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[ "# Difference between pointer to an array and array of pointers\n\n• Difficulty Level : Easy\n• Last Updated : 13 Jun, 2019\n\nPointer to an array:\n\nTake a step-up from those \"Hello World\" programs. Learn to implement data structures like Heap, Stacks, Linked List and many more! Check out our Data Structures in C course to start learning today.\n\nPointer to an array is also known as array pointer. We are using the pointer to access the components of the array.\n\n``` int a = {3, 4, 5 };\nint *ptr = a;\n```\n\nWe have a pointer ptr that focuses to the 0th component of the array. We can likewise declare a pointer that can point to whole array rather than just a single component of the array.\nSyntax:\n\n`data type (*var name)[size of array];`\n\nDeclaration of the pointer to an array:\n\n```// pointer to an array of five numbers\nint (* ptr) = NULL; ```\n\nThe above declaration is the pointer to an array of five integers. We use parenthesis to pronounce pointer to an array. Since subscript has higher priority than indirection, it is crucial to encase the indirection operator and pointer name inside brackets.\n\nExample:\n\n `// C program to demonstrate``// pointer to an array.`` ` `#include `` ` `int` `main()``{`` ` `    ``// Pointer to an array of five numbers``    ``int``(*a);`` ` `    ``int` `b = { 1, 2, 3, 4, 5 };`` ` `    ``int` `i = 0;`` ` `    ``// Points to the whole array b`` ` `    ``a = &b;`` ` `    ``for` `(i = 0; i < 5; i++)`` ` `        ``printf``(``\"%d\\n\"``, *(*a + i));`` ` `    ``return` `0;``}`\nOutput:\n```1\n2\n3\n4\n5\n```\n\nArray of pointers:\n\nArray of pointers” is an array of the pointer variables. It is also known as pointer arrays.\nSyntax:\n\n`int *var_name[array_size];`\n\nDeclaration of an array of pointers:\n\n` int *ptr;`\n\nWe can make separate pointer variables which can point to the different values or we can make one integer array of pointers that can point to all the values.\n\nExample:\n\n `// C program to demonstrate``// example of array of pointers.`` ` `#include `` ` `const` `int` `SIZE = 3;`` ` `void` `main()``{`` ` `    ``// creating an array``    ``int` `arr[] = { 1, 2, 3 };`` ` `    ``// we can make an integer pointer array to``    ``// storing the address of array elements``    ``int` `i, *ptr[SIZE];`` ` `    ``for` `(i = 0; i < SIZE; i++) {`` ` `        ``// assigning the address of integer.``        ``ptr[i] = &arr[i];``    ``}`` ` `    ``// printing values using pointer``    ``for` `(i = 0; i < SIZE; i++) {`` ` `        ``printf``(``\"Value of arr[%d] = %d\\n\"``, i, *ptr[i]);``    ``}``}`\nOutput:\n```Value of arr = 1\nValue of arr = 2\nValue of arr = 3\n```\n\nExample: We can likewise make an array of pointers to the character to store a list of strings.\n\n `#include `` ` `const` `int` `size = 4;`` ` `void` `main()``{`` ` `    ``// array of pointers to a character``    ``// to store a list of strings``    ``char``* names[] = {``        ``\"amit\"``,``        ``\"amar\"``,``        ``\"ankit\"``,``        ``\"akhil\"``    ``};`` ` `    ``int` `i = 0;`` ` `    ``for` `(i = 0; i < size; i++) {``        ``printf``(``\"%s\\n\"``, names[i]);``    ``}``}`\nOutput:\n```amit\namar\nankit\nakhil\n```\n\nMy Personal Notes arrow_drop_up" ]

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[ "# Determining the frequency of a monochromatic light source from measurements of irradiance alone?\n\nLet's say we have a monochromatic source of unknown frequency (say, a laser) and some device/sensor that measures the total photon energy absorbed over a fixed time interval (essentially, irradiance).\n\nSince photon counting follows Poisson statistics, we can determine the light frequency via repeat measurements,\n\n$$\\langle E\\rangle = \\langle N \\rangle h\\nu \\\\ {\\sigma_{E}}^2 = \\langle N \\rangle h^2 \\nu^2$$\n\nwhere $$\\langle E\\rangle$$ is the mean total photon energy across measurements, $${\\sigma_{E}}^2$$ is the variance, $$\\langle N \\rangle$$ is the mean number of photons absorbed across measurements, and $$\\nu$$ is the frequency.\n\nWe can find the frequency,\n\n$$\\nu = \\frac{{\\sigma_{E}}^2}{h \\langle E\\rangle}$$\n\nThis is interesting because you determine the frequency from the variance-to-mean ratio alone without any real calibration required (assuming you can accurately measure the total photon energy).\n\nHas this been meaningfully used in practice or actual research?\n\nMy intuition says you could do this at microwave frequencies (with bolometers used to measure total energy). Could you measure a laboratory hydrogen maser this way?\n\nThis is probably a terrible estimator compared to just doing spectroscopy - but it's interesting how simple it is (in terms of how few variables it depends on).\n\nThe idea that the coefficient-of-variation in a Poisson process:\n\n$$c_V\\equiv \\frac{\\bar x}{\\sigma} = \\frac{\\alpha N}{\\alpha \\sqrt N}=\\sqrt N$$\n\ncan be used to estimate $$N$$ has been used before. For example, in a gas Cherenkov counter, you can expect 6-ish photons per meter in air. One can estimate the actual number in the apparatus from $$c^2_V$$ of the PMT output distribution.\n\nIn your case, you would measure the mean energy and use:\n\n$$E = Nh\\nu = c^2_Vh\\nu$$\n\nNote that $$c_V$$ is measured entirely from the shape of measured data, without calibration of the axes. Of course, you need to calibrate the device to get $$E$$.\n\nSome things to watch out for are:\n\nBiased vs. unbiased estimators of $$c_V$$.\n\nAlso compound distributions: you don't aways have direct access to the photons you're measuring, e.g., they may go through a device with a non-zero efficiency. Thus, you're convolving a Poisson creation process with a Bernoulli detection process. The result is a different Poisson process.\n\nIf the detection involves amplification (e.g., an electron avalanche), it can convolve the Poisson process with a Gaussian process. This leads to Pólya-distributed data (aka: The Dirichlet-multinomial distribution).\n\n• Do you know if that has ever been used to obtain the wavelength of e.g. a laser and what the accuracy is? As based entiely on poisson noise, a perfect measurement device (no readout noise etc.) shoud give an exact measurement... however, this works only for monochromatic light... Sep 18, 2021 at 23:25" ]

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[ "", null, "", null, "Encyclopedia > Computational physics\n\nArticle Content\n\nComputational physics\n\nComputational Physics is the study and implementation of numerical algorithms in order to solve problems in physics for which a quantitative theory already exists.\n\nPhysicists often have a very precise mathematical theory describing exactly how a system will operate. Unfortunately, it is often the case that solving these equations in order to produce a useful prediction is a computationally difficult problem. This is especially true with quantum mechanics, where only a handful of simple models can be solved exactly. Even apparently simple problems, such as calculating the wavefunction of an electron orbiting an atom in a strong electric field, may require great effort to formulate a practical algorithm.\n\nIn addition, quantum mechanical problems are generally exponential in the size of the system (see computational complexity theory).\n\nMany other more general numerical problems fall loosely under the domain of computational physics, although they could easily be considered pure mathematics or part of any number of applied areas. For example:\n\nAll Wikipedia text is available under the terms of the GNU Free Documentation License\n\nSearch Encyclopedia\n Search over one million articles, find something about almost anything!\n\nFeatured Article\n Doetinchem ... water). The municipality of Doetinchem also includes the following towns, villages and townships: Dichteren[?], Gaanderen[?], IJzevoorde, Langerak[?], ...", null, "", null, "" ]

[ null, "http://www.kids.net.au/images/spacer.gif", null, "http://www.kids.net.au/images/spacer.gif", null, "http://www.kids.net.au/images/spacer.gif", null, "http://www.kids.net.au/images/spacer.gif", null ]

[ "org.apache.commons.math4.stat.descriptive\n\n## Class SynchronizedMultivariateSummaryStatistics\n\n• All Implemented Interfaces:\nSerializable, StatisticalMultivariateSummary\n\npublic class SynchronizedMultivariateSummaryStatistics\nextends MultivariateSummaryStatistics\nImplementation of MultivariateSummaryStatistics that is safe to use in a multithreaded environment. Multiple threads can safely operate on a single instance without causing runtime exceptions due to race conditions. In effect, this implementation makes modification and access methods atomic operations for a single instance. That is to say, as one thread is computing a statistic from the instance, no other thread can modify the instance nor compute another statistic.\nSince:\n1.2\nSerialized Form\n• ### Constructor Summary\n\nConstructors\nConstructor and Description\nSynchronizedMultivariateSummaryStatistics(int k, boolean isCovarianceBiasCorrected)\nConstruct a SynchronizedMultivariateSummaryStatistics instance\n• ### Method Summary\n\nAll Methods\nModifier and Type Method and Description\nvoid addValue(double[] value)\nAdd an n-tuple to the data\nvoid clear()\nResets all statistics and storage\nboolean equals(Object object)\nReturns true iff object is a MultivariateSummaryStatistics instance and all statistics have the same values as this.\nRealMatrix getCovariance()\nReturns the covariance matrix of the values that have been added.\nint getDimension()\nReturns the dimension of the data\nStorelessUnivariateStatistic[] getGeoMeanImpl()\nReturns the currently configured geometric mean implementation\ndouble[] getGeometricMean()\nReturns an array whose ith entry is the geometric mean of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\ndouble[] getMax()\nReturns an array whose ith entry is the maximum of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\nStorelessUnivariateStatistic[] getMaxImpl()\nReturns the currently configured maximum implementation\ndouble[] getMean()\nReturns an array whose ith entry is the mean of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\nStorelessUnivariateStatistic[] getMeanImpl()\nReturns the currently configured mean implementation\ndouble[] getMin()\nReturns an array whose ith entry is the minimum of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\nStorelessUnivariateStatistic[] getMinImpl()\nReturns the currently configured minimum implementation\nlong getN()\nReturns the number of available values\ndouble[] getStandardDeviation()\nReturns an array whose ith entry is the standard deviation of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\ndouble[] getSum()\nReturns an array whose ith entry is the sum of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\nStorelessUnivariateStatistic[] getSumImpl()\nReturns the currently configured Sum implementation\ndouble[] getSumLog()\nReturns an array whose ith entry is the sum of logs of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\nStorelessUnivariateStatistic[] getSumLogImpl()\nReturns the currently configured sum of logs implementation\ndouble[] getSumSq()\nReturns an array whose ith entry is the sum of squares of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\nStorelessUnivariateStatistic[] getSumsqImpl()\nReturns the currently configured sum of squares implementation\nint hashCode()\nReturns hash code based on values of statistics\nvoid setGeoMeanImpl(StorelessUnivariateStatistic[] geoMeanImpl)\nSets the implementation for the geometric mean.\nvoid setMaxImpl(StorelessUnivariateStatistic[] maxImpl)\nSets the implementation for the maximum.\nvoid setMeanImpl(StorelessUnivariateStatistic[] meanImpl)\nSets the implementation for the mean.\nvoid setMinImpl(StorelessUnivariateStatistic[] minImpl)\nSets the implementation for the minimum.\nvoid setSumImpl(StorelessUnivariateStatistic[] sumImpl)\nSets the implementation for the Sum.\nvoid setSumLogImpl(StorelessUnivariateStatistic[] sumLogImpl)\nSets the implementation for the sum of logs.\nvoid setSumsqImpl(StorelessUnivariateStatistic[] sumsqImpl)\nSets the implementation for the sum of squares.\nString toString()\nGenerates a text report displaying summary statistics from values that have been added.\n• ### Methods inherited from class java.lang.Object\n\nclone, finalize, getClass, notify, notifyAll, wait, wait, wait\n• ### Constructor Detail\n\n• #### SynchronizedMultivariateSummaryStatistics\n\npublic SynchronizedMultivariateSummaryStatistics(int k,\nboolean isCovarianceBiasCorrected)\nConstruct a SynchronizedMultivariateSummaryStatistics instance\nParameters:\nk - dimension of the data\nisCovarianceBiasCorrected - if true, the unbiased sample covariance is computed, otherwise the biased population covariance is computed\n• ### Method Detail\n\npublic void addValue(double[] value)\nthrows DimensionMismatchException\nAdd an n-tuple to the data\nOverrides:\naddValue in class MultivariateSummaryStatistics\nParameters:\nvalue - the n-tuple to add\nThrows:\nDimensionMismatchException - if the length of the array does not match the one used at construction\n• #### getDimension\n\npublic int getDimension()\nReturns the dimension of the data\nSpecified by:\ngetDimension in interface StatisticalMultivariateSummary\nOverrides:\ngetDimension in class MultivariateSummaryStatistics\nReturns:\nThe dimension of the data\n• #### getN\n\npublic long getN()\nReturns the number of available values\nSpecified by:\ngetN in interface StatisticalMultivariateSummary\nOverrides:\ngetN in class MultivariateSummaryStatistics\nReturns:\nThe number of available values\n• #### getSum\n\npublic double[] getSum()\nReturns an array whose ith entry is the sum of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\nSpecified by:\ngetSum in interface StatisticalMultivariateSummary\nOverrides:\ngetSum in class MultivariateSummaryStatistics\nReturns:\nthe array of component sums\n• #### getSumSq\n\npublic double[] getSumSq()\nReturns an array whose ith entry is the sum of squares of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\nSpecified by:\ngetSumSq in interface StatisticalMultivariateSummary\nOverrides:\ngetSumSq in class MultivariateSummaryStatistics\nReturns:\nthe array of component sums of squares\n• #### getSumLog\n\npublic double[] getSumLog()\nReturns an array whose ith entry is the sum of logs of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\nSpecified by:\ngetSumLog in interface StatisticalMultivariateSummary\nOverrides:\ngetSumLog in class MultivariateSummaryStatistics\nReturns:\nthe array of component log sums\n• #### getMean\n\npublic double[] getMean()\nReturns an array whose ith entry is the mean of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\nSpecified by:\ngetMean in interface StatisticalMultivariateSummary\nOverrides:\ngetMean in class MultivariateSummaryStatistics\nReturns:\nthe array of component means\n• #### getStandardDeviation\n\npublic double[] getStandardDeviation()\nReturns an array whose ith entry is the standard deviation of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\nSpecified by:\ngetStandardDeviation in interface StatisticalMultivariateSummary\nOverrides:\ngetStandardDeviation in class MultivariateSummaryStatistics\nReturns:\nthe array of component standard deviations\n• #### getCovariance\n\npublic RealMatrix getCovariance()\nReturns the covariance matrix of the values that have been added.\nSpecified by:\ngetCovariance in interface StatisticalMultivariateSummary\nOverrides:\ngetCovariance in class MultivariateSummaryStatistics\nReturns:\nthe covariance matrix\n• #### getMax\n\npublic double[] getMax()\nReturns an array whose ith entry is the maximum of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\nSpecified by:\ngetMax in interface StatisticalMultivariateSummary\nOverrides:\ngetMax in class MultivariateSummaryStatistics\nReturns:\nthe array of component maxima\n• #### getMin\n\npublic double[] getMin()\nReturns an array whose ith entry is the minimum of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\nSpecified by:\ngetMin in interface StatisticalMultivariateSummary\nOverrides:\ngetMin in class MultivariateSummaryStatistics\nReturns:\nthe array of component minima\n• #### getGeometricMean\n\npublic double[] getGeometricMean()\nReturns an array whose ith entry is the geometric mean of the ith entries of the arrays that have been added using MultivariateSummaryStatistics.addValue(double[])\nSpecified by:\ngetGeometricMean in interface StatisticalMultivariateSummary\nOverrides:\ngetGeometricMean in class MultivariateSummaryStatistics\nReturns:\nthe array of component geometric means\n• #### toString\n\npublic String toString()\nGenerates a text report displaying summary statistics from values that have been added.\nOverrides:\ntoString in class MultivariateSummaryStatistics\nReturns:\nString with line feeds displaying statistics\n• #### clear\n\npublic void clear()\nResets all statistics and storage\nOverrides:\nclear in class MultivariateSummaryStatistics\n• #### equals\n\npublic boolean equals(Object object)\nReturns true iff object is a MultivariateSummaryStatistics instance and all statistics have the same values as this.\nOverrides:\nequals in class MultivariateSummaryStatistics\nParameters:\nobject - the object to test equality against.\nReturns:\ntrue if object equals this\n• #### hashCode\n\npublic int hashCode()\nReturns hash code based on values of statistics\nOverrides:\nhashCode in class MultivariateSummaryStatistics\nReturns:\nhash code\n• #### getSumImpl\n\npublic StorelessUnivariateStatistic[] getSumImpl()\nReturns the currently configured Sum implementation\nOverrides:\ngetSumImpl in class MultivariateSummaryStatistics\nReturns:\nthe StorelessUnivariateStatistic implementing the sum\n• #### setSumImpl\n\npublic void setSumImpl(StorelessUnivariateStatistic[] sumImpl)\nthrows DimensionMismatchException,\nMathIllegalStateException\n\nSets the implementation for the Sum.\n\nThis method must be activated before any data has been added - i.e., before addValue has been used to add data; otherwise an IllegalStateException will be thrown.\n\nOverrides:\nsetSumImpl in class MultivariateSummaryStatistics\nParameters:\nsumImpl - the StorelessUnivariateStatistic instance to use for computing the Sum\nThrows:\nDimensionMismatchException - if the array dimension does not match the one used at construction\nMathIllegalStateException - if data has already been added (i.e if n > 0)\n• #### getSumsqImpl\n\npublic StorelessUnivariateStatistic[] getSumsqImpl()\nReturns the currently configured sum of squares implementation\nOverrides:\ngetSumsqImpl in class MultivariateSummaryStatistics\nReturns:\nthe StorelessUnivariateStatistic implementing the sum of squares\n• #### setSumsqImpl\n\npublic void setSumsqImpl(StorelessUnivariateStatistic[] sumsqImpl)\nthrows DimensionMismatchException,\nMathIllegalStateException\n\nSets the implementation for the sum of squares.\n\nThis method must be activated before any data has been added - i.e., before addValue has been used to add data; otherwise an IllegalStateException will be thrown.\n\nOverrides:\nsetSumsqImpl in class MultivariateSummaryStatistics\nParameters:\nsumsqImpl - the StorelessUnivariateStatistic instance to use for computing the sum of squares\nThrows:\nDimensionMismatchException - if the array dimension does not match the one used at construction\nMathIllegalStateException - if data has already been added (i.e if n > 0)\n• #### getMinImpl\n\npublic StorelessUnivariateStatistic[] getMinImpl()\nReturns the currently configured minimum implementation\nOverrides:\ngetMinImpl in class MultivariateSummaryStatistics\nReturns:\nthe StorelessUnivariateStatistic implementing the minimum\n• #### setMinImpl\n\npublic void setMinImpl(StorelessUnivariateStatistic[] minImpl)\nthrows DimensionMismatchException,\nMathIllegalStateException\n\nSets the implementation for the minimum.\n\nThis method must be activated before any data has been added - i.e., before addValue has been used to add data; otherwise an IllegalStateException will be thrown.\n\nOverrides:\nsetMinImpl in class MultivariateSummaryStatistics\nParameters:\nminImpl - the StorelessUnivariateStatistic instance to use for computing the minimum\nThrows:\nDimensionMismatchException - if the array dimension does not match the one used at construction\nMathIllegalStateException - if data has already been added (i.e if n > 0)\n• #### getMaxImpl\n\npublic StorelessUnivariateStatistic[] getMaxImpl()\nReturns the currently configured maximum implementation\nOverrides:\ngetMaxImpl in class MultivariateSummaryStatistics\nReturns:\nthe StorelessUnivariateStatistic implementing the maximum\n• #### setMaxImpl\n\npublic void setMaxImpl(StorelessUnivariateStatistic[] maxImpl)\nthrows DimensionMismatchException,\nMathIllegalStateException\n\nSets the implementation for the maximum.\n\nThis method must be activated before any data has been added - i.e., before addValue has been used to add data; otherwise an IllegalStateException will be thrown.\n\nOverrides:\nsetMaxImpl in class MultivariateSummaryStatistics\nParameters:\nmaxImpl - the StorelessUnivariateStatistic instance to use for computing the maximum\nThrows:\nDimensionMismatchException - if the array dimension does not match the one used at construction\nMathIllegalStateException - if data has already been added (i.e if n > 0)\n• #### getSumLogImpl\n\npublic StorelessUnivariateStatistic[] getSumLogImpl()\nReturns the currently configured sum of logs implementation\nOverrides:\ngetSumLogImpl in class MultivariateSummaryStatistics\nReturns:\nthe StorelessUnivariateStatistic implementing the log sum\n• #### setSumLogImpl\n\npublic void setSumLogImpl(StorelessUnivariateStatistic[] sumLogImpl)\nthrows DimensionMismatchException,\nMathIllegalStateException\n\nSets the implementation for the sum of logs.\n\nThis method must be activated before any data has been added - i.e., before addValue has been used to add data; otherwise an IllegalStateException will be thrown.\n\nOverrides:\nsetSumLogImpl in class MultivariateSummaryStatistics\nParameters:\nsumLogImpl - the StorelessUnivariateStatistic instance to use for computing the log sum\nThrows:\nDimensionMismatchException - if the array dimension does not match the one used at construction\nMathIllegalStateException - if data has already been added (i.e if n > 0)\n• #### getGeoMeanImpl\n\npublic StorelessUnivariateStatistic[] getGeoMeanImpl()\nReturns the currently configured geometric mean implementation\nOverrides:\ngetGeoMeanImpl in class MultivariateSummaryStatistics\nReturns:\nthe StorelessUnivariateStatistic implementing the geometric mean\n• #### setGeoMeanImpl\n\npublic void setGeoMeanImpl(StorelessUnivariateStatistic[] geoMeanImpl)\nthrows DimensionMismatchException,\nMathIllegalStateException\n\nSets the implementation for the geometric mean.\n\nThis method must be activated before any data has been added - i.e., before addValue has been used to add data; otherwise an IllegalStateException will be thrown.\n\nOverrides:\nsetGeoMeanImpl in class MultivariateSummaryStatistics\nParameters:\ngeoMeanImpl - the StorelessUnivariateStatistic instance to use for computing the geometric mean\nThrows:\nDimensionMismatchException - if the array dimension does not match the one used at construction\nMathIllegalStateException - if data has already been added (i.e if n > 0)\n• #### getMeanImpl\n\npublic StorelessUnivariateStatistic[] getMeanImpl()\nReturns the currently configured mean implementation\nOverrides:\ngetMeanImpl in class MultivariateSummaryStatistics\nReturns:\nthe StorelessUnivariateStatistic implementing the mean\n• #### setMeanImpl\n\npublic void setMeanImpl(StorelessUnivariateStatistic[] meanImpl)\nthrows DimensionMismatchException,\nMathIllegalStateException\n\nSets the implementation for the mean.\n\nThis method must be activated before any data has been added - i.e., before addValue has been used to add data; otherwise an IllegalStateException will be thrown.\n\nOverrides:\nsetMeanImpl in class MultivariateSummaryStatistics\nParameters:\nmeanImpl - the StorelessUnivariateStatistic instance to use for computing the mean\nThrows:\nDimensionMismatchException - if the array dimension does not match the one used at construction\nMathIllegalStateException - if data has already been added (i.e if n > 0)" ]

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[ "Factors of Cube Root of 83815", null, "Here we will show you how to get the factors of cube root of 83815 (factors of ∛83815). We define factors of cube root of 83815 as any integer (whole number) or cube root that you can evenly divide into cube root of 83815. Furthermore, if you divide ∛83815 by a factor of ∛83815, it will result in another factor of ∛83815.\n\nFirst, we will find all the cube roots that we can evenly divide into cube root of 83815. We do this by finding all the factors of 83815 and add a radical (∛) to them like this:\n\n∛1, ∛5, ∛16763, and ∛83815\n\nNext, we will find all the integers that we can evenly divide into cube root of 83815. We do that by first identifying the perfect cube roots from the list above:\n\n∛1\n\nThen, we take the cube root of the perfect cube roots to get the integers that we can evenly divide into cube root of 83815.\n\n1\n\nFactors of cube root of 83815 are the two lists above combined. Thus, factors of cube root of 83815 (cube roots and integers) are as follows:\n\n1, ∛1, ∛5, ∛16763, and ∛83815\n\nLike we said above, cube root of 83815 divided by any of its factors, will result in another of its factors. Therefore, if you divide ∛83815 by any of factors above, you will see that it results in one of the other factors.\n\nWhat can you do with this information? For one, you can get cube root of 83815 in its simplest form. Cube root of 83815 simplified is the largest integer factor times the cube root of 83815 divided by the largest perfect cube root. Thus, here is the math to get cube root of 83815 in its simplest radical form:\n\n∛83815\n= 1 × (∛83815 ÷ ∛1)\n= ∛83815\n\nCube Root Factor Calculator\nDo you need the factors of another cube root? No problem, please enter your cube root in the box below.\n\nFactors of Cube Root of 83816\nWe hope this information was helpful. Do you want to learn more? If so, go here to get the factors of the next cube root on our list." ]

[ null, "https://factorization.info/images/factors-of-cube-root.png", null ]

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[ "# Least Common Multiple of 45 and 30\n\nFree LCM Calculator determines the least common multiple (LCM) between 45 and 30 the smallest integer that is 90 that is divisible by both numbers.\n\nLeast Common Multiple (LCM) of 45 and 30 is 90.\n\nLCM(45,30) = 90\n\n## LCM of 45 and 30\n\nLeast common multiple or lowest common denominator (LCD) can be calculated in three ways;\n\nLCM of:\nand\n\n### Least Common Multiple (LCM) of 45 and 30 with Primes\n\nLeast common multiple can be found by multiplying the highest exponent prime factors of 45 and 30. First we will calculate the prime factors of 45 and 30.\n\nPrime Factorization of 45\n\n 3 45 3 15 5 5 1\n\nPrime factors of 45 are 3,5. Prime factorization of 45 in exponential form is:\n\n45 = 32×51\n\nPrime Factorization of 30\n\n 2 30 3 15 5 5 1\n\nPrime factors of 30 are 2, 3,5. Prime factorization of 30 in exponential form is:\n\n30 = 21×31×51\n\nNow multiplying the highest exponent prime factors to calculate the LCM of 45 and 30.\n\nLCM(45,30) = 21×32×51\nLCM(45,30) = 90\n\nFactors of 45\n\nList of positive integer factors of 45 that divides 45 without a remainder.\n\n1, 3, 5, 9, 15, 45\n\nFactors of 30\n\nList of positive integer factors of 30 that divides 30 without a remainder.\n\n1, 2, 3, 5, 6, 10, 15, 30\n\n### Least Common Multiple of 45 and 30 with GCF Formula\n\nThe formula of LCM is LCM(a,b) = ( a × b) / GCF(a,b).\nWe need to calculate greatest common factor 45 and 30, than apply into the LCM equation.\n\nGCF(45,30) = 15\nLCM(45,30) = ( 45 × 30) / 15\nLCM(45,30) = 1350 / 15\nLCM(45,30) = 90\n\n### Properties of LCM 45 and 30\n\n(i) The LCM of 30 and 45 is associative\n\nLCM of 45 and 30 = LCM of 30 and 45\n\n### Frequently Asked Questions on LCM of 45 and 30\n\n1. What is the LCM of 45 and 30?\n\nAnswer: LCM of 45 and 30 is 90.\n\n2. What are the Factors of 45?\n\nAnswer: Factors of 45 are 1, 3, 5, 9, 15, 45. There are 6 integers that are factors of 45. The greatest factor of 45 is 45.\n\n3. What are the Factors of 30?\n\nAnswer: Factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30. There are 8 integers that are factors of 30. The greatest factor of 30 is 30.\n\n4. How to Find the LCM of 45 and 30?\n\nLeast Common Multiple of 45 and 30 = 90\n\nStep 1: Find the prime factorization of 45\n\n45 = 3 x 3 x 5\n\nStep 2: Find the prime factorization of 30\n\n30 = 2 x 3 x 5\n\nStep 3: Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the lcm:\n\nLCM = 90 = 2 x 3 x 3 x 5\n\nStep 4: Therefore, the least common multiple of 45 and 30 is 90." ]

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[ "Note the value of R-squared on the graph. collected to make an accurate statistical. You may know that you can make this calculation, often designated by the letter r , How to Calculate Pearson's R (Pearson Correlations) in Microsoft Excel Put the values of the two variables you wish to correlate into two. Calculate R-squared in Microsoft Excel by creating two data ranges to way is to get two data sets and use the built-in R-squared formula.\n\n## linear regression excel\n\nWe usually use correlation coefficient (a value between -1 and 1) to display how strongly In Excel, we also can use the CORREL function to find the correlation . Calculating Pearson's r Correlation Coefficient with Excel Creating a Scatterplot of Correlation Data with Excel. The correlation coefficient (a value between -1 and +1) tells you how strongly two add-in in Excel to find the correlation coefficient between two variables.\n\nWe'll show you how to find and graph the correlation coefficient with I used the CORREL function to see if the model year and value were. Use Excel's RSQ function to compute R^2. The RSQ function has two arguments, the first represents the Y values and the second represents the X values. If an array or reference argument contains text, logical values, or empty cells, those values are ignored; however, cells with the The formula for the Pearson product moment correlation coefficient, r, is: Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. Get instant Excel help.\n\n## how to add r squared value in excel 2016\n\nThe tutorial explains how to find correlation in Excel, calculate a correlation For example, the R2 value in the second graph is How to find Pearson's r by hand or using technology. The correlation coefficient in Excel will always return a value, even if your data is something other. Note: the Analysis TookPak is no longer included in Excel for the Mac. You need to download a find the regression option and click “OK”. Now input the cells the R square value, highlighted at left in the pink cell. You then report the R2. Why am I getting a different r-square value computed from regression analysis and from fitting a trendline in excel? While I was trying to fit a linear relationship. R-squared, also known as the coefficient of determination, is the statistical There are several steps that you need to calculate before you can get to this point . First, you use the line of best fit equation to predict y values on the chart based on. You can use the following steps to calculate the correlation, r, from a data set: Find Find the standard deviation of all the x-values (call it sx) and the standard . fits a particular set of experimental data. In this module, we have relied on the R. 2 value computed in Excel to determine how closely our data conform to a linear . The coefficient of correlation R ranges between -1 and 1, with 0 showing no correlation, 1 showing How to Make One Long Column Into Multiple Columns in Excel and sales may be strongly correlated and have a high R-squared value. And when you squared this result we get R squared which is nothing but the coefficient of determination. You can download this R Squared Formula Excel Template here – R Let's now input the values in the formula to arrive at the figure. Explains that the R-squared value in the chart tool will be incorrect when you select the Set Intercept = 0 box in Excel To work around this." ]

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[ "Sale ends in\nTeacher: Ebru\nCustomers Who Have Viewed This Course: 5785\n\\$299.00\n\\$137.01\n\n#### 0 Introduction 04:37\n\nHere is a brief introduction to the course and the instructor.\n\n#### 1 Lesson 1 - The Complex Number System 28:27\n\nIn this lesson, we have four objectives: 1. To define a complex number 2. To plot a complex number in the complex plane 3. To work with complex numbers and their conjugates 4. To perform arithmetic operations on complex numbers\n\n#### 2 Lesson 1 - Worksheet Review 15:32\n\nLesson 1 - Worksheet Review\n\n#### 3 Lesson 2 - The Zero Product Rule and Factoring 17:48\n\nIn this lesson, we talk about the zero product rule and review the process of factoring. We go through examples where we: 1. factor out the common factor 2. use identities such as difference of squares, difference and sum of cubes, perfect squares 3. use grouping 4. convert the given equation into quadratic form.\n\n#### 4 Lesson 2 - Worksheet Review 11:10\n\nLesson 2 - Worksheet Review\n\n#### 5 Lesson 3 - Radical Equations 10:29\n\nIn this lesson, we learn about solving radical equations. We also cover situations where we need to check for extraneous solutions.\n\n#### 6 Lesson 3 - Worksheet Review 10:45\n\nLesson 3 - Worksheet Review\n\n#### 7 Lesson 4 - Integer and Rational Exponents 15:06\n\nIn this lesson, we learn how to define and evaluate expressions with integer and rational exponents. We also solve some equations.\n\n#### 8 Lesson 4 - Worksheet Review 06:15\n\nLesson 4 - Worksheet Review\n\n#### 9 Lesson 5 - Properties of Exponents and Radicals 13:36\n\nIn this lesson, we discuss properties of exponents and radicals and apply these properties to the task of simplifying algebraic expressions.\n\n#### 10 Lesson 5 - Worksheet Review 09:13\n\nLesson 5 - Worksheet Review\n\n#### 11 Lesson 6 - Problems With Systems of Inequalities in Two Variables 11:44\n\nIn this lesson, we look at problems which can be modelled with systems of inequalities in two variables. We learn to write the inequalities, sketch their graphs and interpret the solution to the system.\n\n#### 12 Lesson 6 - Worksheet Review 10:32\n\nLesson 6 - Worksheet Review\n\n#### 13 Lesson 7 - Linear Systems in Three Variables 12:04\n\nIn this lesson, we look at linear systems in three variables and we solve these systems using substitution and/or elimination. We also talk about the geometric interpretation of these solutions.\n\n#### 14 Lesson 7 - Worksheet Review 10:00\n\nLesson 7 - Worksheet Review\n\n#### 15 Lesson 8 - Arithmetic Operations on Polynomials 19:31\n\nIn this lesson, we introduce basic vocabulary regarding polynomials, such as degree, coefficient, leading term, and we also perform addition, subtraction, multiplication and division on polynomials.\n\n#### 16 Lesson 8- Worksheet Review 11:57\n\nLesson 8- Worksheet Review\n\n#### 17 Lesson 9 - Successive Differences in Polynomials 12:50\n\nIn this lesson, we examine the method of finite differences to find the expression for a polynomial. We also verify the method of finite differences for a linear, quadratic and cubic polynomial.\n\n#### 18 Lesson 9 - Worksheet Review 12:14\n\nLesson 9 - Worksheet Review\n\n#### 19 Lesson 10 - Polynomial Identities and Equations 11:40\n\nIn this lesson, we use polynomial identities to solve equations in complex numbers.\n\n#### 20 Lesson 10 - Worksheet Review 13:14\n\nLesson 10 - Worksheet Review\n\n#### 21 Lesson 11 - Adding and Subtracting Rational Expressions 13:11\n\nIn this lesson, we add and subtract rational expressions where we expand the rational expressions using a least common denominator.\n\n#### 22 Lesson 11 - Worksheet Review 10:34\n\nLesson 11 - Worksheet Review\n\n#### 23 Lesson 12 - Multiplying and Dividing Rational Expressions 12:47\n\nIn this lesson, we multiply or divide rational expressions and simplify the result through factoring and simplifying common factors between the numerator and the denominator. We also work with compound fractions, fractions where, in the numerator and/or the denominator, we have a sum or a difference of rational expressions.\n\n#### 24 Lesson 12 - Worksheet Review 14:23\n\nLesson 12 - Worksheet Review\n\n#### 25 Lesson 13 - Completing the Square, the Quadratic Formula and Types of Roots of a Quadratic Equation 15:38\n\nIn this lesson, we revisit the quadratic equation and use completing the square to solve the equation. Then, we use the quadratic formula to solve some quadratic equations and we learn to use the discriminant to identify the nature of the roots.\n\n#### 26 Lesson 13 - Worksheet Review 09:05\n\nLesson 13 - Worksheet Review\n\n#### 27 Lesson 14- Defining and Graphing a Parabola 11:41\n\nIn this lesson, we look at the main features of a parabola, such as its direction, its line of symmetry, vertex and intercepts. We also sketch the graphs of parabolas using these features.\n\nWorksheet Review\n\n#### 29 Lesson 15 - Graphing and Solving Systems of Equations 10:46\n\nIn this lesson, we are working on solving systems of linear and quadratic equations. We are also graphing the systems and interpreting the solutions geometrically.\n\nWorksheet Review\n\n#### 30 Midterm Exam 1 00:00\n\nThis midterm exam covers lesson 1 through 15. The duration is 80 minutes.\n\n#### 31 Lesson 16 - Logarithms 12:54\n\nIn this lesson, we learn about the definition of a logarithm and how to evaluate a logarithmic value. We also talk about the common logarithm and the natural logarithm.\n\nWorksheet Review\n\n#### 33 Lesson 17 - Properties of Logarithms 17:21\n\nIn this lesson, we explore the properties of logarithms, including the Change of Base Formula, and we evaluate and simplify logarithmic expression with these properties.\n\n#### 34 Lesson 17 - Worksheet Review 06:21\n\nLesson 17 - Worksheet Review\n\n#### 35 Lesson 18 - Logarithmic Equations 12:11\n\nIn this lesson, we apply the properties of logarithms in solving logarithmic equations. We also learn to identify extraneous solutions to these equations.\n\nWorksheet Review\n\n#### 37 Lesson 19 - Exponential Equations 10:14\n\nIn this lesson, we will solve exponential equations. We will express some of the solutions as logarithmic expressions and use factoring techniques in some of the equations.\n\nWorksheet Review\n\n#### 39 Lesson 20 - Choosing a Model and Modeling with Exponential Functions 16:15\n\nIn this lesson, we take a brief look at when polynomial, linear, trigonometric and exponential models are useful for certain real-life situations. In the second half of the lesson, we derive exponential models for given problems.\n\n#### 40 Lesson 20 - Worksheet Review 05:25\n\nLesson 20 - Worksheet Review\n\n#### 41 Lesson 21 -Concept of a Function, Function Notation and Building Functions 15:56\n\nIn this lesson, we are introducing functions, function notation, domain, range and building functions.\n\n#### 42 Lesson 21 - Worksheet Review 05:33\n\nLesson 21 - Worksheet Review\n\n#### 43 Lesson 22 - Creating New Functions from Existing Functions 16:36\n\nIn this lesson, we look at operations on functions and transformations of functions.\n\n#### 44 Lesson 22 - Worksheet Review 05:39\n\nLesson 22 - Worksheet Review\n\n#### 45 Lesson 23 - Even/Odd Functions and Inverse Functions 21:10\n\nIn this lesson, we first decide if a given function is odd, even or neither. We also look at the symmetry of odd and even function graphs. Then, we learn to take the inverse of a given function and to graph the inverse function on the coordinate plane.\n\n#### 46 Lesson 23 - Worksheet Review 07:19\n\nLesson 23 - Worksheet Review\n\n#### 47 Lesson 24 - Rates of Change and Linear Functions 11:22\n\nIn this lesson, we learn to calculate average rates of change for functions over different intervals. Then, we learn to find linear functions whose rate of change is constant over any interval.\n\n#### 48 Lesson 24 - Worksheet Review 08:36\n\nLesson 24 - Worksheet Review\n\n#### 49 Lesson 25 Characteristics of Quadratic Functions and Comparing Properties of Functions 18:05\n\nIn this lesson, we look at characteristics of quadratic functions, such as vertex and the axis of symmetry of a parabola. We find the vertex of a parabola using the intercepts, the quadratic formula and the method of completing the square. Then, we compare properties of linear and quadratic functions.\n\n#### 50 Lesson 25 Worksheet Review 09:39\n\nLesson 25 Worksheet Review\n\n#### 51 Lesson 26 - Exponential Functions 10:49\n\nIn this lesson, we study exponential functions and their graphs. We look at important features such as intercepts, end behavior, asymptotes. We also cover transformations of exponential functions and go over finding the equations for exponential functions.\n\n#### 52 Lesson 26 - Worksheet Review 06:24\n\nLesson 26 - Worksheet Review\n\n#### 53 Lesson 27 - Trigonometric Functions 26:38\n\nThis lesson has four objectives: 1. Applying the idea of a function to trigonometric functions 2. Revisiting the unit circle to sketch the parent sine and cosine curves. 3. Establishing features of trigonometric curves such as period and amplitude. 4. Applying transformations towards sketching graphs of sine and cosine curves.\n\n#### 54 Lesson 27 - Worksheet Review 18:13\n\nLesson 27 - Worksheet Review\n\n#### 55 Lesson 28 - Logarithmic Functions 11:41\n\nIn this lesson, we look at logarithmic functions and graphs. We define logarithmic functions as the inverse of exponential functions, we identify features of their graphs and cover transformations of parent logarithmic graphs.\n\n#### 56 Lesson 28 - Worksheet Review 06:34\n\nLesson 28 - Worksheet Review\n\n#### 57 Lesson 29 - Probability and Venn Diagrams 16:34\n\nIn this lesson, we look at probabilities of simple events and we go over definitions of event, sample space, outcome and complement. We then represent situations with Venn diagrams and calculate probabilities from Venn diagrams.\n\n#### 58 Lesson 29 - Worksheet Review 08:02\n\nLesson 29 - Worksheet Review\n\n#### 59 Lesson 30 - Probability Rules, Conditional Probability, Tree Diagrams 17:51\n\nIn this lesson, we work with the probability rules and learn to display information using tree diagrams. We also calculate conditional probability and learn about independent and mutually exclusive events.\n\n#### 60 Lesson 30 - Worksheet Review 10:30\n\nLesson 30 - Worksheet Review\n\n#### 60 Midterm Exam 2 00:00\n\nThis midterm exam covers lesson 16 through 30. The duration is 80 minutes.\n\n#### 61 Lesson 31 - Center, Spread and Shape of Distributions 11:48\n\nIn this lesson, we learn to calculate the measures of center and spread of distributions. We then interpret the shape of a distribution from its histogram.\n\n#### 62 Lesson 31 - Worksheet Review 04:51\n\nLesson 31 - Worksheet Review\n\n#### 63 Lesson 32 - The Normal Distribution 14:55\n\nIn this lesson, we cover the normal distribution and the standard normal distribution. We calculate probabilities from data values and data values from probabilities. We also find the mean and the standard deviation from given information about percentiles.\n\n#### 64 Lesson 32 - Worksheet Review 11:09\n\nLesson 32 - Worksheet Review\n\n#### 65 Lesson 33 - Types of Statistical Studies, Estimation of Population Characteristic, Sampling Variability 12:10\n\nIn this lesson, we look at three issues in Statistics. We learn about the three types of statistical studies. we look at how we can estimate a population characteristic from a sample and we examine the sampling variability in considering the sample mean as an estimate of the population mean.\n\n#### 66 Lesson 33 - Worksheet Review 08:02\n\nLesson 33 - Worksheet Review\n\n#### 67 Lesson 34 - Square Roots and Proportional Reasoning 13:22\n\nIn this lesson, we cover finding square roots of positive rational numbers and learn to apply proportional reasoning to problems.\n\n#### 68 Lesson 34 - Worksheet Review 06:54\n\nLesson 34 - Worksheet Review\n\n#### 69 Lesson 35 - Properties of Angles in Triangles 13:07\n\nIn this lesson, we look at properties of angles in triangles, polygons and on parallel lines.\n\n#### 70 Lesson 35 - Worksheet Review 04:14\n\nLesson 35 - Worksheet Review\n\n#### 71 Lesson 36 - Theorems About Circles 10:03\n\nIn this lesson, we apply theorems about angles and segment lengths related to circles.\n\n#### 72 Lesson 36 - Worksheet Review 09:20\n\nLesson 36 - Worksheet Review\n\n#### 73 Lesson 37- The Unit Circle 18:41\n\nIn this lesson, we talk about defining angles and trigonometric ratios of angles in the coordinate plane. Then, we introduce the unit circle and learn to use the unit circle to find trigonometric values.\n\n#### 74 Lesson 37 - Worksheet Review 17:31\n\nLesson 37 - Worksheet Review\n\n#### 75 Lesson 38 - Modeling Periodic Data with Trigonometric Functions 10:00\n\nIn this lesson, we look at how we can model a periodic trend with a trigonometric function. We apply the concepts of amplitude, period, phase shift and vertical translation in devising such models.\n\n#### 76 Lesson 38 - Worksheet Review 08:41\n\nLesson 38 - Worksheet Review\n\n#### 77 Lesson 39 - Proving Trigonometric Identities 18:31\n\nIn this lesson, we work with trigonometric identities to prove other trigonometric identities and simplify trigonometric expressions.\n\n#### 78 Lesson 39 - Worksheet Review 08:04\n\nLesson 39 - Worksheet Review\n\n#### 79 Final Exam 00:00\n\nThis exam covers the entire course. The duration is 120 minutes. Good luck!\n\nCourse Description\n\nThis course is a gateway to Precalculus and it provides you with all the knowledge and skills you need in Precalculus.\n\nThe course includes:\n\nThe course comprises 10 units and 39 lessons\n\nOver 15 hours of video-lessons!\n\nEach lesson includes a video lecture, a worksheet, a worksheet review video and an online quiz.\n\n39 video lectures\n\n39 worksheets\n\n39 worksheet review videos with every question solved step-by step by the instructor\n\n39 online quizzes\n\n2 midterms (on paper)\n\n1 final exam (on paper)\n\nCOURSE GOALS\n\nUpon course completion, you will learn how to:\n\n• solve quadratic equations, including complex solutions\n• work with exponents and radicals\n• solve systems of inequalities in two variables and linear systems in three variables\n• perform operations on polynomials and rational expressions\n• define and graph a parabola\n• define logarithms and apply properties of logarithms\n• solve logarithmic and exponential equations\n• choose between models and model with a logarithmic or a trigonometric function\n• apply the concept of a function and function notation\n• perform operations on functions, including function composition and transformations\n• find the inverse of a function and decide if a function is odd, even or neither\n• find the average rate of change of a function over a given interval\n• work with linear, quadratic, trigonometric, exponential and logarithmic functions\n• calculate probabilities and conditional probabilities from probability rules and Venn or tree diagrams\n• examine the center, spread and shape of distributions\n• work with the normal distribution and the standard normal distribution\n• consider types of statistical studies, estimation of a population characteristic and sampling variability in the distribution of sample means\n• work with proportional reasoning\n• apply theorems about angles in triangles and circles\n• use the unit circle in finding trigonometric values\n• prove trigonometric identities and simplify trigonometric expressions\n\nTARGET AUDIENCE\n\nThis video-course is primarily intended for 11th grade students who would like to solidify their mathematical skills and knowledge before they take Precalculus.\n\nCOURSE REQUIREMENTS\n\nStudents taking this course will need to have completed Algebra I, Math for 10th Grade or an equivalent Math course prior to enrolling in this course.\n\nUNIT 1 FACTORING AND SOLVING QUADRATIC EQUATIONS\n\nThe Complex Number System\n\nThe Zero Product Rule and Factoring\n\nInteger and Rational Exponents\n\nUNIT 3 SYSTEMS OF EQUATIONS AND INEQUALITIES\n\nProblems with Systems of Inequalities in Two Variables\n\nLinear Systems in Three Variables\n\nUNIT 4 POLYNOMIALS AND RATIONAL EXPRESSIONS\n\nArithmetic Operations on Polynomials\n\nSuccessive Differences in Polynomials\n\nPolynomial Identities and Equations\n\nMultiplying and Dividing Rational Expressions\n\nUNIT 5 MORE ON SYSTEMS OF EQUATIONS WITH QUADRATIC EQUATIONS\n\nCompleting the Square, the Quadratic Formula and Types of Roots of a Quadratic Equation\n\nDefining and Graphing a Parabola\n\nGraphing and Solving Systems of Equations\n\nUNIT 6 EXPONENTS AND LOGARITHMS\n\nLogarithms\n\nProperties of Logarithms\n\nLogarithmic Equations\n\nExponential Equations\n\nChoosing a Model and Modeling with Exponential Functions\n\nUNIT 7 FUNCTIONS AND TYPES OF FUNCTIONS\n\nConcept of a Function, Function Notation and Building Functions\n\nCreating New Functions from Existing Functions\n\nEven/Odd Functions and Inverse Functions\n\nRates of Change and Linear Functions\n\nCharacteristics of Quadratic Functions and Comparing Properties of Functions\n\nExponential Functions\n\nTrigonometric Functions\n\nLogarithmic Functions\n\nUNIT 8 PROBABILITY\n\nProbability and Venn Diagrams\n\nProbability Rules, Conditional Probability, Tree Diagrams\n\nUNIT 9 STATISTICS\n\nCenter, Spread and Shape of Distributions\n\nThe Normal Distribution\n\nTypes of Statistical Studies, Estimation of Population Characteristic, Sampling Variability\n\nUNIT 10 GEOMETRY AND CIRCULAR FUNCTIONS\n\nSquare Roots and Proportional Reasoning\n\nProperties of Angles in Triangles\n\nThe Unit Circle\n\nModeling Periodic Data with Trigonometric Functions\n\nProving Trigonometric Identities\n\n• Teacher: Ebru\n• Areas of expertise: Mathematics\n• Education: Oberlin College, BA in Mathematics and Anthropology 1993 Texas A&M University MSc in Mathematics 2011\n• Interests: Epistemology, Acting\n• Skills: Learning Languages\n• Associations: National Council of Teachers of Mathematics (NCTM)\n• Issues I care about: Sustainability, Social Responsibility, Giving Back to the Community\n\nI am happy to start working with Lernsys students as I reach out to you all beyond the physical boundaries of the classroom.\n\n#### Midterm 1 Solutions\n\nMidterm 1 Solutions\n\n#### Midterm 2 Solutions\n\nMidterm 2 Solutions\n\n#### Final Exam Solutions\n\nFinal Exam Solutions" ]

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[ "", null, "# Quick Answer: What Is A Multiple Of 3 And 8?\n\n## What is the least common multiple of 3 2 and 8?\n\nThe smallest common multiple of two or more numbers is called the lowest common multiple (LCM).\n\nE.g.\n\nMultiples of 8 are 8, 16, 24, 32, … Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, ….\n\n## What are the five multiples of 3?\n\nMultiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, … Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, … So, the common multiples of 3 and 4 are 12, 24, 36, …\n\n## What is the lowest common multiple of 2 3 and 7?\n\nThe LCM of 2,3,7 2 , 3 , 7 is 2⋅3⋅7=42 2 ⋅ 3 ⋅ 7 = 42 .\n\n## What are the multiples of 8?\n\nThe multiples of 8 are 8,16,24,32,40,48,56,64,72,80,88,96,104,112,120,128,136,144,152,160,168,176,184,192,200,208 and so on.\n\n## What are the factors of 3 and 8?\n\nGreatest common factor (GCF) of 3 and 8 is 1. We will now calculate the prime factors of 3 and 8, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 3 and 8.\n\n## How do you find multiples of 3?\n\nThe multiples of a whole number are found by taking the product of any counting number and that whole number. For example, to find the multiples of 3, multiply 3 by 1, 3 by 2, 3 by 3, and so on. To find the multiples of 5, multiply 5 by 1, 5 by 2, 5 by 3, and so on.\n\n## What is a multiple of 3 and 4?\n\nCommon multiples of 3 and 4: 12, 24, 36, …\n\n## What are the multiples of 3 up to 1000?\n\nAnswer : 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99,102,105,108,111,114,117,120,123,126,129,132,135,138,141,144,147, Related Links : What are the factors of 3?\n\n## What are the first 10 multiples of 3?\n\nThe first ten multiples of 3 are listed below: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.\n\n## What are the common multiples of 3 and 7?\n\nThe common multiples of 3 and 7 up to 70 are 21, 42, and 63.\n\n## Why do multiples of 3 add up to 3?\n\nBecause the digits multiply by ( a multiple of three, plus one), for each digit. The first part of this is obviously divisible by 3. So for the whole thing to be divisible, the sum of the digits has to be as well.\n\n## What is the multiple of 12?\n\nThe multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144,…\n\n## What are the multiples of 3?\n\nHow to list multiples of a number?Multiples of 11, 2, 3, 4, 5, 6, 7, 8, 9, 10, …Multiples of 22, 4, 6, 8, 10, 12, 14, 16, 18, 20, …Multiples of 33, 6, 9, 12, 15, 18, 21, 24, 27, 30, …Multiples of 44, 8, 12, 16, 20, 24, 28, 32, 36, 40, …Multiples of 55, 10, 15, 20, 25, 30, 35, 40, 45, 50, …10 more rows\n\n## What are the first 10 multiples of 8?\n\nThe first ten multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, and 80. The first six multiples of 15 are 15, 30, 45, 60, 75, and 90.\n\n## What are the 5 multiples of 8?\n\nAnswer. five multiples of 8 are 16,24,32,40,48 ……\n\n## What are the multiples of 8 between 100 and 200?\n\nAnswer:104 , 112 , 120 , 128 , 136 , 144 , 152 , 160 , 168 , 176 , 184 , 192.Step-by-step explanation:So, starting from 100 , we are going to check the numbers which are divisible by 8.The next number will be 120 , 128 , 136 …….. 192.\n\n## What do you notice about multiples of 3?\n\nPatterns in Multiplying by 3’s Multiples of 3 have a pattern of 3, 6, 9, 2, 5, 8, 1, 4, 7, 0 in the ones place. Every other multiple of 3 is even. The digits in multiples of 3 add up to a multiple of 3 (36 = 3 + 6, 111 = 1 + 1 + 1, etc.)\n\n## What is the LCM of 3 and 9?\n\nLeast common multiple (LCM) of 3 and 9 is 9.\n\n## What are factors of 3?\n\nTable of Factors and MultiplesFactorsMultiples1, 3391, 2, 44121, 55151, 2, 3, 661841 more rows\n\n## What is the least common multiple of 10 and 15?\n\n30Least common multiple (LCM) of 10 and 15 is 30.\n\n## What is the smallest multiple of 3 and 4?\n\n12The Least Common Multiple (LCM) of some numbers is the smallest number that the numbers are factors of. Like the LCM of 3 and 4 is 12, because 12 is the smallest number that 3 and 4 are both factors for." ]

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[ "# Algorithm for maximum weight matching\n\nI have been given the following algorithm for finding a maximum weight matching in a bipartite graph:\n\n1. $M=\\emptyset$\n\n2. Find augmenting path $C$ with minimal length when looking at length function $l_{ij} = w_{ij}$ if $(v_2, v_j)\\in M$, or $l_{ij} = -w_{ij}$ otherwise. Stop when there is no augmenting path, or every augmenting path has non-negative length.\n\n3. $M = M \\oplus C$ and go back to step 2\n\nNow I am being asked:\n\n1. Prove correctness by showing that in iteration $k$, matching $M$ has maximal weight compared to any other matching of cardinality $k$.\n\n2. Show that the complexity is $\\mathcal{O}(n^4)$.\n\nFor (1), I was thinking of doing it by induction, but this doesn't seem to work: The case for the first iteration is clear, and that the weight of the matching grows every iteration is obvious as well, but how do we know there isn't some way to first choose \"lesser\" matchings and then finally find an augmenting path which gives the best possible matching $M$ (I don't know if induction is the best way by the way, but I wouldn't know how else to approach it).\n\nFor (2), I would think the complexity lies in finding the augmenting path in step 2 of the algorithm. However, I don't see how the $n^4$ would come into play here.\n\nIf anyone has any hints or ways in which I could try to approach this, I would be most grateful.\n\nFirst, $M= M \\oplus C$ means that we add all edges in $C$ to $M$, and if any edge is in $C$ and old $M$, we remove it from $M$.\n\nFor $(1)$, induction is a fine approach. The key observation is that an augmenting path $C$ can 'undo' all previous edge choices, provided we keep the same matched nodes. Fortunately this is always the case.\n\nSuppose for some maximum weight matching with $k$ edges $M_k$ with total weight $w_k$ there is a maximum weight matching with $k+1$ edges with some edge that is disjoint from the matched vertexes of $M_k$. Then there exists $k+1$ edges $e_i$ such that:\n\n1. $e_{k+1} = (u,v) \\not \\in M_k$ and further $u,v$ are unmatched by $M_k$ (by assumption)\n2. $w_k \\geq (\\sum_i w(e_i) ) - e_{k+1} \\implies w_k + w(e_{k+1}) \\geq \\sum_i w(e_i)$ (first part follows from $M_k$ being maximal with $k$ edges)\n\nNote that $C = \\{e_{k+1} \\}$ would be a valid augmenting path, giving a matching with $k+1$ edges. This is because $e_{k+1}$ is disjoint from $M_k$.\n\nMeanwhile, suppose there is a maximum weight matching $M'_{k+1}$ with $k+1$ edges that has no disjoint edges from $M_k$. Then we can follow a path from the first new vertex, alternating between edges in $M_k$ and the edges in $M'_{k+1}$ (and skipping edges both share). Because they are both matchings, and they share matched vertecies, every matched vertex in $M_k$ has an edge from $M'_k$ and from $M_k$. If this happens to be the same edge, the agumenting path need not go through this node. If they are different, then that node can be part of an augmenting path coming in/out from one and out the other (hence alternating being in $M$ and out of $M$). Further, if they are not the same then the vertexes they connect to must also be matched differently by $M_k$ and $M'_k$ (otherwise some node would be matched twice). Therefore these nodes with different matchings in $M$ and $M'$ form a path, and since $M'_k$ is maximal, this path will be a minimal augmenting path.\n\nThus in either case, be there a maximum matching on $k+1$ edges that is disjoint or overlapping, we can find an augmenting path that will make $M$ do just as good.\n\nFor $(2)$, one way to find an augmenting path is via Bellman Ford, which takes $O(|V||E|) = O(n^3)$ time (Assuming one has the maximum of $O(n^2)$ edges). Naively you would have to run bellman from every start node, but you can modify the graph so that Bellman ford finds the shortest augmenting path with only one call. The trick is to modify the graph slightly before giving it to the algorithm. We add 2 new vertexes $s,t$, with $s$ connected to one bipartition by edges with weight $0$, and $t$ connected to the other. You then just need to find the minimal path from $s$ to $t$, which bellman ford does.\n\nThis only adds a linear number of edges and a constant number of vertecies, so the asymptotic runtime is not changed. Check how many calls we need to Bellman Ford (or your favorite $O(n^3)$ shortest paths algorithm), and we're done.\n\n• Are we using the same definition of augmenting path? I'm under the impression your answer is aimed at the \"maximum flow\" meaning of augmenting path, so a positive flow from $s$ to $t$. However, what I mean is an augmenting path in the context of a matching, so a path where edges are alternatively in and not in matching $M$ (See for example the definition here, under paragraph 6.1) – Radical Jul 27 '17 at 8:28\n• @Radical aha! Perfect. Yes we were using different definitions, I'll edit to use yours in a few hours. The good news is these can change choices too, I just need a slightly more careful argument on (1). (2) is still correct, because the graph is bipartite you can make the edges directed st bellman must pick an augmenting path by this def. The trick is not bad. The question was phrased for a hint, do you want the full arguments? – Artimis Fowl Jul 27 '17 at 15:35\n• Thanks! I'm giving you the bounty regardless, but if you have time there's a couple of things I don't totally understand yet. In your second, \"No disjoint edges\", argument for (1), you claim that creating the augmenting path in the way you describe will result in matching $M'_{k+1}$. While I see this happening if I make an example and it sort of makes sense intuitively, I don't see how augmenting $M_k$ along $M'_{k+1}$'s edges guarantees to get you to $M'_{k+1}$ (although I see how you would always get an augmenting path). – Radical Jul 28 '17 at 8:31\n• Then for (2), even if we can find an augmenting path in $\\mathcal{O}(n^3)$, wont we need to find all augmenting paths in each iteration to find the minimal length one? Let's say we can find one augmenting path in $\\mathcal{O}(n^3)$. In the worst case there are $\\frac{n}{2}$ augmenting paths getting us to $\\mathcal{O}(n^4)$ to find the shortest one, and we need to do so for $\\frac{n}{2}$ iterations getting us $\\mathcal{O}(n^5)$ – Radical Jul 28 '17 at 8:31\n• I have updated for your concerns with $(2)$, the trick being that Bellman ford finds this shortest path. It's not immediately clear to me that DFS would do the same, as they suggest in your earlier link. I will update for your concerns on $(1)$ in awhile. – Artimis Fowl Jul 28 '17 at 17:44" ]

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[ "# Changeset 230 for docs/ASPLOS09\n\nIgnore:\nTimestamp:\nDec 10, 2008, 1:16:12 PM (10 years ago)\nMessage:\n\nFurther improvements to population count section.\n\nFile:\n1 edited\n\nUnmodified\nAdded\nRemoved\n• ## docs/ASPLOS09/asplos094-cameron.tex\n\n r229 implementation for a 32-bit integer {\\tt x} adapted from Warren \\cite{HackersDelight}, while Figure \\ref{inductivepopcount} shows the corresponding SWAR implementation for a vector of 32-bit fields using the inductive doubling instruction set architecture.  Each implementation employs Figure \\ref{inductivepopcount} shows the corresponding IDISA implementation for a vector of 32-bit fields.  Each implementation employs five steps of inductive doubling to produce population counts within 32 bit fields.  The traditional implementation employs explicit masking and shifting operations, while these operations are implicit within the semantics of the inductive doubling instructions used in Figure \\ref{inductivepopcount}. doubling instructions shown in Figure \\ref{inductivepopcount}. In each implementation, the first step determines the the population counts within 2-bit fields With the substitution of longer mask constants replicated for four 32-bit fields, the implementation of Figure \\ref{HD-pop} can be easily adapted to SWAR processing using 128-bit registers. Such an implementation requires 10 operations to load or generate directly adapted to SWAR processing using 128-bit registers. Each binary operator is replaced by a corresponding binary SIMD operation.   Without the IDISA features, a straightforward SWAR implementation of population count for 32-bit fields thus employs 10 operations to load or generate mask constants, 10 bitwise-and operations, 5 shifts and 5 adds for a total of 30 operations to complete the task in comparison to a mere   However, Warren further refines the implementation to an optimized version requiring only 20 operations, 5 of which are required to generate mask constants.  At the cost of register pressure, it is possible that these constants could be kept preloaded.  In any case, the IDISA implementation total of 30 operations to complete the task.   Employing optimization identified by Warren, this can be reduced to 20 operations, 5 of which are required to generate mask constants. At the cost of register pressure, it is possible that these constants could be kept preloaded in long vector processing. In any case, the IDISA implementation requires only 5 operations to carry out the work requiring 15 to 20 operations with the optimized version using standard SWAR operations.   IDISA thus offers offers a 3X to 4X improvement in instruction count as well as a reduction in register pressure. The pattern illustrated by population count is typical.  An inductive doubling algorithm of $n$ steps typically applies two mask or shift operations at each step for each of the two operands being combined in the step.   If the combination can be implemented in a single SWAR operation, then a total of $3n$ operations are the minimum required for the SWAR algorithm with IDISA features.   An IDISA implementation typically eliminates the explicit mask and shift operations through appropriate half-operand modifiers, reducing the total instruction count to $n$.   Thus a 3X improvement is typical. \\section{Transposition to Parallel Bit Streams}\nNote: See TracChangeset for help on using the changeset viewer." ]

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[ "# How tall is 178 cm in feet?\n\n CM To Swap FEET\nName Unit\nFeet 0 Feet\nYards 0 Yards\nAcre 0 Acre\nMiles 0 Mile\nMeters 0 Meters\nKilometers 0 Km\nInches 0 Inches\nMillimeters 0 mm\n\n## How tall is 178 cm in feet:\n\n178 centimeters is equal to approximately 5 feet 10 inches. To convert from centimeters to feet and inches, you can use the following conversion factors:\n\n1 inch = 2.54 centimeters 1 foot = 12 inches\n\nSo, to convert from centimeters to feet and inches, you can divide the number of centimeters by 2.54 to get the number of inches, and then divide the number of inches by 12 to get the number of feet.\n\nFor example, to convert 178 centimeters to feet and inches:\n\n178 centimeters / 2.54 = 70 inches 70 inches / 12 = 5 feet 10 inches\n\nSo, 178 centimeters is equal to 5 feet 10 inches." ]

