nHello! I have a function for which i have some variables as input that create output as follows: function(A,B,C,D,E,F,G) - So ...\n\n7 maanden ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nOut of Memory on Function\nHello, I have the below function from Matlab file exchange. I am getting an out of memory alert on RSI every time I run a lar...\n\n7 maanden ago | 0 answers | 0\n\n### 0\n\nQuestion\n\nCumulative Summation down a matrix in loop keeping total per section\nHello! I have 2 matrix, I would like to sum Matrix B values running cummulative given condition. The condition is that it start...\n\n8 maanden ago | 2 answers | 0\n\n### 2\n\nQuestion\n\nHi all, I have the below parfor loop which is good when I am wanting to pass thru negative values to evaluate. I am getting the...\n\n10 maanden ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nFor loop to carry value down depending on another matrix value\nHello! I have the following issue. I have 2 matrices (MatrixA and MatrixB), I want to create a third one(MatrixC) that uses the...\n\n12 maanden ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nCumulative Summation down a matrix in loop\nHello! I have 2 matrix, I would like to sum Matrix B values cummulative given condition. The condition is that it starts to s...\n\n12 maanden ago | 2 answers | 0\n\n### 2\n\nQuestion\n\nFor Loop that Checks 2 Matrices to create a Combined New one based on IF Statement\nGood Morning, I have the following statement that check 3 data sets to combine an answer (1,0,-1) but i am not able to get it t...\n\n12 maanden ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nLoop Rate of Change with Negative Values\nHello, I would like to use a loop to calculate across a table a rate of change ( (Current Value - 4thValue) / 4th Value)). I t...\n\nongeveer een jaar ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nDifference Calculation between current value and a specified prior value\nHello, I use the following to calculate percentage change: B = (diff(VMA)./(VMA(1:end-1))*100; What this does is (x1 - x2) / ...\n\nongeveer een jaar ago | 1 answer | 0\n\n### 1\n\nSending an email from outlook.com using matlab\nTry this mail = '[email protected]'; Tomail = '[email protected]'; password = 'password'; port='587'; setpref('Internet','SMTP_Se...\n\nongeveer een jaar ago | 0\n\nQuestion\n\nparfor loop and decimal step increments in variable\nHello, What is the best way to be able to use a \"parfor\" loop when you have a variable that uses a decimal step as below... y...\n\nongeveer een jaar ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nFractional steps in for loops\nHello! This is the first time i am running fractional steps in for loop and i am not sure wheter this is just comepletely wrong ...\n\nongeveer een jaar ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nPassing multiple Arrays to loop a Function and creating a 1 column array\nHello! I have the following but I am not sure what i am missing: A=ArrayA; (1,000x1) Double C=ArrayC; (1,000x1) Double V=Arra...\n\nongeveer een jaar ago | 2 answers | 0\n\n### 2\n\nQuestion\n\nMultidimensional Array deleting entire max record keeping information across in order\nI have the following loop: sz = 20; m = zeros(size(sz)); for i = 1:sz [max_val, position] = max(sh(:)); %Find Max Pos...\n\nongeveer een jaar ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nFinding position of a multidimensional array given manual value\nHello, I have the following: [max_val, position] = max(sh(:)); %Gets max value of sh and provides position [yAxInd,yCxInd...\n\nongeveer een jaar ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nIF Statement with Multiple Conditions and priority\nI have a set of conditions but one has priority over the other example: Data = zeros(size(V)); for t = 1:numel(Data) if ...\n\nongeveer een jaar ago | 1 answer | 0\n\n### 1\n\nQuestion\n\nNested \"For Loop\" Array Size\nHow can i size my array to the output of a nested for loop? Thanks for the help! DataArray = nan(300,1); % this is the array ...\n\nongeveer een jaar ago | 1 answer | 0" ]

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[ "This notebook gives a short introduction to the Python package Numba, which offers an open source jit (just-in-time) compiler that translates a subset of Python and NumPy code into fast machine code.\n\n## Introduction¶\n\nAs described in the main documentation, Numba translates Python functions to optimized machine code at runtime. The compiled numerical algorithms in Python can then approach the speeds of C or FORTRAN. Without the need of running a separate compilation step, Numba can be simply applied by adding decorators to the Python function. Since there are certain restriction on the Python code, we recommend to only use Numba for compiling functions that are performance-critical. Usually, this is only a small fraction of code. In the following we give a couple of illustrating example and refer to the main documentation for details.\n\nNote: Using numba can really make a difference, since it may easily lead to accelerations of a factor between $10$ and $100$. However, using numba also comes at a cost, since it imposes significant restrictions on the programming. Also, debugging may become much harder, since numba often outputs hard-to-understand error messages. Therefore, we recommend to do the programming in two stages:\n• First write your function without using numba.\n• After the function works as you want, add the jit-decorator and see what you need to change.\n\nIn particular, unknown data types of a function's argument variables may cause unexpected errors when using numba, which tries to infer the types at call time. For example, when initializing a variable with var=None in a function header may cause problems. We come back to this issue later in this notebook.\n\n## Compiling with jit¶\n\nAs a first example, we consider a function that performs an average filtering over a spectrogram. To illustrate the kind of accelerations introduced by Numba, we implement this filtering as a naive double loop.\n\nIn :\nimport os\nimport numpy as np\nimport librosa\nfrom matplotlib import pyplot as plt\n%matplotlib inline\n\nfn_wav = os.path.join('..', 'data', 'B', 'FMP_B_Note-C4_Piano.wav')\nX = librosa.stft(x)\nY = np.abs(X)\n\ndef spectrogram_average_filter_naive(Y, filter_len):\nK, N = Y.shape\nfilter_left = filter_len // 2\nfilter_right = filter_len - filter_left - 1\nnp.zeros((K, filter_right))), axis=1)\nY_new = np.empty_like(Y)\nfor k in range(K):\nfor n in range(N):\nY_new[k, n] = Y_pad[k, n:n+filter_len].sum() / filter_len\nreturn Y_new\n\nfilter_length = 21\nY_filt = spectrogram_average_filter_naive(Y, filter_length)\n\nplt.figure(figsize=(10, 3))\n\nplt.subplot(1, 2, 1)\nplt.imshow(np.log(1 + 100 * Y), aspect='auto', origin='lower', cmap='gray_r')\nplt.title('Original spectrogram')\nplt.colorbar()\nplt.subplot(1, 2, 2)\nplt.imshow(np.log(1 + 100 * Y_filt), aspect='auto', origin='lower', cmap='gray_r')\nplt.title('Smoothed spectrogram')\nplt.colorbar()\nplt.tight_layout()", null, "Using the same function we use the jit-decorator to compile it. In the following code, we check that the outputs of the resulting function is the same as before and then report on the runtime.\n\nIn :\nfrom numba import jit\nimport timeit\n\n@jit(nopython=True)\ndef spectrogram_average_filter_jit(Y, filter_len):\nK, N = Y.shape\n\nfilter_left = filter_len // 2\nfilter_right = filter_len - filter_left - 1\n\nnp.zeros((K, filter_right))), axis=1)\nY_new = np.empty_like(Y)\n\nfor k in range(K):\nfor n in range(N):\nY_new[k, n] = Y_pad[k, n:n+filter_len].sum() / filter_len\n\nreturn Y_new\n\nfilter_length = 21\nY_filt_naive = spectrogram_average_filter_naive(Y, filter_length)\nY_filt_jit = spectrogram_average_filter_jit(Y, filter_length)\nassert np.allclose(Y_filt_naive, Y_filt_jit)\n\nexecuctions = 3\ntime_nai = timeit.timeit(lambda: spectrogram_average_filter_naive(Y, filter_length),\nnumber=execuctions) / execuctions\ntime_jit = timeit.timeit(lambda: spectrogram_average_filter_jit(Y, filter_length),\nnumber=execuctions) / execuctions\nprint('Runtime for naive implementation: %7.5f seconds' % time_nai)\nprint('Runtime for jit implementation: %7.5f seconds' % time_jit)\n\n# An alternative for measuring running time:\n# %timeit spectrogram_average_filter_naive(Y, filter_length)\n# %timeit spectrogram_average_filter_jit(Y, filter_length)\n\nRuntime for naive implementation: 0.25182 seconds\nRuntime for jit implementation: 0.00321 seconds\n\n\n## Parellel Computing¶\n\nOn a standard computer, this the jit-compiled function may be about 100 times faster then the original one. With multiple CPU cores, one can obtain further accelerations by parallelizing the loops.\n\nIn :\nimport numba\nfrom numba import jit, prange\n\n@jit(nopython=True, parallel=True)\ndef spectrogram_average_filter_parallel(Y, filter_len):\nK, N = Y.shape\nfilter_left = filter_len // 2\nfilter_right = filter_len - filter_left - 1\nnp.zeros((K, filter_right))), axis=1)\nY_new = np.empty_like(Y)\nfor k in prange(K):\nfor n in prange(N):\nY_new[k, n] = Y_pad[k, n:n+filter_len].sum() / filter_len\nreturn Y_new\n\nfilter_length = 21\nY_filt_naive = spectrogram_average_filter_naive(Y, filter_length)\nY_filt_jit = spectrogram_average_filter_jit(Y, filter_length)\nY_filt_parallel = spectrogram_average_filter_parallel(Y, filter_length)\nassert np.allclose(Y_filt_naive, Y_filt_parallel)\n\nexecuctions=3\ntime_nai = timeit.timeit(lambda: spectrogram_average_filter_naive(Y, filter_length),\nnumber=execuctions) / execuctions\ntime_jit = timeit.timeit(lambda: spectrogram_average_filter_jit(Y, filter_length),\nnumber=execuctions) / execuctions\ntime_par = timeit.timeit(lambda: spectrogram_average_filter_parallel(Y, filter_length),\nnumber=execuctions) / execuctions\n\nprint('Runtime for naive implementation: %7.5f seconds' % time_nai)\nprint('Runtime for jit implementation: %7.5f seconds' % time_jit)\nprint('Runtime for parallel implementation (using %d threads): %7.5f seconds' % (num_threads, time_par))\n\nRuntime for naive implementation: 0.25911 seconds\nRuntime for jit implementation: 0.00347 seconds\nRuntime for parallel implementation (using 4 threads): 0.00110 seconds\n\n\n## Loop-Based vs. Matrix-Based Implemetation¶\n\nThe function implements the smoothing filter as a naive double-nested look. When using packages like NumPy, it is often much more efficient to use matrix operations while avoiding loop structures. Still the jit-compiled versions may be a bit faster.\n\nIn :\ndef spectrogram_average_filter_matrix(Y, filter_len):\nK, N = Y.shape\nfilter_left = filter_len // 2\nfilter_right = filter_len - filter_left - 1\nY_pad = np.concatenate((np.zeros((K, filter_left)), Y, np.zeros((K, filter_right))), axis=1)\nY_new = np.empty_like(Y)\nfor n in range(N):\nY_new[:, n] = np.mean(Y_pad[:, n:n+filter_len], axis=1)\nreturn Y_new\n\nfilter_length = 21\nY_filt_naive = spectrogram_average_filter_naive(Y, filter_length)\nY_filt_jit = spectrogram_average_filter_jit(Y, filter_length)\nY_filt_parallel = spectrogram_average_filter_parallel(Y, filter_length)\nY_filt_matrix = spectrogram_average_filter_matrix(Y, filter_length)\nassert np.allclose(Y_filt_naive, Y_filt_matrix)\n\nexecuctions = 3\ntime_naive = timeit.timeit(lambda: spectrogram_average_filter_naive(Y, filter_length),\nnumber=execuctions) / execuctions\ntime_jit = timeit.timeit(lambda: spectrogram_average_filter_jit(Y, filter_length),\nnumber=execuctions) / execuctions\ntime_par = timeit.timeit(lambda: spectrogram_average_filter_parallel(Y, filter_length),\nnumber=execuctions) / execuctions\ntime_mat = timeit.timeit(lambda: spectrogram_average_filter_matrix(Y, filter_length),\nnumber=execuctions) / execuctions\n\nprint('Runtime for naive implementation: %7.5f seconds' % time_nai)\nprint('Runtime for jit implementation: %7.5f seconds' % time_jit)\nprint('Runtime for parallel implementation (using %d threads): %7.5f seconds' % (num_threads, time_par))\nprint('Runtime for matrix implementation: %7.5f seconds' % time_mat)\n\nRuntime for naive implementation: 0.25911 seconds\nRuntime for jit implementation: 0.00346 seconds\nRuntime for parallel implementation (using 4 threads): 0.00203 seconds\nRuntime for matrix implementation: 0.00457 seconds\n\n\n## Deviations from Python Semantics¶\n\nNote that not all features of Python and Numpy are available when compiling with Numba, see the list of supported Python features and the list of supported NumPy features for more details. For example, we cannot jit-compile the last version of our function spectrogram_average_filter_matrix due to the keyword axis used in the function mean.\n\nIn :\n@jit(nopython=True)\ndef spectrogram_average_filter_matrixJit(Y, filter_len):\nK, N = Y.shape\n\nfilter_left = filter_len // 2\nfilter_right = filter_len - filter_left\n\nY_pad = np.concatenate((np.zeros((K, filter_left)), Y, np.zeros((K, filter_right))), axis=1)\nY_new = np.empty_like(Y)\n\nfor n in range(K):\nY_new[:, n] = np.mean(Y_pad[:, n:n+filter_len], axis=1)\n\nreturn Y_new\n\ntry:\nY_filt_matrixJit = spectrogram_average_filter_matrixJit(Y, 12)\nexcept Exception as ex:\nprint('Got a %s because of an unsupported numpy feature in numba.' % type(ex).__name__)\n\nGot a TypingError because of an unsupported numpy feature in numba.\n\n\nIn particular, the usage of data types that are determined during runtime may cause unexpected errors. In the following, we give some examples, which work for usual Python functions, but cause problems when using jit:\n\n• The command np.zeros([n, n]) does not work when using jit. It needs to be replaced by np.zeros((n, n)).\n• axis keyword in numpy function $\\leadsto$ loop\n• optional argument used as index (argument arg=None) $\\leadsto$ use different variable name inside function\n• binary masking for multidimensional arrays $\\leadsto$ loop over indexes\nIn :\ndef matrix_zeros_nojit(n=5):\nresult1 = np.zeros([n, n])\nresult2 = np.zeros((n, n), 'int')\nresult3 = np.zeros((n, n)).astype('int')\nreturn result1, result2, result3\n\n@jit(nopython=True)\ndef matrix_zeros_jit(n=5):\nresult1 = np.zeros((n, n))\nresult2 = np.zeros((n, n), np.int64)\nresult3 = np.zeros((n, n)).astype(np.int32)\nreturn result1, result2, result3\n\ndef average_axis0_nojit(x):\nresult = np.mean(x, axis=0)\nreturn result\n\n@jit(nopython=True)\ndef average_axis0_jit(x):\nresult = np.empty(x.shape)\nfor i in range(x.shape):\nresult[i] = np.mean(x[:, i])\nreturn result\n\ndef optional_arg_nojit(x, idx=None):\nx = np.arange(3)\nif idx is None:\nidx = np.argmin(x)\nreturn x[idx]\n\n@jit(nopython=True)\ndef optional_arg_jit(x, idx=None):\nif idx is None:\n_idx = np.argmin(x)\nelse:\n_idx = idx\nreturn x[_idx]\n\ndef treshold_nojit(x, tresh):\nx = x.copy()\nx[x > tresh] = 0\nreturn x\n\n@jit(nopython=True)\ndef treshold_jit(x, tresh):\nx = x.copy()\nfor idx1, idx2 in zip(*np.where(x > tresh)):\nx[idx1, idx2] = 0.0\nreturn x\n\nAcknowledgment: This notebook was created by Frank Zalkow and Meinard Müller." ]

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 ", null ]

[ "# Table 1 The difference between non-invasive cuff and invasive arterial blood pressure measurements\n\nDifference NIBP–IABP SBP DBP MAP\nFor the 66 flush-tested arterial lines*\nOptimal damping (n = 30) 0.8 (12.2); 0.5 (−9, 8.5) −5.2 (8.7); −5.5 (−10.5, 0.3) −4.9 (7.6); −6.5 (−10, −2)\nOver damping (n = 25) −1.5 (11.2); −1 (−7, 6) 2 (12.5); 1 (−5.5, 6.5) −1.3 (8.5); −1.3 (−7, 2)\nUnder damping (n = 11) −2.7 (12.5); −3 (−11, 6) 1.9 (8.4); 1 (−2, 7) −1.4 (9.6); −4 (−10, 6)\nFor all the n = 148 arterial lines\nCombined results −1.7 (11.6); −2 (−8, +5) −2.1 (11.4); −3 (−8, +3) −5.0 (9.2); −5 (−10, −1)\nCombined results r = 0.87 r = 0.78 r = 0.87\nFor the 42 flush tested arterial lines in the d1–3 category\nOptimal damping (n = 19) −3.1 (12.2); −4 (−12, 6) −5.4 (7.4); −6 (−10, 0) −6.5 (5.6); −8 (−10, −3)\nOver damping (n = 16) −2.0 (12.7); −0.5 (−7.5, 5.5) −1.9 (9.7); −1 (−6.8, 3) −3.5 (7.8); −2.5 (−8, 0.5)\nUnder damping (n = 7) −2.9 (13.8); −4 (−9, 1) 1.4 (10.4); −3 (−8, 5.5) −2.0 (10.4); −6 (−14, −1)\nFor all the 83 arterial lines in the d1–3 category\nCombined results −1.5 (11.6); −2.0 (−8, 6) −3.5 (8.4); −4.0 (−8, 1) −5.5 (7.0); −6.0 (−10, −1)\nCombined results r = 0.83 r = 0.84 r = 0.89\n1. Given as mean (SD); median (IQR). *Analysis of variance, diastolic blood pressure (DBP) values are different by damping category (p = 0.024). NIBP non-invasive blood pressure, IABP invasive arterial blood pressure, SBP systolic blood pressure, DBP diastolic blood pressure, MAP mean arterial pressure, r correlation coefficient, d day" ]

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[ "# Pipe network analysis: static pressure of a closed-loop pipe\n\nI've learned fluid dynamics but does not have much knowledge about its real-world applications.\nMy question is about the static pressure of a closed-loop pipe.", null, "[Fig. 1. Pipe network. Source: Wikipedia Pipe_network_analysis]\n\nAccording to the pipe network analysis, two conditions are satisfied for a steady-state closed-loop pipe flow.\n\n1. At any junction, the total flow into a junction equals the total flow out of that junction.\n2. Between any two junctions, the head loss is independent of the path taken. This is equivalent mathematically to the statement that on any closed loop in the network, the head loss around the loop must vanish.\n\nI can understand above fundamental laws for the pipe network analysis.\nHowever, I have difficulty analyzing the static pressure of each junction (not the head loss between two junctions).\n\nLet a pump is located between J3 and J6 of Fig. 1.\nThen following conditions will be hold:\n\n1. $$Q_{in}=Q_{out}$$\n2. $$P_{6} - P_{3} =$$ (actual head loss of the pump)\n3. $$(P_{6} - P_{3})$$ and $$Q_{in}=Q_{out}$$ are related by the pump characteristic curve.\n\nAbove five (2+3) conditions just tell about the pressure difference, not the absolute pressure value itself.\nIf the pump is turned off (i.e. no flow and no head loss) and there is no valve, each joint would have a same absolute pressure value which is determined by the total volume and mass of the fluid filled within the pipe network.\n\nHowever, what will be the absolute value of $$P_{3}$$ and $$P_{6}$$ when the pump is operated?\nDoes the absolute pressure of each joint maitain a steady value for a steady pipe network?\nIf so, which factor determines the steady-absolute pressure value of each joint?\nOr, can the steady-absolute pressure value vary depending on transient fluctuations during the pump turn-on period?\nOr, can the absolute pressure value vary even for a steady pip network?\n\n• The analysis is analogous to an electrical network -- network theorems like Norton or Thevenin (and other more modern ones) can be applied - writing out system of equations. Sometimes it can be made easier by transforming the network into a series of \"voltage dividers\", sometimes some algebraic manipulation is unavoidable. If flow is laminar, the analogy to impedance is just like a resistor. Otherwise the impedances are also nonlinear, but that is completely separate from the network analysis, which is done in terms of a-priori-unknown \"impedance\" of segments: Z_1,2 = (P_2 - P_1) / Q_1,2 etc Aug 7, 2021 at 14:43\n• There is nothing in your math model to determine the absolute pressure, and (within practical limits) the absolute pressure makes no difference to the flow. A real world closed-loop system (e.g. a domestic heating system with a pump, boiler, and radiators) has an external connection that is opened to set the absolute pressure and closed when the system is running - and also a device to monitor the absolute pressure and shut the system down if it goes outside its safe operating limits. Aug 7, 2021 at 15:06\n• @PeteW Thank you for your comment. Yes, this question is somewhat related to an electrical network. In case of a closed-loop electrical circuit without a ground, it is possible to calculate the voltage difference between two junctions. The term 'absolute voltage' is uncertain because the voltage meter requires two measuring point. Aug 7, 2021 at 15:35\n• Then, what if we measure the voltage difference between a junction and ground (like a gauge pressure meter with one-measuring point)? There is no constraint to determine this voltage difference. If so, will this value be continuosly changed? Or will it maintain a fixed value? My question is a fluid-dynamics version of this question. Aug 7, 2021 at 15:35\n• @alephzero Thank you for your comment. I agree that the absolute pressure does not affect the fluid dynamics and only related to the safety problem. This question is originated from this safety problem. In my system, the absolute pressure value fluctuates during the pump operation. I want to identify this is a common operation or a failure that can cause an extreme high pressure situation. Aug 7, 2021 at 15:35\n\nSuppose a simple pipe loop with only one path from the pump outlet to the pump inlet.\n\n[Assumption] Neglect effects of other system components and pipe bending.\n[Def.] Let the stream direction as $$s$$.\n[Def.] Let the total pipe path length as $$L$$.\n[Def.] Let the total mass of the fluid filled within the pipe as $$m$$.\n\nThere are three variables to be determined: $$P_{in}$$, $$P_{out}$$, and $$Q$$.\n($$P_{out}$$ and $$P_{in}$$ are the pressure values of pump outlet and inlet flow, respectively.)\nThree equations are needed to solve this problem.\n\n1. [Pump characteristic] $$P_{out} - P_{in} = f_{pump} ( Q )$$.\n2. [System curve] $$P_{out} - P_{in} = f_{pump} ( Q ) = k L Q^{n}$$ where $$k$$ is the friction factor.\n3. [Mass conservation] $$m = \\int_{0}^{L} \\rho(P,T)Ads = \\int_{0}^{L} \\rho(P_{in} + k s Q^{n},T)Ads$$\n\nThere are two issues related to the Equation 3.\n\n1. Equation 3 assumes a compressible flow (otherwise, it is meaningless). The compressible effect also has to be applied to other two equations.\n2. Bulk modulus of water is quite high. The compressible flow assumption does not make much sense for water.\n• (2) is wrong, P_out = P_in - kLQ^n put another way flow is alsways high pressure to low & flow causes pressure loss, not gain. (1) looks ok, (3) I didn't check en detail, water is incompressible for our purposes (even gases can be treated as incompressible when the pressure loss is low) so it should be far simpler.\n– mart\nSep 7, 2021 at 13:23\n• @mart Thanks, I missed the sign. In case of the compressible flow assumption, I agree that it does not make much sense but there is no other equation I can think. Sep 8, 2021 at 1:13\n• mass conversation is accounted for for all nodes with Q_in = Q_3 + Q_9 (one equation for each node plus one for the network)\n– mart\nSep 8, 2021 at 12:44\n• @mart I agree that the name 'mass conservation' in this answer caused some misunderstanding. Q_in=Q_3+Q_9 stands for the mass conservation for each node; while (3) stands for the mass conservation for the total fluid within the closed-loop pipe. Sep 14, 2021 at 3:43\n• Mass conversation of the whole network is counted for by Q:in = Q_out. What's the use of the very complicated integration at ?\n– mart\nSep 14, 2021 at 6:53" ]

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[ "# What is the Greatest Common Factor of 7 and 12?\n\nGreatest common factor (GCF) of 7 and 12 is 1.\n\nGCF(7,12) = 1\n\nWe will now calculate the prime factors of 7 and 12, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 7 and 12.\n\nGCF Calculator and\nand\n\n## How to find the GCF of 7 and 12?\n\nWe will first find the prime factorization of 7 and 12. After we will calculate the factors of 7 and 12 and find the biggest common factor number .\n\n### Step-1: Prime Factorization of 7\n\nPrime factors of 7 are 7. Prime factorization of 7 in exponential form is:\n\n7 = 71\n\n### Step-2: Prime Factorization of 12\n\nPrime factors of 12 are 2, 3. Prime factorization of 12 in exponential form is:\n\n12 = 22 × 31\n\n### Step-3: Factors of 7\n\nList of positive integer factors of 7 that divides 7 without a remainder.\n\n1\n\n### Step-4: Factors of 12\n\nList of positive integer factors of 12 that divides 7 without a remainder.\n\n1, 2, 3, 4, 6\n\n#### Final Step: Biggest Common Factor Number\n\nWe found the factors and prime factorization of 7 and 12. The biggest common factor number is the GCF number.\nSo the greatest common factor 7 and 12 is 1.\n\nAlso check out the Least Common Multiple of 7 and 12" ]

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[ "• # rdd\n\n• Referenced in 1 article [sw32641]\n• supported. Estimation is accomplished using local linear regression. A provided function will utilize Imbens-Kalyanaraman...\n• # sharpData\n\n• Referenced in 1 article [sw34928]\n• /aos/1015957396>. Capabilities for enhanced local linear regression function and derivative estimation are included, as well ... data sharpening estimator for any degree of local polynomial regression estimation. A cross-validation-based...\n• # npsp\n\n• Referenced in 4 articles [sw31433]\n• methods for multidimensional: linear binning, local polynomial kernel regression, density and variogram estimation. Nonparametric methods...\n• # regtools\n\n• Referenced in 1 article [sw27435]\n• regtools: Regression Tools. Tools for linear, nonlinear and nonparametric regression and classification. Parametric fit assessment ... Nonparametric regression for general dimension, locally-linear option. Nonlinear regression with Eickert-White method...\n• # Evolino\n\n• Referenced in 19 articles [sw36450]\n• still have problems when there are numerous local minima. We introduce a general framework ... sequence learning, EVOlution of recurrent systems with LINear outputs (Evolino). Evolino uses evolution to discover ... weights, while using methods such as linear regression or quadratic programming to compute optimal linear...\n\n• Referenced in 8 articles [sw14077]\n• methods (using a total variance decomposition into local and global components, analogous to Moran ... analysis and within/between groups analyses, many linear regression methods including lowess and polynomial regression, multiple...\n• # COBRA\n\n• Referenced in 9 articles [sw14858]\n• estimators of the regression function is introduced. Instead of building a linear or convex optimized ... training data and a test observation. This local distance approach is model-free and very ... companion R package called (standing for COmBined Regression Alternative) is presented (downloadable on http://cran.r-project.org...\n• # cvplogistic\n\n• Referenced in 5 articles [sw29394]\n• solution surface for concave penalized logistic regression model. The SCAD and MCP (default ... penalty, the package also provides the local linear approximation by coordinate descant...\n• # McSpatial\n\n• Referenced in 2 articles [sw16522]\n• package McSpatial. Locally weighted regression, semiparametric and conditionally parametric regression, fourier and cubic spline functions ... linearized spatial logit and probit, k-density functions and counterfactuals, nonparametric quantile regression and conditional...\n• # MAIC\n\n• Referenced in 8 articles [sw19964]\n• Akaike’s Information Criterion in Linear Regression Analysis via Mixed Integer Nonlinear Program. Akaike ... unreasonable. Instead, stepwise methods, which are local search algorithms, are commonly used to find...\n• # StOpt\n\n• Referenced in 10 articles [sw32903]\n• methods based on Monte Carlo with regressions (global, local and sparse regressors), for underlying states ... Methods are proposed to solve some non linear PDEs. For each method, a framework...\n• # L2WPMA\n\n• Referenced in 9 articles [sw04326]\n• given positive integer. Hence, the piecewise linear interpolant to the fit consists of k monotonic ... This calculation can have about O(nk) local minima, because the positions of the turning ... problem (monotonic fit or isotonic regression) for each set. So it calculates efficiently a global...\n• # ROOT\n\n• Referenced in 5 articles [sw20052]\n• span a large number of files on local disks, the web, or a number ... mathematical and statistical functions, including linear algebra classes, numerical algorithms such as integration and minimization ... various methods for performing regression analysis (fitting). In particular, the RooFit package allows the user...\n• # LDOD\n\n• Referenced in 1 article [sw20245]\n• Locally D-optimal designs for Logistic, Negative Binomial, Poisson, Michaelis-Menten, Exponential, Log-Linear, Emax ... Richards, Weibull and Inverse Quadratic regression models and also functions for auto-constructing Fisher information...\n• # SNPStats\n\n• Referenced in 3 articles [sw20043]\n• association is based on linear or logistic regression according to the response variable (quantitative ... AVAILABILITY: http://bioinfo.iconcologia.net/SNPstats. Source code for local installation is available under GNU license...\n• # SDRcausal\n\n• Referenced in 1 article [sw38424]\n• score) and outcome regression models, are fitted by using semiparametric locally efficient dimension reduction estimators ... large sets of confounding covariates. Techniques including linear extrapolation, numerical differentiation, and truncation have been...\n• # PROC GAM\n\n• Referenced in 1 article [sw12075]\n• Nonparametric regression relaxes the usual assumption of linearity and enables you to uncover structure ... such as the LOESS procedure for local regression and the TPSPLINE procedure for thin-plate ... explored simultaneously; the distributional flexibility of generalized linear models (Nelder and Wedderburn 1972). Thus..." ]