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[ "## VALENCE BOND METHODS Theory and applications\n\nValence Bond Methods Theory and applications Gordon A. Gallup VALENCE BOND METHODS Theory and applications Valence bond theory is one of two commonl...\nAuthor: Julian Wheeler\nValence Bond Methods Theory and applications Gordon A. Gallup\n\nVALENCE BOND METHODS Theory and applications Valence bond theory is one of two commonly used methods in molecular quantum mechanics, the other is molecular orbital theory. This book focuses on the first of these methods, ab initio valence bond theory. The book is split into two parts. Part I gives simple examples of two-electron calculations and the necessary theory to extend these to larger systems. Part II gives a series of case studies of related molecule sets designed to show the nature of the valence bond description of molecular structure. It also highlights the stability of this description to varying basis sets. There are references to the CRUNCH computer program for molecular structure calculations, which is currently available in the public domain. Throughout the book there are suggestions for further study using CRUNCH to supplement discussions and questions raised in the text. The book will be of primary interest to researchers and students working on molecular electronic theory and computation in chemistry and chemical physics. GORDON A. GALLUP\n\nwas born (9 March 1927) and raised in St Louis, Missouri and attended the public schools there. After High School and a short stint in the US Navy, he attended Washington University (St Louis) and graduated with an AB in 1950. He received the PhD degree from the University of Kansas in 1953 and spent two years at Purdue University carrying out post-doctoral research. In 1955 he was appointed to the faculty of chemistry at the University of Nebraska and rose through the ranks, becoming full professor in 1964. He spent a year at the Quantum Theory Project at the University of Florida, and a year in England at the University of Bristol on an SERC fellowship. In 1993 he retired from teaching and since then has spent time as a research professor with the Department of Physics and Astronomy at the University of Nebraska. His research interests over the years include infrared spectroscopy and molecule vibrations, theory of molecular electronic structure, valence bond theory, electron scattering from atoms and molecules, and dissociative electron attachment. During his career he has held grants from the National Science Foundation, the Department of Energy, and others. He has had over 100 articles published in 10–15 different chemistry and physics journals, as well as articles in edited compendia and review books.\n\nVALENCE BOND METHODS Theory and applications GORDON A. GALLUP University of Nebraska\n\nPUBLISHED BY CAMBRIDGE UNIVERSITY PRESS (VIRTUAL PUBLISHING) FOR AND ON BEHALF OF THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE The Pitt Building, Trumpington Street, Cambridge CB2 IRP 40 West 20th Street, New York, NY 10011-4211, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia http://www.cambridge.org © Gordon A. Gallup 2002 This edition © Gordon A. Gallup 2003 First published in printed format 2002\n\nA catalogue record for the original printed book is available from the British Library and from the Library of Congress Original ISBN 0 521 80392 6 hardback\n\nISBN 0 511 02037 6 virtual (netLibrary Edition)\n\nTo my wife Grace, for all her encouragement, and to the memory of our son, Michael, 1956–1995.\n\nContents\n\nPreface List of abbreviations\n\npage xiii xv\n\nI Theory and two-electron systems 1 Introduction 1.1 History 1.2 Mathematical background 1.2.1 Schr¨odinger’s equation 1.3 The variation theorem 1.3.1 General variation functions 1.3.2 Linear variation functions 1.3.3 A 2 × 2 generalized eigenvalue problem 1.4 Weights of nonorthogonal functions 1.4.1 Weights without orthogonalization 1.4.2 Weights requiring orthogonalization 2 H2 and localized orbitals 2.1 The separation of spin and space variables 2.1.1 The spin functions 2.1.2 The spatial functions 2.2 The AO approximation 2.3 Accuracy of the Heitler–London function 2.4 Extensions to the simple Heitler–London treatment 2.5 Why is the H2 molecule stable? 2.5.1 Electrostatic interactions 2.5.2 Kinetic energy effects 2.6 Electron correlation 2.7 Gaussian AO bases 2.8 A full MCVB calculation vii\n\n3 3 4 5 9 9 9 14 16 18 19 23 23 23 24 24 27 27 31 32 36 38 38 38\n\nviii\n\nContents\n\n2.8.1 Two different AO bases 2.8.2 Effect of eliminating various structures 2.8.3 Accuracy of full MCVB calculation with 10 AOs 2.8.4 Accuracy of full MCVB calculation with 28 AOs 2.8.5 EGSO weights for 10 and 28 AO orthogonalized bases 3 H2 and delocalized orbitals 3.1 Orthogonalized AOs 3.2 Optimal delocalized orbitals 3.2.1 The method of Coulson and Fisher 3.2.2 Complementary orbitals 3.2.3 Unsymmetric orbitals 4 Three electrons in doublet states 4.1 Spin eigenfunctions 4.2 Requirements of spatial functions 4.3 Orbital approximation 5 Advanced methods for larger molecules 5.1 Permutations 5.2 Group algebras 5.3 Some general results for finite groups 5.3.1 Irreducible matrix representations 5.3.2 Bases for group algebras 5.4 Algebras of symmetric groups 5.4.1 The unitarity of permutations 5.4.2 Partitions 5.4.3 Young tableaux and N and P operators 5.4.4 Standard tableaux 5.4.5 The linear independence of Ni Pi and Pi Ni 5.4.6 Von Neumann’s theorem 5.4.7 Two Hermitian idempotents of the group algebra 5.4.8 A matrix basis for group algebras of symmetric groups 5.4.9 Sandwich representations 5.4.10 Group algebraic representation of the antisymmetrizer 5.5 Antisymmetric eigenfunctions of the spin 5.5.1 Two simple eigenfunctions of the spin 5.5.2 The \u0002 function 5.5.3 The independent functions from an orbital product 5.5.4 Two simple sorts of VB functions 5.5.5 Transformations between standard tableaux and HLSP functions 5.5.6 Representing θ N PN \u0002 as a functional determinant\n\n40 42 44 44 45 47 47 49 49 49 51 53 53 55 58 63 64 66 68 68 69 70 70 70 71 72 75 76 76 77 79 80 81 81 84 85 87 88 91\n\nContents\n\n6 Spatial symmetry 6.1 The AO basis 6.2 Bases for spatial group algebras 6.3 Constellations and configurations 6.3.1 Example 1. H2 O 6.3.2 Example 2. NH3 6.3.3 Example 3. The π system of benzene 7 Varieties of VB treatments 7.1 Local orbitals 7.2 Nonlocal orbitals 8 The physics of ionic structures 8.1 A silly two-electron example 8.2 Ionic structures and the electric moment of LiH 8.3 Covalent and ionic curve crossings in LiF\n\nix\n\n97 98 98 99 100 102 105 107 107 108 111 111 113 115\n\nII Examples and interpretations 9 Selection of structures and arrangement of bases 9.1 The AO bases 9.2 Structure selection 9.2.1 N2 and an STO3G basis 9.2.2 N2 and a 6-31G basis 9.2.3 N2 and a 6-31G∗ basis 9.3 Planar aromatic and π systems 10 Four simple three-electron systems 10.1 The allyl radical 10.1.1 MCVB treatment 10.1.2 Example of transformation to HLSP functions 10.1.3 SCVB treatment with corresponding orbitals 10.2 The He+ 2 ion 10.2.1 MCVB calculation 10.2.2 SCVB with corresponding orbitals 10.3 The valence orbitals of the BeH molecule 10.3.1 Full MCVB treatment 10.3.2 An SCVB treatment 10.4 The Li atom 10.4.1 SCVB treatment 10.4.2 MCVB treatment 11 Second row homonuclear diatomics 11.1 Atomic properties 11.2 Arrangement of bases and quantitative results\n\n121 121 123 123 123 124 124 125 125 126 129 132 134 134 135 136 137 139 141 142 144 145 145 146\n\nx\n\n12\n\n13\n\n14\n\n15\n\n16\n\nContents\n\n11.3 Qualitative discussion 11.3.1 B2 11.3.2 C2 11.3.3 N2 11.3.4 O2 11.3.5 F2 11.4 General conclusions Second row heteronuclear diatomics 12.1 An STO3G AO basis 12.1.1 N2 12.1.2 CO 12.1.3 BF 12.1.4 BeNe 12.2 Quantitative results from a 6-31G∗ basis 12.3 Dipole moments of CO, BF, and BeNe 12.3.1 Results for 6-31G∗ basis 12.3.2 Difficulties with the STO3G basis Methane, ethane and hybridization 13.1 CH, CH2 , CH3 , and CH4 13.1.1 STO3G basis 13.1.2 6-31G∗ basis 13.2 Ethane 13.3 Conclusions Rings of hydrogen atoms 14.1 Basis set 14.2 Energy surfaces Aromatic compounds 15.1 STO3G calculation 15.1.1 SCVB treatment of π system 15.1.2 Comparison with linear 1,3,5-hexatriene 15.2 The 6-31G∗ basis 15.2.1 Comparison with cyclobutadiene 15.3 The resonance energy of benzene 15.4 Naphthalene with an STO3G basis 15.4.1 MCVB treatment 15.4.2 The MOCI treatment 15.4.3 Conclusions Interaction of molecular fragments 16.1 Methylene, ethylene, and cyclopropane 16.1.1 The methylene biradical\n\n148 149 152 154 157 160 161 162 162 164 166 168 171 173 174 174 175 177 177 177 186 187 189 191 192 192 197 198 200 203 205 208 208 211 211 212 213 214 214 215\n\nContents\n\nxi\n\n16.1.2 Ethylene 16.1.3 Cyclopropane with a 6-31G∗ basis 16.1.4 Cyclopropane with an STO-3G basis 16.2 Formaldehyde, H2 CO 16.2.1 The least motion path 16.2.2 The true saddle point 16.2.3 Wave functions during separation\n\n215 218 224 225 226 227 228\n\nReferences Index\n\n231 235\n\nPreface\n\nOne senses that it is out of style these days to write a book in the sciences all on one’s own. Most works coming out today are edited compilations of others’ articles collected into chapter-like organization. Perhaps one reason for this is the sheer size of the scientific literature, and the resulting feelings of incompetence engendered, although less honorable reasons are conceivable. Nevertheless, I have attempted this task and submit this book on various aspects of what is called ab initio valence bond theory. In it I hope to have made a presentation that is useful for bringing the beginner along as well as presenting material of interest to one who is already a specialist. I have taught quantum mechanics to many students in my career and have come to the conclusion that the beginner frequently confuses the intricacies of mathematical arguments with subtlety. In this book I have not attempted to shy away from intricate presentations, but have worked at removing, insofar as possible, the more subtle ones. One of the ways of doing this is to give good descriptions of simple problems that can show the motivations we have for proceeding as we do with more demanding problems. This is a book on one sort of model or trial wave function that can be used for molecular calculations of chemical or physical interest. It is in no way a book on the foundations of quantum mechanics – there are many that can be recommended. For the beginner one can still do little better than the books by Pauling and Wilson and Eyring, Walter, and Kimbal. A more recent work is by Levine, and for a more “physicsish” presentation the book by Messiah is recommended. These are a little weak on the practice of group theory for which Cotton may serve. A more fundamental work on group theory is by Hammermesh. Some further group theory developments, not to my knowledge in any other book, are in Chapter 5. Some of what we do with the theory of symmetric groups is based fairly heavily on a little book by Rutherford. This is a book on ab initio valence bond (VB) theory. There is a vast literature on “valence bond theory” – much of it devoted to semiempirical and qualitative xiii\n\nxiv\n\nPreface\n\ndiscussions of structure and reactivity of many chemical substances. It is not my purpose to touch upon any of this except occasionally. Rather, I will restrict myself principally to the results and interpretation of the ab initio version of the theory. It must be admitted that ab initio VB applications are limited to smaller systems, but we shall stick to this more limited goal. Within what practitioners call ab initio VB theory there are, in broad terms, two different approaches. r Calculations in which the orbitals used are restricted to being centered on only one atom of\n\nthe molecule. They are legitimately called “atomic orbitals”. Treatments of this sort may have many configurations involving different orbitals. This approach may be considered a direct descendent of the original Heitler–London work, which is discussed in Chapter 2. r Calculations in which the orbitals range over two or more atomic centers in the molecule. Although the resulting orbitals are not usually called “molecular orbitals” in this context, there might be some justification in doing so. Within this group of methods there are subcategories that will be addressed in the book. Treatments of this sort usually have relatively few configurations and may be considered descendents of the work of Coulson and Fisher, which is discussed in Chapter 3.\n\nEach of these two approaches has its enthusiasts and its critics. I have attempted an even-handed description of them. At various places in the text there are suggestions for further study to supplement a discussion or to address a question without a currently known answer. The CRUNCH program package developed by the author and his students is available on the Web for carrying out these studies.1 This program package was used for all of the examples in the book with the exception of those in Sections 2.2–2.6. I wish to thank Jeffrey Mills who read large parts of the manuscript and made many useful comments with regard to both style and clarity of presentation. Lastly, I wish to thank all of the students I have had. They did much to contribute to this subject. As time passes, there is nothing like a group of interested students to keep one on one’s toes. Lincoln, Nebraska November 2001\n\n1\n\nSee http://phy-ggallup.unl.edu/crunch\n\nGordon A. Gallup\n\nList of abbreviations\n\nAO CI CRUNCH DZP EGSO ESE GAMESS GGVB GUGA HLSP LCAO LMP MCVB MO MOCI RHF ROHF SEP SCF SCVB UHF VB\n\natomic orbital configuration interaction computational resource for understanding chemistry double-zeta plus polarization eigenvector guided sequential orthogonalization electronic Schr¨odinger equation general atomic and molecular electronic structure system Goddard’s generalized valence bond graphical unitary group approach Heitler–London–Slater–Pauling linear combination of atomic orbitals least motion path multiconfiguration valence bond molecular orbital molecular orbital configuration interaction spin-restricted Hartree–Fock spin-restricted open-shell Hartree–Fock static-exchange potential self-consistent-field spin coupled valence bond unrestricted Hartree–Fock valence bond\n\nxv\n\nPart I Theory and two-electron systems\n\n1 Introduction\n\n1.1 History In physics and chemistry making a direct calculation to determine the structure or properties of a system is frequently very difficult. Rather, one assumes at the outset an ideal or asymptotic form and then applies adjustments and corrections to make the calculation adhere to what is believed to be a more realistic picture of nature. The practice is no different in molecular structure calculation, but there has developed, in this field, two different “ideals” and two different approaches that proceed from them. The approach used first, historically, and the one this book is about, is called the valence bond (VB) method today. Heitler and London, in their treatment of the H2 molecule, used a trial wave function that was appropriate for two H atoms at long distances and proceeded to use it for all distances. The ideal here is called the “separated atom limit”. The results were qualitatively correct, but did not give a particularly accurate value for the dissociation energy of the H−H bond. After the initial work, others made adjustments and corrections that improved the accuracy. This is discussed fully in Chapter 2. A crucial characteristic of the VB method is that the orbitals of different atoms must be considered as nonorthogonal. The other approach, proposed slightly later by Hund and further developed by Mulliken is usually called the molecular orbital (MO) method. Basically, it views a molecule, particularly a diatomic molecule, in terms of its “united atom limit”. That is, H2 is a He atom (not a real one with neutrons in the nucleus) in which the two positive charges are moved from coinciding to the correct distance for the molecule.1 HF could be viewed as a Ne atom with one proton moved from the nucleus out to the molecular distance, etc. As in the VB case, further adjustments and corrections may be applied to improve accuracy. Although the united atom limit is not often mentioned in work today, its heritage exists in that MOs are universally 1\n\nAlthough this is impossible to do in practice, we can certainly calculate the process on paper.\n\n3\n\n4\n\n1 Introduction\n\nconsidered to be mutually orthogonal. We touch only occasionally upon MO theory in this book. As formulated by Heitler and London, the original VB method, which was easily extendible to other diatomic molecules, supposed that the atoms making up the molecule were in (high-spin) S states. Heitler and Rumer later extended the theory to polyatomic molecules, but the atomic S state restriction was still, with a few exceptions, imposed. It is in this latter work that the famous Rumer diagrams were introduced. Chemists continue to be intrigued with the possibility of correlating the Rumer diagrams with bonding structures, such as the familiar Kekul´e and Dewar bonding pictures for benzene. Slater and Pauling introduced the idea of using whole atomic configurations rather than S states, although, for carbon, the difference is rather subtle. This, in turn, led to the introduction of hybridization and the maximum overlap criterion for bond formation. Serber and Van Vleck and Sherman continued the analysis and introduced symmetric group arguments to aid in dealing with spin. About the same time the Japanese school involving Yamanouchi and Kotani published analyses of the problem using symmetric group methods. All of the foregoing work was of necessity fairly qualitative, and only the smallest of molecular systems could be handled. After WWII digital computers became available, and it was possible to test many of the qualitative ideas quantitatively. In 1949 Coulson and Fisher introduced the idea of nonlocalized orbitals to the VB world. Since that time, suggested schemes have proliferated, all with some connection to the original VB idea. As these ideas developed, the importance of the spin degeneracy problem emerged, and VB methods frequently were described and implemented in this context. We discuss this more fully later. As this is being written at the beginning of the twenty-first century, even small computers have developed to the point where ab initio VB calculations that required “supercomputers” earlier can be carried out in a few minutes or at most a few hours. The development of parallel “supercomputers”, made up of many inexpensive personal computer units is only one of the developments that may allow one to carry out ever more extensive ab initio VB calculations to look at and interpret molecular structure and reactivity from that unique viewpoint. 1.2 Mathematical background Data on individual atomic systems provided most of the clues physicists used for constructing quantum mechanics. The high spherical symmetry in these cases allows significant simplifications that were of considerable usefulness during times when procedural uncertainties were explored and debated. When the time came\n\n1.2 Mathematical background\n\n5\n\nto examine the implications of quantum mechanics for molecular structure, it was immediately clear that the lower symmetry, even in diatomic molecules, causes significantly greater difficulties than those for atoms, and nonlinear polyatomic molecules are considerably more difficult still. The mathematical reasons for this are well understood, but it is beyond the scope of this book to pursue these questions. The interested reader may investigate many of the standard works detailing the properties of Lie groups and their applications to physics. There are many useful analytic tools this theory provides for aiding in the solution of partial differential equations, which is the basic mathematical problem we have before us. 1.2.1 Schr¨odinger’s equation Schr¨odinger’s space equation, which is the starting point of most discussions of molecular structure, is the partial differential equation mentioned above that we must deal with. Again, it is beyond the scope of this book to give even a review of the foundations of quantum mechanics, therefore, we assume Schr¨odinger’s space equation as our starting point. Insofar as we ignore relativistic effects, it describes the energies and interactions that predominate in determining molecular structure. It describes in quantum mechanical terms the kinetic and potential energies of the particles, how they influence the wave function, and how that wave function, in turn, affects the energies. We take up the potential energy term first. Coulomb’s law Molecules consist of electrons and nuclei; the principal difference between a molecule and an atom is that the latter has only one particle of the nuclear sort. Classical potential theory, which in this case works for quantum mechanics, says that Coulomb’s law operates between charged particles. This asserts that the potential energy of a pair of spherical, charged objects is q1 q2 q1 q2 , (1.1) = V (|\u0003r1 − r\u00032 |) = |\u0003r1 − r\u00032 | r12 where q1 and q2 are the charges on the two particles, and r12 is the scalar distance between them. Units A short digression on units is perhaps appropriate here. We shall use either Gaussian units in this book or, much more frequently, Hartree’s atomic units. Gaussian units, as far as we are concerned, are identical with the old cgs system of units with the added proviso that charges are measured in unnamed electrostatic units, esu. The value of |e| is thus 4.803206808 × 10−10 esu. Keeping this number at hand is all that will be required to use Gaussian units in this book.\n\n6\n\n1 Introduction\n\nHartree’s atomic units are usually all we will need. These are obtained by assigning mass, length, and time units so that the mass of the electron, m e = 1, the electronic charge, |e| = 1, and Planck’s constant, h¯ = 1. An upshot of this is that the Bohr radius is also 1. If one needs to compare energies that are calculated in atomic units (hartrees) with measured quantities it is convenient to know that 1 hartree is 27.211396 eV, 6.27508 × 105 cal/mole, or 2.6254935 × 106 joule/mole. The reader should be cautioned that one of the most common pitfalls of using atomic units is to forget that the charge on the electron is −1. Since equations written in atomic units have no m e s, es, or h¯ s in them explicitly, their being all equal to 1, it is easy to lose track of the signs of terms involving the electronic charge. For the moment, however, we continue discussing the potential energy expression in Gaussian units. The full potential energy One of the remarkable features of Coulomb’s law when applied to nuclei and electrons is its additivity. The potential energy of an assemblage of particles is just the sum of all the pairwise interactions in the form given in Eq. (1.1). Thus, consider a system with K nuclei, α = 1, 2, . . . , K having atomic numbers Z α . We also consider the molecule to have N electrons. If the molecule is uncharged \u0001 as a whole, then Z α = N . We will use lower case Latin letters, i, j, k, . . . , to label electrons and lower case Greek letters, α, β, γ , . . . , to label nuclei. The full potential energy may then be written V =\n\n\u0002 e2 Z α Z β α 0. We form the matrix H \u0011 A+C H\u0011 = H − S, 2 \b a b = , b −a\n\n(1.37)\n\nwhere a = A−\n\nA+C 2\n\n(1.38)\n\nand A+C s. (1.39) 2 It is not difficult to show that the eigenvectors of H \u0011 are the same as those of H . Our generalized eigenvalue problem thus depends upon three parameters, a, b, and s. Denoting the eigenvalue by W and solving the quadratic equation, we obtain \u0014 a 2 (1 − s 2 ) + b2 sb ± . (1.40) W =− (1 − s 2 ) (1 − s 2 ) b=B−\n\n16\n\n1 Introduction\n\nWe note the possibility of an accident that cannot happen if s = 0 and b = 0: Should b = ±as, one of the two values of W is either ±a, and one of the two diagonal elements of H \u0011 is unchanged.5 Let us for definiteness assume that b = as and it is a we obtain. Then, clearly the vector C1 we obtain is \b 1 , 0 and there is no mixing between the states from the application of the variation theorem. The other eigenvector is simply determined because it must be orthogonal to C1 , and we obtain √ \b −s/√ 1 − s 2 , C2 = 1/ 1 − s 2 so the other state is mixed. It must normally be assumed that this accident is rare in practical calculations. Solving the generalized eigenvalue problem results in a nonorthogonal basis changing both directions and internal angles to become orthogonal. Thus one basis function could get “stuck” in the process. This should be contrasted with the case when S = I , in which basis functions are unchanged only if the matrix was originally already diagonal with respect to them. We do not discuss it, but there is an n × n version of this complication. If there is no degeneracy, one of the diagonal elements of the H-matrix may be unchanged in going to the eigenvalues, and the eigenvector associated with it is [0, . . . , 0, 1, 0, . . . , 0]† . 1.4 Weights of nonorthogonal functions The probability interpretation of the wave function in quantum mechanics obtained by forming the square of its magnitude leads naturally to a simple idea for the weights of constituent parts of the wave function when it is written as a linear combination of orthonormal functions. Thus, if \u0002 ψi Ci , (1.41) \u0007= i\n\nand \bψi |ψ j = δi j , normalization of \u0007 requires \u0002 |Ci |2 = 1.\n\n(1.42)\n\ni\n\nIf, also, each of the ψi has a certain physical interpretation or significance, then one says the wave function \u0007, or the state represented by it, consists of a fraction 5\n\nNB We assumed this not to happen in our discussion above of the convergence in the linear variation problem.\n\n1.4 Weights of nonorthogonal functions\n\n17\n\n|Ci |2 of the state represented by ψi . One also says that the weight, wi of ψi in \u0007 is wi = |Ci |2 . No such simple result is available for nonorthogonal bases, such as our VB functions, because, although they are normalized, they are not mutually orthogonal. Thus, instead of Eq. (1.42), we would have \u0002\n\nCi∗ C j Si j = 1,\n\n(1.43)\n\nij\n\nif the ψi were not orthonormal. In fact, at first glance orthogonalizing them would mix together characteristics that one might wish to consider separately in determining weights. In the author’s opinion, there has not yet been devised a completely satisfactory solution to this problem. In the following paragraphs we mention some suggestions that have been made and, in addition, present yet another way of attempting to resolve this problem. In Section 2.8 we discuss some simple functions used to represent the H2 molecule. We choose one involving six basis functions to illustrate the various methods. The overlap matrix for the basis is \n\n 1.000 000  0.962 004  1.000 000    0.137 187  0.181 541 1.000 000  , −0.254 383 −0.336 628  0.141 789 1.000 000    0.181 541  0.137 187 0.925 640 0.251 156 1.000 000 0.336 628 0.254 383 −0.251 156 −0.788 501 −0.141 789 1.000 000 and the eigenvector we analyze is \n\n 0.283 129  0.711 721     0.013 795     −0.038 111  .    −0.233 374  0.017 825\n\n(1.44)\n\nS is to be filled out, of course, so that it is symmetric. The particular chemical or physical significance of the basis functions need not concern us here. The methods below giving sets of weights fall into one of two classes: those that involve no orthogonalization and those that do. We take up the former group first.\n\n18\n\n1 Introduction\n\nTable 1.1. Weights for nonorthogonal basis functions by various methods. Chirgwin– Coulson\n\nInverseoverlap\n\nSymmetric orthogon.\n\nEGSOa\n\n0.266 999 0.691 753 –0.000 607 0.016 022 0.019 525 0.006 307\n\n0.106 151 0.670 769 0.000 741 0.008 327 0.212 190 0.001 822\n\n0.501 707 0.508 663 0.002 520 0.042 909 0.051 580 0.000 065\n\n0.004 998 0.944 675 0.000 007 0.002 316 0.047 994 0.000 010\n\na\n\nEGSO = eigenvector guided sequential orthogonalization.\n\n1.4.1 Weights without orthogonalization The method of Chirgwin and Coulson These workers suggest that one use wi = Ci∗\n\n\u0002\n\nSi j C j ,\n\n(1.45)\n\nj\n\nalthough, admittedly, they proposed it only in cases where the quantities were real. As written, this wi is not guaranteed even to be real, and when the Ci and Si j are real, it is not guaranteed to be positive. Nevertheless, in simple cases it can give some idea for weights. We show the results of applying this method to the eigenvector and overlap matrix in Table 1.1 above. We see that the relative weights of basis functions 2 and 1 are fairly large and the others are quite small. Inverse overlap weights Norbeck and the author suggested that in cases where there is overlap, the basis functions each can be considered to have a unique portion. The “length” of this may be shown to be equal to the reciprocal of the diagonal of the S −1 matrix corresponding to the basis function in question. Thus, if a basis function has a unique portion of very short length, a large coefficient for it means little. This suggests that a set of relative weights could be obtained from wi ∝ |Ci |2 /(S −1 )ii ,\n\n(1.46)\n\nwhere these wi do not generally sum to 1. As implemented, these weights are renormalized so that they do sum to 1 to provide convenient fractions or percentages. This is an awkward feature of this method and makes it behave nonlinearly in some contexts. Although these first two methods agree as to the most important basis function they transpose the next two in importance.\n\n1.4 Weights of nonorthogonal functions\n\n19\n\n1.4.2 Weights requiring orthogonalization We emphasize that here we are speaking of orthogonalizing the VB basis not the underlying atomic orbitals (AOs). This can be accomplished by a transformation of the overlap matrix to convert it to the identity N † S N = I.\n\n(1.47)\n\nInvestigation shows that N is far from unique. Indeed, if N satisfies Eq. (1.47), NU will also work, where U is any unitary matrix. A possible candidate for N is shown in Eq. (1.18). If we put restrictions on N , the result can be made unique. If N is forced to be upper triangular, one obtains the classical Schmidt orthogonalization of the basis. The transformation of Eq. (1.18), as it stands, is frequently called the canonical orthogonalization of the basis. Once the basis is orthogonalized the weights are easily determined in the normal sense as \u0005 \u00052 \u0005\u0002 \u0005 \u0005 \u0005 (N −1 )i j C j \u0005 , (1.48) wi = \u0005 \u0005 j \u0005 and, of course, they sum to 1 exactly without modification. Symmetric orthogonalization L¨owdin suggested that one find the orthonormal set of functions that most closely approximates the original nonorthogonal set in the least squares sense and use these to determine the weights of various basis functions. An analysis shows that the appropriate transformation in the notation of Eq. (1.18) is \u0003 \u0004−1/2 † T = S −1/2 = (S −1/2 )† , (1.49) N = T s (n) which is seen to be the inverse of one of the square roots of the overlap matrix and Hermitian (symmetric, if real). Because of this symmetry, using the N of Eq. (1.49) is frequently called a symmetric orthogonalization. This translates easily into the set of weights \u0005 \u00052 \u0005\u0002 \u0005 \u0005 \u0005 (S 1/2 )i j C j \u0005 , (1.50) wi = \u0005 \u0005 j \u0005 which sums to 1 without modification. These are also shown in Table 1.1. We now see weights that are considerably different from those in the first two columns. w1 and w2 are nearly equal, with w2 only slightly larger. This is a direct result of the relatively large value of S12 in the overlap matrix, but, indirectly, we note that the hypothesis behind the symmetric orthogonalization can be faulty. A least squares problem like that resulting in this orthogonalization method, in principle, always has an answer, but that gives no guarantee at all that the functions produced really are close to the original ones. That is really the basic difficulty. Only if the overlap\n\n20\n\n1 Introduction\n\nmatrix were, in some sense, close to the identity would this method be expected to yield useful results. An eigenvector guided sequential orthogonalization (EGSO) As promised, with this book we introduce another suggestion for determining weights in VB functions. Let us go back to one of the ideas behind inverse overlap weights and apply it differently. The existence of nonzero overlaps between different basis functions suggests that some “parts” of basis functions are duplicated in the sum making up the total wave function. At the same time, consider function 2 (the second entry in the eigenvector (1.44)). The eigenvector was determined using linear variation functions, and clearly, there is something about function 2 that the variation theorem likes, it has the largest (in magnitude) coefficient. Therefore, we take all of that function in our orthogonalization, and, using a procedure analogous to the Schmidt procedure, orthogonalize all of the remaining functions of the basis to it. This produces a new set of Cs, and we can carry out the process again with the largest remaining coefficient. We thus have a stepwise procedure to orthogonalize the basis. Except for the order of choice of functions, this is just a Schmidt orthogonalization, which normally, however, involves an arbitrary or preset ordering. Comparing these weights to the others in Table 1.1 we note that there is now one truly dominant weight and the others are quite small. Function 2 is really a considerable portion of the total function at 94.5%. Of the remaining, only function 5 at 4.8% has any size. It is interesting that the two methods using somewhat the same idea predict the same two functions to be dominant. If we apply this procedure to a different state, there will be a different ordering, in general, but this is expected. The orthogonalization in this procedure is not designed to generate a basis for general use, but is merely a device to separate characteristics of basis functions into noninteracting pieces that allows us to determine a set of weights. Different eigenvalues, i.e., different states, may well be quite different in this regard. We now outline the procedure in more detail. Deferring the question of ordering until later, let us assume we have found an upper triangular transformation matrix, Nk , that converts S as follows: \b 0 Ik † (Nk ) S Nk = , (1.51) 0 Sn−k where Ik is a k × k identity, and we have determined k of the orthogonalized weights. We show how to determine Nk+1 from Nk . Working only with the lower right (n − k) × (n − k) corner of the matrices, we observe that Sn−k in Eq. (1.51) is just the overlap matrix for the unreduced portion of the basis and is, in particular, Hermitian, positive definite, and with diagonal\n\n1.4 Weights of nonorthogonal functions\n\nelements equal to 1. We write it in partitioned form as \b 1 s Sn−k = † , s S\u0011\n\n21\n\n(1.52)\n\nwhere [1 s] is the first row of the matrix. Let Mn−k be an upper triangular matrix partitioned similarly, \b 1 q Mn−k = , (1.53) 0 B and we determine q and B so that †\n\n(Mn−k ) Sn−k Mn−k\n\n\b\n\n1 q + sB = , (q + s B)† B † (S \u0011 − s † s)B \b 1 0 , = 0 Sn−k−1\n\nwhere these equations may be satisfied with B the diagonal matrix \u0004 \u0004−1/2 \u0003 \u0004−1/2 \u0003\u0003 ··· 1 − s22 B = diag 1 − s12\n\n(1.54) (1.55)\n\n(1.56)\n\nand q = −s B. The inverse of Mn−k is easily determined: \b 1 −1 (Mn−k ) = 0 and, thus, Nk+1 = Nk Q k , where Qk =\n\n\b\n\nIk 0\n\n(1.57) s , B −1\n\n0 . Mn−k\n\n(1.58)\n\n(1.59)\n\nThe unreduced portion of the problem is now transformed as follows: (Cn−k )† Sn−k Cn−k = [(Mn−k )−1 Cn−k ]† (Mn−k )† Sn−k Mn−k [(Mn−k )−1 Cn−k ]. (1.60) Writing \b C1 , (1.61) Cn−k = C\u0011 we have\n\n\b\n\nC1 + sC \u0011 , [(Mn−k ) Cn−k ] = B −1 C \u0011 \b C1 + sC \u0011 . = Cn−k−1 −1\n\nPutting these together, we arrive at the total N as Q 1 Q 2 Q 3 · · · Q n−1 .\n\n(1.62) (1.63)\n\n22\n\n1 Introduction\n\nWhat we have done so far is, of course, no different from a standard top-down Schmidt orthogonalization. We wish, however, to guide the ordering with the eigenvector. This we accomplish by inserting before each Q k a binary permutation matrix Pk that puts in the top position the C1 + sC \u0011 from Eq. (1.63) that is largest in magnitude. Our actual transformation matrix is N = P1 Q 1 P2 Q 2 · · · Pn−1 Q n−1 .\n\n(1.64)\n\nThen the weights are simply as given (for basis functions in a different order) by Eq. (1.48). We observe that choosing C1 + sC \u0011 as the test quantity whose magnitude is maximized is the same as choosing the remaining basis function from the unreduced set that at each stage gives the greatest contribution to the total wave function. There are situations in which we would need to modify this procedure for the results to make sense. Where symmetry dictates that two or more basis functions should have equal contributions, the above algorithm could destroy this equality. In these cases some modification of the procedure is required, but we do not need this extension for the applications of the EGSO weights found in this book.\n\n2 H2 and localized orbitals\n\n2.1 The separation of spin and space variables One of the pedagogically unfortunate aspects of quantum mechanics is the complexity that arises in the interaction of electron spin with the Pauli exclusion principle as soon as there are more than two electrons. In general, since the ESE does not even contain any spin operators, the total spin operator must commute with it, and, thus, the total spin of a system of any size is conserved at this level of approximation. The corresponding solution to the ESE must reflect this. In addition, the total electronic wave function must also be antisymmetric in the interchange of any pair of space-spin coordinates, and the interaction of these two requirements has a subtle influence on the energies that has no counterpart in classical systems.\n\n2.1.1 The spin functions When there are only two electrons the analysis is much simplified. Even quite elementary textbooks discuss two-electron systems. The simplicity is a consequence of the general nature of what is called the spin-degeneracy problem, which we describe in Chapters 4 and 5. For now we write the total solution for the ESE \u0001(1, 2), where the labels 1 and 2 refer to the coordinates (space and spin) of the two electrons. Since the ESE has no reference at all to spin, \u0001(1, 2) may be factored into separate spatial and spin functions. For two electrons one has the familiar result that the spin functions are of either the singlet or triplet type, \u0001 \u0002\u0003√ 1 φ0 = η1/2 (1)η−1/2 (2) − η−1/2 (1)η1/2 (2) 2, (2.1) φ1 = η1/2 (1)η1/2 (2), \u0001 \u0002\u0003√ 3 φ0 = η1/2 (1)η−1/2 (2) + η−1/2 (1)η1/2 (2) 2,\n\n(2.3)\n\nφ−1 = η−1/2 (1)η−1/2 (2),\n\n(2.4)\n\n3\n\n3\n\n23\n\n(2.2)\n\n24\n\n2 H2 and localized orbitals\n\nwhere on the φ the anterior superscript indicates the multiplicity and the posterior subscript indicates the m s value. The η±1/2 are the individual electron spin functions. If we let Pi j represent an operator that interchanges all of the coordinates of the i th and j th particles in the function to which it is applied, we see that P12 1 φ0 = −1 φ0 ,\n\n(2.5)\n\nP12 3 φm s = 3 φm s ;\n\n(2.6)\n\nthus, the singlet spin function is antisymmetric and the triplet functions are symmetric with respect to interchange of the two sets of coordinates. 2.1.2 The spatial functions The Pauli exclusion principle requires that the total wave function for electrons (fermions) have the property P12 \u0001(1, 2) = \u0001(2, 1) = −\u0001(1, 2),\n\n(2.7)\n\nbut absence of spin in the ESE requires (2S+1)\n\n\u0001m s (1, 2) = (2S+1) ψ(1, 2) ×\n\n(2S+1)\n\nφm s (1, 2),\n\n(2.8)\n\nand it is not hard to see that the overall antisymmetry requires that the spatial function ψ have behavior opposite to that of φ in all cases. We emphasize that it is not an oversight to attach no m s label to ψ in Eq. (2.8). An important principle in quantum mechanics, known as the Wigner–Eckart theorem, requires the spatial part of the wave function to be independent of m s for a given S. Thus the singlet spatial function is symmetric and the triplet one antisymmetric. If we use the variation theorem to obtain an approximate solution to the ESE requiring symmetry as a subsidiary condition, we are dealing with the singlet state for two electrons. Alternatively, antisymmetry, as a subsidiary condition, yields the triplet state. We must now see how to obtain useful solutions to the ESE that satisfy these conditions. 2.2 The AO approximation The only uncharged molecule with two electrons is H2 , and we will consider this molecule for a while. The ESE allows us to do something that cannot be done in the laboratory. It assumes the nuclei are stationary, so for the moment we consider a very stretched out H2 molecule. If the atoms are distant enough we expect each one\n\n2.2 The AO approximation\n\n25\n\nto be a normal H atom, for which we know the exact ground state wave function.1 The singlet wave function for this arrangement might be written 1\n\nψ(1, 2) = N [1sa (1)1sb (2) + 1sb (1)1sa (2)],\n\n(2.9)\n\nwhere 1sa and 1sb are 1s orbital functions centered at nuclei a and b, respectively, and N is the normalization constant. This is just the spatial part of the wave function. We may now work with it alone, the only influence left from the spin is the “+” in Eq. (2.9) chosen because we are examining the singlet state. The function of Eq. (2.9) is that given originally by Heitler and London. Perhaps a small digression is in order on the use of the term “centered” in the last paragraph. When we write the ESE and its solutions, we use a single coordinate system, which, of course, has one origin. Then the position of each of the particles, r\u0007i for electrons and r\u0007α for nuclei, is given by a vector from this common origin. When determining the 1s state of H (with an infinitely massive proton), one obtains the result (in au) 1 (2.10) 1s(r ) = √ exp(−r ), π where r is the radial distance from the origin of this H atom problem, which is where the proton is. If nucleus α = a is located at ra then 1sa (1) is a shorthand for 1 (2.11) 1sa = 1s(|\u0007r1 − r\u0007a |) = √ exp(−|\u0007r1 − r\u0007a |), π and we say that 1sa (1) is “centered at nucleus a”. In actuality it will be useful later to generalize the function of Eq. (2.10) by changing its size. We do this by introducing a scale factor in the exponent and write \u0004 α3 exp(−αr ). (2.12) 1s (α, r ) = π When we work out integrals for VB functions, we will normally do them in terms of this version of the H-atom function. We may reclaim the real H-atom function any time by setting α = 1. Let us now investigate the normalization constant in Eq. (2.9). Direct substitution yields 1 = 1 ψ(1, 2)|1 ψ(1, 2) = |N |2 ( 1sa (1)|1sa (1) 1sb (2)|1sb (2) + 1sb (1)|1sb (1) 1sa (2)|1sa (2) + 1sa (1)|1sb (1) 1sa (2)|1sb (2) + 1sb (1)|1sa (1) 1sb (2)|1sa (2) ), 1\n\n(2.13)\n\n(2.14)\n\nThe actual distance required here is quite large. Herring has shown that there are subtle effects due to exchange that modify the wave functions at even quite large distances. In addition, we are ignoring dispersion forces.\n\n26\n\n2 H2 and localized orbitals\n\nwhere we have written out all of the terms. The 1s function in Eq. (2.10) is normalized, so\n\n1sa (1)|1sa (1) = 1sb (2)|1sb (2) , = 1sb (1)|1sb (1) , = 1sa (2)|1sa (2) , = 1.\n\n(2.15) (2.16) (2.17) (2.18)\n\nThe other four integrals are also equal to one another, and this is a function of the distance, R, between the two atoms called the overlap integral, S(R). The overlap integral is an elementary integral in the appropriate coordinate system, confocal ellipsoidal–hyperboloidal coordinates. In terms of the function of Eq. (2.12) it has the form S(w) = (1 + w + w 2 /3) exp(−w), w = α R,\n\n(2.19) (2.20)\n\nand we see that the normalization constant for 1 ψ(1, 2) is N = [2(1 + S 2 )]−1/2 .\n\n(2.21)\n\nWe may now substitute 1 ψ(1, 2) into the Rayleigh quotient and obtain an estimate of the total energy, E(R) = 1 ψ(1, 2)|H |1 ψ(1, 2) ≥ E 0 (R),\n\n(2.22)\n\nwhere E 0 is the true ground state electronic energy for H2 . This expression involves four new integrals that also can be evaluated in confocal ellipsoidal–hyperboloidal coordinates. In this case all are not so elementary, involving, as they do, expansions in Legendre functions. The final energy expression is (α = 1 in all of the integrals) E = 2h − 2\n\nj2 + k2 1 j1 (R) + S(R)k1 (R) + + , 2 2 1 + S(R) 1 + S(R) R\n\n(2.23)\n\nwhere h j1 k1 j2\n\n= α 2 /2 − α, = −[1 − (1 + w)e−2w ]/R, = −α(1 + w)e−w , = [1 − (1 + 11w/8 + 3w2 /4 + w3 /6)e−2w ]/R,\n\n(2.24) (2.25) (2.26) (2.27)\n\nk2 = {6[S(w)2 (C + ln w) − S(−w)2 E1 (4w) + 2S(w)S(−w)E1 (2w)] + (25w/8 − 23w 2/4 − 3w3 − w4 /3)e−2w }/(5R), l2 = [(5/16 + w/8 + w2 )e−w − (5/16 + w/8)e−3w ]/R.\n\n(2.28) (2.29)\n\n2.4 Extensions to the simple Heitler–London treatment\n\n27\n\nOf these only the exchange integral of Eq. (2.28) is really troublesome to evaluate. It is written in terms of the overlap integral S(w), the same function of −w, S(−w) = (1 − w + w2 /3)ew , the Euler constant, C = 0.577 215 664 901 532 86, and the exponential integral \u0005 ∞ dy E1 (x) = e−y , (2.30) y x which is discussed by Abramowitz and Stegun. In our discussion we have merely given the expressions for the five integrals that appear in the energy. Those interested in the problem of evaluation are referred to Slater. In practice, these expressions are neither very important nor useful. They are essentially restricted to the discussion of this simplest case of the H2 molecule and a few other diatomic systems. The use of AOs written as sums of Gaussian functions has become universal except for single-atom calculations. We, too, will use the Gaussian scheme for most of this book. The present discussion, included for historical reasons, is an exception. 2.3 Accuracy of the Heitler–London function We are now in a position to compare our results with experiment. A graph of E(R) given by Eq. (2.23) is shown as curve (e) in Fig. 2.1. As we see, it is qualitatively correct, showing the expected behavior of having a minimum with the energy rising to infinity at shorter distances and reaching a finite asymptote for large R values. Nevertheless, it misses 34% of the binding energy (comparing with curve (a) of Fig. 2.1), a significant fraction, and its minimum is clearly at too large a bond distance.2 2.4 Extensions to the simple Heitler–London treatment In the last section, our calculation used only the function of Eq. (2.9), what is now called the “covalent” bonding function. According to our discussion of linear variation functions, we should see an improvement in the energy if we perform a two-state calculation that also includes the ionic function, 1\n\nψ I (1, 2) = N [1sa (1)1sa (2) + 1sb (1)1sb (2)].\n\n(2.31)\n\nWhen this is done we obtain the curve labeled (d) in Fig. 2.1, which, we see, represents a small improvement in the energy. 2\n\nThe quantity we have calculated here is appropriately compared to De , the bond energy from the bottom of the curve. This differs from the experimental bond energy, D0 , by the amount of energy due to the zero point motion of the vibration. There is no vibration in our system, since the nuclei are infinitely massive. We use the theoretical result for comparison, since it is today considered more accurate than experimental numbers.\n\n28\n\n2 H2 and localized orbitals\n\nEnergy (hartree)\n\n−1.00\n\n−1.05\n\n−1.10\n\n(e) (d) (c)\n\n−1.15\n\n(b) (a) 0\n\n2\n\n4\n\n6\n\n8\n\n10\n\nH---H distance (bohr)\n\nFigure 2.1. Energies of H2 for various calculations using the H-atom 1s orbital functions; (b) covalent + ionic, scaled; (c) covalent only, scaled; (d) covalent + ionic, unscaled; (e) covalent only, unscaled. (a) labels the curve for the accurate function due to Kolos and Wolniewicz. This is included for comparison purposes.\n\nA much larger improvement is obtained if we follow the suggestions of Wang and Weinbaum and use the 1s (1) of Eq. (2.12), scaling the 1s H function at each internuclear distance to give the minimum energy according to the variation theorem. In Fig. 2.1 the covalent-only energy is labeled (c) and the two-state energy is labeled (b). There is now more difference between the one- and two-state energies and the better binding energy is all but 15% of the total.3 The scaling factor, α, shows a smooth rise from ≈1 at large distances to a value near 1.2 at the energy minimum. Rhetorically, we might ask, what is it about the ionic function that produces the energy lowering, and just how does it differ from the covalent function? First, we note that the normalization constants for the two functions are the same, and, indeed, they represent exactly the same charge density. Nevertheless, they differ in their two-electron properties. 1 ψC (adding a subscript C to indicate covalent) gives a higher probability that the electrons be far from one another, while 1 ψ I gives 3\n\nPerhaps we should note that in this relatively simple case, we will approach the binding energy from below as we improve the wave function. This is by no means guaranteed in more complicated systems where both the equilibrium and asymptotic wave functions will be approximate. One of the most important problems of bond energy calculations is to have a “balanced” treatment. This happens if the accuracies at the equilibrium and asymptotic geometries are sufficiently high to give an accurate difference. More realistically in larger systems, one must have any errors at the two geometries sufficiently close to one another to obtain a useful value of the difference. We will take up this question again later.\n\n2.4 Extensions to the simple Heitler–London treatment\n\n29\n\njust the opposite. This is only true, however, if the electrons are close to one or the other of the nuclei. If both of the electrons are near the mid-point of the bond, the two functions have nearly the same value. In fact, the overlap between these two functions is quite close to 1, indicating they are rather similar. At the equilibrium distance the basic orbital overlap from Eq. (2.19) is S = 1ss |1sb = 0.658 88.\n\n(2.32)\n\nA simple calculation leads to \u0007 \b \u0006 = 1 ψC \u00071 ψ I =\n\n2S = 0.918 86. 1 + S2\n\n(2.33)\n\n(We consider these relations further below.) The covalent function has been characterized by many workers as “overcorrelating” the two electrons in a bond. Presumably, mixing in a bit of the ionic function ameliorates the overage, but this does not really answer the questions at the beginning of this paragraph. We take up these questions more fully in the next section, where we discuss physical reasons for the stability of H2 . At the calculated energy minimum (optimum α) the total wave function is found to be \u0001 = 0.801 981 1 ψC + 0.211 702 1 ψ I .\n\n(2.34)\n\nThe relative values of the coefficients indicate that the variation theorem thinks better of the covalent function, but the other appears fairly high at first glance. If, however, we apply the EGSO process described in Section 1.4.2, we obtain 0.996 50 1 ψC + 0.083 54 1 ψ I , where, of course, the covalent function is unchanged, but 1 ψ I is the new ionic function orthogonal to 1 ψC . On the basis of this calculation we conclude that the the covalent character in the wave function is (0.996 50)2 = 0.993 (99.3%) of the total wave function, and the ionic character is only 0.7%. Further insight into this situation can be gained by examining Fig. 2.2, where a geometric representation of the basis vectors and the eigenfunction is given. The overlap integral is the inner product of the two vectors (basis functions) and is the cosine of the angle between them. Since arccos( ) = 23.24◦ , some care was taken with Fig. 2.2 so that the angle between the vectors representing the covalent and ionic basis functions is close to this value. One conclusion to be drawn is that these two vectors point, to a considerable extent, in the same direction. The two smaller segments labeled (a) and (b) show how the eigenfunction Eq. (2.34) is actually put together from its two components. Now it is seen that the relatively large coefficient of 1 ψ I is required because it is poor in “purely ionic” character, rather than because the eigenvector is in a considerably different direction from that of the covalent basis function.\n\n30\n\n2 H2 and localized orbitals \"Ionic\" orthonormal vector\n\nIonic\n\n(b) (a)\n\nEigenvector Covalent\n\n\"Covalent\" orthonormal vector\n\nFigure 2.2. A geometric representation of functions for H2 in terms of vectors for R = Req . The small vectors labeled (a) and (b) are, respectively, the covalent and ionic components of the eigenvector. The vectors with dashed lines are the symmetrically orthogonalized basis functions for this case.\n\nIt is important to realize that the above geometric representation of the H2 Hilbert space functions is more than formal. The overlap integral of two normalized functions is a real measure of their closeness, as may be seen from \u0006 1\n\n\u0007\n\n\b ψC − 1 ψ I \u0007 1 ψC − 1 ψ I = 2(1 − ),\n\n(2.35)\n\nand, if the two functions were exactly the same, would be 1. As pointed out above, is a dependent upon S, the orbital overlap. Figure 2.3 shows the relation between these two quantities for the possible values of S. In addition, in Fig. 2.2 we have plotted with dashed lines the symmetrically orthogonalized basis functions in this treatment. It is simple to verify that \u00061\n\n\u0007 \b ψC − S 1 ψ I \u00071 ψ I − S 1 ψC = 0,\n\n(2.36)\n\nwhere S is the orbital overlap. Therefore, the vectors given in Fig. 2.2 are just the normalized versions of those in Eq. (2.36). Since they must be at right angles, they must move out 33.38◦ from the vector they are supposed to approximate. Thus, the real basis functions are closer together than their orthogonalized approximations are to the functions they are to represent. Clearly, writing the eigenfunction in terms of the two symmetrically orthogonalized basis functions will require nearly equal coefficients, a situation giving a very overblown view of the importance of\n\n2.5 Why is the H2 molecule stable?\n\n31\n\n1.0\n\nCovalent−ionic overlap\n\n0.8\n\n〈 ψ C ψI〉\n\n0.6\n\n〈1sa1sb 〉 0.4\n\n0.2\n\n0.0 0.0\n\n0.2\n\n0.4 0.6 Orbital overlap\n\n0.8\n\n1.0\n\nFigure 2.3. The relation between the orbital and covalent–ionic function overlaps.\n\n√ ◦ the “ionic” component of the eigenfunction.4 Since cos 30 = 3/2, this√situation √ will be the same for any pair of vectors with S > 3/2. Only when S \u0012 3/2, is it likely that the symmetrical orthogonalization will, in the two-vector case, produce coefficients that are qualitatively useful. Another interesting conclusion can be drawn from Fig. 2.2. The linear combination of AOs-molecular orbital (LCAO-MO) function for H2 is\n\n\u0001 M O = 0.510 46 1 ψC + 1 ψ I , (2.37) and, although it has not been shown in the figure to reduce crowding, the vector corresponding to \u0001 M O would appear halfway between the two for 1 ψC and 1 ψ I . It is thus only a little farther from the optimum eigenfunction than 1 ψC alone. 2.5 Why is the H2 molecule stable? Our discussion so far has not touched upon the origin of the stability of the H2 molecule. Reading from the articles of the early workers, one obtains the impression that most of them attributed the stability to “resonance” between the 1sa (1)1sb (2) form of the wave function and the 1sa (2)1sb (1) form. This phenomenon was new to physicists and chemists at the time and was frequently invoked in “explaining” quantum effects. Today’s workers prefer explanations that use classical language, 4\n\nAlthough the function is more complicated, this is actually what happened in the third column of Table 1.1, where the first two entries are nearly equal compared to those in other columns.\n\n32\n\n2 H2 and localized orbitals\n\ninsofar as possible, and attempt to separate the two styles of description to see as clearly as possible just where the quantum effects are. 2.5.1 Electrostatic interactions Equation (2.23) is not well adapted to looking at the nature of the bonding. We rearrange it so that we see the terms that decrease the energy beyond that of two H atoms. This gives E(R) = 2E H +\n\nJ (R) + K (R) . 1 + T (R)\n\n(2.38)\n\nHere E H is the energy of a normal hydrogen atom, J (R), K (R), and T (R) were called by Heitler and London the Coulomb integral, the exchange integral, and the overlap integral, respectively. The reader should perhaps be cautioned that the terms “Coulomb”, “exchange”, and “overlap” integrals have been used by many other workers in ways that differ from that initiated by Heitler and London. In the present section we adhere to their original definitions, as follows: J (R) = 1sa (1)1sb (2)|V (1, 2)|1sa (1)1sb (2) , K (R) = 1sa (1)1sb (2)|V (1, 2)|1sb (1)1sa (2) , T (R) = 1sa (1)1sb (2)|1sb (1)1sa (2) ,\n\n(2.39) (2.40) (2.41)\n\n= 1sa (1)|1sb (1) 2 , and V (1, 2) = −1/r2a − 1/r1b + 1/r12 + 1/Rab .\n\n(2.42)\n\nThese equations are obtained by assigning electron 1 to proton a and 2 to b, so that the kinetic energy terms and the Coulomb attraction terms −1/r1a − 1/r2b give rise to the 2E H term in Eq. (2.38). V (1, 2) in Eq. (2.42) is then that part of the Hamiltonian that goes to zero for the atoms at long distances. It is seen to consist of two attraction terms and two repulsion terms. As observed by Heitler and London, the bonding in the H2 molecule arises from the way these terms balance in the J and K integrals. We show a graph of these integrals in Fig. 2.4. Before continuing, we discuss modifications of Eq. (2.38) when scaled 1s orbitals are used. With the 1s function of Eq. (2.12), we obtain E(α, R) = 2E H + (α − 1)2 + α\n\nJ (α R) + K (α R) , 1 + T (α R)\n\n(2.43)\n\nwhich reduces to the energy expression of Eq. (2.38) when α = 1. The changes brought by including the scale factor are only quantitative in nature and leave the qualitative conclusions unmodified.\n\n2.5 Why is the H2 molecule stable?\n\n33\n\n3 2 1\n\nEnergy (eV)\n\n0 −1\n\nJ(R)\n\n−2 −3 −4\n\nK(R)\n\n−5 1\n\n2\n\n3\n\n4\n\n5\n\nInternuclear distance (bohr)\n\nFigure 2.4. The relative sizes of the J (R) and K (R) integrals.\n\nIt is important to understand why the J (R) and K (R) integrals have the sizes they do. We consider J (R) first. As we have seen from Eq. (2.42), V (1, 2) is the sum of four different Coulombic terms from the Hamiltonian. If these are substituted into Eq. (2.39), we obtain J (R) = 2 j1 (R) + j2 (R) + 1/R, j1 (R) = 1sa | − 1/rb |1sa = 1sb | − 1/ra |1sb , j2 (R) = 1sa (1)1sb (2)|1/r12 |1sa (1)1sb (2) . The quantity j1 (R) is seen to be the energy of Coulombic attraction between a point charge and a spherical charge distribution, j2 (R) is the energy of Coulombic repulsion between two spherical charge distributions, and 1/R is the energy of repulsion between two point charges. J (R) is thus the difference between two attractive and two repulsive terms that cancel to a considerable extent. The magnitude of the charges is 1 in every case. This is shown in Fig. 2.5, where we see that the resulting difference is only a few percent of the magnitudes of the individual terms. This is to be contrasted with the situation for the exchange integral. In this case we have K (R) = 2k1 (R)S(R) + k2 (R) + S(R)2 /R, k1 (R) = 1sa | − 1/rb |1sb = 1sa | − 1/ra |1sb , k2 (R) = 1sa (1)1sb (2)|1/r12 |1sa (2)1sb (1) .\n\n34\n\n2 H2 and localized orbitals 1.6 1.4 1.2 −2j1(R)\n\nEnergy (hartree)\n\n1.0\n\nj2 (R)+1/R\n\n0.8 0.6 0.4 0.2 0.0\n\nJ(R) −0.2 1\n\n2\n\n3\n\n4\n\n5\n\nInternuclear distance (bohr)\n\nFigure 2.5. Comparison of the sizes of j2 + 1/R and −2 j1 that comprise the positive and negative terms in the Coulomb integral.\n\nThe magnitude of the charge in the overlap distribution, 1sa 1sb , is S(R), and here again the overall result is the difference between the energies of attractive and repulsive terms involving the same sized charges of different shaped distributions. The values are shown in Fig. 2.6, where we see that now there is a considerably greater difference between the attractive and repulsive terms. This leads to a value of about 20% of the magnitude of the individual terms. These values for J (R) and K (R) may be rationalized in purely electrostatic terms involving charge distributions of various sizes and shapes.5 From the point of view of electrostatics, J (R) is the interaction of points and spherical charge distributions. The well-known effect, where the interaction of a point and spherical charge at a distance R is due only to the portion of the charge inside a sphere of radius R, leads to an exponential fall-off J (R), as R increases. The situation is not so simple with K (R). The overlap charge distribution is shown in Fig. 2.7 and is far from spherical. The upshot of the differences is that the k2 (R) integral is the self-energy of the overlap distribution and is more dependent upon its charge than upon its size. In addition, at any distance there is in k1 (R) a portion of the distribution that surrounds the point charge, and, again, the distance dependence is decreased. The overall effect is thus that shown in Fig. 2.4. 5\n\nIt should not be thought that the result |J (R)| \u0012 |K (R)| is peculiar to the 1s orbital shape. It is fairly easy to show that a single spherical Gaussian orbital in the place of the 1s leads to a qualitatively similar result. In addition, two sp hybrid orbitals, oriented toward one another, show the effect, although compared with spherical orbitals, the disparity between J and K is reduced.\n\n2.5 Why is the H2 molecule stable?\n\n35\n\n1.4 1.2\n\nEnergy (hartree)\n\n1.0 0.8 −2S(R)k1(R)\n\n0.6 0.4\n\nk 2 (R)+S(R)2/R 0.2 0.0 K(R) −0.2 1\n\n2\n\n3\n\n4\n\n5\n\nInternuclear distance (bohr)\n\nFigure 2.6. Comparison of the sizes of k2 + S 2 /R and −2k1 S that comprise the positive and negative terms in the exchange integral.\n\nOverlap charge density 0.08 0.06 0.04 0.02 0.00 1.5 1.0 −1.5\n\n−1.0\n\n−0.5 z (bohr)\n\n0.0\n\n0.5\n\n1.0\n\n0.5 0.0 x (bohr) −0.5 −1.0 1.5 −1.5\n\nFigure 2.7. The overlap charge distribution when the H–H distance is near the molecular equilibrium value. We show an altitude plot of the value on the x–z plane.\n\nWe have not yet spoken of the effect of optimizing the scale factor in Eq. (2.43). Wang showed, for the singlet state, that it varies from 1 at R = ∞ to about 1.17 at the equilibrium separation. Since both J and K have relatively small slopes near the equilibrium distance, the principal effect is to increase the potential energy portion of the energy by about 17%. The (α − 1)2 term increases by only 3%. Thus the qualitative picture of the bond is not changed by this refinement.\n\n36\n\n2 H2 and localized orbitals\n\n2.5.2 Kinetic energy effects When we go beyond the simple covalent treatment and include an ionic function in the Heitler–London treatment, we obtain a further lowering of the calculated energy. At first glance, perhaps this is surprising, since the ionic function has more electron repulsion than the covalent. Although we saw in Section 1.3.2 that any additional linear variation term must lower the energy, that does not give any physical picture of the process. We will now give a detailed analysis of how the lowering comes about and its physical origin. In the previous section we examined the variational result of the two-term wave function consisting of the covalent and ionic functions. This produces a 2 × 2 Hamiltonian, which may be decomposed into kinetic energy, nuclear attraction, and electron repulsion terms. Each of these operators produces a 2 × 2 matrix. Along with the overlap matrix these are 1 SI C TI I TI C ; T = ; S= SC I 1 TC I TCC VI I VI C GI I GIC ; Ge = . Vn = VC I VCC G C I G CC As we discussed above, the two functions have the same charge density,6 and this implies that TI I = TCC and VI I = VCC , but we expect G I I > G CC . The normalization of the wave function requires 1 = C I2 + 2S I C C I CC + CC2 .\n\n(2.44)\n\nTwo similar expressions give us the expectation values of T amd Vn ,\n\nT = C I2 TI I + 2TI C C I CC + CC2 TCC ,\n\n(2.45)\n\nVn = C I2 VI I + 2VI C C I CC + CC2 VCC .\n\n(2.46)\n\nMultiplying Eq. (2.44) by TCC and VCC in turn and subtracting the result from the corresponding Eq. (2.45) or Eq. (2.46), we arrive at the equations\n\nT − TCC = 2(TI C − TCC S I C )C I CC ,\n\nVn − VCC = 2(VI C − VCC S I C )C I CC ,\n\n(2.47) (2.48)\n\nand we see that the differences depend on how the off-diagonal matrix elements compare to the overlap times the diagonal elements. A similar expression for G e is more complicated:\n\nG e − G CC = 2(G I C − G CC S I C )C I CC + (G I I − G CC )C I2 . 6\n\nThey have the same first order density matrices.\n\n(2.49)\n\n2.5 Why is the H2 molecule stable?\n\n37\n\nTable 2.1. Numerical values for overlap, kinetic energy, nuclear attraction, and electron repulsion matrix elements in the two-state calculation.\n\nII CI CC\n\nS\n\nT\n\nV\n\nG\n\n1.0 0.933 221 6 1.0\n\n1.146 814 1 0.954 081 4 1.146 814 1\n\n−3.584 134 6 −3.322 881 7 −3.584 134 6\n\n0.705 610 0 0.600 313 7 0.584 097 3\n\nThe numerical values of the matrix elements for R = R min are shown in Table 2.1 Putting in the numbers we see that TI C − TCC S I C = −0.116 15, and, therefore, the kinetic energy decreases as more of the ionic function is mixed in with the covalent. The nuclear attraction term changes in the opposite direction, but by only about one fifth as much, VI C − VCC S I C = 0.021 910. The magnitudes of the numbers in the G column are generally smaller than in the others and we have G I C − G CC S I C = 0.055 221 and G I I − G CC = 0.121 513. Since C I is not very large, the squared term in Eq. (2.49) is not very important. As C I grows from zero the decrease in the energy is dominated by the kinetic energy until the squared term in Eq. (2.49) can no longer be ignored. Therefore, the principal role of the inclusion of the ionic term in the wave function is the reduction of the kinetic energy from the value in the purely covalent wave function. Thus, this is the delocalization effect alluded to above. We saw in the last section that the bonding in H2 could be attributed principally to the much larger size of the exchange integral compared to the Coulomb integral. Since the electrical effects are contained in the covalent function, they may be considered a first order effect. The smaller added stabilization due to the delocalization when ionic terms are included is of higher order in VB wave functions. We have gone into some detail in discussing the Heitler–London treatment of H2 , because of our conviction that it is important to understand the details of the various contributions to the energy. Our conclusion is that the bonding in H2 is due primarily to the exchange effect caused by the combination of the Pauli exclusion principle and the required singlet state. The peculiar shape of the overlap distribution causes the exchange effect to dominate. Early texts (see, e.g., Ref. ) frequently emphasized the resonance between the direct and exchange terms, but this is ultimately due to the singlet state and Pauli principle. Those more familiar with the language of the molecular orbital (MO) picture of bonding may be surprised that the concept of delocalization energy does not arise here. That effect would occur in the VB treatment only if ionic terms were included. We thus conclude that delocalization is less important than the exchange attraction in bonding.\n\n38\n\n2 H2 and localized orbitals\n\n2.6 Electron correlation We have pushed this basis to its limit. In fact, it has a basic defect that does not allow a closer approach to the correct answer. The electrons repel each other, and the variation theorem tries to arrange that they not be too close together on the average. This type of effect is called electron correlation, i.e., if the electrons stay away from each other to some extent, their motion is said to be correlated. This calculation does produce some correlation, since we saw that the covalent function tends to keep the electrons apart. This is, however, only in a direction parallel to the bond. When the molecule forms there is also the possibility for angular correlation around the bond direction. Our simple basis makes no provision for this at all, and a significant fraction of the remaining discrepancy is due to this failing. In addition, Rosen added pz AOs to each center to produce polarization. These, in addition to pπ orbitals, will provide more correlation of the type important when the atoms are close as well as correlation of the type generally called van der Waal’s forces. We will correct some of these defects in the next section.\n\n2.7 Gaussian AO bases We now turn to considering calculations with the AOs represented by sums of Gaussians. This approach was pioneered by Boys, and is used almost universally today. We will settle on a particular basis and investigate its use for a number of different VB-like calculations. A double-ζ + polarization basis We define a ten-function AO basis for the H2 molecule that has two different s-type orbitals and one p-type set on each H atom. It will be recalled that Weinbaum allowed the scale factor of the 1s orbital to adjust at each internuclear distance. Using two “different sized” s-type orbitals on each center accomplishes a similar effect by allowing the variation theorem to “choose” the amount of each in the mixture. Our orbitals are shown in Table 2.2. The s-type orbitals are a split version of the Huzinaga 6-Gaussian H function and the p-type orbitals are adjusted to optimize the energy at the minimum. It will be observed that the pσ and pπ scale factors are different. We will present an interpretation of this below.\n\n2.8 A full MCVB calculation The author and his students have used the term multiconfiguration valence bond (MCVB) to describe a linear variation calculation involving more than one VB structure (function). This practice will be continued in the present book. Other terms have been used that mean essentially the same thing. We defer a fuller\n\n2.8 A full MCVB calculation\n\n39\n\nTable 2.2. A double-zeta plus polarization (DZP) basis for H2 calculations. 1s\n\n1s\n\nexp\n\nc\n\nexp\n\nc\n\nexp\n\nc\n\nexp\n\nc\n\n68.160 0 10.246 5 2.346 48 0.673 320 0.224 660\n\n0.00 255 0.01 938 0.09 280 0.29 430 0.49 221\n\n0.082 217\n\n0.24 260\n\n0.9025\n\n1.0\n\n0.5625\n\n1.0\n\n−0.96 −0.98 −1.00\n\nEnergy (hartree)\n\n−1.02\n\nIonic\n\n−1.04 −1.06 −1.08 −1.10 −1.12 −1.14 −1.16 0\n\n2\n\n4 6 H−−H distance (bohr)\n\n8\n\n10\n\nFigure 2.8. A global view of the adiabatic energies of the H2 molecule with the DZP basis.\n\ndiscussion of these terms until Chapter 7. When the MCVB calculation consists of all of the n-electron functions the basis can support, the treatment is said to be full.7 In the present case we have ten AO basis functions, and these provide a set of 55 symmetric (singlet) spatial functions. Only 27 of these, however, can enter into functions satisfying the spatial symmetry, 1 \u000fg+ , of the ground state of the H2 molecule. Indeed, there are only 14 independent linear combinations for this subspace from the total, and, working in this subspace, the linear variation matrices are only 14 × 14. We show the energy for this basis as the lowest energy curve in Fig. 2.8. We will discuss the other curves in this figure later. 7\n\nCompare with the term “full CI” used in the MO regime.\n\n40\n\n2 H2 and localized orbitals\n\nTable 2.3. Comparison of MCVB coefficients for orthogonalized AOs and “raw” AOs at the equilibrium internuclear distance. The ordering is determined by the orthogonalized AOs. No.\n\nType\n\nSymmetry function\n\nOrth. AOs\n\nRaw AOs\n\n1 2 3 4 5 6 7\n\nC I C C I C I\n\n0.743 681 90 0.144 201 30 −0.104 077 87 0.050 698 84 −0.036 309 53 −0.025 127 35\n\n0.762 703 95 0.064 562 30 −0.036 499 99 0.065 531 68 −0.069 271 56 −0.034 232 23\n\n8 9 10 11 12 13 14\n\nI I I I C C C\n\n(1sa 1sb ) (1sb 1sb ) + (1sa 1sa ) (1sa 1sb ) + (1sa 1sb ) (1sa 1sb ) (1sa 1sa ) + (1sb 1sb ) ( pza 1sb ) − (1sa pzb ) ( pxb pxb ) + ( p yb p yb ) + ( pxa pxa ) + ( p ya p ya ) ( pza pza ) + ( pzb pzb ) (1sb pzb ) − (1sa pza ) (1sa 1sa ) + (1sb 1sb ) (1sa pza ) − (1sb pzb ) (1sa pzb ) − ( pza 1sb ) ( pxa pxb ) + ( p ya p yb ) ( pza pzb )\n\n−0.024 702 92 −0.018 420 72 0.016 544 02 −0.015 623 66 −0.011 973 50 −0.011 494 85 −0.007 198 05 −0.006 660 28\n\n−0.024 702 92 −0.018 420 85 0.026 116 73 0.137 650 60 0.009 689 91 0.009 130 60 −0.007 198 05 −0.006 660 48\n\n2.8.1 Two different AO bases The Gaussian group functions given in Table 2.2 could be used in “raw” form for our calculation, or we could devise two linear combinations of the raw functions that are orthogonal. The most natural choice for the latter would be a linear combination that is the best H1s orbital and the function orthogonal to it. It should perhaps be emphasized that the energies are identical for the two calculations, except for minor numerical rounding differences. We show the MCVB coefficients for each of these in Table 2.3. The p-type orbitals are already orthogonal to the s-type and to each other, of course. It will be observed that we orthogonalize only on the same center, not between centers. This is, of course, the sine qua non of VB methods. Examination of the coefficients shows that, although the numbers are not greatly different, there are some significant equalities and differences between the two sets. Considering the equalities first, we note that this occurs for functions 7 and 13. These contribute to angular correlation around the internuclear axis and are completely orthogonal to all of the other functions. This is the reason that the coefficients are the same for the two bases. At any internuclear separation, the overlap of the raw s-type orbitals at the same center is\n\n1sa |1sa = 0.709 09,\n\n(2.50)\n\nwhich is fairly large. This produces the greatest difference between the two sets,\n\n2.8 A full MCVB calculation\n\n41\n\nTable 2.4. Comparison of EGSO weights for orthogonalized AOs and “raw” AOs at the equilibrium internuclear distance. The ordering is determined by the orthogonalized AOs. No.\n\nType\n\nSymmetry function\n\nOrth. AOs\n\nRaw AOs\n\n1 10 7\n\nC I I\n\n0.946 195 61 0.043 836 16\n\n0.980 762 89 0.002 080 27\n\nC I I C I\n\n0.004 199 01 0.002 147 20 0.002 071 71 0.000 662 56\n\n0.004 199 01\n\n3 11 2 12 5\n\n(1sa 1sb ) (1sa 1sa ) + (1sb 1sb ) ( pxb pxb ) + ( p yb p yb ) + ( pxa pxa ) + ( p ya p ya ) (1s a 1sb ) + (1sa 1sb ) (1sa pza ) − (1sb pzb ) (1sb 1sb ) + (1sa 1sa ) (1sa pzb ) − ( pza 1sb ) (1sa 1sa ) + (1sb 1sb )\n\na\n\n0.003 052 99 0.000 654 57\n\na\n\na\n\na\n\n0.009 689 91 a\n\nThese and the functions not listed contribute < 0.01%.\n\nfunctions 2 and 10. For the orthogonalized set, the 1sa and 1sb are good approximations to the H1s orbital and are important in the ionic function. With the raw AOs the most important ionic term becomes function 10, which mixes the two types. The fact that the function 1 coefficient is larger for the raw AOs than for the other basis should not be considered too important. There is a rather larger amount of overlap8 for the former basis. We now compare the EGSO weights of the important functions between the orthogonalized and raw bases. These are shown in Table 2.4. At first glance, in looking at the covalent function, one might be surprised at how much larger the weight is for the raw orbital. A little reflection will show, however, that this is to be expected and is related to the way these calculations accomplish the effects of orbital scaling. It will be recalled that the orbital scale factor in Eq. (2.12) was optimally ≈1.2 at the equilibrium internuclear separation in the simple calculation. The raw AO called 1s in Table 2.4 does not have the long range component contained in the optimized AO. Therefore, the raw AO is “tighter” and the closer of the two to the AO the molecule desires at Req . On the other hand the orthogonalized 1s AO is the function appropriate for long range. The “ionic” function (1sa 1sa ) + (1sb 1sb ), which contains both short and long range orthogonalized AOs, compensates for the too diffuse character of the orthogonalized 1s. We note that the sum of the first two weights is 0.990 031 77 for orthogonalized AOs and 0.982 843 16 for raw AOs. These are not too far from one another and indicate a similar representation of the total wave function. 8\n\nThis is measured by the determinant of 14 × 14 MCVB overlap matrix. The smaller this is, the larger are all of the coefficients of the VB functions.\n\n42\n\n2 H2 and localized orbitals\n\nTable 2.5. Comparison of MCVB coefficients for orthogonalized AOs and raw AOs at the internuclear distance of 20 bohr. No.\n\nType\n\nSymmetry function\n\nOrth. AOs\n\nRaw AOs\n\n1 2 3\n\nC C C\n\n(1sa 1sb ) (1sa 1sb ) + (1sa 1sb ) (1sa 1sb )\n\n1.000 000 00 0.000 000 00 0.000 000 00\n\n0.661 346 76 0.197 293 67 0.058 856 86\n\nTable 2.6. Comparison of EGSO weights for orthogonalized and raw AOs at an internuclear distance of 20 bohr. No.\n\nType\n\nSymmetry function\n\nOrth. AOs\n\nRaw AOs\n\n1 2 3\n\nC C C\n\n(1sa 1sb ) (1sa 1sb ) + (1sa 1sb ) (1sa 1sb )\n\n1.000 000 00 0.000 000 00 0.000 000 00\n\n0.942 329 44 0.056 814 21 0.000 856 35\n\nWhen we make these same comparisons for an internuclear separation of 20 bohr, we obtain the coefficients shown in Table 2.5 and the weights shown in Table 2.6. Now the orthogonalized AOs give the asymptotic function with one configuration, while it requires three for the raw AOs. The energies are the same, of course. The EGSO weights imply the same situation. A little reflection will show that the three terms in the raw VB function are just those required to reconstruct the proper H1s orbital. It should be clear that coefficients and weights in such calculations as these depend on the exact arrangement of the basis, and that their interpretations also depend upon how much physical or chemical significance can be associated with individual basis functions.\n\n2.8.2 Effect of eliminating various structures As we stated above, there are 14 different symmetry functions in the full MCVB with the present basis we are discussing. It will be instructive to see how the adiabatic energy curve changes as we eliminate these various functions in a fairly systematic way. This is the source of the higher-energy curves in Fig. 2.8. We showed all of them there in spite of the fact that they are not all easily distinguishable on that scale, because that gives a better global view of how they change. We “blow up” the region around the minimum and show this in Fig. 2.9 where the six lowest ones are labeled (a)–(f). In addition, the Kolos and Wolniewicz curve is shown for comparison and marked “K&W”.\n\n2.8 A full MCVB calculation\n\n43\n\n−1.10 −1.11\n\n(f) (e)\n\nEnergy (au)\n\n−1.12 −1.13\n\n(d)\n\n−1.14 (c) −1.15\n\n(b) (a)\n\n−1.16 −1.17 1.0\n\nK&W 1.2\n\n1.4 1.6 H−−H distance (au)\n\n1.8\n\n2.0\n\nFigure 2.9. A detailed view of the adiabatic energies of the H2 molecule with the DZP basis. (a) This is the full calculation with all of the functions of Table 2.3. (b) This has had functions 7 and 13 (see Table 2.3) removed. These give angular correlation around the internuclear axis. (c) This has had functions 4 and 5 eliminated (in addition to 7 and 13). This clearly does not have much effect. We can categorize different structures according to how many electrons are in orbitals that might be considered “excited”. In this case all of the AOs except the two H1s orbitals will be considered excited. Functions 7 and 13 are of a sort we will call “double excitations” and do not contribute much to the energy. (d) Now 6, 8, 9, 11, 12, and 14 have been eliminated. These are the configurations that contain the pσ AO. These are fairly important, it is seen. (e) When we now eliminate 3 and 10 and we are left with only the covalent and ionic configurations of the asymptotic H1s function. (f) This is only function 1, the covalent one. It is seen that the ionic term contributes fairly little at this level.\n\nThe energy increments from omitting selected basis functions are not additive, thus, using the amount that the energy is raised by elimination as a measure of the importance of various configurations is not a unique process, since the result depends upon the order of elimination. Nevertheless, the previous exercise was instructive. Although one would not expect a good answer, we show one more calculation with a limited basis – that is the H1s “ionic” function by itself. It does not fall in the region plotted in Fig. 2.9, but is shown in Fig. 2.8 where it is marked “Ionic”.\n\n44\n\n2 H2 and localized orbitals\n\nTable 2.7. A TZ2P1Da basis for H2 calculations. s, s , s\n\np, p\n\nd\n\nexpb\n\ncc\n\nexpb\n\ncc\n\nexpb\n\ncc\n\n68.160 0 10.246 5 2.346 48 0.673 320\n\n0.002 55 0.019 38 0.092 80 0.294 30\n\n5.027 243 1.190 621\n\n0.092 05 0.474 06\n\n1.140 046\n\n1.0\n\n0.450 098\n\n0.578 60\n\n0.224 660\n\n0.492 21\n\n0.082 217\n\n0.242 60\n\nTriple-zeta + 2 p-functions + 1 d-function. exp = exponential scale factor. c c = coefficient.\n\na\n\nb\n\n2.8.3 Accuracy of full MCVB calculation with 10 AOs The full MCVB calculation gives the best answer we have obtained so far. Compared to the Kolos and Wolniewicz result we now have 91.5% of the binding energy, but a the minimum is at 0.778 instead of 0.741 A , almost 5% too large. One must realize that the difficulty here is not with the VB method, but, rather, with the underlying AO basis. We are evaluating the energy for the full calculation, which would be the same whether we are using the VB method, orthogonal MOs followed by a full configuration interaction (CI), or some combination.9 2.8.4 Accuracy of full MCVB calculation with 28 AOs It is instructive to increase the size of the AO basis to see where we get to in calculating the binding energy of H2 . This is a so-called triple-ζ basis with a split p set and a d set on each center. It is shown in Table 2.7 and is based upon the same six-function Huzinaga orbital as is used in the previous Gaussian basis, Table 2.2. There are 406 singlet functions that can be made from this basis, but only 128 of them can enter into 1 \u000fg+ molecular states, and these give 58 linearly independent 1 \u000fg+ functions. The 1s, 2s, 3s, 1 p, and 2 p orbitals we use are the eigenfunctions of the H-atom Hamiltonian matrix in the s, s , s , p, and p group function basis. There is only one d-function, and it needs no modification. The results are considerably improved over the basis of Table 2.7. We now obtain a 98.6% of the binding energy and the minimum is at 0.7437 A , which is only 0.3% 9\n\nIt is perhaps not too difficult to see that a nonsingular linear transformation of the underlying AO basis produces a nonsingular linear transformation of the n-electron basis. Thus, the H and S matrices imply the same eigenvalues, although the coefficients in the sum giving the wave function will differ. Nevertheless, the actual wave function for a given eigenvalue (nondegenerate ones, at least) will be the same.\n\n2.8 A full MCVB calculation\n\n45\n\nTable 2.8. EGSO weights (>0.001) for two bases. 10 AOs 1 2 3 4 5 6 7\n\n(1sa 1sb ) (1sa 2sa ) + (1sb 2sb ) ( pxa pxa ) + ( p ya p ya ) + ( pxb pxb ) + ( p yb p yb ) (1sa 2sb ) + (2sa 1sb ) (1sa pza ) + (1sb pzb ) Total\n\n28 AOs 0.946 195 61(C) 0.043 836 16(I)\n\n0.947 336 55(C) 0.030 228 93(I) 0.012 119 46(C)\n\n0.004 199 01(I) 0.002 147 20(C)\n\n0.003 736 22(I) 0.002 316 93(C)\n\n(1sa 1sb ) (1sa 2sa ) + (1sb 2sb ) (1sa 3sb ) + (1sb 3sa ) (1 pxa 1 pxa ) + (1 p ya 1 p ya ) + (1 pxb 1 pxb ) + (1 p yb 1 p yb ) (1sa 1 pzb ) + (1sb 1 pza )\n\n0.002 071 71(I) 0.998 449 70\n\n0.995 738 09\n\nTotal\n\ntoo large. We could improve these results further, but for our purposes in discussing VB theory this is not particularly pertinent. Rather, we compare the EGSO weights of the two calculations to ascertain how much they change.\n\n2.8.5 EGSO weights for 10 and 28 AO orthogonalized bases In Table 2.8 we show a comparison of the EGSO weights for the two full MCVB calculations we have made with orthogonalized Gaussian bases. These are quite close to one another. We have only listed functions with weights > 0.001, and in each case there are five. We can interpret the various weights as follows. 1. Covalent The principal function in each case is the conventional Heitler–London covalent basis function with a weight very close to 95%. 2. Ionic The function, in each case, with the next highest weight, 3–4%, is ionic and involves a single excitation into the 2s AO. This contributes to adjusting the electron correlation and also contributes to adjusting the size of the wave function along the lines of the scale adjustment of the Weinbaum treatment. As we have shown, it also contributes to delocalization. 3. Covalent This function at 1.2% appears only with the larger basis set involving, as it does, the higher 3s-function. It will contribute to scaling. 4. Ionic At ≈0.4% the next function type appears in both sets and contributes to the angular correlation around the internuclear line. 5. Covalent At ≈0.2% the next function appears only with the smaller basis. It is the counterpart of the 3s covalent function with the larger basis, but is relatively less important. 6. Covalent At ≈0.2% this function contributes to polarization with the larger basis. 7. Ionic Again, at ≈0.2% this function contributes to polarization with the smaller basis.\n\n46\n\n2 H2 and localized orbitals\n\nThese functions contribute over 99% of the total wave function in both cases. The smaller value for the larger basis reflects the larger number of small contribution basis functions in that case. Although the fairly large number of basis functions that contribute with only minor weights have an important impact on lowering the energy, the large weight of the covalent function indicates that the bond in the H2 molecule is just as chemists always describe it: a covalent bond.\n\n3 H2 and delocalized orbitals\n\nWe now examine VB functions where the orbitals are allowed to take much more general forms, but only one configuration is used. This more general form allows the orbitals to range over more than one atomic center. As we shall see later, the restriction to one configuration is appropriate only to two-electron systems, so we must postpone the discussions of more configurations until we treat the more advanced methods in Chapter 5.\n\n3.1 Orthogonalized AOs Before we examine the more general case, let us look at an unusual result due to Slater. Earlier, in discussing solids Wannier had shown how linear combinations of the AOs could be made that rendered the functions orthogonal while retaining a relatively large concentration on one center. Slater adapted this idea to the H2 molecule. In modern language this is just making a symmetric orthogonalization (see Section 1.4.2) of the basis, which in this case is a H1s function on each of two centers, 1sa and 1sb . We are here again, following Slater, using the correct exponential functions of Eq. (2.10). The overlap matrix for this basis is \u0001 \u0002 1 S S¯ = , (3.1) S 1 and the inverse square root is \n\nS¯−1/2\n\n1 1 + √ √ 2 1+S 2 1−S =  1 1 − √ √ 2 1+S 2 1−S 47\n\n 1 − √ 2 1+S 2 1−S  ,  1 1 + √ √ 2 1+S 2 1−S √\n\n1\n\n(3.2)\n\n48\n\n3 H2 and delocalized orbitals\n\nwhere S = \u00051sa |1sb \u0007, and the signs are appropriate for S > 0. This orthogonalization gives us two new functions (see Eq. (1.48)) |A\u0007 = P|1sa \u0007 + Q|1sb \u0007, |B\u0007 = Q|1sa \u0007 + P|1sb \u0007,\n\n(3.3) (3.4)\n\nwhere 1 1 P= √ + √ , 2 1+S 2 1−S 1 1 − √ . Q= √ 2 1+S 2 1−S\n\n(3.5) (3.6)\n\nWe use these in a single Heitler–London covalent configuration, \u0005orth = A(1)B(2) + B(1)A(2), and calculate❛ the energy. When R → ∞ we obtain E = −1 au, just as we should. At R = 0.741 A, however, where we have seen that the energy should be a minimum, we obtain E = −0.6091 hartree, much higher than the correct value of –1.1744 hartree. The result for this orthogonalized basis, which represents not only no binding but actual repulsion, could hardly be worse. It is interesting to consider this function in terms of the covalent and ionic functions of Chapter 2. If the |A\u0007 and |B\u0007 functions in terms of the basic AOs are substituted into \u0005orth and the result normalized, one sees that √ 1 + S2 \u0005orth = (ψC − Sψ I ), 1 − S2 where, as always, S is the orbital overlap. Thus, this is exactly the symmetrically orthogonalized function closer to ψC discussed above, and its vector representation in Fig. 2.2 is clearly a considerable distance from the optimum eigenvector. Thus we should not be surprised at the poor value for the variational energy corresponding to \u0005orth . The early workers do not comment particularly on this result, but, in light of present understanding, we may say that the symmetric orthogonalization gives very close to the poorest possible linear combination for determining the lowest energy. This results from the added kinetic energy of the orbitals produced by a node that is not needed. Alternatively, one could say that the symmetric orthogonalization yields antibonding orbitals where bonding orbitals are needed. This is a good example of how the orthogonalization between different centers can have serious consequences for obtaining good energies and wave functions. We shall see shortly that there are, however, linear combinations determined in other ways that work quite well.\n\n3.2 Optimal delocalized orbitals\n\n49\n\n3.2 Optimal delocalized orbitals We now investigate orbitals that range over both centers with linear combinations that minimize the calculated energy. For this simple two-electron system these may all be viewed as extensions of the Coulson–Fisher approach we describe next. We use the basis of Table 2.2 and compare these results with the appropriate full MCVB calculations of Section 2.8.\n\n3.2.1 The method of Coulson and Fisher The first calculation of the energy of H2 for optimal delocalized orbitals used A = 1sa + λ1sb , B = λ1sa + 1sb ,\n\n(3.7) (3.8)\n\nand, using the “covalent” function, A(1)B(2) + B(1)A(2), in the Rayleigh quotient, adjusted the value of λ to minimize the energy. We will not duplicate this calculation here, but bring this up, because the methods we will discuss are generalizations of the Coulson–Fisher approach where we use in the orbitals all of the functions of our basis with the appropriate symmetry.\n\n3.2.2 Complementary orbitals It will be observed that the Coulson–Fisher functions satisfy the relations σh A = B\n\n(3.9)\n\nσh B = A,\n\n(3.10)\n\nand\n\nwhere σh is the operation of D∞h that reflects the molecule end for end. If A and B are also of σ symmetry,1 the “covalent” function A(1)B(2) + B(1)A(2) is of 1 g+ symmetry. Thus, the overall state symmetry is correct, although the orbitals do not belong to a single irreducible representation. For our first calculation we take all of the σ-type AOs of the basis and form (the unnormalized) A = 1sa + a1sb + b2sa + c2sb + dpza + epzb\n\n(3.11)\n\nB = σh A,\n\n(3.12)\n\nand\n\n1\n\nThe reader should note carefully the two different uses of the symbol σ here. One is a group operation, the other the state designation of an orbital.\n\n50\n\n3 H2 and delocalized orbitals\n\nTable 3.1. Energies of optimal orbital calculations. Distance ❛ A\n\nComplementary orbitals\n\nUnsymmetric orbitals\n\nFull MCVB\n\n0.741 ∞\n\n−1.143 356 −1.0\n\n−1.147 368 −1.0\n\n−1.148 052 −1.0\n\nOrbital value 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 3 2 −3\n\n1 −2\n\n−1 0 z-coordinate\n\n1\n\n2\n\n3\n\n−3\n\n−2\n\n−1\n\n0\n\nx-coordinate\n\nFigure 3.1. Altitude drawing of the A optimal complementary orbital for values in the x–z plane. The H nuclei are on the z-axis. The two vertical lines point at the nuclei.\n\nwhere a, . . . , e are the variation constants to be optimized.2 These orbitals are inserted in the covalent function, and the Rayleigh quotient minimized with respect to the variation parameters. We show the results for two internuclear distances in the second and fourth columns of Table 3.1 together with the calculation of the full MCVB using the same AO basis, i.e., omitting the pπ AOs. It will be recalled by examining Table 2.3 that there are 12 independent σ -AOonly VB functions in the MCVB. Our complementary orbital function has only five independent parameters, so it certainly cannot duplicate the MCVB energy, but it reproduces 96.8% of the binding energy of the latter calculation. We show a 3D altitude drawing of the amplitude of the A orbital in Fig. 3.1. It is easily seen to be extended over both nuclei, and it is this property that produces in the wave function the adjustment of the correlation and delocalization that is provided by the ionic function in the linear variation treatment with the same AO basis. We point out that these results are obtained without any “ionic” states in the wave function and such are not needed. As we argued in Chapter 2, the principal role of the ionic functions is to provide delocalization of the electrons when the molecule 2\n\nWe note that we cannot introduce the pπ AOs here and retain the 1 g+ state symmetry.\n\n3.2 Optimal delocalized orbitals\n\n51\n\nforms. Since the orbital is itself delocalized, the wave function requires nothing further. We also anticipate the discussion of Chapter 7 by pointing out that the wave function we have obtained here is a simple version of Goddard’s generalized valence bond (GGVB) or the spin coupled valence bond SCVB treatment of Gerratt, Cooper, and Raimondi. The GGVB in general has orthogonality prescriptions that do not, however, arise in the two electron case.\n\n3.2.3 Unsymmetric orbitals Instead of using Eq. (3.12) we might use a B defined as B = 1sb + a 1sa + b 2sb + c 2sa + d pzb + e pza .\n\n(3.13)\n\nOf course, if this is used in A(1)B (2) + B (1)A(2), the result does not have the correct symmetry, therefore we must use a projection operator to obtain the 1 g+ state. Defining A = σh B and B = σh A, we have 1 [I + σh ][A(1)B (2) + B (1)A(2)] 2 1 = [A(1)B (2) + B (1)A(2) + B(1)A (2) + A (1)B(2)], 2\n\n(3.14)\n\nand when this ten-parameter function is optimized with the Rayleigh quotient we obtain the results in the third column of Table 3.1 We now have 99.5% of the full binding energy, which is a credible showing. These orbitals are visibly different from the complementary optimal orbital as can be seen in the plots of A in Fig. 3.2\n\nOrbital amplitude 1\n\n0.5\n\n0 3 2 −3\n\n1 −2\n\n−1 0 z-coordinate\n\n1\n\n2\n\n3\n\n−3\n\n−2\n\n−1\n\n0\n\nx-coordinate\n\nFigure 3.2. Altitude drawing of the A optimal unsymmetric orbital for values in the x–z plane. The H nuclei are on the z-axis. The two vertical lines point at the nuclei.\n\n52\n\n3 H2 and delocalized orbitals\n\nOrbital amplitude 1.0\n\n0.5\n\n0.0 3 2 −3\n\n1 −2\n\n−1 0 z-coordinate\n\n1\n\n2\n\n3\n\n−3\n\n−2\n\n−1\n\n0\n\nx-coordinate\n\nFigure 3.3. Altitude drawing of the A optimal unsymmetric orbital for values in the x–z plane. The H nuclei are on the z-axis. The two vertical lines point at the nuclei.\n\nand A in Fig. 3.3. This A is more evenly distributed over the molecule, and A is less so, being somewhat outside of the region between the nuclei. The question might be asked: can angular correlation be included in an optimal orbital treatment? The answer is yes, but it is somewhat troublesome in general with infinite groups like D∞h . We merely need to generalize the trick we pulled to obtain the wave function of Eq. (3.14). The projection operator there guarantees the g (gerade) subscript on the state symmetry, 1 g+ . If we add pπ orbitals to our unsymmetric optimal orbitals we must also apply a projection operator to guarantee the part of the state symmetry. The appropriate operator in general is then 2π 1 Cφ dφ, (3.15) (I + σh ) 4π 0 where Cφ is a rotation about the z-axis of φ radians. This is not an operation convenient to deal with on a digital computer. We will not pursue these ideas further. As stated, such integral projection operators are troublesome to implement, and, in particular, they are clearly not very useful if there is no symmetry, which is true of most molecules.\n\n4 Three electrons in doublet states\n\nIn Chapter 5 we give an analysis of VB functions that is general for any number of electrons. In order to motivate some of the considerations we discuss there we first give a detailed example of the requirements when one is to construct an antisymmetric doublet eigenfunction of the spin for a three-electron system. Pauncz has written a useful workbook on this subject. We will first give a discussion of some results of general spin-operator algebra; not much is needed. This is followed by a derivation of the requirements spatial functions must satisfy. These are required even of the exact solution of the ESE. We then discuss how the orbital approximation influences the wave functions. A short qualitative discussion of the effects of dynamics upon the functions is also given. 4.1 Spin eigenfunctions The total spin operator and operator for the z-component are S\u00012 = S\u000121 + S\u000122 + S\u000123 , Sz = Sz1 + Sz2 + Sz3 ,\n\n(4.1) (4.2)\n\nwhere we see that both operators are symmetric1 sums of operators for the three identical electrons. Many treatments of spin discuss the raising and lowering operators for the z-component of the total spin. These are symmetric operators we symbolize as S + = Sx + i S y\n\n(4.3)\n\nS − = Sx − i S y\n\n(4.4)\n\nfor raising and\n\n1\n\nThe term symmetric is used in a variety of ways by mathematicians and in this book. The important point here is that the term implies that for n particles these spin operators commute with any permutation of n objects.\n\n53\n\n54\n\n4 Three electrons in doublet states\n\nfor lowering. With these, two alternative forms of S\u00012 are possible: S\u00012 = S − S + + Sz (Sz + 1), = S + S − + Sz (Sz − 1).\n\n(4.5) (4.6)\n\nThese are quite useful for constructing spin eigenfunctions and are easily seen to be true, not only for three electrons, but for n. In Chapter 2 we used η±1/2 to represent individual electron spin functions, but we would now like to use a more efficient notation. Thus we take [+ + +] to represent the product of three m s = +1/2 spin functions, one for electron 1, one for electron 2, and one for electron 3. As part of the significance of the symbol we stipulate that the + or – signs refer to electrons 1, 2, and 3 in that order. Thus, in the notation of Chapter 2, we have, for example, [+ + −] = η1/2 (1)η1/2 (2)η−1/2 (3).\n\n(4.7)\n\nFamiliar considerations show that there are all together eight different [± ± ±], they are all normalized and mutually orthogonal, and they form a complete basis for spin functions of three electrons. The significance of Eqs. (4.5) and (4.6) is that any spin function φ with the properties S+φ = 0\n\n(4.8)\n\nSz φ = M S φ,\n\n(4.9)\n\nand is automatically also an eigenfunction of the total spin with eigenvalue M S (M S + 1). Similarly, if φ satisfies S−φ = 0\n\n(4.10)\n\nSz φ = M S φ,\n\n(4.11)\n\nand\n\nit is automatically also an eigenfunction of the total spin with eigenvalue M S (M S − 1). We may use this to construct doublet eigenfunctions of the total spin for our three electrons. Thus, consider φ = a[− + +] + b[+ − +] + c[+ + −],\n\n(4.12)\n\nwhere, clearly, we have Sz φ = 1/2φ. Applying the operator S + to this gives S + φ = (a + b + c)[+ + +].\n\n(4.13)\n\nAccording to our requirements, this must be zero if φ is to be an eigenfunction of the total spin, therefore, we must have (a + b + c) = 0, since [+ + +] certainly is not\n\n4.2 Requirements of spatial functions\n\n55\n\nzero. We have one (homogeneous) equation in three unknowns so there is more than one solution – in fact, there are an infinite number of solutions. Nevertheless, all of them may be written as linear combinations of (in this case) just two. We observe that we can write three solutions of the form (a, b, c) = (1, −1, 0), (1, 0, −1), and (0, 1, −1), but that any one of these may be written as the difference of the other two. Thus, there are only two linearly independent solutions among our three, and any doublet spin function for three electrons may be written as a linear combination of these two. When dealing with spin functions it is normally convenient to arrange the bases to be orthonormal, and we obtain two functions, 1 2 φ1 = √ (2[+ + −] − [+ − +] − [− + +]) (4.14) 6 and 1 2 (4.15) φ2 = √ ([+ − +] − [− + +]). 2 For simplicity we do not label these functions with the M S value. Our work in VB theory and solving the ESE seldom needs any but the principal spin function with M S = S. The S − operator is always available should other M S values be needed. With the spin eigenfunctions of Eqs. (4.14) and (4.15) we have an example of the spin degeneracy alluded to in Chapter 2. Unlike the single singlet function we arrived at for two electrons in Section 2.1.1 we now obtain two.2 Writing out the equations specifically, \u00022 \u0001 S\u00012 2 φ1 = 1/2 1/2 + 1 φ1 , (4.16) \u0002 \u0001 2 (4.17) S\u00012 2 φ2 = 1/2 1/2 + 1 φ2 , we see that both of the functions have the same eigenvalue, and it is degenerate. In Chapter 5 we shall see that the degree of this degeneracy is related to the sizes of irreducible representations of the symmetric groups. We defer further discussion until that place. 4.2 Requirements of spatial functions We now have a significant difference from the case of two electrons in a singlet state, namely, we have two spin functions to combine with spatial functions for a solution to the ESE rather than only one. For a doublet three-electron system our general solution must be 2\n\n2\n\n\u0006 = 2 ψ1 2 φ1 + 2 ψ2 2 φ2 ,\n\nFour really, considering that each 2 φ1 and 2 φ2 has both m S = ±1/2 forms, also.\n\n(4.18)\n\n56\n\n4 Three electrons in doublet states\n\nand, apparently, we have two spatial functions to determine. We have not yet applied the antisymmetry requirement, however. With this it will develop that 2 ψ1 and 2 ψ2 are not really independent and only one need be determined. We must now investigate the effect of the binary interchange operators, Pi j on the 2 φi functions. We suppress the 2 spin-label superscript for these considerations. It is straightforward to determine P12 φ1 = φ1 , P12 φ2 = −φ2 , √ P13 φ1 = (2[− + +] − [+ − +] − [+ + −])/ 6 √ 1 3 = − φ1 − φ2 , 2 2 √ P13 φ2 = ([+ − +] − [+ + −])/ 2 √ 1 3 =− φ1 + φ2 , 2 2 √ P23 φ1 = (2[+ − +] − [+ + −] − [− + +])/ 6 √ 1 3 = − φ1 + φ2 , 2 2 √ P23 φ2 = ([+ + −] − [− + +])/ 2 √ 1 3 = φ1 + φ2 , 2 2\n\n(4.19) (4.20)\n\n(4.21)\n\n(4.22)\n\n(4.23)\n\n(4.24)\n\nand the results of applying higher permutations may be determined from these. We now apply the Pi j operators to \u0006 and require the results to be antisymmetric. Using the fact that the φi are linearly independent, for P12 we obtain P12 \u0006 = −\u0006 = (P12 ψ1 )φ1 − (P12 ψ2 )φ2 , P12 ψ1 = −ψ1 , P12 ψ2 = ψ2 ,\n\n(4.25) (4.26)\n\nand the others in a similar way give P13 ψ1 P13 ψ2 P23 ψ1 P23 ψ2\n\n√ 1 3 = ψ1 + ψ2 , 2 2 √ 1 3 = ψ1 − ψ2 , 2 2 √ 1 3 = ψ1 − ψ2 , 2 2 √ 3 1 ψ1 − ψ 2 , =− 2 2\n\n(4.27) (4.28) (4.29) (4.30)\n\n4.2 Requirements of spatial functions\n\n57\n\nand, finally,\n\n√ 1 3 ψ2 , (4.31) P12 P13 ψ1 = − ψ1 + 2 2 √ 1 3 ψ1 − ψ2 , (4.32) P12 P13 ψ2 = − 2 2 √ 1 3 ψ2 , P12 P23 ψ1 = − ψ1 − (4.33) 2 2 √ 1 3 (4.34) ψ1 − ψ2 . P12 P23 ψ2 = 2 2 With all of these relations it is not surprising that we can find several that express ψ2 in terms of sums of permuted ψ1 functions. An example is √ (4.35) ψ2 = (P13 − P23 )ψ1 / 3. This allows us to obtain some information about the normalization of the ψi functions, since 1 \u0007ψ2 |ψ2 = \u0007(P13 − P23 )ψ1 |(P13 − P23 )ψ1 , 3 1 = \u0007ψ1 |(I − P13 P23 − P23 P13 + I )ψ1 , 3 1 = \u0007ψ1 |(2I − P12 P13 − P12 P23 )ψ1 , 3 (4.36) = \u0007ψ1 |ψ1 , where we have used Eqs. (4.31) and (4.33).3 Thus, the spin eigenfunction– antisymmetry conditions require that ψ1 and ψ2 have the same normalization, whatever it is. Furthermore, the Pi j operators commute with the Hamiltonian of the ESE, and an argument similar to that leading to Eq. (4.36) yields \u0007ψ2 |H |ψ2 = \u0007ψ1 |H |ψ1 .\n\n(4.37)\n\nThese considerations may now be used to simplify the Rayleigh quotient for \u0006, and we see that \u0007ψ1 |H |ψ1 + \u0007ψ2 |H |ψ2 \u0007\u0006|H |\u0006 = , (4.38) \u0007\u0006|\u0006 \u0007ψ1 |ψ1 + \u0007ψ2 |ψ2 \u0007ψ1 |H |ψ1 , (4.39) = \u0007ψ1 |ψ1 and contrary to what appeared might be necessary above, we need to determine only one function to obtain the energy. We emphasize that Eq. (4.39) is true even 3\n\nWe remind the reader that all permutations are unitary operators. Since binary permutations are equal to their own inverses, they are also Hermitian. Products of commuting binaries are also Hermitian.\n\n58\n\n4 Three electrons in doublet states\n\nfor the exact solution to the ESE. The constraints that our spin eigenfunction– antisymmetry conditions impose on the wave function require that ψ1 and ψ2 be closely related, and, if a method is available for obtaining ψ1 , ψ2 may then be determined using Pi j operators. If we wish to apply the variation theorem to ψ1 , we still need the condition it must satisfy. Reflecting back upon the two-electron systems, we see that the requirement of symmetry for singlet functions could have been written 1 (I + P12 ) 1 ψ(12) = 1 ψ(12). 2\n\n(4.40)\n\nExamining our previous results we see that a corresponding relation for the threeelectron case may be constructed: 1 (2I − P12 P13 − P12 P23 ) 2 ψ(123) = 2 ψ(123). 3\n\n(4.41)\n\nThis has the correct form: it is Hermitian and idempotent, but that it is actually correct will be more easily ascertained after our general discussion of the next chapter. 4.3 Orbital approximation We now specialize our ψ-function, considering it to be a linear combination of products of only three independent orbitals. At the outset we use a, b, and c to represent three different functions that are to be used as orbitals. To keep the notation from becoming too cumbersome, we use an adaptation of the [· · ·] symbols above. Thus we let [abc] = a(1)b(2)c(3), [bca] = b(1)c(2)a(3),\n\n(4.42) (4.43)\n\netc. There are, of course, six such functions, since there are six permutations of three objects. Applying the doublet projector in Eq. (4.41) to each of the six product functions, we obtain the six linear combinations, w1 w2 w3 w4 w5\n\n= {2[abc] − [bca] − [cab]}/3, = {2[bca] − [cab] − [abc]}/3, = {2[cab] − [abc] − [bca]}/3, = {2[acb] − [cba] − [bac]}/3, = {2[cba] − [bac] − [acb]}/3,\n\n(4.44) (4.45) (4.46) (4.47) (4.48)\n\n4.3 Orbital approximation\n\n59\n\nand w6 = {2[bac] − [acb] − [cba]}/3.\n\n(4.49)\n\nIt is easy to see that these are not all linearly independent: in fact, w1 + w2 + w3 = 0\n\n(4.50)\n\nw4 + w5 + w6 = 0.\n\n(4.51)\n\nand\n\nThere are, therefore, apparently four functions based upon these orbitals to be used for doublet states. Again, there seems to be too many, but we now show how these are to be used. To proceed, we dispense with w3 and w6 , since they are not needed. We now construct functions that satisfy Eqs. (4.25) and (4.26). By direct calculation we find that P12 w1 = −w4 − w5 , P12 w2 = w5 , P12 w4 = −w1 − w2 ,\n\n(4.52) (4.53) (4.54)\n\nP12 w5 = w2 .\n\n(4.55)\n\nand\n\nThe ws constitute a basis for a matrix representation of P12 \u0003 \u0004 0 A P12 = , A 0 where\n\n\u0003 A=\n\n−1 0\n\n\u0004 −1 . 1\n\n(4.56)\n\n(4.57)\n\nA has eigenvalues ±1 and is diagonalized by the (nonunitary) similarity transformation: \u0003 \u0004 −1 0 −1 , (4.58) M AM = 0 1 \u0003 \u0004 1 1 M= . (4.59) 0 −2 We now subject P12 to a similarity transformation by N, \u0003 \u0004 M −M N= , M M\n\n(4.60)\n\n60\n\n4 Three electrons in doublet states\n\nand obtain \n\n−1  0 N−1 P12 N =   0 0\n\n0 1 0 0\n\n0 0 1 0\n\n 0 0 . 0 −1\n\n(4.61)\n\nWe may work out P23 in the same way and subject it to the same similarity transformation to obtain   1/2 3/2 0 0  1/2 −1/2 0 0 . N−1 P23 N =  (4.62)  0 0 −1/2 −3/2  0 0 −1/2 1/2 These do not yet quite satisfy the conditions on antisymmetry given in Eqs. (4.25), (4.26), (4.29), and (4.30), but further transformation by \n\n1 0 √0  0 −1/ 3 0 Q= 0 0 √0 0 0 1/ 3\n\n 0 0  1 0\n\n(4.63)\n\n 0 0 0 0  −1 0  0 1\n\n(4.64)\n\nyields \n\n−1  0 Q−1 N−1 P12 NQ =   0 0\n\n0 1 0 0\n\nand \n\n√ 0 1/2 − 3/2 √  −1/2 0  − 3/2 Q−1 N−1 P23 NQ =  0 0 1/2  √ 0 0 − 3/2\n\n 0  √ 0 , − 3/2  −1/2\n\n(4.65)\n\nwhich do agree. Since P13 = P12 P23 P12 the requirements for that matrix will also be satisfied. Putting together all of the transformations we eventually arrive at x = Q−1 N−1 w,\n\n(4.66)\n\n4.3 Orbital approximation\n\n61\n\nwhere w = [w1 w2 w4 w5 ]† with the components defined in Eqs. (4.44), (4.45), (4.47), and (4.48), respectively. Thus we have the four functions,4 x1 = {[abc] − [cab] + [acb] − [bac]}/2, √ x2 = {2[bca] + 2[cba] − [cab] − [acb] − [abc] − [bac]}/ 12, √ x3 = {2[bca] − 2[cba] − [cab] + [acb] − [abc] + [bac]}/ 12,\n\n(4.67) (4.68) (4.69)\n\nx4 = {−[abc] + [cab] + [acb] − [bac]}/2.\n\n(4.70)\n\nand\n\nTherefore, the four linearly independent functions we obtain in the orbital approximation can be arranged into two pairs of linear combinations, each pair of which satisfies the transformation conditions to give an antisymmetric doublet function. The most general total wave function then requires another linear combination of the pair of functions. In this case Eq. (4.18) can be written 2\n\n\u0006 = (x1 + αx3 ) 2 φ1 + (x2 + αx4 ) 2 φ2 ,\n\n(4.71)\n\nwhere α is a new variation parameter that is characteristic of the doublet case when we use orbital product functions. The same value, α, is required in both terms because of Eq. (4.35). In addition, Eq. (4.39) is still valid, of course, so that the energy is calculated from W =\n\n\u0007(x1 + αx3 )|H |(x1 + αx3 ) . \u0007(x1 + αx3 )|(x1 + αx3 )\n\n(4.72)\n\nThus, even without mixing in configurations of different orbitals, determining the energy of a doublet system of three electrons in three different orbitals is a sort of two-configuration calculation. The way this function represents the system is strongly influenced by the dynamics of the problem, as well as the flexibility allowed. If we were to find the set of three orbitals and value of α minimizing W , we obtain essentially the SCVB wave function. What this looks like depends significantly on the potential energy function. If we are treating the π system of the allyl radical, where all three orbitals are nearly degenerate, we obtain one sort of answer. If, on the other hand, we treat a deep narrow potential like the Li atom, we would obtain two orbitals close to one another and like the traditional 1s orbital of self-consistent-field (SCF) theory. The third would resemble the 2s orbital, of course. 4\n\nThese are displayed with an arbitrary overall normalization. This is unimportant in the Rayleigh quotient so long as the functions’ normalizations are correct relative to one another. The real normalization constant depends upon the overlaps, of course.\n\n62\n\n4 Three electrons in doublet states\n\nIf, as a further approximation, we force the two inner orbitals of the Li-atom treatment to be the same, the need for α disappears. In Eqs. (4.67)–(4.70) we put c = b, and the result is x1 = [abb] − [bab], √ x2 = {2[bba] − [bab] − [abb]}/ 3,\n\n(4.73) (4.74)\n\nx3 = x4 = 0,\n\n(4.75)\n\nand\n\nand we no longer have two pairs of functions with the correct properties. There is not an extra variation parameter to determine in this case. The most general wave function of this sort with optimized orbitals is the ordinary spin-restricted open-shell Hartree–Fock (ROHF) function.5 If we have three orbitals the same, c = b = a, we then see that all of the xi vanish identically. This is clearly the familiar answer: the Pauli exclusion principle prohibits three electrons having the same spatial part of their spin orbitals. In Chapter 10, after we have discussed the general n-electron problem, we will illustrate these two three-electron doublet systems with some calculations. We delay these examples because notational problems will be considerably simpler at that time. Above, we commented on the unfortunate increase in complexity in going from a two-electron singlet system to a three-electron doublet system. Unfortunately, the complexity accelerates as the number of electrons increases. 5\n\nThe ordinary unrestricted Hartree–Fock (UHF) function is not written like either of these. It is not a pure spin state (doublet) as are these functions. The spin coupled VB (SCVB) function is lower in energy than the UHF in the same basis.\n\n5 Advanced methods for larger molecules\n\nAs was seen in the last chapter, the effect of permutations on portions of the wave function is important in enforcing their correct character. The permutations of n entities form a group in the mathematical sense that is said to be one of the symmetric groups.1 In particular, when we have all of the permutations of n entities the group is symbolized Sn . In this chapter we give, using the theory of the symmetric groups, a generalization of the special treatment of three electrons discussed above. There are several more or less equivalent methods for dealing with the twin problems of constructing antisymmetric functions that are also eigenfunctions of the spin. Where orbitals are orthogonal the graphical unitary group approach (GUGA), based upon the symmetric group and unitary group representations, is popular today. With VB functions, which perforce have nonorthogonal orbitals, a significant problem centers around devising algorithms for calculating matrix elements of the Hamiltonian that are efficient enough to be useful. In the past symmetric group methods have been criticized as being overcomplicated. Nevertheless, the present author knows of no other techniques for obtaining what appears to be the optimal algorithm for these calculations. This chapter is the most complicated and formal in the book. Looking back to Chapter 4 we can obtain an idea of what is needed in general. In this chapter we: 1. outline the theory of the permutation (symmetric) groups and their algebras. The goal here is the special, “factored” form for the antisymmetrizer of Section 5.4.10, since, in this form the influence of the spin state on the spatial functions is especially transparent; 2. show how the resultant spatial functions allow an optimal algorithm for the evaluation of matrix elements of the Hamiltonian, which is given; 3. show the way to generate HLSP functions from the previous treatment. 1\n\nA word of caution here is in order. Groups describing spatial symmetry are frequently spoken of as symmetry groups. These should be distinguished from the symmetric groups.\n\n63\n\n64\n\n5 Advanced methods for larger molecules\n\nThese are the principal ideas of this chapter although the order is not exactly given by the list, and we start with an outline of permutation groups. 5.1 Permutations The word permutation has two meanings in common usage. The standard dictionary definition “an arrangement of a number of objects” is one of them, but we will reserve it to mean the act of permuting a set of objects. In this work the word “arrangement” will be used to refer to the particular ordering and “permutation” will always refer to the act of changing the arrangement. The set of “acts” that result in a particular rearrangement is not unique, but we do not need to worry about this. We just consider it the permutation producing the rearrangement. In Chapter 4 we used symbols like Pi j to indicate a binary permutation, but this notation is much too inefficient for general use. Another inefficient notation sometimes used is \u0002 \u0001 1 2 3 ··· n , ↓ i1 i2 i3 · · · in where i 1 , i 2 , . . . , i n is a different arrangement of the first n integers. We interpret this to mean that the object (currently) in position j is moved to the position i j (not necessarily different from j). The inverse permutation could be symbolized by reversing the direction of the arrow to ↑. There is too much redundancy in this symbol for convenience, and permutations are most frequently written in terms of their cycle structure. Every permutation can be written as a product, in the group sense, of cycles, which are represented by disjoint sets of integers. The symbol (12) represents the interchange of objects 1 and 2 in the set. This is independent of the number of objects. A cycle of three integers (134) is interpreted as instructions to take the object in position 1 to position 3, that in position 3 to position 4, and that in position 4 to position 1. It should be clear that (134), (341), and (413) all refer to actions with the same result. A permutation may have several cycles, (12)(346)(5789). It should be observed that there are no numbers common between any of the cycles. A unary cycle, e.g., (3), says that the object in position 3 is not moved. In writing permutations unary cycles are normally omitted. The group nature of the symmetric groups arises because the application of two permutations sequentially is another permutation, and the sequential application can be defined as the group multiplication operation. If we write the product of two permutations, (124) × (34) = (1243),\n\n(5.1)\n\n5.1 Permutations\n\n65\n\nwe say (34) is applied first and then (124). Working out results like that of Eq. (5.1) is fairly simple with a little practice. If we decide to write cycles with the smallest number in them first, we would start by searching the product of cycles from the right for the smallest number, which is 1. The rightmost reference to 1 says “1 → 2”, the rightmost reference to 2 says “2 → 4”, the rightmost reference to 4 says “4 → 3”, the rightmost reference to 3 says “3 → 4”, but 4 appears again in the left factor where “4 → 1”, closing the cycle. If two cycles have no numbers in common, their product is just the two of them written side by side. The order is immaterial; thus they commute. It may be shown also that the product defined this way is associative, (ab)c = a(bc). The inverse of a cycle is simply obtained by writing the numbers in reverse order. Thus (1243)−1 = (3421) = (1342), and (1243)(1342) = I, the identity, which corresponds in this case to no action, of course. We have here all the requirements of a finite group. 1. A set of quantities with an associative law of composition yielding another member of the set. 2. An identity appears in the set. The identity commutes with all elements of the set. 3. Corresponding to each member of the set there is an inverse. (The first two laws guarantee that an element commutes with its inverse.)\n\nA cycle can be written as a product of binary permutations in a number of ways. One of these is (i 1 i 2 i 3 · · · i n−1 i n ) = (i 1 i 2 )(i 2 i 3 ) · · · (i n−1 i n ).\n\n(5.2)\n\nThe ternary cycle that is the product of two binary permutations with one number in common can be written in three equivalent ways, (i 1 i 2 )(i 2 i 3 ) = (i 2 i 3 )(i 1 i 3 ) = (i 1 i 3 )(i 1 i 2 ). Clearly, these transformations could be applied to the result of Eq. (5.2) to arrive at a large number of products of different binaries. Nevertheless, each one contains the same number of binary interchanges. A cycle of n numbers is always the product of n − 1 interchanges, regardless of the way it is decomposed. In addition, these decompositions can vary in their efficiency. Thus, e.g., (12)(23)(14)(24)(14) = (23)(13)(12) = (13) all represent the same permutation, but they all have an odd number of interchanges in their representation. In general, a permutation is the product of m 2 binary cycles, m 3 ternary cycles, m 4 quaternary cycles, etc., all of which are noninteracting. If all of these are factored into (now interacting) binaries, the number is σ =\n\njmax \u0003 j=2\n\n( j − 1)m j ,\n\n(5.3)\n\n66\n\n5 Advanced methods for larger molecules\n\nwhich is called the signature of the permutation. A permutation is said to be even or odd according to whether σ is an even or odd number. The value of σ depends upon the efficiency of the decomposition, but its oddness or evenness does not. Therefore, the product of two even or two odd permutations is even, while the product of an even and an odd permutation is odd.\n\n5.2 Group algebras We need to generalize the idea of a group to that of group algebra. The reader has probably already used these ideas without the terminology. The antisymmetrizer we have used so much in earlier discussions is just such an entity for a symmetric group, Sn , 1 \u0003 (−1)σπ π, (5.4) A= n! π ∈Sn = A2 ,\n\n(5.5)\n\nwhere σπ is the signature of the permutation defined in Eq. (5.3). We note that Eq. (5.4) describes an entity in which we have multiplied group elements by scalars (±1) and added the results together. Equation (5.5) implies that we may multiply two such entities together, collect the terms by adding together the coefficients of like permutations, and write the result as an algebra element. Hence, A is idempotent. NB The assumption that we can identify the individual group elements to collect coefficients is mathematically equivalent to assuming the group elements themselves form a linearly independent set of algebra elements.2 The reader may feel that couching our argument in terms of group algebras is unnecessarily abstract, but, unfortunately, without this idea the arguments become excessively tedious. Thus, we define the operations of multiplying a group element by a scalar and adding two or more such entities. In this, we assume the elements of the group to be linearly independent, otherwise the mathematical structure we are dealing with would be unworkable. An element, x, of the algebra associated with Sn can be written \u0003 xρ ρ, (5.6) x= ρ∈Sn\n\nwhere xρ is, in general, a complex number. Two elements of the algebra may be 2\n\nIn most arguments involving spatial symmetry, the group character projections used are implicitly (if not explicitly) elements of the algebra of the corresponding symmetry group.\n\n5.2 Group algebras\n\n\u0003\n\n67\n\n(xρ ± yρ )ρ\n\n(5.7)\n\nρ∈Sn\n\nand xy =\n\n\u0003\u0003\n\nxρ yπ ρπ\n\nρ∈Sn π ∈Sn\n\n=\n\n\u0004 \u0003 \u0003\n\nη∈Sn\n\n\u0005 xρ yρ −1 η η,\n\n(5.8)\n\nρ∈Sn\n\nwhere π = ρ −1 η. The way the product is formed in Eq. (5.8) should be carefully noted. We also note that individual elements of the group necessarily possess inverses, but this is not true for the general algebra element. These considerations make the elements of a group embedded in the algebra behave like a basis for a vector space, and, indeed, this is a normed vector space. Let x be any element of the algebra, and let [[x]] stand for the coefficient of I in x. Also, for all of the groups we consider in quantum mechanics it is necessary that the group elements (not algebra elements) are assumed to be unitary. There will be more on this below in Section 5.4 This gives the relation ρ † = ρ −1 . Thus we have \u0003 |xρ |2 ≥ 0, (5.9) ||x||2 = [[x † x]] = ρ\n\nwhere the equality holds if and only if x = 0. One of the important properties of [[x y]] is [[x y]] = [[yx]]\n\n(5.10)\n\nfor any two elements of the algebra. We will frequently use the “[[· · ·]]” notation in later work. Since the group elements we are working with normally arise as operators on wave functions in quantum mechanical arguments, by extension, the algebra elements also behave this way. Because of the above, one of the important properties of their manipulation is φ|xψ = x † φ|ψ .\n\n(5.11)\n\nThe idea of a group algebra is very powerful and allowed Frobenius to show constructively the entire structure of irreducible matrix representations of finite groups. The theory is outlined by Littlewood, who gives references to Frobenius’s work.\n\n68\n\n5 Advanced methods for larger molecules\n\n5.3 Some general results for finite groups 5.3.1 Irreducible matrix representations Many works[5, 6] on group theory describe matrix representations of groups. That is, we have a set of matrices, one for each element of a group, G, that satisfies D(ρ)D(η) = D(ρη),\n\n(5.12)\n\nfor each pair of group elements.3 Any matrix representation may be subjected to a similarity transformation to obtain an equivalent representation: ¯ D(ρ) = N −1 D(ρ)N ;\n\nρ ∈ G,\n\n(5.13)\n\nwhere N is any nonsingular matrix. Amongst all of the representations, unitary ones are frequently singled out. This means that D(ρ)−1 = D(ρ −1 ) = D(ρ)† ,\n\n(5.14)\n\nand, for a finite group, such unitarity is always possible to arrange. For our work, however, we need to consider representations that are not unitary, so some of the results quoted below will appear slightly different from those seen in expositions where the unitary property is always assumed. The theory of group representation proves a number of results. 1. There is a set of inequivalent irreducible representations. The number of these is equal to the number of equivalence classes among the group elements. If the α th irreducible representation is an f α × f α matrix, then \u0003 f α2 = g, (5.15) α\n\nwhere g is the number of elements in the group. 2. The elements of the irreducible representation matrices satisfy a somewhat complicated law of composition: \u0003 g β D αji (ρ −1 )Dlk (ηρ) = δαβ δ jk Dliα (η). (5.16) fα ρ 3. If we specify in the previous item that η = I , the orthogonality theorem results: \u0003 g β D αji (ρ −1 )Dlk (ρ) = δαβ δ jk δil . (5.17) fα ρ\n\nEquation (5.17) has an important implication. Consider a large table with entries, and the columns labeled by the possible values of α, i, and j. Because of Eq. (5.15) the table is square, and may be considered Diαj (ρ −1 ), and the rows labeled by ρ\n\n3\n\nThe representation property does not imply, however, that ρ \u000e= η implies D(ρ) \u000e= D(η).\n\n5.3 Some general results for finite groups\n\n69\n\na g × g matrix. Because of Eq. (5.17) the matrix is necessarily nonsingular and possesses an inverse. The inverse matrix is then an array with g rows and g columns where the entries4 are ( f α /g)D αji (ρ), the rows are labeled by α, i, and j, and the columns by ρ. In the theory of matrices, it may be proved that matrix inverses commute, therefore we have another relation among the irreducible representation matrix elements: \u0003 fα D αji (η)Diαj (ρ −1 ) = δηρ , (5.18) g αi j where δηρ is 1 or 0, according as η and ρ are or are not the same. 5.3.2 Bases for group algebras The matrices of the irreducible representations provide one with a special set of group algebra elements. We define f α \u0003 α −1 D ji (ρ )ρ, (5.19) eiαj = g ρ and using Eq. (5.16) one can show that β\n\neiαj ekl = δαβ δ jk eilα .\n\n(5.20)\n\nEquation (5.19) gives the algebra basis as a sum over the group elements. Using Eq. (5.18) we may also write the group elements as a sum over the algebra basis, \u0003 ρ= Diαj (ρ)eiαj (5.21) αi j\n\nand, if ρ is taken as the identity,\n\n\u0003\n\neiiα = I.\n\n(5.22)\n\nαi\n\nIn the theory of operators over vector spaces Eq. (5.22) is said to give the resolution of the identity, since by Eq. (5.20) each (eiiα )2 = eiiα , and is idempotent. We note another important property of these bases. Irreducible representation matrices may be obtained from the eiαj by using the relation \u0006\u0006 \u0007\u0007 (5.23) Diαj (ρ) = ρeαji , where we have used the [[· · ·]] notation defined above to obtain the coefficient of the identity operation. 4\n\nNB In the inverse we have interchanged the index labels of the irreducible representation matrix.\n\n70\n\n5 Advanced methods for larger molecules\n\nThe theory of matrix representations of groups is more commonly discussed than the theory of group algebras. The latter are, however, important for our discussion of the symmetric groups, because Young (this theory is discussed by Rutherford) has shown, for these groups, how to generate the algebra first and obtain the matrix representations from them. In fact, we need not obtain the irreducible representation matrices at all for our work; the algebra elements are the operators we need to construct spatial VB basis functions appropriate for a given spin. 5.4 Algebras of symmetric groups 5.4.1 The unitarity of permutations Before we actually take up the subject of this section we must give a demonstration that permutations are unitary. This was deferred from above. The n-particle spatial (or spin) functions we work with are elements of a Hilbert space in which the permutations are operators. If (1, 2, . . . , n) and ϒ(1, 2, . . . , n) are two such functions we generally understand that \b (5.24) |ϒ ≡ (1, 2, . . . , n)∗ ϒ(1, 2, . . . , n) dτ1 dτ2 · · · dτn . If Pop and Q op are operators in the Hilbert space and Q op |ϒ = |Pop ϒ\n\n(5.25)\n\nfor all and ϒ in the Hilbert space, Q op is said to be the Hermitian conjugate of † Pop , i.e., Q op = Pop . Consider the integral \b\n\n(1, 2, . . . , n)∗ πϒ(1, 2, . . . , n) dτ1 dτ2 · · · dτn \b = [π −1 (1, 2, . . . , n)]∗ ϒ(1, 2, . . . , n) dτ1 dτ2 · · · dτn , (5.26) where π is some permutation. Equation (5.26) follows because of the possibility of relabeling variables of definite integrals, and, since it is true for all and ϒ, π † = π −1 .\n\n(5.27)\n\nThis is the definition of a unitary operator. 5.4.2 Partitions The theory of representations of symmetric groups is intimately connected with the idea of partitions of integers. Rutherford gives what is probably the most accessible treatment of these matters. A partition of an integer n is a set of smaller\n\n5.4 Algebras of symmetric groups\n\n71\n\nintegers, not necessarily different, that add to n. Thus 5 = 3 + 1 + 1 constitutes a partition of 5. Partitions are normally written in {· · ·}, and another way of writing the partition of 5 is {3,12 }. We use exponents to indicate multiple occurrences of numbers in the partition, and we will write them with the numbers in decreasing order. The distinct partitions of 5 are {5}, {4,1}, {3,2}, {3,12 }, {22 ,1}, {2,13 }, and {15 }. There are seven of them, and the theory of the symmetric groups says that this is also the number of inequivalent irreducible representations for the group S5 made up of all the permutations of five objects. We have written the above partitions of 5 in the standard order, such that partition i is before partition j if the first number in i differing from the corresponding one in j is larger than the one in j. When we wish to speak of a general partition, we will use the symbol, λ. 5.4.3 Young tableaux and N and P operators Associated with each partition there is a table, called by Young a tableau. In our example using 5, we might place the integers 1 through 5 in a number of rows corresponding to the integers in a partition, each row having the number of entries of that part of the partition, e.g., for {3,2} and {22 ,1} we would have  \n\n1 2 1 2 3 and 3 4 , 4 5 5 respectively. The integers might be placed in another order, but, for now, we assume they are in sequential order across the rows, finishing each row before starting the next. Associated with each tableau, we may construct two elements of the group algebra of the corresponding symmetric group. The first of these is called the row symmetrizer and is symbolized by P. Each row of the tableau consists of a distinct subset of the integers from 1 through n. If we add together all of the permutations involving just those integers in a row with the identity, we obtain the symmetrizer for that row. Thus for the {3,2} tableau, the symmetrizer for the first row is I + (12) + (13) + (23) + (123) + (132) and for the second is I + (45). Thus, the total row symmetrizer, P is P = [I + (12) + (13) + (23) + (123) + (132)][I + (45)].\n\n(5.28)\n\n72\n\n5 Advanced methods for larger molecules\n\nThe second of these is the column antisymmetrizer and is symbolized by N . As might be expected, for the {3,2} tableau the column antisymmetrizer is the product of the antisymmetrizer for each column and is N = [I − (14)][I − (25)].\n\n(5.29)\n\nIn these expressions a symmetrizer is the sum of all of the corresponding permutations and the antisymmetrizer is the sum with plus signs for even permutations and minus signs for odd permutations. In Eqs. (5.28) and (5.29) the specific P and N are given for the arrangement of numbers in the {3,2} tableau above. A different arrangement of integers in this same shape would in many, but not all, cases give different P and N operators. As a further example we give the P and N operators for the above tableau associated with the shape {22 ,1}. For this we have P = [I + (12)][I + (34)], N = [I − (13) − (15) − (35) + (135) + (153)][I − (24)]. Here, again, I is the only operation in common between P and N . A central result of Young’s theory is that the product N P is proportional to an algebra element that will serve as one of the eiiα basis elements discussed above, and the proportionality constant is f α /n!, n! being the value of g in this case. The product PN serves equally well, but is, of course, a different element of the algebra, since N and P do not normally commute. 5.4.4 Standard tableaux In a tableau corresponding to a partition of n, there are, of course, n! different arrangements of the way the first n integers may be entered. Among these there is a subset that Young called standard tableaux. These are those for which the numbers in any row increase to the right and downward in any column. Thus, we have for {3,2}\n\n1 2 3 1 2 4 1 2 5 1 3 4 1 3 5 , , , , and , 4 5 3 5 3 4 2 5 2 4 and among the 120 possible arrangements, only five are standard tableaux. These standard tableaux have been ordered in a particular way called a lexical sequence. We label the standard tableaux, T1 , T2 , . . . and imagine the numbers of the tableau written out in a line, row 1, row 2, . . . . We say that Ti is before T j if the first number of T j that differs from the corresponding one in Ti is the larger of the two. In our succeeding work we express the idea of Ti being earlier than T j with the symbols Ti < T j .\n\n5.4 Algebras of symmetric groups\n\n73\n\nThe operators P i and N i corresponding to Ti are, of course, different and, in fact, have no permutations in common other than the identity. The first important result is that [[Ni Pi ]] = 1,\n\n(5.30)\n\nsince the only permutation N i and P i have in common is I, and the numbers adding to the coefficient of the identity cannot cancel. Thus N i P i is never zero. The second important result here is that Pi N j = N j Pi = 0, if Ti < T j .\n\n(5.31)\n\nThis is so because there is some pair of numbers appearing in the same row of Ti that must appear in the same column of T j , if it is later. To see this suppose that the entries in the tableaux are (Ti )kl and (T j )kl , where k and l designate the row and column in the shape. Let the first difference occur at row m and column n. Thus, (T j )mn > (Ti )mn , but (Ti )mn must appear somewhere in T j . Because of the way standard tableaux are ordered it must be (T j )m \u0014 n \u0014 , where m \u0014 > m and n \u0014 < n. Now, also by hypothesis, (T j )mn \u0014 = (Ti )mn \u0014 , since this is in the region where the two are the same. Therefore, there is a pair of numbers in the same row of Ti that appear in the same column of T j . Calling these numbers p and q, we have Pi = (1/2)Pi [I + ( pq)], = (1/2)[I + ( pq)]Pi , N j = (1/2)N j [I − ( pq)], = (1/2)[I − ( pq)]N j ,\n\n(5.32) (5.33) (5.34) (5.35)\n\nand [I + ( pq)][I − ( pq)] = [I − ( pq)][I + ( pq)] = 0.\n\n(5.36)\n\nOne should not conclude, however, that P j Ni = Ni P j = 0 if Ti < T j . Although true in some cases, we see that it does not hold true for the first and last of the tableaux above. No pair in a row of the last is in a column of the first. In fact, the nonstandard tableau\n\n1 5 3 \u0014 T = 4 2 can be obtained from T5 by permutations within rows and from T1 by permutations within columns. Thus, P \u0014 = P 5 and N \u0014 = ±N 1 , and, therefore, P5 N1 = ±P \u0014 N \u0014 \u000e= 0. This would also be true for the operators written in the other order.\n\n(5.37)\n\n74\n\n5 Advanced methods for larger molecules\n\nWe stated above that there is an inequivalent irreducible representation of Sn associated with each partition of n, and we use the symbol f λ to represent the number of standard tableaux corresponding to the partition, λ. Using induction on n, Young proved the theorem \u0003 f λ2 = n!, (5.38) λ\n\nwhich should be compared with Eq. (5.15). Young also derived a formula for f λ , but, as will be seen, we need only a small number of partitions for our work with fermions like electrons. These are either {n − k,k} or {2k ,1n−2k } for all k = 0, 1, . . . , such that n − 2k ≥ 0. In fact, the shapes of the tableaux corresponding to these two partitions are closely related, being transposes of one another. Letting n = 5 and k = 2, the shape of {3,2} may be symbolized with dots as • • • • • If we interchange rows and columns in this shape, we obtain • • • • • which is seen to be the shape of the partition {22 ,1}. Partition shapes and tableaux related this way are said to be conjugates, and we use the symbol λ˜ to represent the partition conjugate to λ. It should be reasonably self-evident that the conjugate of a standard tableau is a standard tableau of the conjugate shape. Therefore, f λ = f λ˜ , and irreducible representations corresponding to conjugate partitions are the same size. In fact, the irreducible representations are closely related. If D λ (ρ) is one of the irreducible representation matrices for partition λ, one has D λ (ρ) = (−1)σρ D λ (ρ), ˜\n\n(5.39)\n\nwhere σρ is the signature of ρ. As we noted above, Young derived a general expression for f λ for any shape. For the partitions we need there is, however, some simplification of the general expression, and we have for either {n − k,k} or {2k ,1n−2k } \u0001 \u0002 n − 2k + 1 n + 1 fλ = , (5.40) k n+1 \u0001 \u0002 p! p . (5.41) = q q!( p − q)!\n\n5.4 Algebras of symmetric groups\n\n75\n\n5.4.5 The linear independence of N i P i and P i N i The relationships expressed in Eq. (5.31) can be used to prove the very important result that the set of algebra elements, N i P i , is linearly independent. First, from Eq. (5.30) we have seen that they are not zero, so we suppose there is a relation \u0003 ai Ni Pi = 0. (5.42) i\n\nWe multiply Eq. (5.42) on the right, starting with the final one Nf , and, because of Eq. (5.31), we obtain a f N f P f N f = 0.\n\n(5.43)\n\nTherefore, either a f or N f P f N f , or both must be 0. We observe, however, that \u0006\u0006 \u0007\u0007 (5.44) [[Ni Pi Ni ]] = Ni 2 Pi = gN [[Ni Pi ]] = gN ,\n\n(5.45) (5.46)\n\nwhere gN is the order of the subgroup of N , and this is true for any i. Thus, N f P f N f is not zero and a f in Eq. (5.43) must be. Now that we know a f is zero, we may multiply Eq. (5.42) on the right by N f −1 , and see that a f −1 must also be zero. Proceeding this way until we reach N 1 , we see that all of the ai are zero, and the result is proved. Permutations are unitary operators as seen in Eq. (5.27). This tells us how to take the Hermitian conjugate of an element of the group algebra, \u0004 \u0005† \u0003 † xπ π , (5.47) x = π\n\n=\n\n\u0003 π\n\n=\n\n\u0003 π\n\nxπ∗ π −1 ,\n\n(5.48)\n\nxπ∗ −1 π.