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[ "• Select a School\n\n• Language\n\n### Acceleration\n\nLesson on Acceleration\nPosted on 08/25/2016", null, "Mrs. Shelton's Lesson - Acceleration\n\nStandard\n:\n\nSPS8. Students will determine relationships among force, mass, and motion.\n\na. Calculate velocity and acceleration.\n\nLearning Targets:\n\nStudents should know and will be assessed on their ability to:\n\n1. Calculate the motion of object (speed, velocity, and acceleration).\n\n2. Identify how acceleration and velocity are related.\n\nMore photos." ]

[ null, "https://p10cdn4static.sharpschool.com/UserFiles/Servers/Server_63943/Image/2016%202017/news/IMG_5825.JPG", null ]

[ "# Finitely generated module\n\n(Redirected from Finitely presented module)\n\nIn mathematics, a finitely generated module is a module that has a finite generating set. A finitely generated module over a ring R may also be called a finite R-module, finite over R, or a module of finite type.\n\nRelated concepts include finitely cogenerated modules, finitely presented modules, finitely related modules and coherent modules all of which are defined below. Over a Noetherian ring the concepts of finitely generated, finitely presented and coherent modules coincide.\n\nA finitely generated module over a field is simply a finite-dimensional vector space, and a finitely generated module over the integers is simply a finitely generated abelian group.\n\n## Definition\n\nThe left R-module M is finitely generated if there exist a1, a2, ..., an in M such that for any x in M, there exist r1, r2, ..., rn in R with x = r1a1 + r2a2 + ... + rnan.\n\nThe set {a1, a2, ..., an} is referred to as a generating set for M in this case. The finite generators need not be a basis, since they need not be linearly independent over R. What is true is: M is finitely generated if and only if there is a surjective R-linear map:\n\n$R^{n}\\to M$", null, "for some n (M is a quotient of a free module of finite rank.)\n\nIf a set S generates a module that is finitely generated, then the finite generators of the module can be taken from S at the expense of possibly increasing the number of the generators (since only finitely many elements in S are needed to express the finite generators).\n\nIn the case where the module M is a vector space over a field R, and the generating set is linearly independent, n is well-defined and is referred to as the dimension of M (well-defined means that any linearly independent generating set has n elements: this is the dimension theorem for vector spaces).\n\nAny module is the union of the directed set of its finitely generated submodules.\n\nA module M is finitely generated if and only if any increasing chain Mi of submodules with union M stabilizes: i.e., there is some i such that Mi = M. This fact with Zorn's lemma implies that every nonzero finitely generated module admits a maximal submodule. If any increasing chain of submodules stabilizes (i.e., any submodule is finitely generated), then the module M is called a Noetherian module.\n\n## Examples\n\n• If a module is generated by one element, it is called a cyclic module.\n• Let R be an integral domain with K its field of fractions. Then every finitely generated R-submodule I of K is a fractional ideal: that is, there is some nonzero r in R such that rI is contained in R. Indeed, one can take r to be the product of the denominators of the generators of I. If R is Noetherian, then every fractional ideal arises in this way.\n• Finitely generated modules over the ring of integers Z coincide with the finitely generated abelian groups. These are completely classified by the structure theorem, taking Z as the principal ideal domain.\n• Finitely generated (say left) modules over a division ring are precisely finite dimensional vector spaces (over the division ring).\n\n## Some facts\n\nEvery homomorphic image of a finitely generated module is finitely generated. In general, submodules of finitely generated modules need not be finitely generated. As an example, consider the ring R = Z[X1, X2, ...] of all polynomials in countably many variables. R itself is a finitely generated R-module (with {1} as generating set). Consider the submodule K consisting of all those polynomials with zero constant term. Since every polynomial contains only finitely many terms whose coefficients are non-zero, the R-module K is not finitely generated.\n\nIn general, a module is said to be Noetherian if every submodule is finitely generated. A finitely generated module over a Noetherian ring is a Noetherian module (and indeed this property characterizes Noetherian rings): A module over a Noetherian ring is finitely generated if and only if it is a Noetherian module. This resembles, but is not exactly Hilbert's basis theorem, which states that the polynomial ring R[X] over a Noetherian ring R is Noetherian. Both facts imply that a finitely generated commutative algebra over a Noetherian ring is again a Noetherian ring.\n\nMore generally, an algebra (e.g., ring) that is a finitely generated module is a finitely generated algebra. Conversely, if a finitely generated algebra is integral (over the coefficient ring), then it is finitely generated module. (See integral element for more.)\n\nLet 0 → M′MM′′ → 0 be an exact sequence of modules. Then M is finitely generated if M′, M′′ are finitely generated. There are some partial converses to this. If M is finitely generated and M'' is finitely presented (which is stronger than finitely generated; see below), then M′ is finitely generated. Also, M is Noetherian (resp. Artinian) if and only if M′, M′′ are Noetherian (resp. Artinian).\n\nLet B be a ring and A its subring such that B is a faithfully flat right A-module. Then a left A-module F is finitely generated (resp. finitely presented) if and only if the B-module BA F is finitely generated (resp. finitely presented).\n\n## Finitely generated modules over a commutative ring\n\nFor finitely generated modules over a commutative ring R, Nakayama's lemma is fundamental. Sometimes, the lemma allows one to prove finite dimensional vector spaces phenomena for finitely generated modules. For example, if f : MM is a surjective R-endomorphism of a finitely generated module M, then f is also injective, and hence is an automorphism of M. This says simply that M is a Hopfian module. Similarly, an Artinian module M is coHopfian: any injective endomorphism f is also a surjective endomorphism.\n\nAny R-module is an inductive limit of finitely generated R-submodules. This is useful for weakening an assumption to the finite case (e.g., the characterization of flatness with the Tor functor.)\n\nAn example of a link between finite generation and integral elements can be found in commutative algebras. To say that a commutative algebra A is a finitely generated ring over R means that there exists a set of elements G = {x1, ..., xn} of A such that the smallest subring of A containing G and R is A itself. Because the ring product may be used to combine elements, more than just R-linear combinations of elements of G are generated. For example, a polynomial ring R[x] is finitely generated by {1,x} as a ring, but not as a module. If A is a commutative algebra (with unity) over R, then the following two statements are equivalent:\n\n• A is a finitely generated R module.\n• A is both a finitely generated ring over R and an integral extension of R.\n\n## Generic rank\n\nLet M be a finitely generated module over an integral domain A with the field of fractions K. Then the dimension $\\operatorname {dim} _{K}(M\\otimes _{A}K)$", null, "is called the generic rank of M over A. This number is the same as the number of maximal A-linearly independent vectors in M or equivalently the rank of a maximal free submodule of M. (cf. rank of an abelian group.) Since $(M/F)_{(0)}=M_{(0)}/F_{(0)}=0$", null, ", $M/F$", null, "is a torsion module. When A is Noetherian, by generic freeness, there is an element f (depending on M) such that $M[f^{-1}]$", null, "is a free $A[f^{-1}]$", null, "-module. Then the rank of this free module is the generic rank of M.\n\nNow suppose the integral domain A is generated as algebra over a field k by finitely many homogeneous elements of degrees $d_{i}$", null, ". Suppose M is graded as well and let $P_{M}(t)=\\sum \\operatorname {dim} _{k}(M_{n})t^{n}$", null, "be the Poincaré series of M. By the Hilbert–Serre theorem, there is a polynomial F such that $P_{M}(t)=F(t)\\prod (1-t^{d_{i}})^{-1}$", null, ". Then $F(1)$", null, "is the generic rank of M.\n\nA finitely generated module over a principal ideal domain is torsion-free if and only if it is free. This is a consequence of the structure theorem for finitely generated modules over a principal ideal domain, the basic form of which says a finitely generated module over a PID is a direct sum of a torsion module and a free module. But it can also be shown directly as follows: let M be a torsion-free finitely generated module over a PID A and F a maximal free submodule. Let f be in A such that $fM\\subset F$", null, ". Then $fM$", null, "is free since it is a submodule of a free module and A is a PID. But now $f:M\\to fM$", null, "is an isomorphism since M is torsion-free.\n\nBy the same argument as above, a finitely generated module over a Dedekind domain A (or more generally a semi-hereditary ring) is torsion-free if and only if it is projective; consequently, a finitely generated module over A is a direct sum of a torsion module and a projective module. A finitely generated projective module over a Noetherian integral domain has constant rank and so the generic rank of a finitely generated module over A is the rank of its projective part.\n\n## Equivalent definitions and finitely cogenerated modules\n\nThe following conditions are equivalent to M being finitely generated (f.g.):\n\n• For any family of submodules {Ni | i ∈ I} in M, if $\\sum _{i\\in I}N_{i}=M\\,$", null, ", then $\\sum _{i\\in F}N_{i}=M\\,$", null, "for some finite subset F of I.\n• For any chain of submodules {Ni | i ∈ I} in M, if $\\bigcup _{i\\in I}N_{i}=M\\,$", null, ", then Ni = M for some i in I.\n• If $\\phi :\\bigoplus _{i\\in I}R\\to M\\,$", null, "is an epimorphism, then the restriction $\\phi :\\bigoplus _{i\\in F}R\\to M\\,$", null, "is an epimorphism for some finite subset F of I.\n\nFrom these conditions it is easy to see that being finitely generated is a property preserved by Morita equivalence. The conditions are also convenient to define a dual notion of a finitely cogenerated module M. The following conditions are equivalent to a module being finitely cogenerated (f.cog.):\n\n• For any family of submodules {Ni | i ∈ I} in M, if $\\bigcap _{i\\in I}N_{i}=\\{0\\}\\,$", null, ", then $\\bigcap _{i\\in F}N_{i}=\\{0\\}\\,$", null, "for some finite subset F of I.\n• For any chain of submodules {Ni | i ∈ I} in M, if $\\bigcap _{i\\in I}N_{i}=\\{0\\}\\,$", null, ", then Ni = {0} for some i in I.\n• If $\\phi :M\\to \\prod _{i\\in I}N_{i}\\,$", null, "is a monomorphism, where each $N_{i}$", null, "is an R module, then $\\phi :M\\to \\prod _{i\\in F}N_{i}\\,$", null, "is a monomorphism for some finite subset F of I.\n\nBoth f.g. modules and f.cog. modules have interesting relationships to Noetherian and Artinian modules, and the Jacobson radical J(M) and socle soc(M) of a module. The following facts illustrate the duality between the two conditions. For a module M:\n\n• M is Noetherian if and only if every submodule N of M is f.g.\n• M is Artinian if and only if every quotient module M/N is f.cog.\n• M is f.g. if and only if J(M) is a superfluous submodule of M, and M/J(M) is f.g.\n• M is f.cog. if and only if soc(M) is an essential submodule of M, and soc(M) is f.g.\n• If M is a semisimple module (such as soc(N) for any module N), it is f.g. if and only if f.cog.\n• If M is f.g. and nonzero, then M has a maximal submodule and any quotient module M/N is f.g.\n• If M is f.cog. and nonzero, then M has a minimal submodule, and any submodule N of M is f.cog.\n• If N and M/N are f.g. then so is M. The same is true if \"f.g.\" is replaced with \"f.cog.\"\n\nFinitely cogenerated modules must have finite uniform dimension. This is easily seen by applying the characterization using the finitely generated essential socle. Somewhat asymmetrically, finitely generated modules do not necessarily have finite uniform dimension. For example, an infinite direct product of nonzero rings is a finitely generated (cyclic!) module over itself, however it clearly contains an infinite direct sum of nonzero submodules. Finitely generated modules do not necessarily have finite co-uniform dimension either: any ring R with unity such that R/J(R) is not a semisimple ring is a counterexample.\n\n## Finitely presented, finitely related, and coherent modules\n\nAnother formulation is this: a finitely generated module M is one for which there is an epimorphism\n\nf : RkM.\n\nSuppose now there is an epimorphism,\n\nφ : FM.\n\nfor a module M and free module F.\n\n• If the kernel of φ is finitely generated, then M is called a finitely related module. Since M is isomorphic to F/ker(φ), this basically expresses that M is obtained by taking a free module and introducing finitely many relations within F (the generators of ker(φ)).\n• If the kernel of φ is finitely generated and F has finite rank (i.e. F=Rk), then M is said to be a finitely presented module. Here, M is specified using finitely many generators (the images of the k generators of F=Rk) and finitely many relations (the generators of ker(φ)). See also: free presentation. Finitely presented modules can be characterized by an abstract property within the category of R-modules: they are precisely the compact objects in this category.\n• A coherent module M is a finitely generated module whose finitely generated submodules are finitely presented.\n\nOver any ring R, coherent modules are finitely presented, and finitely presented modules are both finitely generated and finitely related. For a Noetherian ring R, finitely generated, finitely presented, and coherent are equivalent conditions on a module.\n\nSome crossover occurs for projective or flat modules. A finitely generated projective module is finitely presented, and a finitely related flat module is projective.\n\nIt is true also that the following conditions are equivalent for a ring R:\n\n1. R is a right coherent ring.\n2. The module RR is a coherent module.\n3. Every finitely presented right R module is coherent.\n\nAlthough coherence seems like a more cumbersome condition than finitely generated or finitely presented, it is nicer than them since the category of coherent modules is an abelian category, while, in general, neither finitely generated nor finitely presented modules form an abelian category." ]

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[ "MRPT  1.9.9\nCPose2DInterpolator.cpp\nGo to the documentation of this file.\n1 /* +------------------------------------------------------------------------+\n2  | Mobile Robot Programming Toolkit (MRPT) |\n3  | http://www.mrpt.org/ |\n4  | |\n5  | Copyright (c) 2005-2018, Individual contributors, see AUTHORS file |\n8  +------------------------------------------------------------------------+ */\n9\n10 #include \"poses-precomp.h\" // Precompiled headers\n11\n13 #include \"CPoseInterpolatorBase.hpp\" // templ impl\n15\n16 using namespace mrpt::poses;\n17\n19\n20 uint8_t CPose2DInterpolator::serializeGetVersion() const { return 0; }\n22 {\n23  out << m_path;\n24 }\n27 {\n28  switch (version)\n29  {\n30  case 0:\n31  {\n32  in >> m_path;\n33  }\n34  break;\n35  default:\n37  };\n38 }\n39\n40 namespace mrpt::poses\n41 {\n42 // Specialization for DIM=2\n43 template <>\n45  const TTimePosePair &p1, const TTimePosePair &p2,\n46  const TTimePosePair &p3, const TTimePosePair &p4,\n47  const TInterpolatorMethod method, const mrpt::Clock::time_point &t, pose_t& out_interp) const\n48 {\n49  using mrpt::math::TPose2D;\n50  using doubleDuration = std::chrono::duration<double>;\n51  doubleDuration durationT(t.time_since_epoch());\n52  double td = durationT.count();\n54  ts = std::chrono::duration_cast<doubleDuration>(p1.first.time_since_epoch()).count();\n55  ts = std::chrono::duration_cast<doubleDuration>(p2.first.time_since_epoch()).count();\n56  ts = std::chrono::duration_cast<doubleDuration>(p3.first.time_since_epoch()).count();\n57  ts = std::chrono::duration_cast<doubleDuration>(p4.first.time_since_epoch()).count();\n58\n60  X = p1.second.x;\n61  Y = p1.second.y;\n62  yaw = p1.second.phi;\n63  X = p2.second.x;\n64  Y = p2.second.y;\n65  yaw = p2.second.phi;\n66  X = p3.second.x;\n67  Y = p3.second.y;\n68  yaw = p3.second.phi;\n69  X = p4.second.x;\n70  Y = p4.second.y;\n71  yaw = p4.second.phi;\n72\n73  unwrap2PiSequence(yaw);\n74\n75  // Target interpolated values:\n76  switch (method)\n77  {\n78  case imSpline:\n79  {\n80  // ---------------------------------------\n81  // SPLINE INTERPOLATION\n82  // ---------------------------------------\n83  out_interp.x = math::spline(td, ts, X);\n84  out_interp.y = math::spline(td, ts, Y);\n85  out_interp.phi = math::spline(td, ts, yaw, true); // Wrap 2pi\n86  }\n87  break;\n88\n89  case imLinear2Neig:\n90  {\n91  out_interp.x =\n92  math::interpolate2points(td, ts, X, ts, X);\n93  out_interp.y =\n94  math::interpolate2points(td, ts, Y, ts, Y);\n95  out_interp.phi = math::interpolate2points(\n96  td, ts, yaw, ts, yaw, true); // Wrap 2pi\n97  }\n98  break;\n99\n100  case imLinear4Neig:\n101  {\n102  out_interp.x =\n103  math::leastSquareLinearFit<double, decltype(ts), 4>(td, ts, X);\n104  out_interp.y =\n105  math::leastSquareLinearFit<double, decltype(ts), 4>(td, ts, Y);\n106  out_interp.phi =\n107  math::leastSquareLinearFit<double, decltype(ts), 4>(\n108  td, ts, yaw, true); // Wrap 2pi\n109  }\n110  break;\n111\n112  case imSSLLLL:\n113  {\n114  out_interp.x = math::spline(td, ts, X);\n115  out_interp.y = math::spline(td, ts, Y);\n116  out_interp.phi =\n117  math::leastSquareLinearFit<double, decltype(ts), 4>(\n118  td, ts, yaw, true); // Wrap 2pi\n119  }\n120  break;\n121\n122  case imSSLSLL:\n123  {\n124  out_interp.x = math::spline(td, ts, X);\n125  out_interp.y = math::spline(td, ts, Y);\n126  out_interp.phi = math::spline(td, ts, yaw, true); // Wrap 2pi\n127  }\n128  break;\n129\n130  case imLinearSlerp:\n131  {\n132  const double ratio = (td - ts) / (ts - ts);\n133  const double Aang = mrpt::math::angDistance(yaw, yaw);\n134  out_interp.phi = yaw + ratio * Aang;\n135\n136  out_interp.x =\n137  math::interpolate2points(td, ts, X, ts, X);\n138  out_interp.y =\n139  math::interpolate2points(td, ts, Y, ts, Y);\n140  }\n141  break;\n142\n143  case imSplineSlerp:\n144  {\n145  const double ratio = (td - ts) / (ts - ts);\n146  const double Aang = mrpt::math::angDistance(yaw, yaw);\n147  out_interp.phi = yaw + ratio * Aang;\n148\n149  out_interp.x = math::spline(td, ts, X);\n150  out_interp.y = math::spline(td, ts, Y);\n151  }\n152  break;\n153\n154  default:\n155  THROW_EXCEPTION(\"Unknown value for interpolation method!\");\n156  }; // end switch\n157 }\n158\n159 // Explicit instantations:\n160 template class CPoseInterpolatorBase<2>;\n161 }\n162\nGLuint GLuint GLsizei count\nDefinition: glext.h:3528\nGLdouble GLdouble t\nDefinition: glext.h:3689\nThis class stores a time-stamped trajectory in SE(2) (mrpt::math::TPose2D poses). ...\nstd::chrono::time_point< Clock > time_point\nDefinition: Clock.h:26\n#define THROW_EXCEPTION(msg)\nDefinition: exceptions.h:41\nstd::pair< mrpt::Clock::time_point, pose_t > TTimePosePair\n#define IMPLEMENTS_SERIALIZABLE(class_name, base, NameSpace)\nThis must be inserted in all CSerializable classes implementation files.\nCArrayNumeric is an array for numeric types supporting several mathematical operations (actually...\nDefinition: CArrayNumeric.h:25\nT angDistance(T from, T to)\nComputes the shortest angular increment (or distance) between two planar orientations, such that it is constrained to [-pi,pi] and is correct for any combination of angles (e.g.\nDefinition: wrap2pi.h:96\ntypename mrpt::poses::SE_traits< DIM >::lightweight_pose_t pose_t\nTPose2D or TPose3D.\nunsigned char uint8_t\nDefinition: rptypes.h:41\n#define MRPT_THROW_UNKNOWN_SERIALIZATION_VERSION(__V)\nFor use in CSerializable implementations.\nDefinition: exceptions.h:90\nvoid serializeTo(mrpt::serialization::CArchive &out) const override\nPure virtual method for writing (serializing) to an abstract archive.\nClasses for 2D/3D geometry representation, both of single values and probability density distribution...\nVirtual base class for \"archives\": classes abstracting I/O streams.\nDefinition: CArchive.h:52\nvoid unwrap2PiSequence(VECTOR &x)\nModify a sequence of angle values such as no consecutive values have a jump larger than PI in absolut...\nDefinition: wrap2pi.h:72\nvoid serializeFrom(mrpt::serialization::CArchive &in, uint8_t serial_version) override\nPure virtual method for reading (deserializing) from an abstract archive.\nNUMTYPE spline(const NUMTYPE t, const VECTORLIKE &x, const VECTORLIKE &y, bool wrap2pi=false)\nInterpolates the value of a function in a point \"t\" given 4 SORTED points where \"t\" is between the tw...\nDefinition: interp_fit.hpp:34\nGLuint in\nDefinition: glext.h:7274\nLightweight 2D pose.\nvoid impl_interpolation(const TTimePosePair &p1, const TTimePosePair &p2, const TTimePosePair &p3, const TTimePosePair &p4, const TInterpolatorMethod method, const mrpt::Clock::time_point &td, pose_t &out_interp) const\ndouble interpolate2points(const double x, const double x0, const double y0, const double x1, const double y1, bool wrap2pi=false)\nLinear interpolation/extrapolation: evaluates at \"x\" the line (x0,y0)-(x1,y1).\nDefinition: math.cpp:436\nTInterpolatorMethod\nType to select the interpolation method in CPoseInterpolatorBase derived classes. ...\n\n Page generated by Doxygen 1.8.14 for MRPT 1.9.9 Git: 7d5e6d718 Fri Aug 24 01:51:28 2018 +0200 at lun nov 2 08:35:50 CET 2020", null, "" ]

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[ "", null, "全文文献 工具书 数字 学术定义 翻译助手 学术趋势 更多", null, "", null, "", null, "3 d function 的翻译结果: 查询用时：0.008秒", null, "在分类学科中查询 所有学科 计算机软件及计算机应用 更多类别查询", null, "历史查询", null, "", null, "d function 三维功能(5)三维函数(1)", null, "", null, "三维功能", null, "三维函数\n The paper introduced the method of operating Matlab with the technology of ActiveX in VB.NET. It realized mixing programming of Matlab and VB NET. One example was given to implement 3D function image drawing based on the method. 文中介绍了在VB.NET中,利用ActiveX自动化技术应用和操作Matlab的方法,实现了Matlab和VB.NET的混合编程,并基于此方法实现了对三维函数图像的绘制。 短句来源", null, "“3 d function”译为未确定词的双语例句\n Implementing 3D function image drawing base on VB.NET and Matlab 基于VB.NET和Matlab实现三维图像的绘制 短句来源 And OpenGL is widely used with its powerful 3D function and platform independent. 而OpenGL 则以其强大的3D 功能及平台独立等特点获得了广泛的应用。 短句来源 A method of building drawing models of potential failure modes by the use of 3D function of AutoCAD software is given in the paper. 本文给出了利用AUTOCAD软件包的三维图形功能来建立滑体破坏模式图形模型的方法。 短句来源 This paper introduces the method of design function building the mathematical model of parts and drawing the section of parts, fulfiling the drawing of section in 2D circumstance, so the TLTBIower CAD parameter design system has the 3D function. 该文介绍设计函数法建立起“中分面法兰”总件数学模型,进而绘制中分面法兰图,实现在二维参数化设计中对剖视图的绘制,使TLT离心风机CAD参数化设计具有三维的功能。 短句来源 By adapting matrix calculation method for using MATLAB drawing graph technology, 2D and 3D function graphs could be drawn, which is widely applied for science research and engineering technology. MATLAB图形技术采用矩阵运算方法,可以方便地绘制出各种二维和三维图形,广泛用于科学研究和工程技术。 短句来源", null, "查询“3 d function”译词为用户自定义的双语例句\n\n我想查看译文中含有：的双语例句", null, "", null, "d function\n In order to reconstruct the shape of the object, an almost arbitrarily initialized 3D function is propagated on a rectangular grid, so that a level set of this function tracks the height contours of the shape. An empirical hydrophobic field-like 3D function has been calculated with the program HINT (hydrophobic interactions) and imported into the SYBYL implementation of CoMFA (Comparative Molecular Field Analysis). This paper proposes a new method which is based on the modification of the integrated 2D or 3D function into a 1D (spherical distance angle) function applied to an optimal quadrature. In this approach, a 3D function is propagated on a grid. Intermediate shapes are simply the zero-valued contours of 2D slices through this 3D function. 更多", null, "In particle physics, it has been shown that elementary strong interaction is not nuclear force, but quark interaction. On the basis of the model of one-gluon exchange with a confining potential (OGEC), we have proposed a new model of one-gluon exchange with a confining potential, the quark-quark intermediate range potential and the form factor spin dependent potential (OGECIF) by introducing the quark-quark intermediate range potential, the linear confining potential and replacing the d function in the... In particle physics, it has been shown that elementary strong interaction is not nuclear force, but quark interaction. On the basis of the model of one-gluon exchange with a confining potential (OGEC), we have proposed a new model of one-gluon exchange with a confining potential, the quark-quark intermediate range potential and the form factor spin dependent potential (OGECIF) by introducing the quark-quark intermediate range potential, the linear confining potential and replacing the d function in the spin-dependent potential by a suitable form factor f(r, r0) of range r(). The nucleon-nucleon interaction has been derived from the pheno-menological quark-quark potential. Our result is an improvement over those of the OGEC model'-'-1, with a stronger short range repulsive central nucleon-nucleon potential than that in the OGEC model in the odd channels and an appreciable increase in intermediate range attraction in the even channels, therefore, with an expectation that there may be a deuteron bound. Finally, the result is compared with the Tamagaki phenomenological potential. 本文从具有禁闭的单胶子交换模型(OGEC)理论出发,通过引入形状因子自旋相关势和中程势以及线性禁闭势,建立了具有特设禁闭势和形状因子自旋相关势以及中程势的单胶子交换模型(OGECIF).从唯象层子势导出了核子—核子相互作用,比较显著地改进了OGEC模型的结果。 In this paper the general analytical solution of lagrange's problem is derived through the hydro-dynamic fundamental equation. It is found that there exists an approximate similar criterion that is adequate for it. Based on the sets of p-t curves actully measured in gun bore the authors have obtained a linear function of l/l_0 to express the ratio of P_t/P_d approximately. The interior ballistic solving method has shown that such a solution using this P_tP_d function gives higher precision than is obtainable... In this paper the general analytical solution of lagrange's problem is derived through the hydro-dynamic fundamental equation. It is found that there exists an approximate similar criterion that is adequate for it. Based on the sets of p-t curves actully measured in gun bore the authors have obtained a linear function of l/l_0 to express the ratio of P_t/P_d approximately. The interior ballistic solving method has shown that such a solution using this P_tP_d function gives higher precision than is obtainable using lagrange's approximation. 本文应用气动力基本方程导出了拉格朗日问题解的一般形式,且得出其近似的相似准则。本文还利用膛内p-t曲线实测结果确定了P_r/P_d与l/l_c的一阶近似表达式。该式应用于解内弹道问题时,与拉格朗日近似解相比,得到了与实验更为一致的结果。 This paper presents a CAMAC system which is an internationally accepted standard interface. It is composed of the standard DATAWAY and the standard modules assembled in the standard DATAWAY.Three kinds of selected standard modules——function modules, interface modules and a control module——may be assembled in the CAMAC standard DATAWAY.In DO 960 hybrid simulation system, the analog computing elements are arranged in 16 modules which are controlled and monitored via a bus. The bus is controlled by the CAMAC... This paper presents a CAMAC system which is an internationally accepted standard interface. It is composed of the standard DATAWAY and the standard modules assembled in the standard DATAWAY.Three kinds of selected standard modules——function modules, interface modules and a control module——may be assembled in the CAMAC standard DATAWAY.In DO 960 hybrid simulation system, the analog computing elements are arranged in 16 modules which are controlled and monitored via a bus. The bus is controlled by the CAMAC DATAWAY with a microprocesser as the main control module via interface module. The computing results of analog computing elements are acquired by the microprocesser via A/D function module assembled in the CAMAC DATAWAY. The computing results of microprocesser are transmitted via CAMAC DATAWAY and bus to the D/A function module which is assembled in the standard CAMAC DATAWAY. The transformed analog signal is then transmitted to analog computing elements. 本文介绍一种国际公认的标准接口(CAMAC)。它由标准总线和装在标准总线上的标准模件组成。在CAMAC标准总线上可以安装标准化的可供选择的下列三种模件:功能模件、接口模件和控制模件。在DO960混合仿真系统中,模拟计算元件被布置在十六块模块上。这些模块受总线控制和监督,而总线受以微处理器为主要控制模件的CAMAC标准总线控制。模拟计算元件的计算结果可以通过装在CAMAC标准总线上的A/D功能模件采集到微处理器上,微处理器的计算结果经过CAMAC标准总线传送到装在CAMAC标准总线上的D/A功能模件上,转换的模拟量传送到模拟计算元件中。 << 更多相关文摘", null, "相关查询\n\n CNKI小工具 在英文学术搜索中查有关3 d function的内容 在知识搜索中查有关3 d function的内容 在数字搜索中查有关3 d function的内容 在概念知识元中查有关3 d function的内容 在学术趋势中查有关3 d function的内容\n\n CNKI主页 |  设CNKI翻译助手为主页 | 收藏CNKI翻译助手 | 广告服务 | 英文学术搜索", null, "2008 CNKI－中国知网", null, "2008中国知网(cnki) 中国学术期刊(光盘版)电子杂志社" ]