\n\n(5.49)\n\nIn passing we note that N and P are Hermitian, since the coefficients are real and equal for inverse permutations. In general Pi Ni is not equal to Ni Pi but is its Hermitian conjugate, since (ρπ)† = † † π ρ . Therefore, it should be reasonably obvious that the Pi Ni operators are also linearly independent. We note that an alternative, but very similar, proof that all ai = 0 in Eq. (5.42) could be constructed by multiplying on the left by P j ; j = 1, 2, . . . , f sequentially. It is now fairly easy to see that we could form a new set of linearly independent quantities xi Ni Pi ;\n\ni = 1, 2, . . . , f,\n\n(5.50)\n\n76\n\n5 Advanced methods for larger molecules\n\nwhere xi is any set of elements of the algebra that do not result in xi Ni Pi = 0. Corresponding results are true for right multiplication (i.e., Ni Pi xi ). As is probably not surprising there are parallel results for right or left multiplication on Pi Ni . An important application of this result (for left multiplication) is an algebra element like xi = ρPi , where ρ is any operation of the group, with corresponding expressions for the other cases. The operators Pi and P j differ only in being based upon a different arrangement of the numbers in the standard tableau they are associated with. Therefore, there exists a permutation, πi j that will interconvert Pi and P j with the relation πi j P j = Pi πi j ,\n\n(5.51)\n\nwith a similar expression for N i and N j . The theorems of this section can thus be stated in a different way. For example, we see that the quantities, P1 N1 π1 j = π1 j P j N j , satisfy the definition of Eq. (5.50), and are thus linearly independent. Three similar results pertain for the other three possible combinations of the ordering of the products of P and N on either side of the equation. Explicitly, for one of these cases, we may write that the relation \u0003 P1 N1 π1i ai = 0 (5.52) i\n\nimplies that all ai = 0, with similar implications for the other cases.\n\n5.4.6 Von Neumann’s theorem Von Neumann proved a very useful theorem for our work (quoted by Rutherford). Using our notation it can be written P xN = [[P xN ]]PN ,\n\n(5.53)\n\nwhere x is any element of the algebra and N and P are based upon the same tableau. A similar expression holds for N xP.\n\n5.4.7 Two Hermitian idempotents of the group algebra We will choose arbitrarily to work with the first of the standard tableaux5 of a given partition. With this we can form the two Hermitian algebra elements u = θPN P 5\n\nAny tableau would do, but we only need one. This choice serves.\n\n(5.54)\n\n5.4 Algebras of symmetric groups\n\n77\n\nand u\u0014 = θ \u0014 N PN ,\n\n(5.55)\n\nwhere θ and θ \u0014 are real. We must work out what to set these values to so that u 2 = u and u \u00142 = u \u0014 . We stated at the end of Section 5.4.3 that ( f /g)N P or ( f /g)PN , g = n!, will serve as an idempotent element of the algebra associated with the partition upon which they are based, although these are, of course, not Hermitian. This means that N PN P =\n\ng N P. f\n\n(5.56)\n\nThus, observing that P 2 = gP P, we have (PN P)2 = PN P 2 N P, = gP PN PN P, ggP PN P, = f\n\n(5.57) (5.58) (5.59)\n\nwhere gP is the order of the subgroup of the P operator. Thus, we obtain u=\n\nf PN P ggP\n\n(5.60)\n\nas an idempotent of the algebra that is Hermitian. A very similar analysis gives u\u0014 =\n\nf N PN , ggN\n\n(5.61)\n\nwhere gN is the order of the subgroup of the N operator. Although portions of the following analysis could be done with the original non-Hermitian Young idempotents, the operators of Eqs. (5.60) and (5.61) are required near the end of the theory and, indeed, simplify many of the intervening steps. 5.4.8 A matrix basis for group algebras of symmetric groups In the present section we will give a construction of the matrix basis only for the u = θPN P operator. The treatment for the other Hermitian operator above is identical and may be supplied by the reader. Consider now the quantities, m i j = πi1 uπ1 j ,\n\n(5.62)\n\n= (m ji )† , =\n\nπi−1 uπ j ;\n\n(5.63) π j = π1 j ,\n\n(5.64)\n\n78\n\n5 Advanced methods for larger molecules\n\nnone of which is zero. Since u is based upon the first standard tableau, from now on we suppress the “1” subscript in these equations. This requires us, however, to use the inverse symbol, as seen. Now familiar methods are easily used to show that m i j \u000e= 0 for all i and j. In fact, the above results show that the m i j constitute f 2 linearly independent elements of the group algebra that, because of Young’s results, completely span the space associated with the irreducible representation labeled with the partition. Thus, because of Eq. (5.38) we have found a complete set of linearly independent elements of the whole group algebra. We now determine the multiplication rule for m i j and m kl , m i j m kl = πi−1 uπ j πk−1 uπl .\n\n(5.65)\n\nExamining the inner factors of this product, we see that uπ j πk−1 u = θ 2 PN Pπ j πk−1 PN P, θ=\n\nf . gg P\n\nWe now apply Eq. (5.53) to some inner factors and obtain \u0006\u0006 \u0007\u0007 Pπ j πk−1 PN = Pπ j πk−1 PN PN , \u0007\u0007 \u0006\u0006 = π j πk−1 PN P PN , \u0007\u0007 \u0006\u0006 uπ j πk−1 u = θ 2 π j πk−1 PN P PN PN P, \u0006\u0006 \u0007\u0007 = θ 2 g −1 π j πk−1 PN P PN PPN P, P = Mk j u, \u0006\u0006 −1 \u0007\u0007 πk PN Pπ j . Mk j = g −1 P\n\n(5.66) (5.67)\n\n(5.68) (5.69) (5.70) (5.71) (5.72) (5.73)\n\nPutting these transformations together, m i j m kl = Mk j m il .\n\n(5.74)\n\nAll of the coefficients in PN P are real and the matrix M is thus real symmetric (and Hermitian). Since the m i j are linearly independent, M must be nonsingular. In addition, g −1 P [[PN P]] is equal to 1, so the diagonal elements of M are all 1. M is essentially an overlap matrix due to the non-orthogonality of the m i j . We note that if the matrix M were the identity, the m i j would satisfy Eq. (5.20). An orthogonalization transformation of M may easily be effected by the nonsingular matrix N N † M N = I,\n\n(5.75)\n\n5.4 Algebras of symmetric groups\n\n79\n\nas we saw in Section 1.4.2. It should be recalled that N is not unique; added conditions are required to make it so. We wish m 11 to be unchanged by the transformation, and an upper triangular N will accomplish both of these goals. If we require all of the diagonal elements of N also to be positive, it becomes uniquely determined. Making the transformations we have \u0003 (N † )ik Nl j m kl , (5.76) ei j = kl\n\nei j ekl = δ jk eil , e11 = m 11 ,\n\n(5.77) (5.78)\n\nas desired. These ei j s constitute a real matrix basis for the symmetric group and, clearly, generate a real unitary representation through the use of Eq. (5.23). 5.4.9 Sandwich representations The reader might ask: “Is there a parallel to Eq. (5.23) for the nonorthogonal matrix basis we have just described?” We answer this in the affirmative and show the results. Clearly, we can define matrices T (ρ)i j = [[ρm i j ]], and it is seen that a normal unitary representation may be obtained from \u0003 (N † )ik T (ρ)kl Nl j , D(ρ)i j =\n\n(5.79)\n\n(5.80)\n\nkl\n\nwhere we have used Eq. (5.76). The upshot of these considerations is that the T (ρ) matrices satisfy T (ρ)M −1 T (π ) = T (ρπ ),\n\n(5.81)\n\nand these have been called sandwich representations, because of a fairly obvious analogy. In arriving at Eq. (5.81) we have used N N † = M −1 , which is a consequence of Eq. (5.75). We may also derive a result analogous to Eq. (5.21), \u0003 (M −1 )λki T (ρ)λkl (M −1 )lλj m iλj , ρ=\n\n(5.82)\n\n(5.83)\n\nλi jkl\n\nwhere we have added a partition label to each of our matrices and summed over it.\n\n80\n\n5 Advanced methods for larger molecules\n\n5.4.10 Group algebraic representation of the antisymmetrizer As we have seen in Eq. (5.21), an element of the group may be written as a sum over the algebra basis. For the symmetric groups, this takes the form, \u0003 Diλj (ρ)eiλj . (5.84) ρ= λi j\n\nWe wish to apply permutations, and the antisymmetrizer to products of spin-orbitals that provide a basis for a variational calculation. If each of these represents a pure spin state, the function may be factored into a spatial and a spin part. Therefore, the whole product, \u0014, may be written as a product of a separate spatial function and a spin function. Each of these is, of course, a product of spatial or spin functions of the individual particles, \u0014 = \u0015 Ms ,\n\n(5.85)\n\nwhere is a product of orbitals and \u0015 Ms is a sum of products of spin functions that is an eigenfunction of the total spin. It should be emphasized that the spin function has a definite Ms value, as indicated. If we apply a permutation to \u0014, we are really applying the permutation separately to the space and spin parts, and we write ρ\u0014 = ρr ρs \u0015 Ms ,\n\n(5.86)\n\nwhere the r or s subscripts indicate permutations affecting spatial or spin functions, respectively. Since we are defining permutations that affect only one type of function, separate algebra elements also arise: eiλj,r and eiλj,s . These considerations provide us with a special representation of the antisymmetrizer6 that is useful for our purposes: 1\u0003 (−1)σρ ρr ρs (5.87) A= g ρ∈Sn \u0003\u0003 \u0003 \u0014 ˜\u0014 = Diλj (ρ)eiλj,r Diλ\u0014 j \u0014 (ρ)eiλ\u0014 j \u0014 ,s (5.88) ρ\n\n=\n\nλi j λ\u0014 i \u0014 j \u0014\n\n\u0003 1 ˜ eiλj,r eiλj,s , f λ λi j\n\n(5.89)\n\nwhere we have used Eq. (5.18) and the symbol for the conjugate partition. In line with the last section we give a version of Eq. (5.89) using the nonorthogonal matrix basis, \u0003 1 λ˜ (M −1 )iλj m λjk,r (M −1 )λkl m il,s , (5.90) A= f λi jkl λ where we need not distinguish between the M −1 matrices for conjugate partitions. 6\n\n√ We use the antisymmetrizer in its idempotent form rather than that with the ( n! )−1 prefactor.\n\n5.5 Antisymmetric eigenfunctions of the spin\n\n81\n\n5.5 Antisymmetric eigenfunctions of the spin In this section we investigate the connections between the symmetric groups and spin eigenfunctions. We have briefly outlined properties of spin operators in Section 4.1. The reader may wish to review the material there. One of the important properties of all of the spin operators is that they are symmetric. The total vector spin operator is a sum of the vector operators for individual electrons n \u0003 (5.91) S\u0017i , S\u0017 = i=1\n\nindicating that the electrons are being treated equivalently in these expressions. This means that every π ∈ Sn must commute with the total vector spin operator. Since all of the other operators, S\u00172 , raising, and lowering operators, are algebraic \u0017 they also commute with every permutation. We functions of the components of S, use this result heavily below. 5.5.1 Two simple eigenfunctions of the spin Consider an n electron system in a pure spin state S. The associated partition is {n/2 + S, n/2 − S}, and the first standard tableau is\n\n1 · · · n/2 − S · · · n/2 + S , n/2 + S + 1 · · · n where we have written the partition in terms of the S quantum number we have targeted. We consider also an array of individual spin functions with the same shape and all η1/2 in the first row and η−1/2 in the second\n\nα ··· α ··· α , β ··· β where we have used the common abbreviations α = η1/2 and β = η−1/2 . Associating symbols in corresponding positions of these two graphical shapes generates a product of αs and βs with specific particle labels, \u0015 = α(1) · · · α(n/2 + S)β(n/2 + S + 1) · · · β(n), Sz \u0015 = M S \u0015, M S = S.\n\n(5.92) (5.93) (5.94)\n\nIf now we operate upon \u0015 with N (corresponding to {n/2 + S, n/2 − S}) we obtain a function with n/2 − S antisymmetric products of the [αβ − βα] sort, N \u0015 = [α(1)β(n/2 + S + 1) − α(n/2 + S + 1)β(1)] · · · α(n/2 + S).\n\n(5.95)\n\n82\n\n5 Advanced methods for larger molecules\n\nThe spin raising operator (see Eq. (4.3)) may now be applied to this result, and we obtain S + N \u0015 = N S + \u0015, = 0,\n\n(5.96)\n\nwhere we have used the commutation of S + with all permutations. The following argument indicates why the zero results. The terms of S + give zero with each α encountered but turn each β encountered into an α. Thus S + \u0015 is n/2 − S terms of products, each one of which has no more than n/2 − S − 1 β functions in it. Considering how these would fit into the tableau shape, we see that there would have to be, for each term, one column in the tableau that has an α in both rows. This column, with its corresponding factor from N , would thus appear as [I − (i j)]α(i)α( j), which is clearly zero. Eq. (5.96) is the consequence. Thus, N \u0015 is an eigenfunction of S\u00172 because of Eq. (4.5), S\u00172 N \u0015 = S(S + 1)N \u0015,\n\n(5.97)\n\nand has total spin quantum number S (also the M S value for this function). Other values of M S are available with S − should they be needed. We now investigate the behavior of \u0015 when we apply our two simple Hermitian idempotents discussed earlier, \u0015 PN P = θ PN P\u0015, = g P θ PN \u0015, \u0015 N PN = θ \u0014 N PN \u0015.\n\n(5.98) (5.99) (5.100)\n\nSince S + and Sz both commute with N and P, both \u0015 PN P and \u0015 N PN are eigenfunctions of the S\u0017 2 operator with total spin S and M S = S. Heretofore in this section we have been working with the partition λ = {n/2 + S, n/2 − S}, but references to it in the equations have been suppressed. We now write \u0015λP N P and \u0015λN PN . Applying the antisymmetrizer to the function of both space and spin that contains \u0015λPN P , λ A\u0014 PN P =\n\n\u0003 1 ˜\u0014 \u0014 eiλj,r eiλj,s \u0015λP N P . \u0014 f λ\u0014 i j λ\n\n(5.101) \u0014\n\nλ If the antisymmetrizer has been conditioned (see Eqs. (5.75)–(5.78)) so that e11,s λ\u0014 is θPN P , we obtain \u0014\n\nλ eiλj,s \u0015λP N P = δ1 j δλλ\u0014 ei1,s \u0015λP N P ,\n\nbecause of the orthogonality of the eiλj for different λs.\n\n(5.102)\n\n5.5 Antisymmetric eigenfunctions of the spin\n\n83\n\nWe make a small digression and note that the spin-degeneracy problem we have alluded to before is evident in Eq. (5.102). It will be observed that i = 1, . . . , f λ λ in the index of ei1,s \u0015λP N P , and these functions are linearly independent since the λ ei j,s are. There are, thus, f λ linearly independent spin eigenfunctions of eigenvalue S(S + 1). Each of these has a full complement of M S values, of course. In view of Eq. (5.40) the number of spin functions increases rapidly with the number of electrons. Ultimately, however, the dynamics of a system governs if many or few of these are important. Returning to our antisymmetrized function, we see it is now 1 \u0003 λ˜ e eλ \u0015λ , (5.103) A\u0014 P N P = f λ i i1,r i1,s P N P and we are in a position to examine its properties with regard to the Rayleigh quotient. Considering first the denominator, we have \u0003 \u000f\u0010 ˜ \u0011 ˜ \u0012 λ\n\n\u0011eλj1,r ei1,r A\u0014 P N P |A\u0014 P N P = f λ−2 ×\n\nij \u0011 \u0012\u0013 λ ei1,s \u0015λP N P \u0011eλj1,s \u0015λP N P ,\n\n\u0010\n\n\u0011 λ \u0012\u0010 \u0010 \u0011 λ˜ \u0012 = f λ−1 \u0011e11,r\n\n\u0015 P N P \u0011e11,s \u0015λP N P , since\n\n\u0011 ˜ \u0012 \u0010 \u0011 λ˜ λ˜ \u0012 λ˜ ei1,r\n\n\u0011eλj1,r = \u0011e1i,r e j1,r , \u0012 \u0010 \u0011 ˜ = δi j \u0011eλ ,\n\n\u0010\n\n11,r\n\n(5.104) (5.105)\n\n(5.106) (5.107)\n\nwith a very similar expression for the spin integral. Since the Hamiltonian of the ESE commutes with all permutations and symmetric group algebra elements, the same reductions apply to the numerator, and we obtain \u0011 λ \u0012\u0010 \u0010 \u0011 \u0011 λ˜ \u0012 A\u0014 P N P |H |A\u0014 P N P = f λ−1 \u0011 H \u0011e11,r\n\n\u0015λP N P \u0011e11,s \u0015λP N P . (5.108) This result should be carefully compared to that of Eq. (4.37), where there were two functions that have the same integral. Here we have f λ of them.7 Our final expression for the Rayleigh quotient is A\u0014 P N P |H |A\u0014 P N P , A\u0014 P N P |A\u0014 P N P \u0012 \u0010 λ˜\n\n|H |e11 = \u0010 \u0012 . λ˜\n\n|e11\n\nE=\n\n7\n\n(5.109) (5.110)\n\nWe may note in passing that the partition for three electrons in a doublet state is {2,1} and f λ for this is 2. That is why we found two functions in our work in Chapter 4.\n\n84\n\n5 Advanced methods for larger molecules\n\nWe are now done with spin functions. They have done their job to select the correct irreducible representation to use for the spatial part of the wave function. Since we no longer need spin, it is safe to suppress the s subscript in Eq. (5.110) and all of the succeeding work. We also note that the partition of the spatial function λ˜ is conjugate to the spin partition, i.e., {2n/2−S , 22S }. From now on, if we have occasion to refer to this partition in general by symbol, we will drop the tilde and represent it with a bare λ. 5.5.2 The function We have so far said little about the nature of the space function, . Earlier we implied that it might be an orbital product, but this was not really necessary in our general work analyzing the effects of the antisymmetrizer and the spin eigenfunction. We shall now be specific and assume that is a product of orbitals. There are many ways that a product of orbitals could be arranged, and, indeed, there are many λ would produce zero. The partition of these for which the application of the e11 corresponding to the spin eigenfunction had at most two rows, and we have seen that the appropriate ones for the spatial functions have at most two columns. Let us illustrate these considerations with a system of five electrons in a doublet state, and assume that we have five different (linearly independent) orbitals, which we label a, b, c, d, and e. We can draw two tableaux, one with the particle labels and one with the orbital labels,     1 4 a b  c d  and 2 5 . 3 e Associating symbols in corresponding positions from these two tableaux we may write down a particular product = a(1)c(2)e(3)b(4)d(5). There are, of course, 5! = 120 different arrangements of the orbitals among the particles, and all of the products are linearly independent. When we operate on them with the idempotent8 e11 , however, the linear independence is greatly reduced and instead of 120 there are only f = 5 remaining.9 This reduction is discussed in general by Littlewood. For our work, however, we note that e11 = m 11 = u, and uπi are linearly independent algebra elements. Therefore, using Eq. (5.64), the set consisting of the functions, uπi a(1)c(2)e(3)b(4)d(5); i = 1, . . . , 5 is linearly independent.10 There are many sets of five that have this property, but we only need a set that spans the vector 8 9 10\n\nWe now suppress the λ superscript. At the beginning of Section 5.4.4 we saw that there were five standard tableau for the conjugate of the current shape. The linear independence of this sort of set is discussed in Section 5.4.5.\n\n5.5 Antisymmetric eigenfunctions of the spin\n\n85\n\nspace of these functions, and the ones given here, based upon standard tableaux, will serve. We saw in Eq. (4.71) that there are two linearly independent orbital functions for the three-electron doublet state in the most general case, this being a consequence of the spin degeneracy of two. The result here is merely an extension. For the partition and tableau above the spin degeneracy is five, and the number of independent orbital functions is the same. We saw in Chapter 4 that the number of independent functions is reduced to one if two of the three electrons are in the same orbital. A similar reduction occurs in general. In our five-electron example, if b is set equal to a and c \u000e= d, there are only two linearly independent functions, illustrating a specific case of the general result that the number of linearly independent functions arising from any orbital product is determined only by the orbitals “outside” the doubly occupied set. This is an important point, for which now we take up the general rules.\n\n5.5.3 The independent functions from an orbital product Assume we have a set of m linearly independent orbitals. In order to do a calculation we must have m ≥ n/2 + S, where n is the number of electrons. Any fewer than this would require at least some triple occupancy of some of the orbitals, and any such product, , would yield zero when operated on by uπi . This is the minimal number; ordinarily there will be more. Any particular product can be characterized by an occupation vector, γ\u0017 = [γ1 γ2 . . . γm ] where γi = 0, 1, or, 2, and m \u0003\n\nγi = n.\n\n(5.111)\n\ni=1\n\nClearly, the number of “2”s among the γi cannot be greater than n/2 − S. It is not difficult to convince oneself that functions with different γ\u0017 s are linearly independent. Therefore, the only cases we have to check are those produced from one occupation vector. Littlewood shows how this is done considering standard tableaux with repeated elements. We choose an ordering for the labels of the orbitals we are using, a1 < a2 < · · · < ak ; n/2 + S ≤ k ≤ n that is arbitrary other than a requirement that the ai with γi = 2 occur first in the ordering.11 We now place these orbitals in a tableau shape with the rule that all symbols are nondecreasing to the right in the rows and definitely increasing downward in the columns. Considering our five-electron case again, assume we have four orbitals a < b < c < d and a is doubly occupied. The rules for standard tableaux with repeated elements then 11\n\nThis ordering can be quite arbitrary and, in particular, need not be related to an orbital’s position in a product with a different γ\u0017 .\n\n86\n\nproduce\n\n5 Advanced methods for larger molecules\n\na b d\n\n  a a c  and b c\n\n a d\n\nas the two possible arrangements for this case. Using only three orbitals, a < b < c, with a and b doubly occupied, we obtain only one such tableau,   a a b b  . c The general result states that the number of linearly independent functions from the set uπi (γ\u0017 ); i = 1, . . . , f is the number of standard tableaux with repeated elements that can be constructed from the labels in the product. As a general principle, this is not so easy to prove as some of the demonstrations of linear independence we have given above. The interested reader might, however, examine the case of two-column tableaux with which we are concerned. Examining the nature of the πi for this class of tableau, it is easy to deduce the result using N PN . This is all that is needed, of course. The number of linearly independent functions cannot depend upon the representation. γ\u0017 We now see that for each γ\u0017 we have f γ\u0017 linearly independent functions, uπi (γ\u0017 ), γ\u0017 where πi ; i = 1, . . . , f γ\u0017 is some subset12 of all of the πi appropriate for (γ\u0017 ). The method for putting together a CI wave function is now clear. After choosing the γ\u0017 s to be included, one obtains \u0003 γ\u0017 Ci γ\u0017 πi (γ\u0017 ), (5.112) u\u0017 = u i γ\u0017\n\nwhere the Ci γ\u0017 are the linear variation parameters to optimize the energy. u\u0017 is thus a function satisfying the antisymmetry and spin conditions we choose and suitable for use with the ESE. We recall that u\u0017 is all that is needed to determine the energies. Minimizing the energy given by the Rayleigh quotient u\u0017|H |u\u0017 u\u0017|u\u0017 \u0017|H |u\u0017 = \u0017|u\u0017\n\nE=\n\n(5.113) (5.114)\n\nleads to a conventional nonorthogonal CI. 12\n\nWe see now why there were relatively few spin functions generated by operators from the symmetric groups. For the partition {n/2 + S, n/2 − S} and an M S = S, there is only one standard tableau with repeated elements for the ordering α < β. Thus only the πi−1 N PN \u0015 are linearly independent. All expressions of the form N PN π j \u0015, π j \u000e= I are zero.\n\n5.5 Antisymmetric eigenfunctions of the spin\n\n87\n\nNow that we know the number of linearly independent n-particle functions for a particular γ\u0017 , we can ask for the total number of linearly independent n-particle functions that can be generated from m orbitals. Weyl gave a general expression for all partitions and we will only quote his result for our two-column tableaux. The total number of functions, D(n, m, S), i.e., the size of a full CI calculation is \u0001 \u0002\u0001 \u0002 2S + 1 m+1 m+1 . (5.115) D(n, m, S) = n/2 − S m + 1 n/2 + S + 1 This is frequently called the Weyl dimension formula. For small S and large m and n, D can grow prodigiously beyond the capability of any current computer.13\n\n5.5.4 Two simple sorts of VB functions We saw in Section 5.4.7 that there were two Hermitian idempotents of the algebra that were easily constructed for each partition. Using these alternatives gives us two different (but equivalent) specific forms for the spatial part of the wave function. Specifically, if we choose u = θN PN , we obtain the standard tableau functions introduced by the author and his coworkers. If, on the other hand, we take u\u0014 = θ \u0014 PN P, we obtain the traditional Heitler–London–Slater–Pauling (HLSP) VB functions as discussed by Matsen and his coworkers. In actual practice γ\u0017 the πi for this case are not usually chosen from among those giving standard tableaux, but rather to give the Rumer diagrams (see Section 5.5.5). We asserted above that the permutations giving standard tableaux were only one possible set yielding linearly independent elements of the group algebra. This is a case in point. For the two-column tableaux the Rumer permutations are an alternative set that can be used, and are traditionally associated with different bonding patterns in the molecule. Of these two schemes, it appears that the standard tableaux functions have properties that allow more efficient evaluation. This is directly related to the occurrence of the N on the “outside” of θ N PN . Tableau functions have the most antisymmetry possible remaining after the spin eigenfunction is formed. The HLSP functions have the least. Thus the standard tableaux functions are closer to single determinants, with their many properties that provide for efficient manipulation.14 Our discussion of evaluation methods will therefore be focused on them. Since the two sets are equivalent, methods for writing the HLSP functions in terms of the others allow us to compare results for weights (see Section 1.1) of bonding patterns where this 13\n\n14\n\nThese considerations are independent of the nature of the orbitals other than their required linear independence. Thus, D is the size of the full Hamiltonian matrix in either a VB treatment or an orthogonal molecular orbital CI. One may compare this difference with Goddard’s discussion of what he termed the G1 and Gf methods.\n\n88\n\n5 Advanced methods for larger molecules\n\nis desired. If only energies and other properties calculated from expectation values are needed, the standard tableaux functions are sufficient. We note finally that if f γ\u0017 = 1 for a particular product function the standard tableaux function and HLSP function are the same.\n\n5.5.5 Transformations between standard tableaux and HLSP functions Since the standard tableaux functions and the HLSP functions span the same vector space, a linear transformation between them is possible. Specifically, it would appear that the task is to determine the ai j s in \u0003 ai j θ \u0014 PN Pρ j , (5.116) θN PN πi = j\n\nwhere the πi are the permutations interconverting standard tableaux, and ρ j similarly interconvert Rumer diagrams. It turns out, however, that Eq. (5.116) cannot be valid. The difficulty arises because on the left of the equal sign the left-most operator is N , while on the right it is P. To see that Eq. (5.116) leads to a contradiction multiply both sides by N . After factoring out some constants, one obtains \u0003 1 ai j N Pρ j , (5.117) θ N PN πi = gN gP j which has a right hand side demonstratively different from that of Eq. (5.116). The left hand sides are, however, the same, so the two together lead to a contradiction. We must modify Eq. (5.116) by eliminating one or the other of the offending factors. It does not matter which, in principle, but the calculations are simpler if we use instead \u0003 f θ N PN πi = ai j N Pρ j . (5.118) g j In order to see why this modified problem actually serves our purpose, we digress to discuss some results for non-Hermitian idempotents. The perceptive reader may already have observed that the functions we use can take many forms. Consider the non-Hermitian idempotent ( f /g)PN . Using the permutations interconverting standard tableaux, one finds that ( f /g)PN πi ; i = 1, . . . , f is a set of linearly independent functions (if has no double occupancy). Defining a linear variation function in terms of these, \u0017=\n\n\u0003 f ai πi , PN g i\n\n(5.119)\n\n5.5 Antisymmetric eigenfunctions of the spin\n\n89\n\none obtains for the matrix system, Hi j = K × πi |H N PN |π j\n\n, Si j = K × πi |N PN |π j\n\n,\n\n(5.120) (5.121)\n\nwhich is easily seen to be the same system as that obtained from the Hermitian idempotent, θN PN . The “K ” is different, of course, but this cancels between the numerator and denominator of the Rayleigh quotient. Thus, \u0014\u0014 =\n\n\u0003 f N PN ai πi gg N i\n\n(5.122)\n\nwill produce the same eigensystem and eigenvectors as the variation function \u0014 of Eq. (5.119), but the resulting spatial functions are not equal, \u0014 \u000e= \u0014 \u0014 . Some considerable care is required in interpreting this result. It must be remembered that the spatial functions under discussion are only a fragment of the total wave function, and are related to expectation values of the total wave function only if the operator involved commutes with all permutations of Sn . There are two important cases that demonstrate the care that must be used in this matter. Consider an operator commonly used to determine the charge density: \u0003 δ(\u0017ri − ρ), \u0017 (5.123) Dop = i\n\nwhere ρ\u0017 is the position at which the density is given and i now labels electrons. This operator commutes with all permutations and is thus satisfactory for determining the charge density from \u0014, \u0014 \u0014 , or the whole wave function. The spatial probability density is another matter. In this case the operator is \u0014 δ(\u0017ri − ρ\u0017i ), (5.124) Pop = i\n\nwhere the ρ\u0017i are the values at which the functions are evaluated. As it stands, this is satisfactory for the whole wave function, but for neither \u0014 nor \u0014 \u0014 . To work with the latter two, we must make it commute with all permutations, and it must be modified to 1 \u0003 −1 \u0014 \u0014 = τ δ(\u0017ri − ρ\u0017i )τ, (5.125) Pop n! τ ∈Sn i \u0014 form gives the same where the permutations do not operate on the ρ\u0017i . The Pop value in all three cases. After this digression we now return to the problem of determining the HLSP functions in terms of the standard tableaux functions. We solve Eq. (5.118) by\n\n90\n\n5 Advanced methods for larger molecules\n\nmultiplying both sides by πk−1 and evaluating both sides for the identity element:15 \u0007\u0007 \u0003 \u0006\u0006 \u0007\u0007 1 \u0006\u0006 (5.126) N PN πi πk−1 = N Pρ j πk−1 a ji gN j \u0003 Bk j a ji , (5.127) Mki = j\n\nand denoting by A the matrix with elements ai j , we obtain A = B −1 M.\n\n(5.128)\n\n\u0007\u0007 1 \u0006\u0006 −1 πk N PN πi , gN\n\n(5.129)\n\nIn Eq. (5.128) Mki =\n\nis the “overlap” matrix for θ N PN (see Eq. (5.73) and following). For singlet systems the bonding patterns for Rumer diagrams are conventionally obtained by writing the symbols for the orbitals in a ring (shown here for six), and drawing all diagrams where all pairs of orbital symbols are connected by a line and no lines cross[2, 13].\n\na\n\na\n\na\n\na\n\na\n\nf\n\nb\n\nf\n\nb f\n\nb f\n\nb f\n\ne\n\nc\n\ne\n\nc\n\nc e\n\nc e\n\nd\n\nd\n\ne d\n\nd\n\nb c d\n\nOur treatment has been oriented towards using tableaux to represent functions rather than Rumer diagrams, and it will be convenient to continue. Thus, corresponding to the five canonical diagrams for a ring of six orbital symbols one can write           a f a b a d a f a b  c d   b c   c f  b c  b f  d e R d e R e f R c d R e f R where the symbols in the same row are “bonded” in the Rumer diagram. We have made a practice in using [ ] around our tableaux, and those that refer to functions where we use PN P will be given “R” subscripts to distinguish them from functions where we have used N PN . This notational device will be used extensively in Part II of the book where many comparisons between standard tableaux functions and HLSP functions are shown. 15\n\nWe commented above that the form of Eq. (5.118) was simpler than the result of removing N from the other side. This arises because determining [[PN Pτ ]] is, in general, much more difficult than evaluating [[N Pτ ]], because simple expressions for PN P are known only for singlet and doublet systems.\n\n5.5 Antisymmetric eigenfunctions of the spin\n\n91\n\nThe tableaux in the last paragraph are, of course, not unique. In any row either orbital could be written first, and any order of rows is possible. Thus, there are 23 × 3! = 48 different possible arrangements for each. We have made them unique by setting a < b < c < d < e < f , making each row increase to the right and the first column increase downward. These are not standard tableaux – the second column is not alway increasing downward. Using   1 4 2 5 3 6 for the particle label tableau, it is seen that the permutations I , (25 364), (365), (254), and (23 564) will generate all five orbital tableaux from the first, and can be used for the ρi of Eq. (5.118). This transformation is tedious to obtain by hand, and computer programs are to be preferred. A few special cases have been given. An example is also given in Section 6.3.2. 5.5.6 Representing θ N P N as a functional determinant For the efficient evaluation of matrix elements, it is useful to have a representation of θN PN as a functional determinant. We consider subgroups and their cosets to obtain the desired form. The operator N consists of terms for all of the permutations of the subgroup G N , and P those for the subgroup G P . Except for the highest multiplicity case, S = n/2, G N is smaller than the whole of Sn . Let ρN ∈ G N and τ1 \u000e∈ G N . Consider all of the permutations ρN τ1 for fixed τ1 as ρN runs over G N . This set of permutations is called a right coset of G N . The designation as “right” arises because τ1 is written to the right of all of the elements of G N . We abbreviate the right coset as G N τ1 . There is also a left coset τ1 G N , not necessarily the same as the right coset. Consider a possibly different right coset G N τ2 , τ2 \u000e∈ G N . This set is either completely distinct from G N τ1 or identical with it. Thus, assume there is one permutation in common between the two cosets, ρ1 τ1 = ρ2 τ2 ; ρ1 , ρ2 ∈ G N ρ3 ρ1 τ1 = ρ3 ρ2 τ2 ; ρ3 ∈ G N ,\n\n(5.130) (5.131)\n\nand, as ρ3 ranges over G N , the right and left hand sides of Eq. (5.131) run over the two cosets and we see they are the same except possibly for order. The test may be stated another way: if τ2 τ1−1 = ρ2−1 ρ1 ∈ G N , τ1 and τ2 generate the same coset.\n\n(5.132)\n\n92\n\n5 Advanced methods for larger molecules\n\nWe conclude that one can find a number of right coset generators giving distinct cosets until the permutations of Sn are exhausted. Symbolizing the right coset generators as τ1 = I, τ2 , . . . , τ p , we have Sn = G N I ⊕ G N τ2 ⊕ · · · ⊕ G N τ p ,\n\n(5.133)\n\nwhere the first coset is G N itself. This leads to the often quoted result that the order of any subgroup must be an integer divisor of the order of the whole group and, in this case, we have \u0001 \u0002 n p= . (5.134) n/2 − S Our goal now is to find a convenient set of right coset generators for G N that gives Sn . Let us now consider specifically the case for the {2k 1n−2k } partition with k = n/2 − S, and the tableau, \n\n1  ..  .   k  k + 1   .  .. n−k\n\n n−k+1 ..   .   n .    \n\nThe order of G N is gN = (n − k)!k!. Now let i 1 , i 2 , . . . , il be l ≤ k of the integers from the first column of our tableau and let j1 , j2 , . . . , jl be the same number from the second column. These two sets of integers define a special permutation [(i)( j)]l = (i 1 j1 ) · · · (il jl ), which is a product of l noninteracting binaries. Since each binary contains a number from each column, none with l > 0 are members of G N . Some, but not all, are members of G P , however. Amongst all of these there is a subset that we call canonical in which i 1 < i 2 < · · · < il and j1 < j2 < · · · < jl . The number of these is \u0001 \u0002\u0001 \u0002 n−k k , l l and it is easily shown that \u0002\u0001 \u0002 k \u0001 \u0003 n−k k l=0\n\nl\n\nl\n\n\u0001 \u0002 n = . k\n\n(5.135)\n\n5.5 Antisymmetric eigenfunctions of the spin\n\n93\n\nThus, if the distinct canonical [(i)( j)] generate distinct cosets, we have all of them, since \u0001 \u0002 n = n!. (5.136) gN k Considering [(i)( j)]l and [(i \u0014 )( j \u0014 )]l \u0014 , we use Eq. (5.132) to test whether they generate the same coset. The [(i)( j)]l are, of course, their own inverses, and for the present test we have ?\n\n[(i)( j)]l [(i \u0014 )( j \u0014 )]l \u0014 ∈ G N .\n\n(5.137)\n\nIf [(i)( j)]l and [(i \u0014 )( j \u0014 )]l \u0014 have any binaries with no numbers in common then these will remain unaffected in the product, and since none of the binaries is a member of G N , neither is the product and the cosets must be different. If there are any binaries in common these cancel and there remain only binaries that have numbers connected in one or more chains. Consider a simple two-member chain, (a b)(c b) = (a c)(a b). The binary (a c) ∈ G N but (a b) is not, so this chain cannot be a member of G N , and, again, the cosets are different. Our simple two-member chain could, however, be the start of a longer one, and proceeding this way we see that we always arrive at the conclusion that the canonical [(i)( j)]l s generate all of the cosets. Going back to θN PN , we write out the N on the right explicitly and carry out a number of transformations. \u0003 (−1)σν ν, (5.138) θ N PN = θ N P ν∈G N\n\n= θN\n\n\u0003\n\n(−1)σν νν −1 Pν,\n\n(5.139)\n\nν∈G N\n\nf N B, n! 1 \u0003 −1 ν Pν, B= gN ν∈G N =\n\n(5.140) (5.141)\n\nand we see that the B operator is a sort of symmetrization of the P operator. We note first that [[B]] = 1. The operator P is a sum of terms P=I+\n\nk \u0003\n\npl ,\n\n(5.142)\n\nl=1\n\nwhere pl is a sum over all of the sort of [(i)( j)]l that correspond to l and have “horizontal” binaries only. There are ( kl ) permutations in pl . Next, we observe that\n\n94\n\n5 Advanced methods for larger molecules\n\neach term in Eq. (5.141) has the form \u0003 ν\n\nν\n\n−1\n\n\u0001 \u0002\u0001 \u0002−1 \u0001 \u0002−1 k k n−k pl ν = (n − k)!k! bl , l l l \u0001 \u0002−1 n−k bl , = gN l\n\n(5.143) (5.144)\n\nwhere bl is the sum of all of the coset generators corresponding to l. Equation (5.143) is obtained merely by the correct counting: the factors on the right are the number of terms in the sum and pl divided by the number of terms in bl . Thus, f N B, g \u0002−1 k \u0001 f \u0003 n−k bl , = N l g l=0\n\nθN PN =\n\n=\n\nf BN , g\n\n(5.145) (5.146) (5.147)\n\nwhere we know N and B commute, since they are both Hermitian and so is N PN . As an example of how N and bl operators work together we observe that the full antisymmetrizer corresponding to Sn may be written with N and the bl operators, A=\n\nk 1 \u0003 N (−1)l bl , n! l=0\n\n(5.148)\n\nsince the right hand side has each permutation once and each will have the correct sign. We emphasize that this is valid for any k. Now consider n functions u 1 , u 2 , . . . , u n and form the n-particle product function\n\n= u 1 (1)u 2 (2) · · · u n (n). Using the form of the antisymmetrizer of Eq. (5.148) we see that \u0011 \u0011 \u0011u 1 (1) · · · u n (1)\u0011 \u0011 1 \u0011\u0011 . .. \u0011 , (5.149) A = \u0011 .. \u0011 . \u0011 n! \u0011 \u0011u 1 (n) · · · u n (n)\u0011 and for each k of Eq. (5.148) we have a way of representing a determinant. These correspond to different Lagrange expansions that can be used to evaluate determinants, and, in particular, the use of k = 1 is closely associated with Cramer’s rule. We now define another operator (group algebra element) using the bl coset generator sums, D(q) =\n\nk \u0003 l=0\n\nq l bl ,\n\n(5.150)\n\n5.5 Antisymmetric eigenfunctions of the spin\n\n95\n\nwhere q could be complex. With N this new operator may be applied to the orbital product . A little reflection will convince the reader that the result may be written as a functional determinant, \u0011 \u0011 \u0011 P Q\u0011 \u0011 \u0011, N D(q) = \u0011 (5.151) −q R S \u0011 where P, Q, R, and S are blocks from the determinant in Eq. (5.149). Their sizes and shapes depend upon k: P is (n − k) × (n − k), Q is (n − k) × k, R is k × (n − k), and S is k × k. The block −q R represents the variable −q multiplying each function in the R-block. We note that if q = −1 the operator D(q) is just the sum of coset generators in Eq. (5.148), and the determinant in Eq. (5.151) just becomes the one in Eq. (5.149). We may now use the β-function integral, \u0001 \u0002−1 \b 1 l n−k−l −1 n − k t (1 − t) dt = (n − k + 1) , (5.152) l 0 and, letting q = t/(1 − t), convert D(q) to B. Thus, one obtains \b 1 (1 − t)(n−k) D[(t/(1 − t))] dt = B. (n − k + 1)\n\n(5.153)\n\n0\n\nPutting together these results, we obtain the expression for θ N PN as the integral of a functional determinant, \u0011 \u0011 \b \u0011 P Q\u0011 (n − k + 1) f 1 (n−k) \u0011 \u0011 θN PN = (1 − t) (5.154) \u0011−q R S \u0011 dt, g 0 t . (5.155) q= 1−t The same sort of considerations allow one to determine matrix elements. Let v1 (1) · · · vn (n) = ϒ be another orbital product. There is a joint overlap matrix between the v- and u-functions:   v1 |u 1 · · · v1 |u n  ..  , ¯ =  ... (5.156) S(¯v , u) .  vn |u 1\n\n···\n\nvn |u n\n\nand we may use it to assemble a functional determinant. Thus, we have \u0011 \u0011 \b \u0011 P \u0014 Q\u0014\u0011 (n − k + 1) f 1 (n−k) \u0011 \u0011 ϒ|θN PN\n\n= (1 − t) \u0011−q R \u0014 S \u0014 \u0011 dt, g 0 t , q= 1−t\n\n(5.157) (5.158)\n\n96\n\n5 Advanced methods for larger molecules\n\n¯ in the same way as where the primed blocks of the determinant come from S(¯v , u) the blocks of Eq. (5.151) were obtained from the determinant of Eq. (5.149). Somewhat more complicated but similar considerations provide for the determination of the matrix elements of the Hamiltonian. When there are doubly occupied orbitals among the u i , simplifications occur in these expressions. In addition, the integrand is a polynomial in t of degree n − k, and may be evaluated exactly in computer applications using an optimal Gauss quadrature formula. For further details on both these points, the reader is referred to the literature. The VB calculations reported on in Part II of the book were all carried out by a computer program implementing the discussion of this section.\n\n6 Spatial symmetry\n\nSpatial symmetry plays a role in a large number of the examples in Part II of this book. This can arise in a number ways, but the two most important are simplification of the calculations and labeling of the energy states. We have devoted considerable time and space in Chapter 5 to the effects of identical particle symmetry and spin. In this chapter we look at some of the ways spatial symmetry interacts with antisymmetrization. We first note that spatial symmetry operators and permutations commute when applied to the functions we are interested in. Consider a multiparticle function φ(\u0001 r 1 , r\u00012 , . . . , r\u0001n ), where each of the particle coordinates is a 3-vector. Applying a permutation to φ gives r π1 , r\u0001π2 , . . . , r\u0001πn ), π φ(\u0001 r 1 , r\u00012 , . . . , r\u0001n ) = φ(\u0001\n\n(6.1)\n\nwhere {π1 , π2 , . . . , πn } is some permutation of the set {1, 2, . . . , n}. Now consider the result of applying a spatial symmetry operator,1 i.e., a rotation, reflection, or rotary-reflection, to φ. Symbolically, we write for a spatial symmetry operation, R, Rr\u0001 = r\u0001\u0005 , \u0001 \u0002 Rφ(\u0001 r 1 , r\u00012 , . . . , r\u0001n ) = φ r\u0001\u0005 1 , r\u0001\u0005 2 , . . . , r\u0001\u0005 n ,\n\n(6.2) (6.3)\n\nπ Rr\u0001 = Rπ r\u0001, \u0001 \u0002 = φ r\u0001\u0005 π1 , r\u0001\u0005 π2 , . . . , r\u0001\u0005 πn .\n\n(6.4) (6.5)\n\nand we see that\n\n1\n\nIn physics and chemistry there are two different forms of spatial symmetry operators: the direct and the indirect. In the direct transformation, a rotation by π/3 radians, e.g., causes all vectors to be rotated around the rotation axis by this angle with respect to the coordinate axes. The indirect transformation, on the other hand, involves rotating the coordinate axes to arrive at new components for the same vector in a new coordinate system. The latter procedure is not appropriate in dealing with the electronic factors of Born–Oppenheimer wave functions, since we do not want to have to express the nuclear positions in a new coordinate system for each operation.\n\n97\n\n98\n\n6 Spatial symmetry\n\n6.1 The AO basis The functions we use are products of AOs, and, to be useful in a calculation, the AOs must be a basis for a representation of the spatial group. Since the spatial operations and permutations commute, the tableau functions we use also provide a basis for a representation of the spatial group. This is generally true regardless of the nature of the representation provided by the AOs themselves. Nevertheless, to work with tableaux on computers it greatly simplifies programs if the AO basis provides a representation of a somewhat special sort we call generalized permutation. If we have an appropriate AO basis, it supports a unitary representation of the spatial group G S = {I, R2 , R3 , . . . , R f }, \u0003 χk D(Ri )k j , (6.6) Ri χ j = k\n\nwhere χ j are the AOs and the D(Ri )k j are, in general, reducible. D(Ri ) is a generalized permutation matrix if every element is either zero or a number of unit magnitude. Because of the unitarity, each row or column of D(Ri ) has exactly one nonzero element, and this one is ±1. As it turns out, this is not an extremely special requirement, but it is not always possible to arrange. The following are some guidelines as to when it is possible.2 r G is abelian. S r G has a principal rotation axis of order >2, and no atoms of the molecule are centered S\n\non it. This frequently requires the coordinate axes for the AOs to be different on different atoms. r G transforms the x-, y-, and z-coordinate axes into ± themselves, and we use S tensorial rather than spherical d, f , . . . functions. That is, our d-set transforms as {x 2 , y 2 , z 2 , x y, x z, yz} with similar sets for the higher l-values.\n\nIn cases where these guidelines cannot be met, one must use the largest abelian subgroup from the true G S of the molecule. We will show some examples later.\n\n6.2 Bases for spatial group algebras Just as we saw with the symmetric groups, groups of spatial operations have associated group algebras with a matrix basis for this algebra, eiαj =\n\n2\n\ng fα \u0003 D(Ri )iα∗j Ri . g i=1\n\n(6.7)\n\nWe emphasize these rules are not needed theoretically. They are merely those that the symmetry analysis in CRUNCH requires to work.\n\n6.3 Constellations and configurations\n\n99\n\nThis should be compared with Eq. (5.19), but in this case we can assume that the irreducible representation is unitary without causing any complications. The law of combination is identical with the earlier Eq. (5.20), β\n\neiαj ekl = δ jk δαβ eilα .\n\n(6.8)\n\nWe use the same symbol for the two kinds of groups. This normally causes no confusion. These operators of course satisfy \u0001 α \u0002† (6.9) ei j = eαji , and, thus, eiiα is Hermitian. All of the eiαj also commute with the Hamiltonian. α is a projector for the first component of the α th irreducible The element e11 representation basis. Using standard tableaux functions we can select a function of a given symmetry and a given spin state with α ψ αj = e11 θ N PN T j ,\n\n(6.10)\n\nwhere T j is a product of AOs associated with the j th standard tableau. When we evaluate matrix elements of either the overlap or the Hamiltonian between two functions of these types we have \u0005 β \u0006 \u0004 α\u0005 β\u0006 \u0004 α (6.11) ψ j \u0005ψk = e11 θ N PN T j \u0005e11 θ N PN Tk , \u0004 \u0005 β \u0006 = δαβ T j \u0005e11 θN PN Tk , (6.12) \u0004 \u0005 \u0005 β \u0006 \u0004 α\u0005 \u0005 β\u0006 (6.13) ψ j \u0005 H \u0005ψk = δαβ T j \u0005 H \u0005e11 θ N PN Tk . 6.3 Constellations and configurations In quantum mechanical structure arguments we often speak of a configuration as a set of orbitals with a particular pattern of occupations. In this sense, if we consider the first of a set of standard tableaux, T1 , we can see that it establishes a configuration of orbitals. The other standard tableaux, T2 , . . . , T f , all establish the same configuration. Consider, however, the result of operating on T1 with an element of G S . It is simple to see why the assumption that the representation D(R) in Eq. (6.6) consists of generalized permutation matrices simplifies the result of this operation: in this case Ri T1 is just another product function ±T \u0005 . It may involve the same configuration or a different one, but it is just a simple product function. We use the term constellation to denote the collection of configurations that are generated by all of the elements operating upon Ri T1 ; i = 1, 2, . . . , g. Putting this another way, a constellation is a set of configurations closed under the operations of G S . It will be useful to illustrate some of these ideas with examples. We give\n\n100\n\n6 Spatial symmetry\n\nTable 6.1. Transformation of H2 O AOs. I\n\nC2\n\nσx z\n\nσzy\n\n2s 2 px 2 py 2 pz 1sa 1sb\n\n2s −2 px −2 p y 2 pz 1sb 1sa\n\n2s 2 px −2 p y 2 pz 1sb 1sa\n\n2s −2 px 2 py 2 pz 1sa 1sb\n\nthree: a C2v system, H2 O; a C3v system, NH3 ; and a D6h system, the π system of benzene. 6.3.1 Example 1. H2 O Consider a water molecule with a minimal basis on the atoms. We have a 1s, 2s, 2 px , 2 p y , 2 pz set on the O atom and 1sa and 1sb on the H atoms. We assume the molecule is oriented in the y–z plane with the O on the z-axis and the center of mass at the origin of a right-handed Cartesian coordinate system. It does not detract from this illustration if we ignore the O1s, and we suppress them from all tableaux. H2 O belongs to the C2v symmetry group, which is abelian and, hence, satisfies one of our guidelines above. Table 6.1 gives the transformation of the AOs under the operations of the group. Consider a configuration 2s 2 2 px2 1sa2 2 p y 2 pz . The identity and σzy operations leave it unchanged and the other two give 2s 2 2 px2 1sb2 2 p y 2 pz , and these configurations comprise one of the constellations for H2 O and this basis. The projector for the A1 symmetry species of C2v is 1 e A1 = (I + C2 + σx z + σzy ), 4 and taking\n\n2s  2 px   1sa 2 py as the defining tableau, we obtain    2s 2s 2s     2 px 2 px  1  2 px e A1   1sa 1sa  = 2  1sa 2 p y 2 pz 2 py\n\n(6.14)\n\n 2s 2 px   1sa  2 pz   2s 2s   2 px   2 px − 1sa   1sb\n\n 2s  2 px   , 1sb \n\n2 pz\n\n2 pz\n\n2 py\n\n(6.15)\n\n6.3 Constellations and configurations\n\n101\n\nas the A1 symmetry function based upon this constellation. If, alternatively, we used the B2 projector, 1 e B2 = (I − C2 − σx z + σzy ), (6.16) 4 we would obtain the same two tableaux as in Eq. (6.15), but with a + sign between them. The other two projectors yield zero. The symmetry standard tableaux functions are not always so intuitive as those in the first case we looked at. Consider, e.g., the configuration 2s2 px2 2 p 2y 2 pz 1sa 1sb , for which there are two standard tableaux and no other members in the constellation,     2 px 2 px 2 px 2 px   2 py 2 py    and  2 p y 2 p y  .    2s  2 pz 2s 1sa  1sa 1sb 2 pz 1sb When we apply e A1 to the first of these, we obtain      2 px 2 px 2 px 2 px 2 px       1 2 p 2 p 2 p 2 p 2 py y y y  y e A1  =  +   2s     2 pz 2s 2 pz 2s 2 1sa 1sb 1sa 1sb 1sb\n\n 2 px  2 py   ,  2 pz  1sa\n\n(6.17)\n\nwhere the second term on the right is not a standard tableau, but may be written in terms of them. Using the methods of Chapter 5 we find that       2 px 2 px 2 px 2 px 2 px 2 px  2 py 2 py   2 py 2 py   2 py 2 py  = − ,  (6.18)  2s 2 pz   2s 2 pz   2s 1sa  1sb 1sa 1sa 1sb 2 pz 1sb and thus\n\n2 px  2 py e A1   2s 1sa\n\n  2 px 2 px  2 py 2 py  = 2 pz   2s 1sb 1sa\n\n  2 px 2 px  2 py 1 2 py  −  2 pz  2  2s 1sb 2 pz\n\n 2 px 2 py  , 1sa  1sb\n\nwhich is a projected symmetry function, although not manifestly so. It is not difficult to show that   2 px 2 px  2 py 2 py   = 0, e A1   2s 1sa  2 pz\n\n(6.19)\n\n(6.20)\n\n1sb\n\nand the second standard tableau does not contribute to 1 A1 wave functions. This result indicates that the first standard tableau is not by itself a pure symmetry\n\n102\n\n6 Spatial symmetry\n\nTable 6.2. Transformation of NH3 AOs. I\n\nC3\n\nC32\n\nσx\n\nσy a\n\nσz a\n\n2s 2 px 2 py 2 pz 1sx b 1s y b 1sz b\n\n2s 2 py 2 pz 2 px 1s y 1sz 1sx\n\n2s 2 pz 2 px 2 py 1sz 1sx 1s y\n\n2s 2 px 2 pz 2 py 1sx 1sz 1s y\n\n2s 2 pz 2 py 2 px 1sz 1s y 1sx\n\n2s 2 py 2 px 2 pz 1s y 1sx 1sz\n\na\n\nEach reflection plane is labeled with the coordinate axis that is contained in it. b Each H-atom orbital is labeled with the reflection plane it resides on.\n\ntype but contains A1 and B2 components, while the second is pure B2 . The linear combination of Eq. (6.19) removes the unwanted part from the first tableau. We emphasize that these results are specific to the way we have ordered the particle numbers in the AOs. Other arrangements could give results that look quite different, but which would, nevertheless, be equivalent as far as giving the same eigenvalues of the ESE is concerned.\n\n6.3.2 Example 2. NH3 C3v is not an abelian group, but it is not difficult to orient a minimal basis involving s and p orbitals to make the representation of the AO basis a set of generalized permutation We orient the C3 -axis of the group along the unit vector √ √ matrices. √ {1/ 3,1/ 3,1/ 3}. The center of mass is at the origin and the N atom is on the C3 -axis in the negative direction from the origin. The three reflection planes of the group may be defined by the rotation axis and the three coordinate axes, respectively. There is an H atom in each of the reflection planes at an N---H bond distance from the N atom and at an angle of ≈76◦ from the rotation axis. In our description we suppress the closed 1s 2 core as before. Table 6.2 shows the transformation properties of the basis. We consider the configuration 2s 2 2 px 2 p y 2 pz 1sx 1s y 1sz , which is the only member of its constellation. Once we have chosen a specific arrangement for the first tableau, the other four standard tableaux may be given       2s 2s 2s 2s 2s 2s 2s 2s 2s 2s  2 px 1sx   2 px 1sx   2 px 2 p y   2 px 2 p y   2 px 1s y         2 p y 1s y   2 p y 2 pz   1sx 1s y   1sx 2 pz   1sx 2 pz  , 2 pz 1sz 1s y 1sz 2 pz 1sz 1s y 1sz 2 p y 1sz\n\n6.3 Constellations and configurations\n\n103\n\nand these will be symbolized by T1 , . . . , T5 in the order given. The A1 projector for C3v is e A1 =\n\n\u0002 1\u0001 I + C3 + C32 + σx + σ y + σz , 6\n\n(6.21)\n\nand using symgenn from the CRUNCH suite, we find that θ N PN T1 is a 1 A1 symmetry function on its own, e A1 θN PN T1 = θ N PN T1 .\n\n(6.22)\n\nApplying e A1 to θ N PN T2 yields 1 e A1 θN PN T2 = θ N PN (2T2 + 2T3 − T4 + 3T5 ). 6\n\n(6.23)\n\nUsing e A1 with T3 , T4 , or T5 does not give a function linearly independent of those we have found already. Thus, there are two linearly independent 1 A1 functions that can be formed from the configuration above. The first of these is not hard to understand when one examines the consequences of the antisymmetry of the columns of the standard tableaux functions. The second, however, is much less obvious and would be very tedious to determine without the computer program. To obtain the symmetry functions in terms of HLSP functions we can transform the standard tableaux functions using the methods of Chapter 5. The transformation matrix is given in Eq. (5.128): \n\n0  −1/3  A=  −1/3  0 −1/3\n\n0 0 0 −1/3 1/3 1/3 1/3 −1/3 1/3 2/3 2/3 2/3 1/3 1/3 −1/3\n\n −2/3 −1/3   −1/3  , −2/3 \n\n(6.24)\n\n−1\n\nand multiplying this by the coefficients of the symmetry functions of Eqs. (6.22) and (6.23), we obtain \n\n  1 0 0 0   1/3   −1/3   =  −1/3 A 0 1/3     0 −1/6   0 0 1/2 −1/3\n\n −1/3 −2/9   −2/9  , 0 −2/9\n\n(6.25)\n\n104\n\n6 Spatial symmetry\n\nas the coefficients of 1 A1 symmetry HLSP functions. The Rumer tableaux, Ti(R) , for the HLSP functions are \n\n2s  2 px   2 py 2 pz\n\n  2s 2s   1sx   2 px 1s y   1s y 1sz R 2 p y\n\n  2s 2s   1sx   1sx 2 pz   2 p x 1sz R 2 pz\n\n  2s 2s   2 p y   1sx 1s y   1s y 1sz R 2 px\n\n  2s 2s   2 py   2 py 2 pz   1sx 1sz R 2 px\n\n 2s 1s y   2 pz  1sz R\n\nThese Rumer tableaux are based upon the following diagrams:\n\n2px\n\n2px\n\n2px\n\n2px\n\n2px\n\n1sz\n\n1sx\n\n1sz\n\n1sx 1s z\n\n1sx\n\n1sz\n\n1sx 1sz\n\n1sx\n\n2pz\n\n2py 2pz\n\n2py 2pz\n\n2py 2pz\n\n2py 2pz\n\n2py\n\n1sy\n\n1sy\n\n1sy\n\n1sy\n\n1sy\n\nwhere we have arranged3 the orbitals below the 2s pair in a circle. T1(R) and T4(R) are the two “Kekul´e”4 diagrams and the others are the “Dewar” diagrams. T1(R) is the HLSP function with three electron pair bonds between the 2 pi orbital and the closest 2si . One sees that the T4(R) Kekul´e structure is completely missing from the 1 A1 functions. We, of course, could have determined the symmetry HLSP functions by examining them directly. Clearly T1(R) is by itself a symmetry function and a sum of the three Dewar structures is also. It is not so obvious that T4(R) does not contribute. One must confess that these A1 symmetry results we have obtained for NH3 are reasonably simple, because we chose the order of the AOs the way we did. One could arrange the orbitals in some other order and obtain valid results, but have symmetry functions that are very nonintuitive. The reader is urged to experiment with symgenn to see this. This is also evident when we consider A2 symmetry, the projector for which is e A2 =\n\n\u0002 1\u0001 I + C3 + C32 − σx − σ y − σz . 6\n\n(6.26)\n\nAs e A2 is applied to the Ti in turn, we obtain zero until 1 e A2 θN PN T4 = θ N PN (T4 − T5 ). 2 3 4\n\n(6.27)\n\nThe order of these Rumer diagrams is determined by the automatic generation routine in the computer program. Although NH3 does not have the spatial symmetry of a hexagon, we still may use this terminology in describing the structures.\n\n6.3 Constellations and configurations\n\n105\n\nT5 gives the same function. This function may be written in terms of Rumer tableaux, Ti(R) also, and we obtain \u0002 \u0001 1 1 θN PN (T4 − T5 ) = N P T1(R) + T2(R) + T3(R) + 2T4(R) + T5(R) , (6.28) 2 9 not a result that is particularly intuitive. T4(R) is present in this sum, and we will see shortly that this Rumer tableau is the only one that has an A2 component. It is not pure, as is seen from Eq. (6.28), but none of the other tableaux have any A2 part at all. There are two linear combinations of the standard tableaux functions that comprise a pair of E symmetry. The E projectors are \u0002 1\u0001 I − 1/2C3 − 1/2C32 + σx − 1/2σ y − 1/2σz , 3 √ √ √ \u0002 1 \u0001√3 = /2C3 − 3/2C32 + 3/2σ y − 3/2σz . 3\n\nE = e11\n\n(6.29)\n\nE e21\n\n(6.30)\n\nThe computations show that 1 E e11 θ N PN T2 = θ N PN (2T2 − T3 − T4 ), 3√ 3 E e21 θ N PN T3 = θ N PN (−T3 + T4 ), 6\n\n(6.31) (6.32)\n\nwhere the energy for either component will be the same. Again, the functions may be expressed in terms of the Rumer structures, and we obtain 1 θN PN (2T2 − T3 − T4 ) = 3 √ 3 θ N PN (−T3 + T4 ) = 6\n\n\u0001 \u0002 2 N P − 2T2(R) + T3(R) + T5(R) , 9 \u0001 \u0002 1 √ N P T3R − T5R . 27\n\n(6.33) (6.34)\n\nAs we commented on above, T4R is missing from all of the functions except for the one of A2 symmetry. 6.3.3 Example 3. The π system of benzene In Chapter 15 we give an extensive treatment of the π system of benzene, but here we outline briefly some of the symmetry considerations. We consider the configuration p1 p2 p3 p4 p5 p6 , where pi stands for a C2 pz orbital at the i th C atom, numbered sequentially and counterclockwise around the ring. The set of five standard tableaux is       p1 p2 p1 p3 p1 p3 p1 p4 p1 p2  p3 p4   p3 p5   p2 p4   p2 p5   p2 p5  p5 p6 p4 p6 p5 p6 p4 p6 p3 p6\n\n106\n\n6 Spatial symmetry\n\nand we label them T1 , . . . , T5 in that order. The e A1g in merely the sum of all of the elements of the D6h symmetry group divided by 24, the value of g. We obtain two linear combinations (6.35) e A1g θN PN T1 = θN PN T1 , 1 e A1g θN PN T2 = θ N PN (3T1 − T2 − T3 + 2T4 − 3T5 ). (6.36) 6 Here again, the second of these is not obviously a symmetry function. The Rumer diagrams for benzene actually mirror the real spatial symmetry, and thus the Kekul´e and Dewar structures emerge,\n\np1\n\np1\n\np1\n\np1\n\np1\n\np6\n\np2\n\np6\n\np2\n\np6\n\np2\n\np6\n\np2\n\np6\n\np2\n\np5\n\np3\n\np5\n\np3\n\np5\n\np3\n\np5\n\np3\n\np5\n\np3\n\np4\n\np4\n\np4\n\nand with these we associate the Rumer tableaux        p1 p2 p1 p2 p1 p4 p1  p3 p4   p3 p6   p2 p3   p2 p5 p6 R p4 p5 R p5 p6 R p4\n\np4\n\n  p6 p1 p3   p2 p5 R p3\n\np4\n\n p6 p5  · p4 R\n\nThe transformation from standard tableaux functions to HLSP functions is independent of the spatial symmetry and so we need the A-matrix in Eq. (6.24) again. This time the results are     1 1/2 0 1/3  0 −1/6   −1/3 2/9       =  −1/3 2/9  · A (6.37) 0 −1/6     0 1/3   0 1/3  0 −1/2 −1/3 2/9 In this case the symmetry functions in terms of the Rumer tableau are fairly obvious, as can be seen by inspection of the Rumer diagrams added together in them. We give more details of symmetry in benzene in Chapter 15.\n\n7 Varieties of VB treatments\n\nThe reader will recall that in Chapter 2 we gave examples of H2 calculations in which the orbitals were restricted to one or the other of the atomic centers and in Chapter 3 the examples used orbitals that range over more than one nuclear center. The genealogies of these two general sorts of wave functions can be traced back to the original Heitler–London approach and the Coulson–Fisher approach, respectively. For the purposes of discussion in this chapter we will say the former approach uses local orbitals and the latter, nonlocal orbitals. One of the principal differences between these approaches revolves around the occurrence of the so-called ionic structures in the local orbital approach. We will describe the two methods in some detail and then return to the question of ionic structures in Chapter 8.\n\n7.1 Local orbitals The use in VB calculations of local orbitals is more straightforward than the alternative. In its simplest form, when atomic AOs are used and considered fixed, the wave function is \u0001 Ci φi , (7.1) \u0001= i\n\nwhere the φi are n-electron basis functions as described in Chapter 5. The wave function presents a linear variation problem, and the only real problem is the practical one of choosing a suitable set of φi functions. We will discuss this latter problem more fully in Chapter 9. A primary characteristic of this approach is that each φi can be interpreted as a representation of the molecule in which each atom has a more-or-less definite state or configuration. In this way the molecule as a whole may be thought of as consisting of a mixture of atomic states including ionic ones, and in ideal circumstances we 107\n\n108\n\n7 Varieties of VB treatments\n\nmay calculate fractional weights for these states. The focus here is thus on the way atoms in a number of states interact to form the molecule. This is, of course, the approach used by all of the early VB workers. In more recent times, after computing machinery allowed ab initio treatments, this is the sort of wave function proposed by Balint-Kurti and Karplus, which they called a multistructure approach. The present author and his students have proposed the multiconfiguration valence bond (MCVB) approach, which differs from the BalintKurti–Karplus wave function principally in the way the φi are chosen. The local approach may be extended, as Hiberty suggests, by allowing the AOs to “breathe”. This is accomplished in modern times by writing the orbitals in φi as linear combinations of more primitive AOs, all at one nuclear center, and optimizing these linear combinations along with the coefficients in Eq. (7.1). The breathing thus contributes a nonlinear component to the energy optimization. This latter is, of course, only a practical problem; it contributes no conceptual difficulty to the interpretation of the wave function. We may summarize the important characteristics of VB calculations with local orbitals. 1. The n-electron basis consists of functions that have a clearcut interpretation in terms of individual atomic states or configurations. 2. Many atomic states in φi are of the sort termed “ionic”. 3. In a highly accurate energy calculation many terms may be required in Eq. (7.1). 4. If Rumer tableaux are used for φi , these may in many cases be put in a one-to-one relation with classical bonding diagrams used by chemists. 5. In its simplest form the energy optimization is a linear variation problem. 6. If a molecule dissociates, the asymptotic wave function has a clear set of atomic states.\n\n7.2 Nonlocal orbitals In all of the various VB methods that have been suggested involving nonlocal orbitals it is obvious that the orbitals must be written as linear combinations of AOs at many centers. Thus one is always faced with some sort of nonlinear minimization of the Rayleigh quotient. Historically, the first of the modern descendents of the Coulson–Fisher method proposed was the GGVB approach. Nevertheless, we will postpone its description, since it is a restricted version of still later proposals. We describe first the SCVB proposal of Gerratt et al. We use here the notation and methods of Chapter 5. These workers originally used the genealogical representations of the symmetric groups, but so long as the irreducible representation space is completely spanned, any representation will give the same energy and wave function. Balint-Kurti and van Lenthe proposed using an equivalent wave\n\n7.2 Nonlocal orbitals\n\n109\n\nfunction. The principal differences between these proposals deal with methods of optimization. We will continue to use the SCVB acronym for this method. Consider a system of n electrons in a spin state S. We know that there are for n linearly independent orbitals \u0002 \u0003 2S + 1 n+1 f = (7.2) n + 1 n − S/2 linearly independent standard tableaux functions or HLSP functions that can be constructed from these orbitals. In the present notation the SCVB wave function is written as the general linear combination of these: \u0001SCVB =\n\nf \u0001\n\nCi φi (u 1 , . . . , u n ),\n\n(7.3)\n\ni=1\n\nwhere the orbitals in φi are, in general, linear combinations of the whole AO basis.1 The problem is to optimize the Rayleigh quotient for this wave function with respect to both the Ci and the linear coefficients in the orbitals. Using familiar methods of the calculus of variations, one can set the first variation of the energy with respect to the orbitals and linear coefficients to zero. This leads to a set of Fock-like operators, one for each orbital. Gerratt et al. use a second order stabilized Newton–Raphson algorithm for the optimization. This gives a set of occupied and virtual orbitals from each Fock operator as well as optimum Ci s. The SCVB energy is, of course, just the result from this optimization. Should a more elaborate wave function be needed, the virtual orbitals are available for a more-or-less conventional, but nonorthogonal, CI that may be used to improve the SCVB result. Thus an accurate result here may also involve a wave function with many terms. The GGVB wave function can have several different forms, each one of which, however, is a restricted version of a SCVB wave function. As originally proposed, a GGVB calculation uses just one of the genealogical irreducible representation functions and optimizes the orbitals in it, under a constraint of some orthogonality. In general, the orbitals are ordered into two sets, orthogonality is enforced within the sets but not between them. Thus, there are f different GGVB wave functions, depending upon which of the genealogical φi functions is used. Goddard designated these as the G1, G2, . . . , Gf methods, the general one being Gi. Each of these, in general, yields a different energy, and one could choose the wave function for the lowest as the best GGVB answer. In actual practice only the G1 or Gf methods have been much used. In simple cases the Gf wave function is a standard tableaux function while the G1 is a HLSP function. For Gf wave functions 1\n\nThe requirements of symmetry may modify this.\n\n110\n\n7 Varieties of VB treatments\n\nwe may show that the above orthogonality requirement is not a real constraint on the energy. From Chapter 5 we have, for n electrons in a singlet state, the unnormalized function \u0001G f = θN PN u 1 (1) · · · u m (m)v1 (m + 1) · · · vm (n),\n\n(7.4)\n\nwhere m = n/2, and the us and vs are the two sets of orbitals. In principle, we could optimize the energy and orbitals corresponding to Eq. (7.4), and afterwards the presence of the N will allow the formation of linear combinations among the us and, in general, different ones among the vs that will render the two sets internally orthonormal. This does not change the value of \u0001G f , of course, except possibly for its overall normalization. On the other hand, no such invariance of G1 or HLSP functions occurs, so the orthogonality constraint has a real impact on the calculated energy. We saw in Chapter 3 how the delocalization of the orbitals takes the place of the ionic terms in localized VB treatments, and this phenomenon is generally true for n electron systems. We now summarize the main characteristics of VB calculations with nonlocal orbitals. 1. The wave function is reasonably compact, normally having no more than f terms. 2. There are no structures in the sum that must be interpreted as “ionic” in character. For many people this is a real advantage to these VB functions. 3. The SCVB function produces a considerable portion of the correlation energy. 4. If Rumer tableaux are used for φi , these may in many cases be put in a one-to-one relation with classical bonding diagrams used by chemists. 5. If a molecule dissociates, the asymptotic wave function has a clear set of atomic states.\n\nIllustrations of both of these classes of VB functions will be given for a number of systems in Part II of this book.\n\n8 The physics of ionic structures\n\nThe existence of many ionic structures in MCVB wave functions has often been criticized by some workers as being unphysical. This has been the case particularly when a covalent bond between like atoms is being represented. Nevertheless, we have seen in Chapter 2 that ionic structures contribute to electron delocalization in the H2 molecule and would be expected to do likewise in all cases. Later in this chapter we will see that they can also be interpreted as contributions from ionic states of the constituent atoms. When the bond is between unlike atoms, it is to be expected that ionic structures in the wave function will also contribute to various electric moments, the dipole moment being the simplest. The amounts of these ionic structures in the wave functions will be determined by a sort of “balancing act” in the variation principle between the “diagonal” effects of the ionic state energies and the “off-diagonal” effect of the delocalization. In this chapter we will also focus on the dipole moment of molecules. With these, some of the most interesting phenomena are the molecules for which the electric moment is in the “wrong” direction insofar as the atomic electronegativities are concerned. CO is probably the most famous of these cases, but other molecules have even more striking disagreements. One of the larger is the simple diatomic BF. We will take up the question of the dipole moments of molecules like BF in Chapter 12. In this chapter we will examine in a more general way how various sorts of structures influence electric moments for two simple cases. For some of the discussion in this chapter we restrict ourselves to descriptions of minimal basis set results, since these satisfactorily describe the physics of the effects. In other cases a more extensive treatment is necessary. 8.1 A silly two-electron example In Chapter 2 we described several treatments of the H2 molecule, and, of course, there the question of dipole moments was irrelevant, although we could have 111\n\n112\n\n8 The physics of ionic structures\n\ncalculated a quadrupole moment. We want here to consider the properties of a covalent VB structure for H2 using a silly basis set consisting of 1s orbitals for the two atoms that have different scale factors. Such a wave function is certain not to have the correct 1 \u0001g+ symmetry for H2 and will not have a credible energy, but an important point emerges. Let the singlet standard tableaux function, ψ, be ψ = N [1sa\u0003 1sb ],\n\n(8.1)\n\nwhere, as indicated by the prime, we use differently scaled AOs at the two centers. The question we ask is: What is the electric dipole moment implied by this wave function? Assuming the molecule is situated along the z-axis, the x- and y-components of the moment are zero. In atomic units the z-component of the moment is z a + z b + 2S\u00051sa\u0003 |z|1sb \u0007 , µz = z a + z b − 1 + S2 S[S(z a + z b ) − 2\u00051sa\u0003 |z|1sb \u0007] = , (8.2) 1 + S2 where S = \u00051sa\u0003 |1sb \u0007. It is clear from Eq. (8.2) that, whatever its value at small distances, µz goes to zero as the interatomic distance goes to infinity, since S also a goes to zero. µz is not zero, however, at 0.7 A, a distance near that at equilibrium in H2 . Taking the scale of 1s to be 1.0 and that of 1s \u0003 to be 1.2, a value close to that which is optimum for the molecule,1 we obtain µz = −0.118 D. STO6Gs were used and z b < z a , i.e., the less diffuse orbital is in the positive z-direction from the other. This calculated moment is not very large, but it arises from a purely covalent function. If we do the same calculation for the triplet function, \u0001 \u0003\u0002 1sa , 1sb we obtain µz = 0.389 D, in the direction opposite to that for the singlet function. In the singlet case the electron distribution is more toward the less extended AO and in the triplet case more toward the more extended AO. It is useful to state this result in different language. In general, we expect more electronegative atoms to have tighter less diffuse orbitals in comparable shells than atoms of lower electronegativity. In our case this means we have a surrogate atom for higher electronegativity in the positive z-direction from the other. Therefore, bonding interactions have the electrons moving toward the more electronegative atom and antibonding interactions have them moving toward the less electronegative atom. The usual sign convention confusion occurs, of course; the dipole moment points in a direction opposite to the electron movement. 1\n\nWe note that if the orbitals were scaled equally 2\u00051sa |z|1sb \u0007 = S(z a + z b ) and µz is correctly zero at all distances.\n\n8.2 Ionic structures and the electric moment of LiH\n\n113\n\n6\n\nElectric moment (au)\n\n5\n\n4\n\n3 +\n\nLi —H\n\n2\n\nLi−—H +\n\nThree-term wave function 1 Li—H 0 0\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\nLi—H distance (bohr)\n\nFigure 8.1. The electric dipole function versus the internuclear distance for LiH using the two ionic and one covalent functions, individually. The sign of the moment for Li− −H+ has been changed to facilitate plotting and comparison of the magnitudes.\n\nIt is clear, when we calculate the electric moment with a more realistic wave function, that even the so-called covalent functions can make nonzero contributions when symmetry allows.\n\n8.2 Ionic structures and the electric moment of LiH There are two rather different questions that arise when considering ionic structures in VB wave functions. The first of these we discuss is the contribution to electric dipole moments. LiH is considered as an example. In the next section we take up ionic structures and curve crossings, using LiF to illustrate the points. LiH is the simplest uncharged molecule that has a permanent electric dipole moment.2 We examine here some of the properties of the simplest VB functions for this molecule. The molecule is oriented along the z-axis with the Li atom in the positive direction. Figure 8.1 shows the expectation values of the electric moment for the Li+ −H− and covalent structures. The graph gives the negative of the moment for Li− −H+ for easy comparison. In addition, the moment for the three-term wave function involving all three of the other functions is given. Although such a simple wave 2\n\nThe small moments in isotopic hydrogen, HD and HT, for example, do not interest us here.\n\n114\n\n8 The physics of ionic structures\n\nfunction does not reproduce the total moment accurately, qualitatively the signs are correct. The behavior of the three expectation values deserves comment. 1. Li− −H+ From the point of view of electrostatics this structure is the simplest. We have a spherical Li− ion and a bare proton. The dipole moment should just be −R in atomic units, and the line giving the dependence of the moment on R should be straight with a slope of −45◦ and through the origin. (NB The sign of this curve has been changed in Fig. 8.1 to facilitate comparison of magnitudes.) 2. Li+ −H− At longer distances the molecule is essentially an undistorted Li+ ion and an H− ion. As such, the moment equals the internuclear distance. As the ions approach one another, the Pauli principle interaction between the Li1s and H1s AOs causes the moment to be larger than the value due only to the distance. It should be noted that the effect of the Pauli principle is in the same direction as is the triplet example we gave earlier for the H2 molecule with unequally scaled AOs, i.e., the charge density is pushed toward the more diffuse orbital. The exchange interaction between two doubly occupied orbital distributions is essentially like the triplet interaction between two singly occupied orbitals of the same sort. 3. Covalent The dipole moment of the covalent structure is never larger than 1 (in atomic units) and is always positive. A simple analysis is not so easy here, but the same Li1s and H1s interaction as appeared above occurs in this case also.\n\nFigure 8.2 shows the coefficients of the three structures in the total three-term wave function. As expected, the covalent term predominates at all distances, but 1.0\n\n0.8 Coefficients in wave function\n\nLi—H 0.6\n\n0.4\n\nLi +—H −\n\n0.2\n\n0.0\n\nLi −—H +\n\n−0.2 0\n\n1\n\n2 3 4 Li—H distance (bohr)\n\n5\n\n6\n\nFigure 8.2. The coefficients of the three structures in the simple three-term wave function.\n\n8.3 Covalent and ionic curve crossings in LiF\n\n115\n\n6\n\n5\n\nDipole moment (au)\n\n4\n\n3\n\n2\n\n1\n\n0 0\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\nLi —H distance (bohr)\n\nFigure 8.3. The dipole function for the full valence wave function of LiH. It is a little difficult to see on the present scale, but the moment is –0.033 au at an internuclear distance of 0.2 bohr.\n\nthe Li+ −H− term is the next largest at all distances. The structure with the “wrong” sign is always the smallest in the range of the graph. Figure 8.2 admittedly does not allow a quantitative analysis of the way the three terms produce the total moment, but it does provide a suggestive picture of how the various terms contribute. When a full valence calculation is done with the present basis, there are 48 standard tableaux functions that produce 45 1 \u0001 + functions. Figure 8.3 shows the dipole moment function for this wave function. The figure also shows the “45◦ ” line that would be the moment if the charges at the ends were unit magnitude. The curve has some interesting structure. As noted in the caption, even this molecule shows the dipole going the “wrong” way to a slight extent at very close internuclear distances. As the distance increases the moment rises above the 45◦ line, indicating an effective charge on the ends greater in magnitude than one au. By the time the equilibrium distance is reached (3.019 bohr), the effective charges have fallen so that the moment is 5.5 D with the positive end at the Li, as the electronegativities predict. 8.3 Covalent and ionic curve crossings in LiF Lithium and fluorine form a diatomic molecule that has a large dipole moment in the gas phase; it has been measured to be 6.3248a D in the ground vibrational state. The equilibrium internuclear distance is 1.564 A, and, therefore, the apparent\n\n116\n\n8 The physics of ionic structures 4 15 Electric moment → ←E 4 Energy (eV)\n\n2\n\n10\n\n←E 3 5\n\n1 ←E 2 0\n\n0\n\n←E 1\n\n−1\n\n−2\n\nDipole moment (D)\n\n3\n\n6\n\n8\n\n−5\n\n10 12 Li—F distance (bohr)\n\n14\n\n16\n\nFigure 8.4. The first four 1 \u0001 states of LiF near the ionic–covalent curve crossing region. The electric dipole moment is also given showing its change in relation to the crossing of the first two states. The energy curves refer to the left vertical axis and the dipole moment to the right.\n\ncharge on the atoms is ±0.84 electrons. When there is a long distance between them, the overall ground state is the pair of atoms, each in its neutral ground state. The first excited overall state has the Li atom excited 2s → 2 p at 1.847 eV. The ionization potential of Li is 5.390 eV and the electron affinity of F is 3.399 eV, so the next state up is the transfer of an electron from Li to F at an energy of 1.991 eV. This is a lower energy than any 2s → 3l states of Li or any 2 p → 3s states of F. If there were no other interaction the energy of the atoms in the Li+ −F− state would fall as they approach each other from ∞. The energy of this state would cross the ground state at about 12 bohr. In actuality the states interact, and there is an avoided crossing. The energies of the first four states of LiF are shown in Fig. 8.4 as a function of internuclear distance. In addition, the electric dipole moment is shown. The point we wish to bring out here is that the dipole moment curve around 6–7 bohr is very nearly the value expected for ± one electronic charge separated by that distance. As one moves outward, the avoided crossing region is traversed, the state of the molecule switches over from Li+ −F− to Li–F and the dipole falls rapidly. The avoided crossing we have discussed occurs between the two curves in Fig. 8.4 labeled E 1 and E 2 . Another avoided crossing, farther out than our graph\n\n8.3 Covalent and ionic curve crossings in LiF\n\n117\n\nshows, occurs between the Li+ −F− state curve and E 3 and E 4 . These latter two states involve Li2 pσ –F2 pσ and Li2 pπ –F2 pπ , both coupled to 1 \u0001 + . The crossing in these cases occurs at such large distances there is very little interaction. The calculations in this illustration were not done with a minimal basis set, since, if such were used, they would not show the correct behavior, even qualitatively. This happens because we must represent both F and F− in the same wave function. Clearly one set of AOs cannot represent both states of F. Li does not present such a difficult problem, since, to a first approximation, it has either one orbital or none. The calculations of Fig. 8.4 were done with wave functions of 1886 standard tableaux functions. These support 1020 1 \u0001 + symmetry functions. We will discuss the arrangement of bases more fully in Chapter 9.\n\nPart II Examples and interpretations\n\n9 Selection of structures and arrangement of bases\n\nSince, for any but the smallest of systems, a full VB calculation is out of the question, it is essential to devise a useful and systematic procedure for the arrangement of the bases and for the selection of a manageable subset of structures based upon these orbitals. These two problems are interrelated and cannot be discussed in complete isolation from one another, but we will consider the basis question first. In our two-electron calculations we have already addressed some of the issues, but here we look at the problems more systematically.\n\n9.1 The AO bases The calculations described in this section of the book have, for the most part, been carried out using three of the basis sets developed by the Pople school. STO3G A minimal basis. This contains exactly the number of orbitals that might be occupied in each atomic shell. 6-31G A valence double-ζ basis. This basis has been constructed for atoms up through Ar. 6-31G∗ A valence double-ζ basis with polarization functions added. Polarization functions are functions of one larger l-value than normally occurs in an atomic shell in the ground state.\n\nAny departures from these will be spelled out at the place they are used. Our general procedure is to represent the atoms in a molecule using the Hartree– Fock orbitals of the individual atoms occurring in the molecule. (We will also consider the interaction of molecular fragments where the Hartree–Fock orbitals of the fragments are used.) These are obtained with the above bases in the conventional way using Roothaan’s RHF or ROHF procedure, extended where necessary. ROHF calculations are not well defined, and the reader is cautioned that this term has meanings that differ among workers. Some computational packages, GAMESS is an example, in doing a single-atom calculation, do not treat the 121\n\n122\n\n9 Selection of structures and arrangement of bases\n\natom in a spherical environment. For example, for N in its 4 S ground state, the three p orbitals are divided into σ and π sets and are not all equal. This is not a matter of any importance in that milieu. An atom in a molecule will not be in a spherically symmetric environment for the Hartree–Fock function to be determined. In all of the calculations described here, however, we use the original Roothaan specifications that produce sets of l-functions that transform into one another under all rotations. This can have an important consequence in our VB calculations, if we treat a problem in which the energies of the system are important as we move to asymptotic geometries. An example will clarify this point. C2 is in a 1 \u0004g+ ground state, but there are two couplings of two C atoms, each in a 3 P e ground state, that have this symmetry. In our calculations these two will have the correct asymptotic degeneracy only if we use “spherical” atoms. Conventional basis set Hartree–Fock procedures also produce a number of virtual orbitals in addition to those that are occupied. Although there are experimental situations where the virtual orbitals can be interpreted physically, for our purposes here they provide the necessary fine tuning of the atomic basis as atoms form molecules. The number of these virtual orbitals depends upon the number of orbitals in the whole basis and the number of electrons in the neutral atom. For the B through F atoms from the second row, the minimal STO3G basis does not produce any virtual orbitals. For these same atoms the 6-31G and 6-31G∗ bases produce four and nine virtual orbitals, respectively. There is a point we wish to make about the orbitals in these double-ζ basis sets. A valence orbital and the corresponding virtual orbital of the same l-value have approximately the same extension in space. This means that the virtual orbital can efficiently correct the size of the more important occupied orbital in linear combinations. As we saw in the two-electron calculations, this can have an important effect on the AOs as a molecule forms. We may illustrate this situation using N as an example. The 6-31G basis for N has three s-type Gaussian groups. In the representation of the normal atom the 1s and 2s occupied orbitals are two linear combinations of the three-function basis and the s-type virtual orbital is the third. For convenience, we will call the last orbital 3s, but it should not be thought to be a good representation of a real orbital of that sort in an excited atom. A typical Hartree–Fock calculation yields 1s = 0.996 224 75g6 + 0.019 984 19g3 − 0.004 639 97g1 , 2s = −0.226 268 21g6 + 0.515 913 17g3 + 0.568 841 00g1 , 3s = 0.091 019 32g6 − 1.5148 075 3g3 + 1.478 256 46g1 ,\n\n(9.1) (9.2) (9.3)\n\nwhere g6 , g3 , and g1 are, respectively, the 6, 3, and 1 Gaussian groups from the basis. The 1s orbital is predominantly the g6 function, but the other two have roughly equal\n\n9.2 Structure selection\n\n123\n\n(in magnitude) parts of g3 and g1 . Therefore, a linear combination, a2s + b3s with a fairly small b can have an extension in space differing from that of 2s itself. If b/a < 0 it is more compact; if b/a > 0 it is less compact. 9.2 Structure selection Our discussion of the structure selection must be somewhat more involved. In part, this is a discussion of a crucial member of the CRUNCH package, the program entitled symgenn. It possesses a number of configuration selection devices, for the details of which the reader is referred to the CRUNCH manual. The present discussion will focus on the desired outcome of the selections rather than on how to accomplish them. Again, it is convenient to describe these by giving an example, that of the N2 molecule, which will be discussed quantitatively in Chapter 11.\n\n9.2.1 N2 and an STO3G basis N2 has 14 electrons and there are 10 orbitals in an STO3G basis. The Weyl dimension formula, Eq. (5.115), gives 4950 configurations1 for a singlet state. Physical arguments suggest that configurations with electrons excited out of the 1s cores should be quite unimportant. If we force a 1sa2 1sb2 occupation at all times, Eq. (5.115) now gives us 1176 configurations, a considerable reduction. These are not just \u0004g+ states, of course. Symgenn will allow us to select linear combinations having this spatial symmetry only. This reduces the size of the linear variation matrix to 102 × 102, a further significant reduction. Another number that symgenn tells us is that, among the 1176 configurations, only 328 appear as any part of a linear combination giving a \u0004g+ state. This number would be difficult to determine by hand.2 At this stage symgenn has done its job and the matrix generator uses the symgenn results to compute the Hamiltonian matrix. Thus, we would call this a full valence calculation of the energy of N2 with an STO3G basis.\n\n9.2.2 N2 and a 6-31G basis We still have 14 electrons, but the larger basis provides 18 orbitals in the basis. The full calculation now has 4 269 359 850 configurations, a number only slightly 1 2\n\nThis is the number of linearly independent standard tableaux or Rumer functions that the entire basis supports. To be precise, we should point out that we have symgenn treat N2 as a D4h system, rather than the completely correct D∞h . In projecting symmetry blocks out of Hamiltonian matrices, it is never wrong to use a subgroup of the full symmetry, merely inefficient. It would be a serious error, of course, to use too high a symmetry. It happens for the STO3G basis that there is no difference between A1g D4h and \u0004g+ D∞h projections.\n\n124\n\n9 Selection of structures and arrangement of bases\n\nsmaller than 232 . Forcing a 1sa2 1sb2 occupation reduces this to 4 504 864, which is a considerable reduction, but still much too large a number in practice. The reduction of these to \u0004g+ states is not known, but the number is still likely to be considerable. Instead, we use physical arguments again to reduce the number of configurations further. Many of the 4 504 864 configurations have mostly virtual orbitals occupied and we expect these to be unimportant. The number of occupied orbitals from the 6-31G basis is the same number as the total number of orbitals from the STO3G basis. Therefore, there are again 102 \u0004g+ functions from the occupied orbitals. These include charge separations as high as ±3. We add to this full valence set those configurations that have one occupied orbital replaced by one virtual orbital in the valence configurations with charge separation no higher than ±1. Symgenn could work out the number of configurations resulting, but we have not done this. If this selection scheme is combined with \u0004g+ symmetry projection, we obtain a 1086×1086 Hamiltonian matrix, an easily manageable size. 9.2.3 N2 and a 6-31G∗ basis When we add d orbitals to the basis on each atom we have the possibility that polarization can occur. Of course, as far as an atom in the second or third rows is concerned, the d orbitals merely increase the number of virtual orbitals and increase the number of possibilities for substitutions from the normally filled set. We do not give any of the numbers here, but will detail them when we discuss particular examples. 9.3 Planar aromatic and π systems In later chapters we give a number of calculations of planar unsaturated systems. Because of the plane of symmetry, the SCF orbitals can be sorted into two groups, those that are even with respect to the symmetry plane, and those that are odd. The former are commonly called σ orbitals and the latter π orbitals. Although it is an approximation, there has been great interest in treating the π parts of these systems with VB methods and ignoring the σ parts. The easiest way of doing this, while still using ab initio methods, is to arrange all configurations to have all of the occupied σ orbitals doubly occupied in the same way. In addition, σ virtual orbitals are simply ignored. The π AOs may then be used in their raw state or in any linear combinations desired. In this sort of arrangement, the π electrons are subjected to what is called the static-exchange potential (SEP) of the nuclei and σ core. The most important molecules of this sort are the aromatic hydrocarbons, but many examples containing oxygen and nitrogen also exist.\n\n10 Four simple three-electron systems\n\nIn this chapter we describe four rather different three-electron systems: the π system of the allyl radical, the He+ 2 ionic molecule, the valence orbitals of the BeH molecule, and the Li atom. In line with the intent of Chapter 4, these treatments are included to introduce the reader to systems that are more complicated than those of Chapters 2 and 3, but simple enough to give detailed illustrations of the methods of Chapter 5. In each case we will examine MCVB results as an example of localized orbital treatments and SCVB results as an example of delocalized treatments. Of course, for Li this distinction is obscured because there is only a single nucleus, but there are, nevertheless, noteworthy points to be made for that system. The reader should refer back to Chapter 4 for a specific discussion of the three-electron spin problem, but we will nevertheless use the general notation developed in Chapter 5 to describe the results because it is more efficient. 10.1 The allyl radical All of the calculations on allyl radicals are based upon a conventional ROHF treatment with a full geometry optimization using a 6-31G∗ basis set. The σ “core” was used to construct an SEP as described in Chapter 9. The molecule possesses C2v symmetry. The C2 symmetry axis is along the z-axis and the nuclei all reside in the x–z plane. Thus the “π ” AOs consist of the p y s, dx y s, and d yz s, of which there are 12 in all for this basis. At each C there is a 2 p y , a 3 p y , a 3dx y , and a 3d yz . The 2 p y is the SCF orbital for the atomic ground state, and the 3 p y is the virtual orbital of the same symmetry. Table 10.1 shows for reference the pertinent portions of the C2v character table. We number the π orbitals from one end of the molecule and use 2 p1 , 2 p2 , and 2 p3 , remembering that they are all of the 2 p y sort. The effect of the σx z and σ yz operations of the group is seen to be σ yz 2 pi = −2 pi , σ x z 2 p1 = 2 p3 , 125\n\n(10.1) (10.2)\n\n126\n\n10 Four simple three-electron systems\n\nTable 10.1. C2v characters. C2v\n\nI\n\nC2\n\nσx z\n\nσ yz\n\nA1 A2 B1 B2\n\n1 1 1 1\n\n1 1 −1 −1\n\n1 −1 1 −1\n\n1 −1 −1 1\n\nTable 10.2. Results of 128-function MCVB calculation. −116.433 248 63 au −116.477 396 60 au 1.201 eV 0.9003\n\nSCF energy MCVB energy Correlation energy \u0001 2 p1 EGSO pop. of 2 p3\n\nσ x z 2 p2 = 2 p2 , σ x z 2 p3 = 2 p1 .\n\n2 p2\n\n\u0002\n\n(10.3) (10.4)\n\nThe effect of the C2 operation is easily determined since C2 = σx z σ yz . There is, of course, a completely parallel set of relations for the 3 p y set of orbitals. Writing out the corresponding relations for the 3d orbitals is left to the interested reader.\n\n10.1.1 MCVB treatment An MCVB calculation with a full set of configurations involving the six 2 p y and 3 p y orbitals with further configurations involving all possible single excitations out of this set into the d-set gives 256 standard tableau functions, which can form 128 2 A2 symmetry functions and a Hamiltonian matrix of the same dimension. Table 10.2 gives several results from the calculation, and we see that there is about 1.2 eV of correlation energy. Because of the static exchange core, all of this is in the π system, of course. In addition we see that the EGSO population suggests that the wave function is 90% of the basic VB function with unmodified AOs. This is true, of course, for either standard tableaux functions or HLSP functions. It is instructive to examine the symmetry of the standard tableaux function of highest EGSO population given in Table 10.2. The effects of the two symmetry\n\n127\n\nplanes of C2v on the 2 pi orbitals are given above, and, consequently, \u0001 \u0002 \u0001 \u0002 2 p1 2 p2 2 p1 2 p2 σ yz =− , 2 p3 2 p3 \u0001 \u0002 \u0001 \u0002 2 p1 2 p2 2 p3 2 p2 σx z = , 2 p3 2 p1 \u0001 \u0002 2 p1 2 p2 =− . 2 p3\n\n(10.5) (10.6) (10.7)\n\nIt is important to recognize why Eq. (10.7) is true. From Chapter 5 we have \u0002 \u0001 2 p1 2 p2 = θN PN 2 p1 (1)2 p3 (2)2 p2 (3), (10.8) 2 p3 except for normalization. Since N is a column antisymmetrizer, if we interchange 2 p1 (1)2 p3 (2), the sign of the whole function changes, and this standard tableaux function has 2 A2 symmetry. The spatial projector for A2 symmetry may be constructed from Table 10.1, e A2 = 1/4[I + C2 − σx z − σ yz ], and we see that\n\n\u0001 e A2\n\n2 p1 2 p3\n\n2 p2\n\n\u0002\n\n\u0001 =\n\n2 p1 2 p3\n\n2 p2\n\n(10.9)\n\n\u0002 .\n\n(10.10)\n\nThe second standard tableaux function \u0002 \u0001 2 p1 2 p3 2 p2 is not a pure symmetry type; in fact, it is a linear combination of 2A2 and 2B2 . Since there cannot be three linearly independent functions from these tableaux, the two 2 A2 functions must be the same, and we do not need the second standard tableaux function for this calculation. The e A2 operator may be applied to this tableau to obtain the result in a less formal fashion, \u0003\u0001 \u0001 \u0002 \u0002 \u0001 \u0002\u0004 1 2 p1 2 p3 2 p3 2 p1 A 2 2 p1 2 p3 e = − , (10.11) 2 p2 2 p2 2 p2 2 where we have a nonstandard tableau in the result. Again, the methods of Chapter 5 come to our aid, and we have \u0002 \u0003\u0001 \u0002 \u0001 \u0002\u0004 \u0001 2 p1 2 p3 2 p1 2 p2 2 p3 2 p1 = − , (10.12) 2 p2 2 p2 2 p3\n\n128\n\n10 Four simple three-electron systems\n\nTable 10.3. Results of smaller VB calculations. −116.433 248 63 au SCF energy 32-function MCVB – d-functions removed −116.470 007 69 au MCVB energy 1.000 eV Apparent correlation energy \u0001 \u0002 2 p1 2 p2 0.9086 EGSO pop. of 2 p3 4-Function MCVB – 2 p1 , 2 p2 , 2 p3 only −116.461 872 28 au MCVB energy 0.779 eV Apparent correlation energy \u0001 \u0002 2 p1 2 p2 0.9212 EGSO pop. of 2 p3 2-function VB – 2 p1 , 2 p2 , 2 p3 covalent only −116.413 426 76\n\nEnergy\n\nand substituting this result into Eq. (10.11), we obtain \u0001 \u0001 \u0002 \u0002 1 2 p1 2 p2 A 2 2 p1 2 p3 e = . 2 p2 2 2 p3\n\n(10.13)\n\nOur ability to represent the wave function for allyl as one standard tableaux function should not be considered too important. If we had ordered our 2 p orbitals differently with respect to particle labels, there are cases where the 2A2 function would require using both standard tableaux functions. This happens when we consider the most important configuration using HLSP functions. The two Rumer diagrams are shown with dots to indicate the extra electron.\n\np2 p1\n\np2 p3 p1\n\np3\n\nTransforming our wave function to the HLSP function basis,1 we obtain \u0003\u0001 \u0001 \u0002 \u0002 \u0004 2 p2 2 p1 2 p2 2 p3 2 A2 = 0.41115 − + ···. 2 p3 2 p1 R R\n\n(10.14)\n\nwhere we have used Rumer tableaux (see Chapter 5). We emphasize that the EGSO populations are the same regardless of the basis. In Table 10.3 we give data for smaller calculations of the allyl π system. As expected, the MCVB energies increase as fewer basis functions are included, the 1\n\nDetails of this sort of calculation are given in the next section.\n\n129\n\napparent correlation energy decreasing by about 0.5 eV. In fact, unlike the case with H2 , the covalent only VB energy is above the SCF energy. This is a frequent occurrence in systems where resonance occurs between equivalent structures. It arises because of the delocalization tendencies of the electrons. We will take this question up in greater detail in Chapter 15 when we discuss benzene. In actuality, the two smaller correlation energies shown in Table 10.3 are not very significant, since the AO basis is really different from that giving the SCF energy. What is significant is the relative constancy of the EGSO weight for the most important configuration. Since there are only four terms, we give the whole wave function for the smallest calculation. In terms of standard tableaux functions one obtains \u0001 \u0002 2 p1 2 p2 2 A2 = 0.730 79 2 p3 \u0003\u0001 \u0002 \u0001 \u0002\u0004 2 p3 2 p3 2 p1 2 p1 − + 0.140 64 2 p3 2 p1 \u0002 \u0001 \u0002\u0004 \u0003\u0001 2 p2 2 p2 2 p2 2 p2 − + 0.139 95 2 p3 2 p1 \u0003\u0001 \u0002 \u0001 \u0002\u0004 2 p1 2 p1 2 p3 2 p3 + 0.061 32 − . (10.15) 2 p2 2 p2 The HLSP function form of this wave function is easily obtained with the method of Section 5.5.5, \u0003\u0001 \u0001 \u0002 \u0002 \u0004 2 p2 2 p1 2 p2 2 p3 2 A2 = 0.411 88 − 2 p3 2 p1 R R + three other terms the same as in Eq. (10.15).\n\n(10.16)\n\nThe reader will recall that a given configuration has different standard tableaux functions and HLSP functions if and only if it supports more than one standard tableaux function (or HLSP function). It will be instructive to detail the calculations leading from Eq. (10.15) to Eq. (10.16). This provides an illustration of the methods of Section 5.5.5.\n\n10.1.2 Example of transformation to HLSP functions The permutations we use are based upon the particle label tableau \u0001 \u0002 1 3 , 2\n\n130\n\n10 Four simple three-electron systems\n\nand, therefore, 1/ N PN 6\n\n\u0005 = 1/3 I − (12) + 1/2(13) + 1/2(23) \u0006 − 1/2(123) − 1/2(132) ,\n\n/3 N P = 1/3[I − (12) + (13) − (132)].\n\n1\n\n(10.17) (10.18)\n\nThe standard tableaux for the present basis are \u0001\n\n2 p1 2 p3\n\n2 p2\n\n\u0002\n\n\u0001 and\n\n2 p3\n\n2 p1 2 p2\n\n\u0002 ;\n\nit should be clear that the permutation yielding the second from the first is (23). Thus, the permutations of the sort defined in Eq. (5.64) are {πi } = {I , (23)}, and we obtain \b \u0007 1 1/2 , (10.19) M= 1 /2 1 where we have used an N PN version of Eq. (5.73), and the numbers are obtained from the appropriate coefficient in Eq. (10.17). The Rumer tableaux may be written \u0001\n\n2 p1 2 p3\n\n2 p2\n\n\u0001\n\n\u0002 and R\n\n2 p2\n\n2 p3 2 p1\n\n\u0002 R\n\nand the {ρi } set is {I , (12)}. Thus the matrix B from Eq. (5.126) is \u0001\n\n\u0002 −1 , −1\n\n1 B= 0\n\n(10.20)\n\nand A from Eq. (5.128) is \u0007 A = B −1 M =\n\n1\n\n/2\n\n−1/2\n\n−1/2\n\n−1\n\n\b .\n\n(10.21)\n\nWe also give the inverse transformation \u0007 A\n\n−1\n\n=\n\n4/ 3 −2/3\n\n−2/3 −2/3\n\n\b .\n\n(10.22)\n\n131\n\nThe results of multiplying Eqs. (10.17) and (10.18) by (23) and (12), respectively, from the right are seen to be \u0006 \u0005 /6 N PN (23) = 1/3 1/2I − 1/2(12) − 1/2(13) + (23) − (123) + 1/2(132) ,\n\n1\n\n1/ N P(12) 3\n\n= 1/3 [−I + (12) + (123) − (23)] ,\n\nand we obtain \u0005\n\n\u0006\n\n1/ N P 1/ I − 1/ (12) 3 2 2 \u0005 \u0006 1/ N P − 1/ I − (12) 3 2\n\n= 1/6 N PN ,\n\n(10.23)\n\n= 1/6 N PN (23).\n\n(10.24)\n\nFor completeness we also give the inverse transformation: \u0005\n\n\u0006 − 2/3(23) = 1/3 N P, \u0005 \u0006 1/ N PN − 2/ I − 2/ (23) = 1/ N P(12). 6 3 3 3 1/ N PN 4/ I 6 3\n\n(10.25) (10.26)\n\nWe now return to the problem, and, using the first row of the matrix in Eq. (10.21), we see that \u0001\n\n2 p2\n\n2 p1 2 p3\n\n\u0002\n\n1 = 2 1 = 2\n\n\u0003\u0001 \u0003\u0001\n\n2 p1 2 p3 2 p1 2 p3\n\n2 p2\n\n\u0002\n\n\u0001\n\n2 p3 − 2 p1 R \u0001 \u0002 2 p2 2 p2 − 2 p1 R\n\n2 p2\n\n\u0002 \u0004 ,\n\n(10.27)\n\n.\n\n(10.28)\n\nR\n\n2 p3\n\n\u0002 \u0004 R\n\nThis result does not quite finish the problem, however, in that it deals with unnormalized functions. The coefficients that we show are given assuming the tableau functions of either sort are individually normalized to 1. We must therefore consider some normalization integrals. The normalization and overlap integrals of the two standard tableaux functions may be written as a matrix\n\nst f Si j = 2 p1 2 p2 2 p3 πi−1 θ N PN π j 2 p1 2 p2 2 p3 , \u0001\n\nS\n\nst f\n\n\u0002 0.365 144 70 0.182 572 36 = . 0.313 656 66\n\n(10.29)\n\n132\n\n10 Four simple three-electron systems\n\nThe corresponding normalization and overlap integrals for the HLSP functions are then obtained with the transformation of Eq. (10.22), \u0001 \u0002 0.463 976 0 −0.266 313 37 R S = . (10.30) 0.463 976 03 It is seen that the two diagonal elements of S R are equal, reflecting the symmetrical equivalence of the two Rumer tableaux and diagrams. The coefficients in the wave functions Eqs. (10.15) and (10.16) are all appropriate for each individual tableau √ st f function’s being normalized to 1. Therefore, (1/ S11 )θN PN 2 p1 2 p2 2 p3 is a normalized standard tableaux function, with a similar expression for the HLSP functions. In terms of normalized tableau functions we have \n\n\u0001 \u0002u \u0001 \u0002u R 1 S 1 2 p1 2 p2 2 p1 2 p2 11  1 = st f 2 p3 2 p3 2 S11 st f R R S11 S11  \u0001 \u0002u 1 2 p2 2 p3  , (10.31) − 2 p1 R R S22 where we have designated unnormalized tableau functions with a superscript “u”. √ st f R We now see that 1/2 S11 /S11 should convert the coefficient of the standard tableaux function in Eq. (10.15) to the coefficient of the HLSP function in Eq. (10.16), i.e., \u0012 1 0.463 976 0 = 0.411 88. (10.32) 0.730 79 × 2 0.365 144 70 For a system of any size, these considerations are tedious and best done with a computer.\n\n10.1.3 SCVB treatment with corresponding orbitals The SCVB method can also be used to study the π system of the allyl radical. As we have seen already, only one of the two standard tableaux functions is required because of the symmetry of the molecule. We show the results in Table 10.4, where we see that one arrives at 85% of the correlation energy from the largest MCVB calculation in Table 10.2. There is no entry in Table 10.4 for the EGSO weight, since it would be 1, of course. The single standard tableaux function is \u0002 \u0001 2 p1 2 p2 , 2 p3\n\n133\n\nTable 10.4. Results of SCVB calculation. −116.433 248 63 au −116.470 933 15 au 1.025 eV\n\nSCF energy SCVB energy Correlation energy\n\nOrbital amplitude 0.4 0.3 0.2 0.1 0.0\n\n−3\n\n−2\n\n−1\n\n0\n\n1 Distance from center (Å)\n\n2\n\n3\n\n4 4\n\n1\n\n2\n\n3\n\n0\n\n−2\n\n−1\n\n−3\n\n−4\n\nDistance from center (Å)\n\nFigure 10.1. The first SCVB orbital fora the allyl radical. The orbital amplitude is given in a plane parallel to the radical and 0.5 A distant.\n\nand the orbitals satisfy σ yz 2 p1 = 2 p3 ,\n\n(10.33)\n\nσ yz 2 p2 = 2 p2 ,\n\n(10.34)\n\nσ yz 2 p3 = 2 p1 ,\n\n(10.35)\n\neach one consisting of a linear combination of all of the π AOs allowed by symmetry. In terms of HLSP functions the wave function has two terms, of course: \u0003\u0001 0.537 602 87\n\n2 p2 2 p3\n\n2 p1\n\n\u0001\n\n\u0002 R\n\n2 p2 − 2 p1\n\n2 p3\n\n\u0002 \u0004 , R\n\nand the overlap between the two HLSP functions is −0.730 003. In Fig. 10.1 we show an altitude drawing of the orbital amplitude of the first of the SCVB orbitals of the allyl π system. The third can be obtained by merely reflecting this one in the y–z plane of the molecule. It is seen to be concentrated at\n\n134\n\n10 Four simple three-electron systems Orbital amplitude 0.4 0.3 0.2 0.1 0.0\n\n−3\n\n−2\n\n−1\n\n0\n\n1 Distance from center (Å)\n\n2\n\n3\n\n4 4\n\n3\n\n1\n\n2\n\n0\n\n−1\n\n−2\n\n−3 −4\n\nDistance from center (Å)\n\nFigure 10.2. The second SCVB orbital for a the allyl radical. The orbital amplitude is given in a plane parallel to the radical and 0.5 A distant.\n\none end of the radical, with the amplitude falling off fairly rapidly as one moves away from that end. The second SCVB orbital is shown in Fig. 10.2. It is seen to be concentrated on the middle carbon atom with lobes symmetrically placed ona either end carbon. Both of these drawings are plotted for amplitudes in a plane 0.5 A from the plane in which the nuclei occur. Since these are π orbitals, the amplitude is, of course, zero in the nuclear plane. 10.2 The He+ 2 ion The He+ 2 ion has the archetype three-electron bond originally described by Pauling , and this section gives a description of MCVB calculation and SCVB treatments for this system. All of these use a Huzinaga 6-G 1s function split (411), a 4-G 2s function and a pz function with the scale set to 0.9605. We take up the MCVB treatment first. 10.2.1 MCVB calculation The basis described was used to generate one 1s occupied and four virtual RHF orbitals. Using these a full calculation yields 250 standard tableaux functions, which may be combined into 125 functions of 2 u+ symmetry. The results for energy, bond distance, and vibrational frequency are shown in Tablea 10.5. We see that the agreement for De is within 0.1 eV, for Re is within 0.01 A, and for ωe is within 20 cm−1 . Even at the equilibrium nuclear separation, the wave function is dominated\n\n10.2 The He+ 2 ion\n\n135\n\nTable 10.5. Dissociation energy, bond distance, and vibrational frequency from MCVB calculation of He+ 2 .\n\nCalc. Exp.\n\na\n\nDe eV\n\nRe A\n\nωe cm−1\n\n2.268 2.365\n\n1.088 8 1.080 8\n\n1715.8 1698.5\n\nTable 10.6. Energy differences E SCVB − E MCVB for He+ 2 . E(Rmin ) eV\n\nE(R∞ ) eV\n\n1.088\n\n1.214\n\nby the first term, and only the second is of further importance,\n\n\u00132\n\nu+\n\n\u0014\n\n\u0003\u0001\n\n\u0002 \u0001 \u0002\u0004 1sa 1sa 1sb 1sb = 0.967 975 − 1sb 1sa \u0003\u0001 \u0002 \u0001 \u0002\u0004 2sa 2sa 2sb 2sb − 0.135 988 − + ···. 1sb 1sa\n\n(10.36)\n\n10.2.2 SCVB with corresponding orbitals The three orbitals we use are two we label 1sa and 1sb that are symmetrically equivalent and one 2 pσ that has the symmetry indicated. Thus if σh is the horizontal reflection from D∞h we have the transformations σh 1sa = 1sb , σh 1sb = 1sa , σh 2 pσ = −2 pσ . When these orbitals are optimized, the energies of the SCVB wave functions are higher, of course, than those of the full MCVB wave functions. We show the differences at the equilibrium and infinite internuclear separations in Table 10.6. The energy curves are parallel within ≈0.1 eV, but the SCVB energy is about 1.1 eV higher. Because of the spatial symmetry there is only one configuration (as with allyl), and in this case the HLSP function function is the simpler of the two. We have for\n\n136\n\n10 Four simple three-electron systems Amplitude 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 2 1 −2\n\n0\n\n−1\n\n0 z-direction (Å)\n\n−1 1\n\nFigure 10.3. The first SCVB orbital for the x–z plane, which contains the nuclei.\n\nthe two forms SCVB\n\n\u0001\n\n1sa = 2 pσ\n\n1sb\n\nHe+ 2\n\nx-direction (Å)\n\n2 −2\n\nion. The orbital amplitude is given in the\n\n\u0002 , R\n\n\u0001\n\n1sa = 1.282 557 82 2 pσ\n\n1sb\n\n\u0002\n\n\u0001\n\n1sa − 0.803 090 63 1sb\n\n2 pσ\n\n\u0002 ,\n\n(10.37)\n\nwhere each of the tableaux functions is individually normalized. The second standard tableaux function on the right hand side of Eq. (10.37) is of pure 2 g+ symmetry, as can be seen by methods we have used above. Thus the other tableau is of mixed symmetry, and the second term subtracts out the “wrong” part from the first.2 The 1sa orbital is shown in Fig. 10.3, and it is seen to be located predominately on one of the nuclei. We may compare this orbital to that for H2 given in Section 3.2.2. The present one is seen to be more localized near the nuclei, reflecting the larger nuclear charge for He. The 1sb orbital is obtained by reflecting with σh . The 2 pσ orbital is shown as an altitude drawing in Fig. 10.4, where it is seen to have the symmetry indicated by its symbol. 10.3 The valence orbitals of the BeH molecule In this section we give the results of MCVB and SCVB treatments of BeH using a conventional 6-31G∗∗ basis.3 Although there are some similarities to the He+ 2 ion, the lack of g–u symmetry in this case introduces a number of interesting 2 3\n\nThe relative values of the coefficients in Eq. (10.37) are not determined by the variation theorem, but are imposed by the symmetry and overlaps. That is, there is a set of d orbitals on Be and a set of p orbitals on H.\n\n10.3 The valence orbitals of the BeH molecule\n\n137\n\nTable 10.7. Dissociation energy, bond distance, vibrational frequency, and electric dipole moment from full MCVB calculation of BeH.\n\nCalc. Exp.\n\na\n\nDe eV\n\nRe A\n\nωe cm−1\n\nµD\n\n1.963 2.034\n\n1.348 6 1.342 6\n\n211 7.1 206 0.8\n\n0.262\n\nAmplitude 1.5 1.0 0.5 0.0 −0.5 −1.0 −1.5 2 1 −2\n\n0\n\n−1\n\n0 z-direction (Å)\n\n1\n\n−1\n\nx-direction (Å)\n\n2 −2\n\nFigure 10.4. The third SCVB orbital for the He+ 2 ion. The orbital amplitude is given in the x–z plane.\n\ndifferences. The use of corresponding orbitals does not arise, since there is no symmetry requirement to be satisfied.\n\n10.3.1 Full MCVB treatment A full MCVB calculation on BeH with the above basis yields 504 doublet standard tableaux functions, and these combine into 344 2 + symmetry functions. In Table 10.7 we give some details of the results with experimental values for comparison. The calculated De is within 0.1 eV of the experimental value, the values of Re are quite close, and the vibrational frequency is within 60 cm−1 . An experimental value for the dipole moment is apparently not known. The principal configurations in the wave function are shown as HLSP functions in Table 10.8 and as standard tableaux functions in Table 10.9. Considering the HLSP functions, the first is the ground state configuration of the separated atoms, the next two are bonding functions with the s–p hybrid of Be and the fourth contributes polarization to the Be2 pz component. The corresponding entries in the third and fourth columns of Table 10.9 do not include the tableau function with the 3 pz orbital,\n\n138\n\n10 Four simple three-electron systems\n\nTable 10.8. The principal HLSP function configurations in BeH at R = Re . Coef.\n\nTableau \u0002 2s 2s 1s \u0002R \u0001 1s 2 pz 2s \u0001 \u0002R 2s 1s 2 pz \u0001 \u0002R 1s 3 pz 2s R \u0001\n\n0.332 76 0.295 79 −0.280 28 0.126 09\n\nTable 10.9. Large components of the wave function for distances on either side of the cross-over from negative to positive dipole moments. The n = 1 orbitals are on H and the n = 2 or 3 orbitals are on Be. The Be1s orbitals from the core are omitted in the tableaux. a\n\nR = 1.0 A Coef.\n\nTableau \u0001\n\n0.552 01 −0.272 34 0.187 00 0.107 65 −0.079 44 0.059 29\n\nR = Re Coef. \u0002\n\n2s 1s 2 pz \u0001 \u0002 2s 2s 1s \u0002 \u0001 2s 1s 3 pz \u0002 \u0001 2 pz 2 pz 1s \u0001 \u0002 1s 3 pz 2 pz \u0001 \u0002 1s 1s 2 pz\n\nTableau \u0001\n\n0.530 40 −0.332 76 0.152 74 0.117 30 0.103 57 0.089 61\n\n\u0002 2s 1s 2 pz \u0001 \u0002 2s 2s 1s \u0001 \u0002 2s 1s 3 pz \u0001 \u0002 1s 1s 2s \u0002 \u0001 2 pz 2 pz 1s \u0001 \u0002 1s 1s 2 pz\n\nbut do include the ionic functions involving 1s 2 2s. Overall, therefore the two different sorts of functions give similar pictures of the bonding in this system. We now consider the dipole moment, which we will analyze in terms of standard tableaux functions only. Our calculated value for the moment is fairly small, but it is\n\n10.3 The valence orbitals of the BeH molecule\n\n139\n\n1.0\n\n0.8\n\nDipole moment (debye)\n\n0.6\n\n0.4 0.2 0.0\n\n1.349\n\n−0.2 −0.4 0.5\n\n1.0\n\n1.5 2.0 Be—H distance (Å)\n\n2.5\n\n3.0\n\nFigure 10.5. The dipole moment function from the MCVB calculation of BeH. The vertical dotted line marks the calculated equilibrium internuclear distance.\n\nin the “right” direction according to the electronegativity difference. The molecule is oriented with the Be atom in the positive z-direction in these calculations. We show the dipole as a function of distance in Fig. 10.5. One sees that the Re of BeH is fairly close to a point at which the moment changes sign to the “wrong” direction. In Table 10.9 we give the major tableau in the wave function on either side of the crossover to see how the dipole moment depends upon distance. At both distances the first three tableaux are covalent and do not have large moments. We saw, however, in Chapter 8 that covalent functions could have small moments in the direction of less diffuse orbitals. The biggest difference here is the overlap of the H1s with the Be1s, and we expect the covalent functions to have small negative moments. The H1s and Be2s orbitals are not so different in size and will not contribute so much. The fourth function at Re is ionic with a large moment in the positivea direction. This sort of function does not come in until the sixth place at R = 1.0 A and then with a coefficient only half the size. Thus the main contribution to a positive moment recedes as the distance gets smaller. 10.3.2 An SCVB treatment The allyl radical and the He+ 2 ion both have end-for-end symmetry and thus the corresponding orbital SCVB treatment is applied. Consequently, there was only one tableau function in each of those cases. BeH is different in this regard. In the\n\n140\n\n10 Four simple three-electron systems Orbital amplitude\n\n0.2 0.0 −0.2 −0.4 −0.6 2 1 −2\n\n−1\n\n0 z-distance (Å)\n\n1\n\n2\n\n−2\n\n−1\n\n0 x-distance (Å)\n\nFigure 10.6. The first SCVB orbital for the BeH molecule and associated with the Be nucleus. This has the general appearance of an s– p hybrid pointed toward the H atom, and we denote it the inner hybrid, h i . The orbital amplitude is given in the x–z plane, which contains the nuclei.\n\nOrbital amplitude 0.2 0.0 −0.2 −0.4 −0.6 2 1 −2\n\n−1\n\n0 z-distance (Å)\n\n1\n\n2\n\n−2\n\n−1\n\n0\n\nx-distance (Å)\n\nFigure 10.7. The second SCVB orbital for the BeH molecule and associated with the Be nucleus. This has the general appearance of an s– p hybrid pointed away from the H atom, and we denote it the outer hybrid, h o . The orbital amplitude is given in the x–z plane, which contains the nuclei.\n\nwave function there are three different orbitals and, consequently, two independent tableaux (of either sort) and an extra variation parameter associated with their mixing. To a considerable extent we may associate two of the orbitals with the Be nucleus and one with the H nucleus. Altitude drawings of the three orbitals are shown in Figs. 10.6, 10.7, and 10.8. They are all orthogonal to the Be1s core orbitals and this results in the sharp negative peak at the Be nucleus. The two Be orbitals in Figs. 10.6 and 10.7 have the general characters of s– p hybrids\n\n10.4 The Li atom\n\n141\n\nTable 10.10. Energy differences between SCVB and MCVB treatment of BeH. E(Re ) eV\n\nE(R∞ ) eV\n\n0.653\n\n0.772\n\nOrbital amplitude\n\n0.6 0.4 0.2 0.0 −0.2 −0.4 2 1 −2\n\n−1\n\n0 z-distance (Å)\n\n1\n\n2\n\n−2\n\n−1\n\n0\n\nx-distance (Å)\n\nFigure 10.8. The SCVB orbital for the BeH molecule associated with the H nucleus. The orbital amplitude is given in the x–z plane, which contains the nuclei.\n\npointed toward and away from the H atom, respectively. We denote these the inner and outer hybrids, h i and h o . The third orbital shown in Fig. 10.8 and located mainly on the H atom we will denote simply by 1s. The energy differences at Re and ∞ geometries are in Table 10.10, where it is seen that the two curves are parallel within about 0.12 eV. The extra variation parameter with the two tableaux that occur here is shown by the coefficients in Table 10.11, where we use the orbital symbols defined above. The representation of the total wave function is rather similar with the two different sorts of tableau functions. Nevertheless, the HLSP functions have a slight edge in that the perfect pairing function between the inner hybrid and the H1s is a better single-function approximation to the wave function than any of the other tableaux. This is very clear from the EGSO weights that are given. 10.4 The Li atom As we stated in Chapter 4, the Li atom has a much deeper and narrower potential for three electrons than does the allyl radical. One consequence is that the nuclear attraction part of the dynamical effects is relatively more important in Li. Because\n\n142\n\n10 Four simple three-electron systems\n\nTable 10.11. Coefficients and tableaux for standard tableaux functions and HLSP functions for SCVB treatment of BeH. The orbital symbols are defined in the text. Standard tableaux functions Coef.\n\nEGSO Wt\n\nTableau \u0001\n\n1.003 58\n\n0.877 5 \u0001\n\n−0.235 80\n\n0.122 5\n\nHLSP functions\n\nhi ho\n\n1s\n\nhi 1s\n\nho\n\nCoef.\n\nEGSO Wt\n\n\u0002\n\nTableau \u0002 h i 1s ho R \u0002 \u0001 1s h o hi R\n\n\u0001 1.009 08\n\n0.9991\n\n0.011 28\n\n0.0009\n\n\u0002\n\nit resembles the SCF result, we, in this case, take up the SCVB wave function first. These Li atom calculations used Huzinaga’s (10/73) basis set, further split to (10/73/5221) to yield four basis functions. This is an “s” only basis, so our treatments will not produce any angular correlation, but the principles are well illustrated, nevertheless.\n\n10.4.1 SCVB treatment There is no added symmetry in this example to cause one of the standard tableaux functions to disappear. Thus, the SCVB wave function is \u0002 \u0001 \u0002 \u0001 1s 2s 1s 1s +B , (10.38) =A 2s 1s where 1s, 1s , and 2s are three different linear combinations of the four basis functions. In this case the tableaux in Eq. (10.38) can be interpreted as either the standard or the Rumer sort. The energies and wave functions obtained are shown in Table 10.12. We observe that the wave function in terms of HLSP functions is a little simpler in that the function with 1s and 1s coupled to singlet is very nearly all of it. It has been observed that correlation energies are frequently close to 1 eV per pair of electrons, particularly in atoms. The value in Table 10.12 is only a third of that. This is to be expected since we have included the possibility of only radial correlation in our wave function.4 4\n\nWe do not go into this, but only observe that there are three directions in which electrons may avoid one another. In many cases each direction contributes approximately 1/3 of the correlation energy.\n\n10.4 The Li atom\n\n143\n\nTable 10.12. Results of SCVB calculation. The A and B in line 4 are defined in Eq. (10.38). −7.432 300 22 au −7.444 280 860 8 au 0.326 eV Standard tableaux Rumer tableaux\n\nSCF energy SCVB energy Correlation energy A B 1.012 989 45 −0.156 041 34 0.993 746 95 −0.006 921 54\n\n0.40 1s\n\n0.35\n\nr × (orbital amplitude)\n\n0.30 1s'\n\n0.25 0.20\n\n2s 0.15 0.10 0.05 0.00 −0.05 0.0\n\n0.5\n\n1.0\n\n1.5 2.0 2.5 Radial distance (bohr)\n\n3.0\n\n3.5\n\n4.0\n\nFigure 10.9. The three SCVB orbitals for the Li atom. The orbital amplitudes times the radial coordinate are shown.\n\nIn terms of the primitive split Gaussian basis, we obtain for the three SCVB orbitals 1s = 0.670 361 39s1 + 0.421 010 56s2 + 0.094 469 35s3 − 0.094 700 67s4 , 1s = 0.196 626 76s1 + 0.861 563 17s2 − 0.088 741 86s3 + 0.089 095 19s4 , 2s = −0.071 164 88s1 − 0.102 382 30s2 + 0.683 095 99s3 + 0.365 043 02s4 , where s1 , . . . , s4 are the four functions in increasing order of diffuseness. These SCVB orbitals are plotted in Fig. 10.9 where we see that the inner shell orbitals are split and show radial correlation. The outer orbital has one radial node like the familiar 2s orbital of SCF theory, but in this case it is exactly orthogonal to neither of the inner orbitals.\n\n144\n\n10 Four simple three-electron systems\n\nTable 10.13. Results of MCVB calculation for Li. −7.432 300 22 au −7.446 822 77 au 0.395 eV 0.9746\n\nSCF energy MCVB energy Correlation energy \u0001 1s EGSO pop. of 2s\n\n1s\n\n\u0002\n\n10.4.2 MCVB treatment We now describe the full MCVB treatment of the Li atom using a basis consisting of the three SCVB orbitals 1s, 1s , and 2s to which we add the primitive s4 for completeness. The Hamiltonian matrix is 20 × 20, and the energy is of course the same regardless of the sort of CI performed, so long as it is “full”. The results are shown in Table 10.13, where we see that the SCVB calculation arrived at about 83% of the correlation energy available from this basis. The EGSO population of the principal SCVB standard tableaux function is very high. The additional correlation energy from the MCVB is principally from intershell correlation and is produced by the accumulation of a number of configurations with fairly small coefficients.\n\n11 Second row homonuclear diatomics\n\nFor many years chemists have considered that an understanding of the theory of the bonding of the homonuclear diatomic molecules from the second row of the periodic table is central to understanding all of bonding, and we consider these stable molecules first from our VB point of view. The stable molecules with interesting multiple bonds are B2 , C2 , N2 , O2 , and F2 . Of course, F2 has only a single bond by ordinary bonding rules, but we include it in our discussion. Li2 is stable, but, qualitatively, is similar to H2 . The question of the existence of Be2 is also interesting, but is really a different sort of problem from that of the other molecules. Of the five molecules we do consider, B2 and C2 are known only spectroscopically, while the other three exist at room temperature all around us or in the laboratory.\n\n11.1 Atomic properties Before we launch into the discussion of the molecules, we examine the nature of the atoms we are dealing with. As we should expect, this has a profound effect on the structure of the molecules we obtain. We show in Fig. 11.1 the first few energy levels of B through F with the ground state taken at zero energy. The L-S term symbols are also shown. The ground configurations of B and F each support just one term, 2 P, but the other three support three terms. All of these are at energies below ≈4.2 eV (relative to their ground state energies). The states from the ground configuration are connected with lines marked with a G in Fig. 11.1. We also consider two sorts of excited states. r Those arising from configurations that differ from the ground configuration by one\n\n2s →2p transition. This may be called a valence excited configuration. The lines connecting these states are marked with a “V”. r Those arising from a 2 p → 3s transition. This is the first configuration of a Rydberg-like series. The lines connecting these states are marked with an “R”. 145\n\n146\n\n11 Second row homonuclear diatomics 20\n\n4\n\nP\n\n3\n\nP\n\n15 2\n\nS\n\n4\n\nEnergy (eV)\n\nP\n\n3\n\n10\n\nP\n\n3\n\nV S\n\nP S\n\n5 1\n\nG S 1\n\nG D 2 G P\n\nB\n\n2\n\nP\n\n4\n\n0\n\nS\n\n3\n\nV 2D R 2S V P\n\nP\n\nD\n\n2\n\n5\n\n4\n\n5\n\n1\n\nS\n\n1 2\n\nD\n\nD\n\n3\n\nP\n\n4\n\nS\n\n3\n\n2\n\nC\n\nN\n\nO\n\nF\n\nP\n\nP\n\nFigure 11.1. Energy levels for five atoms.\n\nIn the cases of B and C three terms have been observed from the valence excited configuration. Apparently only one has been observed for the others. Only one state has been plotted for the Rydberg state. We note that the energy of the lowest valence excited state crosses the lowest Rydberg state energy between C and N. As two identical atoms from the group approach one another to form a molecule, the ground and several excited states mix together in the final structure. The general point we make is that we expect valence states to be more important than Rydberg states in bonding,1 and the atoms at the start of the list have much lower valence state energies than those later, and the valence states should mix in more easily upon molecule formation. The energy necessary for the excitation of the atomic states must come from the bonding energy. We analyze the bonding of these molecules using the above general ideas. 11.2 Arrangement of bases and quantitative results A goal of this chapter is to show, for the diatomic molecules under discussion, both the capability of the VB method in providing quantitative estimates of molecular properties and its capability of giving qualitative pictures of the bonding. The quantitative results are illustrated in Table 11.1, where we give values for Re , the equilibrium bond distance, and De ,2 determined theoretically with STO3G, 6-31G, 1 2\n\nAmong other things, the 3s orbital is too diffuse to overlap other orbitals well. D0 + ωe /2, see Huber and Herzberg.\n\n11.2 Arrangement of bases and quantitative results\n\n147\n\nTable 11.1. Bond distances and De for some second row homonuclear diatomics. Source a\n\nB2\n\nC2\n\nN2\n\nO2\n\nF2\n\nExp.\n\nRe A De eV\n\n1.590 3.085\n\n1.243 6.325\n\n1.098 9.905\n\n1.208 5.214\n\n1.412 1.659\n\nSTO3G\n\nRe De\n\n1.541 3.778\n\n1.261 6.851\n\n1.198 6.452\n\n1.263 4.529\n\n1.392 2.082\n\n6-31G\n\nRe De\n\n1.623 2.781\n\n1.243 6.164\n\n1.137 7.666\n\n1.290 3.830\n\n1.502 1.565\n\n6-31G∗\n\nRe De\n\n1.610 3.053\n\n1.261 6.331\n\n1.129 8.768\n\n1.257 4.580\n\n1.436 1.867\n\nTable 11.2. Number of symmetry functions for 6-31G and 6-31G∗ basis sets. Basis\n\nB2\n\nC2\n\nN2\n\nO2\n\nF2\n\n6-31G 6-31G∗\n\n777 1268\n\n991 1864\n\n1086 1812\n\n697 1188\n\n109 131\n\nand 6-31G∗ bases compared with those from experiment. All of the VB calculations here have the two 1s 2 shells in each atom completely closed, and, in order to save space, none of the tableaux shown in tables later in this chapter include these orbitals explicitly. In Chapter 9 we gave a somewhat generic description of the way we arrange basis sets. More details are given here. STO3G Each of the treatments may be described as a full valence calculation. The details of numbers of VB structures is given below in Table 11.4. 6-31G The VB structure basis is a full valance set augmented by structures involving a single excitation from one valence orbital to one virtual orbital, using all possible combinations of the excitation (outside the 1s shells). Table 11.2 shows the number of symmetry functions (the dimension of the H and S matrices) for each case. 6-31G∗ The AO basis in this case is the same as that for the 6-31G set with a set of d orbitals added. In these calculations the dσ and the dπ orbitals are included in the virtual orbital set in which single excitations are included in generating structures. The dδ orbitals were not used. The inclusion of these d orbitals provides polarization when the molecule is formed.\n\nAn examination of the values in Table 11.1 shows a variety of results for the different molecules and bases. In general, the values of Re are too large except for\n\n148\n\n11 Second row homonuclear diatomics a\n\nB2 and F2 values using the STO3G basis. The 6-31G∗ values average about 0.03 A too large. The calculated De values vary somewhat more randomly. In general the values for B2 and C2 are the closest to the experimental ones, followed by those for F2 . The values of De for N2 are the hardest to obtain followed by those for O2 . With the 6-31G∗ basis set, the calculated De values for B2 , C2 , and F2 are all within 0.2 eV of experiment, while N2 is off by more than 1 eV. It is not clear why N2 presents such a challenge.\n\n11.3 Qualitative discussion If one wishes a qualitative picture of the bonding and structure of a molecule it has become evident that this is most easily determined from a reasonably minimal basis calculation. As one increases the size of the basis, the set of important structures remains reasonably stable, but there is frequently some jockeying around. As we argued earlier, the STO3G set was historically optimized to be appropriate for molecular geometries, therefore it is, perhaps, not surprising that it gives a reasonable picture of molecular structure, even when taken over to the VB method. In spite of this bias toward the molecular state, the STO3G basis also gives a good account of the states that the system migrates into as the separation between the atoms goes to ∞. In the present section we therefore examine the wave functions obtained with this basis for the molecules we are discussing to determine the VB picture of their bonding. In Table 11.3 we show the ground states of the atoms and the ground state of the diatomic molecules they form. Except possibly for B2 , all of these are well established spectroscopically. This same table shows the total degeneracy for two infinitely separated atoms. For example, atomic boron is in a 2 P o state, which, ignoring spin-orbit coupling (i.e., using the ESE), is six-fold degenerate. Each of these states can couple with each in another atom, so, all together, we expect 6 × 6 = 36 different states to come together at ∞. This will include singlets, triplets, \u0006, \u0007, and \b states with various g and u and + and − labels, but the number will add up to 36. We are not interested in discussing most of these but the interested reader can make calculations for each of the symmetries with CRUNCH. As we stated earlier, all of the configurations we use have the two 1s 2 shells occupied. Thus if we allow all possible occupations of the remaining eight valence orbitals in the STO3G basis, we may speak of a full valence VB. We have this same number of valence orbitals in all of the molecules we treat this way. As we pass from B2 to F2 , the number of electrons that the orbitals must hold increases, however, causing a considerable variation in the number of allowed states. We show the\n\n11.3 Qualitative discussion\n\n149\n\nTable 11.3. Atomic ground states and asymptotic molecular symmetries. Molecular states Atom\n\nDeg. at ∞\n\nAtomic state 2\n\nPo Pe 4 o S 3 e P 2 o P\n\nB C N O F\n\n36 81 16 81 36\n\n3\n\nBonding state \u0006g− \u0006g+ 1 + \u0006g 3 − \u0006g 1 + \u0006g 3 1\n\nTable 11.4. Statistics of full valence calculations for STO3G basis. Molecule\n\n2S + 1\n\nNumber of electronsa\n\nNumber of constellations\n\nNumber of symmetry functions\n\nB2 (3 \u0006g− ) C2 (1 \u0006g+ ) N2 (1 \u0006g+ ) O2 (3 \u0006g− ) F2 (1 \u0006g+ )\n\n3 1 1 3 1\n\n6 8 10 12 14\n\n18 35 76 18 8\n\n41 84 102 30 8\n\na\n\nOutside a 1sa2 1sb2 core.\n\nnumbers in Table 11.4, where it is seen that the total size of the variational problem is a maximum for the case of N2 . We now proceed to a discussion of the individual molecules. 11.3.1 B2 The very first question that we might ask is: From our knowledge of the properties of VB functions and knowing that the atom is in a 2 P o state, can we predict the likely ground state symmetry of the molecule? With B2 this may be tricky. We list some conjectures. 1. The very first guess might be that, outside of the two 2s 2 closed subshells, a single σ bond is formed from the two p orbitals in the σ orientation. A singlet state is expected. 2. A more intricate situation arises if the excited configuration, 2s2 p 2 , can come into play. Then the two 2s and two 2 pσ electrons can each form an electron pair bond, but there are still two 2 pπ electrons hanging around.\n\n150\n\n11 Second row homonuclear diatomics\n\nTable 11.5. Principal standard tableaux function structures for B2 at an asymptotic bond distance.\n\nNum.a Tab.b Ci (∞)\n\n1\n\n2\n\n2  2sa 2sa  2sb 2sb  p  xa p yb 0.665 111 24\n\n8  2sa 2sa  pxb pxb  p  xa p yb −0.118 177 21\n\n3 8  pxa pxa  p yb p yb  p  xb p ya 0.020 997 77 \n\na\n\nThe number of terms in the symmetry function that is generated from the tableau shown. (See text.) b These tableau symbols exclude the core orbitals.\n\nTable 11.6. Principal HLSP function structures for B2 at an asymptotic bond distance. 1 Num.a Tab.b Ci (∞)\n\n2  2sa 2sa  2sb 2sb  p  xa p yb R 0.665 111 24 \n\n2 8  2sa 2sa  pxb pxb  p  xa p yb R −0.118 177 21 \n\n3 8  pxa pxa  p yb p yb  p  xb p ya R 0.020 997 77 \n\na\n\nThe number of terms in the symmetry function that is generated from the tableau shown. (See text.) b These tableau symbols exclude the core orbitals.\n\n(a) The two 2 pπ orbitals could form an electron pair bond. Thus we would expect a 1 \bg state with three bonds. (b) The two 2 pπ electrons could arrange themselves in two one-electron bonds, one for the x-direction and one for the y-direction. i. If the two electrons are singlet coupled, we have a 1 \u0006g+ state. ii. If the two electrons are triplet coupled, we have a 3 \u0006g− state.\n\nWe have collected some results for standard tableaux functions and HLSP functions in Tables 11.5 and 11.6. The structure of these tables will be repeated in several later sections, and we describe it here. 1. The unlabeled row gives the ordinal number of the following entries. 2. The “Num.” row gives the number of tableau functions in a symmetry function. Thus the “2” for column one indicates that the tableau function below it is the first of a two function sum that has the correct 1\u0006g+ symmetry. 3. The “Tab.” row gives the actual tableaux in terms of AO symbols.\n\n11.3 Qualitative discussion\n\n151\n\nTable 11.7. Principal standard tableaux function structures for B2 at energy minimum bond distance.\n\nNum. Tab. Ci (Rmin )\n\n1\n\n2\n\n2  2sa 2sb  pza pzb   p xa p yb 0.215 367 05\n\n2  2sa 2sa  2sb pza  p  xb p yb −0.206 418 77\n\n3\n\n4\n\n2  2sa 2sa  2sb 2sb  p  xa p yb 0.198 518 19 \n\n2  2sa 2sb  pza p yb  p  zb pxa 0.115 676 19 \n\n4. The Ci are the coefficients in the wave function corresponding to the tableaux. The values assume that the actual tableau function is normalized to 1 as well as the overall wave function. The “∞” indicates the values in this case are for large R-values. Elsewhere, other arguments appear.\n\nWe return to consideration of the entries in these tables, where we give the principal structures for two B atoms at long distance. It can be seen that we did not need to give two tables since they are the same. The reader should recall that the two sorts of VB functions are the same when there is only one standard tableau, as is the case here. Focusing on Table 11.5 we see that the principal structure involves the orbitals of the atomic configuration 2s 2 2 p on each atom. The relatively small coefficient is caused by the fact that the principal structure is really \n\n2sa  2sb  0.665 111 24   pxa p yb\n\n  2sa 2sa   2sb   2sb  +  pxb\n\n 2sa  2sb   , \n\np ya\n\ni.e., there are two terms in the symmetry function. If the symmetry function were √ normalized in the form ( f + g)/ 2 the coefficient would be ≈0.941. We also emphasize that the footnotes in Tables 11.5 and 11.6 apply equally well to all of the tables in this chapter that show tableaux and coefficients. The second and third terms involve excited states that produce electron correlation (particularly of the angular sort) in the closed 2s 2 shells of the atoms. Therefore, the wave function for the asymptotic geometry is essentially the product of two atomic wave functions. When the two atoms are at the geometry of the energy minimum the results are as shown in Tables 11.7 and 11.8, where, as before, we give both the standard tableaux function and HLSP function results. It is clear that the wave function is now a more complicated mixture of many structures. In addition, the apparent importance of the structures based upon the values of the coefficients is somewhat different for the\n\n152\n\n11 Second row homonuclear diatomics\n\nTable 11.8. Principal HLSP function structures for B2 at the energy minimum bond distance. 