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[ "# اعداد و ارقام و تاریخ ها در آیلتس، چند کلمه حساب می شوند؟\n\nHow words are counted in IELTS\n\nExample: NOT MORE THAN THREE WORDS AND/OR A NUMBER\n\n1. Numbers, dates and time are counted as words in writing. For example 30,000 = one word / 55 = one word / 9.30am = one word / 12.06.2016 = one word. In listening, 30,000 is counted as one number and 9.30AM is also counted as one number.\n2. Dates written as both words and numbers are counted in this way: 12th July = one number and one word.\n3. Symbols with numbers are not counted. For example, 55% = one number (the symbol “%” is not counted as a word). However, if you write “۵۵ percent” it is counted as one word and one number.\n4. Small words such as “a” or “an” are counted as one word. All prepositions, such as “in” or “at” are also counted. All words are counted.\n5. Hyphenated words like “up-to-date” are counted as one word.\n6. Compound nouns which are written as one word are also counted as one word. For example, blackboard = one word.\n7. Compound nouns which are written as two separate words, are counted as two words. For example, university bookshop = two words.\n8. All words are counted, including words in brackets. For example in IELTS writing, “The majority of energy was generated by electricity (55%).”. This sentence is counted as 9 words. The number in brackets is counted.\nتماس با پشتیبانی\nارسال به واتس اپ" ]

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[ "# 4. Kern E. Kenyon, Newton's Hypothetical Orbits Independently Derived\n\n\\$25.00 each\n\n + Add to cart –\n\nFor purchase of this item, please read the instructions.\n\nVolume 12: Pages 39-43, 1999\n\n# Newton's Hypothetical Orbits Independently Derived\n\nKern E. Kenyon\n\n4623 North Lane, Del Mar, California 920144134 U.S.A.\n\nThe mathematical results of four hypothetical orbital problems from the Principia are confirmed by an independent physical method. Each orbital problem that Newton posed and solved is characterized as follows. Given the shape of the orbit and the position of the force center, find the functional form of the central attractive force that will keep a body moving around the orbit. None of Newton's hypothetical orbital problems has so far found any apparent practical application, whereas the Kepler problem, also solved by Newton in the Principia, is of great importance to physics. The Kepler problem too can be derived easily by the present method. Newton used primarily geometrical constructions and logical deductions to arrive at his force functions. In contrast to this, the present (inverse) approach is based on a force balance: as a body moves along a curved path the outward centrifugal force always balances the component of the inward attractive force that is perpendicular to the orbit. Taking the functional form for the central force derived by Newton and inserting it into the force balance, the orbital shape can be derived by solving an ordinary secondorder differential equation—the forced harmonic oscillator equation. Two of Newton's four force functions examined in this way lead to (different) fully nonlinear differential equations, which, surprisingly, can both be solved analytically and in closed form by means of the elementary functions that describe the shapes of the orbits.\n\nReceived: February 4, 1997; Published online: December 15, 2008" ]

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[ "# How many ways can $2$ different history books, $5$ different math books, and $4$ different novels be arranged on a shelf if ...?\n\nThis is my first class in probability so I just wanted verification as to my attempted solution.\n\nQuestion:\n\nIn how many ways can $$2$$ different history books, $$5$$ different math books, and $$4$$ different novels be arranged on a shelf if the books of each type must be together?\n\nSolution:\n\n$$2! \\cdot 5! \\cdot 4! \\cdot 3! = 34560$$\n\nThere are $$34560$$ ways the books can be arranged.\n\n• Looks fine to me... These days people put too much stress on just the numeric result being right, without the \"show your works\" bit: I have a guess that you probably understand this matter and can explain where those $2!$, $5!$, $4!$ and $3!$ are coming from, but it would be better if you explicitly spelled them in your solution. (\"First, we have $3!$ ways to decide the order of the groups of the books of the same type. Then, for every one of those ways, we can order the history books in their group in $2!$ ways etc. etc.\") Sep 6, 2019 at 20:31\n• Thank you for the constructive feedback. I will definitely do that in the future.\n– Mark\nSep 6, 2019 at 20:32\n\nQuestion: \"In how many ways can 2 different history books, 5 different math books, and 4 different novels be arranged on a shelf if the books of each type must be together?\"\n\nIn this question, sequence of the books is not important, therefore:\n\n• For the 2 history books: 2 ways to arrange them (AB and BA), or $$2!$$\n• For the 5 math books: $$5*4*3*2*1 = 5!$$ ways to arrange them, or 120\n• For the 4 novels: $$4*3*2*1 =$$4!\\$ ways to arrange them, or 24\n\nThink like this:\n\n• For the history books (assuming we only look at the history books): 2 options for the first slot, and 1 for the last\n• For the math books (again, only look at the math books): 5 options for the first slot, $$5-1=4$$ for the second slot, $$5-2=3$$ for the third and so on\n• The same for the novels\n\nWe also have three types of books, so, the order of first-to-appear is, by the same logic, 3!\n\nTherefore, in the the end you have $$2!*5!*4!*3!=34560$$ ways to arrange those books" ]

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[ "# Asset Risk Score\n\nEnso Standalone prioritizes assets mitigation according to the asset's importance, related vulnerabilities and other risk factors. In this article we explain how the scoring calculation works.\nThe asset risk is based on:\n\n### Security Gap\n\n• the sum of all defects’ cost and coverage gaps.\n• Defects cost\n• The cost is based on severity per security control and the mitigation status. The cost can be configured separately per each security control and severity according to your definitions.\n• Mitigation status - when a defect's status is moved to in progress, a 50% discount is applied to the defect’s cost.\n• Coverage gap - when an asset wasn't scanned as defined in the coverage policy (see policy), the asset gets a coverage gap penalty that increases the security gap", null, "### Impact Deduction\n\n• Each impact class determines a different level of the asset’s business criticality.\n• Defining the asset class can be done using an automated policy (see policy) or by specifying it manually in the inventory (see inventory).\n• The formula to calculate the impact’s deduction value is: Security gap X ((100-Impact)/100)\n• The impact's value itself is based on:\n• The class which determines the minimum/maximum thresholds\n• A specific set of parameters such as risk factors (appear also as tags), number of commits in repositories, number of hits in hosts, etc. The mentioned parameters determine the asset’s position within the impact class range.", null, "### Risk Calculation\n\nSummarizing the above, below is the high-level risk formula:\nRisk =\n• Security gap\n• Sum of (defect cost - mitigation discount)\n• + coverage gap\n• - Impact deduction", null, "The asset's severity is determined according to the position of the Risk in the risk scale", null, "" ]

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[ "# Loading...\n\nThere was an error loading the page.\n\nTry to answer the following questions here:\n\n• What is this question used for?\n• What does this question assess?\n• What does the student have to do?\n• How is the question randomised?\n• Are there any implementation details that editors should be aware of?\n\n### Metadata", null, "Multiple linear regression - decide which variable to exclude by", null, "Newcastle University Mathematics and Statistics\n\nStats\n\n### Feedback\n\nFrom users who are members of Stats :\n\n### History\n\n#### Checkpoint description\n\nDescribe what's changed since the last checkpoint.\n\n#### Mario Orsi1 year, 3 months ago\n\nGave some feedback: Ready to use\n\nPublished this.\n\n#### Mario Orsi1 year, 3 months ago\n\nSaved a checkpoint:\n\nBefore changin randomization\n\n#### Mario Orsi1 year, 3 months ago\n\nCreated this as a copy of Multiple linear regression - decide which variable to exclude.\nName Status Author Last Modified\nMultiple linear regression - decide which variable to exclude draft Newcastle University Mathematics and Statistics 20/11/2019 14:51\nMultiple linear regression - decide which variable to exclude draft Lauren Frances Desoysa 08/08/2018 10:32\nMultiple linear regression draft Lauren Frances Desoysa 09/08/2018 11:04\nMultiple linear regression - model selection Ready to use Mario Orsi 31/03/2020 22:23\njack's copy of Multiple linear regression - model selection draft jack simons 06/01/2021 16:02\nMLR model selection based on adjusted R^2 draft Richard Pymar 02/03/2021 22:17\n\nGive any introductory information the student needs.\n\nNo variables have been defined in this question.\n\n#### Warning\n\n• When applying the function,\n##### Suggestions\n• Change the signature of parameter from to .\n\nThis variable is an HTML node. HTML nodes can not be relied upon to work correctly when resuming a session - for example, attached event callbacks will be lost, and mathematical notation will likely also break.\n\nIf this causes problems, try to create HTML nodes where you use them in content areas, instead of storing them in variables.\n\nDescribe what this variable represents, and list any assumptions made about its value.\n\n← Depends on:\n→ Used by:\n\nThis variable doesn't seem to be used anywhere.\n\nName Type Generated Value\n\n#### Error in variable testing condition\n\nThere's an error in the condition you specified in the Variable testing tab. Variable values can't be generated until it's fixed.\n\nError:\n\nfor seconds\n\nRunning for ...\n\nName Limit\n\n### Penalties\n\nName Limit\n\nNo parts have been defined in this question.\n\nSelect a part to edit.\n\nAsk the student a question, and give any hints about how they should answer this part.\n\nThe correct answer is an equation. Use the accuracy tab to generate variable values satisfying this equation so it can be marked accurately.\n\n#### Checking accuracy\n\nDefine the range of points over which the student's answer will be compared with the correct answer, and the method used to compare them.\n\n#### Variable value generators\n\nGive expressions which produce values for each of the variables in the expected answer. Leave blank to pick a random value from the range defined above, following the inferred type of the variable.\n\n#### Advanced settings\n\nFor each choice, specify the number of marks to add or subtract when the student picks it.\n\nFor each choice, write 1 if the student should tick it, or 0 if they should leave it unticked.\n\nYou must set a maximum number of marks in order to use this marking method.\n\n#### Advanced settings\n\nYou must set a maximum number of marks in order to use this marking method.\n\nFor each combination of answer and choice, specify the number of marks to add or subtract when the student picks it.\n\nFor each combination of answer and choice, write 1 if the student should tick it, or 0 if they should leave it unticked.\n\nBoth choices and answers must be defined for this part.\n\n#### Advanced settings\n\nHelp with this part type\n\n#### Test that the marking algorithm works\n\nCheck that the marking algorithm works with different sets of variables and student answers using the interface below.\n\nCreate unit tests to save expected results and to document how the algorithm should work.\n\nThere's an error which means the marking algorithm can't run:\n\nGap Expected answer Student's answer\n\n#### Question variables\n\nThese variables are available within the marking algorithm.\n\nName Value\n\n#### Marking parameters\n\nThese values are available as extra variables in the marking algorithm.\n\nName Value\n\n#### Part settings\n\nThese values are available as entries in the settings variable.\n\nName Value\n\nAlternative used:\n\nNote\nValue Feedback\n\nClick on a note's name to show or hide it. Only shown notes will be included when you create a unit test.\n\n#### Unit tests\n\nNo unit tests have been defined. Enter an answer above, select one or more notes, and click the \"Create a unit test\" button.\n\nThe following tests check that the question is behaving as desired.\n\n### This test has not been run yet This test produces the expected output This test does not produce the expected output\n\nThis test is not currently producing the expected result. Fix the marking algorithm to produce the expected results detailed below or, if this test is out of date, update the test to accept the current values.\n\nOne or more notes in this test are no longer defined. If these notes are no longer needed, you should delete this test.\n\nName Value\n\n This note produces the expected output\n\nThis test has not yet been run.\n\nWhen you need to change the way this part works beyond the available options, you can write JavaScript code to be executed at the times described below.\n\nRun this script the built-in script.\n\nThis script runs after the built-in script.\n\nTo account for errors made by the student in earlier calculations, replace question variables with answers to earlier parts.\n\nIn order to create a variable replacement, you must define at least one variable and one other part.\n\nVariable Answer to use Must be answered?\n\nThe variable replacements you've chosen will cause the following variables to be regenerated each time the student submits an answer to this part:\n\nThese variables have some random elements, which means they're not guaranteed to have the same value each time the student submits an answer. You should define new variables to store the random elements, so that they remain the same each time this part is marked.\n\nMore information about this problem\n\nThis part can't be reached by the student.\n\nAdd a \"next part\" reference to this part from another part.\n\nNone of the parts which can lead to this part are reachable either.\n\n### Next part options\n\nDefine the list of parts that the student can visit after this one.\n\n• #### Variable replacements\n\nNo variable replacements have been defined for this next part option.\nVariable Value\n\n### Previous parts\n\nThis part can follow on from:\n\nThis part doesn't follow on from any others.\n\n### Parts\n\n#### Alternative answers\n\nGive a worked solution to the whole question.\n\nSelect extensions to use in this question.\n\n• There was an error loading this extension.\n\nDefine rulesets for simplification and display of mathematical expressions.\n\nDefine functions to use in JME expressions.\n\n• Parameters\n• : of\n\n## Built-in constants\n\nTick the built-in constants you wish to include in this question.\n\n## Custom constants\n\nYou can define constants in terms of the built-in constants, even if they're disabled.\nNames Value LaTeX\n\nAdd styling to the question's display and write a script to run when the question is created.\n\nThis script will run after the question's variable have been generated but before the HTML is attached.\nApply styling rules to the question's content.\n\nThis question is used in the following exam:" ]

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[ "+88 01714 700026 [email protected]\nSelect Page\n\n#### Digital Logic Gate\n\n AND Gate        OR Gate        NOT Gate        NAND Gate        NOR Gate        Ex-OR Gate          Ex-NOR Gate\n\n## Logic OR Gate Tutorial\n\nIf the two input values are both 0, the output value is 0; instead, the output is 1.", null, "#### Two-input OR GATE\n\nAn OR gate performs associate degree ORing operation on 2 or over 2 logic variables. The OR operation on 2 freelance logic variables A and B is written as Y =A+B and reads as Y equals A OR B and not as A and B. associate degree OR gate could be a logic circuit with 2 or additional inputs and one output. The output of associate degree OR gate is LOW only all of its inputs square measure LOW. For all alternative potential input combos, the output is HIGH. This statement once understood for a positive logic system means that the subsequent. The output of associate degree OR gate could be a logic ‘0’ only all of its inputs square measure at logic ‘0’. For all alternative potential input combos, the output could be a logic ‘1’. Figure 4.3 shows the circuit image and also the truth table of a two-input OR gate. The operation of a two-input OR gate is explained by the logic expression.\n\nY=A+B\n\n#### Two-input Transistor OR Gate\n\nA simple 2-input inclusive logic gate may be created victimization RTL Resistor-transistor switches connected along as shown below with the inputs connected on to the semiconductor unit bases. Either semiconductor unit should be saturated “ON” for Associate in Nursing output at Y.\n\n#### Fig-6: 3-Input OR gate IC configuration", null, "#### Three-input Transistor OR Gate\n\nA simple 3-input inclusive logic gate may be created victimization RTL Resistor-transistor switches connected along as shown below with the inputs connected on to the semiconductor unit bases. Either semiconductor unit should be saturated “ON” for Associate in Nursing output at Y.\n\n#### Commonly available digital logic OR gate IC Number:\n\nTTL Logic AND Gate\n\n74LS32 Quad 2-input IC Pin number 14\n\nHCF4075 Triple 3-input IC Pin number 14\n\nTTL Logic AND Gate\n\nCD4072 Dual 4-input IC Pin number 14\n\nCD4071 Quad 2-input IC Pin number 14\n\nCD4075 Triple 3-input IC Pin number 14" ]

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[ "# 40d:Yellow sand\n\n `▓` `▓` `▓` `▓` `▓` `▓` `▓` `≈` `=` `=` `=` `▓` `▓` `▓` `≈` `≈` `=` `=` `=` `▓` `▓` `≈` `≈` `≈` `≈` `=` `=` `▓` `≈` `≈` `≈` `≈` `≈` `≈` `=`" ]

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[ "# Aggregate Demand Curve\n\n(redirected from Aggregate Demand Curves)\n\n## Aggregate Demand Curve\n\nA graph representing demand for goods and services in an economy at different prices. If prices are increasing while demand remains constant, this indicates the economy's aggregate supply is inadequate to meet demand. One calculates the aggregate demand curve by combining and properly weighting the demand curves for individual goods and services.\nMentioned in ?\nReferences in periodicals archive ?\nThe second assumption is that any difference in the price level elasticity of aggregate demand for this paper arises only from either the real balance or interest rate effect, with the corollary being that the price level sensitivity of the balance of trade is assumed equal for aggregate demand curves of different price level elasticities.\nPanel A of Figure 1 shows the standard aggregate demand, short-run aggregate supply model expanded to include two aggregate demand curves of different price level elasticities.\nNow assume that a positive aggregate demand shock occurs and shifts both aggregate demand curves horizontally by the distance [E.sub.0]X.\nPanel A of Figure 2 again shows the standard aggregate demand, aggregate supply model expanded with two aggregate demand curves of different elasticities, labeled as before.\nFor any given increase of aggregate supply with a negatively sloped aggregate demand curve, the price level will decrease, real GDP will increase, and the net effect on the balance of trade is ambiguous In this case the price level elasticity of demand, by governing the relative strengths of the price level and real GDP changes, determines both the direction of change of the balance of trade and the size of the resultant trade deficit or surplus.\n\"A Note on Macroeconomics Textbooks: The Use of the Aggregate Demand Curve,\" Journal of Economic Literature, 12: 896-897.\n\"A Comment on 'The Use of the Aggregate Demand Curve,'\" Journal of Economic Literature, 13: 472-473.\nThe intersection of these aggregate demand curves and the aggregate supply curve produce the goods market-clearing equilibrium values of [P.sup.H.sub .t], [[Y.sup.H.sub.t], and [P.sup.L.sub.t], [Y.sup.L.sub.t], respectively, portrayed in Figure 2.\nFor the sake of illustration, the three aggregate demand curves ([AD.sub.1, AD.sub.2, and AD.sub.3]) are assumed to represent the aggregate demand curves of either Ricardo or Keynes.\n\nSite: Follow: Share:\nOpen / Close" ]

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[ "", null, "", null, "", null, "\\$6.00\n\n# 5th Grade Math Centers | 10 Numerical Expressions Centers | Order of Operations\n\n\\$6.00\n\n## Description\n\nEverything you need to practice and reteach numerical expressions can be found inside of this set of 5th grade math centers. There are 10 different numerical expressions centers included in this set.\n\nAlong with 10 engaging math centers, this set also includes a teacher guide that walks you through prepping each center, storing your centers, and implementing the centers.\n\nActivities Included:\n\nThere are 5 different types of activities included in this set of centers. Each activity includes 2 different centers. Check out the included preview to see an example of each of the math activities below.\n\n• Task Cards: generating numerical patterns\n• Board Games: multiplying by powers of 10\n• Error Analysis: using the order of operations\n• Secret Picture: multiplying and dividing by powers of 10\n• Square Puzzles: using the order of operations\n\nHow to Use These Math Centers:\n\n• Each math center has an answer sheet that can be turned in and graded for student accountability.\n• The centers can be reused over and over again. Just prep them once and use them for years to come!\n• If you prefer, you can also create single-use centers with the included black and white, printable options.\n• The teacher guide includes a printable packet cover page that you can use if you prefer to create weekly math center work packets for your students.\n\nSkills Covered:\n\n• CCSS.MATH.CONTENT.5.OA.A.1: Use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols.\n• CCSS.MATH.CONTENT.5.OA.A.2: Write simple expressions that record calculations with numbers, and interpret numerical expressions without evaluating them. For example, express the calculation “add 8 and 7, then multiply by 2” as 2 √ó (8 + 7). Recognize that 3 √ó (18932 + 921) is three times as large as 18932 + 921, without having to calculate the indicated sum or product.\n• CCSS.MATH.CONTENT.5.OA.B.3: Generate two numerical patterns using two given rules. Identify apparent relationships between corresponding terms. Form ordered pairs consisting of corresponding terms from the two patterns, and graph the ordered pairs on a coordinate plane. For example, given the rule “Add 3” and the starting number 0, and given the rule “Add 6” and the starting number 0, generate terms in the resulting sequences, and observe that the terms in one sequence are twice the corresponding terms in the other sequence. Explain informally why this is so.\n• CCSS.MATH.CONTENT.5.NBT.A.1: Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left.\n• CCSS.MATH.CONTENT.5.NBT.A.2: Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole-number exponents to denote powers of 10.\n\nAll of the centers are included in one ZIP download. After unzipping this resource, you will see 6 PDFs: the teacher guide, the task card centers, the board game centers, the error analysis centers, the secret picture centers, and the square puzzle centers. All of the teacher instructions can be found inside of the teacher guide. You will find everything you need for each center inside of the corresponding PDFs.\n\n***************************************************************************\n\n## Preview", null, "Loading...", null, "Taking too long?", null, "Reload document\n|", null, "Open in new tab\n\n## Reviews\n\nThere are no reviews yet.\n\nOnly logged in customers who have purchased this product may leave a review.\n\nSelect" ]