1 Num. Tab. Ci (Rmin )\n\n2  2sa 2sa  2sb 2sb   p xa p yb R 0.198 518 19 \n\n2 2  2sa 2sa  2sb pza  p  xb p yb R −0.134 559 43 \n\n3 2  2sb pza  2sa pzb  p  xa p yb R −0.118 190 06 \n\n4 2  2sa 2sa  2sb 2sb  p  xa p ya R 0.097 154 09 \n\nstandard tableaux function and HLSP function bases. Considering Table 11.7 first, we see that structure 1 consists of orbitals from the excited valence configuration of each atom, 2s2 p 2 . Structure 4 is another of the nine standard tableaux from this arrangement. Structure 2 has the orbitals of one atom in the ground state and one in the excited valence state. Structure 3 is from the two atoms each in their ground states. Thus, the VB picture of the B2 molecule consists of roughly equal parts of these three atomic configurations. There are, of course, many smaller terms leading to electron correlation. The picture from the results of Table 11.8 is not significantly different. Structure 3 from before is now 1 (with the same coefficient, of course), but we have a mixture of the same atomic configurations. The new structures 2 and 3 show standard twoelectron bonds involving 2s and 2 pz orbitals on opposite atoms. This feature is not so clear from the standard tableaux functions. 11.3.2 C2 The ground state of the C atom is 3P from a 2s 2 2 p 2 configuration. In the case of B we saw that the excited valence configuration played an important role in the structures describing the B2 molecule. C2 has more electrons with the possibility of more bonds, and, thus, there may be more tendency for the valence excited configuration to be important in this molecule than in B2 . Our conjectures concerning the lowest state of C2 are as follows. 1. The two 3 P atoms could form a pσ−pσ bond with the two remaining pπ orbitals coupled: (a) 3 \u0006 − as in B2 ; (b) 1 \u0007 to give a doubly degenerate ground state. 2. The two 3 P atoms could form two pπ bonds to produce a 1 \u0006 + ground state. If the valence excited state is important as argued above, these two 5 S states could also couple to 1 \u0006 + interacting strongly with the two pπ bonds.\n\nIt is this last situation that pertains.\n\n11.3 Qualitative discussion\n\n153\n\nTable 11.9. Principal standard tableaux function structures for C2 at an asymptotic bond distance.\n\nNum.\n\n1\n\n2\n\n2  2sa 2sa  2sb 2sb  p pzb  za pxa pxb 0.690 603 71\n\n4  2sa 2sa  p yb p yb  p pzb  za pxa pxb −0.106 757 10 \n\nTab. Ci (∞)\n\n3 2  pxa pxa  pxb pxb  p pzb  za p ya p yb 0.016 503 07 \n\nTable 11.10. Principal HLSP function structures for C2 at an asymptotic bond distance. 1 Num. Tab. Ci (∞)\n\n2  2sa 2sa  2sb 2sb  p pzb  za pxa pxb R 0.398 720 24 \n\n2 2  2sa 2sa  2sb 2sb  p pxa  zb pza pxb R −0.398 720 24 \n\n3 8  2sa 2sa  p yb p yb  p pzb  za pxa pxb R −0.061 636 24 \n\n4 4  2sa 2sa  p yb p yb  p pzb  za pxa pxb R 0.009 528 05 \n\nFirst, however, we examine the asymptotic geometry. The principal structures are shown in Tables 11.9 and 11.10. In the standard tableaux function case structure 1 is one of the possible 1 \u0006 + couplings of two 3 P atoms, and structures 2 and 3 produce electron correlation in the closed 2s shell. The results with HLSP functions are essentially the same with some differences in the coefficients. The apparently smaller coefficients in the latter case result mainly from the larger number of terms in the symmetry functions. We show the standard tableaux function results for the energy minimum geometry in Table 11.11. Here we see that the valence excited configuration has become the dominant structure, and the 1 \u0006g+ coupling of the 3 P atomic ground states is structure 2. Structure 3 is a mixture of the 5S and 3 P states while structure 4 is another of the standard tableau associated with structure 1. We call attention to a significant similarity between structure 1 and the tableau for the 5 S atomic state, which is   2s  2 pz     2 px  . 2 py\n\n154\n\n11 Second row homonuclear diatomics\n\nTable 11.11. Principal standard tableaux function structures for C2 at the energy minimum bond distance. 1 Num. Tab. Ci (Rmin )\n\n1  2sa 2sb  pza pzb  p pxb  xa p ya p yb 0.438 636 13 \n\n2 1  2sa 2sa  2sb 2sb  p pxb  xa p ya p yb 0.293 039 91 \n\n3\n\n4\n\n4  2sb 2sb  p ya p ya   2s pzb  a pxa pxb 0.158 969 69\n\n1  2sa 2sb  pza pxb  p p ya  zb pxa p yb −0.131 203 05\n\nTable 11.12. Principal HLSP function structures for C2 at the energy minimum bond distance. 1 Num. Tab. Ci (Rmin )\n\n1  2sa 2sa  2sb 2sb  p pxb  xa p ya p yb R 0.240 701 83 \n\n2 1  2sb pza  2sa pzb  p pxb  xa p ya p yb R −0.179 769 81 \n\n3 2  2sa 2sa  p yb p yb   2s pza  b pxa pxb R −0.165 030 55 \n\n4 2  2sb 2sb  p ya p ya   2s pzb  a pxa pxb R 0.127 710 28 \n\nThis function is antisymmetric with respect to the interchange of any pair of orbitals. The same pertains to structure 1 of Table 11.11 with respect to either of the columns. Thus the dominant structure is very much two 5S atoms. The results for HLSP functions in Table 11.12 show a somewhat different picture. In this case the dominant (but not by much) structure is the one with two π bonds and structures 3 and 4 provide a σ bond. Structure 2 is the double 5S structure, but, since HLSP functions do not have a close relationship to the actual 5S state as above, there is less importance to just one Rumer coupling scheme.\n\n11.3.3 N2 We commented above that the energies of the first excited valence states of B and C are fairly low and there is a large jump between C and N. The reason for this is principally the Coulomb repulsion energy in the states. For B and C the excited valence state has one less paired orbital than the corresponding ground state, while for N, O, and F the numbers are the same. Since the Coulomb repulsion energy tends to be largest between two electrons in the same orbital, this trend is not surprising.\n\n11.3 Qualitative discussion\n\n155\n\nTable 11.13. Principal standard tableaux function structure for N2 at an asymptotic bond distance. 1 Num.\n\n1  2sa 2sa  2sb 2sb     pza pzb  p pxb  xa p ya p yb 1.000 001 54 \n\nTab.\n\nCi (∞)\n\nTable 11.14. Principal HLSP function structures for N2 at an asymptotic bond distance. 1 Num. Tab.\n\nCi (∞)\n\n1  2sa 2sa  2sb 2sb     pzb pxa  p pxb  za p ya p yb R 0.471 403 79 \n\n2 1  2sa 2sa  2sb 2sb     pxa pxb  p p ya  zb pza p yb R 0.471 403 79 \n\n3 1  2sa 2sa  2sb 2sb     pza pzb  p p ya  xb pxa p yb R 0.471 403 79 \n\nIn addition, we are comparing these molecules with a minimal basis. With eight valence orbitals and ten electrons, configurations that produce some angular correlation in the 2s shell cannot occur in the asymptotic region. The upshot is that there is just one principal standard tableaux function at long distance, and this is shown in Table 11.13. Because of the antisymmetry in the columns of standard tableaux functions, we see that this function represents two noninteracting 4 S N atoms. The situation is not so simple with HLSP functions. They do not have the antisymmetry characteristic mentioned above, and the asymptotic state requires a sum of three of them as shown in Table 11.14. When two 4S N atoms form a molecule we have the possibility that there could be three bonds, one from the two pσ orbitals, and two from the four pπ orbitals. Some mixing of the 2s with the pσ orbitals might lead to hybridization. No other possibilities seem likely. We show the principal configurations in the HLSP function and standard tableaux function cases in Tables 11.15 and 11.16, respectively. We see that the same orbitals are present in both main structures. The situation with\n\n156\n\n11 Second row homonuclear diatomics\n\nTable 11.15. Principal HLSP function structures for N2 at the energy minimum bond distance. 1 Num.\n\n2\n\n1  2sa 2sa  2sb 2sb     pza pzb  p pxb  xa p ya p yb R 0.207 439 81 \n\nTab.\n\nCi (Rmin )\n\n3\n\n1  2sa 2sa  pzb pzb     2sb pza  p pxb  xa p ya p yb R 0.103 862 35 \n\n4\n\n2  2sa 2sa  2sb 2sb     pxa pxa  p p yb  yb pza pzb R 0.081 907 16 \n\n4  2sa 2sa  2sb 2sb     pzb pzb  p p ya  ya pxa pxb R −0.075 261 88 \n\nTable 11.16. Principal standard tableaux function structures for N2 at the energy minimum bond distance. 1 Num.\n\n1  2sa 2sa  2sb 2sb    Tab.  pza pzb  p pxb  xa p ya p yb Ci (Rmin ) 0.329 868 28 \n\n2 1  2sa 2sa  2sb 2sb     pza pxb  p p ya  zb pxa p yb −0.158 776 96 \n\n3\n\n4\n\n5\n\n2  2sa 2sa  pzb pzb     2sb pza  p pxb  xa p ya p yb 0.112 116 70\n\n1  2sa 2sa  2sb 2sb     pza pxa  p pxb  zb p ya p yb −0.111 073 50\n\n1  2sa 2sa  2sb 2sb     pza pzb  p p ya  xa pxb p yb −0.110 745 84\n\nthe HLSP functions is somewhat simpler. The main structure has three electron pair bonds involving the 2 p orbitals, and structure 2 involves one atom in the first excited valence state with an electron pair bond between 2s and 2 pσ orbitals. This latter occurrence, of course, indicates a certain amount of s–p hybridization in the σ bond. Structures 3 and 4 represent ionic contributions to the π and σ bonds, respectively. The results for the standard tableaux functions at the energy minimum are shown in Table 11.16. Structures 1, 2, 4, and 5 are different standard tableaux corresponding to two ground state atoms and represent mixing in different states from the ground configurations. The standard tableaux functions are not so simple here since they do not represent three electron pair bonds as a single tableau. Structure 3 represents one of the atoms in the first excited valence state and contributes to s–p hybridization in the σ bond as in the HLSP function case. It is clear that, regardless of the sort of basis function we use, our results give the bonding picture of N2 as a triple bond. There is, in addition, some indication\n\n11.3 Qualitative discussion\n\n157\n\nthat the excited valence configuration is less important compared to the ground configuration than was the case with B2 and C2 . Two properties of the atoms could contribute to this. r As already mentioned, the excited valence state is of higher energy and is less likely to\n\nmix as strongly.\n\nr Exciting the atom in this case does not change the number of paired electrons, and, thus,\n\na no greater opportunity for bonding presents itself than in the ground state.\n\n11.3.4 O2 From Table 11.3 we see that the ground state of O is 3P, and there are only two unpaired orbitals in the ground configuration. Since the L shell is more than half full, valence excitations will not reduce the number of double occupations. We can make the following conjectures. 1. The two free 2 p orbitals from each atom could combine to form two π bonds to give a 1 + \u0006g state. 2. One of the 2 p orbitals on each atom could join with the other to form a σ bond. (a) The other 2 p orbitals could combine as a pair of π bonds to give a 1 \u0007u state. (b) The other single 2 p orbitals could combine with the doubly occupied orbital on the other atom to form two three-electron bonds, giving a 3 \u0006g− state.\n\nIt is, of course, the last case that occurs, and we consider first the nature of a three-electron bond. Any elementary inorganic structure book will describe, in MO terms, the π bonds in O2 as each having a doubly occupied bonding orbital and a singly occupied antibonding orbital. (This is the MO description of a three-electron bond.) We may analyze this description, using the properties of tableau functions, to see how it relates to the VB picture. We take a very simple case of a pair of orbitals a and b that can bond. We assume the orbitals are at two different centers. The simplest LCAO approximation to the bonding orbital is σ = A(a + b), and the antibonding counterpart is σ ∗ = √ √ B(a − b). Here A = 1/ 2(1 + S) and B = 1/ 2(1 − S), where S is the overlap integral, are the normalization constants. Consider the simple three-electron doublet wave function\n\n\u000e σ σ ψ= , (11.1) σ∗\n\n\u000e a+b a+b 2 . (11.2) =A B a−b\n\n158\n\n11 Second row homonuclear diatomics\n\nTable 11.17. Principal standard tableaux function structures for O2 at an asymptotic bond distance. 1 Num.\n\nTab.\n\nCi (∞)\n\n2  2sa 2sa  2sb 2sb     2 pxa 2 pxa     2 p yb 2 p yb    2p  za 2 pzb    2 pxb 2 p ya 0.612 374 06 \n\n2 2  2sa 2sa  2sb 2sb     2 pxa 2 pxa     2 p yb 2 p yb    2p  za 2 pxb    2 pzb 2 p ya −0.612 374 06 \n\nIt will be recalled from our discussion of Chapter 5 that the tableau in Eq. (11.2) is a shorthand for the result of operating upon a particular orbital product with the operator θ N PN , and N is the column antisymmetrizer. Thus, our function contains a 2 × 2 functional determinant involving a ± b and two particles in all terms. Any row or column operations legal in a determinant may be used to simplify our function, and the determinant may be converted to the equal one involving just 2a and −b. Equation (11.2) becomes\n\n\u000e 2a a + b , (11.3) ψ = A2 B −b \u000f \u000e \u000e\u0010 b b a a 2 − = 2A B , (11.4) a b and shows us how the three-electron bond is represented in the VB scheme. We also emphasize that the tableaux of Eqs. (11.2) and (11.4) are of the sort where the standard tableaux functions and the HLSP functions are the same. Thus, that distinction does not affect our picture. Now, let us consider the principal structures for the asymptotic geometry shown in Tables 11.17 and 11.18. Both forms of the wave function correspond to 3 \u0006g− couplings of the two atoms in their 3P ground states. When we consider the principal structures at the energy minimum geometry we see the three-electron bonds discussed above. These are shown in Tables 11.19 and 11.20. Considering the principal tableaux of either sort, we see there are two three2 pxb and p 2yb p ya . There is, of course, a normal two-electron electron sets present, pxa σ bond present also. When we move to the second structure, there are differences.\n\n11.3 Qualitative discussion\n\n159\n\nTable 11.18. Principal HLSP function structures for O2 at an asymptotic bond distance. 1 Num.\n\nTab.\n\nCi (∞)\n\n2\n\n2  2sa 2sa  2sb 2sb     2 pxa 2 pxa     2 p yb 2 p yb    2p  za 2 pzb    2 pxb 2 p ya R 0.500 000 0\n\n2  2sa 2sa  2sb 2sb     2 pxa 2 pxa     2 p yb 2 p yb   2p   ya 2 pxb   2 pzb  2 pza R 0.500 000 0 \n\nTable 11.19. Principal standard tableaux function structures for O2 at the energy minimum bond distance. 1 Num.\n\n2\n\n3\n\n4\n\n5\n\n2 2 2 2 2      2sa 2sa 2sa 2sa 2sa 2sa 2sa 2sa 2sa 2sa  2sb     2sb  2sb  2sb  2sb  2sb    2sb   2sb   2sb   2sb    2 pxa 2 pxa   2 pxa 2 pxa   2 pxa 2 pxa   2 pza 2 pza   2 pxb 2 pxb        Tab.  2 p yb 2 p yb   2 p ya 2 p ya   2 p yb 2 p yb   2 pxb 2 pxb   2 p ya 2 p ya   2p           za 2 pzb   2 pza 2 pzb   2 pza 2 pxb   2 p yb 2 p yb   2 pza 2 pxa   2 pxb   2 pxb   2 pzb   2 pxa   2 pzb  2 p ya 2 p yb 2 p ya 2 p ya 2 p yb Ci (Rmin ) 0.385 676 56 −0.190 603 83 −0.189 268 76 0.175 541 11 −0.164 445 74 \n\nTable 11.20. Principal HLSP function structures for O2 at the energy minimum bond distance. 1 Num.\n\n2  2sa 2sa  2sb 2sb     2 pxa 2 pxa    Tab.  2 p yb 2 p yb    2p  za 2 pzb    2 pxb 2 p ya R Ci (Rmin ) 0.327 172 36 \n\n2 2  2sa 2sa  2sb 2sb     2 pza 2 pza     2 pxb 2 pxb   2p   yb 2 p yb   2 pxa  2 p ya R 0.175 541 11 \n\n3 2  2sa 2sa  2sb 2sb     2 pxa 2 pxa     2 p ya 2 p ya   2p   za 2 pzb   2 pxb  2 p yb R −0.171 075 55 \n\n4 4  2sa 2sa  2sb 2sb     2 pza 2 pza     2 pxa 2 pxa   2p   yb 2 p yb   2 pxb  2 p ya R −0.113 001 07 \n\n160\n\n11 Second row homonuclear diatomics\n\nTable 11.21. Principal standard tableaux and HLSP function structures for F2 at an asymptotic bond distance. 1 Num.\n\nTab.\n\nCi (∞)\n\n2  2sa 2sa  2sb 2sb     pza pza     pzb pzb   p  xa pxa   pxb pxb  p ya p yb 0.707 106 78 \n\n1. With standard tableaux functions: (a) structure 2 is ionic, having the two three-electron bonds pointed the same way; (b) structures 3 and 4 are the other standard tableau associated with structure 1; (c) structure 5 makes ionic contributions to all bonds, but in such a way that the net charge on the atoms is zero. The charge in a three-electron bond is one way, and the charge in the σ bond is opposite. 2. With HLSP functions: (a) structure 2 is ionic with a zero net atomic charge. This is similar to structure 5 in terms of the standard tableaux functions; (b) structure 3 is ionic with a net charge. The two three-electron bonds point in the same direction; (c) structure 4 is ionic with respect to the σ bond.\n\n11.3.5 F2 As we pass to F2 , with a minimal basis the amount of flexibility remaining is small. The only unpaired orbital in the atom is a 2 p one, and these are expected to form a σ electron pair bond and a 1 \u0006g+ molecular state. In fact, with 14 electrons and 8 orbitals (outside the core) there can be, at most, one unpaired orbital set in any structure. Therefore, in this case there is no distinction between the standard tableaux and HLSP function representations of the wave functions, and we give only one set of tables. As is seen from Table 11.21, there is only one configuration present at asymptotic distances. That shown is one of the 1 \u0006g+ combinations of two 2P atoms. Table 11.22 shows the principal structures at the energy minimum bond distance. Structure 1 is a σ bond comprising the two pσ orbitals, and structure 2 is\n\n11.4 General conclusions\n\n161\n\nTable 11.22. Principal standard tableaux and HLSP function structures for F2 at the energy minimum bond distance.\n\nNum.\n\nTab.\n\nCi (Rmin )\n\n1\n\n2\n\n1  2sa 2sa  2sb 2sb     pxa pxa     pxb pxb   p  ya p ya   p yb p yb  pza pzb 0.779 221 33\n\n2  2sa 2sa  2sb 2sb     pzb pzb     pxa pxa  p   xb pxb   p ya p ya  p yb p yb −0.232 134 50\n\n3 2  2sa 2sa  pzb pzb     pxa pxa     pxb pxb  p   ya p ya   p yb p yb  2sb pza 0.053 534 27 \n\n4 2  2sa 2sa  pza pza     pxa pxa     pxb pxb  p   ya p ya   p yb p yb  2sb pzb 0.044 702 64 \n\nionic, contributing to correlation in the bond. Structures 3 and 4 contribute to s–p hybridization in the bond. 11.4 General conclusions In Section 11.1 we pointed out that B and C atoms have relatively low-lying valence excited states compared to the other atoms considered. It is seen that these valence excited states comprise the principal structures in the bonded state of B2 and C2 , but not in the other molecules where they contribute less than the ground configuration. We shall discuss these effects in further detail for C atoms in Chapter 13. If we treat the one- and three-electron bonds as one-half a bond we see that B2 , C2 , N2 , O2 , and F2 have two, three, three, two, and one bond(s) in the molecule, respectively. Were it not for the low-lying valence excited states in B and C, the molecules corresponding to these might be expected to have one and two bonds, respectively. Nevertheless, the more open structure of the valence excited states allows more bonding between the atoms. The two molecules that have one- or three-electron π bonds show triplet ground states. This conforms to Hund’s rule in atoms where one has unpaired electrons distributed among degenerate orbitals to produce the highest possible multiplicity. The other molecules all have electron pair bonds or unshared pairs and are in singlet states.\n\n12 Second row heteronuclear diatomics\n\nThe consideration of isoelectronic sequences can provide considerable physical understanding of structural details. We here give details of the calculation of a series of isoelectronic diatomic molecules from the second row of the periodic table, N2 , CO, BF, and BeNe. By studying this sequence we see how the competition between nuclear charges affects bonding. All of these are closed-shell singlet systems, and, at least in the cases of the first two, conventional bonding arguments say there is a triple bond between the two atoms. We expect, at most, only a Van der Waals type of bond between Be and Ne, of course. Our calculations should predict this. The three polar molecules in the series are interesting because they all have anomalous directions to their dipole moments, i.e., the direction is different from that predicted by an elementary application of the idea of electronegativity, accepting the fact that there may be ambiguity in the definition of electronegativity for Ne. We will see how VB ideas interpret these anomalous dipole moments. We do the calculations with a 6-31G∗ basis in the same way as was done in Chapter 11 and for three arrangements of STO3G bases. This will allow us both to judge the stability of the qualitative predictions to the basis and to assess the ability of the calculations to obtain quantitative answers. We have already treated N2 in Chapter 11, but will look at it here from a somewhat different point of view. 12.1 An STO3G AO basis Results of calculations carried out with three different selection schemes and an STO3G AO will be described. The reader will recall that the scale factors for this basis are traditionally adjusted to give molecular geometries, and this must be remembered when interpreting the results. By now the reader should suspect that such a basis will not produce very accurate energies. Nevertheless, we see that the qualitative trends of the quantities match the experimental values. 162\n\n12.1 An STO3G AO basis\n\n163\n\nTable 12.1. Dissociation energies and equilibrium distances for isoelectronic series with an STO3G basis and a full calculation. Energies are in electron volts and distances are in a˚ ngstroms. Basis\n\nSTO3G\n\n6-31G∗\n\na\n\nExp. De Rm (±1, grouped) De Rm (±1, ungrouped) De Rm (full) Ed Rm Full valence+Sa De Rm\n\nN2\n\nCO\n\nBF\n\nBeNe\n\n9.905 1.098\n\n11.226 1.128\n\n7.897 1.263\n\n6.101 1.199\n\n8.988 1.195\n\n6.917 1.257\n\n0.023 2.583\n\n6.448 1.198\n\n9.444 1.196\n\n7.162 1.265\n\n0.123 2.159\n\n6.452 1.265\n\n9.460 1.196\n\n7.181 1.264\n\n0.125 2.151\n\n8.768 1.129\n\n11.053 1.155\n\n7.709 1.278\n\n0.053 3.066\n\nS is an abbreviation for ‘single excitations’.\n\nThe three ways in which the structures are selected for the calculation follow, and in all cases the 1s orbitals of the atoms are doubly occupied. r “±1, grouped” This indicates that the structures included in the VB calculation are re-\n\nstricted to those in which there is at most only one electron transferred from one atom to the other and in which there are six σ, two πx , and two π y electrons. r “±1, ungrouped” This indicates that the structures included in the VB calculation are restricted to those in which there is at most only one electron transferred from one atom to the other. r “full” This is the full (valence) VB calculation.\n\nDissociation energies and minimum energy atomic separations from the STO3G bases are given in Table 12.1 along with those for the 6-31G∗ basis, which we will discuss later. We note that the restriction to ±1 ionicities has an effect on the energy of at most 10–20 meV for this basis. We give tables of the important structures in the full wave function using spherical AOs and using the s– p hybrids, 2s ± 2 pz . The energies are, of course, the same for these alternatives, but the apparent importance of the standard tableaux functions or HLSP functions differs. We also discuss EGSO results for the series. Again we see that the De of N2 is the most poorly predicted in this series. We have no clear explanation for this at present.\n\n164\n\n12 Second row heteronuclear diatomics\n\nTable 12.2. N2 : The most important terms in the wave function when spherical AOs are used as determined by the magnitudes of the coefficients. Results for standard tableaux and HLSP functions are given. See text. 1 Num.\n\n2\n\n3\n\n4\n\n1  2sa 2sa  2sb 2sb     2 pza 2 pzb   2p 2 pxb  xa 2 p ya 2 p yb 0.329 86\n\n1  2sa 2sa  2sb 2sb     2 pza 2 pxb   2p 2 p ya  zb 2 pxa 2 p yb −0.158 78\n\n2  2sb 2sb  2 pza 2 pza    2 pzb   2sa  2p 2 pxb  xa 2 p ya 2 p yb −0.112 12\n\n1  2sa 2sa  2sb 2sb    HLSP  2 pza 2 pzb   2p 2 pxb  xa 2 p ya 2 p yb R Ci (min) 0.207 44\n\n1  2sa 2sa  2 pzb 2 pzb    2 pza   2sb  2p 2 pxb  xa 2 p ya 2 p yb R 0.103 86\n\n2  2sa 2sa  2sb 2sb     2 pxa 2 pxa   2p 2 p yb  yb 2 pza 2 pzb R 0.081 91\n\nSTF Ci (min) Num.\n\n1  2sa 2sa  2sb 2sb     2 pza 2 pxa   2p 2 pxb  zb 2 p ya 2 p yb −0.111 07 1  2sa 2sa  2sb 2sb     2 pza 2 pza   2p 2 p yb  yb 2 pxa 2 pxb R –0.075 26\n\n12.1.1 N2 In Table 12.2 we show the four most important structures in the wave function as determined by the magnitude of the coefficients for standard tableaux functions and for HLSP functions. Table 12.3 shows the same information for the σ AOs formed into s– p hybrids. The symbols “h ox ” or “h i x ” represent the outward or inward pointing hybrids, respectively. Using the size of the coefficients as a measure of importance we see that the expected structure involving one σ and two π bonds is the largest in the wave function. It appears that the hybrid orbital arrangement is slightly preferred for standard tableaux functions while the spherical orbital arrangement is slightly preferred for HLSP functions, but the difference is not great. These results suggest that an intermediate rather than one-to-one hybridization might be preferable, but a great difference is not expected. Nevertheless, it is clear that the VB method predicts a triple bond between the two atoms in N2 . The layout of Tables 12.3 and 12.4 is similar to that of Tables 11.5 and 11.6 described in Section 11.3.1. There is, nevertheless, one point concerning the “Num.” row that merits further comment. In Chapter 6 we discussed how the symmetric group projections interact with spatial symmetry projections. Functions 1, 2, and 4 are members of one constellation, and the corresponding coefficients may not be entirely independent. There are three linearly independent 1 \u0003g+ symmetry functions from the five standard tableaux of this configuration. The 1, 2, and 4 coefficients are thus possibly partly independent and partly connected by group theory. In none\n\n12.1 An STO3G AO basis\n\n165\n\nTable 12.3. N2 : The most important terms in the wave function when s–p hybrid AOs are used as determined by the magnitudes of the coefficients. Results for standard tableaux and HLSP functions are given. 1 Num.\n\n2\n\n3\n\n4\n\n1  h oa h oa  h ob h ob    h ib   h ia  2p 2 pxb  xa 2 p ya 2 p yb 0.337 43\n\n1  h oa h oa  h ob h ob    2 pxb   h ia h 2 p ya  ib 2 pxa 2 p yb −0.144 76\n\n1  h oa h oa  h ob h ob    h ib   h ia  2p 2 p ya  xa 2 pxb 2 p yb −0.107 25\n\n1  h oa h oa  h ob h ob    h ib  HLSP  h ia  2p 2 pxb  xa 2 p ya 2 p yb R Ci (min) 0.189 12\n\n4  h oa h oa  h ob h ob    h ia   h ia  2p 2 pxb  xb 2 p ya 2 p yb R −0.089 64\n\n1  h oa h oa  h ia h ia    h ib   h ob  2p 2 pxb  xa 2 p ya 2 p yb R 0.086 69\n\nSTF Ci (min) Num.\n\n2  h ob h ob  h ia h ia    h ib   h oa  2p 2 pxb  xa 2 p ya 2 p yb 0.099 41 1  h oa h oa  h ob h ob     2 pxb 2 pxb  h h ib  ia 2 p ya 2 p yb R 0.086 31\n\nof the tables do we attempt to elucidate this sort of question. It really requires a detailed examination of the output of the symgenn segment of the CRUNCH suite. Table 12.4 shows spherical and hybrid AO results when subjected to the EGSO weight analysis. Unlike the coefficients, the EGSO analysis for these results shows that an ionic function is the single structure that contains the largest fraction of the wave function in both of these cases. This is a common result in molecules with multiple bonding. We have seen that the ionic structures contribute to delocalization of the electrons (see Chapter 2) and thereby reduce the kinetic energy of the structure. In a complicated symmetry function involving the sum of several terms the mixing of the improved correlation energy of the covalent functions and the improved kinetic energy of the ionic functions can produce a symmetry constellation that has the highest weight. It is thus important to interpret these results correctly. We may comment that the principal configuration with spherical AOs measured by coefficient in the wave function,   2sa 2sa  2sb 2sb     2 pza 2 pzb  ,    2 pxa 2 pxb  2 p ya 2 p yb is not present among the first four functions measured by the EGSO weights. In fact its weight is 0.012 69, a little lower than those in the table. The situation is similar\n\n166\n\n12 Second row heteronuclear diatomics\n\nTable 12.4. N2 : EGSO weights (standard tableaux functions) for spherical AOs, upper group, and s–p hybrids, lower group. These are weights for whole symmetry functions rather than individual tableaux. It should be recalled from Chapter 6 that the detailed forms of symmetry functions are dependent on the particular arrangement of the orbitals in the tableaux and are frequently nonintuitive. 1 Num. STF Wt Num. STF Wt\n\n4  2sa 2sa  2sb 2sb     2 pxa 2 pxa   2p 2 pzb  za 2 p ya 2 p yb 0.558 62 \n\n4  h oa h oa  h ob h ob     2 pxa 2 pxa  h h ib  ia 2 p ya 2 p yb 0.611 66 \n\n2\n\n3\n\n4\n\n2  2sa 2sa  2sb 2sb     2 pza 2 pza   2p 2 pxb  xa 2 p ya 2 p yb 0.200 53\n\n6  2sa 2sa  2 pzb 2 pzb     2 pxa 2 pxa   2s 2 pza  b 2 p ya 2 p yb 0.066 58\n\n2  h oa h oa  h ob h ob     h ia h ia   2p 2p  xa xb 2 p ya 2 p yb 0.156 66\n\n6  h oa h oa  h ia h ia     2 p yb 2 p yb  h h ib  ob 2 pxa 2 pxb 0.057 14\n\n4  2sa 2sa  2 pza 2 pza     2 pxa 2 pxa   2p 2 p yb  yb 2sb 2 pzb 0.040 89 2  h oa h oa  h ob h ob     2 pxa 2 pxa   2p 2 p yb  yb h ia h ib 0.038 70\n\nwith the hybrid orbitals. In this case the standard tableaux function   h oa h oa  h ob h ob     h ia h ib     2 pxa 2 pxb  2 p ya\n\n2 p yb\n\nhas an EGSO weight of 0.005 32, rather smaller than the value in the case of spherical functions. The reader should not find these small contributions too unexpected. The ionic structures singled out by the EGSO can be looked at as one-third ionic and two-thirds covalent. When the orthogonalization inherent in the method works, the effect of the purely covalent functions is considerably depressed and is already taken care of by the mixed functions.\n\n12.1.2 CO The set of tables we give for CO follows the pattern given for N2 in the last section. Table 12.5 shows the four most important structures in the wave function of CO\n\n12.1 An STO3G AO basis\n\n167\n\nTable 12.5. CO: The most important terms in the wave function when spherical AOs are used as determined by the magnitudes of the coefficients. Results for standard tableaux and HLSP functions are given. 1 Num.\n\n2\n\n3\n\n4\n\n2  2sa 2sa  2sb 2sb     2 pxb 2 pxb   2p 2 pzb  za 2 p ya 2 p yb 0.205 59\n\n1  2sa 2sa  2sb 2sb     2 pzb 2 pzb   2p 2 pxb  xa 2 p ya 2 p yb −0.192 23\n\n1  2sa 2sa  2sb 2sb     2 pza 2 pzb   2p 2 pxb  xa 2 p ya 2 p yb 0.149 09\n\n2  2sa 2sa  2sb 2sb    HLSP  2 pxb 2 pxb   2p 2 pzb  za 2 p ya 2 p yb R Ci (min) 0.179 15\n\n1  2sa 2sa  2sb 2sb     2 pzb 2 pzb   2p 2 pxb  xa 2 p ya 2 p yb R −0.157 016\n\n2  2sa 2sa  2sb 2sb     2 pzb 2 pzb   2p 2 pxb  xb 2 p ya 2 p yb R −0.094 20\n\nSTF Ci (min) Num.\n\n1  2sb 2sb  2 pzb 2 pzb    2 pza   2sa  2p 2 pxb  xa 2 p ya 2 p yb −0.120 44 1  2sa 2sa  2sb 2sb     2 pxb 2 pxb   2p 2 p yb  yb 2 pza 2 pzb R 0.093 19\n\nas determined by the magnitude of the coefficients for standard tableaux functions and HLSP functions. Table 12.6 shows the same information for the σ AOs formed into s– p hybrids. The symbols “h ox ” or “h i x ” are used as before. Using the size of the coefficients as a measure of importance, we see that VB theory predicts CO to have only two covalent bonds between the atoms. We saw in Section 11.1 that C and O are both in 3P ground states, thus elementary considerations suggest that there is one σ covalent bond and one π covalent bond cylindrically averaged to achieve 1 \u0003 + symmetry. This view, although too simplistic, is different from that often seen where CO is written like N2 with a triple bond. The latter must also be too simplistic, since, if CO had anything close to an evenly shared triple bond, its dipole moment would be large, although in the experimentally correct direction. We will discuss the dipole moments of the polar molecules all together in Section 12.3. The triple bond structure appears in the third place with spherical AOs and standard tableaux functions, but is not among the first four with HLSP functions. This is actually misleading due to the arbitrary cutoff at four functions in the table. The HLSP function triple bond has a coefficient of 0.09182, only slightly smaller that function 4 in the table. The appearance of the triple bond structure in this wave function is the quantum mechanical manifestation of the “π back-bonding” phenomenon invoked in qualitative arguments concerning bonding. We thereby have a quantitative approach to the concept.\n\n168\n\n12 Second row heteronuclear diatomics\n\nTable 12.6. CO: The most important terms in the wave function when s–p hybrid AOs are used as determined by the magnitudes of the coefficients. Results for standard tableaux and HLSP functions are given. 1 Num.\n\n2\n\n3\n\n4\n\n2  h oa h oa  h ob h ob     2 p yb 2 p yb  h h ib  ia 2 pxa 2 pxb 0.274 04\n\n1  h oa h oa  h ob h ob    h ib   h ib  2p 2 pxb  xa 2 p ya 2 p yb 0.222 04\n\n1  h oa h oa  h ob h ob    h ib   h ia  2p 2 pxb  xa 2 p ya 2 p yb 0.188 09\n\n2  h oa h oa  h ob h ob    HLSP  2 pxb 2 pxb  h h ib  ia 2 p ya 2 p yb R Ci (min) 0.240 94\n\n1  h oa h oa  h ob h ob    h ib   h ib  2p 2 pxb  xa 2 p ya 2 p yb R 0.181 73\n\n1  h oa h oa  h ob h ob     2 pxb 2 pxb   2p 2 p yb  yb h ia h ib R 0.138 44\n\nSTF Ci (min) Num.\n\n1  h ob h ob  h ib h ib    h ia   h oa  2p 2 pxb  xa 2 p ya 2 p yb 0.146 23 2  h oa h oa  h ob h ob    h ib   h ib  2p 2 pxb  xb 2 p ya 2 p yb R 0.125 86\n\nComparing Tables 12.5 and 12.6 with those for N2 , Tables 12.2 and 12.3, we see that CO prefers hybrid orbitals to a somewhat greater extent. The differences are not great, however. The EGSO weights shown in Table 12.7 display a behavior rather different from those for N2 . In this case the principal configuration is the same for both sorts of measure. The smaller EGSO weights are different, however.\n\n12.1.3 BF The pattern of tables for BF follows the earlier treatments in the chapter. Table 12.8 shows the four most important structures in the wave function of BF as determined by the magnitude of the coefficients for standard tableaux functions and HLSP functions. Table 12.9 shows the same information for the σ AOs formed into s– p hybrids. We use the “h ox ” or “h i x ” symbols as before. In this case, all of the principal structures except number 4 for the hybrid AOs have no more than one unpaired set of orbitals. Therefore, the coefficients for the standard tableaux functions and the HLSP functions differ in that case only. The hybrid orbital arrangement is again preferred, but the difference is only somewhat greater than that for CO. The principal configuration is definitely a single σ bond between the two atoms. When interpreted as a configuration of BF, the most important one from CO changes to the ionic sort.\n\n12.1 An STO3G AO basis\n\n169\n\nTable 12.7. CO: EGSO weights (standard tableaux functions) for spherical AOs, upper group, and s–p hybrids, lower group. These are weights for whole symmetry functions rather than individual tableaux. It should be recalled from Chapter 6 that the detailed forms of symmetry functions are dependent on the particular arrangement of the orbitals in the tableaux and are frequently nonintuitive. 1 Num. STF Wt Num. STF Wt\n\n2\n\n3\n\n4\n\n2  2sa 2sa  2sb 2sb     2 p yb 2 p yb   2p 2 pzb  za 2 pxa 2 pxb 0.550 208\n\n1  2sa 2sa  2sb 2sb     2 pzb 2 pzb   2p 2 pxb  xa 2 p ya 2 p yb 0.140 467\n\n2  2sb 2sb  2 pzb 2 pzb     2 p yb 2 p yb   2s 2 pza  a 2 pxa 2 pxb 0.105 266\n\n2  h oa h oa  h ob h ob     2 p yb 2 p yb  h h ib  ia 2 pxa 2 pxb 0.637 66\n\n1  h oa h oa  h ob h ob    h ib   h ib  2p 2 pxb  xa 2 p ya 2 p yb 0.099 84\n\n2  h ob h ob  h ib h ib     2 p yb 2 p yb  h h ia  oa 2 pxa 2 pxb 0.041 74\n\n2  2sa 2sa  2sb 2sb     2 pzb 2 pzb   2p 2 pxa  xa 2 p yb 2 p yb 0.033 16 1  h oa h oa  h ob h ob     2 pxb 2 pxb   2p 2 p yb  yb h ia h ib 0.031 57\n\nTable 12.8. BF: The most important terms in the wave function when spherical AOs are used as determined by the magnitudes of the coefficients. Results for standard tableaux and HLSP functions are the same. 1 Num. STF or HLSP Ci (min)\n\n1  2sa 2sa  2sb 2sb     2 pxb 2 pxb   2p 2 p yb  yb 2 pza 2 pzb 0.277 78 \n\n2 2  2sa 2sa  2sb 2sb     2 pzb 2 pzb   2p 2 p yb  yb 2 pxa 2 pxb −0.224 278 \n\n3 1  2sa 2sa  2sb 2sb     2 pzb 2 pzb   2p 2 pxb  xb 2 p yb 2 p yb −0.217 808 \n\n4 1  2sb 2sb  2 pzb 2 pzb     2 pxb 2 pxb   2p 2 p yb  yb 2sa 2 pza −0.135 735 \n\nFor spherical AOs it is not among the first four, but appears in the eighth position with a coefficient of 0.114 16, and for hybrids it is the eighth one down with a coefficient of 0.120 80. We therefore predict that, quantitatively, there is less π back-bonding in BF than in CO. For neither arrangement of orbitals is the triply bonded structure of N2 important for BF.\n\n170\n\n12 Second row heteronuclear diatomics\n\nTable 12.9. BF: The most important terms in the wave function when hybrid AOs are used as determined by the magnitudes of the coefficients. Results for standard tableaux and HLSP functions are the same. 1 Num.\n\n2\n\n1  h oa h oa  h ob h ob     2 pxb 2 pxb   2p 2 p yb  yb h ia h ib 0.335 49 0.335 49 \n\nSTF or HLSP st f\n\nCi (min) hlsp Ci (min)\n\n1  h oa h oa  h ob h ob    h ib   h ib  2p 2 pxb  xb 2 p yb 2 p yb 0.260 92 0.260 92 \n\n3 2  h oa h oa  h ob h ob    h ib   h ib  2p 2 p yb  yb 2 pxa 2 pxb 0.228 79 0.228 79 \n\n4 2  h ob h ob  h ib h ib     2 p yb 2 p yb  h h ia  oa 2 pxa 2 pxb 0.160 75 0.147 62 \n\nTable 12.10. BF: EGSO weights (standard tableaux functions) for spherical AOs, upper group, and hybrid AOs, lower group. These are weights for whole symmetry functions. 1 Num. STF Wt Num. STF Wt\n\n2\n\n3\n\n4\n\n2  2sa 2sa  2sb 2sb     2 pzb 2 pzb   2p 2 p yb  yb 2 pxa 2 pxb 0.464 83\n\n1  2sa 2sa  2sb 2sb     2 pxb 2 pxb   2p 2 p yb  yb 2 pza 2 pzb 0.302 50\n\n1  2sb 2sb  2 pzb 2 pzb     2 pxb 2 pxb   2p 2 p yb  yb 2sa 2 pza 0.082 13\n\n1  h oa h oa  h ob h ob     2 pxb 2 pxb   2p 2 p yb  yb h ia h ib 0.486 57\n\n2  h oa h oa  h ob h ob    h ib   h ib  2p 2 p yb  yb 2 pxa 2 pxb 0.252 67\n\n2  h ob h ob  h ib h ib     2 p yb 2 p yb  h h ia  oa 2 pxa 2 pxb 0.078 54\n\n2  2sb 2sb  2 pzb 2 pzb     2 p yb 2 p yb   2s 2 pza  a 2 pxa 2 pxb 0.028 11 1  h ob h ob  h ia h ia     2 pxb 2 pxb   2p 2 p yb  yb h oa h ib 0.036 12\n\nThe EGSO weights shown in Table 12.10 for the two orbital arrangements display an interesting switch. For the spherical AOs a cylindrically averaged π bond is the principal configuration and the σ bond is the second one. The hybrid AOs show the opposite order with the σ bond structure relatively more strongly favored than in the spherical case. In both cases there is considerable competition between the two bond types, and the VB prediction is that they are strongly mixed in the molecule. We defer a discussion of the dipole moment until later.\n\n12.1 An STO3G AO basis\n\n171\n\nTable 12.11. BeNe: The most important terms in the wave function when spherical AOs are used, as determined by the magnitudes of the coefficients. Results for standard tableaux and HLSP functions are the same for these terms in the wave function. 1 Num. STF or HLSP Ci (min)\n\n1  2sa 2sa  2sb 2sb     2 pzb 2 pzb   2p 2 pxb  xb 2 p yb 2 p yb 0.879 46 \n\n2 2  2sb 2sb  2 pzb 2 pzb     2 pxa 2 pxa   2p 2 pxb  xb 2 p yb 2 p yb −0.188 65 \n\n3 1  2sa 2sa  2sb 2sb     2 pxb 2 pxb   2p 2 p yb  yb 2 pza 2 pzb −0.186 14 \n\n4 1  2sb 2sb  2 pza 2 pza     2 pzb 2 pzb   2p 2 pxb  xb 2 p yb 2 p yb −0.144 05 \n\nTable 12.12. BeNe: The most important terms in the wave function when s–p hybrid AOs are used as determined by the magnitudes of the coefficients. Results for standard tableaux and HLSP functions are the same for these terms in the wave function. 1 Num. STF or HLSP Ci (min)\n\n1  h ob h ob  h ib h ib     2 pxb 2 pxb   2p 2 p yb  yb h oa h ia 0.718 54 \n\n2 1  h oa h oa  h ob h ob    h ib   h ib  2p 2 pxb  xb 2 p yb 2 p yb 0.429 47 \n\n3 1  h ob h ob  h ia h ia    h ib   h ib  2p 2 pxb  xb 2 p yb 2 p yb 0.298 39 \n\n4 2  h ob h ob  h ib h ib     2 pxb 2 pxb   2p 2 p ya  ya 2 p yb 2 p yb –0.188 64 \n\n12.1.4 BeNe When we arrive at BeNe in our series we expect no real electron pair bond between the two atoms, but we provide the same sorts of tables as before. Table 12.11 shows the four most important structures in the wave function of BF as determined by the magnitude of the coefficients for standard tableaux functions or HLSP functions. Table 12.12 shows the same information for the σ AOs formed into s– p hybrids. The symbols “h ox ” or “h i x ” are used as before. In this case, where none of the principal structures has more than one pair of unpaired orbitals, there is no difference in the coefficients between the standard tableaux functions and the HLSP functions. For BeNe the spherical AO arrangement is definitely preferred. Examination of Table 12.12 shows an unusual inner–outer hybrid pairing on Be in the principal configuration. This pairing is not a good substitute for the 2s 2 ground\n\n172\n\n12 Second row heteronuclear diatomics\n\nTable 12.13. BeNe: EGSO weights (standard tableaux functions) for spherical AOs. These are weights for whole symmetry functions. 1 Num. STF Wt Num. STF Wt\n\n2\n\n3\n\n4\n\n1  2sa 2sa  2sb 2sb     2 pzb 2 pzb   2p 2 pxb  xb 2 p yb 2 p yb 0.845 66\n\n2  2sb 2sb  2 pzb 2 pzb     2 pxa 2 pxa   2p 2 pxb  xb 2 p yb 2 p yb 0.078 54\n\n1  2sa 2sa  2sb 2sb     2 pxb 2 pxb   2p 2 p yb  yb 2 pza 2 pzb 0.035 88\n\n1  h ob h ob  h ib h ib     2 pxb 2 pxb   2p 2 p yb  yb h oa h ia 0.573 15\n\n1  h oa h oa  h ob h ob    h ib   h ib  2p 2 pxb  xb 2 p yb 2 p yb 0.208 39\n\n1  h ob h ob  h ia h ia    h ib   h ib  2p 2 pxb  xb 2 p yb 2 p yb 0.093 72\n\n1  2sb 2sb  2 pza 2 pza     2 pzb 2 pzb   2p 2 pxb  xb 2 p yb 2 p yb 0.020 73 2  h ob h ob  h ib h ib     2 pxa 2 pxa   2p 2 pxb  xb 2 p yb 2 p yb 0.078 54\n\nstate of Be and leads to the somewhat smaller value of the coefficient compared to the primary structure in the case of spherical AOs. Because of the column antisymmetry of the standard tableaux functions the hybrids on Ne do not cause a similar difficulty. Comparing the two sets of weights in Table 12.13 shows the same phenomenon. The principal spherical AO structure represents over 80% of the total wave function while the weights for the hybrid structures fall off more slowly. It addition it will be observed that the principal terms in the wave functions and the EGSO weights are completely parallel in the case of BeNe. This is in contrast to other members in this series of molecules. We see that the third function in any of the spherical AO series is an ionic structure equivalent to the principal configuration for BF and thus represents one σ bond. This is a relatively minor constituent of the wave function, but, nevertheless, has a surprisingly large coefficient. It is possible that this sort of term is overemphasized in the STO3G basis, since it predicts an improbably short bond between Be and Ne, judged by the value obtained with the higher-quality 6-31G∗ basis.1 As a test of this conjecture, a recalculation of the STO3G structures at the 6-31G∗ equilibrium distance reduces the importance of this ionic structure to the fourth place with an EGSO weight of 0.4%. The π back-bonding 1\n\nThis is very likely a manifestation of basis set superposition error that occurs frequently in MO calculations, also.\n\n12.2 Quantitative results from a 6-31G∗ basis\n\n173\n\nTable 12.14. Statistics on 6-31G∗ calculations for N2 , CO, BF, and BeNe. The dipole moments are also given. See Section 12.3 for a discussion of the signs of the moments. Moment (D) Molecule\n\nNI a\n\nNS b\n\nCalc.\n\nExp.\n\nN2 CO BF BeNe\n\n6964 5736 3166 1210\n\n1812 2986 1680 672\n\n0.0 −0.087 −1.084 −0.312\n\n0.0 −0.122 −0.5c ?\n\nThe number of basis functions involved in 1 \u0003g+ or 1 \u0003 + symmetry functions. The number of symmetry functions supported. c For ν = 0.\n\na\n\nb\n\nstructure at ≈ 0.01% is even less important in this molecule. It appears that VB theory predicts there to be no electron pair bonds between the two atoms here. The minimum in the internuclear potential curve is due to Van der Waals interactions. In spite of this the molecule has a small dipole moment, which we discuss below. 12.2 Quantitative results from a 6-31G∗ basis In Chapter 11 we described calculations using the occupied AOs in a full MCVB with added configurations involving single excitations into all of the atomic virtual orbitals excepting the dδ . We give the values for De and Rm in Table 12.1. The results for N2 are the same, of course, as those in Table 11.1. As was the case with the homonuclear molecules in Chapter 11, we again see that the calculated energy for N2 is the farthest from experiment for the known values. There seems at the moment no good explanation for this. Nevertheless, the higher-quality basis gives closer agreement with experiment. In Table 12.14 we present statistics for the number of n-electron basis functions involved in the calculations. Apparently there are no experimental data on BeNe. If we fit a Morse function to the parameters we obtain for the dissociation curve, it is estimated that there would be 14–15 bound vibrational states for this Van der Waals molecule. Thus, VB theory predicts the existence of stable gaseous BeNe, if it is cold enough, since De is only 2kT for room temperature. As stated above, we consider the dipole moments of the heteronuclear molecules in the next section, but we give in Table 12.14 the dipole moments at the equilibrium geometry and determined with the 6-31G∗ basis.\n\n174\n\n12 Second row heteronuclear diatomics\n\n12.3 Dipole moments of CO, BF, and BeNe Elementary discussions define the electronegativity of an atom as a measure of its ability to attract electrons to itself. Several authors, Pauling, Mulliken, and Allen, have devised quantitative values as a measure of this ability. Such elementary discussions usually emphasize the connection between the dipole moments of heteronuclear bonds and the comparative electronegativities of the atoms involved. In particular the expectation is that the electronegativity difference should tell the direction of the moment. In general, this idea works well with many diatomic molecules that have single bonds between the atoms. Examples are hydrogen halides and (gaseous) alkali halides. Discussions of LiH and LiF representing this sort of system are in Chapter 8. There are, nevertheless, a number of diatomic molecules that have an anomalous direction of the dipole moment between different atoms. CO is probably the most notorious of these anomalies but others are known. Huzinaga et al. have examined a number of these and describe the effects in terms of MO theories. The interested reader is referred to the article for the details, since this work stresses VB analyses of chemical phenomena. 12.3.1 Results for 6-31G∗ basis Figure 12.1 shows the dipole moment functions in terms of internuclear distance of CO, BF, and BeNe, calculated with our conventional 6-31G∗ basis arrangement. 4 3\n\nElectric dipole moment (D)\n\nBF 2 CO\n\n1 0\n\nBeNe\n\n−1 −2 −3 0\n\n1\n\n2\n\n3\n\n4\n\n5\n\nInternuclear distance (Å)\n\nFigure 12.1. The dipole moment functions for CO, BF, BeNe calculated at a number of distances with the conventional 6-31G∗ basis arrangement.\n\n12.3 Dipole moments of CO, BF, and BeNe\n\n175\n\nThe curves for CO and BF show the form typical of these systems as is emphasized in Ref. . In our discussion of LiF in Section 8.3 we emphasized how the nature of a wave function could change from ionic to covalent with a change in internuclear distance. Here again we appear to have the “signature” of this sort of phenomenon: a the change of sign of the moment at internuclear distances around 1.0–1.5 A strongly suggests the interplay of two effects where the winning one changes fairly rapidly with distance. On the other hand the sign of BeNe does not change and this suggests that one of these effects is absent in this molecule. From the signs on the moments and our work in Chapters 2 and 8 we interpret these curves as follows (for the interpretation of the signs the reader is reminded that all three of our systems are oriented with the less electronegative atom in the positive z-direction). 1. At internuclear distances intermediate, but greater than equilibrium, the familiar ideas of electronegativity win out, and the more electronegative atom has an excess of negative charge. At the maxima the charge on O in CO is around −0.29|e| and on F in BF −0.70|e|. It is not surprising that in BF the effect is larger. No legitimate argument would suggest that Ne has any sort of negative ion propensity, and we do not see a maximum in that curve. 2. When systems are pushed together, nonbonded electrons, on the other hand, tend to retreat toward the system that has the more diffuse orbitals. In this case that is C, B, or Be. Since the nonbonded electrons are generally in orbitals less far out, this effect occurs at closer distances and, according to our calculations, wins out at equilibrium distances for CO and BF. This is the only effect for BeNe, and the moment is in the same direction at all of the distances we show. This retreat of electrons is definitely a result of the Pauli exclusion principle.\n\nBoth sorts of physical effects tend to fall off exponentially as the distance between the atoms increases – the dipole moment must go to zero asymptotically. A close examination of the CO results shows that the moment goes to very small negative a values again around 4.0 A. Whether this is real is difficult to decide without further calculations. It might be that the Pauli exclusion effect wins again at these distances, the result might be different for a still larger basis. Also, Gaussian basis functions can cause troubles at larger distances because individually they really fall off much too rapidly with distance.\n\n12.3.2 Difficulties with the STO3G basis We also calculated the dipole moment functions for CO, BF, and BeNe with an STO3G basis, and it can be seen in Fig. 12.2 that there are real difficulties with the minimal basis. We have argued that the numerical value and sign of the electric\n\n176\n\n12 Second row heteronuclear diatomics 1 BF\n\nElectric dipole moment (D)\n\n0.5\n\n0 −0.5\n\nCO\n\nBeNe\n\n−1 −1.5 −2 −2.5 0\n\n1\n\n2\n\n3\n\n4\n\n5\n\nInternuclear distance (Å)\n\nFigure 12.2. The dipole moment functions for CO, BF, BeNe calculated at a number of distances with the full valence minimal STO3G basis.\n\ndipole moment in these molecules is the result of a balance between two opposing effects. When we pass to the STO3G basis, this balance is thrown off.2 It is instructive, nevertheless, to see that STO3Gs reproduce the Pauli exclusion effect better than the formation of a partial negative ion at the more electronegative end of the molecule. This is expected. The more diffuse 6-31G∗ basis, with its capability for allowing the AOs to breathe and polarize can much better represent the negative ions. 2\n\nThis is in spite of the qualitatively reasonable energies the basis yields. Such an outcome is a familiar one, however – the energy is the result of a variational calculation and is expected to be produced to higher order than quantities like the electric moment. In addition, the minimal basis does better in the region of the minimum and asymptotically than elsewhere. Thus, De may not suffer too greatly.\n\n13 Methane, ethane and hybridization\n\nIn Chapter 11 we discussed the properties of the atoms in the second row of the periodic table and how these might influence molecules formed from them. We focus on carbon in this chapter and examine how the bonding changes through the series CH, CH2 , CH3 , and CH4 . The first three of these are known only spectroscopically, in matrix isolation, or as reaction intermediates, but many of their properties have been determined. The reader will recall that carbon exhibits relatively low-energy excited valence configurations. For carbon the excitation energy is around 4 eV, and among the atoms discussed in Chapter 11, only boron has a lower excitation energy. If this excited configuration is to have an important role in the bonding, the energy to produce the excitation must be paid back by the energy of formation of the bond or bonds. We shall see that VB theory predicts this happens between CH and CH2 . After our discussion of these single carbon compounds, we will consider ethane, CH3 CH3 , as an example for dealing with larger hydrocarbons.\n\n13.1 CH, CH2 , CH3 , and CH4 13.1.1 STO3G basis We first give calculations of these four molecules with an STO3G basis. The total energies and first bond dissociation energies are collected in Table 13.1. We see that, even with the minimal basis, the bond energies are within 0.4 eV of the experimental values except for CH3 , which has considerable uncertainty. The calculated values tend to be smaller, as expected for a minimal VB treatment. We now give a discussion of each of the molecules, first considering the atomic structure of the carbon atom and attempting to predict the bonding pattern. The predictions are followed by the results of the STO3G calculations.\n\n177\n\n178\n\n13 Methane, ethane and hybridization\n\nTable 13.1. CHn STO3G energies. n\n\nEnergy (au)\n\nDissociation\n\nEnergy (eV)\n\n−37.438 66 −38.050 28 −38.693 63 −39.338 38 −39.989 73\n\nDC−H DCH−H DCH2 −H DCH3 −H\n\n3.065 3.901 3.939 4.118\n\n0 1 2 3 4\n\nC (3 P) CH (2 \u0001) CH2 (3 B1 ) CH3 (2 A\u0002\u00022 ) CH4 (1 A1 )\n\na\n\nSee Refs. [49, 54].\n\nExp. (eV)a 3.465 4.33 ≤4.90 4.406\n\nTable 13.2. Principal standard tableaux functions for CH at the equilibrium internuclear distance. This is the x-component of a π-pair.\n\nNum.a Tab.b Ci (min)\n\n1\n\n2\n\n1 \u0002 2s 2s 1s 2 pz 2 px 0.729 684 53\n\n1 \u0002 2s 2s 1s 2 px 2 pz −0.320 400 21\n\n\u0001\n\n3\n\n\u0001\n\n1 \u0002 2s 2s 2 pz 2 pz 2 px 0.227 852 03\n\n\u0001\n\n4 1 \u0002 2 pz 2 pz 2s 1s 2 px 0.168 832 87\n\n\u0001\n\na\n\nThe number of terms in the symmetry function that is generated from the tableau shown. (See text.) b These tableau symbols exclude the core orbitals.\n\nCH 3\n\n3\n\nThe P (we call it P(1)) ground state of the C atom has two unpaired p electrons. When an H atom approaches, it should be able to form an electron pair bond with one of these orbitals, while the other would remain unpaired. This scenario leads to the expectation that CH should have a 2 \u0001 ground state. We have commented on the possible involvement of the excited C 5 S state, but symmetry prohibits such mixing here. There is a higher energy 3 P(2) valence state that is allowed to interact through symmetry. There are 75 standard tableaux functions in a full valence treatment, but only 36 are \u0001 states, half being x-components and half y-components. The variation problem therefore has two 18 × 18 matrices. The principal standard tableaux functions in the wave function are shown in Table 13.2. The predominant term in the wave function clearly involves the C atom in its 3 P(1) state. The calculated dipole\n\n13.1 CH, CH2 , CH3 , and CH4\n\n179\n\nz 6\n\nHb\n\nx\n\n\u0011 Q Q \u0011 Q \b\u0011 C \b \b\b \b \u0019\b \b\n\nHa y\n\nFigure 13.1. Orientation of the CH2 diradical.\n\nmoment at the equilibrium separation for this basis is 1.4983 D with H the positive end. CH2 The methylene radical has enjoyed a certain notoriety concerning the nature of the ground state. It is now known to be in a triplet state with a bent geometry. This is perhaps not what is expected if we just think of an H atom interacting with the remaining unpaired p orbital of CH, an outcome that should lead to a singlet state of some geometry. At this stage in our series we will see, however, that the excited 5 S state becomes dominant in the wave function. A quintet state coupled with two doublet H atoms can have no lower multiplicity than triplet. In Fig. 13.1 we show the orientation of the CH2 diradical in a Cartesian coordinate system and assume C2v symmetry. With six electrons and six orbitals in a full valence calculation we expect 189 standard tableaux functions. These support 51 3 B1 symmetry functions that, however, involve a total of only 97 standard tableaux functions out of the possible 189. Table 13.3 shows the principal terms in the wave function for the equilibrium geometry. The principal standard tableaux function is \n\nT1AO\n\n 2s 1sa  2 px 1sb  , =  2 py  2 pz\n\n(13.1)\n\nwhere the subscripts on the 1s orbitals are associated with the corresponding subscripts on the H atoms in Fig. 13.1, and the 1sa orbital is on the positive y side of the x–z plane with 1sb on the other side. We add a superscript “AO” to the tableau\n\n180\n\n13 Methane, ethane and hybridization\n\nTable 13.3. Principal standard tableaux functions for CH2 at the equilibrium internuclear geometry. 1 Num.a Tab.b Ci (min)\n\n1  2s 1sa  2 px 1sb   2p  y 2 pz 0.379 324 12 \n\n2\n\n3\n\n2  2 py 2 py 1sa   2s  2p  x 2 pz 0.092 437 13\n\n2  2s 2s  2 px 1sb   2p  z 1sa –0.091 619 94\n\n4 1  2s 2s  2 px 2 p y   1s  a 1sb –0.088 823 52 \n\na\n\nThe number of terms in the symmetry function that is generated from the tableau shown. (See text). b These tableau symbols exclude the core orbitals.\n\nsymbol to distinguish it from tableaux we introduce later that have hybrid orbitals in them. As we have pointed out many times previously, the columns of the standard tableaux functions are antisymmetrized, and the orbitals in a column may be replaced by any linear combination of them with no more than a change of an unimportant overall constant. In this case, consider a linear combination that has two hybrid orbitals that point directly at the H atoms in accord with Pauling’s principle of maximum overlap. Using the parameter φ we have three orthonormal hybrids  h a = cos(θ )(2s) + sin(θ)[sin(φ/2)(2 p y ) + cos(φ/2)(2 pz )],     h b = cos(θ )(2s) + sin(θ)[−sin(φ/2)(2 p y ) + cos(φ/2)(2 pz )],  cos(φ/2) sin(θ)(2s) −cos(θ )(2 pz )   , hz =  2 cos2 (θ) + cos2 (φ/2) sin (θ )\n\n(13.2)\n\nwhere φ (>π/2)1 is the angle between these two hybrids, and θ = arccot\n\n\u000f \u000e − cos(φ) .\n\nWe keep the 2 px orbital unchanged. This set is a variant of the canonical sp2 hybrid set in which, however, all three orbitals are not symmetrically equivalent. If φ is the H–C–H angle, h a and h b point directly at the H atoms. Because of the invariance 1\n\nHybrids consisting of s and p orbitals without this angle restriction can be complex. We are not interested in such cases.\n\n13.1 CH, CH2 , CH3 , and CH4\n\n181\n\nTable 13.4. Energies for T1AO and T1h R as a function of hybrid angle. \u0010 Hybrid angle\n\n\u0011 T1AO |H |T1AO \u0010 AO AO \u0011 au T1 |T1\n\n120.0 130.0 140.0 150.0\n\n38.479 436 −38.479 436 −38.479 436 −38.479 436\n\nof the function in Eq. (13.1), we obtain  T1AO\n\nha  hb =  hz 2 px\n\n\u0010\n\n\u0011 T1h R |H |T1h R \u0010 h R h R \u0011 au T1 |T1 −38.595 532 −38.599 825 −38.597 295 −38.590 391\n\n 1sa 1sb  . \n\n(13.3)\n\nWe have written out these hybrids, but the reader should realize that the eightelectron wave function (including the 1s 2 ) based upon the standard tableaux function of Eq. (13.3) has an energy expectation value independent2 of the angle parameter used in the hybrids, so long as φ > π/2. There are, however, nine standard tableaux functions for the orbital configuration in Eq. (13.3). These may be combined into five other combinations of 3 B1 symmetry. The T1AO above is the only one that shows the invariance to hybrid angle. When we combine all five in a wave function, the energy does depend upon the hybrid orbital directions. Nevertheless, as we add more and more structures to the wave function, we eventually arrive at a full calculation, and the energy is again invariant to the hybrid orbital directions. Thus the principal of maximum overlap has a meaning only for wave functions that do not involve a linear combination of all possible structures for the underlying AO basis. Some further numerical examples are illuminating when we compare the standard tableaux function results with those of HLSP functions. We define   1sa ha  hb 1sb   . T1h R =  (13.4)  2 px  hz R Consider the energies in Table 13.4, where we see that the energy of T1h R varies up and down around 100–200 meV in the angle range shown, while that of T1AO is 2\n\nThis does not mean that the energy of this T1h is independent of the actual angle in the molecule. Among other things, the nuclear repulsion energy depends upon the distance between the H atoms.\n\n182\n\n13 Methane, ethane and hybridization\n\nconstant. Using the methods of Chapter 5, we may write T1h R in terms of standard tableaux functions: \n\nT1h R\n\nha 12   hb = 5  hz 2 px  ha 6 1s a −  5 h z 2 px  ha  9 1sa −  5  hb 2 px\n\n  1sa ha   6 1sb  h b  − 5 1sb 2 px   hb ha  1sa 3 1sb  +   5  1sb 2 px   1sb ha   h z  9  1sa  + 5  hb hz\n\n  1sa ha   6 hz  h b  + 5 1sb hz   hb ha  1sa 3 hz  −   5  1sb hz   1sb ha   2 px  6  1sa  − 5  hb 1sb\n\n 1sa 2 px     hb 2 px     hz 2 px   . (13.5) \n\nWe see immediately that T1AO , the standard tableaux function invariant to the hybrid angles, is actually the largest term in T1h R , but not overwhelmingly so. The others all depend on the hybrid angles and, therefore, so does T1h R . We may also note that using hybrid orbitals T1h R has a lower energy by ≈3 eV, but, as seen from Table 13.1, the full calculation, with either sort of basis, is still more stable by another ≈3 eV. The calculated value of the dipole moment is 0.6575 D for this basis with the charge positive at the H-atom end of the bonds. CH3 Adding an H atom to CH2 might be expected to do little more than regularize the hybrids we gave in Eq. (13.2), converting them to a canonical sp2 set. With this we expect a planar doublet system. Whether the molecule is really planar is difficult to judge from qualitative considerations. Calculations and experiment bear out the planarity, however. A full valence orbital VB calculation in this basis involves 784 standard tableaux functions, of which only 364 are involved in 68 2 A2 \u0002\u0002 symmetry functions. For CH3 we present the results in terms of sp2 hybrids. This has no effect on the energy, of course. We show the principal standard tableaux functions in Table 13.5. The molecule is oriented with the C3 -axis along the z-axis and one of the H atoms on the x-axis. The three trigonal hybrids are oriented towards the H atoms. The “x” subscript on the orbital symbols in Table 13.5 indicates the functions on the x-axis, the “a” subscript those 120◦ from the first set, and the “b” subscript those 240◦ from the first set.\n\n13.1 CH, CH2 , CH3 , and CH4\n\n183\n\nTable 13.5. Principal standard tableaux function structures for CH3 at equilibrium bond distances.\n\nNum.a Tab.b Ci (min)\n\n1\n\n2\n\n3\n\n1  h x 1sx  h a 1sa  h 1sb  b 2 pz 0.521 922 72\n\n1  h x 1sa  1sx 1sb  h 2 pz  a hb −0.320 281 42\n\n1  h x 1sx  h a 1sb   1s 2 p  a z hb −0.211 957 78\n\n4\n\n3  hx hx  h a 1sa  h 1sb  b 2 pz 0.175 124 90 \n\na\n\nThe number of terms in the symmetry function that is generated from the tableau shown. (See text.) b These tableau symbols exclude the core orbitals.\n\nTable 13.6. Second moments of the charge for CH3 .\n\na\n\na\n\nComponent\n\nValue (D A )a\n\n(x x + yy + zz) (2zz − x x − yy)/2 (x x − yy)/2 xz yz xy\n\n−24.160 72 −1.829 70 0.0 0.0 0.0 0.0\n\na\n\nUnits of debye angstroms.\n\nReturning to the entries in Table 13.5, we see that the principal standard tableaux function is based upon the C 5 S state in line with our general expectations for this molecule with three C—H bonds. We considered in some detail the invariance of this sort of standard tableaux function to hybrid angle in our CH2 discussion. We do not repeat such an analysis here, but the same results would occur. As we have seen in Chapter 6, standard tableaux functions frequently are not simply related to functions of definite spatial symmetry. The second and third standard tableaux functions are members of the same constellation as the first, but are part of pure 2 A2 \u0002\u0002 functions only when combined with other standard tableaux functions with smaller coefficients that do not show at the level to appear in the table. These other standard tableaux functions are associated with L S-coupled valence states of carbon at higher energies than that of 5 S. The fourth term is ionic and associated with a negative C atom and partly positive H atoms. The dipole moment of CH3 is zero, of course, but the second moments of the charge have been determined, and r 2 and the quadrupole moments are given in Table 13.6. The sign of the z-axial quadrupole term indicates the distribution of\n\n184\n\n13 Methane, ethane and hybridization\n\nFigure 13.2. Drawing of positive axial quadrupole.\n\ncharge in the wave function. Figure 13.2 shows the general shape of the z-axial quadrupole with the signs of the regions. Since the moment of the molecule is negative, we see that its signs are reversed compared to those in the figure, and the individual C—H bonds are relatively positive at the H-atom ends. We may arrive at this conclusion another way. In Table 13.6 the components (x x − yy)/2, x y, x z, and yz are zero indicating that the quadrupole field is cylindrically symmetric about the z-axis. The axial moment around the x- or y-axis is (2x x − yy − zz)/2 = 3(x x − yy)/4 − (2zz − x x − yy)/4,\n\n(13.6)\n\n= (2yy − x x − zz)/2,\n\n(13.7)\n\n= 0.91485 D A,\n\n(13.8)\n\na\n\nfor which the positive sign again indicates the positive nature of the H end of the C—H bonds. This direction of the dipole moment is the same as that of CH and CH2 , and is again expected because of the relative predominance in the wave function of the ionic term shown in Table 13.5. CH4 We expect methane to be formed by the combination of an H-atom with the remaining unpaired pz orbital of CH3 . If the principal configuration is still that involving the C 5 S state and its nondirectional character predominates, we expect methane to be tetrahedral, thereby minimizing the repulsion energy between pairs of H atoms. This is borne out by the calculations as we see in Table 13.7.\n\n13.1 CH, CH2 , CH3 , and CH4\n\n185\n\nTable 13.7. Principal standard tableaux function structures for CH4 at equilibrium bond distances.\n\nNum.a Tab.b Ci (min)\n\n1\n\n2\n\n3\n\n4\n\n1  h a 1sa  h b 1sb   h 1s  c c h d 1sd 0.372 037 96\n\n1  h a 1sb  1sa 1sc  h hd  b h c 1sd −0.155 900 88\n\n4  hd hd  h a 1sa   h 1s  b b h c 1sc 0.129 792 00\n\n1  h a 1sa  h b 1sc   1s h  b d h c 1sd −0.105 365 69\n\na\n\nThe number of terms in the symmetry function that is generated from the tableau shown. (See text.) b These tableau symbols exclude the core orbitals.\n\nTable 13.8. Apparent partial electronic charge on H atoms based upon lowest nonzero moment and the corresponding calculated bond lengths, STO3G basis. a\n\nMolecule\n\nCharge\n\nRCH A\n\nCH CH2 CH3 CH4\n\n0.254|e| 0.137 0.194 0.188\n\n1.231 1.163a 1.145 1.150\n\na\n\nThe H—C—H bond angle is 129.1◦ .\n\nA full valence calculation on CH4 gives 1764 standard tableaux functions, and all of these are involved in the 164 1 A1 symmetry functions. The second and fourth tableaux are also present in the principal constellation and, as with the earlier cases, these are not simple symmetry functions alone. The third tableau is ionic with the negative charge at the C atom. As before, this contributes to the relative polarity of the C—H bonds. This is seen from a calculation of the electric moments. Methane has no nonzero a dipole or quadrupole moments, but the x yz component of the octopole is 1.144 D A2 . All of the others are zero if the molecule has the orientation in the coordinate system that is used here. The value is positive, showing the same qualitative electronic distribution in C—H bonds as was seen for the other CHn molecules we have examined. Quantitatively, the octopole moment is equivalent to a charge of 0.204|e| at the H-atom nuclei. For easy comparison we show in Table 13.8 the apparent charge on the H atom in each of our molecules. The trend in these charges is broken between one and\n\n186\n\n13 Methane, ethane and hybridization\n\nTable 13.9. Statistics for 6-31G∗ calculations of CHn .\n\nCH CH2 CH3 CH4\n\nState\n\nNum. symm. funcs.\n\nNumber of tableaux\n\n\u0001 B1 2 \u0002\u0002 A2 1 A1\n\n213 828 1597 2245\n\n546 1651 9375 26 046\n\n2\n\n3\n\ntwo H atoms. The likely interpretation here is that this is the place where the most important atomic configuration changes as one progresses through the list. This is seen clearly in Tables 13.2, 13.3, 13.5, and 13.7, where the principal configuration in the wave functions is shown. In the early days of VB theory workers were concerned with the “valence state” of carbon. Our calculations cannot really address this question because it is well defined only within a perfect pairing single tableau wave function.3 The notion was contrived to explain the relatively constant bond energies through the CHn series, while there is a requirement to pay back the energy loss in having the principal configuration change to higher energy. In the context of a full valence calculation we may only give a somewhat more qualitative argument. The 5 S state of C is about 4 eV above the ground state. This suggests that each of the actual C—H bond energies in CH2 with respect to some hypothetical frozen carbon state is about 2 eV higher than the apparent calculated or measured value. We attribute this to the greater effectiveness for bonding when spn hybrids are involved. 13.1.2 6-31G∗ basis After our discussion of the STO3G results we, in this section, compare some of these obtained with a 6-31G∗ basis arranged as described in Chapter 9. As before, we find that the larger basis gives more accurate results, but the minimal basis yields more useful qualitative information concerning the states of the atoms involved and the bonding. The statistics on the number of symmetry functions and standard tableaux functions for the various calculations are given in Table 13.9. From Table 13.10 we see that the bond distances are reproduced better in this case than with the STO3G basis. We see that the break in the trend between CH and CH2 again appears, and we continue to attribute it to the change in the important atomic configuration at this juncture in the list. The calculated bond distances are about 4.2% high. The success in calculating bond energies is more difficult to assess, since there is considerably more uncertainty in the experimental results. 3\n\nEven then, it is a purely theoretical concept. There appears to be no experimental approach to the energy of this state.\n\n13.2 Ethane\n\n187\n\nTable 13.10. Energies, bond distances, and bond energies of CHn for 6-31G∗ bases. a\n\nBond length (A )\n\nC CH CH2 CH3 CH4 a\n\nDe (eV)\n\nEnergy (au)\n\nCalc.\n\nExp.\n\nDissociation\n\nCalc.\n\nExp.\n\n−37.712 51 −38.321 54 −38.983 55 −39.624 08 −40.295 47\n\n1.169 1.104a 1.109 1.119\n\n1.1190 1.029 1.079 1.094\n\nDC−H DCH−H DCH2 −H DCH3 −H\n\n2.978 4.502 3.918 4.758\n\n3.462 4.33 ≤4.90 4.406\n\nThe H—C—H bond angle is 130.5◦ .\n\nTable 13.11. Various multipole moments and the apparent charges on H atoms from 6-31G∗ calculations. Momenta CH CH2 CH3 CH4 a\n\nD D AQ O\n\nValue 1.20030 0.53033 −1.33895 0.58764\n\nCharge 0.214|e| 0.119 0.152 0.022\n\nD, dipole; AQ, axial quadrupole; O, octopole.\n\nThe apparent charges on the H atoms in this basis are shown in Table 13.11. These may be compared to the similar values in Table 13.8. We see that the larger basis yields smaller values, particularly for methane. Nevertheless, we still predict that the H atoms in these small hydrocarbons are more positive than the C atom.\n\n13.2 Ethane Ethane presents a considerably greater challenge for calculation than the single carbon molecules above. Even if we continue the practice of putting 1s electrons in the “core” we have seven bonds and 14 electrons. A full minimal basis calculation, such as with STO3Gs, will produce 2 760 615 standard tableaux functions or HLSP functions for the total 14-electron basis. Not all of these are 1 A1g (assuming D3d symmetry) but the number would be considerable. With 14 electrons and 14 orbitals, none doubly occupied, there are 429 possible Rumer diagrams or standard tableaux functions. We will not attempt any “full” calculations with ethane, but rather focus on basis set arrangements that are designed to yield useful results with greater efficiency.\n\n188\n\n13 Methane, ethane and hybridization\n\nTable 13.12. Energies for covalent only calculations of D3d and D3h ethane.\n\nTreatment Cartesian AO Hybrid AO Perfect pairing (hybrid) a\n\nEnergy (hartree)\n\nNum. symm. funcs.\n\nD3d\n\nD3h\n\n52 52 1\n\n−78.367 895 −78.577 391 −78.565 885\n\n−78.575 229 −78.563 937\n\na\n\nThis was not run.\n\nWe first contrast using a Cartesian basis with sp3 hybrids on the C atoms for a covalent-only calculation.4 Table 13.12 shows these along with the perfect pairing energy. We see that there is a considerable lowering of the energy at E = 5.7 eV from using hybrid orbitals on the C atom instead of the original Cartesian basis. The hybrids are arranged to be pointing at the H atoms and the other C atom. We also see that the perfect pairing wave function is not a great deal higher in energy than the full covalent-only energy at E = 0.313 or 0.307 eV for the D3d or D3h geometry, respectively. The perfect pairing function is the only Rumer tableau that is a symmetry function by itself. We saw earlier that a perfect pairing function with Cartesian AOs is frequently not sensible, and this is another such case. Because they have no ionic states, the previous covalent-only results have too high a kinetic energy contribution, as discussed in Chapter 2. Adding all possible ionic states would lead to the very large number of basis functions quoted in the first paragraph of the discussion of ethane. We will consider the following physical arguments that may be used to limit the number of ionic state functions. This will all be done in the context of hybrid orbitals on the C atoms. 1. Adjacent ionic structures are the most important. This is expected since reductions in the kinetic energy will only occur if the overlap between the orbitals is fairly sizable. This is accomplished by assigning two electrons to each pair of orbitals that are arranged to bond in the molecule, and then requiring that this pair always have two electrons occupying them. 2. Only a few ionic bonds are required. We accomplish this by restricting the number of doubly occupied orbitals in a structure. 3. Highly charged atoms are unlikely. We accomplish this by preventing the charge depletion or build-up on either C atom from being outside ±1.\n\nTable 13.13 shows the energies for several treatments of ethane using these arguments. The first addition of one set of ionic structures per basis function produces 4\n\nThe reader is reminded that different linear combinations of the AOs yield different energies for less than full treatments.\n\n13.3 Conclusions\n\n189\n\nTable 13.13. Energies for various hybrid orbital calculations of D3d and D3h ethane. Energy (hartree)\n\nIonic structures\n\nNum. symm. funcs.\n\nTableaux\n\nD3d\n\nD3h\n\n0 1 2\n\n52 214 448\n\n429 2277 4797\n\n−78.577 391 −78.731 700 −78.742 547\n\n−78.575 228 −78.730 171 −78.741 195\n\nTable 13.14. Internal rotation barrier in ethane. Energy (eV) Ionic structures\n\nTheory\n\n0 1 2\n\n0.059 0.042 0.037\n\nExp.\n\n0.127a a\n\nSee Ref. .\n\na lowering of ≈4.2 eV, or nearly 0.6 eV per bond. The second ionic structure produces only 0.04 eV more per bond. In Chapter 2 the lowering of the energy in H2 when the ionic states are added is nearly 1 eV. The overlap there is rather greater at ≈0.9 than the values here, which are around 0.7 for either a C—H or a C—C bond. We have calculated ethane in both D3d and D3h geometries. From Table 13.13 we obtain the calculated barriers to internal rotation given in Table 13.14. It is seen that the calculated barrier height is falling as the number of ionic states increases. It is not yet converged, but we do not give the result obtained by including three ionic structures in the basis functions. The interested reader can work this out. The trend here with the addition of ionic states runs counter to predictions using another method published by Pophristic and Goodman. In addition it appears that this minimal basis calculation is unable to give a result close to the experimental value for the rotation barrier. We do not pursue this further here, but leave it as an open question.\n\n13.3 Conclusions In its original form VB theory was proposed using only states of atoms like the 5 S for C that we have invoked in describing our results. These are produced by standard\n\n190\n\n13 Methane, ethane and hybridization\n\ntableaux functions of the particular sort that is antisymmetric with respect to the interchange of any of the four C orbitals 2s, 2 px , 2 p y , and 2 pz . The functions based upon the other standard tableaux of the constellation correspond to the inclusion of other L S states of the same configuration. Although not as important in the wave function, these functions do enter and allow one to infer that the step suggested by Slater and Pauling, the inclusion of all states of a configuration, was an important addition to the VB method. Our results in this chapter also show that using hybrid orbitals with restricted bases can make an important improvement in the wave functions, at least when the criterion is energy lowering. We also see that the number of basis functions grows rapidly with the number of electrons. In Chapter 16 we will discuss another method for dealing with the escalation of basis size with greater numbers of atoms and electrons.\n\n14 Rings of hydrogen atoms\n\nIn this chapter we examine some results for four model systems consisting of rings of H atoms. These calculations show how the number of atoms in a complex reaction may influence rates of reaction, particularly through the activation energy. The systems are as follows. r Four H atoms in a rectangular geometry of D symmetry. The rectangle is characterized 2h\n\nby two distances, RA and RB. We map out a region of the ground state energy for this four-electron system as a function of the two distances. r Six H atoms in a hexagonal geometry of D symmetry. This is not a regular hexagon, in 3h general, but, like the system of four H atoms, is characterized by two distances we also label RA and RB. These two distances alternate around the ring. We also calculate the map of the ground state energy for this six-electron system. r Eight H atoms in an octagonal geometry of D symmetry and the specific shape char4h acterized by the RA and RB variables as above. For this larger system we only determine the saddle point with respect to the same sort of variables. r Ten H atoms in a decagonal geometry of D symmetry and the RA and RB variables. 5h Again, we determine only the saddle point.\n\nSince the geometries of these systems are in most regions not regular polygons, we will symbolize them as (H2 )n , emphasizing the number of H2 molecules rather than the total number of atoms. a For any of these, if RA = 0.7 A and RB is quite large, the rings represent 2–5 normal H2 molecules well separated from one another. If the roles of RA and RB are reversed, the H2 molecules have executed a metathesis in which the molecules transform into an equivalent set. These are, without doubt, somewhat artificial systems. For real systems, one could not tell if anything happened, unless isotopic labeling could be arranged. An even greater problem would occur in the gas phase, since the entropy penalty required for these peculiar geometries would be expected to make them very improbable. 191\n\n192\n\n14 Rings of hydrogen atoms\n\nTable 14.1. Number of symmetry functions of three types for H-ring calculations of (H2 )2 and (H2 )3 .\n\n(H2 )2 (H2 )3\n\nBase configs.\n\nSingle exc.\n\nDouble exc.\n\nTotal\n\n8 13\n\n17 130\n\n33 411\n\n58 554\n\nState 1\n\nA1g A\u00021\n\n1\n\nTable 14.2. Number of symmetry functions for saddle point calculations of (H2 )4 and (H2 )2 .\n\n(H2 )4 (H2 )5\n\nNum. Symm. Funcs.\n\nNum. tab.\n\n146 768\n\n1134 7602\n\nState 1\n\nA1g A\u00021\n\n1\n\nNevertheless, the results have considerable interest, bearing, as they do, on the same sort of considerations as the Woodward–Hoffman rules. 14.1 Basis set The calculations were all performed with an “s”-only basis of a 1s and a “2s” at each center. These are written in terms of the Huzinaga 6-Gaussian function as (6/42). This is the s part of the basis used in Chapter 2 for the H2 molecule and is shown in Table 2.2. It will be recalled that the “2s” orbital is not a real H2s orbital, but the second eigenfunction for this basis. As such it provides orbital breathing flexibility in the wave function. We show some statistics for these calculations in Table 14.1. Ionic states are restricted to ±1 at any center. The saddle point calculations for the larger two systems were carried out with more restricted bases involving valenceonly covalent and single-, and double-ionic structures. The statistics for these are shown in Table 14.2. 14.2 Energy surfaces The energy surface for (H2 )2 , divided by 2, is shown in Fig. 14.1, and that for (H2 )3 , divided by 3, is in Fig. 14.2. Because of the division by the number of H2 molecules, the energy goes to −1 hartree as RA and RB both grow large. Examination of the two surfaces shows clearly that they are quite different. The (H2 )2 energy surface has a fairly sharp ridge between the two stable valleys. This is completely missing in the (H2 )3 case. The difference between the energies E H4 /2 − E H6 /3\n\n14.2 Energy surfaces\n\n193\n\nEnergy (hartree) −0.6 −0.7 −0.8 −0.9 −1.0 −1.1 −1.2\n\n0.4 0.6 0.8 1.0 1.2 2.0 2.2 1.4 RA (Å) 1.6 1.6 1.8 1.4 1.8 1.0 1.2 RB (Å) 2.0 2.2 0.4 0.6 0.8\n\nFigure 14.1. The energy surface for the conversion of 2H2 → 2H2 in D2h geometries. The energy is per H2 molecule. Energy (hartree) −0.6 −0.7 −0.8 −0.9 −1.0 −1.1 −1.2\n\n0.4 0.6 0.8 1.0 1.2 2.0 2.2 1.4 RA (Å) 1.6 1.6 1.8 1.4 1.8 1.0 1.2 RB (Å) 2.0 2.2 0.4 0.6 0.8\n\nFigure 14.2. The energy surface for the conversion of 3H2 → 3H2 in D3h geometries. The energy is per H2 molecule.\n\nfor the two surfaces is plotted in Fig. 14.3. This is everywhere >−0.001 eV within the region of the plot. Thus E H4 /2 is always relatively higher. In Table 14.3 we show the saddle points and activation energies of the four systems. It is seen that there is a tendency for the quantities to alternate between higher and lower values as the number of H2 molecules is either even or odd. The differences decrease, however, as the rings become larger, and it appears that further calculations might show that the effect levels out. Nevertheless, the activation energy for the (H2 )2 system is almost three times higher than that of the\n\n194\n\n14 Rings of hydrogen atoms\n\nTable 14.3. Properties of the saddle points for the four hydrogen rings. RA A\n\nRB A\n\nEnergy au\n\nActivation energy eV\n\n1.310 0.998 1.203 1.107\n\n1.310 0.998 1.203 1.107\n\n–1.0367 –1.1067 –1.0475 –1.0819\n\n3.04 1.14 2.75 1.81\n\na\n\n(H2 )2 (H2 )3 (H2 )4 (H2 )5\n\na\n\nEnergy (eV) 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 −0.5 0.4 0.6 0.8 1.0 1.2 0.4 0.6 1.4 0.8 1.0 RA (Å) 1.6 1.2 1.4 1.8 1.6 2.0 1.8 2.0 RB (Å) 2.2 2.2\n\nFigure 14.3. The difference between the energy surfaces for the conversion of 2H2 → 2H2 in D2h geometries and 3H2 → 3H2 in D3h geometries. The energy is per H2 molecule. NB We have changed the aspect from that of Figs. 14.1 and 14.2. a\n\n(H2 )3 system, the saddle point is 0.3 A farther out and therefore more difficult to reach. We can gain further insight into the differences between these two systems from an examination of the two 2 × 2 matrix systems based upon the Kekul´e-like HLSP functions. These are, of course, the only structures for (H2 )2 , but for this comparison we ignore the long-bond functions in the other rings. We show the elements of these two matrix systems in Table 14.4. These systems are particularly simple since the diagonal elements of H are equal, giving simple expressions for the eigenvalues of the problem, viz., H11 ± H12 , 1 ± S12 H12 − S12 H11 = H11 ± . 1 ± S12\n\nE=\n\n(14.1) (14.2)\n\n14.2 Energy surfaces\n\n195\n\nTable 14.4. Values of elements in the two-state HLSP function matrix systems for four hydrogen rings, all evaluated at the saddle points.\n\n(H2 )2 (H2 )3 (H2 )4 (H2 )5\n\n2H11 /n(= 2H22 /n)\n\n2H12 /n\n\nS12\n\n2(H12 − S12 H11 )/n\n\n−1.0113 −1.0279 −1.0422 −1.0452\n\n0.7058 −0.6308 0.3284 −0.2474\n\n−0.6763 0.5752 −0.3015 0.2242\n\n0.02196 −0.03955 0.01418 −0.01307\n\nAs we see from Table 14.4, there is a peculiar alternation in sign in passing through the series. This results in the lower energy arising from the upper or lower sign in Eq. (14.2) for n odd or even in (H2 )n , respectively. If we represent the two Kekul´e structure functions by the symbols K 1(n) and K 2(n) , the wave function at the saddle point is \u0001 \u0002 (n) = Nn K 1(n) − (−1)n K 2(n) . \u0005sad\n\n(14.3)\n\nThe immediate consequence of this is a tendency for the electrons to stay away from the center of the ring for the even n systems. For example, consider the n = 2 case: \u0003\u0004 \u0004 \u0005 \u0005 \u0006 a b b c (2) = 0.54615 − , (14.4) \u0005sad c d R d a R where a, b, c, and d are the four orbitals around the ring, in order. This function would certainly be zero if the electrons were at locations such that all of the orbitals were of equal value. Because of the Pauli principle each of the Rumer tableaux functions is also zero at such a point, but there is an extra tendency toward zero because of the difference in Eq. (14.4). These do not occur in the odd n cases. Such a point is the center of the ring. Therefore, we interpret the in-and-out alternation of the saddle point as a result of the extra tendency of the electrons in the even systems to avoid the center. As the ring becomes larger and the center farther away, the effect would be expected to decrease. We also note that the values of the energy and overlap elements vary monotonically with n, contrary to the alternating characteristic so far emphasized. Their specific values give a maximum at n = 3 in the last column, however. The maximum remains after division by (1 ± S12 ). Thus the (H2 )3 system has the largest interaction at the saddle point, as measured by the energy decrease when the two structures interact. It is also interesting that H11 decreases as n increases. An immediate explanation for this is not available.\n\n196\n\n14 Rings of hydrogen atoms\n\nThe real reactions that most resemble these are the production of cyclobutane from two ethylene molecules ((H2 )2 ) and the Diels–Alder reaction between butadiene and ethylene ((H2 )3 ). Even these cannot be made to react in the bare forms, but fairly simple activation by substituents will allow the (H2 )3 analog to proceed. Apparently, no form of the (H2 )2 analog has ever been observed. Our analysis suggests that there is a fundamental difference between the four-electron and six-electron systems that produces the effect. The book by Woodward and Hoffman may be consulted for a rationalization of these results based upon MO theory. As we continue to larger rings, the results are not so clearcut. There is a tendency for the saddle points to alternate in and out somewhat, but the interaction energy appears to be a maximum at the ring of six atoms. These last more-or-less qualitative comments have been based upon just the two simple Kekul´e-like structures of the rings, and may not be able to show the proper behavior. The actual surface calculations of (H2 )2 and (H2 )3 included many more structures and show strikingly different qualitative behavior.\n\n15 Aromatic compounds\n\nBenzene is the archetypal aromatic hydrocarbon and its study has been central to the understanding of aromaticity and resonance from the early times. In addition, it has the physical property of having its π electrons reasonably independent from those in σ bonds, leading early quantum mechanics workers to treat the π electrons alone. Since benzene is a ring and the rules for forming Rumer diagrams have one draw noncrossing lines between orbital symbols written in a circle, the Rumer diagrams correspond to the classical Kekul´e and Dewar bond schemes that chemists had postulated far earlier than the VB treatments occurred. This parallel has intrigued people since its first observation and led to many discussions concerning its significance. It has also led to considerable work in more qualitative “graphical methods” for which the reader is directed to the literature. (See, inter alia, Randi´c.) We will examine benzene with different bases and also discuss some of the ideas that consideration of this molecule has led to, such as resonance and resonance energy. We show again the traditional five covalent Rumer diagrams for six electrons and six orbitals in a singlet coupling and emphasize that the similarity between the ring of orbitals and the shape of the molecule considerably simplifies the understanding of the symmetry for benzene.\n\na f e\n\nf\n\nc\n\ne\n\na\n\na\n\na b\n\nb\n\nf\n\nb\n\nc e\n\na b f\n\nf\n\nc e\n\nb c\n\nc e\n\nd\n\nd\n\nd\n\nd\n\nd\n\nK1\n\nK2\n\nD1\n\nD2\n\nD3\n\nMost of the discussion we give here on the nature of the wave function will focus on HLSP functions. An early ab initio study by Norbeck and the present author 197\n\n198\n\n15 Aromatic compounds\n\nTable 15.1. The four most important categories of HLSP functions in the wave function for benzene. 1 Num.a HLSPb Ci c\n\n\u0001\n\n2 \u0002 2 pa 2 pb 2 pc 2 pd 2 pe 2 p f R 0.160 88\n\n2 \u0001\n\n3 \u0002 2 pb 2 pc 2 pa 2 pd 2 pe 2 p f R −0.057 63\n\n3 \u0001\n\n12 \u0002 2 pa 2 pa 2 pc 2 pd 2 pe 2 p f R 0.051 62\n\n4 \u0001\n\n12 \u0002 2 pa 2 pa 2 pc 2 pc 2 pe 2 p f R 0.027 44\n\na\n\nThe number of terms in the symmetry function that is generated from the tableau shown. (See text.) b These tableau symbols exclude the core orbitals. c In this case all of the terms in a symmetry function have the same sign as well as magnitude for the coefficient.\n\non benzene focused more on interpretation of the standard tableaux function representation of the wave functions. Thus, the present discussion now differs from that earlier in some respects.\n\n15.1 STO3G calculation The Weyl dimension formula (Eq. (5.115)) tells us that six electrons in six orbitals in a singlet state yield 175 basis functions. These may be combined into 22 1 A1g symmetry functions. Table 15.1 shows the important HLSP functions for a π-only calculation of benzene for the SCF optimum geometry in the same basis. The σ orbitals are all treated in the “core”, as described in Chapter 9, and the π electrons are subjected to its SEP. We discuss the nature of this potential farther in the next section. The functions numbered in the first row of Table 15.1 have the following characteristics. 1. The two functions of this type are the classical Kekul´e structures for benzene. One might expect the coefficient to be larger, but we will see below why it is not. 2. These three functions are the classical Dewar structures. 3. The third set of functions, 12 in number, are all of the possible singly ionized structures where the charges are adjacent, and there are no long bonds. 4. The fourth set of functions are all of the doubly charged structures with the + and − charges adjacent and no long bonds.\n\nWhen there is a relatively high degree of symmetry as in benzene, the interpretation of the parts of the wave function must be carried out with some care. This arises from an apparent enhancement of the magnitude of the coefficient of a structure in the wave function when whole symmetry functions are used. Let us consider the\n\n15.1 STO3G calculation\n\n199\n\nKekul´e structures and denote them by K 1 and K 2 . As discussed in Chapter 5, when these HLSP functions are projected from the appropriate product of p orbitals K 1 = θ \u0004 PN Pρ1 , K 2 = θ \u0004 PN Pρ2 , they are not normalized to 1. The “raw” 2 × 2 overlap matrix is \u0003 \u0004 2.246 364 9 , 0.799 515 5 2.246 364 9 and, hence, the true overlap between a normalized K 1 and K 2 is 0.355 915 2. Thus, if we consider the 1 A1g symmetry function involving K 1 and K 2 , we obtain in its normalized form 1\n\nA1g = 0.607 251 7(K 1 + K 2 ),\n\nand, if the wave function is written in terms of this symmetry function, its coefficient would be 0.264 931 3 instead of the number listed in Table 15.1 for the individual Kekul´e structures. In these terms, the Kekul´e structures appear to have a larger coefficient. A similar analysis for the Dewar structures leads to an apparent enhancement of the coefficient magnitude to −0.133 825 9. The apparent enhancement we are discussing here is more pronounced, in general, the greater the number of terms in the symmetry function. We now consider the third sort of function from Table 15.1. These are the 12 short-bond singly ionic functions, and in this case the enhancement of the coefficient is a factor of 5.0685, i.e., the reciprocal of the normalization constant for the symmetry function that is the sum of the individually normalized HLSP functions. The resulting coefficient would then be 0.261 637, a number essentially the same as the coefficient of the Kekul´e symmetry function. Are the Kekul´e functions and the short-bond singly ionic functions really of nearly equal importance in the wave function? This appears to be the only possible conclusion and may be rationalized as follows. We have seen that the covalent-only structures provide for a considerable electron correlation, lowering their potential energies, but constrain the space available to the electrons, thereby raising their kinetic energies. Ionic structures allow delocalization that lowers the kinetic energy while not raising the potential energy enough to prevent an overall decrease in energy. When there are six bonds to be delocalized we expect the effect in the singly ionic structures to be roughly six times that for only one bond. Those we discuss are the adjacent only ionic structures and are expected to be the most important. We also observe that the diagonal element of the Hamiltonian for a single Dewar structure is about 1.8 eV higher than a diagonal element of a single Kekul´e structure. This is a very reasonable number for the difference in energies between a long and a short bond. The short-bond singly ionic structures have a diagonal element of the\n\n200\n\n15 Aromatic compounds\n\nHamiltonian that is about 8.6 eV above the single Kekul´e structure. This number is not exactly comparable to the ionic structures in H2 discussed in Chapter 2. Consider the two ionic structures,\n\na− f e d I1\n\na+\n\nb+\n\nf\n\nc\n\ne\n\nb− c\n\nd I2\n\nwhich are two of the 12 short-bond adjacent singly ionic structures in benzene. The 2 × 2 secular equation corresponding to these two functions is \u0005 \u0005 \u0005 8.562 − \u0005 E \u0005 \u0005 \u0005 −1.442 − 0.0995E 8.562 − E \u0005 = 0, where we have converted the energies to electron volts and have reset the zero to the energy of the single Kekul´e structure, K 1 . The lower root of this equation is 6.476 eV, which is ≈2 eV lower than the energy of I1 alone. We should compare this with the corresponding value for H2 , obtained with methods of Section 2.4, wherein 5.82 eV is obtained. Thus the effect on the diagonal energy of forming the ionic structure pair is in the same direction for the two systems, but much larger in the more compact H2 . 15.1.1 SCVB treatment of π system We have so far emphasized the nature of the wave function. We now examine the energies of some different arrangements of the bases. In Table 15.2 we show energies for five levels of calculation, Kekul´e-only, Kekul´e plus Dewar, SCF, SCVB, and full π structures, where energies are given as the excess energy due to the π system over that from the core. Cooper et al. gave the SCVB treatment of benzene. We note first that the covalent-only calculations give a higher energy than the SCF wave function. We noted this effect with the allyl radical in Chapter 10, and it happens again here with benzene. This is again a manifestation of the delocalization provided by ionic structures in the wave function and the concomitant decrease in the kinetic energy of the electrons. Since this phenomenon does not occur in cases where resonance is absent, we expect it to be greater where there are possibilities for greater numbers of more or less equivalent resonance structures. There is only one equivalent orbital in a highly symmetric π system like that in benzene. This is shown as an altitude plot in Fig. 15.1. We see that each orbital is\n\n15.1 STO3G calculation\n\n201\n\nTable 15.2. Comparison of different calculations of the π system of benzene. All energies are in hartrees. E − E core a −6.711 44 −6.755 53 −6.760 79 −6.834 10 −6.904 88 −6.911 87\n\nK1 K1 & K2 Full cov. SCF SCVB Full π a\n\nE core = −222.142 48.\n\nOrbital amplitude 0.4 0.3 0.2 0.1 0.0\n\n2\n\n−4\n\n−3\n\n−2\n\n−1\n\n0 y-direction (Å)\n\n1\n\n2\n\n3\n\n4 −4\n\n−3\n\n−2\n\n3\n\n4\n\n1 0 −1 x-direction (Å)\n\nFigure 15.1. Altitude plot of the SCVB orbital for the π system of benzene. There are six symmetrically equivalent versions of this around the ring. The amplitude is given in a plane a 0.5 A in the positive z-direction from the plane of the nuclei.\n\npredominantly a single 2 pz with smaller “satellites” in the ortho positions, essentially nothing at the meta positions, and a negative contribution at the para position. The coefficients in the two covalent-only wave functions, \u0007(pure 2 pz ) and \u0007(SCVB 2 pz ) are not very different: \u0007(pure 2 pz ) = 0.402 88(K 1 + K 2 ) − 0.150 26(D1 + D2 + D3 ), \u0007(SCVB 2 pz ) = 0.403 53(K 1 + K 2 ) − 0.122 03(D1 + D2 + D3 ). We may, however, examine the 2 × 2 Kekul´e-only matrices for these two cases. For the pure 2 pz orbitals we have, in hartrees, the secular equation \u0005 \u0005 \u0005 −6.711 44 − \u0005 E \u0005 \u0005 \u0005 −2.448 49 − 0.355 92E −6.711 44 − E \u0005 = 0,\n\n202\n\n15 Aromatic compounds\n\nTable 15.3. Lowering of the energy from resonant mixing for pure and SCVB 2 pz orbitals.\n\nK − only K+D\n\nPure eV\n\nSCVB eV\n\n−1.199 −1.343\n\n−0.870 −0.878\n\nTable 15.4. Comparison of some one- and two-electron matrix elements for pure and SCVB 2 pz orbitals. All energies are in hartrees.\n\nTa Vb [11|11]c [11|22]d\n\nPure\n\nSCVB\n\n1.2299 −2.9613 0.5795 0.3242\n\n1.0891 −2.8750 0.5125 0.3352\n\na\n\nKinetic energy. Nuclear and core potential energy. c Orbital self-repulsion energy. d Adjacent orbital repulsion energy. b\n\nand for the SCVB orbitals, \u0005 \u0005 −6.872 56 − E \u0005 \u0005 −4.616 57 − 0.664 00E\n\n\u0005 \u0005 \u0005 = 0. −6.872 56 − E \u0005\n\nWe see that the K 1 –K 1 diagonal element for the SCVB orbitals is already about 4.4 eV below that for the pure 2 pz orbitals. This is the most immediate explanation for the lower energy of the SCVB result. In fact, this is the larger effect. As seen in Eq. (14.2), the amount of energy lowering in 2 × 2 systems like these is H12 − S12 H11 . 1 + S12 In this case we have 1.20 eV and 0.87 eV for the pure 2 pz and SCVB orbitals, respectively, and the resonance appears somewhat more beneficial for the localized orbital. These results are included in Table 15.3. We may obtain more information from a comparison of some of the oneand two-electron integrals for the individual orbitals. The values are shown in Table 15.4. It is seen that the changes in the potential energy terms nearly cancel with the repulsive self-energy and the nuclear and core potential energies changing in opposite directions. The change in the adjacent repulsion energy is also not\n\n15.1 STO3G calculation\n\n203\n\nTable 15.5. Comparison of different calculations of the π system of linear 1,3,5-hexatriene. All energies are in hartrees. a E − E core\n\nK1 Full cov. SCF SCVB Full π a\n\n−6.129 74 −6.133 78 −6.164 40 −6.251 49 −6.252 41\n\nE core = −223.91056.\n\nlarge. There remains only the kinetic energy term, for which the difference is nearly 3.8 eV, with the SCVB orbital lower. This is easily interpreted to be the result of the delocalization in that orbital, and it thus makes the same sort of contribution as do ionic structures in MCVB wave functions. The values of the n-electron matrix elements are the result of an interplay of considerable complexity among the simpler one- and two-electron matrix elements, and it is not really possible to say much more about the effects of these latter quantities upon the total energies. We show another aspect of these numbers in Table 15.3, where we detail the effects of resonance between the two Kekul´e structures and among all of the covalent structures for the two sorts of 2 pz orbitals. The results suggest that within one structure the SCVB orbitals duplicate, to some extent, the effect of multiple structures, and the configurational mixing produces less energy lowering with them. Those familiar with the long history of the attacks on the question of the resonance energy of benzene may be somewhat surprised at the small numbers in Table 15.3. The energy differences that are given there are for just the sort of process that might be expected to yield a theoretical value for the resonance energy, but experimental determinations yield numbers in the range 1.7–2.3 eV. This is an important question, which we will take up in Section 15.3, where it will turn out that some subtleties must be dealt with. 15.1.2 Comparison with linear 1,3,5-hexatriene In order to put the structure of benzene into better perspective, we give a similar calculation of 1,3,5-hexatriene for comparison. Table 15.5 shows the energies for a set of calculations parallel to those in Table 15.2 for benzene. The most obvious difference is the smaller total spread in the energies, about 3.3 eV rather than the 5.5 eV for benzene, and the SCVB energy is closer to the full π than in benzene. Standard arguments say that there is only a rather small amount of resonance\n\n204\n\n15 Aromatic compounds Orbital amplitude 0.4 0.3 0.2 0.1 0.0\n\n2\n\n−5 −4 −3 −2 −1\n\n0 1 y-direction (Å)\n\n2\n\n3\n\n4\n\n5 −5\n\n−4\n\n3\n\n4\n\n5\n\n1 0 −1 x-direction (Å) −2 −3\n\nFigure 15.2. Altitude plot of the first SCVB orbital for the π system of 1,3,5-hexatriene. There are two symmetrically equivalent versions of this at each end of the molecule. The a amplitude is given in a plane 0.5 A in the positive z-direction from the plane of the nuclei. Orbital amplitude 0.4 0.3 0.2 0.1 0.0\n\n3\n\n−5 −4 −3 −2 −1\n\n0\n\ny-direction (Å)\n\n1\n\n2\n\n3\n\n4\n\n5\n\n−5\n\n−4\n\n−3\n\n−2\n\n−1\n\n0\n\n4\n\n5\n\n2 1 x-direction (Å)\n\nFigure 15.3. Altitude plot of the second SCVB orbital for the π system of 1,3,5-hexatriene. There are two symmetrically equivalent versions of this at each end of the molecule. The a amplitude is given in a plane 0.5 A in the positive z-direction from the plane of the nuclei.\n\nin linear hexatriene, since only the one Kekul´e structure has only short bonds, nevertheless, the SCF energy is lower than the full covalent energy. There are three inequivalent SCVB orbitals for hexatriene, and these are given in Figs. 15.2, 15.3, and 15.4. The first of these shows a principal peak at the first 2 pz orbital and a small satellite at the adjacent position. The second is more interesting with the principal peak at the second C atom, but showing a larger satellite at position 1 than at position 3. This is consistent with having essentially a double bond between atoms 1 and 2 or 5 and 6, with single bonds between 2 and 3 and 4 and 5.\n\n15.2 The 6-31G∗ basis\n\n205\n\nTable 15.6. Comparison of different calculations of the π system of benzene with a 6-31G∗ basis. All energies are in hartrees. E − E core a −6.416 20 −6.464 30 −6.479 44 −6.496 50\n\nSCF Full valence π SCVB Full valence π + Sb a b\n\nE core = −224.275 51. Single excitation.\n\nOrbital amplitude 0.4 0.3 0.2 0.1 0.0\n\n−5 −4 −3 −2 −1\n\n0 y-direction (Å)\n\n1 2\n\n1\n\n2\n\n3\n\n4\n\n−4 5 −5\n\n−3\n\n3\n\n4\n\n5\n\n−1 0 x-direction (Å) −2\n\nFigure 15.4. Altitude plot of the third SCVB orbital for the π system of 1,3,5-hexatriene. There are two symmetrically equivalent versions of this at each end of the molecule. The a amplitude is given in a plane 0.5 A in the positive z-direction from the plane of the nuclei.\n\nThe third orbital has a similar interpretation with a larger satellite at position 4 than at position 2. The overall conclusion is that there is considerably less resonance in hexatriene than in benzene, and the bond lengths and types alternate along the chain unlike the equivalence in benzene. 15.2 The 6-31G∗ basis The SCF, SCVB, full valence π , and full valence π + S results of using a 6-31G∗ basis on benzene are given in Table 15.6. The geometry used is that of the minimum SCF energy of the basis. In this case the SCVB energy is lower by 0.4 eV than the full valence π energy. This is principally due to the 3d polarization orbitals present in the SCVB orbital, but absent in the valence calculation. The SCVB orbital is\n\n206\n\n15 Aromatic compounds\n\nTable 15.7. The centroids of charge implied by the second moment of the charge distribution of the nuclear and σ framework. The C and H nuclear positions are those of the 6-31G∗ SCF equilibrium geometry. a\n\nCharge\n\nC+6 −5|e| H+1\n\n1.3862 1.8730 2.4618\n\nquantitatively so close to that from STO3G orbitals shown in Fig. 15.1 that the eye cannot detect any difference, and we do not draw a 6-31G∗ version of the orbital. The SEP is used again to represent the core, and the following analysis may be made to get a crude picture of its nature. The second moment of the x x(= yy) charge distribution of the core is 0.481 10 bohr2 . At each apex of the hexagon there is a C nucleus and an H nucleus farther out. There are also five electrons per apex contributed by the σ system. The (quadratic) centroid of this charge may be calculated from the second moment, and is shown in Table 15.7. The overlap and kinetic energy one-electron matrix elements of the π AOs are unaffected by the SEP. In addition, our centroid picture does not include any of the exchange effects present.1 The main point is that arguments using a nuclear charge effect of +1 for each π AO may be too simplistic for many purposes. We saw in Chapter 14 that a ring of six H atoms does not want to be in a regular hexagonal geometry at the minimum energy. A question concerning benzene arises then: Is benzene a regular hexagon because of or in spite of the resonance in the π system? The previous calculations have all been done with the regular hexagon geometry forced on the molecule. We now relax that constraint to test the stability of the ring against distortion into an alternating bond length geometry. In Table 15.8 we show the values and first and second derivatives of the SCF, core, valence, and total energies with respect to two distortion directions in the molecule. The first direction we call “a1g ” is a symmetric breathing motion involving the change in length of only the C—C bonds. (The C—H bonds are unchanged.) The second we call “b2u ”, and it involves an alternating increase and decrease in the lengths of the C—C bonds around the rings.2 (C—C—C, C—C—H angles, and C—H distances are not changed.) 1 2\n\nThese are not expected to be very large in a system like benzene where there is a natural symmetry-based orthogonality between the π and σ systems. These group species symbols correspond to the symmetry of vibrational normal modes of the same sort.\n\n15.2 The 6-31G∗ basis\n\n207\n\nTable 15.8. Energies and derivatives of energies relating to the stability of the regular hexagonal structure of benzene.\n\nEnergya ∂ E/∂( R)b ∂ 2 E/∂( R)2c\n\na1g b2u a1g b2u\n\nSCF\n\nCore\n\nπ\n\nTotal\n\n−230.691 71 −0.013 8 0.0 3.492 7 5.513 7\n\n−224.275 51 −2.050 6 0.0 5.287 8 16.742 9\n\n−6.496 50 1.981 1 0.0 −1.878 2 −10.517 4\n\n−230.772 02 −0.069 5 0.0 3.409 6 6.225 5\n\na\n\nHartrees. Hartrees/bohr. R is the change in the C—C distance in the ring in all cases. c Hartrees/bohr2 . b\n\nAt first glance the numbers in Table 15.8 suggest that the answer to the question in the last paragraph is the “in spite of” alternative. The last row of the table shows that the b2u distortion has a stable minimum in the core but a maximum in the π energy. The values are such that the total has a stable minimum. The calculations thus correctly predict that there is no force tending to distort the regular hexagon at the point represented by the regular geometry. Nevertheless, there is a difficulty with the interpretation in the last paragraph. There has developed over the years a considerable literature on this question, with many opinions on both sides. Shaik et al. have written articles on this subject[62, 63]. Such a situation frequently indicates the existence of ambiguities in the definition, and that certainly applies to this case. We may describe the situation according to our current terms. As stated above, we have used the SEP to obtain separate energies for the σ core and the π system. Conventionally, this consists of attributing the whole of the nuclear repulsion energy to the core. We called attention, however, to the nature of the energy surface for six H atoms in a ring (see Chapter 14), which, unlike the benzene π system in our treatment, does have nuclear repulsion included. It might be expected that we would need to make some sort of partitioning of the nuclear repulsion energy to make the H atom and the π systems comparable. One way to do this would be to imagine the effect of one unit charge from the C nuclei being subtracted from the core energy and added to the π energy. None of the totals is affected, of course. The result for the repulsion energy for six |e| charges at the positions of the C atoms is 4.185 55 hartrees, but more important for our purposes is that the second derivative is 0.8067 hartrees/bohr2 in terms of the same R coordinate used in Table 15.8. It is seen that it would take the switch of many more charges from the core to the π system than are actually present to reverse its tendency to distort the regular hexagon. Thus, we need not revise the earlier conclusion.\n\n208\n\n15 Aromatic compounds\n\nTable 15.9. Energies and derivatives of energies relating to the stability of the square geometry of cyclobutadiene.\n\nEnergya ∂ E/∂( R)b ∂ 2 E/∂( R)2c\n\nb2u b2u\n\nCore\n\nπ\n\n−150.522 61 0.0 5.9223\n\n−3.080 46 0.0 −6.9506\n\nTotal −153.603 07 0.0 −0.7283\n\na\n\nHartrees. Hartrees/bohr. c Hartrees/bohr2 . b\n\nThe size of the ring is seen to be a compromise between attraction due to the π system and repulsion due to the core. Again, the sizes of the second derivatives add to give a stable minimum. If we assume that the system behaves harmonically, the derivatives in the last column imply that the ring C—C bond distances are about a 0.0204 bohr (0.0108 A) longer than the value used for the first row of the table, which are the SCF minimum distances. 15.2.1 Comparison with cyclobutadiene It is illuminating to compare the behavior of benzene3 upon the b2u type of distortion with a similar calculation in cyclobutadiene. We do not repeat all of the calculations of the last section, but do include the results in Table 15.9, where we give the energies and the derivatives for the b2u distortion from the geometry of a square. It will be recalled from Chapter 14 that the energy surface for the ring of four H atoms has a higher peak at the square geometry than the six-atom ring does at the regular hexagon. The same situation applies here. The second derivative for the π system is larger in magnitude than the core derivative, making the square an unstable structure for C4 H4 . Thus we predict that cyclobutadiene in a singlet state has no tendency to form a molecule with equal ring bond lengths as happens in benzene. In these discussions of benzene and cyclobutadiene we have compared MCVB level calculations of the π system with SCF level calculations of the core. We do not expect that using correlated wave functions for core energies would change the results enough to give a different qualitative picture. 15.3 The resonance energy of benzene Once Kekul´e had deduced the correct structure of benzene, chemists soon realized that the double bonds in it were considerably more stable than isolated double 3\n\nThe same species symbol can serve in D4h symmetry.\n\n15.3 The resonance energy of benzene\n\n209\n\nbonds in aliphatic hydrocarbons. The principal evidence is that reaction conditions leading to addition to an aliphatic double bond with the removal of the multiple bond do not normally affect benzene, and more vigorous conditions cause an attack that removes a ring H atom and leaves the double bonds unchanged. The conclusion was that the “conjugated” ring double bonds possess an added stability due to their environment. It remained for quantum mechanics to explain this effect in terms of what has come to be called “resonance” among a number of bond structures. Experimental approaches to determining the resonance energy (called stabilization energy by some) have involved comparing thermodynamic measurements of benzene with those of three cyclohexenes. Heats of combustion and heats of hydrogenation have been used. Most feel the hydrogenation method to be superior, since it is expected to involve smaller differencing errors in the determination. The energies and processes are C6 H6 + 3H2 → C6 H12 ; H = −2.13eV, C6 H10 + H2 → C6 H12 ; H = −1.23eV. The difference,4 −1.54 eV, corresponds to the lower energy the three double bonds in benzene have than if they were isolated. This is not much larger than the “pure 2 pz ” entry in the second row of Table 15.4. It was pointed out by Mulliken and Parr, however, that this precise comparison is not what should be done. The number in the table from our calculation does not involve any change in the bond lengths whereas the experiment certainly does. Changes in energy due to bond length change come from both the π bonds and the σ core. It is possible to make a successful comparison of theory with experiment for the resonance energy modified according to the Mulliken and Parr prescription, but there are still many assumptions that must be made that have uncertain consequences. A better approach is to attempt calculations that match more closely what experiment gives directly. This still requires making calculations on what is a nonexistent molecule, but the unreality pertains only to geometry, not to restricted wave functions. Following these ideas, Table 15.10 shows results of 6-31G∗ calculations of the π system of normal benzene and benzene distorted to have alternating bond lengths matching standard double and single bonds, which we will call cyclohexatriene. The cyclohexatriene molecule has a wave function considerably modified from that of benzene. The first few terms are shown in Table 15.11, where the two 4\n\nMost workers change the sign of this to make it positive, but logically an energy corresponding to greater stability should be negative.\n\n210\n\n15 Aromatic compounds\n\nTable 15.10. Calculations of π energies for normal and distorted benzene.\n\nBenzene Cyclohexatriene\n\nEeV\n\nSCF\n\nMCVB\n\n−6.416 20 −6.363 07 −1.446\n\n−6.496 50a −6.442 91b −1.458\n\na\n\nThis gives 3211 tableaux functions formed into 280 symmetry functions. b This gives 3235 tableaux functions formed into 545 symmetry functions.\n\nTable 15.11. The first terms in the MCVB wave function for cyclohexatriene. 1 Num. HLSP Ci\n\n\u0001\n\n1 \u0002 2 pa 2 pb 2 pc 2 pd 2 pe 2 p f R 0.275 450\n\n2 \u0001\n\n6 \u0002 2 pa 2 pa 2 pc 2 pd 2 pe 2 p f R 0.082 38\n\n3 \u0001\n\n1 \u0002 2 pb 2 pc 2 pd 2 pe 2 pa 2 p f R 0.053 11\n\n4 \u0001\n\n3 \u0002 2 pc 2 pd 2 pb 2 pe 2 pa 2 p f R −0.040 78\n\n“Kekul´e” structures have quite different coefficients. We interpret the terms as follows. 1. This is the standard Kekul´e structure with the π bonds principally at the short distance. 2. The second group of functions, six in number, are adjacent single ionic structures corresponding to bonds in the position marked in function 1. 3. Function 3 is the other Kekul´e structure. Its importance in the wave function is low, indicating little π bonding at the long positions. 4. The fourth group consists of the three “Dewar” structures and is also relatively unimportant.\n\nWhen we look at the energies from Table 15.10, perhaps the most striking fact is that the correlation energy in the π system makes so little difference in the E values. As we indicated above, the experimental value for the resonance energy from heats of hydrogenation is −1.54 eV, in quite satisfactory agreement with the result in Table 15.10. The fact that our value is a little lower than the experimental one may be attributed to the small amount of residual resonance remaining in the cyclohexatriene, whereas the isolated double bonds in the experiment are truly isolated in separate molecules.5 5\n\nThere is still interest in the resonance energy of benzene. Beckhaus et al. have synthesized a molecule with a strained benzene ring in it and measured heats of hydrogenation. This is an experimental attempt to assay what we did theoretically. They found similar results.\n\n15.4 Naphthalene with an STO3G basis\n\n211\n\nTable 15.12. Core, π SCF, and π MCVB energies for various calculations of naphthalene. An STO3G basis is used, and all energies are in hartrees.\n\n“Kekul´e” Covalent Covalent + singleionic Covalent + singleand double-ionic Full π\n\nEnergy\n\nNum. symm. funcs.\n\nCore SCF π VB MCVB MCVB\n\n−366.093 70 −14.398 73 −14.221 75 −14.277 12 −14.476 32\n\n1 16 334\n\nMCVB\n\n−14.524 33\n\n1948\n\nMCVB\n\n−14.529 93\n\n4936\n\n15.4 Naphthalene with an STO3G basis We now consider naphthalene, which possesses 42 covalent Rumer diagrams. Many of these, however, will have long bonds between the two rings and are probably not very important. To the author’s knowledge no systematic ab initio study has been made of this question. The molecule has D2h symmetry, and these 42 covalent functions are combined into only 16 1 A g symmetry functions. As with benzene we study only the π system using the SEP to account for the presence of the σ orbitals. It is not the purpose of this book to compare MCVB with molecular orbital configuration interaction (MOCI) results, but we do it in this case.\n\n15.4.1 MCVB treatment We first give the MCVB results in Table 15.12, which shows energies for several levels of calculations with an STO3G basis. A full π calculation for naphthalene consists of 19 404 singlet tableau functions, which may be combined into 4936 1 A g symmetry functions. The covalent plus single ionic calculation involves 1302 singlet tableau functions, which may be combined into 334 symmetry functions, and the covalent, single-, and double-ionic treatment produces 7602 singlet tableau functions, which may be combined into 1948 symmetry functions. The results in Table 15.12 show again that the SCF function has a lower energy than the covalent-only VB. Although a thorough study has not been made, it appears that this difference increases with the size of the system. Certainly, the decrease in energy upon adding the single-ionic structures to the basis is greater here than in benzene, 2 or 3) to be considered nonsingular by standard 16-place accuracy calculations. Therefore, for each high-overlap pair the sum and difference were formed. These are orthogonal, and do not cause any problems.\n\n16.2.1 The least motion path We first comment on the so-called least motion path (LMP), in which the two H atoms move away from the CO atoms, maintaining a C2v symmetry, as shown in Fig. 16.6. Earlier calculations of all sorts indicate that this path does not cross\n\n16.2 Formaldehyde, H2 CO\n\n227\n\nH O\n\nH\n\nC\n\nH H\n\nFigure 16.6. A representation of the LMP for the dissociation of H2 CO. H H O\n\nC\n\nFigure 16.7. The true saddle point for the dissociation of H2 CO. This figure is drawn to scale as accurately as possible.\n\nthe lowest saddle point for the reaction. In fact, there is no real saddle point in geometries constrained to be C2v . Earlier workers have, however, imposed further constraints and produced a pseudo saddle point of this sort. This is done because it illustrates a typical four-electron rearrangement similar to the process discussed in Chapter 14 for four H atoms. This is classified by Woodward and Hoffman as a “forbidden” process, which means, of course, that the energy required for it is relatively high compared to the energy for other geometries that may break the symmetry giving the orbital crossing. In any event the forces on the nuclei along restricted paths such as this tend to lead to separation of all three parts of the molecule rather than the formation of CO and H2 .\n\n16.2.2 The true saddle point Calculations using both MCVB and MOCI wave functions predict a very different geometry at the saddle point for the H2 CO dissociation. The molecule is still planar, but otherwise has no elements of symmetry. We do not describe calculations here that search out the saddle point, but we do show the nature of the wave function there, which will make clear why it has the relatively peculiar geometry shown in Fig. 16.7. This position is such that the tendency of the molecule is to form a H2 molecule. Depending upon the method of calculation the barrier height is estimated to be 4.05–4.06 eV, approximately the energy of one H—H bond. Theoretically2 the exothermicity of the process is very close to 0.0 so the parts separate with at least the activation energy. 2\n\nThe process we are discussing is a so-called isodesmic reaction. This means that the number of bonds is constant. It has been argued that calculations of this sort of process using changes in SCF energies are useful because the correlation energies tend to cancel when taking the difference. See Ref. .\n\n228\n\n16 Interaction of molecular fragments\n\nTable 16.9. The leading Rumer tableaux for the asymptotic state of formaldehyde, H2 + CO. In this case the standard tableaux functions are the same for these terms. 1 Num. Tab. Ci (inf )\n\n2\n\n1  5σ 5σ  1πx 1πx   1π 1π  y y 1sa 1sb R 0.709 68 \n\n2  1sa 1sa 5σ   5σ  1π 1π  x x 1π y 1π y R 0.168 56 \n\n3 2  5σ 5σ  1πx 1πx   1π 1π  y y 1sa 1sb\b R −0.134 08 \n\n4 2  5σ 5σ  1πx 1πx   2π 2π  y y 1sa 1sb R −0.098 04 \n\nTable 16.10. The leading Rumer tableaux in the wave function for the saddle point state of formaldehyde dissociation. 1 Num. Tab. Ci (sad )\n\n2\n\n1  1πx 1πx  1π y 1π y     1sb 5σ \b  1sa 2π y\b R 1.306 47 \n\n1  1πx 1πx  1π y 1π y     1sa 1sb  5σ \b 2π y\b R 0.461 32 \n\n3 1  1πx 1πx  1sb 5σ \b     1sa 1π y  2π y 2π y\b R 0.403 68 \n\n4 1  1πx 1πx  2π y\b 2π y\b     1sb 5σ \b  1sa 1π y R 0.209 90 \n\n16.2.3 Wave functions during separation The wave functions change character, of course, during the dissociation process. The asymptotic region is the simplest and we start with that. Table 16.9 shows the most important Rumer tableaux when CO and H2 are well separated from one another. 1. The leading term is clearly the closed shell 1 A1 state of CO in combination with the HLSP function for H2 . 2. These two terms give the closed-shell CO with the ionic term of H2 . 3. These two terms give the closed-shell CO with a breathing term for the H1s orbitals. 4. The last two terms shown give the electron correlation in the π shell of CO and the leading HLSP function term of H2 .\n\nThe wave function for this geometry is very simple to interpret. In Table 16.10 we show the principal Rumer tableaux for the wave function at the saddle point.\n\n16.2 Formaldehyde, H2 CO\n\n229\n\nTable 16.11. The leading terms in the wave function for the equilibrium geometry of formaldehyde. 1 Standard tableaux functions\n\nNum. Tab.\n\n1πx  1π y   1sa 1sb\n\nCi (min) HLSP functions\n\nNum. Tab. Ci (min)\n\n1\n\n2  1πx 1π y   5σ \b  \b 2π y\n\n5σ \b  1πx   1π y 1sa\n\n0.296 23 \n\n1πx  1π y   1sb 1sa\n\n1  1πx 1π y   5σ \b  2π y\b R\n\n0.310 08\n\n2\n\n3  5σ \b 1πx   1π y  \b 2π y\n\n1πx  1π y  \b  2π y 1sb\n\n0.178 83 \n\n5σ \b  1πx   1π y 1sb\n\n2  5σ \b 1πx   1π y  2π y\b R\n\n0.178 83\n\n2\n\n4  1πx 1π y   2π y\b  5σ \b\n\n1πx  1π y   1sb 2π y\b\n\n0.136 65 \n\n1πx  1π y   1sa 2π y\b\n\n2  1πx 1π y   5σ \b  1sb\b R\n\n0.159 18\n\n2\n\n 1πx 1π y   5σ \b  \b 1sa\n\n0.119 13 \n\n1πx  1π y  \b  2π y 1sb\n\n2  1πx 1π y   2π y\b  5σ \b R\n\n−0.136 65\n\n1. The leading term represents a structure with an electron pair bond between one H and the 5σ \b orbital and another between the other H and the 2π y\b orbital. 2. This term together with the first provides the two Rumer diagrams for the bonding scheme. That this has such a large coefficient indicates that electron pair bonds are not near perfect pairing. 3. This term involves correlation and polarization on the CO portion of the system with the electron pair bonds to the Hs still in place. 4. The fourth term involves a further rearrangement of the electrons on the CO portion of the system. In this case one H is now bonded to the 1π y instead of the 2π y\b orbital. These terms provide correlation, polarization, and also give a combination of both bonding and antibonding π y orbitals so that this sort of bond will disappear and C—H bonds and O nonbonding orbitals will appear as the molecule forms.\n\nTable 16.11 shows the leading terms in the wave function at the equilibrium geometry of H2 CO in both standard tableaux function and HLSP function form. 1. The first standard tableaux function term is essentially triplet H2 (much elongated, of course) coupled with the 3 y state of CO. The HLSP function has the same interpretation. 2. The second terms are the same and are an ionic type associated with the first term. These provide delocalization. 3. The third standard tableaux function and fourth HLSP function terms are the same. These are both ionic and provide antibonding character in the y-direction to “remove” that part of the original triple bond in CO. 4. The fourth standard tableaux function term and the third HLSP function term are the same configuration but not the same function. In both cases, however, the terms involve breathing for the 1s orbitals in the H atoms.\n\n230\n\n16 Interaction of molecular fragments\n\nIn summary we see that the barrier to dissociation in H2 CO can be ascribed to an avoided crossing of the same sort as we described in the dissociation of cyclopropane. The two fragments in triplet couplings bond as they approach and that state crosses the state where they are separately in singlet states. At the saddle point position the triplet fragment states still dominate to some extent, but asymptotically the two fragments are certainly in their respective singlet states.\n\nReferences\n\n L. Pauling and Jr E. B. Wilson. Introduction to Quantum Mechanics. McGraw–Hill Book Co., New York (1935). H. Eyring, J. Walter, and G. E. Kimball. Quantum Chemistry. John Wiley and Sons, Inc., New York (1944). I. N. Levine. Quantum Chemistry (Rev.). Allyn and Bacon, Boston (1974). A. Messiah. Quantum Mechanics, Vols. I and II. North–Holland Publishing Co., Amsterdam (1966). F. A. Cotton. Chemical Applications of Group Theory, 2nd Ed. John Wiley and Sons, Inc., New York (1969). M. Hammermesh. Group Theory. Addison–Wesley, Reading, Mass (1962). D. E. Rutherford. Substitutional Analysis. Edinburgh University Press, reprinted by Hafner, New York (1968). W. Heitler and F. London. Z. Physik, 44, 619 (1927). F. Hund. Z. Physik, 40, 742; 42, 93 (1927). R. S. Mulliken. Phys. Rev., 32, 186, 761 (1928). G. Rumer. G¨ottinger Nachr., 377 (1932). R. Serber. Phys. Rev., 45, 461 (1934). J. H. Van Vleck and A. Sherman. Rev. Mod. Phys., 7, 167 (1935). M. Kotani, A. Amemiya, E. Ishiguro, and T. Kimura. Table of Molecular Integrals. Maruzen Co. Ltd., Tokyo (1955). C. A. Coulson and I. Fisher. Phil. Mag., 40, 386 (1949). H. B. G. Casimir and D. Polder. Phys. Rev., 73, 360 (1948). P. A. M. Dirac. Proc. Roy. Soc. (London), 123, 714 (1929). M. Born and J. R. Oppenheimer. Ann. Physik, 84, 457 (1927). M. Born and K. Huang. Dynamical Theory of Crystal Lattices. Clarendon Press, Oxford (1954). R. Englman. The Jahn Teller Effect in Molecules and Crystals. Wiley-Interscience, New York (1972). L. Brillouin, Jour. de Physique, 3, 379 (1932). J. K. L. MacDonald, Phys. Rev., 43, 830 (1933). B. H. Chirgwin and C. A. Coulson. Proc. Roy. Soc. (London), A201, 196 (1950). G. A. Gallup and J. M. Norbeck. Chem. Phys. Lett., 21, 495 (1973). P.-O. L¨owdin. Ark. Mat. Astr. Fysk., 35A, 9 (1947). C. Herring. Rev. Mod. Phys., 34, 631 (1964). 231\n\n232\n\nReferences\n\n J. C. Slater. Quantum Theory of Molecules and Solids, Vol. 1. McGraw-Hill Book Co., New York (1963). M. Abramowitz and I. A. Stegun. Handbook of Mathematical Functions. National Bureau of Standards, US Government Printing Office (1970). S. C. Wang. Phys. Rev., 31, 579 (1928). S. Weinbaum. J. Chem. Phys, 1, 593 (1933). W. Kolos and L. Wolniewicz. J. Chem. Phys., 41, 3663 (1964). N. Rosen. Phys. Rev., 38, 2099 (1931). S. F. Boys. Proc. Roy. Soc. (London), A200, 542 (1950). G. G. Balint-Kurti and M. Karplus. J. Chem. Phys., 50, 478 (1968). G. Wannier. Phys. Rev., 52, 191 (1937). R. Pauncz. The Construction of Spin Eigenfunctions. Kluwer Academic/Plenum, New York (2000). D. E. Littlewood. The Theory of Group Characters. Oxford University Press, London (1950). H. Weyl. The Theory of Groups and Quantum Mechanics. Dover Publications, Inc., London (1931). G. A. Gallup, R. L. Vance, J. R. Collins, and J. M. Norbeck. Ad. Quantum Chem., 16, 229 (1982). F. A. Matsen. Ad. Quantum Chem., 1, 60 (1964). W. A. Goddard III. Phys. Rev., 157, 81 (1967). A. C. Aitken. Determinants and Matrices. Interscience Publishers, New York (1956). G. A. Gallup. Intern. J. Quantum Chem., 16, 267 (1974). P. C. Hiberty. THEOCHEM, 398–399, 35 (1997). C. C. J. Roothaan. Rev. Mod. Phys., 32, 179 (1960). M. W. Schmidt, K. K. Baldridge, J. A. Boatz, S. T. Elbert, M. S. Gordon, J. H. Jensen, S. Koseki, N. Matsunaga, K. A. Nguyen, S. J. Su, T. L. Windus, M. Dupuis, and J. A. Montgomery. J. Comput. Chem., 14, 1347 (1993). D. Chen and G. A. Gallup. J. Chem. Phys., 93, 8893 (1990). S. Huzinaga. Gaussian Basis Sets for Molecular Calculations. Elsevier Science Publishing Co., New York (1984). K. P. Huber and G. Herzberg. Molecular Spectra and Molecular Structure. IV. Constants of Diatomic Molecules. Van Nostrand Reinhold, New York (1979). L. Pauling. The Nature of the Chemical Bond. Cornell University Press, Ithica (1960). R. S. Mulliken. J. Chem. Phys., 2, 782 (1934). L. Allen. J. Amer. Chem. Soc., 111, 9003 (1989). S. Huzinaga, E. Miyoshi, and M. Sekiya. J. Comp. Chem., 14, 1440 (1993). G. Herzberg. Molecule Spectra and Molecular Structure. III. Electronic Spectra and Electronic Structure of Polyatomic Molecules. Van Nostrand Reinhold, New York (1966). H. H. Voge. J. Chem. Phys., 4, 581 (1936). A. Veillard. Theoret. Chim. Acta, 18, 21 (1970). V. Pophristic and L. Goodman. Nature, 411, 565 (2001). R. B. Woodward and R. Hoffman. The Conservation of Orbital Symmetry. Academic Press, New York (1970). M. Randi´c. in Valence Bond Theory and Chemical Structure. Ed. by D. J. Klein and N. Trinajsti´c. Elsevier, Amsterdam (1990). J. M. Norbeck and G. A. Gallup. J. Amer. Chem. Soc., 96, 3386 (1974). D. L. Cooper, J. Gerratt, and M. Raimondi. Nature, 323, 699 (1986).\n\nReferences\n\n233\n\n S. Shaik, A. Shurki, D. Danovich, and P. C. Hiberty. J. Mol. Struct. (THEOCHEM), 398, 155 (1997). S. Shaik, A. Shurki, D. Danovich, and P. C. Hiberty. Chem. Rev. 101, 1501 (2001). R. S. Mulliken and R. G. Parr. J. Chem. Phys., 19, 1271 (1951). H.-D. Beckhaus, R. Faust, A. J. Matzger, D. L. Mohler, D. W. Rogers, C. R¨uchart, A. K. Sawhney, S. P. Verevkin, K. P. C. Vollhardt, and S. Wolff. J. Amer. Chem. Soc., 122, 7819 (2000). J. R. Collins and G. A. Gallup. J. Amer. Chem. Soc., 104, 1530 (1982). S. Shaik and A. Shurki. Angew. Chem. Int. Ed., 38, 586 (1999). R. L. Vance and G. A. Gallup, Chem. Phys. Lett., 81, 98 (1981). S. S. Shaik, H. B. Schlegel, and S. Wolfe. Theoretical Aspects of Physical Organic Chemistry. The S N 2 Mechanism. Wiley and Sons Inc., New York (1992). T. H. Dunning. J. Chem. Phys., 65, 2823 (1970). W. J. Hehre, L. Radom, P. v.R. Schleyer, and J. A. Pople. Ab Initio Molecular Orbital Theory. Wiley and Sons, New York (1986).\n\nIndex\n\nalternation of saddle points, 193 antisymmetrizer, 66 column, 72 factored, 80 AO bases, 121 ROHF, using, 121 spherical vs. non-spherical environment, 122 AO basis group representation, 98 AOs sizes of, 123 atomic charges in CHn , 185, 187 atomic properties, 145 Rydberg configurations, 145 valence configurations, 145 B2 qualitative results, 149 quantitative results, 147 VB structure statistics, 147 BeH dipole moment, 138 full valence MCVB wave function, 137 SCVB wave function, 139 extra variation parameter in, 141 BeNe quantitative results, 6-31G*, 173 quantitative results, STO3G, 163 BF quantitative results, 6-31G*, 173 quantitative results, STO3G, 162 binary interchanges, 65 bond distances, see quantitative results under individual molecules bonding in B2 , 152 BeNe, 171 BF, 168 C2 , 153 C2 H4 , 216 C3 H6 , 220 CO, 166 “cyclohexatriene”, 209\n\nF2 , 160 N2 , 154, 164 O2 , 157 Born–Oppenheimer approximation, 7 breathing orbital, 108 C2 qualitative results, 152 quantitative results, 147 VB structure statistics, 147 C2 H4 , 214 breathing orbitals, 215 C2 H 6 hybrid vs. Cartesian AOs, 188 ionic structures in, 188 perfect pairing wave function, 188 statistics, 189 C3 H 5 allyl radical, 125 correlation energy, 126 HLSPs, 128 MCVB wave function, 126 SCVB wave function and corresponding orbitals, 132 small basis, 128 standard tableaux functions, 126 symmetry of standard tableaux functions, 126 transformation to HLSPs, 129 normalization and, 131 C3 H 6 6-31G*, 218 change in geometry on dissociation, 220 curve crossing on dissociation, 219 hybrid orbital in, 223 reaction barrier and avoided curve crossing, 219 STO3G, 225 triplet–triplet coupling of tableaux, 222 C4 H 4 electronic structure and equilibrium geometry, 208 C6 H6 , 197 6-31G*, 205 bonding in distorted benzene, 209 electronic structure and equilibrium geometry, 206\n\n235\n\n236\n\nIndex\n\nC6 H6 (cont.) Kekul´e and Dewar structures, 197 resonance energy, 208 SCVB wave function, 200 STO3G, 198 symmetry enhancement of wave function terms, 199 C6 H8 , 1,3,5-hexatriene comparison with C6 H6 , 203 C10 H8 MCVB, 211 MOCI, 212 STO3G, 211 CH 6-31G*, 186 dipole moment, 179 STO3G, 178 CH2 , 214 6-31G*, 186 dipole moment, 182 hybridization, 180 ROHF treatment, 215 STO3G, 178 CH3 6-31G*, 186 quadrupole moment, 184 STO3G, 178 CH4 6-31G*, 186 octopole moment, 185 STO3G, 178 charge separation limits, see VB structure selection CO quantitative results, 6-31G*, 173 quantitative results, STO3G, 163 computer program, 8 configuration, 99 confocal ellipsoidal–hyperboloidal coordinates, 26 constellation, 99 core repulsion in fragment interaction, 218, 219 correlation angular, 38 bond parallel, 38 over-correlation, 29 targeted, 214 Coulomb integral, 32 Coulomb’s law, 5, 6 covalent functions, 27 CRUNCH, xiv, 8, 98, 103, 123, 165 De , see quantitative results under individual molecules dipole moment anomalous direction, 111, 162 BeNe, 174 BF, 174 CH, 179 CH2 , 182 CO, 174 covalent function, from, 112 effect of electronegativity, 111 effects of overlap and spin, 111\n\ninadequacy of STO3G basis, 175 ionic structures and, 111 dipole moment function BeNe, 174 BF, 174 CO, 174 direct transformations, 97 EGSO weights, STO3G BeNe, 172 BF, 170 CO, 169 N2 , 166 eigenvalue problem generalized, 10 2×2, 14 electron correlation B2 , 151 C2 , 153 H2 , 29, 40, 50, 52 energy levels, second row atoms, 145 equilibrium geometry C6 H6 compared with C4 H4 , 208 exchange integral, 32 F2 qualitative results, 160 quantitative results, 147 VB structure statistics, 148 full MCVB, 39 GAMESS, 121 Gaussian units, 5 GGVB, 108 Gi method, 109 group algebra, 66 general bases, 69 Hermitian idempotents of, 76 matrix basis, 77 spatial symmetry groups, 98 groups irreducible matrix representations, 68 matrix representations, 68 H2 , 24 bonding Coulombic interactions in, 32 kinetic energy in, 37 polarization, 38 resonance in, 37 complementary orbitals, 49 delocalization energy, 37 delocalized orbitals, 47 double-ζ basis, 38 exponential orbitals, 25, 47 Gaussian orbitals, 38 GGVB and SCVB treatments, 51 localized orbitals, 23 optimal delocalized orbitals, 49 optimal unsymmetric orbitals, 51 orthogonalized orbitals, 47\n\nIndex triple-ζ basis, 44 vector form of wave function, 29 “wrong” scale in AO, 111 (H2 )2 potential energy surface, 192 (H2 )3 potential energy surface, 192 (H2 )n matrix element sign alternation, 195 (H2 )n -ring statistics, 192 H2 CO, 225 barrier height, 227 change in wave function on dissociation, 228 LMP, a Woodward–Hoffman forbidden process, 226 reaction barrier and avoided crossing, 230 saddle point, 227 H2 O symmetry analysis, 100 Hartree’s atomic units, 5 He+ 2 MCVB wave function, 134 SCVB wave function and corresponding orbitals, 135 HLSP functions, 87 in terms of standard tableaux functions, 88 example, see C3 H5 indirect transformations, 97 ionic functions, 27 ionic structures importance in C6 H6 , 199 physical significance, 111 isoelectronic series, 162 kinetic energy, 6 Li MCVB wave function, 144 nodes in SCVB AOs, 143 SCVB wave function, 142 LiF covalent–ionic curve crossing, 115 dipole moment and covalent–ionic curve crossing, 116 LiH dipole moment full valence wave function, 115 three-state wave function, 113 linear variation functions, 9 local orbitals general, 107 maximum overlap principle, 180 MCVB method, 107 characteristics, 108 metathesis reaction, 191 multi-structure VB, 108 N operator, 71 N2 qualitative results, 154 quantitative results, 147 quantitative results, 6-31G*, 173\n\nquantitative results, STO3G, 163 VB structure statistics, 147 NH3 Rumer tableaux, 103 symmetry analysis, 102 nonlocal orbitals characteristics, 110 general, 108 O2 qualitative results, 157 quantitative results, 147 VB structure statistics, 147 octopole moment CH4 , 185 optimal unsymmetric orbitals H2 , 51 orthogonality theorem, 68 orthogonalization, 19 canonical, 11, 19 Schmidt, 19, 20 symmetric, 19, 47 overlap integral, 32 P operator, 71 partitions, 70 conjugate, 74 Pauli exclusion principle, 24 permutation, 64 cycle structure, 64 inverse, 65 signature, 66 unitarity, 70 potential energy surfaces, 191 principal tableaux B2 , 151 BeNe, 171 BF, 169 C2 , 153 C2 H4 , 217 C3 H6 , 221, 223, 224 C6 H6 , 198 C6 H8 , 210 CH, 178 CH2 , 180 CH3 , 183 CH4 , 185 CO, 167 F2 , 160 H2 CO, 228 N2 , 155, 164 O2 , 158 quadrupole moment CH3 , 183 Rayleigh quotient, 9 reaction surfaces, 191 representation generalized permutation, 98 ROHF, see AO bases, 121\n\n237\n\n238 rotation barrier in C2 H6 , 189 Rumer diagrams, 4 and Rumer tableaux, 90 sandwich representations, 79 Schr¨odinger equation, 7 electronic, 8 SCVB method, 109 characteristics, 110 SCVB orbitals BeH, 139 C3 H5 , 132 C6 H6 , 200 C6 H8 , 204 H2 , 51 He+ 2 , 135 Li, 143 secular equation, 10 SEP, 124 separated atom limit, 3 σ core, 124 signature, see permutation size escalation of basis sets, 190 spatial function symmetry n-electron, 84 orbital approximation n-electron, 85 three-electron, 58 orbital double occupancy n-electron, 86 three-electron, 62 three-electron, 55 two-electron, 24 spatial symmetry, see spatial function symmetry spin degeneracy three-electron, 55 spin eigenfunctions general operators for, 54 n-electron, 81 three-electron, 53, 55 two-electron, 23\n\nIndex standard tableaux functions, 87 static exchange approximation, see SEP supercomputers, 4 symgenn, 103, 104, 123, 165 symmetric groups, 63 symmetrizer row, 71 tableau functions nonintuitive symmetry property, 101, 103, 104 θ N PN \u0004 matrix element evaluation, 91 θP N P\u0004 matrix element evaluation, 88 three-electron bond O2 , 157 united atom limit, 3 units, physical, 5 valence state of C, 186 variation theorem, 9 VB structure selection, 123 orbital excitation, 123 planar molecules, 124 virtual AOs, role of, 122 Von Neumann’s theorem, 76 weights EGSO, 20 Chirgwin and Coulson, 18 inverse overlap, 18 nonorthogonal functions, 18 orthogonal functions, 16 symmetric orthogonalization, 19 Weyl dimension formula, 87 Woodward–Hoffman rules, 192 Young tableau, 71 standard, 72" ]