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[ "# #013c7b Color Information\n\nIn a RGB color space, hex #013c7b is composed of 0.4% red, 23.5% green and 48.2% blue. Whereas in a CMYK color space, it is composed of 99.2% cyan, 51.2% magenta, 0% yellow and 51.8% black. It has a hue angle of 211 degrees, a saturation of 98.4% and a lightness of 24.3%. #013c7b color hex could be obtained by blending #0278f6 with #000000. Closest websafe color is: #003366.\n\n• R 0\n• G 24\n• B 48\nRGB color chart\n• C 99\n• M 51\n• Y 0\n• K 52\nCMYK color chart\n\n#013c7b color description : Dark blue.\n\n# #013c7b Color Conversion\n\nThe hexadecimal color #013c7b has RGB values of R:1, G:60, B:123 and CMYK values of C:0.99, M:0.51, Y:0, K:0.52. Its decimal value is 81019.\n\nHex triplet RGB Decimal 013c7b `#013c7b` 1, 60, 123 `rgb(1,60,123)` 0.4, 23.5, 48.2 `rgb(0.4%,23.5%,48.2%)` 99, 51, 0, 52 211°, 98.4, 24.3 `hsl(211,98.4%,24.3%)` 211°, 99.2, 48.2 003366 `#003366`\nCIE-LAB 25.766, 9.818, -40.461 5.203, 4.668, 19.365 0.178, 0.16, 4.668 25.766, 41.635, 283.639 25.766, -13.979, -51.321 21.605, 5.176, -38.018 00000001, 00111100, 01111011\n\n# Color Schemes with #013c7b\n\n• #013c7b\n``#013c7b` `rgb(1,60,123)``\n• #7b4001\n``#7b4001` `rgb(123,64,1)``\nComplementary Color\n• #01797b\n``#01797b` `rgb(1,121,123)``\n• #013c7b\n``#013c7b` `rgb(1,60,123)``\n• #03017b\n``#03017b` `rgb(3,1,123)``\nAnalogous Color\n• #797b01\n``#797b01` `rgb(121,123,1)``\n• #013c7b\n``#013c7b` `rgb(1,60,123)``\n• #7b0301\n``#7b0301` `rgb(123,3,1)``\nSplit Complementary Color\n• #3c7b01\n``#3c7b01` `rgb(60,123,1)``\n• #013c7b\n``#013c7b` `rgb(1,60,123)``\n• #7b013c\n``#7b013c` `rgb(123,1,60)``\n• #017b40\n``#017b40` `rgb(1,123,64)``\n• #013c7b\n``#013c7b` `rgb(1,60,123)``\n• #7b013c\n``#7b013c` `rgb(123,1,60)``\n• #7b4001\n``#7b4001` `rgb(123,64,1)``\n• #00172f\n``#00172f` `rgb(0,23,47)``\n• #012348\n``#012348` `rgb(1,35,72)``\n• #013062\n``#013062` `rgb(1,48,98)``\n• #013c7b\n``#013c7b` `rgb(1,60,123)``\n• #014894\n``#014894` `rgb(1,72,148)``\n• #0155ae\n``#0155ae` `rgb(1,85,174)``\n• #0261c7\n``#0261c7` `rgb(2,97,199)``\nMonochromatic Color\n\n# Alternatives to #013c7b\n\nBelow, you can see some colors close to #013c7b. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #015b7b\n``#015b7b` `rgb(1,91,123)``\n• #01507b\n``#01507b` `rgb(1,80,123)``\n• #01467b\n``#01467b` `rgb(1,70,123)``\n• #013c7b\n``#013c7b` `rgb(1,60,123)``\n• #01327b\n``#01327b` `rgb(1,50,123)``\n• #01287b\n``#01287b` `rgb(1,40,123)``\n• #011e7b\n``#011e7b` `rgb(1,30,123)``\nSimilar Colors\n\n# #013c7b Preview\n\nThis text has a font color of #013c7b.\n\n``<span style=\"color:#013c7b;\">Text here</span>``\n#013c7b background color\n\nThis paragraph has a background color of #013c7b.\n\n``<p style=\"background-color:#013c7b;\">Content here</p>``\n#013c7b border color\n\nThis element has a border color of #013c7b.\n\n``<div style=\"border:1px solid #013c7b;\">Content here</div>``\nCSS codes\n``.text {color:#013c7b;}``\n``.background {background-color:#013c7b;}``\n``.border {border:1px solid #013c7b;}``\n\n# Shades and Tints of #013c7b\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, #000306 is the darkest color, while #f2f8ff is the lightest one.\n\n• #000306\n``#000306` `rgb(0,3,6)``\n• #000d1a\n``#000d1a` `rgb(0,13,26)``\n• #00162d\n``#00162d` `rgb(0,22,45)``\n• #012041\n``#012041` `rgb(1,32,65)``\n• #012954\n``#012954` `rgb(1,41,84)``\n• #013368\n``#013368` `rgb(1,51,104)``\n• #013c7b\n``#013c7b` `rgb(1,60,123)``\n• #01458e\n``#01458e` `rgb(1,69,142)``\n• #014fa2\n``#014fa2` `rgb(1,79,162)``\n• #0158b5\n``#0158b5` `rgb(1,88,181)``\n• #0262c9\n``#0262c9` `rgb(2,98,201)``\n• #026bdc\n``#026bdc` `rgb(2,107,220)``\n• #0275f0\n``#0275f0` `rgb(2,117,240)``\n• #087ffd\n``#087ffd` `rgb(8,127,253)``\n• #1c89fd\n``#1c89fd` `rgb(28,137,253)``\n• #2f93fd\n``#2f93fd` `rgb(47,147,253)``\n• #439dfd\n``#439dfd` `rgb(67,157,253)``\n• #56a7fe\n``#56a7fe` `rgb(86,167,254)``\n• #6ab1fe\n``#6ab1fe` `rgb(106,177,254)``\n• #7dbbfe\n``#7dbbfe` `rgb(125,187,254)``\n• #91c6fe\n``#91c6fe` `rgb(145,198,254)``\n• #a4d0fe\n``#a4d0fe` `rgb(164,208,254)``\n• #b7dafe\n``#b7dafe` `rgb(183,218,254)``\n• #cbe4ff\n``#cbe4ff` `rgb(203,228,255)``\n• #deeeff\n``#deeeff` `rgb(222,238,255)``\n• #f2f8ff\n``#f2f8ff` `rgb(242,248,255)``\nTint Color Variation\n\n# Tones of #013c7b\n\nA tone is produced by adding gray to any pure hue. In this case, #3a3e42 is the less saturated color, while #013c7b is the most saturated one.\n\n• #3a3e42\n``#3a3e42` `rgb(58,62,66)``\n• #353e47\n``#353e47` `rgb(53,62,71)``\n• #313e4b\n``#313e4b` `rgb(49,62,75)``\n• #2c3d50\n``#2c3d50` `rgb(44,61,80)``\n• #273d55\n``#273d55` `rgb(39,61,85)``\n• #223d5a\n``#223d5a` `rgb(34,61,90)``\n• #1e3d5e\n``#1e3d5e` `rgb(30,61,94)``\n• #193d63\n``#193d63` `rgb(25,61,99)``\n• #143d68\n``#143d68` `rgb(20,61,104)``\n• #0f3c6d\n``#0f3c6d` `rgb(15,60,109)``\n• #0b3c71\n``#0b3c71` `rgb(11,60,113)``\n• #063c76\n``#063c76` `rgb(6,60,118)``\n• #013c7b\n``#013c7b` `rgb(1,60,123)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #013c7b 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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[ "# How do you simplify (-4 - 3n) times -8?\n\n= $32 + 24 n$\nThis would more commonly be written as $- 8 \\left(- 4 - 3 n\\right)$ but as multiplication is commutative, it means the same either way.\n= $32 + 24 n \\text{ be careful of the signs!}$" ]