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[ "# Did the universe need the presence of matter and radiation to start expanding?\n\nHence it is not possible that photons generated by stars is contributing to dark energy.\n\nCould photons generated from the many trillions of stars be some how contributing energy to dark energy? Where do they go and what becomes of them?\n\nThe expansion of the universe is the increase of the distance between two distant parts of the universe with time. It is an intrinsic expansion whereby the scale of space itself changes. The universe does not expand \"into\" anything and does not require space to exist \"outside\" it. Technically, neither space nor objects in space move. Instead it is the metric governing the size and geometry of spacetime itself that changes in scale.\n\nhttps://en.wikipedia.org/wiki/Expansion_of_the_universe\n\nIt's not possible for the universe to be in a non-expanding phase and then enter an inflationary phase.\n\nWas the universe already expanding before inflation occured?\n\nSo basically, in other words, does space expansion require the presence of matter of radiation in any way, or is space expansion independent of what exists in space?\n\nThe first answer says that radiation does not contribute to dark energy. The wiki article says that technically, neither space nor objects in space move, it is rather the metric that changes in scale.\n\nSo space expansion does not require radiation energy, it does not require matter to move, and creating more space does not require energy at all (in fact to our knowledge, space expansion just creates more dark energy). Does this mean that an empty universe, with no matter and radiation in it would expand too? The last answer says that the universe probably was always expanding.\n\nNow my question is about this, if the universe was always expanding, then does this require the presence of any radiation or matter, or could a void universe expand without the presence of any matter or radiation? In other words, do we need the presence of matter and radiation to explain how the universe started expanding, or is expansion independent of those?\n\nAs far as I understand, space expansion is a widely accepted fact, that is independent of everything else, but I do not know whether explaining it needs the presence of matter and radiation.\n\nQuestion:\n\n1. Did the universe need the presence of matter and radiation to start expanding?\n\nThe expansion of the universe is the relative motion of the stuff in it. It's meaningless to talk about expansion of a vacuum.\n\nThere are vacuum FLRW solutions: they are Minkowski space (for $$Λ=0$$), de Sitter space ($$Λ>0$$) and anti de Sitter space ($$Λ<0$$). All of them are maximally symmetric—every spacetime point looks the same as every other—and so there's no generally covariant way to define a time-varying scale factor, or any other quantity that varies with spacetime position. There are many different FLRW parametrizations of those spacetimes, or portions of them, that assign different cosmological times and scale factors to each point. They're just coordinate systems, no one more correct than any other, like inertial reference frames in special relativity.\n\nDuring inflation, spacetime is not actually de Sitter. The inflaton field has a time-varying energy density (the slow roll) which you can use to define a time coordinate and a scale factor. If you don't count that as matter, then the universe doesn't need matter to expand. What it needs, regardless of nomenclature, is something that resembles a Hubble flow, i.e., something that has the defining symmetry of FLRW geometries and no larger symmetry. A random spacetime is not symmetric enough, and a FLRW vacuum is too symmetric, to be unambiguously expanding.\n\nLets look at the data the General Relativity (GR) model is describing. GR is a theory as far as mathematics goes. In order to be relevant to physics it has to model the data of the universe(in this case).\n\nThe data of the universe consist of numbers and observations of radiation and matter, if these are not there , there is no universe, only mathematics and science fiction.\n\nLets see the basic GR relativity mathematical relation", null, "On the left hand side we have the space variables defining a tensor, on the right hand side is the energy momentum tensor. For a specific problem with energy and radiation on the right hand side, it is a complicated equation that has to be calculated for the specific problem of matter and radiation. Because of the mathematical equality between space and energy/momentum the mathematical solutions create what is called curvature of space.\n\nIf the right hand side is zero, there can be no solutions that show curvature of the space coordinates . Here this is mentioned\n\nIn general relativity, the symmetric stress–energy tensor acts as the source of spacetime curvature,..\n\nI found this elaborate answer in Quora that may interest you.\n\nIn my opinion, mathematical equations can have any number of solutions, the job of physics analysis is to establish where those solutions describe the data and observations, and, important, are predictive . When there is no matter and energy one should pick to describe the physical situation those solution that are compatible with that, which have no curvature.\n\nIt could be that the initial virtual matter fields were constrained to our 3d space, while our space was embedded in a highly negatively curved 4d space (somewhat as described in pyrotechnical approaches) The virtual character could induce this negative curvature. It pulled (or pushed) the virtual stuff into real existence in a bang. Thereby imparting positive curvature and stopping the acceleration. We see the universe expanding in the 3d space to which matter is confined, but this 3d space expands in a 4d space. So it took only virtual matter.\n\nIf the matter in our universe, all distances between charged particles expanded at the same rate, the moving away from each other wouldn't be noticeable. All photons would maintain their wavelength, when measured to distances between interacting particles. But particles do interact. So thanks to this interaction we notice expansion. But for the initial speed spurt, inflation, only virtual matter was needed.\n\nYour question addresses deep issues about the beginning of the universe as described by 'Concordance Cosmology' and you are right that there are contradictions.\n\nThe answer is that Concordance Cosmology is wrong.\n\nGeneral Relativity and Big Bang are correct, but the extra additions that have been added since about 1990, inflation and dark energy are wrong.\n\nIt would make the answer too long and subjective to go into further details about how the current cosmology could be improved." ]