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[ "# 3 Interactive Discovery and Inspection of Subpopulations\n\n#### Brief Chapter Summary\n\nAnalysis of population-based cohort data has been mostly hypothesis-driven. We present a workflow and an interactive application for data-driven analysis of population-based cohort data using hepatic steatosis as an example. Our mining workflow includes steps\n\n1. to discover subpopulations that have different distributions with respect to the target variable,\n2. to classify each subpopulation taking class imbalance into account, and\n3. to identify variables associated with the target variable.\n\nWe show that our workflow is suited (a) to build subpopulations before classification to reduce class imbalance and (b) to drill-down on the derived models to identify predictive variables and subpopulations worthy of further investigation.\n\nThis chapter is partly based on:\n\n• Uli Niemann, Henry Völzke, Jens-Peter Kühn, and Myra Spiliopoulou. “Learning and inspecting classification rules from longitudinal epidemiological data to identify predictive features on hepatic steatosis.” In: Expert Systems with Applications 41.11 (2014), pp. 5405-5415. DOI: 10.1016/j.eswa.2014.02.040.\n• Uli Niemann, Myra Spiliopoulou, Henry Völzke, and Jens-Peter Kühn. “Interactive Medical Miner: Interactively exploring subpopulations in epidemiological datasets.” In: ECML PKDD 2014, Part III, LNCS 8726. Springer, 2014, pp. 460-463. DOI: 10.1007/978-3-662-44845-8_35.\n\nThis chapter is organized as follows. In Section 3.1, we motivate for classification and interactive subpopulation discovery in epidemiological cohort studies and review related work. We present our workflow and the interactive assistant in Section 3.2. In Section 3.3, we report our results and main findings. The chapter closes with a summary and a discussion of the main contributions in Section 3.4.\n\n## 3.1 Motivation and Comparison to Related Work\n\nMedical decisions about the diagnosis and treatment of multifactorial conditions such as diseases and disorders are based on clinical and epidemiological studies . The latter contain information about participants with and without a disease and allow learning of discriminatory models and, in longitudinal designs, understanding disease progression. For example, several studies identified risk factors (such as obesity and alcohol consumption) and comorbidities (such as cardiovascular disease) associated with hepatic steatosis . However, these studies identified risk factors and associated outcomes that relate to the entire population. Our work arose from the need to identify such factors for subpopulations to promote personalized diagnosis and treatment, as expected in personalized medicine .\n\n### 3.1.1 Role of Subpopulations in Classifier Learning for Cohorts\n\nClassification on subpopulations was studied by Zhanga and Kodell , who pointed out that classifier performance on the whole dataset may be low if the entire population is very heterogeneous. Therefore, they first trained an ensemble of classifiers and then used each ensemble member’s predictions to create a new feature space. They performed hierarchical clustering to partition the instances into three subpopulations: one where the prediction accuracy is high, one where it is in the intermediate range, and one where it is low. Using this approach, Zhanga and Kodell partition the original data set into subpopulations that are easy or hard to classify. While the method seems appealing in general, it appears inappropriate for the three-class problem of the SHIP data, which has a highly skewed distribution, so it is clear that the low classification accuracy is caused (in part) by the class imbalance. Therefore, we exploratively examined the data set before classification to identify less skewed subpopulations and after classification to determine – within each subpopulation – variables strongly associated with the target.\n\nPinheiro et al. performed association rule discovery in patients with liver cancer . The authors pointed out that early detection of liver cancer reduces the mortality rate. Early detection is still difficult because patients often do not show symptoms in the early stages of liver cancer . Pinheiro et al. used the association rule algorithm FP-growth to discover high-confidence association rules and high-confidence classification rules related to liver cancer mortality. We also considered association rules promising for medical data analysis because they are easy to compute and produce results that are understandable to humans. Therefore, we used association rules as the primary method, but for epidemiological data and classification rather than for mortality prediction. To use association rules for classification, we specified that the rule’s consequence is the target variable.\n\n### 3.1.2 Workflows for Expert-Machine Interaction for Cohort Construction and Analysis\n\nZhang et al. addressed the increasing technical challenges of medical expert-driven subpopulation discovery due to increasingly large and complex medical data, often including information from hundreds of variables for thousands of patients in the form of tables, images, or text. In the past, it was sufficient for a physician to have some basic knowledge of statistics and spreadsheet software such as Microsoft Excel to analyze a small table of patient data. Today, more effective and efficient approaches for managing, analyzing, and summarizing extensive medical data are available . However, domain experts typically rely on technical experts to help them perform these tasks. This back-and-forth is often slow, tedious, and expensive. Therefore, it would be better to provide the domain expert with a technical tool that allows them to perform exploratory analysis by themselves quickly. Zhang et al. presented CAVA, a system that includes various subgroup visualizations (called “views”) and analytical components (called “analytics”) for subgroup comparison. The main panel in Figure 3.1 shows one of the views: a flowchart of patient subgroups with the same sequence of symptoms. The user can obtain additional summaries by interacting with the visualization, for example, by dragging and dropping one of the boxes in the flowchart onto one of the entries in the analysis panel. The user can also expand the selected cohort by having the tool search for patients who do not strictly meet the current inclusion criteria but are somewhat similar to the selected patient subpopulation of interest .", null, "Figure 3.1: CAVA’s graphical user interface. The flowchart visualizes subgroups of cardiac patients organized by the common occurrence of symptoms. Arc color represents the hospitalization risk. The user can switch between graphical representations and data processing methods by dragging and dropping. The upper right panel contains detailed information about the currently selected patients. The lower right panel contains a provenance graph that allows the user to undo operations and revisit previous interaction steps. The figure is taken from .\n\nKrause et al. argued that model selection should not be based only on global performance metrics such as accuracy, as these statistics do not contribute to a better understanding of the model’s reasoning. Moreover, a complex but highly accurate model does not automatically guarantee actionable insights. Krause et al. propose Prospector , a system that provides diagnostic components for complex classification models based on the concepts of partial dependence (PD) plots . PD plots are a popular tool for visualizing marginal effects of features on the predicted probability of the target. Briefly, each point on a PD plot represents the model’s average prediction over all observations, assuming that these observations had a fixed value for a feature of interest. A feature whose PD curve exhibits a high range or high variability is considered more influential on model prediction than a feature with a flat PD curve. Closely related to PD plots are individual conditional expectation (ICE) plots, which display a curve for each observation, helping to reveal contrasting subpopulations that might “average out” in a PD plot. Prospector combines PD and ICE curves to show the relationship between a feature and model prediction at a (global) model level and a (local) patient-individual level. Besides, a custom color bar is provided as a more compact alternative to ICE curves (Figure 3.2 (a)). A stacked bar graph shows the distribution of predicted risk scores for each study group (Figure 3.2 (b)). The user can click on a specific decile to obtain a list of individual patients with their exact predicted score and label. In this way, patients whose prediction scores are close to the decision threshold can be further investigated. For each feature, the authors calculate the “most impactful feature change”: given a patient’s current feature values, they identify a near-counterfactual value that leads to a large change in the predicted risk score by minimizing the difference from the original feature value and maximizing the predicted risk score. The top 5 of these so-called “suggested changes” are displayed – separately for increasing and decreasing disease risk - in a table (cf. Figure 3.2 (c)) and integrated as interactive elements into the IC color bars (cf. Figure 3.2 (d)).", null, "Figure 3.2: Selected model diagnostics of Prospector. (a) The upper plot shows two curves for the characteristic “age”: the gray partial dependence (PD) curve represents the marginal prediction of the model over all patients, while the black individual conditional expectation (ICE) curve illustrates the effect of counterfactual ages on the predicted risk of diabetes for an example patient. The histogram shows the age distribution. The color bar below is a compact representation of the ICE curve above; the circled value represents the selected patient’s feature value. (b) Stacked bars show the distribution of predicted risk scores for each study group. Clicking on one of the bars opens a table showing the ID, predicted risk, and true label for all patients belonging to the selected decile of predicted risk. (c) Summary table of “most impactful feature changes” for a decreasing (upper group) and an increasing (lower group) predicted risk: each row shows the actual feature value and the “suggested change,” i.e., a similar but counterfactual value that would lead to a significant change in predicted risk. (d) Multiple PD color bars augmented with suggested changes (labels outlined in white). The figure is adapted from .\n\nPahins et al. presented COVIZ, a system for cohort construction in large spatiotemporal datasets. COVIZ includes components for exploratory data analysis of treatment pathways and event trajectories, visual cohort comparison, and visual querying. One of the design goals of COVIZ was to be fast, e.g., by using efficient data structures such as Quantile Data Structure to ensure low latency for all computational operations and thus suitability for large data sets. Bernard et al. proposed a system for cohort construction in temporal prostate cancer cohort data that included visualizations for subpopulations and individual patients. To guide users during exploration, visual markers indicate interesting relationships between attributes derived from statistical tests. Recently, Corvo et al. presented a comprehensive visual analytics system for pathological high-throughput data, which encompasses all major steps of a typical data analysis pipeline, such as preprocessing raw histopathology images by interactive segmentation, components for exploratory data analysis, and interactive cohort construction in a high-dimensional feature space, feature engineering which includes extraction of potentially predictive biomarker features, modeling, as well as visualization and summarization of the modeling results. Preim and Lawonn provided comprehensive reviews of visual analytics methods and applications in public health and epidemiology in particular.\n\n### 3.1.3 Previous Work on Subpopulation Discovery with the SHIP Data\n\nSince we carry out the proof-of-concept for our workflow on the SHIP data, we list major scientific preparatory work hereafter. Preim et al. provided an overview of the research that developed data mining and visual analytics methods to gain insights into the SHIP data. Among them is the “3D Regression Cube” of Klemm et al. , a system that allows interactive exploration of feature correlations in epidemiological datasets. The system generates many multiple linear regression models from different combinations of one dependent and up to three independent variables and displays their goodness of fit in a three-dimensional heat map. The system allows the user to modify the regression equation, for example, by changing the number of independent variables, specifying wild cards and interaction terms, fixing one of the variables to reduce computational complexity, or focusing specifically on a variable of interest. Our approach is also able to identify variables that are strongly associated with the target variable. However, we search for subpopulation-specific relationships rather than generating a global model for the entire dataset, and we additionally provide predictive value ranges. Klemm et al. presented a system that combines visual representations of non-image and image data. They identify clusters of back pain patients for the SHIP data. Since we specify hepatic steatosis as the target variable, we instead build supervised models and classification rules that directly capture the relationships between predictors and the target variable. Alemzadeh et al. presented S-ADVIsED, a system for interactive exploration of subspace clusters that incorporates various visualization types such as donut diagrams, correlation heatmaps, scatterplot matrices, mosaic diagrams, and error bar graphs. While S-ADVIsED requires the user to input mining results obtained in advance outside the system, our tool enables expert-driven interactive subpopulation discovery instead of expert-driven interactive result exploration. Hielscher et al. developed a semi-supervised constrained-based subspace clustering algorithm to find diverse sets of interesting feature subsets using the SHIP data. To guide the search for predictive feature subsets, the expert can provide their domain knowledge in the form of a small number of instance-level constraints, forcing pairs of instances (i.e., study participants) to be assigned either to the same or a different cluster. Hielscher et al. extended their work and introduced a mechanism to validate subpopulations on independent cohorts.\n\n## 3.2 Subpopulation Discovery Workflow and Interactive Mining Assistant\n\nIn this section, we present our subpopulation discovery workflow. We build classification models on the whole dataset and different partitions, as described in Section 3.2.1. In Section 3.2.2, we introduce relevant underpinnings of classification rule discovery, followed by a description of the primarily used HotSpot algorithm in Section 3.2.3. We present our interactive mining assistant in Section 3.2.4. The dataset used for population partitioning and class separation on the target variable hepatic steatosis comes from the “Study of Health in Pomerania” (SHIP), recall Section 2.2.1. In Section 3.2.5, we describe the origin and availability of the target variable. In Section 3.2.6, the motivation for data partitioning and the partitioning steps are presented.\n\n### 3.2.1 Classification\n\nFor the classification of cohort participants, we focus on algorithms that provide interpretable models, as we aim to identify predictive conditions, i.e., variables and values/ranges in the models. Therefore, we consider decision trees, classification rules, and regression trees. We use the J4.8 decision tree classification algorithm (equivalent to the C4.5 algorithm ) from the Waikato Environment for Knowledge Analysis (Weka) Workbench . This algorithm builds a tree successively by partitioning each node (a subset of the dataset) to the variable that maximizes the information gain within that node. The original algorithm works only with variables that take categorical values, creating one child node per value. However, the Weka implementation also provides an option that forces the algorithm always to create exactly two child nodes: one for the best separating value and one for all other values. We use this option in our experiments because it yields better quality trees. The Weka algorithm also supports variables that take numeric values: A node is split into two child nodes by partitioning the variable’s range of values into two intervals.\n\nTo deal with the skewed distribution, we consider the following classification variants:\n\n• Naive: the problem of imbalanced data is ignored.\n• InfoGain: we keep only the top 30 of the 66 variables by sorting the variables on information gain towards the target variable.\n• Oversampling: We use SMOTE to resample the dataset with minority-oversampling: for class B, 100% new instances are generated; for class C, 300% new instances are generated, resulting in the following distribution A: 438, B: 216, C: 128.\n• CostMatrix: We prefer to misclassify a negative case rather than not detecting a positive case, so we penalize false negatives (FN) more than false positives (FP). We use the cost matrix depicted in Table 3.1.\nTable 3.1: Cost matrix. Cost matrix to penalize misclassification under class imbalance.\nPredicted\nA B C\nTrue A 0 1 2\nB 2 0 1\nC 3 2 0\n\n### 3.2.2 Classification Rule Discovery\n\nClassification rules can reveal interesting relationships between one or more features and the target variable . Compared to model families such as deep neural networks, support vector machines and random forests, classification rules usually achieve lower accuracy. However, they are easier to interpret and infer and are therefore more suitable for interactive subpopulation discovery. In epidemiological research, interesting subpopulations could subsequently be used to formulate and validate a small set of hypotheses or investigate associations between risk factors for a particular target variable. A subpopulation of interest could be formulated as follows: “In the sample of this study, the prevalence of goiter is 32%, whereas the probability in the subpopulation described by thyroid-stimulating hormone less than or equal to 1.63 mU/l and body mass index greater than 32.5 kg/m2 is 49%.”\n\nClassification rule algorithms induce descriptions of interesting subpopulations where interestingness is quantified by a quality function. A classification rule is an association rule whose consequent is fixed to a specific class value. Consider the exemplary classification rule $$r_1$$: $\\begin{equation} r_1: \\underbrace{som\\_waist\\_s2 < 80 \\wedge age\\_ship\\_s2 > 59 \\left(\\wedge \\ldots \\right)}_{\\text{Antecedent}} \\longrightarrow \\underbrace{\\vphantom{som\\_waist\\_s2 < 80 \\wedge age\\_ship\\_s2 > 59 \\left(\\wedge \\ldots \\right)}hepatic\\_steatosis = pos}_{\\text{Consequent}} \\tag{3.1} \\end{equation}$\n\nClassification rules are expressed in the form of $$r: \\text{antecedent} \\longrightarrow T=v$$. The conjunction of conditions (i.e., feature - feature value pairs) left to the arrow constitutes the rule’s $$\\text{antecedent}$$ (or left-hand side). In the $$\\text{consequent}$$ (or right-hand side), $$v$$ is the requested value for the target variable $$T$$.\n\nWe define $$s(r)$$ as the subpopulation or cover set of $$r$$, i.e., the set of instances that satisfy the antecedent of $$r$$. The coverage of $$r$$, which is the fraction of instances covered by $$r$$, is then defined as $$Cov(r)=|s(r)|/N$$, where $$N$$ is the total number of instances. The support of $$r$$ quantifies the percentage of instances covered by $$r$$ that additionally have $$T=v$$, calculated as $$Sup(r)=|s(r)_{T=v}|/N$$. The confidence of $$r$$ (also referred to as precision or accuracy) is defined as $$Conf(r)= |s(r)_{T=v}|/|s(r)|$$ and expresses the relative frequency of instances satisfying the complete rule (i.e., both the antecedent and the consequent) among those satisfying only the antecedent. The recall or sensitivity of $$r$$ with respect to $$T=v$$ is defined as $$Recall(r)=Sensitivity(r)=\\frac{|s(r)_{T=v}|}{n_{T=v}}$$. The Weighted Relative Accuracy of a rule is an interestingness measure that balances coverage and confidence gain and is often used as an internal criterion for candidate generation . It is defined as\n\n$\\begin{equation} WRA(r) = Cov(r)\\cdot \\left(Conf(r)-\\frac{n_{T=v}}{N} \\right). \\tag{3.2} \\end{equation}$\n\nThe odds ratio of $$r$$ with respect to $$T=v$$ is defined as $\\begin{equation} OR(r) = \\frac{ |s(r)_{T=v}| }{|s(r)_{T\\neq v}|} / \\frac{n_{T=v} - |s(r)_{T=v}| }{ n_{T\\neq v} - |s(r)_{T\\neq v}|}. \\tag{3.3} \\end{equation}$\n\nAs an example, Figure 3.3 illustrates an exemplary rule $$r_2$$ in a dataset with 10 instances and a binary target, where circles in cyan color represent instances from the negative class and red circles are positive instances. The cover set of $$r_2$$ contains instances 7, 8, 9 and 10, hence $$Cov(r_2)$$ = 0.40. Further, $$Sup(r_2)$$ = 0.30, $$Conf(r_2)$$ = 0.75, $$WRA(r_2)$$ = 0.40 $$\\cdot$$ (0.75 - 0.40) = 0.14 and $$OR(r_2)$$ = (3/1) / (1/5) = 15.", null, "Figure 3.3: Exemplary classification rule. The gray area represents the data space of the covered instances.\n\n### 3.2.3 HotSpot\n\nFor classification rule discovery, we use the HotSpot algorithm provided for Weka . HotSpot is a beamwidth search algorithm that implements a general-to-specific approach to rule extraction. A single rule is constructed by successively adding the condition to the antecedent that locally maximizes confidence. Unlike general hill-climbing, which considers only the best rule candidate at each iteration, HotSpot’s beam search retains the b highest-ranked candidates and refines them in later steps. Consequently, HotSpot reduces the “myopia” from which Hill-Climbing search typically suffers. Briefly, hill-climbing approaches consider only the locally optimal candidate at each iteration. As a result, a globally optimal rule will not be found if it is not locally optimal in each iteration. It is also desirable to generate more than one rule from an application perspective since alternative descriptions of subpopulations can facilitate hypothesis generation. The beamwidth can be specified as a maximum branching factor, i.e., the maximum number of conditions that can be added to a candidate rule. In each iteration, the rule candidates must satisfy the minimum value count, the sensitivity threshold. To avoid adding a condition only leads to a marginal improvement of the confidence, the parameter minimum improvement, i.e., the minimum relative improvement of the confidence by adding another condition, can be specified. The rule search’s computational complexity can be reduced by specifying a maximum rule length, i.e., the number of conditions in the antecedent. In our experiments we set the parameters as follows: maximum branching factor = 20, maximum value count = 1/3, minimum improvement = 0.1, maximum rule length = 3.\n\n### 3.2.4 Interactive Medical Miner\n\nClassification rules can provide valuable insights into potentially prevalent conditions for different subpopulations of the cohort under study. However, when the number of rules created is large, as is usually the case with large epidemiological data, the rules’ conditions overlap. Hence, some conditions are present under each of the classes of the target variable. Therefore, the medical expert needs inspection tools to decide which rules are informative and which features should be investigated further. Our Interactive Medical Miner (IMM) allows the expert to\n\n• discover classification rules,\n• inspect the frequency of these rules (a) against each class and (b) against the unlabeled subset of the cohort, and\n• examine the statistics of each rule for the values of selected variables.\n\nWe describe these functionalities below, referring to the screenshot in Figure 3.4.", null, "Figure 3.4: The user interface of the Interactive Medical Miner. Classification rules are discovered for class B and shown in the bottom left panel. For the selected rule som_huef_s2 > 109 & crea_u_s2 > 5.38 $$\\longrightarrow$$ mrt_liverfat_s2 = B, the distribution of the participants covered by the rule among all three classes is shown in absolute values (top middle panel) and as a histogram (bottom right panel) with respect to age (top right panel).\n\nThe user interface consists of six panels. In the “Settings” panel (top left), the medical expert can set the parameters for rule induction before pressing the “Build Rules” button. Below this panel, the discovered rules are displayed. In the “Sorting preference” panel, the expert can specify whether the rules should be sorted by confidence, by coverage, or rather alphabetically for a better overview of overlapping rules.\n\nBefore rule generation, the user can specify a sub-cohort of the dataset. By clicking on the button Select Subpopulation, a popup window appears, where multiple filter queries in the form of <variable> <operator> <value> can be added, e.g., som_bmi_s2 >= 30. The defined constraints are displayed in a table and can be undone. Furthermore, the user can select variables for model creation, e.g., exclude a variable that is already known to be highly correlated with another variable that is already considered for model learning.\n\nMining criteria include the dataset (choose between the whole dataset and one of the partitions), the class for which rules are to be generated (drop-down list “Class”), and the constraints related to this class, i.e., “Minimum number of values” (which can also be specified as a relative value), “Maximum rule length,” “Maximum branching factor” and “Minimum improvement.” As an example of how these parameters affect rule search, consider the selected rule in Figure 3.4, som_huef_s2 > 109 & crea_u_s2 > 5.38 $$\\longrightarrow$$ mrt_liverfat_s2 = B, which has a coverage of 0.12 and a confidence of 0.56. The sensitivity of 38/108 = 0.352 satisfies the minimum value count threshold of 0.33. From the Apriori property, it is evident that each of the two conditions in the antecedent of the rule, namely som_huef_s2 > 109 and crea_u_s2 > 5.38, must also exceed this threshold. The position of a condition within the antecedent indicates at which refinement step the condition was added to the rule candidate. For example, the first condition som_huef_s2 > 109 with a confidence of 44/107 = 0.41 was extended by the second condition crea_u_s2 > 5.38 because the confidence gain exceeds the minimum improvement threshold, i.e., 38/68 - 44/107 = 0.15 > 0.05. However, this rule cannot be extended further because the maximum rule length is set to 2. The maximum branching factor was conservatively set to 1000 to prevent potentially interesting rules from not being generated due to a small beamwidth. The expert can lower this parameter interactively if the number of rules found is too high or rule induction takes too long.\n\nThe output list of an execution run (area below the “Settings”) is scrollable and interactive. When the expert clicks on a rule, the upper-middle area “Summary Statistics” is updated. The first row shows the distribution of cohort participants across classes for the entire dataset. The second row shows how the participants covered by the rule (column “Total” in the second row) are distributed across classes. Thus, the expert can specify the discovery of classification rules for one of the classes and then examine how often each rule’s antecedent occurs among participants in the other classes. For example, a rule that covers most of the participants in the selected class (class B in Figure 3.4) is not necessarily interesting if it also covers a high number of participants in the other classes. The rule som_huef_s2 > 109 & crea_u_s2 > 5.38 $$\\longrightarrow$$ mrt_liverfat_s2 = B covers a total of 68 participants, of which 38 are of class B. To reduce the number of covered participants from other classes, i.e., to increase the confidence, the user can decrease the minimum value count threshold to allow generating rules with a lower sensitivity but higher homogeneity with respect to the selected class.\n\nSome of the data may be incomplete. For example, not all participants in the cohort underwent liver MRI. Therefore, it is also of interest to know the distribution of unlabeled participants who support a given rule’s antecedent. For this purpose, the “Histogram” panel can be used: The expert selects another feature from the interactive “Variable selection” area in the upper right panel and can then see how the values of this variable are distributed among the study participants – both labeled and unlabeled; the latter are marked as “Missing” in the color legend. For plotting the histograms, we use the free Java chart library JFreeChart . Numerical variables are discretized using “Scott’s rule” as follows: let $$X_{s(r)}$$ be the set of values for a numeric variable $$X$$ with respect to the cover set $$s(r)$$. The bin width $$h$$ is then calculated as $$h(X_{s(r)})=\\frac{\\max{X_{s(r)}}-\\min{X_{s(r)}}}{3.49\\sigma_{s(r)}}\\cdot |s(r)|^{\\frac{1}{3}}$$.\n\nIf the expert does not select a variable, the target variable is used by default, and only the distribution of labeled participants is visible. The histogram in Figure 3.4 shows the age distribution of both labeled and unlabeled participants covered by our example rule som_huef_s2 > 109 & crea_u_s2 > 5.38 $$\\longrightarrow$$ mrt_liverfat_s2 = B. The distribution of values among the labeled participants indicates that age may be a risk factor for the indicated subpopulation, as the probability of class B increases with age. This visual finding suggests adding the condition age_ship_s2 > 56.8 to the antecedent of the rule. Indeed, the confidence of this more specific rule increases from 38/68 = 0.56 to 27/40 = 0.675. However, as the sensitivity decreases from 38/108 = 0.352 to 27/108 = 0.250, the minimum value count threshold is no longer met. Thus, visualizing participant statistics for selected rules can provide clues to subpopulations that should be monitored more closely and clues to how to modify algorithm parameters for subsequent runs, in our example, to decrease the minimum value count to 0.25 and increase the maximum rule length to 3.\n\n### 3.2.5 The Target Variable\n\nThe target variable is derived from participants’ liver fat concentration calculated by magnetic resonance imaging (MRI). At the time of writing the original manuscript, MRI results were only available for 578 (from a total of 2333; ca. 24.7%) SHIP-2 participants. We use the data from these participants for classifier learning, while our Interactive Medical Miner also contrasts these data with data from the remaining 1755 participants for whom MRI scans were not made available.\n\nAfter discussions with domain experts, we decided to assign participants with a liver fat concentration of 10% or less to class A (“negative” class, i.e., absence of the disorder); values greater than 10% and less than 25% represent class B (increased liver fat/fatty liver tendency) and values greater than 25% class C (high liver fat). We consider classes B and C as “positive.” The cutoff value of 10% is intentionally higher than the value of 5% proposed by Kühn et al. to separate subjects with and without hepatic steatosis because the primary interest from a medical perspective was to identify predictive variables for subjects likely to be ill. Selecting a high cutoff value exacerbates class imbalance and makes data analysis more difficult. Figure 3.5 depicts the class distribution stratified by sex. Of the 578 participants, 438 belong to class A (approximately 76%), 108 to B (19%), and 32 to C (6%). Men were more likely to have elevated or high liver fat concentration than women (30.7% vs. 18.8% in classes B or C).", null, "Figure 3.5: Sex-specific distribution of the target variable. The boxes’ relative sizes depict the number of female and male participants for each of the classes.\n\nIn addition to the target variable, the data set contains 66 variables extracted from participants’ questionnaire responses and medical tests (cf. ). These are variables on socio-demographics (e.g., sex and age), self-reported lifestyle indicators (e.g., alcohol and cigarette consumption), single-nucleotide polymorphism (SNP), laboratory measurements (e.g., serum concentrations), and liver ultrasound. The two available variables of liver ultrasound are stea_s2 and stea_alt75_s2. Both take symbolic values reflecting the probability that the participant has a fatty liver; the latter is a combination of the former and the participant’s alanine transaminase (ALAT) concentration; details are in the caption of Table 3.3 and in . Almost all variables mentioned below have the suffix _s2 indicating SHIP-2 follow-up measurements, in contrast to SHIP-0 (_s0) and SHIP-1 (_s1). Exceptions are sex, highest school degree, and the 10 SNP variables.\n\n### 3.2.6 Partitioning the Dataset into Subpopulations\n\nBecause the dataset is imbalanced with respect to sex (314 females, 264 males), we decided to partition the dataset before classification. First, we examined the class distributions for each sex. We observed that the distributions were very different, especially for class B (see Figure 3.5). Second, we examined the class distribution by sex and age. We found that age was associated with the female subpopulation, but not with the male subpopulation. Third, we identified a cutoff point for age by introducing a heuristic that determines the age value that minimizes the target variable’s standard deviation. We then performed supervised learning separately on the partitions of female and male participants, referred to as PartitionF and PartitionM hereafter. We also created an additional learner for the subpopulation of older female participants aged above the cutoff point of 52 (Partition F:age>52).\n\nTo understand how age affects the class distribution, we introduced a heuristic that determines the cutoff age value at which PartitionF splits into two bins so that the standard deviations of the liver fat concentration in each bin are minimized. Let $$splitAge$$ denote the cutoff value and $$X_y=\\{x\\in\\mathtt{PartitionF}|\\text{age of } x \\leq splitAge\\}$$, $$X_z=\\{x\\in\\mathtt{PartitionF}|\\text{age of } x > splitAge\\}$$ denote the bins. Further, let $$n$$ be the cardinality of $$X_y\\cup{}X_z$$, i.e., of PartitionF. Then, we define the Sum of Weighted Standard Deviations ($$SWSD$$) as\n\n$\\begin{equation} SwSD\\left(X_y,X_z\\right) = \\frac{|X_y|}{n}\\sigma({X_y})+\\frac{|X_z|}{n}\\sigma({X_z}) \\tag{3.4} \\end{equation}$\n\nwhere $$|X_i|$$ is the cardinality of $$X_i$$ and $$\\sigma(X_i)$$ the standard deviation of the original liver fat values. Our heuristic selects the $$\\mathsf{splitAge}$$ such that $$SwSD$$ is minimal. For PartitionF, the minimum value was 7.44 at the age of 52, i.e., close to the onset of menopause.", null, "Figure 3.6: Distribution of liver fat concentration for each partition. Distribution of liver fat concentration in male participants (PartitionM), and females younger and older than 52 years. The horizontal axis shows the liver fat concentration in bins of 5%, while the vertical axis indicates the number of participants in each bin.\n\nThe histograms in Figure 3.6 depict the differences in the liver fat concentration distributions at the age cutoff value of 52. Next to PartitionM (n=264), we show the subpartitions $$\\mathsf{F:age\\leq{}52}$$ (n=131) and F:age>52 (n=183) of PartitionF. Most of the female participants in $$\\mathsf{F:age\\leq{}52}$$ have no more than 5% liver fat concentration, and ca. 95% have no more than 10%, i.e., they belong to the negative class A. In contrast, ca. 28% of F:age>52 have a liver fat concentration of more than 10%; they belong to the positive classes B and C.\n\n## 3.3 Experiments and Findings\n\n### 3.3.1 Results of Decision Tree Classifiers\n\nFor the evaluation of decision tree classifiers, we consider accuracy, i.e., the ratio of correctly classified participants, sensitivity and specificity, and the F-measure, i.e., the harmonic mean between precision and recall. We consider the two classes B and C together as the positive class for specificity, precision, and recall.\n\nOversampling achieved the best performance with an accuracy of about 80% and an F-measure score of 62%. We found the best decision trees for F:age>52, followed by those for PartitionF, then PartitionM. The large discrepancy between the accuracy and F-measure scores also appears in the partitions’ models, suggesting that the accuracy scores are unreliable in such a skewed distribution. Therefore, we do not report on accuracy below.\n\nOn partition F:age>52, the overall best decision tree is achieved by the oversampling variant. On the larger PartitionF, the best performance was achieved by the decision tree created with the InfoGain variant. In contrast, the best decision tree on PartitionM was created with the CostMatrix variant. The sensitivity and specificity values for these trees are given in Table 3.2, while the trees themselves are shown in Figures 3.7 - 3.9 and discussed in Section 3.3.3.\n\nTable 3.2: Best decision trees for the three partitions. Best separation is achieved in F:age>52; PartitionM is the most heterogeneous one, the performance values are lowest.\nPartition Variant Sensitivity (%) Specificity (%) F-measure (%)\nF:age>52 Oversampling 63.5 93.9 81.5\nPartitionF InfoGain 52.4 94.9 69.7\nPartitionM CostMatrix 38.3 86.3 53.0\n\nTable 3.2 indicates that the decision tree variants perform differently on different partitions. Oversampling is beneficial for F:age>52 because it partially compensates for the class imbalance problem. As PartitionM has the most heterogeneous class distribution out of all partitions, all variants perform relatively poorly on it. Hence, we expected most insights from the decision trees on F:age>52 and PartitionF, where better separation is achieved.", null, "Figure 3.7: Best decision tree for F:age>52, achieved by the variant Oversampling.", null, "Figure 3.8: Best decision tree for PartitionF, achieved by the variant InfoGain.", null, "Figure 3.9: Best decision tree for PartitionM, achieved by the variant CostMatrix.\n\n### 3.3.2 Discovered Classification Rules\n\nWhile the classification rules found by HotSpot on the whole dataset were conclusive for class A but not for the positive classes B and C, we omit to report these rules as they are not useful for diagnostic purposes. The classification rules found on the partitions were more informative. However, classification rules with only one feature in the antecedent had low confidence. To ensure high confidence, we restricted the output on rules with at least two features in the antecedent. To ensure still high coverage, we allowed for at most three features. A selection of high confidence and high coverage rules for each partition and class are shown in Tables 3.3 - 3.5, respectively. We describe the most important features in the antecedent of these rules in the next subsection, together with the most important features of the best decision trees.\n\nTable 3.3: Classification rules (PartitionF). Best HotSpot classification rules (maxLength = 3) for PartitionF (excerpt). Cov: coverage; Sup: support; Conf: confidence. age_ship_s2: age; blt_beg_s2: time of blood sampling; ggt_s_s2: serum Gamma-glutamyltransferase (GGT; $$\\mu$$mol/sl); gluc_s_s2: serum glucose (mmol/l); gx_rs11597390: genetic marker; hrs_s_s2: serum uric acid concentration (µmol/l); ldl_s_s2: serum low-density lipoprotein (LDL; mmol/l); sleeph_s2: sleep hours; sleepp_s2: sleep problems; som_bmi_s0: body mass index; som_huef_s0: hip circumference (cm); som_waist_s2: waist circumference (cm); stea_alt75_s2: hepatic steatosis (ultrasound diagnosis) and alanine aminotransferase (ALAT) concentration $$\\geq$$ 0.55 µmol/sl – 0 = normal, 1 = hypoechogenic, 2 = hyperechogenic, 3 = questionable; stea_s2: hepatic steatosis (ultrasound diagnosis); tg_s_s2: serum triglycerides (mmol/l); tsh_s2: thyroid-stimulating hormone (TSH; mu/l).\nRule antecedent\nCov\nSup\nConf\nVariable 1 Variable 2 Variable 3 Abs Abs Rel (%) Rel (%)\nTarget class: A\nsom_waist_s2 $$\\leq$$ 80 132 132 52 100\nsom_bmi_s2 $$\\leq$$ 24.82 109 109 43 100\nsom_huef_s2 $$\\leq$$ 97.8 118 117 46 99\nstea_s2 = 0 218 214 84 98\nstea_alt75_s2 = 0 202 198 78 98\nTarget class: B\nstea_s2 = 1 gx_rs11597390 = 1 age_ship_s2 > 59 20 17 40 85\nstea_alt75_s2 = 1 hrs_s_s2 > 263 age_ship_s2 > 59 20 17 40 85\nstea_alt75_s2 = 1 hrs_s_s2 > 263 ldl_s_s2 > 3.22 20 17 40 85\nstea_s2 = 1 age_ship_s2 > 66 tg_s_s2 > 1.58 17 14 33 82\nstea_s2 = 1 age_ship_s2 > 64 hrs_s_s2 > 263 17 14 33 82\nTarget class: C\ngluc_s_s2 > 7 tsh_s2 > 0.996 6 6 35 100\nsom_bmi_s2 > 38.42 age_ship_s2 $$\\leq$$ 66 asat_s_s2 > 0.22 6 6 35 100\nsom_bmi_s2 > 38.42 sleeph_s2 > 6 blt_beg_s2 $$\\leq$$ 38340 6 6 35 100\nsom_bmi_s2 > 38.42 sleeph_s2 > 6 stea_s2 = 1 6 6 35 100\nhrs_s_s2 > 371 sleepp_s2 = 0 ggt_s_s2 > 0.55 6 6 35 100\nTable 3.4: Classification rules (F:age>52). Best HotSpot classification rules (maxLength = 3) for F:age>52 (excerpt). Cov: coverage; Sup: support; Conf: confidence. age_ship_s2: age; crea_u_s2: urine creatinine (mmol/l); fib_cl_s2: fibrinogen (Clauss) (g/l); gluc_s_s2: serum glucose (mmol/l); ggt_s_s2: serum Gamma-glutamyltransferase (GGT; $$\\mu$$mol/sl); gx_rs11597390: genetic marker; hdl_s_s2: high-density lipoprotein (mmol/l); hrs_s_s2: serum uric acid concentration (µmol/l); som_bmi_s0: body mass index; som_huef_s0: hip circumference (cm); som_waist_s2: waist circumference (cm); stea_alt75_s2: hepatic steatosis (ultrasound diagnosis) and alanine aminotransferase (ALAT) concentration $$\\geq$$ 0.55 µmol/sl – 0 = normal, 1 = hypoechogenic, 2 = hyperechogenic, 3 = questionable; stea_s2: hepatic steatosis (ultrasound diagnosis).\nRule antecedent\nCov\nSup\nConf\nVariable 1 Variable 2 Variable 3 Abs Abs Rel (%) Rel (%)\nTarget class: A\ncrea_u_s2 $$\\leq$$ 5.39 stea_s2 = 0 75 75 57 100\ncrea_u_s2 $$\\leq$$ 5.39 stea_alt75_s2 = 0 72 72 55 100\nsom_waist_s2 $$\\leq$$ 80 54 54 41 100\nsom_bmi_s2 $$\\leq$$ 24.82 50 50 38 100\ncrea_u_s2 $$\\leq$$ 5.39 ggt_s_s2 $$\\leq$$ 0.43 50 50 38 100\nTarget class: B\nstea_s2 = 1 ggt_s_s2 > 0.48 ggt_s_s2 $$\\leq$$ 0.63 15 15 38 100\nstea_s2 = 1 gx_rs11597390 = 1 hdl_s_s2 $$\\leq$$ 1.53 20 19 48 95\nstea_s2 = 1 gx_rs11597390 = 1 fib_cl_s2 > 3.4 15 14 35 93\ncrea_s_s2 $$\\leq$$ 61 som_waist_s2 > 86 stea_s2 = 1 15 14 35 93\nstea_s2 = 1 gx_rs11597390 = 1 hrs_s_s2 > 261 20 18 45 90\nTarget class: C\nsom_bmi_s2 > 38.42 age_ship_s2 $$\\leq$$ 66 4 4 33 100\nsom_bmi_s2 > 38.42 stea_alt75_s2 = 3 4 4 33 100\nsom_huef_s2 > 124 stea_alt75_s2 = 3 4 4 33 100\nsom_waist_s2 > 108 gluc_s_s2 > 6.2 4 4 33 100\nstea_alt75_s2 = 3 som_bmi_s2 > 37.32 4 4 33 100\nTable 3.5: Classification rules (PartitionM). Best HotSpot classification rules (maxLength = 3) for PartitionM (excerpt). Cov: coverage; Sup: support; Conf: confidence. age_ship_s2: age; ATC_C09AA02_s2: enalapril intake; chol_s_s2: serum cholesterol (mmol/l); crea_u_s2: urine creatinine (mmol/l); crea_s_s2: serum creatinine (µmol/l); fig_cl_s2: Fibrinogen (Clauss) (g/l); ggt_s_s2: serum Gamma-glutamyltransferase (GGT; $$\\mu$$mol/sl); gout_s2: treated gout (self-report); hdl_s_s2: high-density lipoprotein (mmol/l); hgb_s2: haemoglobin (g/l); hrs_s_s2: serum uric acid concentration (µmol/l); jodid_u_s2: urine iodide (µg/dl); quick_s2: thromboplastin time Quick test (%); sleeph_s2: sleep hours; stea_alt75_s2: hepatic steatosis (ultrasound diagnosis) and alanine aminotransferase (ALAT) concentration $$\\geq$$ 0.55 µmol/sl – 0 = normal, 1 = hypoechogenic, 2 = hyperechogenic, 3 = questionable; stea_s2: hepatic steatosis (ultrasound diagnosis); som_bmi_s0: body mass index; som_huef_s0: hip circumference (cm); som_waist_s2: waist circumference (cm); tg_s_s2: serum triglycerides (mmol/l).\nRule antecedent\nCov\nSup\nConf\nVariable 1 Variable 2 Variable 3 Abs Abs Rel (%) Rel (%)\nTarget class: A\nstea_alt75_s2 = 0 106 101 55 95\nstea_s2 = 0 138 131 72 95\nggt_s_s2 $$\\leq$$ 0.52 79 73 40 92\nhrs_s_s2 $$\\leq$$ 310 ggt_s_s2 $$\\leq$$ 0.77 81 74 40 91\nsom_waist_s2 $$\\leq$$ 90.8 79 72 39 91\nTarget class: B\nsom_huef_s2 > 108.1 age_ship_s2 > 39 crea_u_s2 > 7.59 28 22 33 79\nsom_bmi_s2 > 32.29 hdl_s_s2 > 0.94 ATC_C09AA02_s2 = 0 29 22 33 76\nsom_bmi_s2 > 32.29 hgb_s2 > 8.1 gout_s2 = 0 29 22 33 76\nsom_waist_s2 > 109 sleeph_s2 $$\\leq$$ 8 jodid_u_s2 > 9.44 29 22 33 76\nsom_huef_s2 > 108.1 hdl_s_s2 > 0.97 crea_u_s2 > 5.38 29 22 33 76\nTarget class: C\nggt_s_s2 > 1.9 crea_s_s2 $$\\leq$$ 90 quick_s2 > 59 6 6 40 100\nggt_s_s2 > 1.9 crea_s_s2 $$\\leq$$ 90 chol_s_s2 > 4.3 6 6 40 100\nggt_s_s2 > 1.9 crea_s_s2 $$\\leq$$ 90 fib_cl_s2 > 1.9 6 6 40 100\nggt_s_s2 > 1.9 crea_s_s2 $$\\leq$$ 90 crea_u_s2 > 4.74 6 6 40 100\nggt_s_s2 > 1.9 tg_s_s2 > 2.01 som_waist_s2 > 93.5 6 6 40 100\n\n### 3.3.3 Important Features for Each Subpopulation\n\nThe most important features in the decision trees of Figures 3.7 - 3.9 are those closer to the root. For readability, the tree nodes in the figures contain short descriptions instead of the original variable names. In all three decision trees, the root node is the ultrasound diagnosis variable stea_s2. A negative ultrasound diagnosis points to negative class A, but a positive ultrasound diagnosis does not directly lead to the positive classes B and C. The decision trees of the three partitions differ in the nodes placed near the root.\n\nImportant features for PartitionF. In the best decision tree of PartitionF (cf. Figure 3.8), it can be observed that if the ultrasound report is positive and the HbA1C concentration is more than 6.8%, the class is C. The classification rules with high coverage and confidence in Table 3.3) point to further interesting features: a waist circumference of at most 80 cm, a BMI of no more than 24.82 kg/m2, a hip circumference of 97.8 cm or less characterize participants of the negative class. All 6 participants with a serum glucose concentration greater than 7 mmol/l and a TSH concentration greater than 0.996 mu/l belong to class C. Further, severe obesity (a BMI value of more than 38.42 kg/m2 points to class C with high confidence – but only in combination with other variables.\n\nImportant features for F:age>52 In contrast to the best tree for PartitionF, the best decision tree for the subpartition F:age>52 (cf. Figure 3.7) also contains nodes with SNPs, indicating potentially genetic associations to fatty liver for these participants. Classification rules with high coverage and confidence for class B also contain SNPs, as shown in Table 3.4. Similar to PartitionF, high BMI values point to a positive class when combined with other features. Table 3.4 shows that all four participants with stea_alt75_s2 = 3 (i.e., a positive ultrasound diagnosis combined with a critical ALAT value) and a BMI larger than 38.42 kg/m2 belong to class C. A similar association holds for stea_alt75_s2 = 3 combined with a high waist circumference (> 124 cm). 19 out of 20 participants in class B with a positive ultrasound diagnosis, a genetic marker gx_rs11597390 = 1, and a high-density lipoprotein (HDL) serum concentration of at most 1.53 mmol/l.\n\nImportant features for PartitionM. The role of the ultrasound report in predicting the negative class is the same for PartitionM (cf. Figure 3.9 as for PartitionF). As with the best tree for F:age>52, the best tree for PartitionM contains nodes with SNPs and serum Gamma-glutamyltransferase (GGT) value ranges. Such features are also in the antecedent of top Hotspot rules (cf. Table 3.5): a Serum GGT concentration of more than 1.9 $$\\mu$$mol/sl in combination with creatinine concentration of at most 90 mmol/l or a thromboplastin time ratio (quick_s2) of more than 59% points to class C. Similarly, positive ultrasound diagnosis and a serum HDL concentration not exceeding 0.84 mmol/l point to class C.\n\nThe decision trees and classification rules provide insights into features that appear diagnostically important. However, the medical expert needs additional information to decide whether a feature is worth further investigation. Decision trees highlight the importance of a feature only in the context of the subtree in which it is found; a subtree describes a typically very small subpopulation. In contrast, classification rules provide information about larger subpopulations. However, these subpopulations may overlap; for example, the first four rules on class C for PartitionM (cf. Table 3.5) may refer to the same 6 participants.\n\nFurthermore, unless a classification rule has a confidence value close to 100%, participants in the other classes may also support it. Therefore, to decide whether the features in the antecedent of the rule deserve further investigation, the expert also needs knowledge about the statistics of the rule for the other classes. To assist the expert in this task, we have proposed the Interactive Medical Miner. This tool discovers classification rules for each class and provides information about the statistics of these rules for all classes.\n\n## 3.4 Conclusion\n\nTo date, analysis of population-based cohort data has been mostly hypothesis-driven. We have presented a workflow and an interactive application for data-driven analysis of population-based cohort data using hepatic steatosis as an example. Our mining workflow includes steps\n\n1. to discover subpopulations that have different distributions with respect to the target variable,\n2. to classify each subpopulation taking class imbalance into account, and\n3. to identify variables associated with the target variable.\n\nOur workflow has shown that it is appropriate (a) to build subpopulations before classification to reduce class imbalance and (b) to drill-down on the derived models to identify important variables and subpopulations worthy of further investigation.\n\nTo assist the domain expert with the latter objective (b), we have developed the Interactive Medical Miner, an interactive application that allows the user to explore classification rules further and understand how the cohort participants supporting each rule are distributed across the three classes. This exploration step is essential for identifying not-yet-known associations between some variables and the target. These variables must then be further investigated – in hypothesis-driven studies. Therefore, our workflow and Interactive Medical Miner carry the potential of data-driven analysis to provide insights into a multifactorial disease and generate hypotheses for hypothesis-driven studies. Our Interactive Medical Miner has been extended by Schleicher et al. , who added panels that include tables showing additional rule statistics such as lift and p-value. Besides, a mosaic plot contrasts the class distributions of a subpopulation and its “complements,” i.e., subsets of participants who do not meet one or both conditions of a length-2 rule describing that subpopulation.\n\nIn terms of the multifactorial disorder of interest, our results confirm the potential of our data-driven approach because most of the variables in the top positions of our decision trees and classification rules have been previously shown to be associated with hepatic steatosis in independent studies. In particular, indices of fat accumulation in the body (BMI, waist circumference) and the liver enzyme GGT were proposed by Bedogni et al. as a reliable “Fatty Liver Index.” According to Yuan et al. , the SNPs rs11597390, rs2143571, and rs11597086 are among the “Independent SNPs Associated with Liver-Enzyme Levels with Genome-wide Significance in Combined GWAS Analysis of Discovery and Replication Data Sets.” Regarding the effects of alcohol consumption, toxic effects of alcohol on the liver are well established and ascribe an even more significant role to obesity than heavy alcohol consumption concerning fat accumulation in the liver . Indeed, a variable related to alcohol consumption appears only in our decision tree on F:age>52 (see Figure 3.7) and not among our top classification rules, where we tend to see variables associated with a person’s weight and obesity (cf. variables som_bmi_s2, som_huef_s2, som_waist_s2 in all figures and tables in Section 3.3). The subpopulation F:age>52 itself was identified without prior knowledge of this subpopulation’s semantics. Still, it is noteworthy that the age of 52 years is close to the onset of menopause – Völzke et al. showed that menopausal status is associated with hepatic steatosis. Our results also verify another fact that was known to medical experts by independent observation: the sonographic variables (cf. stea_s2, stea_alt75_s2 in all figures and tables in Section 3.3) is associated with liver fat concentration found on MRI, but ultrasound alone does not predict hepatic steatosis .\n\nOur algorithms not only provide variables associated with the target but also identify the value intervals related to a specific class, see, for example, the value intervals of BMI associated with class B for PartitionM (Table 3.5) and with classes A and C for F:age>52 (Table 3.4). These intervals do not imply that a person with a BMI within the specific interval actually belongs to the corresponding class but can serve as a starting point for hypothesis-driven analyses.\n\nOur approach allows us to study subpopulations at two points in time. Before the modeling step, we identify subpopulations that have different class distributions. During the modeling step, our Interactive Medical Miner highlights the subpopulation that supports each classification rule; these are overlapping subpopulations. Overlapping subpopulations are not necessarily a disadvantage, especially for very small subpopulations. However, working with overlapping data sets can be unintuitive and tedious for a domain expert. In Chapter 4, we explore the potential of clustering to identify and reorganize overlapping rules to reduce the user’s cognitive load by displaying only semantically unique and representative rules.\n\nAlthough the workflow’s main application is a longitudinal cohort study, it does not exploit temporal characteristics in the data. However, indicators of lifestyle change are potentially predictive for the later occurrence of a disease. Chapter 6 presents a follow-up method that expands the feature space by extracting temporal variables that describe how a study participant changes over time to derive new informative variables for hypothesis generation." ]

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[ "Enter the weight of asphalt (lbs) and the density of compacted asphalt (lbs/ft^3) into the Calculator. The calculator will evaluate the Asphalt Yield.\n\n## Asphalt Yield Formula\n\nAY = WA / D\n\nVariables:\n\n• AY is the Asphalt Yield (ft^3)\n• WA is the weight of asphalt (lbs)\n• D is the density of compacted asphalt (lbs/ft^3)\n\nTo calculate Asphalt Yield, divide the weight of the asphalt by the density of the compacted asphalt.\n\n## How to Calculate Asphalt Yield?\n\nThe following steps outline how to calculate the Asphalt Yield.\n\n1. First, determine the weight of asphalt (lbs).\n2. Next, determine the density of compacted asphalt (lbs/ft^3).\n3. Next, gather the formula from above = AY = WA / D.\n4. Finally, calculate the Asphalt Yield.\n5. After inserting the variables and calculating the result, check your answer with the calculator above.\n\nExample Problem :\n\nUse the following variables as an example problem to test your knowledge.\n\nweight of asphalt (lbs) = 50,000\n\ndensity of compacted asphalt (lbs/ft^3) = 146" ]