[ null, "https://i.stack.imgur.com/2zpqY.png", null ]

[ "The community is working on translating this tutorial into Indonesian, but it seems that no one has started the translation process for this article yet. If you can help us, then please click \"More info\".\n\nClasses:\n\n# Static members\n\nAs we saw in a previous chapter, the usual way to communicate with a class is to create a new instance of the class and then work on the resulting object. In most cases, this is what classes are all about; the ability to instantiate multiple copies of the same class and then use them differently in some way. However, in some cases, you might like to have a class which you may use without instantiating it, or at least a class where you can use members of it without creating an object for it. For instance, you may have a class with a variable that always remains the same no matter where and how it's used. This is called a static member, because it remains the same.\n\nA class can be static, and it can have static members, both functions and fields. A static class can't be instantiated, so in other words, it will work more as a grouping of related members than an actual class. You may choose to create a non-static class instead, but let it have certain static members. A non-static class can still be instantiated and used like a regular class, but you can't use a static member on an object of the class. A static class may only contain static members.\n\nFirst, here is an example of a static class:\n\n``````public static class Rectangle\n{\npublic static int CalculateArea(int width, int height)\n{\nreturn width * height;\n}\n}``````\n\nAs you can see, we use the static keyword to mark the class as static, and then we use it again to mark the method, CalculateArea, as static as well. If we didn't do that, the compiler would complain, since we can't have a non-static member of a static class.\n\nTo use this method, we call it directly on the class, like this:\n\n``Console.WriteLine(\"The area is: \" + Rectangle.CalculateArea(5, 4));``\n\nWe could add other helpful methods to the Rectangle class, but perhaps you are wondering why we are passing on width and height to the actual method, instead of storing it inside the class and then pulling them from there when needed? Because it's static! We could store them, but only one set of dimensions, because there is only one version of a static class. This is very important to understand.\n\nInstead, we can make the class non-static, and then have the CalculateArea as a utility function on this class:\n\n``````public class Rectangle\n{\nprivate int width, height;\n\npublic Rectangle(int width, int height)\n{\nthis.width = width;\nthis.height = height;\n}\n\npublic void OutputArea()\n{\nConsole.WriteLine(\"Area output: \" + Rectangle.CalculateArea(this.width, this.height));\n}\n\npublic static int CalculateArea(int width, int height)\n{\nreturn width * height;\n}\n}``````\n\nAs you can see, we have made the class non-static. We have also added a constructor, which takes a width and a height and assigns it to the instance. Then we have added an OutputArea method, which uses the static method to calculate the area. This is a fine example of mixing static members with non-static members, in a non-static class.\n\nA common usage of static classes, although frowned upon by some people, are utility/helper classes, where you collect a bunch of useful methods, which might not belong together, but don't really seem to fit elsewhere either." ]