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[ "Opened 8 years ago\n\nClosed 8 years ago\n\n# Problem with points at infinity in hyperelliptic curves\n\nReported by: Owned by: gaudry AlexGhitza minor sage-5.0 algebraic geometry ecc2011, sd35, hyperelliptic curve, conic nestibal, jpflori sage-5.0.beta11 David Eklund Marco Streng N/A\n\nThe function that lists the points on a hyperelliptic curve assumes that the point [0, 1, 0] is always valid. This is wrong and creates the following bug:\n\n```sage: R.<x> = GF(7)[]\nsage: H = HyperellipticCurve(3*x^2 + 5*x + 1)\nsage: H.points()\n...\nTypeError: Coordinates [0, 1, 0] do not define a point on Hyperelliptic Curve over Finite Field of size 7 defined by y^2 = 3*x^2 + 5*x + 1\n```\n\nApply\n\n### comment:1 Changed 8 years ago by leif\n\nThe Author(s) field refers to authors of attached code, e.g. patches fixing a bug, not (necessarily the same as) those who opened a ticket.\n\nIn short, if there's no code, the Author(s) field should be empty.\n\n### comment:2 Changed 8 years ago by zimmerma\n\n• Authors Pierrick Gaudry, Paul Zimmermann deleted\n\n### Changed 8 years ago by davideklund\n\nThe patch bans defining polynomials of degree two for hyperelliptic curves, and corrects two typos.\n\n### comment:5 Changed 8 years ago by davideklund\n\n• Authors set to David Eklund\n• Status changed from new to needs_review\n\nThe problem arises exactly when `F=y^2+h*y-f` has degree 2. But then F defines a conic! Even though the definition of hyperelliptic curve seems to vary (sometimes requiring genus > 1 and sometimes including the genus 1 case) I think that the case where `y^2+h*y-f` has degree 2 should raise an error since it is rather confusing to call a rational curve hyperelliptic. Banning this case also seems to get rid of some additional weirdness such as for example the TypeError? when `h=0` and `f=x+1`.\n\nAs far as I can tell, it is not checked at the moment whether the curve defined by `y^2+h*y-f` has any singularities away from infinity, or whether it is irreducible or reduced.\n\n### comment:7 Changed 8 years ago by mstreng\n\n• Cc mstreng removed\n• Dependencies set to #11930\n• Work issues set to rebase\n\nA big rewrite of the hyperelliptic curve constructor has happened at #11930 and is not completely compatible with the patch here. It is easy to rebase the current patch to apply after #11930, but not the other way around. I'll upload a rebase.\n\n### Changed 8 years ago by mstreng\n\nrebase to apply after #11930\n\n### comment:8 Changed 8 years ago by mstreng\n\n• Description modified (diff)\n• Work issues rebase deleted\n\n### comment:9 Changed 8 years ago by mstreng\n\n• Keywords sd35 hyperelliptic curve conic added\n\n### comment:10 Changed 8 years ago by mstreng\n\n• Status changed from needs_review to needs_info\n\nActually, I don't see any reason to ban conics from being hyperelliptic curves in Sage. Mathematically, I would say that a hyperelliptic curve has genus >= 2, but I don't want to forbid people to use the HyperelllipticCurve? classes for lower-genus curves. For example, a user may have some loop going on where curves sometimes have genus 0, but sometimes higher.\n\nThe point `(0:1:0)` is not a reason to change things: for genus >=2 it is a singular point that you are not really interested in anyway! Indeed, the 0, 1, or 2 rational points that you get when you resolve that singularity are much more interesting.\n\nRather than disallowing conics, we could make a separate case in the `points` function instead.\n\n### comment:11 Changed 8 years ago by davideklund\n\nOk thanks, I will watch out for consistency with #11930.\n\nI will start working on the points method so that it allows conics, that is the case where (0:1:0) does not lie on the curve. This appears to be the better option, since changing the points method seems perfectly safe and banning conics is controversial (at least to some degree).\n\nThe problem when f is of degree 1 and h=0 (or more generally h=constant) stems from issues in the homogenization of `y^2+h*y-f` in the init method of HyperellipticCurve_generic. I will try to deal with these issues, probably opening a new ticket.\n\n### comment:12 Changed 8 years ago by davideklund\n\n• Description modified (diff)\n• Status changed from needs_info to needs_review\n\nI changed the description to a simpler example (to include in the documentation) which illustrates the same issue.\n\n### comment:14 Changed 8 years ago by mstreng\n\nNot related to the example in the ticket description or to the patch, but very much related to the title of this ticket and the first line of the description: what is \"points\" supposed to mean? I assume it means points over the base field, but of which model?\n\n• The affine model in `_repr_` means no points at infinity.\n• The plane projective model (currently underlying the implementation) is not the natural thing to look at at infinity, it always has one point `(0:1:0)` at infinity, which is singular (assuming genus > 1).\n• The smooth projective model (the actual hyperelliptic curve) is what you get when you desingularize `(0:1:0)`. It has 0, 1 or 2 points at infinity defined over the base field. Is there a framework in Sage that would make it possible to have 2 point objects representing the different points at infinity?\n\nSee #11980\n\n### comment:15 Changed 8 years ago by mstreng\n\n• Reviewers set to Marco Streng\n• Status changed from needs_review to needs_work\n\nPoints at infinity are counted incorrectly for degree 2 by this patch.\n\n```sage: C = Conic(GF(7), [1, 0, 0, -1, 0, 1])\nsage: R.<x> = GF(7)[]\nsage: H = HyperellipticCurve(x^2+1)\nsage: C\nProjective Conic Curve over Finite Field of size 7 defined by x^2 - y^2 + z^2\nsage: H\nHyperelliptic Curve over Finite Field of size 7 defined by y^2 = x^2 + 1\nsage: C.is_smooth()\nTrue\nsage: H.points()\n[(0 : 6 : 1), (0 : 1 : 1), (1 : 4 : 1), (1 : 3 : 1), (6 : 4 : 1), (6 : 3 : 1)]\nsage: H([1,1,0])\n(1 : 1 : 0)\nsage: H([1,-1,0])\n(6 : 1 : 0)\n```\n\nHere C and H represent the same curve. It is a smooth conic over a finite field of order 7, hence has 8 rational points, but only 6 are found.\n\nIn general, if H has degree 2, there are 0 or 2 rational points at infinity (as in my previous comment). Since H has degree < 4, the plane model equals the smooth model, so these points can be represented and returned correctly in Sage.\n\n### comment:16 follow-up: ↓ 17 Changed 8 years ago by davideklund\n\nIn the present patch, the points() method returns all the rational points on the plane projective model, including those on the line at infinity.\n\nI agree that when the projective plane model is singular (the typical case) we really want the points() method to spit out points on a smooth model. Maybe the issue should be raised in a separate ticket. A comment on the matter by David Kohel:\n\nThe approach of Magma is to use weighted projective space which allows one to represent two points at infinity (when the degree of `f + h^2/4` is even). In this case one gets two points at infinity rather than one (singular when g > 0).\n\n### comment:17 in reply to: ↑ 16 Changed 8 years ago by mstreng\n\nThe approach of Magma is to use weighted projective space which allows one to represent two points at infinity (when the degree of `f + h^2/4` is even). In this case one gets two points at infinity rather than one (singular when g > 0).\n\nYes, this is a standard textbook approach to hyperelliptic curves. It is requested in the documentation of rational_points as a way of representing two points at infinity. I don't think weighted projective spaces exist in Sage, so a minimal implementation of weighted projective spaces (like Magma's `WeightedProjectiveSpace`) would be a first step.\n\nAnyway, your patch looks good. I'll test it now.\n\n### comment:18 follow-up: ↓ 19 Changed 8 years ago by mstreng\n\nI see that the new patch isn't set to \"needs review\" yet, but I'll comment on it anyway: I think some explanation of the situation at infinity is in order in the documentation. How about the following?\n\n\"This method currently lists points in the plane projective model, which means that one point (0:1:0) at infinity is returned if the degree of the curve is at least 4. Later implementations may consider the more mathematically correct desingularisation at infinity, replacing (0:1:0) by 0 or 2 smooth rational points if the degree is even.\" + EXAMPLE of degree 6.\n\nAlso, your method of finding points at infinity is linear in the field order, this can be improved by taking (1:y:0) and solving for y (i.e., extracting 1 square root), exactly as currently the affine points are found by taking (x:y:1) for each x and solving for y.\n\n### comment:19 in reply to: ↑ 18 Changed 8 years ago by davideklund\n\nI see that the new patch isn't set to \"needs review\" yet, but I'll comment on it anyway: I think some explanation of the situation at infinity is in order in the documentation. How about the following?\n\n\"This method currently lists points in the plane projective model, which means that one point (0:1:0) at infinity is returned if the degree of the curve is at least 4. Later implementations may consider the more mathematically correct desingularisation at infinity, replacing (0:1:0) by 0 or 2 smooth rational points if the degree is even.\" + EXAMPLE of degree 6.\n\nAlso, your method of finding points at infinity is linear in the field order, this can be improved by taking (1:y:0) and solving for y (i.e., extracting 1 square root), exactly as currently the affine points are found by taking (x:y:1) for each x and solving for y.\n\nThank you for your comments! I hoped to attend to this earlier. I think I will get some time to work on it soon.\n\nYes, some comment like the one you supply would be in order in the documentation. About the method for finding points: I used brute force since I deliberately choose simple code before efficiency. But I will implement the refined approach you suggest.\n\n### Changed 8 years ago by davideklund\n\nPatch updated. More efficient method to find the points at infinity and added documentation.\n\n### comment:20 Changed 8 years ago by davideklund\n\n• Status changed from needs_work to needs_review\n\n### comment:21 Changed 8 years ago by mstreng\n\nfor patchbot:\n\napply trac_11800_allow_conics.patch\n\n### Changed 8 years ago by mstreng\n\nbreak at 79 characters, shortened polynomial coefficient access\n\n### comment:22 Changed 8 years ago by mstreng\n\n• Dependencies #11930 deleted\n• Description modified (diff)\n\n### comment:23 follow-up: ↓ 24 Changed 8 years ago by mstreng\n\nPython style guides ask for breaking of lines at 79 characters. Aside from that, the patch is good. If you agree with my reviewer patch, then feel free to set the ticket to \"positive review\".\n\n### comment:24 in reply to: ↑ 23 Changed 8 years ago by davideklund\n\n• Status changed from needs_review to positive_review\n\nPython style guides ask for breaking of lines at 79 characters. Aside from that, the patch is good. If you agree with my reviewer patch, then feel free to set the ticket to \"positive review\".\n\nOk, thanks! I agree with your changes.\n\n### comment:25 Changed 8 years ago by jdemeyer\n\n• Merged in set to sage-5.0.beta11\n• Resolution set to fixed\n• Status changed from positive_review to closed\n\n### comment:26 Changed 7 years ago by mstreng\n\nSee #15115 for getting the correct points at infinity in general.\n\nNote: See TracTickets for help on using tickets." ]

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[ "# Fast Block LMS Filter\n\nCompute output, error, and weights using LMS adaptive algorithm\n\n## Library\n\n`dspadpt3`\n\n•", null, "## Description\n\nThe Fast Block LMS Filter block implements an adaptive least mean-square (LMS) filter, where the adaptation of the filter weights occurs once for every block of data samples. The block estimates the filter weights, or coefficients, needed to convert the input signal into the desired signal. Connect the signal you want to filter to the Input port. The input signal can be a scalar or a column vector. Connect the signal you want to model to the Desired port. The desired signal must have the same data type, complexity, and dimensions as the input signal. The Output port outputs the filtered input signal. The Error port outputs the result of subtracting the output signal from the desired signal.\n\nThe block calculates the filter weights using the Block LMS Filter equations. For more information, see Block LMS Filter. The Fast Block LMS Filter block implements the convolution operation involved in the calculations of the filtered output, y, and the weight update function in the frequency domain using the FFT algorithm used in the Overlap-Save FFT Filter block. See Overlap-Save FFT Filter (Obsolete) for more information.\n\nUse the Filter length parameter to specify the length of the filter weights vector.\n\nThe Block size parameter determines how many samples of the input signal are acquired before the filter weights are updated. The input frame length must be a multiple of the Block size parameter.\n\nThe Step-size (mu) parameter corresponds to µ in the equations. You can either specify a step-size using the input port, Step-size, or enter a value in the Block Parameters: Block LMS Filter dialog box.\n\nUse the Leakage factor (0 to 1) parameter to specify the leakage factor, $0<1-\\mu \\alpha \\le 1$, in the leaky LMS algorithm shown below.\n\n`$w\\left(k\\right)=\\left(1-\\mu \\alpha \\right)w\\left(k-1\\right)-f\\left(u\\left(n\\right),e\\left(n\\right),\\mu \\right)$`\n\nEnter the initial filter weights, $w\\left(0\\right)$, as a vector or a scalar in the Initial value of filter weights text box. When you enter a scalar, the block uses the scalar value to create a vector of filter weights. This vector has length equal to the filter length and all of its values are equal to the scalar value.\n\nWhen you select the Adapt port check box, an Adapt port appears on the block. When the input to this port is nonzero, the block continuously updates the filter weights. When the input to this port is zero, the filter weights remain at their current values.\n\nWhen you want to reset the value of the filter weights to their initial values, use the Reset input parameter. The block resets the filter weights whenever a reset event is detected at the Reset port. The reset signal rate must be the same rate as the data signal input.\n\nFrom the Reset input list, select `None` to disable the Reset port. To enable the Reset port, select one of the following from the Reset input list:\n\n• `Rising edge` — Triggers a reset operation when the Reset input does one of the following:\n\n• Rises from a negative value to a positive value or zero\n\n• Rises from zero to a positive value, where the rise is not a continuation of a rise from a negative value to zero (see the following figure)", null, "• `Falling edge` — Triggers a reset operation when the Reset input does one of the following:\n\n• Falls from a positive value to a negative value or zero\n\n• Falls from zero to a negative value, where the fall is not a continuation of a fall from a positive value to zero (see the following figure)", null, "• `Either edge` — Triggers a reset operation when the Reset input is a `Rising edge` or `Falling edge` (as described above)\n\n• `Non-zero sample` — Triggers a reset operation at each sample time that the Reset input is not zero\n\nSelect the Output filter weights check box to create a Wts port on the block. For each iteration, the block outputs the current updated filter weights from this port.\n\n## Parameters\n\nFilter length\n\nEnter the length of the FIR filter weights vector. The sum of the Block size and the Filter length must be a power of 2.\n\nBlock size\n\nEnter the number of samples to acquire before the filter weights are updated. The number of rows in the input must be an integer multiple of the Block size. The sum of the Block size and the Filter length must be a power of 2.\n\nSpecify step-size via\n\nSelect `Dialog` to enter a value for mu, or select `Input port` to specify mu using the Step-size input port.\n\nStep-size (mu)\n\nLeakage factor (0 to 1)\n\nEnter the leakage factor, $0<1-\\mu \\alpha \\le 1$. Tunable (Simulink).\n\nInitial value of filter weights\n\nSpecify the initial values of the FIR filter weights.\n\nSelect this check box to enable the Adapt input port.\n\nReset input\n\nSelect this check box to enable the Reset input port.\n\nOutput filter weights\n\nSelect this check box to export the filter weights from the Wts port.\n\n## References\n\nHayes, M.H. Statistical Digital Signal Processing and Modeling. New York: John Wiley & Sons, 1996.\n\n## Supported Data Types\n\nPortSupported Data Types\n\nInput\n\n• Double-precision floating point\n\n• Single-precision floating point\n\nDesired\n\n• Must be the same as Input\n\nStep-size\n\n• Must be the same as Input\n\n• Double-precision floating point\n\n• Single-precision floating point\n\n• Boolean\n\n• 8-, 16-, and 32-bit signed integers\n\n• 8-, 16-, and 32-bit unsigned integers\n\nReset\n\n• Double-precision floating point\n\n• Single-precision floating point\n\n• Boolean\n\n• 8-, 16-, and 32-bit signed integers\n\n• 8-, 16-, and 32-bit unsigned integers\n\nOutput\n\n• Same as Input\n\nError\n\n• Same as Input\n\nWts\n\n• Same as Input" ]

[ null, "https://es.mathworks.com/help/dsp/ref/fast_block_lms_block.gif", null, "https://es.mathworks.com/help/dsp/ref/rising_edge.png", null, "https://es.mathworks.com/help/dsp/ref/falling_edge.png", null ]

[ "# Present Value of $260 in 71 Years When you have a single payment that will be made to you, in this case$260, and you know that it will be paid in a certain number of years, in this case 71 years, you can use the present value formula to calculate what that $260 is worth today. Below is the present value formula we'll use to calculate the present value of$260 in 71 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 $260 • n: This is the number of periods, which is 71 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$260 paid in 71 years for interest rates from 2% to 30%.\n\nAs you will see, the present value of $260 paid in 71 years can range from$0.00 to $63.73. Discount Rate Future Value Present Value 2%$260 $63.73 3%$260 $31.88 4%$260 $16.05 5%$260 $8.14 6%$260 $4.15 7%$260 $2.13 8%$260 $1.10 9%$260 $0.57 10%$260 $0.30 11%$260 $0.16 12%$260 $0.08 13%$260 $0.04 14%$260 $0.02 15%$260 $0.01 16%$260 $0.01 17%$260 $0.00 18%$260 $0.00 19%$260 $0.00 20%$260 $0.00 21%$260 $0.00 22%$260 $0.00 23%$260 $0.00 24%$260 $0.00 25%$260 $0.00 26%$260 $0.00 27%$260 $0.00 28%$260 $0.00 29%$260 $0.00 30%$260 $0.00 As mentioned above, the discount rate is highly subjective and will have a big impact on the actual present value of$260. A 2% discount rate gives a present value of $63.73 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 \\$260 over the next 71 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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[ "# Backtesting\n\n## Introduction\n\nA common practice in evaluating backtests of trading strategies is to discount the reported Sharpe ratios by 50%. There are good economic and statistical reasons for reducing the Sharpe ratios. The discount is a result of data mining. This mining may manifest itself by academic researchers searching for asset pricing factors to explain the behavior of equity returns or by researchers at firms that specialize in quantitative equity strategies trying to develop protable systematic strategies.\n\nThe 50% haircut is only a rule of thumb. The goal of our paper is to develop an analytical way to determine the magnitude of the haircut.\n\nOur framework relies on the statistical concept of multiple testing. Suppose you have some new data, Y, and you propose that variable X explains Y. Your statistical analysis finds a significant relation between Y and X with a t-ratio of 2.0 which has a probability value of 0.05. We refer to this as a single test. Now consider the same researcher trying to explain Y with variables X1;X2,.....,X100. In this case, you cannot use the same criteria for significance. You expect by chance that some of these variables will produce t-ratios of 2.0 or higher. What is an appropriate cut-o for statistical significance?\n\nIn Harvey and Liu (HL, 2015), we present three approaches to multiple testing. We answer the question in the above example. The t-ratio is generally higher as the number of tests (or X variables) increases. Consider a summary of our method. Any given strategy produces a Sharpe ratio. We transform the Sharpe ratio into a t -ratio. Suppose that t-ratio is 3.0. While a t-ratio of 3.0 is highly significant in a single test, it may not be if we take multiple tests into account. We proceed to calculate a p-value that appropriately reflects multiple testing. To do this, we need to make an assumption on the number of previous tests.\n\nFor example, Harvey, Liu and Zhu (HLZ, 2015) document that at least 316 factors have been tested in the quest to explain the cross-sectional patterns in equity returns. Suppose the adjusted p-value is 0.05. We then calculate an adjusted t-ratio which, in this case, is 2.0. With this new t -ratio, we determine a new Sharpe ratio. The percentage difference between the original Sharpe ratio and the new Sharpe ratio is the \"haircut\".\n\nThe haircut Sharpe ratio that obtains as a result of multiple testing has the following interpretation. It is the Sharpe ratio that would have resulted from a single test, that is, a single measured correlation of Y and X.\n\nWe argue that it is a serious mistake to use the rule of thumb 50% haircut. Our results show that the multiple testing haircut is nonlinear. The highest Sharpe ratios are only moderately penalized while the marginal Sharpe ratios are heavily penalized.\n\nThis makes economic sense. The marginal Sharpe ratio strategies should be thrown out. The strategies with very high Sharpe ratios are probably true discoveries. In these cases, a 50% haircut is too punitive.\n\nOur method does have a number of caveats { some of which apply to any use of the Sharpe ratio. First, high observed Sharpe ratios could be the results of non-normal returns, for instance an option-like strategy with high ex ante negative skew.\n\nIn this case, Sharpe ratios need to be viewed in the context of the skew. Dealing with these non-normalities is the subject of future research. Second, Sharpe ratios do not necessarily control for risk. That is, the volatility of the strategy may not reflect the true risk. Importantly, our method also applies to Information ratios which use residuals from factor models. Third, it is necessary in the multiple testing framework to take a stand on what qualifies as the appropriate significance level, e.g., is it 0.10 or 0.05? Fourth, a choice needs to made on the multiple testing method. We present results for three methods as well as the average of the methods. Finally, some judgment is needed specifying the number of tests.\n\nGiven choices (3)-(5), it is important to determine the robustness of the haircuts to changes in these inputs. We provide a program at: http://faculty.fuqua.duke.edu/charvey/backtesting that allows the user to vary the key parameters to investigate the impact on the haircuts. We also provide a program that determines the minimal level of protability for a trading strategy to be considered \"significant\"." ]

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[ "11,053 Pages\n\nConway's Game of Life (GOL for short) is a game created by John Horton Conway in the late-middle 20th century and is an example of a very complex system arising from a set of simple rules. The rules of GOL dictate how squares are turned \"on\" or \"off\" in an infinite arena of squares over time, where one set of changes is called a \"generation\".\n\nEvery square in this grid of squares is surrounded by eight squares. The number of switched on squares in this eight-square ring is called the \"count\". There are three rules that dictate the cells' status:\n\n1. Death: if the count is less than 2 or greater than 3, the current square is switched off.\n2. Survival: if the count is exactly 2 or 3 and the current square is on, the current square is left unchanged.\n3. Birth: if the current square is off and the count is exactly 3, the current square is switched on.\n\nThese simple rules allow a great variety of structures and behaviors to manifest in the Game of Life, which are being extensively researched. These patterns include a prime-number calculator as well as a simulation of GOL itself in GOL. The Game of Life is also Turing-complete, which means that any function computable on a finite-state Turing machine can also be computed in GOL.\n\nIn the Game of Life, there is a fastest speed activated squares may spread, referred to as the \"speed of light\" and is used to describe the speed of periodic structures such as the glider or spaceships in general. For example, a glider move one cell diagonally every 4 generations, so its speed is denoted c/4.\n\n## Sources\n\nCommunity content is available under CC-BY-SA unless otherwise noted." ]

[ null ]

[ "# Writing a circle equation examples\n\nQuadratic equation is the second degree equation in one variable contains the variable with an exponent of 2. Let the center be the point h, k and the radius be r.\n\nIt is important to keep in mind that LaTeX has its own way of handling spacing in mathematics mode. The majority of chemical reactions fall into five categories. If we add the adult tickets and children tickets together, we will have a total. Moreover, if you know how to use the symbolic algebra package Maple V, you can get it to produce LaTeX code.\n\nStep 2 above is the most important to remember. An almost inconceivable number of chemical reactions are possible. Left Ball in circular motion — rope provides centripetal force to keep ball in circle Right Rope is cut and ball continues in straight line with velocity at the time of cutting the rope, in accord with Newton's law of inertia, because centripetal force is no longer there.", null, "In a chemical reaction atoms are neither created nor destroyed; they are merely rearranged. A circle can be formed by slicing a right circular cone with a plane traveling parallel to the base of the cone. Solution Think about what you know.", null, "I know it's hard to imagine when you will need to use this skill, or even how to solve problems written in this form. It is hard and unfortunately, it just takes practice.\n\nThe identities of the reactants and products. The function value changes as 'x' changes, so function is dependent upon 'x'. However, we cannot consider all of these equations as Functions. Often times, text book word problems are pretty \"far out\"; however Algebra Class word problems are more realistic and easy to relate to every day living.\n\nThe equation of circles cannot be considered as function as here one input can give you two outputs. We will now extend our discussion and look at the trigonometric functions.\n\nWhen you are evaluating the value of a function, place the value of variable in parenthesis rather than variable. What is a Chemical Equation. Quantify the description by balancing the equation.", null, "This package provides features for aligning several equations, for handling multiline equations, compound symbols and certain kinds of diagrams. The equation also explains the energy terms whether it is absorbed or evolved.\n\nIf you are struggling with this concept, then check out the Algebra Class E-course. Linear equations and quadratic equations are the example of the algebraic equations. Where did all the extra numbers come from.", null, "Later on we will see how to create a label for an equation so that we can refer to the equation in the paper without having to know which number will be assigned to it during typesetting.\n\nCompleting the Square: Circle Equations The technique of completing the square is used to turn a quadratic into the sum of a squared binomial and a number: (x – a) 2 + b.\n\nThe center-radius form of the circle equation is in the format (x – h) 2 + (y – k) 2 = r 2, with the center being at the point (h, k) and the radius being \" r \". In mathematics, a parametric equation defines a group of quantities as functions of one or more independent variables called parameters.\n\nwhich is the standard equation of a circle centered at the origin. Examples in two dimensions Parabola. The simplest equation for a parabola. Solved example to find the equation of a concentric circle: Find the equation of the circle which is concentric with the circle 2x\\(^{2}\\) + 2y\\(^{2}\\) + 3x - 4y + 5 = 0 and whose radius is 2√5 units.\n\nASSIGNMENT BY. SHADRECK S CHITSONGA. PARAMETRIC CURVES.", null, "Let us start by doing a quick review of the ordinary equations. y = mx + c is the general form of a linear function. We know that when we plot this function in the Cartesian plane we get a straight line. \"Starting at \\$ \\ (5, \\ 0) \\$ \" means you are on the circumference of the circle and tracing along the circle.\n\nThe center of the circle is not the starting point. The center of the circle is not the starting point. Explore math with gabrielgoulddesign.com, a free online graphing calculator.\n\nWriting a circle equation examples\nRated 5/5 based on 23 review\nSolving Word Problems in Algebra" ]

[ null, "http://slideplayer.com/696028/2/images/22/Example 1 Writing an Equation of a Circle.jpg", null, "http://slideplayer.com/3431818/12/images/2/Example 1 Write the Equation of a circle with center (2,4) and containing the point (8,12) Use the Distance Formula to find the radius..jpg", null, "http://slideplayer.com/3431207/12/images/6/Notes Over 10.3 Writing an Equation of a Circle.jpg", null, "http://slideplayer.com/5999078/20/images/8/Ex. 1: Writing a Standard Equation of a Circle centered at (-4, 0) and radius 7.1.jpg", null, "http://slideplayer.com/3431588/12/images/27/Example of a Standard Equation of a Circle.jpg", null ]

[ "## The Radius and Centre of a Circle in Apollonian Form\n\nLet", null, "be the generalized circle with  equation", null, "with", null, "and", null, "If", null, "then", null, "is the circle with centre", null, "and radius", null, "where", null, "and", null, "Also", null, "lies on the line through", null, "and", null, "and", null, "If", null, "then", null, "is the line through", null, "perpendicular to", null, "Proof\n\nLet", null, "be the extended mobius transformation defined by", null, "", null, "maps the circle defined by", null, "to the unit circle and so", null, "maps the unit circle to", null, "so that", null, "and", null, "Since", null, "points on the extended real axis are mapped by", null, "to points on the extended line", null, "through", null, "and", null, "If", null, "then the diametrically opposite points -1 and 1 are mapped by", null, "to the diametrically opposite points", null, "and", null, "of", null, "on", null, "because", null, "is conformal at 1 and -1.", null, "It follows that", null, "has centre", null, "on", null, "given by", null, "and radius", null, "given by", null, "Furthermore", null, "", null, "If", null, "then", null, "so", null, "must be a line that meets", null, "in a right angle at", null, "", null, "" ]

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[ "# 1989 USAMO Problems/Problem 2\n\n## Problem\n\nThe 20 members of a local tennis club have scheduled exactly 14 two-person games among themselves, with each member playing in at least one game. Prove that within this schedule there must be a set of 6 games with 12 distinct players\n\n## Solution 1\n\nConsider a graph with", null, "$20$ vertices and", null, "$14$ edges. The sum of the degrees of the vertices is", null, "$28$; by the Pigeonhole Principle at least", null, "$12$ vertices have degrees of", null, "$1$ and at most", null, "$8$ vertices have degrees greater than", null, "$1$. If we keep deleting edges of vertices with degree greater than", null, "$1$ (a maximum of", null, "$8$ such edges), then we are left with at least", null, "$6$ edges, and all of the vertices have degree either", null, "$0$ or", null, "$1$. These", null, "$6$ edges represent the", null, "$6$ games with", null, "$12$ distinct players.\n\n## Solution 2\n\n\\indent Let a slot be a place we can put a member in a game, so there are two slots per game, and 28 slots total. We begin by filling exactly 20 slots each with a distinct member since each member must play at least one game. Let there be", null, "$m$ games with both slots filled and", null, "$n$ games with only one slot filled, so", null, "$2m+n=20$. Since there are only 14 games,", null, "$m+n \\leq 14 \\Longrightarrow 2m+n \\leq 14+m \\Longleftrightarrow 20 \\leq 14+m \\Longrightarrow m \\geq 6$, so there must be at least 6 games with two distinct members each, and we must have our desired set of 6 games." ]

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[ "# Display Result Of Formula In Cell\n\nNov 15, 2006\n\nSub startup\n\nDim numwords As Integer\n\nnumwords = WorksheetFunction. CountA(\"H5:H64\")\nRange(\"H1\").Select\nSelection.FormulaR1C1 = numwords\n\nI have a quick question regarding this simple code thats been drviing me nuts. There x distinct pieces of Data in the range speciifed, but when i run this code, the reply posted in cell H1 is 1, instead of x.\n\nEventually the code will be built to check if the result is odd or even and then add an entry into the first blank cell if the result is odd.\n\n## Define Cell To Display Formula Result In?\n\nJun 5, 2014\n\nI am trying to find a way to display text that is the product of a concatenate function in a defined cell, but not to have the function itself in that cell. Basically, I want to have the below function in cell A1. I want to add a command to it to take the result and display it in cell A2.\n\n=CONCATENATE(C5,\"_\",'Attachment 3-A'!C9:E9,\"_\",IF(C5=\"\",\"\",IF(C6=\"MPL\",IF(VLOOKUP(C5,MPL!B:F,5,FALSE)=\"Lease\",\"Lease_Contributable\",\nIF(VLOOKUP(C5,MPL!B:F,5,FALSE)=\"Pre-Lease\",\"Managed_Pre-Lease\",\nIF(VLOOKUP(C5,MPL!B:F,5,FALSE)=\"Non-Contributable\",\"Managed_Non-Contributable\",\"Error\"))),\nIF(VLOOKUP(C5,Sale!B:F,5,FALSE)=\"Assignable\",\"Lease_Assignable\",\"Managed_Non-Assignable\"))))\n\nThe reason for this is that I need the cell to be selectable (it is generating a file name that needs to be selected and copied), but I don't want the code behind it to be seen. I can't find a way to make the cell selectable AND hide the formula from being seen when the cell is selected. This is because our people keep copying the formula rather than the resulting text.\n\nI was hoping for something like a DISPLAYIN(target) function, but it doesn't seem to exist.\n\n## Display Formula In A1, Result In A2\n\nSep 26, 2007\n\nI have a formula in A1 that I would like to execute from A2. I want the A1 to display the formula (so no \"=\"), but I would like cell A2 to execute the formula in A1.\n\nIf A1 contains: \"SUM(1+2)\", how can I get A2 to display \"3\" without reproducing the formula?\n\nA1: SUM(1+2)\nA2: ????\n\nI suppose I'm looking for something like this:\nA2: ==A1\nBut, of course, that doesn't work.\n\nIf I do this it gets close:\nA2: =\"=\"&A1\n\nBut that returns \"=SUM(1+2)\" instead of executing the formula.\n\n## How To Display Result From Left Of Max Formula\n\nMar 6, 2014\n\nI am using the below formula to find the latest date (column N) based on fund (column G) and Vendor (column O) reference. How can I find (column I) where the balance of that payment is? I know there are formulas like INDIRECT RC[-1] but how to add them to my formula.\n\n{=MAX(IF(Payments!G:G=B4,IF(Payments!O:O=C4,Payments!N:N)))}\n\n## If / Vlookup Display Result Not Formula\n\nDec 8, 2009\n\nHopefully you will be able to help again. Is it possible to do a vlookup that references data on other tabs within the worksheet so that the result of the formula is in the cell not the formula.\n\nSo if my vlookup was =vlookup(a2,\\$a\\$1:\\$b\\$12,3,0) and the result was john smith i want just john smith in the cell. I know about copy and paste values but i was looking for a more automatic way. One that doesn't need intervention.\n\n## Display Result And Hide Macro Formula\n\nMar 17, 2009\n\nExcel 2003.\nI have been struggling for an hour how to hide a number to text macro in a way that only macro result is displayed in a cell.\n\n(I have one 2-3 years old xls, where I have managed with task, but now can't figure out how and how to unhide the macro\nCan it be password protected somehow?\n\n## Stop Displaying Formula - Display Result\n\nJun 10, 2006\n\nI entered a simple formula in a cell =B14*B16 but it won't calculate. The cell only displays the formula and not the result. tell me what I did wrong. I've attached a sample.\n\n## VLOOKUP Formula Dragdown Copies Previous Cell Result Instead Of Unique Result\n\nJun 10, 2014\n\nWhen I drag my VLOOKUP formula down a column in Excel 2010, the return value copies the formula result from the original VLOOKUP formula result. For example, if the first VLOOKUP returns a value of 0.5, I expect to see 0.5 or 1 in the cell below that one. However, I get 0.5 which is not the expected result for the cell below.\n\nWhen, I click the fx on the cells below, the expected return values appear in the formula result. After I click OK, the expected formula results updates and now appears in the cell.\n\nI'm not sure what is causing this issue. My computer was updated recently from an old machine to a new one. I have never experienced this issue before.\n\n## Display Formula Not Formula Result?\n\nMay 2, 2012\n\nI have a s/s which is built on IF functions and references other sheets and I have used the s/s in the past without issue. However now I want to ammend the formula a little and excel will only display the formula itself, not the result of the formula. I have tried CTRL ~ to turn on/off the show formula function but this makes no difference.\n\nMy reason for changeing the current formula is that I need to turn a 2 (numeric) to 02 (which can be either numeric or text). This is the new formula.\n\n=IF(\\$A4=\"\",\"\",IF(VLOOKUP(\\$A4,HCGTH1_DATA,22)=2,\"02\",VLOOKUP(\\$a4,HCGTH1_DATA,22)))\n\nIs there an issue with the formula or some setting in excel?\n\n## Result Using If: Cell To Display The Word OK\n\nFeb 11, 2009\n\nI have a formula on cell A1, if the result of this: =SUM(COUNTIF(F:F;M40);COUNTIF(M:M;M40)) is equal to 1, I want the cell to display the word OK.\n\nI have tryed this but it didn't work:\n=IF((SUM(COUNTIF(F:F;M40);COUNTIF(M:M;M40)))=1;\"ok\";\"error\")\n\n## Compute And Display Result In Cell\n\nSep 26, 2009\n\nI have 15*2,14*2,14.5 in cell A1 and would like this to be computed and its result (i.e 72.5) displayed in cell A2. See below for clearer picture:\n\nA1 A2\n15*2,14*2,14.5 72.5\n\n## Display Many Cell Result In Message Box\n\nNov 9, 2006\n\ni have a routine which loops through a range looking for past dates and when it finds one display the result in a message box in my testbook i only use 25 rows but it gets annoying having to click ok for every find....is it possible to collect all results and display them in the message box at one time?\n\nPrivate Sub Workbook_Open()\nDim Mycell\nDim Rng\nSet Rng = Sheets(\"Sheet1\").Range(\"B1:B25\")\nFor Each Mycell In Rng\nIf Mycell.Value < Date Then\nMsgBox Mycell.Offset(0, -1).Value & \" Is Overdue By \" & Date - Mycell.Value & \" Days, Take Action Now!\", vbOKOnly, \"Tasks Overdue\"\nEnd If\nNext Mycell\nEnd Sub\n\n## Display Cell Address Based On Result Of Combobox\n\nJan 29, 2013\n\nI have a combobox that returns me the names that are in a spreadsheet.\n\nI need a return label, the index (address of that cell that the combobox returned), how do I do that?\n\n## Formula To Display Column Header As Result When Any Value Exists In Cells Under Column\n\nJul 25, 2014\n\nI am trying to write a formula where the column header of the row in which a value other than 0 exists, will display for each instance (row) where a value exists in an array spanning 3 columns. So the result cell could be any of the three column headers, or a combination thereof.\n\nI started the formula in P2 of the GL Detail-2012 tab. File attached.\n\nHere is what I started: =INDEX(\\$M\\$1:\\$O\\$1,SUMPRODUCT(COUNTIF(\\$M\\$2:\\$O\\$67756,))). Not working.\n\n## Formula Cell Showing Formula Not Result\n\nAug 24, 2006\n\nI know this is an easy one but I'm unsure why I can't easily modify a formula. The key event in this formula is \"+190\". Every time I change the value the formula no longer works. I've copied to another cell and the formula no longer works. My question is when you have a formula like this one if you have to modify how do you do it and keep the formula working.\n\n## Display Only One Of Each Result?\n\nMay 2, 2014\n\nI've got results in row 3 - 16 386 and sometimes these results are exactly the same in some rows. What I now would need is row AB to display only one of each result and row AH to display how many percentages of the time this particular result appears.\n\n## Formula In Same Cell As Result\n\nDec 15, 2009\n\nI have a value in Cell A1. Now, I need to be able to enter a number into Cell A3 and upon hitting Enter (or clicking elsewhere), the number I just entered into Cell A3 needs to be replaced with the number I just entered multiplied by the number in Cell A1. For Example: Cell A1 contains the number 1.05. I type 2.50 into Cell A3 and upon hitting Enter Cell A3 reads 2.625.\n\n## IF Formula That References A Cell And Returns A Different Result Dependant On The Number In The Cell Being Referenced\n\nOct 2, 2009\n\nI'm trying to do a formula that references a cell and returns a different result dependant on the number in the cell being referenced.\n\nFor example I've said if A1 has a 3 in it then put the word TEST as the result, plus if it has a 4 put the word RESULT.\n\nWhat I wrote as my formula is as follows-\n\n=IF(A1=3,\"TEST\")+IF(A1=4,\"RESULT\")\n\nIt works fine when I only use one result but goes wrong when I add two. If I change the words I want to show to numbers it comes up fine but with words it just returns a Value error.\n\n## If Number =1 Display The Result\n\nFeb 10, 2009\n\nI have a count for each site for certain cloumn headers.\nBut i want to collate these so that if there is a 1 in the column the it will output it with the column header. But there are 10 column headers and I would like to get a result that has all the columns with 1 in.\n\neg:\na b c d e f g h i j k l m n o p q (Organic Suites) (Inorganic Suites)\nsitea 1 0 1 0 1 1 0 1 0 1 1 1 0 0 0 0 (O1 O3 O5 O6 O8 OS) (I1 I2)\n\n## Sum A Column And Then Display The Result\n\nJan 15, 2010\n\nI want to sum a column and then display the result but in the same cell put some text:-.....\n\n## Formula Result To Cell As Text?\n\nFeb 5, 2014\n\nI have the formula in O62 cell :\n\n[Code]....\n\nIn formula bar when I sellect complete formula and press F9 key, I can see:\n\n[Code] .........\n\nresults.\n\nHow can I write this visible result to \"O62\" cell or, \" another cell example\"P62\" as a text value.(without to copy & paste).\n\nI want to make this with a formula ( if impossible by macro).\n\n## Formula Result As Cell Reference\n\nOct 1, 2009\n\nI need to use the result of a simple calculation in one cell on Sheet1 (=A1+2) as the row in a cell reference on a different sheet. eg =Sheet2!A\"n\" where \"n\" is the result of formula.\n\n## Cell Displays Formula Instead Of Result?\n\nJul 24, 2014\n\nI'm working on someone else's spreadsheet, when I type in a formula the cell shows the actual formula as typed, rather than the expected result of the formula, how do Ii correct this?\n\n## Formula Showing In The Cell But Not The Result\n\nAug 23, 2002\n\nI have a formula in a cell and when I use the formula bar it shows me the correct answer but it doesn't show me it in the worksheet. Only the formula shows up.\n\nThe same formula could be another place in my worksheet and work fine.\n\nI have tried changing format. Copying formula from another location and changing the information to fit my needs it won't show me the result.\n\n## Cell Displays Formula Instead Of Result\n\nNov 6, 2007\n\n=K18+Sheet117!K18\n\nK18 contains the number 54.00, and on Sheet117 cell K18 contains the number 404.00.\n\nCell A1 displays the formula rather than the result, which should be 458.00.\n\nOther similar formulas I have in this workbook return a result of 0 when I know there should be a significantly higher number. Then, the same formula used to refer to different cells will return what appears to be a proper result.\n\n## Having Macro Act Upon Formula Result In Cell\n\nNov 10, 2009\n\nI have created a visual schedule for my team using Excel (2003, SP3 if that matters). Essentially, the user puts in pre determined 1-3 character codes in individual cells, and the macros I have act like a complex Conditional Formatting to keep the formatting neat and consistent throughout the sheet. The actual values are inputted directly into the cells though (this is pertinent to my question), and are things like \"A\" \"M\" \"\\$\" and \"TR\"\n\nI have a sheet for every day in a week. Since there are multiple team members on any given day, I have recently made another sheet which pulls a single Team Member's schedule Sunday through Saturday and displays their schedule for the entire week. I have used formulas (specifically VLOOKUP) to do this.\n\nThe problem that I am having though, is that the macros that I made to format the days of the week sheets, do not seem to recognize the result of the formulas in the individual Team Member's sheet, and thus do not format them as desired.\n\nMy macros are written to evaluate a cell's value via [ Range(\"example\").value ] and will act upon it accordingly with more code. I am assuming that a [ Range(\"example\").value ] would see a cells value as the text of the inputted formula, and not the result of that formula. Is there any way around this? or do I need to avoid the formulas all together and write in code to just copy over what I need?\n\n(I hope this makes sense)\n\n## Send Formula Result To Another Cell\n\nDec 30, 2006\n\nHow is this possible,\n\nin cell A3, (''if'' cell A1 has text \"transfers'')\nthen cell A2=7\n\n## Formula Shows In Cell, Not Result\n\nJan 3, 2008\n\nWhy is it that when I edit some cell's formulas and press enter the result is not the changed formula but the formula itself complete with the '=' sign infront of the fuormula. The work around for me is to cut the formula and paste it into a new cell then drag the old cell over the previous one I tried to edit.\n\n## Refernceing Cells: Formula That Looks At A Cell On A Front Sheet, And Then Returns The Contents Of That Cell As The Result If It Meets The Criteria\n\nApr 5, 2009\n\nI have a formula that looks at a cell on a front sheet, and then returns the contents of that cell as the result if it meets the criteria. So for example this formula would be in Cell A1 on Sheet2 IF(SHEET1!A1,\"New\",Sheet1!A1,\"-\")\n\nThis formula is always in the same cell (different sheet) as the cell that it is looking at, down 1500 rows. Instead of having the formula named for each cell is there anyway to ask excel to 'look at this cell but on this other sheet'.\ne.g IF(Sheet1!\"This Cell\" etc). That way no matter what cell you put the formula in it is always referencing the correct cell for the formula?\n\n## Copy Formula Result & Paste Value/Result Only\n\nSep 3, 2006\n\nI created a simple auto numbering function whereby Cell A7 contained =Row()-6, and Cell A8 contained =(A7+1). I then shift, and pasted the contents of cell A8 until cell A600. My aim is to simply copy the increments of 1 - 600 into another column. However when i copy and paste i'm also copying the initial underlying formula ie: =( A?+1), Is there a way to copy the results, not the formula?" ]

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[ "# Algebra mcqs Test 09", null, "Algebra mcqs test 09 consist of 10 most important multiple choice questions. Prepare these questions for better results and also you can prepareof algebra.\n\n1. The symmetries of rectangle form a\n\n2. Let  be a dihedral group of order 8. Then which of the following is a subgroup of D4\n\n3. Let An be the set of all even permutations of Sis a subgroup of Sn. Then order of Ais\n\n4. The set of cube roots of unity is a subgroup of\n\n5. Any group G van be embedded in a group of bijective mappings of certain sets is a statement of\n\n6. The union of all positive even and all positive odd integers is\n\n7. If X and Y are two sets, then X∩(XUY)’=0\n\n8. Let G be a finite group. Let H be a subgroup of G . Then which of the following divides the order of G\n\n9. Let G be a cyclic group of order 24. Then order of  is\n\n10. Which of the following is abelian" ]

[ null, "https://www.pakmath.com/wp-content/uploads/2019/03/aljebra-mcqs-test-09-300x225.png", null ]

[ "I was asked the following question the other day on White’s reality check.\n\nQUESTION:\n\nI was reading White’s (2000) paper (http://www.ssc.wisc.edu/~bhansen/718/White2000.pdf). It seems to suggest that to examine the pnl characteristic of a strategy, we should do bootstrapping on the P&L. I am wondering why this makes sense. Why scramble the P&L time series? Why not scramble the original returns series, e.g., returns of S&P 500, paste the segments together, and feed the “fake” returns series into the strategy again? Then you can generate as many P&L time series as you like. Then you can use them to compute the expected P&L, Sharpe-ratio, etc. Would the 2nd approach be more realistic than the approach suggested in the paper?\n\n1) In the statistical literature, there are basically two approaches to tackling the data snooping bias problem. The first and more popular approach focuses on data. It tries to avoid re-using the same data set and can be done by testing a particular model on randomly chosen subsamples of the original time series or bootstrapping the original time series (e.g. return). However, one may argue that this sampling approach is somewhat arbitrary and may lack desired objectivity. See Chan, Karceski, and Lakonishok (1998) and Brock, Lakonishok, and Lebaron (1992) for examples of this sampling approach. Your proposed method basically falls into this category.\n\nThe second and more “formal” approach focuses on models by considering all relevant models (trading strategies) and constructing a test with properly controlled type I error (test size). Unfortunately, this method is not feasible when the number of strategies being tested is large. White’s paper basically follows the latter approach but he did in a smart way which does not suffer from the aforementioned problem.\n\n2) However, there’s also a problem associated with White’s reality check. It may reduce rejection probabilities of the test under the null by the inclusion of poor and irrelevant alternative models (trading strategies) because it doesn’t satisfy a relevant similarity condition that is necessary for a test to be unbiased. I did some research and found the following paper by Hansen, which attempts to amend this problem by modifying White’s reality check.\nHansen, P. R. (2005), “A Test for Superior Predictive Ability”, Journal of Business & Economic Statistics, 23, pp. 365-380.\n\n3) Basically, the emphasis of using these two approaches is different. Your proposed method tries to answer the following question: will this particular strategy be profitable on a different data set that exhibits similar characteristics to the original one? White’s (2000) reality check, on the other hand, tries to answer the following question: is at least one trading strategy (out of a pool of many seemingly profitable strategies) really profitable for this particular data set? I think a good and conservative approach would be: (a) answer White’s question first and if the answer is YES, then proceed to (b) answer your question by bootstrapping the return time series (and other covariates).\n\n1.", null, "I had exactly the same question. Thanks for the answer. I would like to ask two little questions regarding the first approach (bootstrapping the timeseries and feeding the fake returns series into the strategy again).\n\n1.\nIf, for example, 10000 bootstrapped timeseries are fed into the strategy, is it “allowed” to use the t-test to find out if the true mean is different for 0 for a given alpha?\n\n2.\nAnother example: I bootstrap 10000 new stock timeseries of stock returns (and can therefore create 10000 new timeseries of stockprices). Now I use a strategy (for example moving averages) on my new 10000 timeseries and get 10000 results (return p.a.) generated by the strategy. Also, with every timeseries the buy and hold strategy is performed. Now I have the pairs:\n.\n.\n.\n\nWould it be allowed to use a paired t-test on the differences of the buy and hold and moving averages strategy returns to find out, if the mean is zero?\n\nAll the best,\nDominik/Germany\n\n2.", null, "Hi Dominik,\n\n1. I guess you meant to test the true mean of the PnL of a particular strategy. Yes, one sample t-test can usually be used in this case.\n\nHowever, t-test, like many other parametric test, is not distribution free. Since t-test is based on the assumption of normality, it should NOT be used if this assumption is seriously violated. We may test normality by using Shapiro-Wilk or Kolmogorov–Smirnov test, though it is controversial whether testing distribution before deciding whether to use a test statistic is a good practice; see Zimmerman (2004) for example. If the distribution of PnL is highly skewed, I would recommend to use transformed data instead.\n\n2. Yes, as mentioned in the preceding paragraph, paired t-test can be used here if the normality assumption isnot seriously violated. Alternatively, we may use non-parametric tests (aka distribution-free tests), which have the obvious advantage of not requiring the assumption of normality or the assumption of homogeneity of variance. The non-parametric analogue of paired t-test is the Wilcoxon t-test (http://en.wikipedia.org/wiki/Wilcoxon_signed-rank_test).\n\nThe following table gives the non-parametric analogue for the paired sample t-test and the independent samples t-test:\n————————————————————-\nParametric Test (Non-Parametric Test Analogue)\n————————————————————-\nPaired sample t-test (Wilcoxon signed-rank test)\nIndependent samples t-test (Mann-Whitney U test)\nPearson’s correlation test (Spearman’s correlation test)\n————————————————————-\nI guess all of them can be found on Wikipedia 🙂\n\nKevin\n\n3.", null, "4.", null, "" ]

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[ "# Applied Algebra/Trigonometry 1\n\nMATH 118\n3 CR\n\nThis course includes the following topics: scientific notation, review of basic algebra, solution of linear equations, graphing of algebraic functions, introduction to trigonometry, solution of right triangles, vectors, graphs of trigonometric functions, solution of oblique triangles. Laboratory experiences will be used in this course to show direct applications. Students are required to have a graphing calculator. Specifications will be made by the instructor. Designed for students in technical, occupational fields. [48-16-64] Lab Fee\n\nPrerequisites: ACCUPLACER reading score of 60, or a \"C\" in TSRE 55; and an ACCUPLACER elementary algebra 85, or a \"C\" in MATH 102 or higher" ]

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[ "# New symbol for an exact sequence\n\nA sequence $A\\to B \\to C$ is exact if $\\operatorname{im} f = \\ker g$, where $f:A\\to B$ and $g:B\\to C$.\n\nWhy is there not a symbol to denote such a sequence? Which one would you suggest?\n\n• What is f? What is g? Sep 19, 2018 at 21:45\n• also please consider using MathJax Sep 19, 2018 at 21:45\n• Because nobody thought it necessary. Sep 19, 2018 at 22:02\n• I just write $(ex)$ at the beginning/end of the exact row/column.. Sep 19, 2018 at 22:22\n• I prefer using the words \"is exact\" after the diagram. Or \"the exact sequence\" before the diagram. Sep 20, 2018 at 0:16\n\n@egreg's comment is totally right — because there's no need in such symbol.\n\nAlso, there are at least two reasons to not introduce it. First one is that heavily symbolized mathematical writing is usually unreadable. Second is high probability of ambiguity in case when diagram is somewhat complicated.\n\nSo I would suggest to just write «this sequence is exact at $B$», or that «pair of morphisms $f, g$ is exact» (which is better, in my opinion).\n\n• \"...heavily symbolized mathematical writing is usually unreadable.\" I would upvote this a dozen times if I could. Sep 20, 2018 at 14:49\n• Not to rant, but I HATE seeing arguments written out in logical symbols. \"Oh, the proof is clear. Just note that $\\neg((\\nexists x \\in P \\forall y \\in Z \\ni x \\notin W(\\alpha) \\vee y) \\Rightarrow (J \\Leftrightarrow K \\wedge Q)$.\" Sep 20, 2018 at 14:56\n\nHere are three ideas. I prefer the last one.", null, "• These seem to indicate more about the individual maps themselves and not the sequence as a whole. At least one is ambiguous (what if we have zigzags?) and the others are already in use for different things. Sep 20, 2018 at 4:29" ]

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[ "Cody\n\n# Problem 627. Compute a dot product of two vectors x and y\n\nSolution 1282243\n\nSubmitted on 6 Oct 2017 by Informaton\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\nans = 7\n\n2   Pass\nx = [1 -1]; y= [1 1]; d_correct=0; assert(isequal(vector_dot(x,y),d_correct))\n\nans = 0" ]

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[ "## Formula Calculator System\n\nFormula Calculator System is a simple project developed in HTML5, CSS, and JavaScript. Basically, the system includes mathematical calculations using different formulas. However, the user simply has to enter the number in the text field to get the result clicking on the equal button. Plus, the front-end of this system is pretty simple so that the user won’t get any problem while working on it.  Also, users can modify or add the function in the project according to their needs.\n\nIn addition, the user can add, subtract, divide, multiply, even check the remainder of any two numbers which are the features of the calculator. Also, the user can check and use the formula given in the function options and click the button with a “=” sign for the result. Moreover, this system used for educational purposes only. he user can extract a zip file containing the source code and then can import into Sublime Text 3 for application development. Besides, this project would be helpful for beginner students where they can use this as a demonstration and can submit in college. The students of BTech, MCA, BCA, Engineering, B.Scs, IT, Software engineering, etc. will find it helpful. They will get more knowledge regarding Javascript and HTML.\n\nHow To Run the Project?\n\n• No need to have any kind of local server." ]

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[ "# Python | Masked Scatter Plot\n\nSubmitted by Anuj Singh, on July 16, 2020\n\nMasking a scatter plot refers to differentiating a data point with respect to a function as shown in the following example. It is one of the most important aspects of plotting.\n\n```area1 = np.ma.masked_where(r < r0, area)\narea2 = np.ma.masked_where(r >= r0, area)\nplt.scatter(x, y, s=area1, marker='^', c=c)\nplt.scatter(x, y, s=area2, marker='o', c=c)\n```", null, "", null, "## Python code for masked scatter plot\n\n```import matplotlib.pyplot as plt\nimport numpy as np\n\nN = 100\nr0 = 0.6\nx = 0.9 * np.random.rand(N)\ny = 0.9 * np.random.rand(N)\narea = (20 * np.random.rand(N))**2 # 0 to 10 point radii\nc = np.sqrt(area)\nr = np.sqrt(x ** 2 + y ** 2)\narea1 = np.ma.masked_where(r < r0, area)\narea2 = np.ma.masked_where(r >= r0, area)\n\nplt.figure()\nplt.scatter(x, y, s=area1, marker='^', c=c)\nplt.scatter(x, y, s=area2, marker='o', c=c)\nplt.title('Show the without boundary between the regions')\nplt.show()\n\nplt.figure()\nplt.scatter(x, y, s=area1, marker='^', c=c)\nplt.scatter(x, y, s=area2, marker='o', c=c)\nplt.title('Show the boundary between the regions')\ntheta = np.arange(0, np.pi / 2, 0.01)\nplt.plot(r0 * np.cos(theta), r0 * np.sin(theta))\nplt.show()\n```\n\nOutput:\n\n```Output is as figure\n```\n\nLanguages: » C » C++ » C++ STL » Java » Data Structure » C#.Net » Android » Kotlin » SQL\nWeb Technologies: » PHP » Python » JavaScript » CSS » Ajax » Node.js » Web programming/HTML\nSolved programs: » C » C++ » DS » Java » C#\nAptitude que. & ans.: » C » C++ » Java » DBMS\nInterview que. & ans.: » C » Embedded C » Java » SEO » HR\nCS Subjects: » CS Basics » O.S. » Networks » DBMS » Embedded Systems » Cloud Computing\n» Machine learning » CS Organizations » Linux » DOS\nMore: » Articles » Puzzles » News/Updates" ]

[ null, "https://www.includehelp.com/python/images/masked-scatter-plot-1.jpg", null, "https://www.includehelp.com/python/images/masked-scatter-plot-2.jpg", null ]