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[ "", null, "# Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be\n\na. 2: 1\n\nb. 1: 4\n\nc. 4: 1\n\nd. 1: 2\n\nGiven that first connection", null, "Second connection\n\nC34 = C + C = 2 C\n\nNow, the ratio of equivalent capacities in the two cases will be –\n\n⇒ C12 / C34 = C2 / 2C ⇒ C12 / C34 = ¼", null, "(0)", null, "(0)" ]

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[ "# Moving Average\n\nMoving averages are used to smooth trends. TC2000 offers four different types of moving averages.\n\nSimple\n\nA simple moving average gives equal weight to each data point for the period. If the period is 3 and the last three data points are 3, 4 and 5 the most recent average value would be (3+4+5)/3=4 (divide by three because there are three data points).\n\nExponential\n\nAn exponential moving average (EMA), sometimes also called an exponentially weighted moving average (EWMA), applies weighting factors which decrease exponentially. The weighting for each older data point decreases exponentially, giving much more importance to recent observations while still not discarding older observations entirely\n\nFront Weighted\n\nA front-weighted average, like an exponential average, allows the most recent data being averaged to impact the average value more than older data. It is calculated differently than exponential averages but it also gives recent data more weight. A 5 period front weighted average is calculated as follows (C is the most recent bar, C4 is 4 bars ago):\n\nFront Weighted Average = ((C*5) + (C1*4) + (C2*3) + (C3*2) + C4) / 15\n\nHull Moving Average - The Hull Moving Average solves the age old dilemma of making a moving average more responsive to current price activity whilst maintaining curve smoothness. In fact the HMA almost eliminates lag altogether and manages to improve smoothing at the same time. Read more at AlanHull.com.\n\nYou can see how the different averaging types produce different results. All four averages are plotted using a period of 21: simple (yellow), exponential (cyan), front-weighted (red) and Hull moving average (orange).\n\nCustom PCF Formula\n Function Version Base `tAVG(w, x)` `w`=Numeric, `x`=Period Simple `AVG(w, x)` Exponential `XAVG(w, x)` Front Weighted `FAVG(w, x)` Hull `HAVG(w, x)` Indicator Version of Price or Volume Base `tAVGpx.z` `t`=AverageType, `p`=PriceVolume, `x`=Period, `z`=Offset Simple `AVGpx.z` Exponential `XAVGpx.z` Front Weighted `FAVGpx.z` Hull `HAVGpx.z`\n\nWhere `w` is any formula returning a numeric value.\n\nWhere `x` is the period which must be an integer.\n\nWhere `t` is the Average Type. Leave blank for Simple, X for Exponential, F for Front Weighted, or H for Hull.\n\nWhere `p` is the price or volume component being averaged. O for Open, H for High, L for Low, C for Close, or V for Volume.\n\nWhere `z` is the offset. An offset of 1 returns the value for the previous bar instead of the current bar.\n\n### Examples\n\nA 21 period simple moving average for the current bar can be written as follows.\n\n`AVGC21.0`\n\nOr as follows.\n\n`AVGC21`\n\nOr as follows.\n\n`AVG(C, 21)`\n\nConverting this to exponential adds an X to the beginning of any of the previous forms.\n\n`XAVGC21`\n\nWhile front weighted uses an F.\n\n`FAVGC21`\n\nAnd Hull uses an H.\n\n`HAVGC21`", null, "In addition, you can choose what element of price to use in the calculation of the average: Last, Open, High, Low, or Typical Price. (1)\n\nMoving averages have an Offset parameter (2) that allows you to shift the average plot forward or backwards (negative offset value). This allows you to plot what are commonly referred to as \"displaced\" moving averages.", null, "" ]

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[ "Limsup of independent exponential random variables\n\nIn the lecture, we are given an example on independent and identically distributed random variables, but I am not quite sure what is the idea of this question. The exercise is as following:\n\nLet (Xn) be a sequence of independent and identically distributed random variables defined on a probability space (Ω,F,P). Assume that X1 is exponentially distributed with parameter λ>0. The distribution of X1 is absolutely continuous with density function $f(x)=λe^{-λx}1_{(0,∞)}(x)$, x∈R.\n\nShow that $P(limsup_{(n→∞)}X_n/logn=1/λ)=1$.\n\nI know that for the exponential distribution, the mean E[x]=1/λ, it says X1 is exponentially distributed, is that mean all of the Xn are also exponentially distributed? $limsup_{(n→∞)}(X_n)$ means that Xn happens for infinitely often times as n close to infinity. so $limsup_{(n→∞)}(X_n)=⋂_{n∈N}⋃_{m>n}X_n$, and also as $n→∞$, $log(n)→∞$ . I am not sure about what this question is asking about? Is it something with 0/1 law? like if an event A happens infinitely times, then P(A)∈{0,1}\n\nCan any one please help me with this question? details are preferred. Thanks!\n\n• please use Latex – Alex May 4 '14 at 6:49\n• I am not sure how to use Latex in here, sorry – xz330 May 4 '14 at 7:00\n• @xz330 I have submitted a revision to your question that fixes the latex formatting. Please check that I haven't altered the meaning of your question. – senshin May 4 '14 at 7:49\n• @senshin thanks! any idea about this question? – xz330 May 4 '14 at 7:54\n• Changed the title to something more accurate. Do you understand why it fits better the content of the question? – Did May 8 '14 at 9:09\n\nShow that $P(lim(n→∞)supXn/logn=1/λ)=1$.\n\n(...) $lim(n→∞)(supXn)$ means that Xn happens for infinitely often times as n close to infinity. so $lim(n→∞)(supXn)=⋂_n⋃_mXn$, where m>n and n∈N.\n\nThis is confusing limsup of events and limsup of random variables. Here, each $X_n$ is a random variable hence $$\\limsup\\limits_{n\\to\\infty}X_n/\\log n$$ is the random variable $Y$ such that, for every $\\omega$ in $\\Omega$, $$Y(\\omega)=\\limsup\\limits_{n\\to\\infty}X_n(\\omega)/\\log n.$$ And the task is to prove that the event $[Y=1/\\lambda]$ has probability $1$.\n\nNote that the question most probably does not reproduce faithfully the text of the exercise asked. For example, $$\\text{lim(n→∞)supXn/logn}$$ should read $$\\limsup\\limits_{n\\to\\infty}X_n/\\log n,$$ where $\\limsup\\limits_{n\\to\\infty}$ acts as a single operation, or, equivalently, $$\\limsup\\limits_{n\\to\\infty}\\frac{X_n}{\\log n}.$$\n\n• oh right, I got what it means now, but how do you prove that $P[Y=1/λ]=1$? – xz330 May 4 '14 at 7:58\n• Yeah, I have changed all the equations to the proper formatting now. – xz330 May 4 '14 at 8:00\n• That is a different question. – Did May 4 '14 at 8:01\n• isn't it what the question is asking about? – xz330 May 4 '14 at 8:01\n• No it is not. And if you want an answer to that other question, you should add your thoughts on it. – Did May 4 '14 at 8:02" ]

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[ "# GATE | GATE-CS-2006 | Question 85\n\n• Last Updated : 28 Jun, 2021\n\nIn the correct grammar of above question, what is the length of the derivation (number of steps starring from S) to generate the string albm with l ≠ m?\n(A) max(l,m) + 2\n(B) l + m + 2\n(C) l + m + 3\n(D) max(l, m) + 3\n\nExplanation:\nCorrect grammar of the last question was (D), which is:\n\n```S -> AC|CB\nC -> aCb|epsilon\nA -> aA|a\nB -> Bb|b\n```\n\nNow, the most optimal and intuitive way to generate a string of the form albm would be to first use “C -> aCb|epsilon” production rule to get as many a and b as we can, which would be min(l,m). To get the rest of the string, we could just use latter two production rules accordingly. Formally deriving the string of the general format albm from the above grammar –\n\n```1. S -> AC\n2. -> A(aCb)\n3. -> ....\n4. -> ....\n5. -> A(am C bm)\n6. -> A(am bm)\n7. -> aA(am bm)\n8. -> ....\n9. -> ....\n10. -> a(l-m-1)A(am bm)\n11. -> al bm\n\n```\n\nFrom above set of derivation steps we can count the total steps as follows:\n\n```\nProduction 1 took 1 step : 1 [using S->AC]\nProduction 2-5 took steps : min(l,m) [using C->aCb]\nProduction 6 took 1 step : 1 [using C->epsilon]\nProduction 7-11 took steps : max(l,m)-min(l,m) [using A -> aA|a or B -> Bb|b]\n\nTotal steps : max(l,m) + 2\n```\n\nHence, answer should be (A): max(l,m) + 2\n\nThis explanation is contributed by Vineet Purswani.\n\nQuiz of this Question\n\nMy Personal Notes arrow_drop_up" ]

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[ "", null, "Fixed Flow Rate - MapleSim Help\n\nFixed Flow Rate\n\nFixed flow-rate source", null, "", null, "Description The Fixed Flow Rate component generates a fixed flow rate through its ports.", null, "Equations $q={q}_{A}=-{q}_{B}=Q$ $p={p}_{A}-{p}_{B}$", null, "Variables\n\n Name Units Description Modelica ID $p$ $\\mathrm{Pa}$ Pressure across component p $q$ $\\frac{{m}^{3}}{s}$ Flow rate through component q", null, "Connections\n\n Name Description Modelica ID $\\mathrm{portA}$ Upstream hydraulic port portA $\\mathrm{portB}$ Downstream hydraulic port portB", null, "Parameters\n\n Name Default Units Description Modelica ID $Q$ $1$ $\\frac{{m}^{3}}{s}$ Flow rate from port A to port B Q" ]

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[ "+0\n\n# hyperbola help\n\n0\n125\n1\n\nhttps://prnt.sc/pbwlek\n\nthank you very much\n\nSep 27, 2019\n\n#1\n+2\n\n7y^2  - 4x^2  + 24x + 14y - 57  =  0        rearrange as\n\n7y^2  + 14y   - 4x^2 + 24x   =   57         factor  as\n\n7(y^2 +2y)  -4(x^2 - 6x)  =  57            complete the square on x and y\n\n7(y^2 + 2y + 1) - 4(x^2 -6 + 9)  =  57 + 7 - 36      simplify\n\n7(y + 1)^2  - 4(x - 3)^2  =   28          divide through by 28\n\n(y + 1)^2         (x - 3)^2\n\n_______   -   _________    =      1\n\n4                  7", null, "", null, "", null, "Sep 27, 2019" ]

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[ "# Expand and factorise\n\n3(n+r)(n+r-1) + (n+r)\nto (n+r)(3n+3r-2)?\n\nI forgot how to factorise, so can someone help me do this please?\n\nCRGreathouse\nHomework Helper\nThe n+r is already factored out. Just write everything in terms of that and you'll see.\n\n(It's actually easier to understand by doing, in this case, than to have it explained!)\n\nMentallic\nHomework Helper\nExplain slowly you say? Maybe that can be expressed through example.\n\nab+ac=a(b+c)\nxy+x=x(y+1)\npqr+r=r(pq+1)\n\nIf you understand how to do the opposite of factorising which is expanding, then these results would be easy to confirm.\n\nSo for 3(n+r)(n+r-1) + (n+r)\nlet n+r=x\nSo we now have 3x(x-1)+x\nFactorising x out: x(3(x-1)+1)\nExpand 3(x-1) and collect like terms, then substitute n+r=x back into the result.\n\nOh wow i get it now. Thank you!!" ]

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[ "Message-ID: <1567396536.29503.1657128138671.JavaMail.confluence@2nwiki> Subject: Exported From Confluence MIME-Version: 1.0 Content-Type: multipart/related; boundary=\"----=_Part_29502_1764001853.1657128138671\" ------=_Part_29502_1764001853.1657128138671 Content-Type: text/html; charset=UTF-8 Content-Transfer-Encoding: quoted-printable Content-Location: file:///C:/exported.html SMS programming - How to configure 2N=C2=AE Lift8 over SMS messa= ges\n\n# SMS programming - How to configure 2N=C2=AE Lift8 over SMS messages=\n\n=20\n=20\n=20\n=20\n=20\n\nInfo\n\n=20 Icon=20\n=20\n\nThis FAQ describes how you can configure 2N® Lift8 via SMS messages. You wil= l learn commands which you can use for SMS configuration and also possible = responds to this commands.\n\n=20\n=20\n=20", null, "=20\n=20\n=20\n=20\n=20\n=20\n\n# Configuration SMS messages\n\nBelow you can find list of possible SMS configuration commands for = 2N® Lift8.  In list below, you can see exact form= at of available messages. You can use up to 5x160 characters per SMS messag= e as 2N® Lift8 supports long SMS messages.\n\n1. Configuration:\nL8 CNF 12345 011=3D00420222222222 012= =3D00420111111111 (You can use any available parameter from parameter chart= )\n2. Put 2N® Lift8 into the factory defaul= t:\nL8 DEF 12345\n3. Restart:\nL8 RST 12345\n\n# Possible responses\n\n### Confirmation messages\n\nL8 CNF OK&= nbsp;- configuration was applied\n\nL8 DEF OK&= nbsp;- reset to factory default was successful\n\nL8 SET OK&= nbsp;- switch of profile was successful\n\nL8 RST OK&= nbsp;- restart of the device was successful\n\n### Error messages\n\nL8 ERR Invalid Message  -If L8 sign is missin= g or is different (L2, L4 WHATEVER ELSE)\n\nL8 ERR Unknown Command  -If another command t= hen CNF, DEF, SET a RST was applied (DFE FNC ETC ...)\n\nL8 ERR Invalid Password  - Incorrect password= for programming (default is 12345)\n\nL8 ERR Invalid Parameters  - releated to CNF = and SET.\n\nL8 ERR Invalid Syntax  -Syntax error, when th= ere are some spaces or chars where shouldn=E2=80=99t be\n\nL8 ERR Does not Respond  - doesn=E2=80=99t an= swer the call after 60 sec\n&= nbsp;                    =              - when the call is answere= d but doesn=E2=80=99t react for CPC programming\n&n= bsp;                     =                      = ;- hang up during programming\n= ;                     &nb= sp;              - doesnt react on WRITE= _START, WRITE_CONFIRM, or CRC_REQUEST\n\n=20\n=20\n=20" ]

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[ "# Concentration of Cobalt diiodide, beta form\n\n## cobalt diiodide, beta form: mass and molar concentration\n\n### Molar concentration per milliliter\n\n 0.01 mmol/ml 10 µmol/ml 10 000 nmol/ml 10 000 000 pmol/ml\n\n### Molar concentration per deciliter\n\n 1 mmol/dl 1 000 µmol/dl 1 000 000 nmol/dl 1 000 000 000 pmol/dl\n\n### Molar concentration per liter\n\n 10 mmol/l 10 000 µmol/l 10 000 000 nmol/l 10 000 000 000 pmol/l\n\n### Mass concentration per milliliter\n\n 0 g/ml 3.13 mg/ml 3 127.42 µg/ml 3 127 420 ng/ml 3 127 420 000 pg/ml\n\n### Mass concentration per deciliter\n\n 0.31 g/dl 312.74 mg/dl 312 742 µg/dl 312 742 000 ng/dl 312 742 000 000 pg/dl\n\n### Mass concentration per liter\n\n 3.13 g/l 3 127.42 mg/l 3 127 420 µg/l 3 127 420 000 ng/l 3 127 420 000 000 pg/l\n\n### Equivalent molar concentration per milliliter\n\n 0.01 meq/ml 10 µeq/ml 10 000 neq/ml 10 000 000 peq/ml\n\n### Equivalent molar concentration per deciliter\n\n 1 meq/dl 1 000 µeq/dl 1 000 000 neq/dl 1 000 000 000 peq/dl\n\n### Equivalent molar concentration per liter\n\n 10 meq/l 10 000 µeq/l 10 000 000 neq/l 10 000 000 000 peq/l\n• The units of  amount of substance (e.g. mole) per milliliter,  liter and deciliter are SI units of measurements of molar concentrations.\n• The units of molar concentration per deciliter:\n• millimole per deciliter [mm/dl],  micromole per deciliter [µm/dl],  nanomole per deciliter [nm/dl]  and  picomole per deciliter [pm/dl].\n• The units of molar concentration per milliliter:\n• millimole per milliliter [mm/ml],  micromole per milliliter [µm/ml],  nanomole per milliliter [nm/ml]  and  picomole per milliliter [pm/ml].\n• The units of molar concentration per liter:\n• millimole per liter [mm/l],  micromole per liter [µm/l],  nanomole per liter [nm/l]  and  picomole per liter [pm/l].\n• The units of  mass  per milliliter,  liter and deciliter are non-SI units of measurements of mass concentrations still used in many countries.\n• The units of mass concentration per deciliter:\n• gram per deciliter [g/dl],  milligram per deciliter [mg/dl],  microgram per deciliter [µg/dl],  nanogram per deciliter [ng/dl]  and  picogram per deciliter [pg/dl].\n• The units of mass concentration per milliliter:\n• gram per milliliter [g/ml],  milligram per milliliter [mg/ml],  microgram per milliliter [µg/ml],  nanogram per milliliter [ng/ml]  and  picogram per milliliter [pg/ml].\n• The units of mass concentration per liter:\n• gram per liter [g/l],  milligram per liter [mg/l],  microgram per liter [µg/l],  nanogram per liter [ng/l]  and  picogram per liter [pg/l].\n• The  equivalent  per milliliter,  liter and deciliter are obsolete, non-SI units of measurements of molar concentrations still used in many countries. An equivalent is the number of moles of an ion in a solution, multiplied by the valence of that ion.\n• The units of equivalent concentration per deciliter:\n• milliequivalent per deciliter [meq/dl],  microequivalent per deciliter [µeq/dl],  nanoequivalent per deciliter [neq/dl]  and  picoequivalent per deciliter [peq/dl].\n• The units of equivalent concentration per milliliter:\n• milliequivalent per milliliter [meq/ml],  microequivalent per milliliter [µeq/ml],  nanoequivalent per milliliter [neq/ml]  and  picoequivalent per milliliter [peq/ml].\n• The units of equivalent concentration per liter:\n• milliequivalent per liter [meq/l],  microequivalent per liter [µeq/l],  nanoequivalent per liter [neq/l]  and  picoequivalent per liter [peq/l].\n\n#### Foods, Nutrients and Calories\n\n70% DARK CHOCOLATE, UPC: 810048017287 contain(s) 475 calories per 100 grams (≈3.53 ounces)  [ price ]\n\n1180 foods that contain Iodine, I.  List of these foods starting with the highest contents of Iodine, I and the lowest contents of Iodine, I, and Recommended Dietary Allowances (RDAs) for Iodine\n\n#### Gravels, Substances and Oils\n\nCaribSea, Freshwater, Instant Aquarium, Moonlight Sand weighs 1 601.85 kg/m³ (100.00023 lb/ft³) with specific gravity of 1.60185 relative to pure water.  Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical  or in a rectangular shaped aquarium or pond  [ weight to volume | volume to weight | price ]\n\nCalcium biphosphate [Ca(H2PO4)2] weighs 2 220 kg/m³ (138.59007 lb/ft³)  [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]\n\nVolume to weightweight to volume and cost conversions for Refrigerant R-123, liquid (R123) with temperature in the range of -28.89°C (-20.002°F) to 93.34°C (200.012°F)\n\n#### Weights and Measurements\n\nA cubic picometer (pm³) is a derived metric SI (System International) measurement unit of volume with sides equal to one picometer (1pm)\n\nElectric current is a motion of electrically charged particles within conductors or space.\n\npm/min² to dm/h² conversion table, pm/min² to dm/h² unit converter or convert between all units of acceleration measurement.\n\n#### Calculators\n\nEstimate the total volume of blood, blood plasma and blood cells" ]

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[ "# Calculate Velocity of Alpha Particle - Nuclear Decay Calculator\n\nAlpha (α) decay is a kind of radioactive decay. During nuclear emission the alpha particles at high energy shows changes in its mass number and atomic number. Calculate here the velocity of α-particle using Geiger law at high energy levels.\n\n## Calculate Velocity of Alpha Particle using Geiger Law in Radioactive Decay\n\nlength\nρ\n\nAlpha (α) decay is a kind of radioactive decay. During nuclear emission the alpha particles at high energy shows changes in its mass number and atomic number. Calculate here the velocity of α-particle using Geiger law at high energy levels.\n\nCode to add this calci to your website", null, "", null, "#### Formula:\n\nv = ( a * ( R - x ) ) 1/3 Where, v = Velocity of Particle a = Constant R = Range x = Distance from Source\n\nAlpha particles are positive helium nuclei." ]

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[ "# 91. Decode Ways (Medium)\n\nhttps://leetcode.com/problems/decode-ways/\n\nA message containing letters from `A-Z` is being encoded to numbers using the following mapping:\n\n```'A' -> 1\n'B' -> 2\n...\n'Z' -> 26\n```\n\nGiven a non-empty string containing only digits, determine the total number of ways to decode it.\n\nExample 1:\n\n```Input: \"12\"\nOutput: 2\nExplanation: It could be decoded as \"AB\" (1 2) or \"L\" (12).\n```\n\nExample 2:\n\n```Input: \"226\"\nOutput: 3\nExplanation: It could be decoded as \"BZ\" (2 26), \"VF\" (22 6), or \"BBF\" (2 2 6).```\n\n## Solutions\n\n``````class Solution {\n\n// For such kind of problems, you always need a map to keep track of visited situations so that you can\n// speed up the execution efficiency.\nMap<String, Integer> countMap = new HashMap<>();\n\npublic int numDecodings(String s) {\nif (s == null || s.length() == 0) {\nreturn 1;\n}\n\nif (countMap.containsKey(s)) {\nreturn countMap.get(s);\n}\n\n// if s starts with '0', there is no substitute letters for \"0\" or \"0?\"\nif (s.charAt(0) == '0') {\ncountMap.put(s, 0);\n\nreturn 0;\n}\n\n// There is always substitute for single digit except \"0\"\nint ans1 = numDecodings(s.substring(1));\n\nint ans2 = 0;\n\n// There is always substitute for 2 digit numbers in range [11, 26]\nif (s.length() >= 2 && Integer.valueOf(s.substring(0, 2)) <= 26) {\nans2 = numDecodings(s.substring(2));\n}\n\nint ans = ans1 + ans2;\n\ncountMap.put(s, ans);\n\nreturn ans;\n}\n}\n``````" ]

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[ "# Half the perimeter of a rectangular garden\n\nBy Rebecca\n\nHalf the perimeter of a rectangular garden, whose length is 4 M more than its width, is 36 M. Find the dimensions of the garden.\n\nGiven\n\n• Length = 4 m\n• Width = 36 m\n\nLet\n\n• l be the length of the rectangular garden\n• b be the width of the rectangular garden\n\nAs given in the question:\n\nLength is 4m more than its width\n\nl = b + 4\n\nl – b = 4————–\n\nHalf Parameter of the rectangle = 36\n\nl + b = 36————\n\nNow, by adding both the equations and , we get,\n\nl + b + l – b = 4 + 36\n\n2l = 40\n\nl =frac402\n\nl = 20\n\nNow by substituting the value of l in the equation we get,\n\nl – b = 4\n\n20 – b = 4\n\nb = 20 – 4\n\nb = 16\n\nTherefore, the length (l) of the rectangle is 20m and the width (b) of the rectangle is 16m.\n\nHalf the perimeter of a rectangular garden\n\nRead more: Isobars have same number of\n\nPosted by Rebecca\nFollow:\nRebecca is an Independent content writer for breldigital, She writes content on any given topic. She loves to write a case study article or reviews on a brand, Be it any topic, she nails it" ]

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[ "## Signal Definition and Classification of Signals:\n\nA signal is a single valued function of one or more variables that conveys some information. Signals are represented mathematically as the functions of one or more independent variables. A signal may be a function of time, temperature, position, pressure, distance etc. In electrical sense, the signal can be voltage or current. The voltage or current is the function of time as an independent variable.  The Classification of Signals are namely\n\n1. The signals can be one-dimensional or multidimensional.\n\n• One-Dimensional Signals: When the function depends on a single variable, the signal is said to be one dimensional. Speech signal, whose amplitude varies with time, is one-dimensional signal.\n• Multidimensional Signals: When the function depends on two or more variables, the signal is said to be multidimensional signal. Image is an multidimensional signal because it is a two-dimensional signal with horizontal and vertical coordinates.\n\n2. Based upon their nature and characteristics in the time domain, the signals may be broadly classified as given below:\n\n• Continuous-Time Signals: This function is defined continuously in the time domain. A continuous-time signal is represented by x(t) where x represents the shape of the signal and t shows that the variable is time. This signal will have some value at every instant of time. Figure 22.4 shows a continuous-time signal.", null, "• Discrete-Time Signals: A discrete-time signal is defined only at certain time-instants. For discrete-time signals, the amplitude between two time instants is not defined. It is represented by x(n), here time n is the independent variable. Figure 22.5 shows a discrete-time signal. Mathematically, a discrete-time signal is denoted as under:", null, "Here arrow indicates the value of x(n) at n = 0.", null, "3. Real and Complex Signals: A signal x(t) is a real signal if its value is a real number. Similarly, a signal x(t) is a complex signal if its value is a complex number.\n\n4. Deterministic and Non-Deterministic (Random) Signals: Deterministic signals are those signals which can be completely specified in time. The pattern of this type of classification of signals is regular and can be characterized mathematically. Also, the nature and amplitude of such a signal at any time can be predicted. So these signals are called deterministic signals.\n\nA non-deterministic signal is one whose occurrence is always random in nature. The pattern of such a signal is quite irregular so these signals are also called random signals. For example, a thermal noise generated in an electric circuit is a non-deterministic signal.\n\n5. Periodic and Non-Periodic Signals:\n\n• Periodic Signal: A signal which repeats itself after a fixed time period is called a periodic signal. It can be defined mathematically as\n\nx(t) = x(t + T0). It is a condition of periodicity.\n\nHere T0 is called as the period of signal x(t). Example of these signals are sine wave, cosine wave, square wave etc.\n\n• Periodic Discrete-Time Signal: For the discrete-time signal, the condition of periodicity is\n\nx(n) = x(n + N)\n\nHere, N is the period of signal and its smallest value is called as fundamental period.\n\n• Non-Periodic Signal: A signal which does not repeat itself after a fixed time period or does not repeat at all is called as non-periodic or aperiodic signal. Mathematical expression for this signal is\n\nx(t) ≠ x(t + T0)\n\nSometimes value of period T0 = ∞ then it is an exponential signal and mathematically can be expressed as\n\nx(t) = e-αt\n\n• Non-Periodic Discrete-Time Signal: Mathematical expression for this classification of signals is,\n\nx(n) ≠ x(n + N)\n\n6. Symmetrical (Even) or Anti symmetrical (Odd) Signals:\n\n• Symmetrical Continuous Time Signal: A signal x(t) is symmetrical or even if it satisfies the following condition:\n\nx(t) = x(-t)", null, "Here x(t) is value of signal for positive t and x(-t) is value of signal for negative t. Cosine wave is an example of symmetrical continuous time signal.\n\n• Antisymmetrical Continuous Time Signal: A signal x(t) is antisymmetrical or odd if it satisfies the following condition:\n\nx(t) = -x(-t)\n\nSine wave is an example of antisymmetrical continuous time signal.\n\n• Symmetrical Discrete-Time Signal: This type of classification of signals satisfies the following condition and is shown in Fig. 22.7(a).\n\nx(n) = x(-n)\n\n• Unsymmetrical Discrete-Time Signal: This type of classification of signals satisfies the following condition and is shown in Fig. 22.7(b).\n\nx(n) = -x(-n)", null, "7. Energy and Power Signals: Signals may be classified as energy and power signals. However, there are some signals which can neither be classified as energy signals nor power signals.\n\nThe energy signal is one which has finite energy and zero average power.\n\nx(t) is an energy signal if it satisfies following condition:\n\n0 < E < ∞ and P = 0\n\nHere, E is the energy and P is the power of signal x(t). The power signal is one which has finite average power and infinite energy.\n\nx(t) is a power signal if it satisfies following condition:\n\n0 < P < ∞ and E = ∞\n\n8. Single Channel and Multichannel Signals: Multichannel signals are generated by multiple sources or multiple sensors. The resultant signal is the vector sum of signals from all channels. It is expressed as:", null, "A common example of multichannel signal is ECG waveform.\n\n9. One-Dimensional and Multidimensional Signals: If a signal is a function of single independent variable, the signal is called as one-dimensional signal. On the other hand, if the signal is a function of multi independent variables then it is called as multidimensional signal.\n\nFor example, to locate a pixel on the TV screen, two coordinates X and Y are required and this pixel is a function of time too. This it is a multidimensional signal and can be expressed mathematically as P(x, y, t).\n\nScroll to Top" ]

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[ "# Sylow\n\noriginal sillier post\n\nNote on notation: I’m going to use $$\\text{Stab}(x)$$ instead of $$G_x$$ for the stabilizer subgroup and $$\\text{Cl}(x)$$ instead of $$^Gx$$ for the conjugacy classes. For the orbit of $$x$$ I’ll stick with the norm and use $$Gx$$, although it’s only used in confusing summation notation that I’ll explain with words too.\n\nWe keep using this silly counting argument which I thought was something like Burnside’s lemma but actually is a lot simpler, just partitioning the set into orbits and slapping the orbit-stabilizer theorem on.\n\nIf $$G$$ is the group and $$S$$ is the set then\n\n$\\vert S\\vert = \\sum_{Gx} \\vert Gx\\vert = \\sum_{Gx} \\frac{\\vert G\\vert}{\\vert\\text{Stab}(x)\\vert}$\n\nwhere in the sum, $$x$$ ranges over a set of representatives of the orbits, i.e. exactly one $$x$$ is taken from each orbit, i.e. $$Gx$$ is equal to each orbit exactly once. Anyway, that sum is a terrible abuse of notation so I’ll just leave it out and explain things in English.\n\nBelow, let $$p$$ denote some fixed prime.\n\n### Cauchy’s Lemma\n\nLet $$G$$ be a finite abelian group. Then if $$p \\mid \\vert G\\vert$$ then $$G$$ has a subgroup of order $$p$$.\n\nProof: Induct on the order of $$G$$.\n\nIf we have an element $$a$$ with order $$pr$$ for some positive integer $$r$$ then $$\\langle a^r \\rangle$$ (the subgroup generated by $$a^r$$) works.\n\nOtherwise, we can at least pick an element $$a \\neq 1$$ with order not divisible by $$p$$. As $$G$$ is abelian, we can take the factor group $$G/\\langle a \\rangle$$, whose order is divisible by $$p$$. By induction, this factor group contains a subgroup with order $$p$$; taking a generator of it, we obtain a coset $$b\\langle a\\rangle$$ with order $$p$$, where $$b \\in G$$. Then $$b$$ has order divisible by $$p$$, because if $$b^s = 1$$ then $$(b\\langle a\\rangle)^s = \\langle a\\rangle$$, the identity of the factor group, and $$p \\mid s$$. As before, let $$b$$’s order be $$pr$$ and take $$\\langle b^r \\rangle$$.\n\n(Alternatively: this is entirely trivial if you know the structure of all finitely generated abelian groups.)\n\n— end Cauchy’s lemma —\n\n### Sylow I\n\nLet $$G$$ be a finite group, let $$p$$ be a prime and let $$k$$ be a positive integer. If $$p^k \\mid \\vert G\\vert$$ then there exists a subgroup $$H$$ of $$G$$ with order $$p^k$$.\n\nProof:\n\nConsider the group action $$G$$ acting via conjugation ($$g: x \\mapsto gxg^{-1}$$) on the set $$G$$ itself.\n\nThen\n\n$\\vert G\\vert = \\sum \\vert\\text{Cl}(x)\\vert = \\sum \\frac{\\vert G\\vert}{\\vert C(x)\\vert}$\n\nwhere in the sum, $$x$$ ranges over a set of representatives of the conjugacy classes, or orbits under conjugation.\n\nConsider the center $$C(G)$$. It’s just the set of $$x$$ for which $$\\vert C(x)\\vert = \\vert G\\vert$$, so $$\\frac{\\vert G\\vert}{\\vert C(x)\\vert} = 1$$ and the conjugacy class of $$x$$ contains only $$x$$ itself. We separate them from the summation like so:\n\n$\\vert G\\vert = \\vert C(G)\\vert + \\sum \\frac{\\vert G\\vert}{\\vert C(x)\\vert}$\n\nwhere in the sum, $$x$$ ranges over a set of representatives of the conjugacy classes with at least two elements.\n\nI.1. If $$\\vert C(G)\\vert$$ is divisible by $$p$$:\n\nOf course, $$C(G)$$ is abelian. So, by Cauchy’s Lemma above, there exists a subgroup $$L < C(G)$$ with order $$p$$. As $$L$$ is a subgroup of the center, it commutes with every element of $$G$$, so it’s normal. Then we can compute the factor group $$G/L$$, whose order is divisible by $$p^{k-1}$$. Thus by induction $$G/L$$ contains a subgroup $$H$$ with order $$p^{k-1}$$. The union of all the cosets in $$H$$ is a subgroup with order $$p^k$$, so we’re done.\n\nI.2. If $$\\vert C(G)\\vert$$ is not divisible by $$p$$:\n\nSince $$\\vert G\\vert$$ is divisible by $$p$$, there would have to be another term on the RHS of the equation not divisible by $$p$$. Thus, for some $$x \\in G$$, the value $$\\frac{\\vert G\\vert}{\\vert C(x)\\vert}$$ is not divisible by $$p$$. This is only possible if $$p^k \\mid \\vert C(x)\\vert$$. But $$C(x) \\lneqq G$$ (if the equality sign held $$x$$ should have been moved out into $$C(G)$$), hence by the induction hypothesis on $$C(x)$$ we have a subgroup $$L < C(X) < G$$ with $$\\vert L\\vert = p^k$$ and we’re done.\n\n— end Sylow I —\n\n### Sylow II\n\nLet $$G$$ be a finite group; suppose $$p^k \\parallel \\vert G\\vert$$ (i.e. $$p^k \\mid \\vert G\\vert, p^{k+1} \\nmid \\vert G\\vert$$). Consider the collection $$\\mathcal{S} = \\{H < G \\mid \\vert H\\vert = p^k \\}$$ of all subgroups of $$G$$ with order $$p^k$$ (which are called Sylow subgroups).\n\nThen:\n\n1. for any two Sylow subgroups $$H_1, H_2 \\in \\mathcal{S}$$, they are conjugate, i.e. $$\\exists x \\in G$$ such that $$xH_1x^{-1} = H_2$$;\n2. $$\\vert\\mathcal{S}\\vert$$ is a divisor of $$\\frac{\\vert G\\vert}{p^k}$$, the index of any Sylow subgroup;\n3. $$\\vert\\mathcal{S}\\vert\\equiv 1 \\bmod p$$;\n4. for any subgroup $$L < G$$ with order $$p^r$$ for some nonnegative integer $$r$$, there is some Sylow subgroup $$H \\in \\mathcal{S}$$ with $$L < H$$.\n\nProof.\n\nFirst, we prove this statement: (d’) given a fixed Sylow subgroup $$H$$, for any subgroup $$L < G$$ with order $$p^r$$ for some nonnegative integer $$r$$, there exists a conjugation $$xHx^{-1}$$ of $$H$$ such that $$L < xHx^{-1}$$.\n\nConsider the group action $$L$$ acting via left translaion ($$\\ell: gH \\mapsto \\ell gH$$) on the set of left cosets of $$H$$, which we’ll denote $$\\mathcal{T} = \\{gH \\mid g \\in G\\}$$.\n\nAgain:\n\n$\\vert\\mathcal{T}\\vert = \\sum \\frac{\\vert L\\vert}{\\vert\\text{Stab}(gH)\\vert}$\n\nwhere $$gH$$ ranges over a set of representatives of the orbits in $$\\mathcal{T}$$.\n\nNow, we certainly know that $$\\vert\\mathcal{T}\\vert = \\frac{\\vert G\\vert}{\\vert H\\vert = p^k}$$ is not divisible by $$p$$. Hence again there’s a term on the RHS that’s not divisible by $$p$$. As $$\\vert L\\vert = p^r$$ this is only possible if, for some coset $$gH$$, we have $$\\text{Stab}(gH) = L$$.\n\nThus, that coset $$gH$$ satisfies: for any $$\\ell \\in L$$, we have $$\\ell gH = gH$$, i.e. $$\\ell \\in gHg^{-1}$$. Then $$L \\subset gHg^{-1}$$, and as both are subgroups of $$G$$ we’re done.\n\nWith this we have proved (d), since (by Sylow I) at least one $$H$$ exists, and the $$gHg^{-1}$$ produced is a subgroup with the same order as $$H$$ and is thus another Sylow subgroup. It also implies (a), just by setting $$L$$ to be another Sylow group.\n\nNow we prove (c). Fix a Sylow subgroup $$H_1$$ and consider the group action $$H_1$$ acting via conjugation ($$x: H \\mapsto xHx^{-1}$$) on the set of Sylow subgroups $$\\mathcal{S}$$. Then\n\n$\\vert\\mathcal{S}\\vert = \\sum \\frac{\\vert H_1\\vert}{\\vert\\text{Stab}(H_j)\\vert}$\n\nwhere $$H_j$$ ranges over a set of representatives of the orbits in $$\\mathcal{S}$$.\n\nIf $$H_j = H_1$$ it’s obvious that every element of $$H_1$$ is in the stabilizer, hence the summand is $$1$$.\n\nOn the other hand, if $$H_j \\neq H_1$$, we claim that $$\\text{Stab}(H_j) \\neq H_1$$. Suppose otherwise; then $$xH_jx^{-1} = H_j$$ for any $$x \\in H_1$$, so $$H_j$$ is a normal subgroup of the subgroup generated by these two Sylow subgroups, $$M = \\langle H_1, H_j \\rangle$$.\n\nConsider the factor group $$M/H_j$$; we claim that its order is also a power of $$p$$. This is because if we take the intersection of each coset with $$H_1$$, we actually get a factor group of $$H_1$$. (I think it’s the “first isomorphism theorem” or something.) That is, $$M/H_j \\cong H_1/(H_1 \\cap H_j)$$ so its order is a divisor of $$\\vert H_1\\vert = p^k$$.\n\nThus $$\\vert M\\vert = p^k\\vert M/H_j\\vert$$ is also a power of $$p$$, and $$\\vert M\\vert > \\vert H_1\\vert = p^k$$. But as $$M$$ is a subgroup of $$G$$, we know $$\\vert M\\vert$$ (a higher power of $$p$$ than $$p^k$$) divides $$\\vert G\\vert$$, contradicting our selection of $$k$$.\n\nTherefore, $$\\text{Stab}(H_j) \\neq H_1$$ and therefore $$\\frac{\\vert H_1\\vert}{\\vert\\text{Stab}(H_j)\\vert}$$ is greater than 1. As $$\\vert H_1\\vert = p^k$$, the result is divisible by $$p$$. Therefore, all terms vanish $$\\bmod p$$ except for the orbit of $$H_1$$ whose summand is $$1$$, so $$\\vert\\mathcal{S}\\vert\\equiv 1 \\bmod p$$.\n\nNow, (b) is easy. Consider the group action $$G$$ acting via conjugation ($$g: H \\mapsto gHg^{-1}$$) on the set of Sylow subgroups $$\\mathcal{S}$$. Since there’s only one orbit,\n\n$\\vert\\mathcal{S}\\vert = \\frac{\\vert G\\vert}{\\vert\\text{Stab}(H)\\vert}$\n\nfrom which it’s obvious $$\\vert G\\vert$$ is divisible by $$\\vert\\mathcal{S}\\vert$$. And since $$\\vert\\mathcal{S}\\vert$$ is not divisible by $$p$$, we know that $$\\vert G\\vert /p^k$$ is divisible by $$\\vert\\mathcal{S}\\vert$$ too.\n\n— end Sylow II —\n\n(note: the commenting setup here is experimental and I may not check my comments often; if you want to tell me something instead of the world, email me!)" ]

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[ "•\n•\n\n#### Article Title\n\nTheorem on the Zeros of Polynomials\n\nResearch\n\n#### Abstract\n\nA simple method of finding limits for the absolute values of the zeros of polynomials will be given in this paper. The radius of a circle about the origin in the complex plane will be found from the coefficient of the polynomial such that all the zeros will lie on or within this circle. For certain polynomials a second circle will be found such that the zeros will lie on or outside this circle. In finding these circles the following well-known theorem of Rouché is used. Rouché's Theorem is very useful. By means of it the fundamental theorem of algebra and other important results may be established. The proof of Rouché's Theorem is ordinarily based on residue theory and may be found in most books on the theory of functions of a complex variable. Rouché, however, used series expansions to prove his theorem.\n\n1951\n\n#### Journal Title\n\nProceedings of the Iowa Academy of Science\n\n58\n\n1\n\n311\n\n312\n\nen\n\napplication/pdf\n\nCOinS" ]

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[ "# #1f6f1f Color Information\n\nIn a RGB color space, hex #1f6f1f is composed of 12.2% red, 43.5% green and 12.2% blue. Whereas in a CMYK color space, it is composed of 72.1% cyan, 0% magenta, 72.1% yellow and 56.5% black. It has a hue angle of 120 degrees, a saturation of 56.3% and a lightness of 27.8%. #1f6f1f color hex could be obtained by blending #3ede3e with #000000. Closest websafe color is: #336633.\n\n• R 12\n• G 44\n• B 12\nRGB color chart\n• C 72\n• M 0\n• Y 72\n• K 56\nCMYK color chart\n\n#1f6f1f color description : Very dark lime green.\n\n# #1f6f1f Color Conversion\n\nThe hexadecimal color #1f6f1f has RGB values of R:31, G:111, B:31 and CMYK values of C:0.72, M:0, Y:0.72, K:0.56. Its decimal value is 2060063.\n\nHex triplet RGB Decimal 1f6f1f `#1f6f1f` 31, 111, 31 `rgb(31,111,31)` 12.2, 43.5, 12.2 `rgb(12.2%,43.5%,12.2%)` 72, 0, 72, 56 120°, 56.3, 27.8 `hsl(120,56.3%,27.8%)` 120°, 72.1, 43.5 336633 `#336633`\nCIE-LAB 40.83, -40.523, 36.111 6.496, 11.759, 3.223 0.302, 0.547, 11.759 40.83, 54.278, 138.295 40.83, -33.375, 43.145 34.291, -26.191, 18.43 00011111, 01101111, 00011111\n\n# Color Schemes with #1f6f1f\n\n• #1f6f1f\n``#1f6f1f` `rgb(31,111,31)``\n• #6f1f6f\n``#6f1f6f` `rgb(111,31,111)``\nComplementary Color\n• #476f1f\n``#476f1f` `rgb(71,111,31)``\n• #1f6f1f\n``#1f6f1f` `rgb(31,111,31)``\n• #1f6f47\n``#1f6f47` `rgb(31,111,71)``\nAnalogous Color\n• #6f1f47\n``#6f1f47` `rgb(111,31,71)``\n• #1f6f1f\n``#1f6f1f` `rgb(31,111,31)``\n• #471f6f\n``#471f6f` `rgb(71,31,111)``\nSplit Complementary Color\n• #6f1f1f\n``#6f1f1f` `rgb(111,31,31)``\n• #1f6f1f\n``#1f6f1f` `rgb(31,111,31)``\n• #1f1f6f\n``#1f1f6f` `rgb(31,31,111)``\nTriadic Color\n• #6f6f1f\n``#6f6f1f` `rgb(111,111,31)``\n• #1f6f1f\n``#1f6f1f` `rgb(31,111,31)``\n• #1f1f6f\n``#1f1f6f` `rgb(31,31,111)``\n• #6f1f6f\n``#6f1f6f` `rgb(111,31,111)``\nTetradic Color\n• #0e330e\n``#0e330e` `rgb(14,51,14)``\n• #144714\n``#144714` `rgb(20,71,20)``\n• #195b19\n``#195b19` `rgb(25,91,25)``\n• #1f6f1f\n``#1f6f1f` `rgb(31,111,31)``\n• #258325\n``#258325` `rgb(37,131,37)``\n• #2a972a\n``#2a972a` `rgb(42,151,42)``\n• #30ab30\n``#30ab30` `rgb(48,171,48)``\nMonochromatic Color\n\n# Alternatives to #1f6f1f\n\nBelow, you can see some colors close to #1f6f1f. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #336f1f\n``#336f1f` `rgb(51,111,31)``\n• #2c6f1f\n``#2c6f1f` `rgb(44,111,31)``\n• #266f1f\n``#266f1f` `rgb(38,111,31)``\n• #1f6f1f\n``#1f6f1f` `rgb(31,111,31)``\n• #1f6f26\n``#1f6f26` `rgb(31,111,38)``\n• #1f6f2c\n``#1f6f2c` `rgb(31,111,44)``\n• #1f6f33\n``#1f6f33` `rgb(31,111,51)``\nSimilar Colors\n\n# #1f6f1f Preview\n\nText with hexadecimal color #1f6f1f\n\nThis text has a font color of #1f6f1f.\n\n``<span style=\"color:#1f6f1f;\">Text here</span>``\n#1f6f1f background color\n\nThis paragraph has a background color of #1f6f1f.\n\n``<p style=\"background-color:#1f6f1f;\">Content here</p>``\n#1f6f1f border color\n\nThis element has a border color of #1f6f1f.\n\n``<div style=\"border:1px solid #1f6f1f;\">Content here</div>``\nCSS codes\n``.text {color:#1f6f1f;}``\n``.background {background-color:#1f6f1f;}``\n``.border {border:1px solid #1f6f1f;}``\n\n# Shades and Tints of #1f6f1f\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, #010401 is the darkest color, while #f3fcf3 is the lightest one.\n\n• #010401\n``#010401` `rgb(1,4,1)``\n• #051305\n``#051305` `rgb(5,19,5)``\n• #0a220a\n``#0a220a` `rgb(10,34,10)``\n• #0e320e\n``#0e320e` `rgb(14,50,14)``\n• #124112\n``#124112` `rgb(18,65,18)``\n• #165016\n``#165016` `rgb(22,80,22)``\n• #1b601b\n``#1b601b` `rgb(27,96,27)``\n• #1f6f1f\n``#1f6f1f` `rgb(31,111,31)``\n• #237e23\n``#237e23` `rgb(35,126,35)``\n• #288e28\n``#288e28` `rgb(40,142,40)``\n• #2c9d2c\n``#2c9d2c` `rgb(44,157,44)``\n• #30ac30\n``#30ac30` `rgb(48,172,48)``\n• #34bc34\n``#34bc34` `rgb(52,188,52)``\nShade Color Variation\n• #3bc83b\n``#3bc83b` `rgb(59,200,59)``\n• #4bcd4b\n``#4bcd4b` `rgb(75,205,75)``\n• #5ad15a\n``#5ad15a` `rgb(90,209,90)``\n• #69d569\n``#69d569` `rgb(105,213,105)``\n• #79d979\n``#79d979` `rgb(121,217,121)``\n• #88de88\n``#88de88` `rgb(136,222,136)``\n• #97e297\n``#97e297` `rgb(151,226,151)``\n• #a7e6a7\n``#a7e6a7` `rgb(167,230,167)``\n• #b6ebb6\n``#b6ebb6` `rgb(182,235,182)``\n• #c5efc5\n``#c5efc5` `rgb(197,239,197)``\n• #d5f3d5\n``#d5f3d5` `rgb(213,243,213)``\n• #e4f7e4\n``#e4f7e4` `rgb(228,247,228)``\n• #f3fcf3\n``#f3fcf3` `rgb(243,252,243)``\nTint Color Variation\n\n# Tones of #1f6f1f\n\nA tone is produced by adding gray to any pure hue. In this case, #454945 is the less saturated color, while #048a04 is the most saturated one.\n\n• #454945\n``#454945` `rgb(69,73,69)``\n• #404e40\n``#404e40` `rgb(64,78,64)``\n• #3a543a\n``#3a543a` `rgb(58,84,58)``\n• #355935\n``#355935` `rgb(53,89,53)``\n• #2f5f2f\n``#2f5f2f` `rgb(47,95,47)``\n• #2a642a\n``#2a642a` `rgb(42,100,42)``\n• #246a24\n``#246a24` `rgb(36,106,36)``\n• #1f6f1f\n``#1f6f1f` `rgb(31,111,31)``\n• #1a741a\n``#1a741a` `rgb(26,116,26)``\n• #147a14\n``#147a14` `rgb(20,122,20)``\n• #0f7f0f\n``#0f7f0f` `rgb(15,127,15)``\n• #098509\n``#098509` `rgb(9,133,9)``\n• #048a04\n``#048a04` `rgb(4,138,4)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #1f6f1f 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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[ "arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.\n\n# Hadron production in heavy ion collisions: Fragmentation and recombination from a dense parton phase\n\nR. J. Fries Department of Physics, Duke University, Durham, NC 27708    S. A. Bass Department of Physics, Duke University, Durham, NC 27708 RIKEN BNL Research Center, Brookhaven National Laboratory, Upton, NY 11973, USA    B. Müller Department of Physics, Duke University, Durham, NC 27708    C. Nonaka Department of Physics, Duke University, Durham, NC 27708\nAugust 9, 2020\n###### Abstract\n\nWe discuss hadron production in heavy ion collisions at RHIC. We argue that hadrons at transverse momenta GeV are formed by recombination of partons from the dense parton phase created in central collisions at RHIC. We provide a theoretical description of the recombination process for GeV. Below GeV our results smoothly match a purely statistical description. At high transverse momentum hadron production is well described in the language of perturbative QCD by the fragmentation of partons. We give numerical results for a variety of hadron spectra, ratios and nuclear suppression factors. We also discuss the anisotropic flow and give results based on a flow in the parton phase. Our results are consistent with the existence of a parton phase at RHIC hadronizing at a temperature of 175 MeV and a radial flow velocity of 0.55.\n\npreprint: DUKE-TH???\n\n## I Introduction\n\nRecent data from the Relativistic Heavy Ion Collider (RHIC) have shown a strong nuclear suppression of the pion yield at transverse momenta larger than 2 GeV/ in central Au + Au collisions, compared to interactions PHENIX . This is widely seen as the experimental confirmation of jet quenching, the phenomenon that high energy partons lose energy when they travel through the hot medium created in a heavy ion collision GyulWang:94 ; BDMS:01 ; Muller:02 , entailing a suppression of intermediate and high hadrons.\n\nHowever, the experiments at RHIC have provided new puzzles. The amount of suppression seems to depend on the hadron species. In fact, in the production of protons and antiprotons between 2 and 4 GeV/ the suppression seems to be completely absent. Generally, pions and kaons appear to suffer from a strong energy loss while baryons and antibaryons do not. Two stunning experimental facts exemplify this PHENIX-B ; STAR-B ; STAR-L ; PHENIX-L . First, the ratio of protons over positively charged pions is equal or above one for and is approximately constant up to 4 GeV/. Second, the nuclear suppression factor below 4 GeV/ is close to one for protons and lambdas, while it is about 0.3 for pions.\n\nThere have been recent attempts to describe the different behavior of baryons and mesons through the existence of gluon junctions GyulVit:01 or alternatively through recombination as the dominant mechanism of hadronization FMNB:03 ; GreKoLe:03 . The recombination picture has attracted additional attention due to the observation that the elliptic flow pattern of different hadron species can be explained by a simple recombination mechanism LinKo:02 ; Voloshin:02 ; MoVo:03 ; LinMol:03 . The anisotropies for the different hadrons are compatible with a universal value of in the parton phase, related to the hadronic flow by factors of two and three depending on the number of valence quarks Sorensen:03 .\n\nIn this work we elaborate on the arguments presented in FMNB:03 . We will present the formalism as well as new numerical results. The competition between recombination and fragmentation delays the onset of the perturbative/fragmentation regime to relatively high transverse momentum of 4–6 GeV/, depending on the hadron species. This is the explanation for several key observations at RHIC:\n\n• the two component form of hadron transverse momentum spectra, including an exponential part and a power law tail with a transition between 4 and 6 GeV/.\n\n• the very different behavior of the nuclear suppression factors of mesons and baryons.\n\n• the particle dependence of the elliptic flow.\n\n• the unusually large baryon/meson ratios.\n\nIn addition, our calculation clarifies the range of applicability of perturbative calculations including energy loss.\n\nThe paper is organized as follows. In the next section we explain why fragmentation might not be the dominant mechanism of hadronization at intermediate transverse momenta of a few GeV/ in heavy ion collisions. We discuss the fundamental principles of recombination and fragmentation. In Sec. III we present the theoretical framework of recombination in more detail and also discuss its shortcomings. In particular, we address the question of applicability at low transverse momentum. In Sec. IV we introduce our parametrization of the parton spectrum and discuss further calculational details, and in Sec. V we present numerical results on spectra, hadron ratios, nuclear suppression and elliptic flow. Sec. VI summarizes our work.\n\n## Ii Fragmentation vs. Recombination\n\n### ii.1 Fragmentation of partons\n\nInclusive hadron production at sufficiently large momentum transfer can be described by perturbative quantum chromodynamics (pQCD). The invariant cross section for a hadron with momentum can be given in factorized form Owens:86\n\nThe sum runs over all parton species and is the cross section for the production of parton with momentum . Thus the parton production cross section has to be convoluted with the probability that parton fragments into hadron . The probabilities are called fragmentation functions CoSo:81 . Like parton distributions they are non-perturbative quantities. However they are universal and once measured, e.g. in annihilations, they can be used to describe hadron production in other hard QCD processes.\n\nUsing (1) we can estimate the ratio of protons and pions. Taking, e.g., the common parametrization of Kniehl, Kramer and Pötter (KKP) KKP:00 , the ratio is always smaller than 0.2 for each parton . This reflects the well known experimental fact that pions are much more abundant than protons in the domain where pQCD is applicable. The excess of pions over protons even holds down to very low , smaller than 1 GeV, where perturbative calculations are no longer reliable. In that domain one can argue that the difference in mass, , lays a huge penalty on proton production. The small value of the ratio predicted by these calculations over the entire range of is the reason why the ratio measured at RHIC is so surprising.\n\nIt has been suggested that the fragmentation functions can be altered by the environment GW:00 ; WaWa:02 . The energy loss of the propagating parton in the surrounding medium leads, in first approximation, to a rescaling of the variable . This would affect all produced hadrons in the same way, and thus cannot explain the observations at RHIC. In a picture with perturbative hadron production and jet quenching alone, the different behavior of hadrons can not be described by one consistent sets of energy loss parameters. To save the validity of the purely perturbative approach species dependent non-perturbative contributions to the fragmentation functions have to be introduced ad hoc to explain the data Wang:03 .\n\nPerturbative hadron production consists of three steps: production of a parton in a hard scattering, propagation and interaction with a medium, and finally hadronization of the parton. Only modifications in our understanding of hadronization are able to provide an explanation of the experimental observations, since the other steps are blind to the hadron species that will eventually be created.\n\n### ii.2 From fragmentation to recombination\n\nFor the production of a hadron with momentum via fragmentation we need to start with a parton with momentum . The fragmentation functions favor very small values of , i.e. the situation where the energy of the fragmenting parton is not concentrated in one hadron. On the other hand, the transverse momentum spectrum of partons is steeply falling with . This makes it clear that fragmentation is a rather inefficient mechanism for the production of high hadrons, since it has to overcome the limited availability of partons at even higher transverse momentum. As a result, the average is larger than what is expected from the shape of the fragmentation functions. For pion production, is about 0.6 for the production from a valence quark, 0.4 for a sea quark and 0.5 for a gluon in the range for leading order KKP fragmentation functions.\n\nAn outgoing high energy parton is not a color singlet and will therefore have a color string attached. The breaking of the string will initiate the creation of quark antiquark pairs until there is a entire jet of partons, which have to share the energy of the initial parton. They will finally turn into many hadrons. The creation of several hadrons from one fragmenting parton is the reason why fragmentation functions prefer small values of . If phase space is already filled with partons, a single parton description might not be valid anymore. Instead one would have to introduce higher twist (multiple parton) fragmentation functions. In the most extreme case, if partons are abundant in phase space, they might simply recombine into hadrons. This means that a and a quark that are “close” to each other in phase space can bind together to form a . The scale of being close will be set by the width of the pion wave function. In this scenario the total pion momentum will be just the sum of the individual quark momenta. We immediately notice that this recombination mechanism is very efficient for steeply falling spectra: in order to produce a 5 GeV pion we can start with two quarks having (on average) about 2.5 GeV/ transverse momentum and each being therefore far more abundant (on average) than a 10 GeV/ parton that could produce the pion via fragmentation. Of course the recombining partons must be close in phase space, i.e. recombination will be suppressed if the phase space density is low.\n\nRecombination can be interpreted as the most “exclusive” form of hadronization, the endpoint of a hypothetical resummation of fragmentation processes to arbitrary twist. We cannot achieve a quantitative understanding of this at the moment. However, we can try to find an effective description which can be tested against observable consequences. In this work we will advocate a simple model for recombination and compare it with single parton fragmentation. These two mechanisms of hadron production compete differently depending on the phase space density of partons. From the above we understand that the competition between fragmentation and recombination is dominated by the slope and the absolute value of the phase space distribution of partons. Below we show that recombination always wins over fragmentation for an exponentially falling parton spectrum, but that fragmentation takes over if the spectrum has the form of a power law, as it is provided by pQCD. We will apply this insight to hadron production in relativistic heavy ion collisions at midrapidity and transverse momenta of a few GeV/ where we expect a densely populated phase space. For the recombination of three quarks into a proton the momenta of three partons have to be added up, but only two momenta in the case of a pion. Assuming an exponential parton spectrum this implies for a proton a distribution and for pions , predicting a constant ratio where the value is determined by simple counting of quantum numbers FMNB:03 . We will show that some of the surprising experimental results from RHIC can be explained in this way.\n\n### ii.3 The recombination concept\n\nThe idea of quark recombination was proposed long ago to describe hadron production in the forward region of collisions DasHwa:77 . This was later justified by the discovery of the leading particle effect, the phenomenon that, in the forward direction of a beam of hadrons colliding with a target, the production of hadrons sharing valence quarks with the beam hadrons are favored. E.g. in the Fermilab E791 fixed target experiment with a 500 GeV beam E791:96 the asymmetry\n\n α(xF)=dσD−/dxF−dσD+/dxFdσD−/dxF+dσD+/dxF (2)\n\nbetween and mesons grows nearly to unity when the Feynman variable , measuring the longitudinal momentum relative to the beam momentum, approaches 1. Fragmentation would predict this asymmetry to be very close to zero. However, recombination of the quark from a pair produced in a hard interaction with a valence quark from the , propagating in forward direction with large momentum, is highly favored compared to the recombination of the with a which is only a sea quark of the . This leads to the enhancement of over mesons in the forward region.\n\nThe leading particle effect is a clear signature for the existence of recombination as a hadronization mechanism and has been addressed in several publications recently AnMaHe:01 ; BraJiaMe:02 . In this case recombination is favored over fragmentation only in a certain kinematic situation (the very forward direction), which is a only a small fraction of phase space.\n\nIn central heavy ion collisions many more partons are produced than in collisions of single hadrons. The idea that recombination may then be important for a wide range of rapidities – and at least up to moderate transverse momenta – was advocated before recomb1 ; recomb2 ; recomb3 . However, it was only recently that RHIC data indicated that recombination could indeed be a valid approach up to surprisingly high transverse momenta of a few GeV/.\n\nCharm and heavy hadron production have the advantage that the heavy quark mass provides a large scale that permits a more rigorous treatment of the recombination process BraJiaMe:02 . The description of recombination into pions and protons seems to be theoretically less rigorous. However, a simple counting of quantum numbers in a picture where the structure of hadrons is dominated by their valence quarks often provides surprisingly good results. This has been pointed out for particle spectra and ratios HwaYa:02 ; FMNB:03 ; GreKoLe:03b and for elliptic flow LinKo:02 ; Voloshin:02 ; MoVo:03 . Most of the work so far has stayed on a qualitative level without quantitative predictions. Recombination of mesons in heavy ion collisions has been investigated in RappShu:03 .\n\nWe will argue below that the counting of quantum numbers is a good description of the recombination process for intermediate momenta. The formalism will set the stage to obtain quantitative results in this regime by recombining quarks from a possibly thermalized phase. We know that the parton phase will not behave like a perturbative plasma near the hadronization point KaKaLaLu:99 . Instead quarks at hadronization will be effective degrees of freedom exhibiting a mass and gluons will disappear as dynamical degrees of freedom. We will assume here that the effective quarks behave like constituent quarks and that there are no dynamical gluons.\n\nThis is different from the work of Greco et al. GreKoLe:03 who suggested to recombine one perturbative quark with thermal quarks, leading to an additional contribution at the transition region between the pure thermal phase dominating below 5 GeV/ and the pure fragmentation regime dominating above 5 GeV/. A similar form of this pick-up reaction of a perturbative parton was recently proposed in the context of the sphaleron model SolShu:03 .\n\nOne can also attempt to extend the recombination concept to low , however, the theoretical situation is much more ambiguous there. The main reasons are that the simple counting of quantum numbers violates energy and entropy conservation at low , where the bulk of the hadron yield resides. Nevertheless, once recombination is recognized to be the dominant hadronization mechanism at intermediate from 2 to 5 GeV/, it is quite reasonable to expect that this mechanism extends down to very small transverse momentum, where quarks are even more abundant. However the theoretical description will only be on solid ground once the problems of energy and entropy conservation are addressed properly.\n\n### ii.4 A non-relativistic model\n\nFor a first estimate, let us consider a simple static model for the recombination process. We start with a system of quarks and antiquarks which are homogeneously distributed in a fixed three-dimensional volume with phase space distributions , so that the number of quarks or antiquarks with a particular set of quantum numbers (color, spin, isospin) is\n\n Na=V∫d3p(2π)3wa(p) (3)\n\nHadronization is assumed to occur instantaneously throughout the volume. The distributions are supposed not to change during the hadronization process, i.e. the quarks are assumed to be quasi-free.\n\nThe spatial wave functions for a two particle quark/antiquark state with momenta and and a meson bound state with momentum are\n\n ⟨x|q,p1p2⟩ =V−1ei(p1x1+p2x2) (4) ⟨x|M,P⟩ =V−1/2eiP⋅RφM(y) (5)\n\nrespectively, with the center of mass and relative coordinates and for the two quarks in the meson system. To keep our notation simple we omit the proper antisymmetrization of multi fermion states, since all combinatorial factors will cancel in the final result. The internal meson wave functions is normalized as\n\n ∫d3y|φM(y)|2=1. (6)\n\nThe overlap amplitude is given by\n\n ⟨q,p1p2|π,P⟩=(2π)3V3/2δ3(P−p1−p2)^φM(q). (7)\n\nHere, we have introduced the relative momentum , conjugate to , and is the Fourier transformed wave function. The squared amplitude is\n\n |⟨q,p1p1|M,P⟩|2=(2π)3V2δ3(P−p1−p2)∣∣^φM(q)∣∣2. (8)\n\nWe conclude that the total number of pions found in the quark/antiquark distribution is\n\n NM=CMV3∫d3P(2π)3d3p1(2π)3d3p2(2π)3×w(p1)w(p2)∣∣⟨q,p1p1|M,P⟩∣∣2 (9)\n\nwith a degeneracy factor . The momentum distribution of the pions is given by\n\n dNMd3P=CπV(2π)3∫d3q(2π)3×w(P2+q)w(P2−q)∣∣^φM(q)∣∣2 (10)\n\nFrom above equation the spectra of mesons can be calculated for given quark distributions. This will require knowledge of . The wave function has some width , and we assume that it drops rapidly for . Let us study the kinematic region where . The integral over the relative momentum is dominated by values and thus we can assume that . We apply a Taylor expansion\n\n (11)\n\nwhere the first order term in the expansion vanishes.\n\nFrom the lowest order term in the expansion we get a contribution to the meson spectrum which is independent of the shape of the wave function. Only the normalization (6) enters and leads to a universal term\n\n CπV(2π)3w(P/2)2. (12)\n\nThe second order term (like all higher order terms) in the expansion generates a correction depending on the shape of the wave function. For the sake of simplicity, let us assume a Gaussian shape\n\n ^φM(q)=NMe−q2/2Λ2M. (13)\n\nThe normalization factor is determined by (6) as . For this example we have the second order correction\n\n CπV(2π)3Λ2M2[w(P/2)△w(P/2)−(∇w(P/2))2]. (14)\n\nWe want to emphasize once more that in the limit the exact shape of the wave function is not important. In a relativistic framework this statement is softened by the fact that we consider the hadron formation in a boosted frame, where is large. Therefore , which is of order in the rest frame of the hadron, will be dilated.\n\nAs an example, let us assume an exponential parton distribution of the form : The meson spectrum at large is then given by\n\n dNπd3P=CπV(2π)3e−P/T[1−2Λ2MTP], (15)\n\nusing (12),(14). The second order term introduces a power correction of order to the universal result, depending both on the width of the wave function and the slope of the parton distribution.\n\nFor nucleons we start with three quarks at coordinates . We introduce center of mass and relative coordinates , and .\n\nThe overlap amplitude between a 3 quark state and a baryon with momentum is\n\n ⟨q,p1p2p3|B,P⟩=(2π)3V2δ3(P−p1−p2−p3)^φB(q,s). (16)\n\nHere is the baryon wave function in momentum space, depending on the relative momenta and conjugate to and respectively.\n\nHence the baryon distribution is given by\n\n dNBd3P=CBV(2π)3∫d3q(2π)3d3s(2π)3∣∣^φB(q,s)∣∣2×w(P3+q2+s)w(P3+q2−s)w(P3−q). (17)\n\nis the appropriate degeneracy factor. In the region where the nucleon momentum is large compared to the intrinsic width of the wave function, we can again expand the product of the quark distribution functions. The leading term is universal and contributes\n\n CπV(2π)3w(P/3)3 (18)\n\nto the nucleon distribution. The second order term provides a correction\n\n CπV(2π)312(Λ2B2+34Λ2B1)[w(P/3)2△w(P/3)−w(P/3)(∇w(P/3))2], (19)\n\nassuming a normalized Gaussian wave function\n\n ^φB(q,s)=NBe−q2/2Λ2B1e−s2/2Λ2B2. (20)\n\n### ii.5 First conclusions\n\nLet us summarize and analyze our first results. The transverse spectrum of mesons from recombination is proportional to whereas fragmentation would generate a distribution proportional to . For an exponential parton spectrum the ratio of recombination to fragmentation is\n\n RF=CM⟨D⟩e−PTT(1−1⟨z⟩) (21)\n\nwhere and are average values of the fragmentation function and the scaling variable. Obviously, for large one always gets . In other words, recombination always wins over fragmentation at “high” for an exponential parton spectrum. The same is true for baryons as well as mesons.\n\nNow let us consider a parton distribution given by a power law spectrum with a scale and . Then the ratio of recombination over fragmentation is\n\n RF=CMA⟨D⟩(4⟨z⟩)α(PTQ)−α (22)\n\nand fragmentation ultimately dominates at high . Again, this holds both for mesons and baryons. This implies that fragmentation from an exponential spectrum and recombination from a power law spectrum are suppressed. We will thus omit these contributions in our work.\n\nOne might ask whether these considerations are still valid in the case of more realistic formulation of recombination. It turns out that the basic formulae obtained above are still valid in a relativistic description. Deviations are less than 20% for GeV/.\n\nGiven an exponential parton spectrum we note that recombination yields a a constant baryon-to-meson ratio. The ratio is then only determined by the degeneracy factors\n\n dNRBdNRM=CBCM. (23)\n\nJust counting the hadron degeneracies, the direct ratio (neglecting protons and pions from secondary hadronic decays) would be , in contrast to from fragmentation in pQCD. We will later see that finite mass effects and superposition with the fragmentation process will bring down the ratio from 2 to approximately 1 in the range between 2 and 4 GeV/ transverse momentum.\n\n## Iii The recombination formalism\n\nIn this section, we turn to a better description of recombination. This will require a more realistic model of the parton phase including longitudinal and transverse expansion as well as an improved space-time picture.\n\nLet us consider a system of quarks and antiquarks evolving in Minkowski space. We choose a spacelike hypersurface on which recombination of these partons into hadrons occurs. In the simplest scenario, that could be just a slice of Minkowski space with fixed time , leading back to the case we described in the previous subsection.\n\nIt has been discussed in the literature WieToHei:98 ; ScheiHei:99 , how the freeze-out can be smeared around the hypersurface to account for a finite hadronization time. However, RHIC experiments suggest a very rapid freeze-out. The measured two-particle correlation functions are consistent with an extremely short emission time in the local rest frame, suggesting a sudden transition after which the individual hadrons interact only rarely RHIC-HBT .\n\nThis can be understood by the fact that the hadronization time in the laboratory frame is Lorentz contracted. The Lorentz factors are 3.35 for a 3 GeV/ proton and 22 for a pion of the same . The hadronization time — even if it is more than 1 fm/ in the rest frame of the hadron — will be experienced as considerably shorter by fast hadronizing particles.\n\n### iii.1 Wigner function formalism\n\nRecombination has already been considered before in a covariant form utilizing Wigner functions for the process of baryons coalescing into light nuclei and clusters in nuclear collisions DHSZ:91 ; ScheiHei:99 . Here, we will provide a derivation for the recombination of a quark antiquark pair into a meson. The generalization to a three quark system recombining into a baryon is straightforward.\n\nBy introducing the density matrix for the system of partons, the number of quark-antiquark states that we will interpret as mesons is given by\n\n NM=∑ab∫d3P(2π)3⟨M;P|^ρab|M;P⟩ (24)\n\nHere is a meson state with momentum and the sum is over all combinations of quantum numbers — flavor, helicity and color — of valence partons that contribute to the given meson . We insert complete sets of coordinates\n\n NM=∫d3P(2π)3d3^r1d3^r′1d3^r2d3^r′2⟨M;P|^r1,^r2⟩×⟨^r1,^r2|^ρ|^r′1,^r′2⟩⟨^r′1,^r′2|M;P⟩. (25)\n\nand change the variables to and . We define the 2-parton Wigner function as\n\n ⟨r1−r′12,r2−r′22∣∣∣^ρ∣∣∣r1+r′12,r2+r′22⟩=∫d3p1(2π)3d3p2(2π)3e−ip1⋅r′1e−ip2⋅r′2Wab(r1,r2;p1,p2) (26)\n\nand introduce a notation for the meson wave function\n\n ⟨r1+r′12,r2+r′22∣∣∣π;P⟩=e−iP⋅(R+R′/2)φM(r−r′2). (27)\n\nIt is convenient to change coordinates again to\n\n R(′)=(r(′)1+r(′)2)/2, (28) r(′)=r(′)1−r(′)2 (29)\n\nwith conjugated momenta\n\n ~P =p1+p2, (30) q =(p1−p2)/2. (31)\n\nWe arrive at\n\n NM=∑ab∫d3P(2π)3d3q(2π)3∫d3Rd3rd3r′×Wab(R+r2,R−r2;P2+q,P2−q)×eiq⋅r′φM(r+r′2)φ∗M(r−r′2). (32)\n\nThe integration over has been carried out and provides the three-momentum conservation .\n\nWe define the Wigner function of the meson as\n\n ΦWM(r,q)=∫d3r′e−iq⋅r′φM(r+r′2)φ∗M(r−r′2). (33)\n\nThen\n\n dNMd3P=(2π)−3∑ab∫d3R∫d3qd3r(2π)3×Wab(R+r2,R−r2;P2+q,P2−q)ΦWM(r,q). (34)\n\nTo evaluate this expression, we have to model the Wigner functions of the parton system and of the meson. We assume that the 2-parton Wigner function can be factorized into a product of classical one-particle phase space distributions . Furthermore the color and helicity states for each flavor will be degenerate. We can therefore replace the sum over quantum numbers and by a degeneracy factor .\n\nWe introduce the following simplifications: the spatial width of the hadron wave function, translating into a width of , will be small compared to the nuclear size of the system at hadronization. The phase space distributions are steeply falling functions of the momentum , but vary much less with the spatial coordinate within the typical size of a hadron. We shall assume that the spatial variation of the phase space distribution is small on this scale, replacing with .\n\nIn our derivation we implicitly chose an equal time formalism by introducing the spatial coordinate states at a fixed time. However, in our result the integration over the final state phase space\n\n d3Pd3R=d3Pd3RP⋅u(R)/E (35)\n\nis manifestly Lorentz invariant. Here is the energy of the four vector and is the future oriented unit vector orthogonal to the hypersurface defined by the hadronization volume. This form can be easily generalized for an arbitrary hadronization hypersurface CoFr:74 ; ScheiHei:99 . We have\n\n EdNMd3P=CM∫Σd3RP⋅u(R)(2π)3∫d3q(2π)3×wa(R;P2−q)ΦWM(q)wb(R;P2+q). (36)\n\n, now only denote the flavors of the valence quarks in meson and\n\n ΦWM(q)=∫d3rΦWM(r,q) (37)\n\nis the spatially integrated Wigner function of the meson.\n\n### iii.2 Local light cone coordinates\n\nThe structure of hadrons is best known in the infinite momentum frame which is described in light cone coordinates. If we let the hadron momentum define the axis of the hadron light cone (HLC) frame, we can introduce the light cone coordinates e.g. , for the relative momentum . Note that we denote transverse momenta in the HLC frame — i.e. the component orthogonal to — by the label , but transverse momenta in the center of mass (CM) frame of the heavy ion collision — orthogonal to the beam axis — by the label . The labels and always refer to light cone coordinates in the HLC frame for a given . We fix the HLC frame by a simple rotation from the CM frame, i.e. . We reintroduce the momentum of parton in the meson. Assuming a mass shell condition with arbitrary but fixed virtuality , we can rewrite the integral\n\n d3k=d4k2k0δ(k2−m2a)=dk+d2k⊥k0k+ (38)\n\nin HLC coordinates. will be of order , or more precisely, of the order of a constituent quark mass. In the HLC frame, where we assume that formally , we parametrize with . Since , we have . We end up with\n\n EdNMd3P=CM∫Σd3RP⋅u(R)(2π)3∫dxP+d2k⊥√2(2π)3wa(R;xP+,k⊥)ΦM(x,k⊥)wb(R;(1−x)P+,−k⊥). (39)\n\nHere we have rewritten the spatially integrated Wigner function of the meson in terms of light cone coordinates for quark and approximated it by a squared light cone wave function of the meson\n\n ΦM(x,k⊥)=∣∣¯ϕM(x,k⊥)∣∣2. (40)\n\nThe light cone wave functions we introduce here do not necessarily coincide with the light cone wave functions used for exclusive processes in QCD ERBL:old . There, a hadron is decomposed into a series of Fock states of perturbative partons, starting from the valence structure. In terms of this expansion we know that the valence Fock state has only a small contribution at scales . is given by the momentum transfer in the hard reaction which is described by perturbative QCD (pQCD). However, because the momentum transfer in a hard exclusive reaction has to be spread over all partons in the hadron state, the exclusive process itself acts like a filter, weighting the lower Fock states more strongly. In other words, the contributions of higher Fock states, though more likely in the wave function, are generally suppressed by inverse powers of the momentum transfer (higher twist). This usually permits a fairly good description of hard exclusive processes in terms of the lowest Fock state.\n\nAs we will discuss below, we expect the parton spectrum of a heavy ion collision at freeze out to be composed of an exponential part at small transverse momentum and a power law tail, given by pQCD, at high transverse momentum. We will study recombination in the pQCD domain in a forthcoming publication.\n\nHere, we want to focus on recombination from the exponential part of the parton spectrum. It will be given by a slope with a temperature-like parameter . also sets the scale for the typical momentum transfer in the parton medium before hadronization. If is an effective blue shifted temperature in an expanding medium with physical temperature and flow velocity , then the typical scale will even be smaller. Here we use the phase transition temperature at zero baryon density Karsch:01 .\n\nWe cannot expect that perturbative QCD will work as a description of partons at the phase transition KaKaLaLu:99 . What are the quanta that recombine? We know that in pQCD for decreasing scales the non-perturbative (long-range) matrix elements describing hadrons get more “valence like”, though we cannot seriously extend this study to scales below 1 GeV. We will assume here that we recombine effective constituent partons, taking into account only the valence structure of the hadron. Gluons are no dynamic degrees of freedom in this picture, and the quarks and antiquarks will have an effective mass.\n\nThis picture is supported by the recent discovery of “magical factors” of 2 and 3 in measurements of spectra and the elliptic flow of mesons and baryons respectively at RHIC Sorensen:03 . Later we will apply our assumptions to partons in the exponential spectrum having as much as 2 GeV/ of transverse momentum. This might raise doubts about the validity of an effective description. However, we have to keep in mind that the momentum is in principle meaningless, only the momentum transfer experienced by a particle in a reaction sets the scale at which we resolve its structure.\n\nFrom the normalization condition\n\n ⟨M;P|M;P′⟩=(2π)3δ(P−P′) (41)\n\nwe infer and finally\n\n ∫dk+d2k⊥√2(2π)3|¯ϕM(x,k⊥)|2=1. (42)\n\nThis is different from the light cone wave functions in exclusive processes which involve a dimensionful quantity connected to the weight of the Fock state, e.g. the pion decay constant in case of the pion.\n\nWe further utilize the fact that we work in a kinematic regime where is large compared to all non-perturbative quantities. Of course, our main concern here is that we do not really know the shape of the wave function . We can choose a factorized ansatz\n\n ¯ϕM(x,k⊥)=ϕM(x)Ω(k⊥) (43)\n\nwith a longitudinal distribution amplitude and a transverse part. We know that the transverse shape should be quite narrow, e.g. given by a Gaussian with a width . Considering hadrons at mid rapidity, will mainly be pointing in the longitudinal and azimuthal directions in the CM frame, where the variation of the parton distributions is small. This implies for typical transverse momenta . We can then integrate the dependence of the wave function. This leaves us with\n\n EdNMd3P=CM∫Σd3RP⋅u(R)(2π)31∫0dxwa(R;xP+)|ϕM(x)|2wb(R;(1−x)P+). (44)\n\nThe amplitude encodes the remaining QCD dynamics. We expect it to be peaked around , meaning that the two quarks will carry roughly the same amount of momentum. But the width of the distribution, since it is formulated in terms of momentum fractions, could be quite broad in momentum space. Thus we cannot use the same argument as for the transverse coordinates in order to integrate out this degree of freedom. However, for an exponential parton spectrum we have\n\n wa(R;xP+)wb(R;(1−x)P+)∼e−xP+/Te−(1−x)P+/T=e−P+/T. (45)\n\nHence the product of parton distributions is independent of and we can perform the integral over , which just gives the trivial normalization of the wave function from (42) FMNB:03 . There will be corrections to that from momentum components other than which are not additive because energy is not conserved (see next subsection). Where we want to take into account wave functions, we adopt the asymptotic form of the perturbative pion distribution amplitude\n\n ϕM(x)=√30x(1−x) (46)\n\nas a model.\n\n### iii.3 Energy conservation\n\nEnergy conservation is not manifest in the recombination approach. Since we are dealing with a 21 process, one of the particles in general needs to be off mass shell ScheiHei:99 . This does not pose a problem in the physical environment where recombination takes place. Both in the quark phase and in the hadronic phase we expect interactions with the surrounding medium to occur. Since recombination, as described in this work, has been simplified to a counting of quantum numbers and momenta, without real QCD dynamics, we neglect effects of these additional interactions that ensure energy conservation. This is tolerable due to the small time scale of hadronization and that changes in the parton distributions by interactions during this time are negligible. We also neglect effects that final state interactions between the hadrons could have on the spectra – these are expected to contribute mainly in the low momentum domain,  GeV.\n\nFor large momenta , the energies of the particles are dominated by the kinetic energies and not by the masses. The light cone wave functions have a transverse width which is a non-perturbative momentum scale. Therefore the momenta of the recombining quarks will be collinear up to transverse momenta of order . This implies that the energy is conserved up to terms of order and , where stands for the masses of the participating particles.\n\nEnergy conservation is a problem at low transverse momentum and can only be overcome if one takes further interactions between the partons into account. This would require a much more sophisticated formulation that includes nonperturbative initial- and final-state effects.\n\nRestricting ourselves to large transverse momentum finally permits to use light cone fractions for the spatial momentum in the CM frame instead of the momentum in the HLC frame. The relation between the quark momentum and the hadron momentum in light cone coordinates translates to\n\n p=xP+O(xM2|P|)+O(Λ⊥). (47)\n\nThis is a leading order expansion in where is the hadron momentum. At mid rapidity this further translates to the simple formula for the transverse momenta.\n\n### iii.4 Low transverse momentum and hadron thermodynamics\n\nIt is well known that total hadron yields at RHIC can be described very accurately by a purely statistical model, using only hadronic properties, such as masses, chemical potentials and spin degeneracies BroFlo:01 ; PBmMaReSt ; XuKa . The hadron yields are given by the -integrated spectra and are naturally dominated by particles with less than 2 GeV/ transverse momentum. The picture of thermal hadron production does not necessarily require the existence of a parton phase.\n\nHowever, one can show that recombination of a thermalized parton phase is consistent with thermal hadron production in the limit . Although pQCD will eventually dominate over both mechanisms at large , this nevertheless suggests that the recombination mechanism connects a thermal parton phase with the observed thermal hadron phase. Therefore it is justified to call recombination from a thermal parton phase the microscopic manifestation of statistical hadron production.\n\n### iii.5 Summary of the formalism\n\nWe want to summarize what we have so far. From (44) the meson spectrum is given by\n\n ENMd3P=CM∫ΣdσRP⋅u(R)(2π)31∫0dxwa(R;xP)|ϕM(x)|2wb(R;(1−x)P). (48)\n\nwhere measures the volume of the hypersurface and . For baryons, the same steps result in the expression\n\n ENBd3P=CB∫ΣdσRP⋅u(R)(2π)3∫Dxiwa(R;x1P)wb(R;x2P)wc(R;x3P)|ϕB(x1,x2,x3)|2. (49)\n\n, and are the valence partons and is the effective wave function of the baryon in light cone coordinates. We use the short notation\n\n ∫Dxi=1∫0dx1dx2dx3δ(x1+x2+x3−1) (50)\n\nfor the integration over three light cone fractions. Inspired by the asymptotic form of the light cone distribution amplitudes for pions and nucleons we choose\n\n ϕM(x) =√30x(1−x), (51) ϕB(x1,x2,x3) =12√35x1x2x3. (52)\n\nThese wave functions are broad in momentum space. In order to study the effect of the width of the wave functions on our results it will be interesting to alternatively explore the case of narrow wave functions in the spirit of Sec.  II.4. The limiting case are -shaped wave functions\n\n |ϕM(x)|2 =δ(x−12) (53) |ϕB(x1,x2,x3)|2 =δ(x1−13)δ(x2−13). (54)\n\nThe spectra are then given by\n\n ENMd3P=CM∫ΣdσRP⋅u(R)(2π)3wa(R;P2)wb(R;P2) (55)\n ENBd3P=CB∫ΣdσRP⋅u(R)(2π)3×wa(R;P3)wb(R;P3)wc(R;P3). (56)\n\nIt is an important observation that in the case of exponential parton distributions, the shape of the wave function is almost negligible. We have to be aware that there will be corrections to the above equations of order and , where is the mass of the hadron or the partons, reducing their range of applicability to large .\n\nOn the other hand the spectrum of hadrons from fragmentation is given by (1)\n\nThe number of partons can be obtained from the cross section via the impact parameter dependent nuclear thickness function . We take for central collisions. is the radius of the nucleus.\n\n## Iv Hadron and parton spectra\n\nIn the following we want to discuss the parton phase created in collisions of gold nuclei at RHIC with GeV per nucleon pair. We will then proceed to calculate the hadron spectra emerging from recombination and fragmentation of this parton phase.\n\n### iv.1 Modeling the parton phase\n\nAssuming longitudinal boost invariance, we fix the hypersurface by choosing for\n\n Rμ=(t,x,y,z)=(τcoshη,ρcosϕ,ρsinϕ,τsinhη). (58)\n\nIt is convenient to introduce the space time rapidity and the radial coordinate , since the measure for the hypersurface then takes the simple form and the normal vector is given by\n\n uμ(R)=(coshη,0,0,sinhη). (59)\n\nUsing a similar parametrization of the parton momentum\n\n pμ=(mTcoshy,pTcosΦ,pTsinΦ,mTsinhy) (60)\n\nwith rapidity and transverse mass , we obtain . We remind the reader that we will use capitalized variables (, etc.) for hadrons.\n\nNow we have to specify the spectrum of partons. As already discussed above we assume that the parton spectrum consists of two domains. At large , the distribution of partons is given by perturbative QCD and follows a power law. For the transverse momentum distribution at midrapidity for central collisions (b=0) we use the parametrization FMS:03 ; SGF:03\n\nThe parameters , , and are taken from a leading order (LO) pQCD calculation and can be found in SGF:03 for the three light quark flavors and gluons. A constant factor of 1.5 is included to roughly account for higher order corrections in EskTuo:01 . The calculation includes nuclear shadowing of the parton distributions, but no higher twist initial state effects. Higher twist effects, like the Cronin effect, will fade like for high transverse momentum LQS:92 . Since we will show that fragmentation and pQCD are only dominant for transverse momenta above 5 GeV/ in the hadron spectrum for RHIC, it is safe to omit the Cronin effect.\n\nEnergy loss of partons, resulting in a shift of the transverse momentum spectrum BDMS:01 ; Muller:02 , is taken into account and parameterized as\n\n ΔpT(b,pT)=ϵ(b)√pT⟨L⟩RA. (62)\n\nFor central collisions we take and therefore (we postpone the discussion of the impact parameter dependence to Subsection IV.C). The choice of neglects the fact that for the strong quenching observed at RHIC energies, jet emission becomes a surface effect Muller:02 ; Shuryak:02 . However, we note that only the product as a whole is a parameter. We have no ambition here to make a connection to the microscopic parameters of jet quenching, therefore we will not disentangle and . This would require a more sophisticated model of the emission geometry. We also do not use a radial profile for the emission and the density of the medium. This can be found discussed elsewhere in the literature Muller:02 ; WaWa:02 ; Wang:03 . Nevertheless, our “minimal” description of energy loss is quite successful to describe the available data on high- spectra ( GeV/) for , and charged hadrons in central Au+Au collisions. From a fit to these data we find for central Au+Au collisions. This value corresponds to an average energy loss of 3 GeV for a 10 GeV parton in a central Au+Au collision. The perturbative spectrum for up and strange quarks at midrapidity is shown in Fig. 1. For fragmentation of pions, kaons, protons and antiprotons we use LO KKP fragmentation functions with the scale set to the hadron transverse momentum KKP:00 . fragmentation is calculated with the LO fragmentation functions of de Florian, Stratmann and Vogelsang dFSV:97 .", null, "Figure 1: Spectrum of u and s quarks at hadronization in a central Au+Au collision at RHIC. Perturbative partons from hard QCD processes with subsequent energy loss (dashed lines) and the thermal phase with T=175 MeV and radial flow vT=0.55c (solid lines) are shown.\n\nBesides the perturbative tail of the parton spectrum that will turn into hadrons via fragmentation, we assume the existence of a spectrum of thermalized partons that are recombining at hadronization and dominate at low and intermediate values of . In this phase we assume the effective degrees of freedom to be constituent quarks without dynamical gluons. We take the spectrum to be exponential with a given temperature\n\n wa(R;p)=γae−p⋅v(R)/Te−η2/2Δ2f(ρ,ϕ). (63)\n\nis a fugacity factor for each parton species . We also include longitudinal and radial flow through the velocity vector\n\n vμ(R)=(coshηLcoshηT,sinhηTcosϕ,sinhηTsinϕ,sinhηLcoshηT). (64)\n\nand are the rapidities of the longitudinal and radial flow which still could depend on the space time point . For the longitudinal expansion we choose a Bjorken scenario where the longitudinal rapidity is simply fixed by the space time rapidity\n\n ηL(R)=η. (65)\n\nThe transverse flow is given by a velocity with . For practical purposes we will not work with a radial profile but assume to be independent of and . However, for collisions with finite impact parameter we will later allow a dependence of on the azimuthal angle in order to describe the measured elliptic asymmetry in the spectra. The space-time structure of the parton source in (63) is given by a transverse distribution and a wide Gaussian rapidity distribution with a width .\n\nWe assume that hadronization occurs at fm at a temperature MeV in the parton phase. This is consistent with predictions of the phase transition temperature at vanishing baryon chemical potential from lattice QCD Karsch:01 . For the spread of the parton distribution in longitudinal direction we choose . The constituent quark masses are taken to be 260 MeV for and quarks and 460 MeV for quarks.\n\nThe two component model of the parton spectrum with an exponential bulk and a power law tail is also predicted by parton cascades like VNI/BMS BMS:02 , although the interactions in that case are purely perturbative. This implies that an exponential shape of the spectrum does not necessarily mean that the parton system is in thermal equilibrium.\n\nIn the region where contributions from recombination and fragmentation are of the same size we expect other mechanisms to play a role, which interpolate between the two pictures. This could include partial recombination and higher twist fragmentation. In the absence of a consistent description of these mechanisms we simply add both contributions to the hadron spectrum – recombination from the exponential part and the fragmentation from the pQCD part – for GeV/.\n\n### iv.2 Degeneracy factors\n\nIt is not a priori clear from QCD what the degeneracy factors for each hadron are. In principle every quark has 3 color and 2 spin degrees or freedom. On could argue that 3 quarks of any color and spin can form a proton and that quantum numbers can be ”corrected” at no cost by the emission of soft gluons. That would lead to degeneracy factors and . On the other hand there are no dynamical gluons in our picture and it would be consistent to require recombining partons to have the right quantum numbers at the beginning.\n\nSurprisingly this is supported from work on recombination in pQCD, where the contributions from color octets and spin-flip states to the recombination of mesons were found to be small BraJiaMe:02 . Using this assumption, the degeneracies are only determined by the degrees of freedom of the hadron, e.g. , etc. These are exactly the degeneracies used in the statistical thermal model. We will not take into account feeddown from decays of resonances, except for the , where the is too close in mass to be suppressed. Hence we use .\n\nThis is different from FMNB:03 where we counted resonances to give nucleons with weight 1. However this overestimates the correction from decays. Nevertheless the degeneracy of given in FMNB:03 for the proton was of the right size due to a mistake in the normalization of the baryon states, that gave an additional factor of . Therefore the numerical results given in FMNB:03 are still valid.\n\nWe should add that due to the small but probably non-vanishing color octet and spin flip contributions and due to feeddown corrections we expect all degeneracy factors to have an error of at least 20%.\n\n### iv.3 Central collisions\n\nFor the momentum spectrum of quarks" ]