[ "", null, "OpenCV  4.2.0 Open Source Computer Vision\nImplementing a face beautification algorithm with G-API\n\n# Introduction\n\nIn this tutorial you will learn:\n\n• Basics of a sample face beautification algorithm;\n• How to infer different networks inside a pipeline with G-API;\n• How to run a G-API pipeline on a video stream.\n\n## Prerequisites\n\nThis sample requires:\n\n• PC with GNU/Linux or Microsoft Windows (Apple macOS is supported but was not tested);\n• OpenCV 4.2 or later built with Intel® Distribution of OpenVINO™ Toolkit (building with Intel® TBB is a plus);\n• The following topologies from OpenVINO™ Toolkit Open Model Zoo:\n• face-detection-adas-0001;\n• facial-landmarks-35-adas-0002.\n\n## Face beautification algorithm\n\nWe will implement a simple face beautification algorithm using a combination of modern Deep Learning techniques and traditional Computer Vision. The general idea behind the algorithm is to make face skin smoother while preserving face features like eyes or a mouth contrast. The algorithm identifies parts of the face using a DNN inference, applies different filters to the parts found, and then combines it into the final result using basic image arithmetics:\n\nBriefly the algorithm is described as follows:\n\n• Input image $$I$$ is passed to unsharp mask and bilateral filters ( $$U$$ and $$L$$ respectively);\n• Input image $$I$$ is passed to an SSD-based face detector;\n• SSD result (a $$[1 \\times 1 \\times 200 \\times 7]$$ blob) is parsed and converted to an array of faces;\n• Every face is passed to a landmarks detector;\n• Based on landmarks found for every face, three image masks are generated:\n• A background mask $$b$$ – indicating which areas from the original image to keep as-is;\n• A face part mask $$p$$ – identifying regions to preserve (sharpen).\n• A face skin mask $$s$$ – identifying regions to blur;\n• The final result $$O$$ is a composition of features above calculated as $$O = b*I + p*U + s*L$$.\n\nGenerating face element masks based on a limited set of features (just 35 per face, including all its parts) is not very trivial and is described in the sections below.\n\n# Constructing a G-API pipeline\n\n## Declaring Deep Learning topologies\n\nThis sample is using two DNN detectors. Every network takes one input and produces one output. In G-API, networks are defined with macro G_API_NET():\n\nG_API_NET(FaceDetector, <cv::GMat(cv::GMat)>, \"face_detector\");\nG_API_NET(LandmDetector, <cv::GMat(cv::GMat)>, \"landm_detector\");\n\nTo get more information, see Declaring Deep Learning topologies described in the \"Face Analytics pipeline\" tutorial.\n\n## Describing the processing graph\n\nThe code below generates a graph for the algorithm above:\n\ncv::GComputation pipeline([=]()\n{\ncv::GMat gimgIn; // input\ncv::GMat faceOut = cv::gapi::infer<custom::FaceDetector>(gimgIn);\nGArrayROI garRects = custom::GFacePostProc::on(faceOut, gimgIn, config::kConfThresh); // post-proc\ncv::GArray<cv::GMat> landmOut = cv::gapi::infer<custom::LandmDetector>(garRects, gimgIn);\ncv::GArray<Landmarks> garElems; // |\ncv::GArray<Contour> garJaws; // |output arrays\nstd::tie(garElems, garJaws) = custom::GLandmPostProc::on(landmOut, garRects); // post-proc\ncv::GArray<Contour> garElsConts; // face elements\ncv::GArray<Contour> garFaceConts; // whole faces\nstd::tie(garElsConts, garFaceConts) = custom::GGetContours::on(garElems, garJaws); // interpolation\ncv::GMat mskSharp = custom::GFillPolyGContours::on(gimgIn, garElsConts); // |\ncv::GMat mskSharpG = cv::gapi::gaussianBlur(mskSharp, config::kGKernelSize, // |\nconfig::kGSigma); // |\ncv::GMat mskBlur = custom::GFillPolyGContours::on(gimgIn, garFaceConts); // |\ncv::GMat mskBlurG = cv::gapi::gaussianBlur(mskBlur, config::kGKernelSize, // |\n// The first argument in mask() is Blur as we want to subtract from // |\n// BlurG the next step: // |\ncv::GMat mskBlurFinal = mskBlurG - cv::gapi::mask(mskBlurG, mskSharpG); // |\ncv::GMat mskFacesGaussed = mskBlurFinal + mskSharpG; // |\ncv::GMat mskFacesWhite = cv::gapi::threshold(mskFacesGaussed, 0, 255, cv::THRESH_BINARY); // |\ncv::GMat mskNoFaces = cv::gapi::bitwise_not(mskFacesWhite); // |\ncv::GMat gimgBilat = custom::GBilatFilter::on(gimgIn, config::kBSize,\nconfig::kBSigmaCol, config::kBSigmaSp);\nconfig::kUnshStrength);\n// as mask() provides CV_8UC1 source only (and we have CV_8U3C)\nreturn cv::GComputation(cv::GIn(gimgIn), cv::GOut(gimgBeautif,\ncv::gapi::copy(gimgIn),\ngarFaceConts,\ngarElsConts,\ngarRects));\n});\n\nThe resulting graph is a mixture of G-API's standard operations, user-defined operations (namespace custom::), and DNN inference. The generic function cv::gapi::infer<>() allows to trigger inference within the pipeline; networks to infer are specified as template parameters. The sample code is using two versions of cv::gapi::infer<>():\n\n• A frame-oriented one is used to detect faces on the input frame.\n• An ROI-list oriented one is used to run landmarks inference on a list of faces – this version produces an array of landmarks per every face.\n\nMore on this in \"Face Analytics pipeline\" (Building a GComputation section).\n\nThe unsharp mask $$U$$ for image $$I$$ is defined as:\n\n$U = I - s * L(M(I)),$\n\nwhere $$M()$$ is a median filter, $$L()$$ is the Laplace operator, and $$s$$ is a strength coefficient. While G-API doesn't provide this function out-of-the-box, it is expressed naturally with the existing G-API operations:\n\nconst int sigma,\nconst float strength)\n{\ncv::GMat blurred = cv::gapi::medianBlur(src, sigma);\ncv::GMat laplacian = custom::GLaplacian::on(blurred, CV_8U);\nreturn (src - (laplacian * strength));\n}\n\nNote that the code snipped above is a regular C++ function defined with G-API types. Users can write functions like this to simplify graph construction; when called, this function just puts the relevant nodes to the pipeline it is used in.\n\n# Custom operations\n\nThe face beautification graph is using custom operations extensively. This chapter focuses on the most interesting kernels, refer to G-API Kernel API for general information on defining operations and implementing kernels in G-API.\n\n## Face detector post-processing\n\nA face detector output is converted to an array of faces with the following kernel:\n\nusing VectorROI = std::vector<cv::Rect>;\nGAPI_OCV_KERNEL(GCPUFacePostProc, GFacePostProc)\n{\nstatic void run(const cv::Mat &inDetectResult,\nconst cv::Mat &inFrame,\nconst float faceConfThreshold,\nVectorROI &outFaces)\n{\nconst int kObjectSize = 7;\nconst int imgCols = inFrame.size().width;\nconst int imgRows = inFrame.size().height;\nconst cv::Rect borders({0, 0}, inFrame.size());\noutFaces.clear();\nconst int numOfDetections = inDetectResult.size;\nconst float *data = inDetectResult.ptr<float>();\nfor (int i = 0; i < numOfDetections; i++)\n{\nconst float faceId = data[i * kObjectSize + 0];\nif (faceId < 0.f) // indicates the end of detections\n{\nbreak;\n}\nconst float faceConfidence = data[i * kObjectSize + 2];\n// We can cut detections by the conf field\n// to avoid mistakes of the detector.\nif (faceConfidence > faceConfThreshold)\n{\nconst float left = data[i * kObjectSize + 3];\nconst float top = data[i * kObjectSize + 4];\nconst float right = data[i * kObjectSize + 5];\nconst float bottom = data[i * kObjectSize + 6];\n// These are normalized coordinates and are between 0 and 1;\n// to get the real pixel coordinates we should multiply it by\n// the image sizes respectively to the directions:\ncv::Point tl(toIntRounded(left * imgCols),\ntoIntRounded(top * imgRows));\ncv::Point br(toIntRounded(right * imgCols),\ntoIntRounded(bottom * imgRows));\noutFaces.push_back(cv::Rect(tl, br) & borders);\n}\n}\n}\n};\n\n## Facial landmarks post-processing\n\nThe algorithm infers locations of face elements (like the eyes, the mouth and the head contour itself) using a generic facial landmarks detector (details) from OpenVINO™ Open Model Zoo. However, the detected landmarks as-is are not enough to generate masks — this operation requires regions of interest on the face represented by closed contours, so some interpolation is applied to get them. This landmarks processing and interpolation is performed by the following kernel:\n\nGAPI_OCV_KERNEL(GCPUGetContours, GGetContours)\n{\nstatic void run(const std::vector<Landmarks> &vctPtsFaceElems, // 18 landmarks of the facial elements\nconst std::vector<Contour> &vctCntJaw, // 17 landmarks of a jaw\nstd::vector<Contour> &vctElemsContours,\nstd::vector<Contour> &vctFaceContours)\n{\nsize_t numFaces = vctCntJaw.size();\nCV_Assert(numFaces == vctPtsFaceElems.size());\nCV_Assert(vctElemsContours.size() == 0ul);\nCV_Assert(vctFaceContours.size() == 0ul);\n// vctFaceElemsContours will store all the face elements' contours found\n// in an input image, namely 4 elements (two eyes, nose, mouth) for every detected face:\nvctElemsContours.reserve(numFaces * 4);\n// vctFaceElemsContours will store all the faces' contours found in an input image:\nvctFaceContours.reserve(numFaces);\nContour cntFace, cntLeftEye, cntRightEye, cntNose, cntMouth;\ncntNose.reserve(4);\nfor (size_t i = 0ul; i < numFaces; i++)\n{\n// The face elements contours\n// A left eye:\n// Approximating the lower eye contour by half-ellipse (using eye points) and storing in cntLeftEye:\ncntLeftEye = getEyeEllipse(vctPtsFaceElems[i], vctPtsFaceElems[i]);\n// Pushing the left eyebrow clock-wise:\ncntLeftEye.insert(cntLeftEye.cend(), {vctPtsFaceElems[i], vctPtsFaceElems[i],\nvctPtsFaceElems[i]});\n// A right eye:\n// Approximating the lower eye contour by half-ellipse (using eye points) and storing in vctRightEye:\ncntRightEye = getEyeEllipse(vctPtsFaceElems[i], vctPtsFaceElems[i]);\n// Pushing the right eyebrow clock-wise:\ncntRightEye.insert(cntRightEye.cend(), {vctPtsFaceElems[i], vctPtsFaceElems[i],\nvctPtsFaceElems[i]});\n// A nose:\n// Storing the nose points clock-wise\ncntNose.clear();\ncntNose.insert(cntNose.cend(), {vctPtsFaceElems[i], vctPtsFaceElems[i],\nvctPtsFaceElems[i], vctPtsFaceElems[i]});\n// A mouth:\n// Approximating the mouth contour by two half-ellipses (using mouth points) and storing in vctMouth:\ncntMouth = getPatchedEllipse(vctPtsFaceElems[i], vctPtsFaceElems[i],\nvctPtsFaceElems[i], vctPtsFaceElems[i]);\n// Storing all the elements in a vector:\nvctElemsContours.insert(vctElemsContours.cend(), {cntLeftEye, cntRightEye, cntNose, cntMouth});\n// The face contour:\n// Approximating the forehead contour by half-ellipse (using jaw points) and storing in vctFace:\n// The ellipse is drawn clock-wise, but jaw contour points goes vice versa, so it's necessary to push\n// cntJaw from the end to the begin using a reverse iterator:\nstd::copy(vctCntJaw[i].crbegin(), vctCntJaw[i].crend(), std::back_inserter(cntFace));\n// Storing the face contour in another vector:\nvctFaceContours.push_back(cntFace);\n}\n}\n};\n\nThe kernel takes two arrays of denormalized landmarks coordinates and returns an array of elements' closed contours and an array of faces' closed contours; in other words, outputs are, the first, an array of contours of image areas to be sharpened and, the second, another one to be smoothed.\n\nHere and below Contour is a vector of points.\n\n### Getting an eye contour\n\nEye contours are estimated with the following function:\n\ninline int custom::getLineInclinationAngleDegrees(const cv::Point &ptLeft, const cv::Point &ptRight)\n{\nconst cv::Point residual = ptRight - ptLeft;\nif (residual.y == 0 && residual.x == 0)\nreturn 0;\nelse\n}\ninline Contour custom::getEyeEllipse(const cv::Point &ptLeft, const cv::Point &ptRight)\n{\nContour cntEyeBottom;\nconst cv::Point ptEyeCenter((ptRight + ptLeft) / 2);\nconst int angle = getLineInclinationAngleDegrees(ptLeft, ptRight);\nconst int axisX = toIntRounded(cv::norm(ptRight - ptLeft) / 2.0);\n// According to research, in average a Y axis of an eye is approximately\n// 1/3 of an X one.\nconst int axisY = axisX / 3;\n// We need the lower part of an ellipse:\nstatic constexpr int kAngEyeStart = 0;\nstatic constexpr int kAngEyeEnd = 180;\ncv::ellipse2Poly(ptEyeCenter, cv::Size(axisX, axisY), angle, kAngEyeStart, kAngEyeEnd, config::kAngDelta,\ncntEyeBottom);\nreturn cntEyeBottom;\n}\n\nBriefly, this function restores the bottom side of an eye by a half-ellipse based on two points in left and right eye corners. In fact, cv::ellipse2Poly() is used to approximate the eye region, and the function only defines ellipse parameters based on just two points:\n\n• The ellipse center and the $$X$$ half-axis calculated by two eye Points;\n• The $$Y$$ half-axis calculated according to the assumption that an average eye width is $$1/3$$ of its length;\n• The start and the end angles which are 0 and 180 (refer to cv::ellipse() documentation);\n• The angle delta: how much points to produce in the contour;\n• The inclination angle of the axes.\n\nThe use of the atan2() instead of just atan() in function custom::getLineInclinationAngleDegrees() is essential as it allows to return a negative value depending on the x and the y signs so we can get the right angle even in case of upside-down face arrangement (if we put the points in the right order, of course).\n\nThe function approximates the forehead contour:\n\nconst cv::Point &ptJawRight,\nconst cv::Point &ptJawLower)\n{\n// The point amid the top two points of a jaw:\nconst cv::Point ptFaceCenter((ptJawLeft + ptJawRight) / 2);\n// This will be the center of the ellipse.\n// The angle between the jaw and the vertical:\nconst int angFace = getLineInclinationAngleDegrees(ptJawLeft, ptJawRight);\n// This will be the inclination of the ellipse\n// Counting the half-axis of the ellipse:\nconst double jawWidth = cv::norm(ptJawLeft - ptJawRight);\n// A forehead width equals the jaw width, and we need a half-axis:\nconst int axisX = toIntRounded(jawWidth / 2.0);\nconst double jawHeight = cv::norm(ptFaceCenter - ptJawLower);\n// According to research, in average a forehead is approximately 2/3 of\n// a jaw:\nconst int axisY = toIntRounded(jawHeight * 2 / 3.0);\n// We need the upper part of an ellipse:\nstatic constexpr int kAngForeheadStart = 180;\nstatic constexpr int kAngForeheadEnd = 360;\n}\n\nAs we have only jaw points in our detected landmarks, we have to get a half-ellipse based on three points of a jaw: the leftmost, the rightmost and the lowest one. The jaw width is assumed to be equal to the forehead width and the latter is calculated using the left and the right points. Speaking of the $$Y$$ axis, we have no points to get it directly, and instead assume that the forehead height is about $$2/3$$ of the jaw height, which can be figured out from the face center (the middle between the left and right points) and the lowest jaw point.\n\nWhen we have all the contours needed, we are able to draw masks:\n\ncv::GMat mskSharp = custom::GFillPolyGContours::on(gimgIn, garElsConts); // |\ncv::GMat mskSharpG = cv::gapi::gaussianBlur(mskSharp, config::kGKernelSize, // |\nconfig::kGSigma); // |\ncv::GMat mskBlur = custom::GFillPolyGContours::on(gimgIn, garFaceConts); // |\ncv::GMat mskBlurG = cv::gapi::gaussianBlur(mskBlur, config::kGKernelSize, // |\n// The first argument in mask() is Blur as we want to subtract from // |\n// BlurG the next step: // |\ncv::GMat mskBlurFinal = mskBlurG - cv::gapi::mask(mskBlurG, mskSharpG); // |\ncv::GMat mskFacesGaussed = mskBlurFinal + mskSharpG; // |\ncv::GMat mskFacesWhite = cv::gapi::threshold(mskFacesGaussed, 0, 255, cv::THRESH_BINARY); // |\ncv::GMat mskNoFaces = cv::gapi::bitwise_not(mskFacesWhite); // |\n\nThe steps to get the masks are:\n\n• fill the contours that should be sharpened;\n• blur that to get the \"sharp\" mask (mskSharpG);\n• fill all the face contours fully;\n• blur that;\n• subtract areas which intersect with the \"sharp\" mask — and get the \"bilateral\" mask (mskBlurFinal);\n• set all non-zero pixels of the result as 255 (by cv::gapi::threshold())\n• revert the output (by cv::gapi::bitwise_not) to get the background mask (mskNoFaces).\n\n# Configuring and running the pipeline\n\nOnce the graph is fully expressed, we can finally compile it and run on real data. G-API graph compilation is the stage where the G-API framework actually understands which kernels and networks to use. This configuration happens via G-API compilation arguments.\n\n## DNN parameters\n\nThis sample is using OpenVINO™ Toolkit Inference Engine backend for DL inference, which is configured the following way:\n\n{\n/*std::string*/ faceXmlPath,\n/*std::string*/ faceBinPath,\n/*std::string*/ faceDevice\n};\n{\n/*std::string*/ landmXmlPath,\n/*std::string*/ landmBinPath,\n/*std::string*/ landmDevice\n};\n\nEvery cv::gapi::ie::Params<> object is related to the network specified in its template argument. We should pass there the network type we have defined in G_API_NET() in the early beginning of the tutorial.\n\nNetwork parameters are then wrapped in cv::gapi::NetworkPackage:\n\nauto networks = cv::gapi::networks(faceParams, landmParams);\n\nMore details in \"Face Analytics Pipeline\" (Configuring the pipeline section).\n\n## Kernel packages\n\nIn this example we use a lot of custom kernels, in addition to that we use Fluid backend to optimize out memory for G-API's standard kernels where applicable. The resulting kernel package is formed like this:\n\nauto customKernels = cv::gapi::kernels<custom::GCPUBilateralFilter,\ncustom::GCPULaplacian,\ncustom::GCPUFillPolyGContours,\ncustom::GCPUPolyLines,\ncustom::GCPURectangle,\ncustom::GCPUFacePostProc,\ncustom::GCPULandmPostProc,\ncustom::GCPUGetContours>();\ncustomKernels);\n\n## Compiling the streaming pipeline\n\nG-API optimizes execution for video streams when compiled in the \"Streaming\" mode.\n\ncv::GStreamingCompiled stream = pipeline.compileStreaming(cv::compile_args(kernels, networks));\n\nMore on this in \"Face Analytics Pipeline\" (Configuring the pipeline section).\n\n## Running the streaming pipeline\n\nIn order to run the G-API streaming pipeline, all we need is to specify the input video source, call cv::GStreamingCompiled::start(), and then fetch the pipeline processing results:\n\nif (parser.has(\"input\"))\n{\nstream.setSource(cv::gapi::wip::make_src<cv::gapi::wip::GCaptureSource>(parser.get<cv::String>(\"input\")));\n}\nauto out_vector = cv::gout(imgBeautif, imgShow, vctFaceConts,\nvctElsConts, vctRects);\nstream.start();\navg.start();\nwhile (stream.running())\n{\nif (!stream.try_pull(std::move(out_vector)))\n{\n// Use a try_pull() to obtain data.\n// If there's no data, let UI refresh (and handle keypress)\nif (cv::waitKey(1) >= 0) break;\nelse continue;\n}\nframes++;\n// Drawing face boxes and landmarks if necessary:\nif (flgLandmarks == true)\n{\ncv::polylines(imgShow, vctFaceConts, config::kClosedLine,\nconfig::kClrYellow);\ncv::polylines(imgShow, vctElsConts, config::kClosedLine,\nconfig::kClrYellow);\n}\nif (flgBoxes == true)\nfor (auto rect : vctRects)\ncv::rectangle(imgShow, rect, config::kClrGreen);\ncv::imshow(config::kWinInput, imgShow);\ncv::imshow(config::kWinFaceBeautification, imgBeautif);\n}\n\nOnce results are ready and can be pulled from the pipeline we display it on the screen and handle GUI events.\n\nSee Running the pipeline section in the \"Face Analytics Pipeline\" tutorial for more details.\n\n# Conclusion\n\nThe tutorial has two goals: to show the use of brand new features of G-API introduced in OpenCV 4.2, and give a basic understanding on a sample face beautification algorithm.\n\nThe result of the algorithm application:", null, "Face Beautification example\n\nOn the test machine (Intel® Core™ i7-8700) the G-API-optimized video pipeline outperforms its serial (non-pipelined) version by a factor of 2.7 – meaning that for such a non-trivial graph, the proper pipelining can bring almost 3x increase in performance." ]

[ null, "https://docs.opencv.org/4.2.0/opencv-logo-small.png", null, "https://docs.opencv.org/4.2.0/example.jpg", null ]

[ "# Tagged: intersection of subspaces\n\n## Problem 595\n\nLet $U$ and $V$ be subspaces of the $n$-dimensional vector space $\\R^n$.\n\nProve that the intersection $U\\cap V$ is also a subspace of $\\R^n$.", null, "Add to solve later" ]

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[ "i3geek.com\n\n# 二叉树——二叉查找树的增、删、查\n\n#### 类型\n\n1. 完全二叉树——若设二叉树的高度为h，除第 h 层外，其它各层 (1～h-1) 的结点数都达到最大个数，第h层有叶子结点，并且叶子结点都是从左到右依次排布，这就是完全二叉树。\n2. 满二叉树——除了叶结点外每一个结点都有左右子叶且叶子结点都处在最底层的二叉树。\n3. 平衡二叉树——平衡二叉树又被称为AVL树（区别于AVL算法），它是一棵二叉排序树，且具有以下性质：它是一棵空树或它的左右两个子树的高度差的绝对值不超过1，并且左右两个子树都是一棵平衡二叉树。\n\n#### 二叉查找树", null, "#### 定义\n\n1. 若左子树不空，则左子树上所有结点的值均小于它的根结点的值；\n2. 若右子树不空，则右子树上所有结点的值均大于或等于它的根结点的值；\n3. 左、右子树也分别为二叉排序树；\n4. 没有键值相等的节点。\n\n#### 特点", null, "#### 查找\n\n1. 二叉树若根结点的关键字值等于查找的关键字，成功\n2. 否则，若小于根结点的关键字值，递归查左子树。若大于根结点的关键字值，递归查右子树。若子树为空，查找不成功\n```pnode search_BST(pnode p, int x)\n{\nbool solve = false;\nwhile(p && !solve){\nif(x == p->val){\nsolve = true;\n}\nelse if(x < p->val){\np = p->lchild;\n}\nelse{\np = p->rchild;\n}\n}\nif(p == NULL){\ncout << \"没有找到\" << x << endl;\n}\nreturn p;\n}```\n\n#### 增加\n\n1. 若当前的二叉查找树为空，则插入的元素为根节点，\n2. 若插入的元素值小于根节点值，则将元素插入到左子树中，\n3. 若插入的元素值不小于根节点值，则将元素插入到右子树中。\n```struct node\n{\nint val;\npnode lchild;\npnode rchild;\n};\n\npnode BT = NULL;\n\n//递归方法插入节点\npnode insert(pnode root, int x)\n{\npnode p = (pnode)malloc(LEN);\np->val = x;\np->lchild = NULL;\np->rchild = NULL;\nif(root == NULL){\nroot = p;\n}\nelse if(x < root->val){\nroot->lchild = insert(root->lchild, x);\n}\nelse{\nroot->rchild = insert(root->rchild, x);\n}\nreturn root;\n}\n\n//非递归方法插入节点\nvoid insert_BST(pnode q, int x)\n{\npnode p = (pnode)malloc(LEN);\np->val = x;\np->lchild = NULL;\np->rchild = NULL;\nif(q == NULL){\nBT = p;\nreturn ;\n}\nwhile(q->lchild != p && q->rchild != p){\nif(x < q->val){\nif(q->lchild){\nq = q->lchild;\n}\nelse{\nq->lchild = p;\n}\n}\nelse{\nif(q->rchild){\nq = q->rchild;\n}\nelse{\nq->rchild = p;\n}\n}\n}\nreturn;\n}```\n\n#### 删除\n\n1. p为叶子节点，直接删除该节点，再修改其父节点的指针（注意分是根节点和不是根节点），如图a。\n2. p为单支节点（即只有左子树或右子树）。让p的子树与p的父亲节点相连，删除p即可；（注意分是根节点和不是根节点）；如图b。\n3. p的左子树和右子树均不空。找到p的后继y，因为y一定没有左子树，所以可以删除y，并让y的父亲节点成为y的右子树的父亲节点，并用y的值代替p的值；或者方法二是找到p的前驱x，x一定没有右子树，所以可以删除x，并让x的父亲节点成为y的左子树的父亲节点。如图c。", null, "", null, "", null, "```bool delete_BST(pnode p, int x) //返回一个标志，表示是否找到被删元素\n{\nbool find = false;\npnode q;\np = BT;\nwhile(p && !find){ //寻找被删元素\nif(x == p->val){ //找到被删元素\nfind = true;\n}\nelse if(x < p->val){ //沿左子树找\nq = p;\np = p->lchild;\n}\nelse{ //沿右子树找\nq = p;\np = p->rchild;\n}\n}\nif(p == NULL){ //没找到\ncout << \"没有找到\" << x << endl;\n}\n\nif(p->lchild == NULL && p->rchild == NULL){ //p为叶子节点\nif(p == BT){ //p为根节点\nBT = NULL;\n}\nelse if(q->lchild == p){\nq->lchild = NULL;\n}\nelse{\nq->rchild = NULL;\n}\nfree(p); //释放节点p\n}\nelse if(p->lchild == NULL || p->rchild == NULL){ //p为单支子树\nif(p == BT){ //p为根节点\nif(p->lchild == NULL){\nBT = p->rchild;\n}\nelse{\nBT = p->lchild;\n}\n}\nelse{\nif(q->lchild == p && p->lchild){ //p是q的左子树且p有左子树\nq->lchild = p->lchild; //将p的左子树链接到q的左指针上\n}\nelse if(q->lchild == p && p->rchild){\nq->lchild = p->rchild;\n}\nelse if(q->rchild == p && p->lchild){\nq->rchild = p->lchild;\n}\nelse{\nq->rchild = p->rchild;\n}\n}\nfree(p);\n}\nelse{ //p的左右子树均不为空\npnode t = p;\npnode s = p->lchild; //从p的左子节点开始\nwhile(s->rchild){ //找到p的前驱，即p左子树中值最大的节点\nt = s;\ns = s->rchild;\n}\np->val = s->val; //把节点s的值赋给p\nif(t == p){\np->lchild = s->lchild;\n}\nelse{\nt->rchild = s->lchild;\n}\nfree(s);\n}\nreturn find;\n}```\n\n### 评论 2\n\n1. #1\n\n二叉树，大学学的东东，全忘完了 😉\n\nhd5年前 (2015-02-25)回复\n• 最近在准备面试。。。又捡起来看了。。唉\n\nyan5年前 (2015-02-25)回复" ]

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[ "", null, "# Projecting EPS (number of years)\n\nI am given a current EPS and a growth rate - the question asks for a projected EPS in 2 years. However, the formula used in the answer is as follows:\n\nEPS3 = EPS0 × (1 + g )3\n\nWhy is it using 3 periods? I am pretty sure I am missing some basics here.\n\nThe 3 should be to the 3rd power to account for 3 years of compounding.\n\nBut why 3 years though if the question is needs a projected EPS in 2 years?\n\nIts got something to do with the difference between projected forward P/E and trailing P/E. The intention is to find the Price using EPS, growth rate in EPS and forward projected P/E after 2 years.\n\nIt looks like a type since the answer is using EPS3.\n\nIs it using the EPS at time 0 and looking for the projected value at time 3?\n\nIts using EPS at time 0 and growth rate to get EPS at time 2.\n\nThe reason for not using EPS2 is “because it uses the forecasted trailing EPS (or EPS2), thus calculating the wrong future price of \\$66.99.”" ]

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