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[ "# NAG FL InterfaceF01 (Matop)Matrix Operations, Including Inversion\n\n## ▸▿ Contents\n\nSettings help\n\nFL Name Style:\n\nFL Specification Language:\n\n## 1Scope of the Chapter\n\nThis chapter provides facilities for four types of problem:\n1. (i)matrix inversion;\n2. (ii)matrix factorizations;\n3. (iii)matrix arithmetic and manipulation;\n4. (iv)matrix functions.\nSee Sections 2.1, 2.2, 2.3 and 2.4 where these problems are discussed.\n\n## 2Background to the Problems\n\n### 2.1Matrix Inversion\n\n1. (i)Nonsingular square matrices of order $n$.\nIf $A$, a square matrix of order $n$, is nonsingular (has rank $n$), then its inverse $X$ exists and satisfies the equations $AX=XA=I$ (the identity or unit matrix).\nIt is worth noting that if $AX-I=R$, so that $R$ is the ‘residual’ matrix, then a bound on the relative error is given by $‖R‖$, i.e.,\n $‖X-A-1‖ ‖A-1‖ ≤‖R‖.$\n2. (ii)General real rectangular matrices.\nA real matrix $A$ has no inverse if it is square ($n×n$) and singular (has rank $\\text{}), or if it is of shape ($m×n$) with $m\\ne n$, but there is a Generalized or Pseudo-inverse ${A}^{+}$ which satisfies the equations\n $AA+A=A, A+AA+=A+, (AA+)T=AA+, (A+A)T=A+A$\n(which of course are also satisfied by the inverse $X$ of $A$ if $A$ is square and nonsingular).\n1. (a)if $m\\ge n$ and $\\mathrm{rank}\\left(A\\right)=n$ then $A$ can be factorized using a $\\mathbit{Q}\\mathbit{R}$ factorization, given by\n $A=Q ( R 0 ) ,$\nwhere $Q$ is an $m×m$ orthogonal matrix and $R$ is an $n×n$, nonsingular, upper triangular matrix. The pseudo-inverse of $A$ is then given by\n $A+=R-1Q~T,$\nwhere $\\stackrel{~}{Q}$ consists of the first $n$ columns of $Q$.\n2. (b)if $m\\le n$ and $\\mathrm{rank}\\left(A\\right)=m$ then $A$ can be factorized using an RQ factorization, given by\n $A=(R 0)QT$\nwhere $Q$ is an $n×n$ orthogonal matrix and $R$ is an $m×m$, nonsingular, upper triangular matrix. The pseudo-inverse of $A$ is then given by\n $A+ = Q~R-1 ,$\nwhere $\\stackrel{~}{Q}$ consists of the first $m$ columns of $Q$.\n3. (c)if $m\\ge n$ and $\\mathrm{rank}\\left(A\\right)=r\\le n$ then $A$ can be factorized using a $QR$ factorization, with column interchanges, as\n $A=Q ( R 0 ) PT,$\nwhere $Q$ is an $m×m$ orthogonal matrix, $R$ is an $r×n$ upper trapezoidal matrix and $P$ is an $n×n$ permutation matrix. The pseudo-inverse of $A$ is then given by\n $A+=PRT(RRT)−1Q~T,$\nwhere $\\stackrel{~}{Q}$ consists of the first $r$ columns of $Q$.\n4. (d)if $\\mathrm{rank}\\left(A\\right)=r\\le k=\\mathrm{min}\\phantom{\\rule{0.125em}{0ex}}\\left(m,n\\right)$ then $A$ can be factorized as the singular value decomposition\n $A=UΣVT,$\nwhere $U$ is an $m×m$ orthogonal matrix, $V$ is an $n×n$ orthogonal matrix and $\\Sigma$ is an $m×n$ diagonal matrix with non-negative diagonal elements $\\sigma$. The first $k$ columns of $U$ and $V$ are the left- and right-hand singular vectors of $A$ respectively and the $k$ diagonal elements of $\\Sigma$ are the singular values of $A$. $\\Sigma$ may be chosen so that\n $σ1≥σ2≥⋯≥σk≥0$\nand in this case, if $\\mathrm{rank}\\left(A\\right)=r$ then\n $σ1≥σ2≥⋯≥σr>0, σr+1=⋯=σk=0.$\nIf $\\stackrel{~}{U}$ and $\\stackrel{~}{V}$ consist of the first $r$ columns of $U$ and $V$ respectively and $\\stackrel{~}{\\Sigma }$ is an $r×r$ diagonal matrix with diagonal elements ${\\sigma }_{1},{\\sigma }_{2},\\dots ,{\\sigma }_{r}$ then $A$ is given by\n $A=U~Σ~V~T$\nand the pseudo-inverse of $A$ is given by\n $A+=V~Σ~-1U~T.$\nNotice that\n $ATA=V(ΣTΣ)VT$\nwhich is the classical eigenvalue (spectral) factorization of ${A}^{\\mathrm{T}}A$.\n5. (e)if $A$ is complex then the above relationships are still true if we use ‘unitary’ in place of ‘orthogonal’ and conjugate transpose in place of transpose. For example, the singular value decomposition of $A$ is\n $A=UΣVH,$\nwhere $U$ and $V$ are unitary, ${V}^{\\mathrm{H}}$ the conjugate transpose of $V$ and $\\Sigma$ is as in (ii)(d) above.\n\n### 2.2Matrix Factorizations\n\nThe routines in this section perform matrix factorizations which are required for the solution of systems of linear equations with various special structures. A few routines which perform associated computations are also included.\nOther routines for matrix factorizations are to be found in Chapters F07, F08 and F11.\nThis section also contains a few routines associated with eigenvalue problems (see Chapter F02). (Historical note: this section used to contain many more such routines, but they have now been superseded by routines in Chapter F08.)\nFinally, this section contains routines for computing non-negative matrix factorizations, which are used for dimensional reduction and classification in data analysis. Given a rectangular $m×n$ matrix, $A$, with non-negative elements, a non-negative matrix factorization of $A$ is an approximate factorization of $A$ into the product of an $m×k$ non-negative matrix $W$ and a $k×n$ non-negative matrix $H$, so that $A\\approx WH$. Typically $k$ is chosen so that $k\\ll \\mathrm{min}\\phantom{\\rule{0.125em}{0ex}}\\left(m,n\\right)$. The matrices $W$ and $H$ are then computed to minimize ${|A-WH|}_{F}$. The factorization is not unique.\n\n### 2.3Matrix Arithmetic and Manipulation\n\nThe intention of routines in this section (f01c, f01d, f01v, and f01z) is to cater for some of the commonly occurring operations in matrix manipulation, i.e., transposing a matrix or adding part of one matrix to another, and for conversion between different storage formats,such as conversion between rectangular band matrix storage and packed band matrix storage. For vector or matrix-vector or matrix-matrix operations refer to Chapters F06 and F16.\n\n### 2.4Matrix Functions\n\nGiven a square matrix $A$, the matrix function $f\\left(A\\right)$ is a matrix with the same dimensions as $A$ which provides a generalization of the scalar function $f$.\nIf $A$ has a full set of eigenvectors $V$ then $A$ can be factorized as\n $A = V D V-1 ,$\nwhere $D$ is the diagonal matrix whose diagonal elements, ${d}_{i}$, are the eigenvalues of $A$. $f\\left(A\\right)$ is given by\n $f(A) = V f(D) V-1 ,$\nwhere $f\\left(D\\right)$ is the diagonal matrix whose $i$th diagonal element is $f\\left({d}_{i}\\right)$.\nIn general, $A$ may not have a full set of eigenvectors. The matrix function can then be defined via a Cauchy integral. For $A\\in {ℂ}^{n×n}$,\n $f(A) = 1 2π i ∫ Γ f(z) (zI-A)-1 dz ,$\nwhere $\\Gamma$ is a closed contour surrounding the eigenvalues of $A$, and $f$ is analytic within $\\Gamma$.\nSome matrix functions are defined implicitly. A matrix logarithm is a solution $X$ to the equation\n $eX=A .$\nIn general, $X$ is not unique, but if $A$ has no eigenvalues on the closed negative real line then a unique principal logarithm exists whose eigenvalues have imaginary part between $\\pi$ and $-\\pi$. Similarly, a matrix square root is a solution $X$ to the equation\n $X2=A .$\nIf $A$ has no eigenvalues on the closed negative real line then a unique principal square root exists with eigenvalues in the right half-plane. If $A$ has a vanishing eigenvalue then $\\mathrm{log}\\left(A\\right)$ cannot be computed. If the vanishing eigenvalue is defective (its algebraic multiplicity exceeds its geometric multiplicity, or equivalently it occurs in a Jordan block of size greater than $1$) then the square root cannot be computed. If the vanishing eigenvalue is semisimple (its algebraic and geometric multiplicities are equal, or equivalently it occurs only in Jordan blocks of size $1$) then a square root can be computed.\nAlgorithms for computing matrix functions are usually tailored to a specific function. Currently, Chapter F01 contains routines for calculating the exponential, logarithm, sine, cosine, sinh, cosh, square root and the general real power of both real and complex matrices. In addition, there are routines to compute a general function of real symmetric and complex Hermitian matrices and a general function of general real and complex matrices.\nThe Fréchet derivative of a matrix function $f\\left(A\\right)$ in the direction of the matrix $E$ is the linear function mapping $E$ to ${L}_{f}\\left(A,E\\right)$ such that\n $f(A+E) - f(A) - Lf(A,E) = O(‖E‖) .$\nThe Fréchet derivative measures the first-order effect on $f\\left(A\\right)$ of perturbations in $A$. Chapter F01 contains routines for calculating the Fréchet derivative of the exponential, logarithm and real powers of both real and complex matrices.\nThe condition number of a matrix function is a measure of its sensitivity to perturbations in the data. The absolute condition number measures these perturbations in an absolute sense and is defined by\n $condabs (f,A) ≔ lim ε→0 sup {‖E‖→0} ‖f(A+E)-f(A)‖ ε .$\nThe relative condition number, which is usually of more interest, measures these perturbations in a relative sense and is defined by\n $condrel (f,A) = condabs (f,A) ‖A‖ ‖f(A)‖ .$\nThe absolute and relative condition numbers can be expressed in terms of the norm of the Fréchet derivative by\n $condabs (f,A) = max E≠0 ‖L(A,E)‖ ‖E‖ ,$\n $condrel (f,A) = ‖A‖ ‖f(A)‖ max E≠0 ‖L(A,E)‖ ‖E‖ .$\nChapter F01 contains routines for calculating the condition number of the matrix exponential, logarithm, sine, cosine, sinh, cosh, square root and the general real power of both real and complex matrices. It also contains routines for estimating the condition number of a general function of a real or complex matrix.\n\n## 3Recommendations on Choice and Use of Available Routines\n\n### 3.1Matrix Inversion\n\nNote:  before using any routine for matrix inversion, consider carefully whether it is needed.\nAlthough the solution of a set of linear equations $Ax=b$ can be written as $x={A}^{-1}b$, the solution should never be computed by first inverting $A$ and then computing ${A}^{-1}b$; the routines in Chapters F04 or F07 should always be used to solve such sets of equations directly; they are faster in execution, and numerically more stable and accurate. Similar remarks apply to the solution of least squares problems which again should be solved by using the routines in Chapters F04 and F08 rather than by computing a pseudo-inverse.\n1. (a)Nonsingular square matrices of order $n$\nThis chapter describes techniques for inverting a general real matrix $A$ and matrices which are positive definite (have all eigenvalues positive) and are either real and symmetric or complex and Hermitian. It is wasteful and uneconomical not to use the appropriate routine when a matrix is known to have one of these special forms. A general routine must be used when the matrix is not known to be positive definite. In most routines, the inverse is computed by solving the linear equations $A{x}_{\\mathit{i}}={e}_{\\mathit{i}}$, for $\\mathit{i}=1,2,\\dots ,n$, where ${e}_{i}$ is the $i$th column of the identity matrix.\nRoutines are given for calculating the approximate inverse, that is solving the linear equations just once, and also for obtaining the accurate inverse by successive iterative corrections of this first approximation. The latter, of course, are more costly in terms of time and storage, since each correction involves the solution of $n$ sets of linear equations and since the original $A$ and its $LU$ decomposition must be stored together with the first and successively corrected approximations to the inverse. In practice, the storage requirements for the ‘corrected’ inverse routines are about double those of the ‘approximate’ inverse routines, though the extra computer time is not prohibitive since the same matrix and the same $LU$ decomposition is used in every linear equation solution.\nDespite the extra work of the ‘corrected’ inverse routines, they are superior to the ‘approximate’ inverse routines. A correction provides a means of estimating the number of accurate figures in the inverse or the number of ‘meaningful’ figures relating to the degree of uncertainty in the coefficients of the matrix.\nThe residual matrix $R=AX-I$, where $X$ is a computed inverse of $A$, conveys useful information. Firstly $‖R‖$ is a bound on the relative error in $X$ and secondly $‖R‖<\\frac{1}{2}$ guarantees the convergence of the iterative process in the ‘corrected’ inverse routines.\nThe decision trees for inversion show which routines in Chapter F04 and Chapter F07 should be used for the inversion of other special types of matrices not treated in the chapter.\n2. (b)General real rectangular matrices\nFor real matrices f08aef and f01qjf return $QR$ and $RQ$ factorizations of $A$ respectively and f08bff returns the $QR$ factorization with column interchanges. The corresponding complex routines are f08asf, f01rjf and f08btf respectively. Routines are also provided to form the orthogonal matrices and transform by the orthogonal matrices following the use of the above routines. f01qgf and f01rgf form the $RQ$ factorization of an upper trapezoidal matrix for the real and complex cases respectively.\nf01blf uses the $QR$ factorization as described in Section 2.1(ii)(a) and is the only routine that explicitly returns a pseudo-inverse. If $m\\ge n$ then the routine will calculate the pseudo-inverse ${A}^{+}$ of the matrix $A$. If $m then the $n×m$ matrix ${A}^{\\mathrm{T}}$ should be used. The routine will calculate the pseudo-inverse $Z={\\left({A}^{\\mathrm{T}}\\right)}^{+}={\\left({A}^{+}\\right)}^{\\mathrm{T}}$ of ${A}^{\\mathrm{T}}$ and the required pseudo-inverse will be ${Z}^{\\mathrm{T}}$. The routine also attempts to calculate the rank, $r$, of the matrix given a tolerance to decide when elements can be regarded as zero. However, should this routine fail due to an incorrect determination of the rank, the singular value decomposition method (described below) should be used.\nf08kbf and f08kpf compute the singular value decomposition as described in Section 2 for real and complex matrices respectively. If $A$ has rank $r\\le k=\\mathrm{min}\\phantom{\\rule{0.125em}{0ex}}\\left(m,n\\right)$ then the $k-r$ smallest singular values will be negligible and the pseudo-inverse of $A$ can be obtained as ${A}^{+}=V{\\Sigma }^{-1}{U}^{\\mathrm{T}}$ as described in Section 2. If the rank of $A$ is not known in advance it can be estimated from the singular values (see Section 2.4 in the F04 Chapter Introduction). In the real case with $m\\ge n$, f08aef followed by f02wuf provide details of the $QR$ factorization or the singular value decomposition depending on whether or not $A$ is of full rank and for some problems provides an attractive alternative to f08kbf. For large sparse matrices, leading terms in the singular value decomposition can be computed using routines from Chapter F12.\n\n### 3.2Matrix Factorizations\n\nMost of these routines serve a special purpose required for the solution of sets of simultaneous linear equations or the eigenvalue problem. For further details, you should consult Sections 3 or 4 in the F02 Chapter Introduction or Sections 3 or 4 in the F04 Chapter Introduction.\nf01brf and f01bsf are provided for factorizing general real sparse matrices. A more recent algorithm for the same problem is available through f11mef. For factorizing real symmetric positive definite sparse matrices, see f11jaf. These routines should only be used when $A$ is not banded and when the total number of nonzero elements is less than 10% of the total number of elements. In all other cases, either the band routines or the general routines should be used.\nf01mdf and f01mef compute the Cheng–Higham modified Cholesky factorization of a real symmetric matrix and the positive definite perturbed input matrix from the factors.\nThe routines f01saf (for dense matrices) and f01sbf (sparse matrices, using a reverse communication interface) are provided for computing non-negative matrix factorizations.\n\n### 3.3Matrix Arithmetic and Manipulation\n\nThe routines in the f01c section are designed for the general handling of $m×n$ matrices. Emphasis has been placed on flexibility in the argument specifications and on avoiding, where possible, the use of internally declared arrays. They are, therefore, suited for use with large matrices of variable row and column dimensions. Routines are included for the addition and subtraction of sub-matrices of larger matrices, as well as the standard manipulations of full matrices. Those routines involving matrix multiplication may use additional-precision arithmetic for the accumulation of inner products. See also Chapter F06.\nThe routines in the f01d section perform arithmetic operations on triangular matrices.\nThe routines in the f01v (LAPACK) and f01z section are designed to allow conversion between full storage format and one of the packed storage schemes required by some of the routines in Chapters F02, F04, F06, F07 and F08.\n\n#### 3.3.1NAG Names and LAPACK Names\n\nRoutines with NAG name beginning f01v may be called either by their NAG names or by their LAPACK names. When using the NAG Library, the double precision form of the LAPACK name must be used (beginning with D- or Z-).\nReferences to Chapter F01 routines in the manual normally include the LAPACK double precision names, for example, f01vef.\nThe LAPACK routine names follow a simple scheme (which is similar to that used for the BLAS in Chapter F06). Most names have the structure XYYTZZ, where the components have the following meanings:\n– the initial letter, X, indicates the data type (real or complex) and precision:\n• S – real, single precision (in Fortran, 4 byte length REAL)\n• D – real, double precision (in Fortran, 8 byte length REAL)\n• C – complex, single precision (in Fortran, 8 byte length COMPLEX)\n• Z – complex, double precision (in Fortran, 16 byte length COMPLEX)\n– the fourth letter, T, indicates that the routine is performing a storage scheme transformation (conversion)\n– the letters YY indicate the original storage scheme used to store a triangular part of the matrix $A$, while the letters ZZ indicate the target storage scheme of the conversion (YY cannot equal ZZ since this would do nothing):\n• TF – Rectangular Full Packed Format (RFP)\n• TP – Packed Format\n• TR – Full Format\n\n### 3.4Matrix Functions\n\nf01ecf and f01fcf compute the matrix exponential, ${e}^{A}$, of a real and complex square matrix $A$ respectively. If estimates of the condition number of the matrix exponential are required then f01jgf and f01kgf should be used. If Fréchet derivatives are required then f01jhf and f01khf should be used.\nf01edf and f01fdf compute the matrix exponential, ${e}^{A}$, of a real symmetric and complex Hermitian matrix respectively. If the matrix is real symmetric, or complex Hermitian then it is recommended that f01edf, or f01fdf be used as they are more efficient and, in general, more accurate than f01ecf and f01fcf.\nf01ejf and f01fjf compute the principal matrix logarithm, $\\mathrm{log}\\left(A\\right)$, of a real and complex square matrix $A$ respectively. If estimates of the condition number of the matrix logarithm are required then f01jjf and f01kjf should be used. If Fréchet derivatives are required then f01jkf and f01kkf should be used.\nf01ekf and f01fkf compute the matrix exponential, sine, cosine, sinh or cosh of a real and complex square matrix $A$ respectively. If the matrix exponential is required then it is recommended that f01ecf or f01fcf be used as they are, in general, more accurate than f01ekf and f01fkf. If estimates of the condition number of the matrix function are required then f01jaf and f01kaf should be used.\nf01elf and f01emf compute the matrix function, $f\\left(A\\right)$, of a real square matrix. f01flf and f01fmf compute the matrix function of a complex square matrix. The derivatives of $f$ are required for these computations. f01elf and f01flf use numerical differentiation to obtain the derivatives of $f$. f01emf and f01fmf use derivatives you have supplied. If estimates of the condition number are required but you are not supplying derivatives then f01jbf and f01kbf should be used. If estimates of the condition number of the matrix function are required and you are supplying derivatives of $f$ then f01jcf and f01kcf should be used.\nIf the matrix $A$ is real symmetric or complex Hermitian then it is recommended that to compute the matrix function, $f\\left(A\\right)$, f01eff and f01fff are used respectively as they are more efficient and, in general, more accurate than f01elf, f01emf, f01flf and f01fmf.\nf01gaf and f01haf compute the matrix function ${e}^{tA}B$ for explicitly stored dense real and complex matrices $A$ and $B$ respectively while f01gbf and f01hbf compute the same using reverse communication. In the latter case, control is returned to you. You should calculate any required matrix-matrix products and then call the routine again. See Section 7 in How to Use the NAG Library for further information.\nf01enf and f01fnf compute the principal square root ${A}^{1/2}$ of a real and complex square matrix $A$ respectively. If $A$ is complex and upper triangular then f01fpf should be used. If $A$ is real and upper quasi-triangular then f01epf should be used. If estimates of the condition number of the matrix square root are required then f01jdf and f01kdf should be used.\nf01eqf and f01fqf compute the matrix power ${A}^{p}$, where $p\\in ℝ$, of real and complex matrices respectively. If estimates of the condition number of the matrix power are required then f01jef and f01kef should be used. If Fréchet derivatives are required then f01jff and f01kff should be used.\n\n## 4Decision Trees\n\nThe decision trees show the routines in this chapter and in Chapter F04, Chapter F07 and Chapter F08 that should be used for inverting matrices of various types. They also show which routine should be used to calculate various matrix functions.\n(i) Matrix Inversion:\n\n### Tree 1\n\n Is $A$ an $n×n$ matrix of rank $n$? Is $A$ a real matrix? see Tree 2 yes yes no no see Tree 3 see Tree 4\n\n### Tree 2: Inverse of a real n by n matrix of full rank\n\n Is $A$ a band matrix? See Note 1. yes no Is $A$ symmetric? Is $A$ positive definite? Do you want guaranteed accuracy? (See Note 2) f01abf yes yes yes no no no Is one triangle of $A$ stored as a linear array? f07gdf and f07gjf yes no f01adf or f07fdf and f07fjf Is one triangle of $A$ stored as a linear array? f07pdf and f07pjf yes no f07mdf and f07mjf Is $A$ triangular? Is $A$ stored as a linear array? f07ujf yes yes no no f07tjf Do you want guaranteed accuracy? (See Note 2) f07abf yes no f07adf and f07ajf\n\n### Tree 3: Inverse of a complex n by n matrix of full rank\n\n Is $A$ a band matrix? See Note 1. yes no Is $A$ Hermitian? Is $A$ positive definite? Is one triangle of $A$ stored as a linear array? f07grf and f07gwf yes yes yes no no no f07frf and f07fwf Is one triangle $A$ stored as a linear array? f07prf and f07pwf yes no f07mrf and f07mwf Is $A$ symmetric? Is one triangle of $A$ stored as a linear array? f07qrf and f07qwf yes yes no no f07nrf and f07nwf Is $A$ triangular? Is $A$ stored as a linear array? f07uwf yes yes no no f07twf f07anf or f07arf and f07awf\n\n### Tree 4: Pseudo-inverses\n\n Is $A$ a complex matrix? Is $A$ of full rank? Is $A$ an $m×n$ matrix with $m? f01rjf and f01rkf yes yes yes no no no f08asf and f08auf or f08atf f08kpf Is $A$ of full rank? Is $A$ an $m×n$ matrix with $m? f01qjf and f01qkf yes yes no no f08aef and f08agf or f08aff Is $A$ an $m×n$ matrix with $m? f08kbf yes no Is reliability more important than efficiency? f08kbf yes no f01blf\nNote 1: the inverse of a band matrix $A$ does not, in general, have the same shape as $A$, and no routines are provided specifically for finding such an inverse. The matrix must either be treated as a full matrix or the equations $AX=B$ must be solved, where $B$ has been initialized to the identity matrix $I$. In the latter case, see the decision trees in Section 4 in the F04 Chapter Introduction.\nNote 2: by ‘guaranteed accuracy’ we mean that the accuracy of the inverse is improved by the use of the iterative refinement technique using additional precision.\n(ii) Matrix Factorizations: see the decision trees in Section 4 in the F02 and F04 Chapter Introductions.\n(iii) Matrix Arithmetic and Manipulation: not appropriate.\n(iv) Matrix Functions:\n\n### Tree 5: Matrix functions $f\\left(A\\right)$ of an n by n real matrix $A$\n\n Is ${e}^{tA}B$ required? Is $A$ stored in dense format? f01gaf yes yes no no f01gbf Is $A$ real symmetric? Is ${e}^{A}$ required? f01edf yes yes no no f01eff Is $\\mathrm{cos}\\left(A\\right)$ or $\\mathrm{cosh}\\left(A\\right)$ or $\\mathrm{sin}\\left(A\\right)$ or $\\mathrm{sinh}\\left(A\\right)$ required? Is the condition number of the matrix function required? f01jaf yes yes no no f01ekf Is $\\mathrm{log}\\left(A\\right)$ required? Is the condition number of the matrix logarithm required? f01jjf yes yes no no Is the Fréchet derivative of the matrix logarithm required? f01jkf yes no f01ejf Is $\\mathrm{exp}\\left(A\\right)$ required? Is the condition number of the matrix exponential required? f01jgf yes yes no no Is the Fréchet derivative of the matrix exponential required? f01jhf yes no f01ecf Is ${A}^{1/2}$ required? Is the condition number of the matrix square root required? f01jdf yes yes no no Is the matrix upper quasi-triangular? f01epf yes no f01enf Is ${A}^{p}$ required? Is the condition number of the matrix power required? f01jef yes yes no no Is the Fréchet derivative of the matrix power required? f01jff yes no f01eqf $f\\left(A\\right)$ will be computed. Will derivatives of $f$ be supplied by the user? Is the condition number of the matrix function required? f01jcf yes yes no no f01emf Is the condition number of the matrix function required? f01jbf yes no f01elf\n\n### Tree 6: Matrix functions $f\\left(A\\right)$ of an n by n complex matrix $A$\n\n Is ${e}^{tA}B$ required? Is $A$ stored in dense format? f01haf yes yes no no f01hbf Is $A$ complex Hermitian? Is ${e}^{A}$ required? f01fdf yes yes no no f01fff Is $\\mathrm{cos}\\left(A\\right)$ or $\\mathrm{cosh}\\left(A\\right)$ or $\\mathrm{sin}\\left(A\\right)$ or $\\mathrm{sinh}\\left(A\\right)$ required? Is the condition number of the matrix function required? f01kaf yes yes no no f01fkf Is $\\mathrm{log}\\left(A\\right)$ required? Is the condition number of the matrix logarithm required? f01kjf yes yes no no Is the Fréchet derivative of the matrix logarithm required? f01kkf yes no f01fjf Is $\\mathrm{exp}\\left(A\\right)$ required? Is the condition number of the matrix exponential required? f01kgf yes yes no no Is the Fréchet derivative of the matrix exponential required? f01khf yes no f01fcf Is ${A}^{1/2}$ required? Is the condition number of the matrix square root required? f01kdf yes yes no no Is the matrix upper triangular? f01fpf yes no f01fnf Is ${A}^{p}$ required? Is the condition number of the matrix power required? f01kef yes yes no no Is the Fréchet derivative of the matrix power required? f01kff yes no f01fqf $f\\left(A\\right)$ will be computed. Will derivatives of $f$ be supplied by the user? Is the condition number of the matrix function required? f01kcf yes yes no no f01fmf Is the condition number of the matrix function required? f01kbf yes no f01flf\n\n## 5Functionality Index\n\n Action of the matrix exponential on a complex matrix f01haf\n Action of the matrix exponential on a complex matrix (reverse communication) f01hbf\n Action of the matrix exponential on a real matrix f01gaf\n Action of the matrix exponential on a real matrix (reverse communication) f01gbf\n Inversion (also see Chapter F07),\n real $m×n$ matrix,\n pseudo-inverse f01blf\n real symmetric positive definite matrix,\n accurate inverse f01abf\n Matrix Arithmetic and Manipulation,\n complex matrices f01cwf\n real matrices f01ctf\n matrix multiplication,\n rectangular matrices,\n update,\n real matrices f01ckf\n triangular matrices,\n in-place,\n complex matrices f01duf\n real matrices f01dgf\n update,\n complex matrices f01dtf\n real matrices f01dff\n matrix storage conversion,\n full to packed triangular storage,\n complex matrices f01vbf\n real matrices f01vaf\n full to Rectangular Full Packed storage,\n complex matrix f01vff\n real matrix f01vef\n packed band $↔$ rectangular storage, special provision for diagonal\n complex matrices f01zdf\n real matrices f01zcf\n packed triangular to full storage,\n complex matrices f01vdf\n real matrices f01vcf\n packed triangular to Rectangular Full Packed storage,\n complex matrices f01vkf\n real matrices f01vjf\n packed triangular $↔$ square storage, special provision for diagonal\n complex matrices f01zbf\n real matrices f01zaf\n Rectangular Full Packed to full storage,\n complex matrices f01vhf\n real matrices f01vgf\n Rectangular Full Packed to packed triangular storage,\n complex matrices f01vmf\n real matrices f01vlf\n matrix subtraction,\n real matrices f01ctf\n matrix transpose f01crf\n Matrix function,\n complex Hermitian $n×n$ matrix,\n matrix exponential f01fdf\n matrix function f01fff\n complex $n×n$ matrix,\n condition number for a matrix exponential f01kgf\n condition number for a matrix exponential, logarithm, sine, cosine, sinh or cosh f01kaf\n condition number for a matrix function, using numerical differentiation f01kbf\n condition number for a matrix function, using user-supplied derivatives f01kcf\n condition number for a matrix logarithm f01kjf\n condition number for a matrix power f01kef\n condition number for the matrix square root, logarithm, sine, cosine, sinh or cosh f01kdf\n Fréchet derivative\n matrix exponential f01khf\n matrix logarithm f01kkf\n matrix power f01kff\n general power\n matrix f01fqf\n matrix exponential f01fcf\n matrix exponential, sine, cosine, sinh or cosh f01fkf\n matrix function, using numerical differentiation f01flf\n matrix function, using user-supplied derivatives f01fmf\n matrix logarithm f01fjf\n matrix square root f01fnf\n upper triangular\n matrix square root f01fpf\n real $n×n$ matrix,\n condition number for a matrix exponential f01jgf\n condition number for a matrix function, using numerical differentiation f01jbf\n condition number for a matrix function, using user-supplied derivatives f01jcf\n condition number for a matrix logarithm f01jjf\n condition number for a matrix power f01jef\n condition number for the matrix exponential, logarithm, sine, cosine, sinh or cosh f01jaf\n condition number for the matrix square root, logarithm, sine, cosine, sinh or cosh f01jdf\n Fréchet derivative\n matrix exponential f01jhf\n matrix logarithm f01jkf\n matrix power f01jff\n general power\n matrix exponential f01eqf\n matrix exponential f01ecf\n matrix exponential, sine, cosine, sinh or cosh f01ekf\n matrix function, using numerical differentiation f01elf\n matrix function, using user-supplied derivatives f01emf\n matrix logarithm f01ejf\n matrix square root f01enf\n upper quasi-triangular\n matrix square root f01epf\n real symmetric $n×n$ matrix,\n matrix exponential f01edf\n matrix function f01eff\n Matrix Transformations,\n complex $m×n\\left(m\\le n\\right)$ matrix,\n $RQ$ factorization f01rjf\n complex matrix, form unitary matrix f01rkf\n complex upper trapezoidal matrix,\n $RQ$ factorization f01rgf\n eigenproblem $Ax=\\lambda Bx$, $A$, $B$ banded,\n reduction to standard symmetric problem f01bvf\n modified Cholesky factorization, form positive definite perturbed input matrix f01mef\n modified Cholesky factorization of a real symmetric matrix f01mdf\n non-negative matrix factorization f01saf\n non-negative matrix factorization, reverse communication f01sbf\n real almost block-diagonal matrix,\n $LU$ factorization f01lhf\n real band symmetric positive definite matrix,\n $ULD{L}^{\\mathrm{T}}{U}^{\\mathrm{T}}$ factorization f01buf\n variable bandwidth, $LD{L}^{\\mathrm{T}}$ factorization f01mcf\n real $m×n\\left(m\\le n\\right)$ matrix,\n $RQ$ factorization f01qjf\n real matrix,\n form orthogonal matrix f01qkf\n real sparse matrix,\n factorization f01brf\n factorization, known sparsity pattern f01bsf\n real upper trapezoidal matrix,\n $RQ$ factorization f01qgf\n tridiagonal matrix,\n $LU$ factorization f01lef\n\nNone.\n\n## 7 Withdrawn or Deprecated Routines\n\nNone.\nGolub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore\nHigham N J (2008) Functions of Matrices: Theory and Computation SIAM, Philadelphia, PA, USA\nWilkinson J H (1965) The Algebraic Eigenvalue Problem Oxford University Press, Oxford\nWilkinson J H (1977) Some recent advances in numerical linear algebra The State of the Art in Numerical Analysis (ed D A H Jacobs) Academic Press\nWilkinson J H and Reinsch C (1971) Handbook for Automatic Computation II, Linear Algebra Springer–Verlag" ]

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[ "# Formulas of reduced multiplication\n\n## The basic laws of algebra\n\nThe square of the sum of two numbers is equal to the square of the first number plus twice the product of the first number by the second, plus the square of the second number.", null, "The squared difference of two numbers is equal to the square of the first minus twice the product of the first number by the second, plus the square of the second number.", null, "The difference between the squares of two numbers is equal to the product of the sum of these numbers is to their difference.", null, "The cube is the sum of two numbers is equal to waste the first day plus triple the square of the product of the first number by the second, tripled plus the product of the first number by the square of the second plus the cube of the second number.", null, "Cube of the difference is equal to the boiler of the first number minus tripled the product of the square of the first number by the second, tripled plus the product of the first number by the square of the second minus the cube of the second number.", null, "The sum of the cubes of two numbers is equal to the product of the sum of these numbers on the incomplete square of the difference of these numbers.", null, "The difference between the cubes of two numbers is equal to the product of the difference of the numbers on the incomplete square of the sum of these numbers.", null, "The square of the sum of multiple expressions is equal to the sum of the squares of all the summands plus all the double works of each expression for each of the following.", null, "## The rules of disclosure brackets", null, "", null, "", null, "", null, "1. Multiplication of a single term to the polynomial\n2.", null, "3. Multiplication of polynomial by polynomial\n4.", null, "5. The imposition of a General polynomial of the brackets\n6.", null, "7. The grouping method\n8.", null, "9. The allocation of square binomials square trinomial\n10.", null, "## Decomposition of algebraic expressions into factors.\n\n### Square trinomial", null, "### The difference of two numbers the n-th degree", null, "### The difference between two numbers of the fourth degree", null, "### The difference between two numbers of the fifth degree", null, "### For odd integer n", null, "## The Binomial Theorem", null, "where the coefficients of this decomposition", null, "can be derived from Pascal's triangle.\n\nTags:\nChapter:\nVersions in other languages:\nShare with friends:" ]

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[ "+0\n\n# Let S be the set of points (a,b) with $0 \\le a,b \\le 1$ such that the equation $x^4 + ax^3 - bx^2 + ax + 1 = 0$has at least one rea\n\n0\n156\n1\n\nLet S be the set of points (a,b) with $$0\\leq a,b \\leq 1$$  such that the equation $$x^4+ax^3-bx^2+ax+1=0$$\nhas at least one real root. Determine the area of the graph of S.\n\nDec 4, 2018\nedited by Guest  Dec 4, 2018\nedited by Guest  Dec 4, 2018" ]

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[ "#", null, "#### [ decimal->nextafter ]\n\nMethod\n\nReturns the next representable double-precision floating-point value following x in the direction of y. Thus, if y is less than x, [decimal->nextafter] returns the largest representable floating-point number less than x. x is the decimal object and y is the required decimal parameter.\n\n• Syntax\n``Decimal Object->nextafter(Decimal Value)``\n• Beginner\n\n### To return the next representable double-precision floating-point number following a decimal in the direction of another decimal:\n\nUse the [decimal->nextafter] method. The following examples illustrate the next numbers which can be represented when followed in two different directions, relative to the original decimal.\n\nCode\n\n``````1.0000005->nextafter(2.0)\n'<br>'\n1.0000005->nextafter(1.0)\n``````\n\nResult\n\n```1.000001\n1.000000```\n##### Resources\n• Articles", null, "Please note that periodically LassoSoft will go through the notes and may incorporate information from them into the documentation. Any submission here gives LassoSoft a non-exclusive license and will be made available in various formats to the Lasso community." ]

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[ "# How to get the size of a JavaScript object?\n\n## How to get the size of a JavaScript object?\n\nI want to know the size occupied by a JavaScript object.\nTake the following function:\nfunction Marks(){\nthis.maxMarks = 100;\n}\n\nfunction Student(){\nthis.firstName = “firstName”;\nthis.lastName = “lastName”;\nthis.marks = new Marks();\n}\n\nNow I instantiate the student:\nvar stud = new Student();\n\nso that I can do stuff like\nstud.firstName = “new Firstname”;\n\nstud.marks.maxMarks = 200;\n\netc.\nNow, the stud object will occupy some size in memory. It has some data and more objects.\nHow do I find out how much memory the stud object occupies? Something like a sizeof() in JavaScript? It would be really awesome if I could find it out in a single function call like sizeof(stud).\nI’ve been searching the Internet for months—couldn’t find it (asked in a couple of forums—no replies).\n\n### Solution 1:\n\nI have re-factored the code in my original answer. I have removed the recursion and removed the assumed existence overhead.\n\n``````function roughSizeOfObject( object ) {\n\nvar objectList = [];\nvar stack = [ object ];\nvar bytes = 0;\n\nwhile ( stack.length ) {\nvar value = stack.pop();\n\nif ( typeof value === 'boolean' ) {\nbytes += 4;\n}\nelse if ( typeof value === 'string' ) {\nbytes += value.length * 2;\n}\nelse if ( typeof value === 'number' ) {\nbytes += 8;\n}\nelse if\n(\ntypeof value === 'object'\n&& objectList.indexOf( value ) === -1\n)\n{\nobjectList.push( value );\n\nfor( var i in value ) {\nstack.push( value[ i ] );\n}\n}\n}\nreturn bytes;\n}\n``````\n\n### Solution 2:\n\nThe Google Chrome Heap Profiler allows you to inspect object memory use.\n\nYou need to be able to locate the object in the trace which can be tricky. If you pin the object to the Window global, it is pretty easy to find from the “Containment” listing mode.\n\nIn the attached screenshot, I created an object called “testObj” on the window. I then located in the profiler (after making a recording) and it shows the full size of the object and everything in it under “retained size”.", null, "In the above screenshot, the object shows a retained size of 60. I believe the unit is bytes here.\n\n### Solution 3:\n\nI just wrote this to solve a similar (ish) problem. It doesn’t exactly do what you may be looking for, ie it doesn’t take into account how the interpreter stores the object.\n\nBut, if you are using V8, it should give you a fairly ok approximation as the awesome prototyping and hidden classes lick up most of the overhead.\n\n``````function roughSizeOfObject( object ) {\n\nvar objectList = [];\n\nvar recurse = function( value )\n{\nvar bytes = 0;\n\nif ( typeof value === 'boolean' ) {\nbytes = 4;\n}\nelse if ( typeof value === 'string' ) {\nbytes = value.length * 2;\n}\nelse if ( typeof value === 'number' ) {\nbytes = 8;\n}\nelse if\n(\ntypeof value === 'object'\n&& objectList.indexOf( value ) === -1\n)\n{\nobjectList[ objectList.length ] = value;\n\nfor( i in value ) {\nbytes+= 8; // an assumed existence overhead\nbytes+= recurse( value[i] )\n}\n}\n\nreturn bytes;\n}\n\nreturn recurse( object );\n}\n``````\n\n### Solution 4:\n\nSometimes I use this to flag really big objects that might be going to the client from the server. It doesn’t represent the in memory footprint. It just gets you approximately what it’d cost to send it, or store it.\n\nAlso note, it’s slow, dev only. But for getting an ballpark answer with one line of code it’s been useful for me.\n\n``````roughObjSize = JSON.stringify(bigObject).length;\n``````\n\n### Solution 5:\n\nThere is a NPM module to get object sizeof, you can install it with `npm install object-sizeof`\n\n`````` var sizeof = require('object-sizeof');\n\n// 2B per character, 6 chars total => 12B\nconsole.log(sizeof({abc: 'def'}));\n\n// 8B for Number => 8B\nconsole.log(sizeof(12345));\n\nvar param = {\n'a': 1,\n'b': 2,\n'c': {\n'd': 4\n}\n};\n// 4 one two-bytes char strings and 3 eighth-bytes numbers => 32B\nconsole.log(sizeof(param));\n``````\n\n### Solution 6:\n\nA little late to the party, but here’s a slightly more compact solution to the problem:\n\n``````const typeSizes = {\n\"undefined\": () => 0,\n\"boolean\": () => 4,\n\"number\": () => 8,\n\"string\": item => 2 * item.length,\n\"object\": item => !item ? 0 : Object\n.keys(item)\n.reduce((total, key) => sizeOf(key) + sizeOf(item[key]) + total, 0)\n};\n\nconst sizeOf = value => typeSizes[typeof value](value);\n``````" ]

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[ "A ball is thrown eastward into the air from the origin (in the direction of the positive x-axis). The initial velocity is 40 i + 64 k,?\n\nwith speed measured in feet per second. The spin of the ball results in a southward acceleration of 8 ft/s2, so the acceleration vector is a = −8 j − 32 k. Where does the ball land? (Round your answers to one decimal place.) . ft from the origin at an angle of ° from the eastern direction toward the south. With... show more with speed measured in feet per second. The spin of the ball results in a southward acceleration of 8 ft/s2, so the acceleration vector is a = −8 j − 32 k. Where does the ball land? (Round your answers to one decimal place.)\n. ft from the origin at an angle of ° from the eastern direction toward the south.\n\nWith what speed does the ball hit the ground? (Round your answer to one decimal place.)\n\nI have:\na(t)=<0,18,-32>\nv(t)=<40,-8t,64-32t>\nr(t)=<40t,-8t^2,64t-32t^s>\nwhen the z component is zero t=0 and t=2\n\nI plugged in t=2 into my position function and tried to do inverse tan to get the first two parts, but my answers are wrong. Can you help me see where I went wrong?" ]

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[ "# Normalizing a PDF\n\nProbability Level 2\n\nA certain continuous random variable has a probability density function (PDF) given by:\n\n$f(x) = C x (1-x)^2,$\n\nwhere $x$ can be any number in the real interval $[0,1]$. Compute $C$ using the normalization condition on PDFs.\n\n×" ]

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[ "# Doppler Effect Wavelength Front Calculation English", null, "Español\n\nDoppler Effect is the change in frequency and wavelength of a wave for an observer moving relative to the source of the waves. Here we can calculate Wavelength in Front of a Moving Source.\n\n## Doppler Effect Wavelength Front Calculator\n\nm/s\nm/s\nHz\nm\n\nDoppler Effect is the change in frequency and wavelength of a wave for an observer moving relative to the source of the waves. Here we can calculate Wavelength in Front of a Moving Source.\n\nCode to add this calci to your website", null, "", null, "#### Wavelength in Front of a Moving Source:", null, "#### Wave Velocity:", null, "#### Source Velocity:", null, "#### Source Frequency:", null, "where, λf = Wavelength in Front of a Moving Source v = Wave Velocity us = Source Velocity f0 = Source Frequency." ]

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[ "# Limits problem\n\nSolve:\nlim x->0 (tan 3(x+h)-tan(3x))/h\n\ni hv no clue where to start =(\n\nerr.. tan (3h)/h?\n\nsimply plugging in 0 for x..\n\nedit: are you sure x is approaching 0?\n\ninitially, I thought it's h..\n\nthen the answer would have been 3+3tan(3x)^2\nand you had to do some mess with identites..\nhttp://www.clarku.edu/~djoyce/trig/identities.html\n\nLast edited:\nDick\nHomework Helper\nDid you mean lim h->0??\n\nDid you mean lim h->0??\nsry, yes i mean lim h->0\n\nDick\nHomework Helper\nLook up a formula for tan(A+B) and apply it to tan(3x+3h). Then simplify, take your limit and use some trig.\n\nIt is actually very simple.. don't even need to any trig after simplifying tan(3x+3h) ..\n\nshould factor out things.. and they would cancel out nicely.\n\nAnd, one more thing tan(x)/x = 1 .. (which is simple to prove is you know sin(x)/x =1 as x-->0)\n\nDick\nHomework Helper\nIt is actually very simple.. don't even need to any trig after simplifying tan(3x+3h) ..\n\nshould factor out things.. and they would cancel out nicely.\n\nAnd, one more thing tan(x)/x = 1 .. (which is simple to prove is you know sin(x)/x =1 as x-->0)\nYep. You don't have to use any trig. But using sec^2(A)=1+tan^2(A) would put it in the simpler form listed in books.\n\nkk thanks i got it =)" ]

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[ "", null, "# 《消失的设计模式》系列之策略模式\n\n## 面向对象的方式\n\n### UML\n\n• Strategy 接口定义了算法的行为\n• 一系列的算法 ConcreteStrategyAConcreteStrategyB 分别实现了该接口\n• 策略被包含在一个 Context 中，拥有私有的 strategy 实例，从而执行具体的算法行为\n• 通常 Context 中还会包含一个 setStrategy 方法，从而动态改变策略\n\n### 代码\n\ninterface IStrategy {\ncalculate(a: number, b: number): number\n}\n\nclass AddStrategy implements IStrategy {\ncalculate(a: number, b: number): number {\nreturn a + b\n}\n}\n\nclass SubtractStrategy implements IStrategy {\ncalculate(a: number, b: number): number {\nreturn a - b\n}\n}\n\nclass Context {\nprivate strategy: IStrategy\n\nconstructor(strategy: IStrategy) {\nthis.strategy = strategy\n}\n\npublic setStrategy(strategy: IStrategy) {\nthis.strategy = strategy\n}\n\npublic calculate(a: number, b: number) {\nreturn this.strategy.calculate(a, b)\n}\n}\n\nconst context = new Context(new AddStrategy())\nconsole.log(context.calculate(1, 2)) // 3\ncontext.setStrategy(new SubtractStrategy())\nconsole.log(context.calculate(1, 2)) // -1\n\ncontext.setStrategy(new SubtractStrategy())\nconsole.log(context.calculate(1, 2)) // -1\n\n## 带上函数式的思考帽\n\ninterface IStrategy {\ncalculate(a: number, b: number): number\n}\n\ninterface IStrategy {\n(a: number, b: number): number\n}\n\ntype IStrategy = (a: number, b: number) => number\n\nconst AddStrategy: IStrategy = (a, b) => a + b\nconst SubtractStrategy: IStrategy = (a, b) => a - b\n\nclass Context {\nprivate strategy: IStrategy\n\nconstructor(strategy: IStrategy) {\nthis.strategy = strategy\n}\n\npublic setStrategy(strategy: IStrategy) {\nthis.strategy = strategy\n}\n\npublic calculate(a: number, b: number) {\nreturn this.strategy(a, b)\n}\n}\n\nconsole.log(context.calculate(1, 2)) // 3\ncontext.setStrategy(SubtractStrategy)\nconsole.log(context.calculate(1, 2)) // -1\n\ncontext.setStrategy(SubtractStrategy)\nconsole.log(context.calculate(1, 2)) // -1\n\n## 数组的排序算法与策略模式\n\n;[1, 5, 3, 4, 2].sort((a, b) => a - b)\n\nclass CustomComparator implements Comparator<X> {\n\npublic int compare(X a, x b) {\nreturn ...\n}\n}\n\nCustomComparator comparator = new CustomComparator()\nArrays.sort(arr, new CustomComparator())\n\n## 总结\n\n#### 参考", null, "" ]

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[ "14 May by\n\n# Free Math Drill Worksheets Unique Best 2 Times Table Worksheet Ideas On Grade 4 Sheets Math Drills Worksheets Answers 5 Free Addition Drill Answers To Math Drills Worksheets", null, "free math drill worksheets unique best 2 times table worksheet ideas on grade 4 sheets math drills worksheets answers 5 free addition drill answers to math drills worksheets.\n\nanswers to math drills worksheets graphing linear equations, worksheets order math drills algebra worksheet answers to,math drills addition and subtraction answers to worksheets , answers to math drills worksheets fractions,math drill sheets addition facts etc preschool numbers drills answers to worksheets, answers to math drills worksheets maths mean median mode range,addition of exponents math medium to large size grade worksheets answers drills , addition answers to math drills worksheets,answers to math drills worksheets improper fractions for kids dividing and multiplying, answers to math drills worksheets." ]

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[ "## Peirce’s 1870 “Logic Of Relatives” • Selection 8\n\nWe continue with §3. Application of the Algebraic Signs to Logic.\n\n### The Signs for Multiplication (cont.)\n\nThus far, we have considered the multiplication of relative terms only. Since our conception of multiplication is the application of a relation, we can only multiply absolute terms by considering them as relatives.\n\nNow the absolute term “man” is really exactly equivalent to the relative term “man that is ──”, and so with any other. I shall write a comma after any absolute term to show that it is so regarded as a relative term.\n\nThen “man that is black” will be written:", null, "$\\mathrm{m},\\!\\mathrm{b}.$\n\nBut not only may any absolute term be thus regarded as a relative term, but any relative term may in the same way be regarded as a relative with one correlate more. It is convenient to take this additional correlate as the first one.\n\nThen:", null, "$\\mathit{l},\\!\\mathit{s}\\mathrm{w}$\n\nwill denote a lover of a woman that is a servant of that woman.\n\nThe comma here after", null, "$\\mathit{l}$ should not be considered as altering at all the meaning of", null, "$\\mathit{l}\\,,$ but as only a subjacent sign, serving to alter the arrangement of the correlates.\n\nIn point of fact, since a comma may be added in this way to any relative term, it may be added to one of these very relatives formed by a comma, and thus by the addition of two commas an absolute term becomes a relative of two correlates.\n\nSo:", null, "$\\mathrm{m},\\!,\\!\\mathrm{b},\\!\\mathrm{r}$\n\ninterpreted like", null, "$\\mathfrak{g}\\mathit{o}\\mathrm{h}$\n\nmeans a man that is a rich individual and is a black that is that rich individual.\n\nBut this has no other meaning than:", null, "$\\mathrm{m},\\!\\mathrm{b},\\!\\mathrm{r}$\n\nor a man that is a black that is rich.\n\nThus we see that, after one comma is added, the addition of another does not change the meaning at all, so that whatever has one comma after it must be regarded as having an infinite number.\n\nIf, therefore,", null, "$\\mathit{l},\\!,\\!\\mathit{s}\\mathrm{w}$ is not the same as", null, "$\\mathit{l},\\!\\mathit{s}\\mathrm{w}$ (as it plainly is not, because the latter means a lover and servant of a woman, and the former a lover of and servant of and same as a woman), this is simply because the writing of the comma alters the arrangement of the correlates.\n\nAnd if we are to suppose that absolute terms are multipliers at all (as mathematical generality demands that we should}, we must regard every term as being a relative requiring an infinite number of correlates to its virtual infinite series “that is ── and is ── and is ── etc.”\n\nNow a relative formed by a comma of course receives its subjacent numbers like any relative, but the question is, What are to be the implied subjacent numbers for these implied correlates?\n\nAny term may be regarded as having an infinite number of factors, those at the end being ones, thus:", null, "$\\mathit{l},\\!\\mathit{s}\\mathrm{w} ~=~ \\mathit{l},\\!\\mathit{s}\\mathrm{w},\\!\\mathit{1},\\!\\mathit{1},\\!\\mathit{1},\\!\\mathit{1},\\!\\mathit{1},\\!\\mathit{1},\\!\\mathit{1}, ~\\text{etc.}$\n\nA subjacent number may therefore be as great as we please.\n\nBut all these ones denote the same identical individual denoted by", null, "$\\mathrm{w};$ what then can be the subjacent numbers to be applied to", null, "$\\mathit{s}\\,,$ for instance, on account of its infinite “that is” ’s? What numbers can separate it from being identical with", null, "$\\mathrm{w}?$ There are only two. The first is zero, which plainly neutralizes a comma completely, since", null, "$\\mathit{s},_0\\!\\mathrm{w} ~=~ \\mathit{s}\\mathrm{w}$\n\nand the other is infinity; for as", null, "$1^\\infty$ is indeterminate in ordinary algebra, so it will be shown hereafter to be here, so that to remove the correlate by the product of an infinite series of ones is to leave it indeterminate.\n\nAccordingly,", null, "$\\mathrm{m},_\\infty$\n\nshould be regarded as expressing some man.\n\nAny term, then, is properly to be regarded as having an infinite number of commas, all or some of which are neutralized by zeros.\n\n“Something” may then be expressed by:", null, "$\\mathit{1}_\\infty.$\n\nI shall for brevity frequently express this by an antique figure one", null, "$(\\mathfrak{1}).$\n\n“Anything” by:", null, "$\\mathit{1}_0.$\n\nI shall often also write a straight", null, "$1$ for anything.\n\n(Peirce, CP 3.73)\n\n### References\n\n• Peirce, C.S. (1870), “Description of a Notation for the Logic of Relatives, Resulting from an Amplification of the Conceptions of Boole’s Calculus of Logic”, Memoirs of the American Academy of Arts and Sciences 9, 317–378, 26 January 1870.  Reprinted, Collected Papers 3.45–149, Chronological Edition 2, 359–429.  Online (1) (2) (3).\n• Peirce, C.S., Collected Papers of Charles Sanders Peirce, vols. 1–6, Charles Hartshorne and Paul Weiss (eds.), vols. 7–8, Arthur W. Burks (ed.), Harvard University Press, Cambridge, MA, 1931–1935, 1958.\n• Peirce, C.S., Writings of Charles S. Peirce : A Chronological Edition, Peirce Edition Project (eds.), Indiana University Press, Bloomington and Indianapolis, IN, 1981–.\nThis entry was posted in C.S. Peirce, Logic, Logic of Relatives, Mathematics, Peirce, Relation Theory, Semiotics and tagged , , , , , , . Bookmark the permalink.\n\n### 6 Responses to Peirce’s 1870 “Logic Of Relatives” • Selection 8\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]

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[ "# How to Tell if a Number if Divisible by 3\n\nHow to Tell if a Number if Divisible by 3", null, "", null, "• A number is divisible by 3 if the sum of the digits is also divisible by 3.\n• If a number is divisible by 3 it means that the number is in the 3 times table.\n• Firstly, add the individual digits of a number.\n• 7 + 7 + 1 = 15.\n• Then check to see if this answer is in the 3 times table.\n• 15 is in the 3 times table and so 771 is also in the 3 times table.\n• 771 is divisible by 3.\n• This divisibility by 3 rule works for every number.\n• The rule can be applied to the result of sum of the digits to check that too.\n• In this example, we know that 15 is divisible by 3 because 1 + 5 = 6 and 6 is divisible by 3.\nAdd the digits of the number.\n\nIf this answer is divisible by 3, the original number is divisible by 3.", null, "", null, "• The rule for divisibility by 3 works for all numbers no matter how large.\n• Add the digits of the number and check if this result is also divisible by 3.\n• We add the individual digits of the number 7, 749, 984.\n• 7 + 7 + 4 + 9 + 9 + 8 + 4 = 48.\n• We might not be sure if 48 is divisible by 3.\n• The rule can be applied again to 48 to see if 48 is divisible by 3.\n• 4 + 8 = 12, which is in the 3 times table.\n• 12 is in the three times table, therefore so is 48 and so is 7, 749, 984.\n• 7, 749, 984 is divisible by 3, which means that 7, 749, 984 is in the 3 times table.", null, "Supporting Lessons", null, "# How to Tell if a Number is Divisible by 3\n\n## How to Tell if a Number is Divisible by 3\n\nTo test if a number is divisible by 3, follow these steps:\n\n1. Add the individual digits of the number to make a total.\n2. If this total is divisible by 3, the original number is divisible by 3.\n3. If you are not sure if the total is divisible by 3, apply the first two steps to that number.\n\nIf a number is divisible by 3, it means that the number is in the 3 times table. A number that is divisible by 3 is a multiple of 3. The number can be divided exactly by 3 to leave no remainder.\n\nIn this example, we will use the divisibility by 3 rule to test 5502. Is 5502 in the 3 times table?\n\nA number is divisible by 3 if the sum of its digits is also divisible by 3. For example, 5502 is divisible by 3 because 5 + 5 + 0 + 2 = 12. 12 is divisible by 3 and so, 5502 is divisible by 3.", null, "The first step is to add the digits of the number.\n\n5 + 5 + 0 + 2 = 12.\n\nThe next step is to check if this new, smaller number is divisible by 3.\n\n12 is 4 × 3. 12 is in the three times table and so, 5502 is also in the three times table.", null, "We can also check that each number in the working out is a multiple of 3 by adding its digits.\n\n12 is a multiple of 3 because 1 + 2 = 3.\n\nHere is another example of using the rule for divisibility by 3 to test if a number is a multiple of 3.\n\nIs the number 409 a multiple of 3?", null, "The first step is to add the digits of the number.\n\n4 + 0 + 9 = 13.\n\nThe next step is to decide if the sum of the digits is a multiple of 3.\n\n13 is not a multiple of 3. It is not in the 3 times table. We can also see that 1 + 3 = 4 and 4 is not a multiple of 3, therefore 13 is not a multiple of 3.", null, "13 is an example of a number that is not divisible by 3 and so, 409 is not divisible by 3.\n\nA number is not divisible by 3 if the sum of its digits is not divisible by 3. 409 is not divisible by 3 because 4 + 0 + 9 = 13 and 13 is not divisible by 3.\n\nPrime numbers are not divisible by 3 because they are ony divisible by 1 and themselves. For example, 13 is a prime number and so it not divisible by 3.\n\nThe rule for divisibility by 3 works for all numbers no matter how large.\n\nFor example, here is the number 529, 943.\n\nThe first step is to add the digits of the number.\n\n5 + 2 + 9 + 9 + 4 + 3 = 32", null, "The next step is to test if the sum of the digits is divisible by 3.\n\n32 is not divisible by 3. We know this because 30 and 33 are multiples of 3 and 32 is in between these numbers.\n\nWe can also use the same test on 32 to show that it is not divisible by 3. We add the digits. 3 + 2 = 5 and 5 is not a multiple of 3.\n\n32 is not a multiple of 3 and therefore 529, 943 is not a multiple of 3 either. 529, 943 is an example of a number that is not divisible by 3.\n\nHere is an example of testing a large number to see if it is in the 3 times table.\n\nIs 7, 749, 984 a multiple of 3?\n\n7 + 7 + 4 + 9 + 9 + 8 + 4 = 48", null, "48 is 16 × 3 and so, it is a multiple of 3.\n\nIf we were not sure if the number was a multiple of 3, add the digits and see if the total is a multiple of 3.\n\n4 + 8 = 12, which is 4 × 3. 12 is a multiple of 3 and so, 48 is a multiple of 3 and 7, 749, 984 is also a multiple of 3\n\n## Why Does the Divisibility Rule for 3 Work?\n\nThe divisibility rule for 3 works because the number represented by each digit can be written as a multiple of 9 plus that digit. 9 is divisible by 3 so if the sum of the digits is divisible by 3, the number itself is too.\n\nHere is the proof that 3174 is divisible by 3.\n\nThe digit 3 stands for 3000, which is 1000 × 3.\n\nThis is the same as 999 × 3 plus one more 3.\n\nThe digit 1 stands for 100, which is 100 × 1.\n\nThis is the same as 99 × 1 plus one more 1.\n\nThe digit 7 stands for 70, which is 10 × 7.\n\nThis is the same as 9 × 7 plus one more 7.\n\nThe digit 4 stands for one lot of 4.", null, "The multiples of 9, 99 and 999 are all divisible by 3.", null, "3174 is equal to 999 × 3 + 99 × 1 + 9 × 7 plus 3 + 1 + 7 + 4.\n\nThe multiples of 9, 99 and 999 are all divisible by 3, so we only need to check the sum of the digits: 3 + 1 + 7 + 4.", null, "3 + 1 + 7 + 4 = 15, which is divisible by 3.\n\nSince the multiples of 9, 99 and 999 along with the sum of the digits are all divisible by 3, the entire number is divisible by 3.\n\nAll numbers are written in base 10. This means that the digits of each number represent a multiple of 9 plus that digit. The multiples of 9 are divisible by 3, so we simply need to test the sum of the digits to see if the whole number is divisible by 3.\n\nNo matter what the number, add the digits to check if it is divisible by 3.\n\n## List of Numbers Divisible by 3\n\nHere is a list of 2-digit numbers less than 100 that are divisible by 3:\n\n3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96 and 99.\n\nThere are 33 2-digit numbers that are divisible by 3. The largest 2 digit number divisible by 3 is 99, which is 33 × 3.\n\nThis list can help us to identify some common multiples of 3, which helps us to identify if a larger number is divisible by 3.\n\nPrime numbers are not be divisible by 3. This is because prime numbers can only be divided by 1 and themselves.\n\nEven numbers can be divisible by 3. For example, the even number of 12 is divisible by 3.\n\nAll numbers that are divisible by 9 are also divisible by 3. This is because 3 divides exactly into 9.", null, "Now try our lesson on Rule for Divisibility by 9 where we learn how to test if a number is divisible by 9.", null, "error: Content is protected !!" ]

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[ "# Exponential Decay Test Quiz\n\nApproved & Edited by ProProfs Editorial Team\nThe editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.\n| By Gregorynaomi\nG\nGregorynaomi\nCommunity Contributor\nQuizzes Created: 1670 | Total Attempts: 728,158\nQuestions: 10 | Attempts: 132", null, "", null, "Settings", null, "", null, "Exponential decay is an abstract mathematical phenomenon which explains why some numbers decrease at a rate directly proportional to their present value. This can be expressed using Calculus. It can also be represented symbolically. This phenomenon is widely applied. Especially in natural sciences, the principle of exponential decay is useful. Do you understand it?\n\n• 1.\n\n### Which of these does not involve the application of exponential decay?\n\n• A.\n\nElectrostatics\n\n• B.\n\nChemical reactions\n\n• C.\n\nHeat transfer\n\n• D.\n\nFriction\n\nD. Friction\nExplanation\nFriction does not involve the application of exponential decay. Exponential decay refers to a decrease in a quantity over time, where the rate of decrease is proportional to the current value. In electrostatics, the charge on an object can decay exponentially over time. In chemical reactions, the concentration of reactants can decrease exponentially as the reaction progresses. In heat transfer, the temperature difference between two objects can decrease exponentially over time. However, friction is a force that opposes motion and does not exhibit exponential decay.\n\nRate this question:\n\n• 2.\n\n### What type of reactions follows exponential decay?\n\n• A.\n\nFirst-order decay\n\n• B.\n\nThird-order decay\n\n• C.\n\nExothermic reactions\n\n• D.\n\nEndothermic reactions\n\nA. First-order decay\nExplanation\nExponential decay is a type of decay in which the rate of decay is proportional to the amount of substance remaining. This type of decay is characteristic of first-order reactions, where the rate of reaction is directly proportional to the concentration of the reactant. In first-order decay, the amount of substance decreases exponentially over time, with a constant half-life. Therefore, the correct answer is \"First-order decay\".\n\nRate this question:\n\n• 3.\n\n### How is exponential decay applicable in geophysics?\n\n• A.\n\nBy quantification of decreasing atmospheric oxygen\n\n• B.\n\nBy quantification of increasing atmospheric oxygen\n\n• C.\n\nBy quantification of fluctuating atmospheric oxygen\n\n• D.\n\nBy quantification of constant atmospheric oxygen\n\nA. By quantification of decreasing atmospheric oxygen\nExplanation\nExponential decay is applicable in geophysics through the quantification of decreasing atmospheric oxygen. This is because the concentration of atmospheric oxygen gradually decreases over time, following an exponential decay pattern. Geophysicists can measure and analyze this decay to understand various processes and phenomena, such as changes in atmospheric composition, climate change, and the impact of human activities on the environment.\n\nRate this question:\n\n• 4.\n\n### How is exponential decay applicable in pharmacology?\n\n• A.\n\n• B.\n\nBioavailability\n\n• C.\n\nTherapeutics\n\n• D.\n\nDrug metabolism\n\nD. Drug metabolism\nExplanation\nExponential decay is applicable in pharmacology through drug metabolism. Drug metabolism refers to the process by which the body breaks down and eliminates drugs. This process follows an exponential decay pattern, where the drug concentration decreases over time as it is metabolized and eliminated from the body. Understanding drug metabolism is crucial in pharmacology as it helps determine the dosage and frequency of drug administration to maintain effective therapeutic levels in the body.\n\nRate this question:\n\n• 5.\n\n### What variable undergoes exponential decay in heat transfer?\n\n• A.\n\nTemperature difference\n\n• B.\n\nTemperature medium\n\n• C.\n\nHeat source\n\n• D.\n\nHeat conduction\n\nA. Temperature difference\nExplanation\nTemperature difference undergoes exponential decay in heat transfer. As heat is transferred from a hotter object to a cooler object, the temperature difference between them decreases over time. This is because the rate of heat transfer is directly proportional to the temperature difference according to Newton's Law of Cooling. As the temperature difference decreases, the rate of heat transfer also decreases exponentially. Therefore, the temperature difference is the variable that undergoes exponential decay in heat transfer.\n\nRate this question:\n\n• 6.\n\n### How is exponential decay applicable in finance?\n\n• A.\n\nTax evasion\n\n• B.\n\nTax enforcement\n\n• C.\n\nPayment of retirement fund\n\n• D.\n\nBudgeting\n\nC. Payment of retirement fund\nExplanation\nExponential decay is applicable in finance for the payment of retirement funds because it represents the gradual decrease in the value of money over time. Retirement funds are typically invested in various financial instruments such as stocks, bonds, and mutual funds, which generate returns over time. However, these returns may decrease over time due to factors like inflation or market fluctuations. Therefore, the value of retirement funds tends to decay exponentially, and it is important to plan and budget accordingly to ensure a steady income during retirement.\n\nRate this question:\n\n• 7.\n\n### What is the duration of exponential decay in Polonium?\n\n• A.\n\n123 days\n\n• B.\n\n145 days\n\n• C.\n\n155 days\n\n• D.\n\n138 days\n\nD. 138 days\nExplanation\nThe correct answer is 138 days. Polonium undergoes exponential decay, which means that its quantity decreases over time. The duration of this decay in Polonium is 138 days, indicating that after this time period, the amount of Polonium will have significantly decreased.\n\nRate this question:\n\n• 8.\n\n### What is the length of exponential decay in Uranium?\n\n• A.\n\n234 days\n\n• B.\n\n189 days\n\n• C.\n\n177 days\n\n• D.\n\n23 days\n\nD. 23 days\nExplanation\nThe correct answer is 23 days. Exponential decay refers to the time it takes for a substance to decrease by a certain factor. In the case of Uranium, it has a half-life of 23 days, meaning that it takes 23 days for half of the Uranium atoms to decay. This indicates that after 23 days, the amount of Uranium will be reduced by half, and the process continues in a similar manner.\n\nRate this question:\n\n• 9.\n\n### What kind of quantity is influenced by exponential decay?\n\n• A.\n\nScalar quantity\n\n• B.\n\nVector quantity\n\n• C.\n\nLinear quantity\n\n• D.\n\nUniform quantity\n\nA. Scalar quantity\nExplanation\nExponential decay is a process in which a quantity decreases exponentially over time. Scalar quantity refers to a physical quantity that has only magnitude and no direction. In the context of exponential decay, the quantity being influenced is only changing in magnitude, not direction. Therefore, the correct answer is scalar quantity.\n\nRate this question:\n\n• 10.\n\n### What mathematical phenomenon is most relevant to exponential decay in all contexts?\n\n• A.\n\nCombination\n\n• B.\n\nProbability\n\n• C.\n\nPermutation\n\n• D.\n\nPoisson process\n\nD. Poisson process\nExplanation\nThe mathematical phenomenon most relevant to exponential decay in all contexts is the Poisson process. The Poisson process is a stochastic process that models the occurrence of events over time. It is often used to describe the decay of radioactive substances, the arrival of customers in a queue, or the occurrence of rare events. In the context of exponential decay, the Poisson process provides a mathematical framework to model the random and independent nature of decay events, where the rate of decay remains constant over time.\n\nRate this question:\n\nRelated Topics", null, "Back to top" ]

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[ "# Proving the reflection principle of Brownian motion\n\nThe reflection principle of Brownian motion states that Brownian motion reflected at some stopping time $\\tau$ is still a Brownian motion.\n\nThe proof found in Mörters & Peres (as well as in several other books), rests on the fact that the concatenation mapping, which takes a continuous path $\\{g(t) : t \\geq 0\\}$ and glues it to the end point of a finite continuous path$\\{f(t) : 0 \\leq t \\leq \\tau\\}$ to form a new continuous path, is measurable.\n\nI don't see how to prove that the concatenation mapping is measurable.\n\nFormally, the reflection principle states (I follow Schilling & Partzsch):\n\nLet $(B(t), \\mathcal{F}_t)_{t \\geq 0}$ be a standard, $d$-dimensional Brownian motion and let $\\tau$ be an a.s. finite stopping time. Consider the process $$W(t,\\omega) := \\begin{cases} B(t,\\omega), & 0 \\leq t < \\tau(\\omega) \\leq +\\infty \\\\ 2B(\\tau(\\omega), \\omega) - B(t,\\omega), & \\tau(\\omega) \\leq t < \\infty \\end{cases}$$ Then $(W(t))_{t \\geq 0}$ is again a standard Brownian motion.\n\nTo prove this theorem we may assume, without loss of generality, that $\\tau(\\omega) < \\infty$ for all $\\omega \\in \\Omega$, and we define the mapping $\\Phi : \\mathbf{C}_{(0)}\\times[0,\\infty)\\times\\mathbf{C}_{(0)} \\rightarrow \\mathbf{C}_{(0)}$, where $\\mathbf{C}_{(0)} := \\{f \\in \\mathbf{C}[0,\\infty) : f(0) = 0\\}$, thus $$\\Phi(f,t,g) := \\begin{cases} f(s), & 0 \\leq s < t \\\\ f(t) + g(s - t), & t \\leq s < \\infty \\end{cases}$$\n\nIt is claimed that $\\Phi$ is $\\mathcal{B}(\\mathbf{C}_{(0)})\\otimes\\mathcal{B}[0,\\infty)\\otimes\\mathcal{B}(\\mathbf{C}_{(0)})/\\mathcal{B}(\\mathbf{C}_{(0)})$ measurable.\n\nSchilling and Partzsch offer the following hint for proving this fact: equip $\\mathbf{C}_{(0)}\\times[0,\\infty)\\times\\mathbf{C}_{(0)}$ with the topology of locally uniform convergence (in $\\mathbf{C}_{(0)}$) and pointwise convergence (in $[0,\\infty)$).\n\nI'd appreciate help in proving why $\\Phi$ is measurable, either by following Schilling & Partzsch's cue, or by another method. Additionally, if it makes life easier, it can be assumed that $d = 1$.\n\nWorks cited\n\n• Mörters, Peter and Peres, Yuval. Brownian Motion. The version freely available from Peres's home page. Accessed 2014-02-18. (Theorem 2.19, pp. 44-45)\n• Schilling, René L. and Partzsch, Lothar. Brownian Motion - An introduction to stochastic processes. De Gruyter. 2012. (Theorem 6.12, pp. 71-72)\n\nTHE MAIN RESULT\n\nTheorem Employing the notation and terminology introduced in the original post, we have: $\\Phi$ is $\\mathcal{B}(\\mathbf{C}_{(0)})\\otimes\\mathcal{B}[0,\\infty)\\otimes\\mathcal{B}(\\mathbf{C}_{(0)})/\\mathcal{B}(\\mathbf{C}_{(0)})$ measurable.\n\nProof (For a listing of pertinent results, notations and definitions consult the next section, \"Auxiliary facts and definitions\". References cited below - e.g. [K], [M] - are listed in full at the bottom of this post in the section \"Works cited\".)\n\nI will prove only the case $d = 1$.\n\nAs suggested in [S] p. 72, it suffices to show that $\\Phi$ is continuous w.r.t. the product topology $\\mathcal{T}$ on $\\mathbf{C}_{(0)} \\times [0,\\infty) \\times \\mathbf{C}_{(0)}$ where $\\mathbf{C}_{(0)}$ is equipped with the $\\mathcal{O}_\\rho$ topology, and $[0,\\infty)$ is equipped with the standard, Euclidean topology, since if $\\Phi$ is $\\mathcal{T}/\\mathcal{O}_\\rho$-continuous, then by lemma 16 (\"continuity and measurability\"), $\\Phi$ is $\\sigma(\\mathcal{T})/\\sigma(\\mathcal{O}_\\rho)$-measurable, but by lemma 15 (\"product and Borel $\\sigma$-fields) and in light of proposition 8 (\"$\\mathbf{C}_{(0)}$ is a Polish space\") we have $\\sigma(\\mathcal{T}) = \\sigma(\\mathcal{O}_\\rho)\\otimes\\mathcal{B}[0,\\infty)\\otimes\\sigma(\\mathcal{O}_\\rho)$ and by definition 7, $\\sigma(\\mathcal{O}_p) = \\mathcal{B}(\\mathbf{C}_{(0)})$.\n\nNow, to prove that $\\Phi$ is $\\mathcal{T}/\\mathcal{O}_\\rho$-continuous, let $(f^*, t^*, g^*) \\in \\mathbf{C}_{(0)} \\times [0,\\infty) \\times \\mathbf{C}_{(0)}$ and let $\\varepsilon \\in (0,\\infty)$. We shall define some $\\delta \\in (0,\\infty)$, such that for all $(f, t, g) \\in \\mathbf{C}_{(0)} \\times [0,\\infty) \\times \\mathbf{C}_{(0)}$ with $\\rho(f, f^*), |t - t^*|, \\rho(g, g^*) < \\delta$, we will have $$\\rho(\\Phi(f,t,g), \\Phi(f^*, t^*, g^*)) < \\varepsilon$$\n\nChoose $k, n \\in \\mathbb{N}_1$ such that\n\na) $t^* + 1/8 \\leq k$,\n\nb) $(k + 1)2^{-k} \\leq \\varepsilon$\n\nc) $k \\leq n$\n\nd) Whenever $s, s' \\in [0, k]$ are such that $|s - s'| < 2^{- (n + 2)}$, $|g^*(s) - g^*(s')| < 2^{- (k + 2)}$\n\nAnd define $\\delta := 2^{- (n + 2)}$.\n\nLet $(f, t, g) \\in \\mathbf{C}_{(0)} \\times [0,\\infty) \\times \\mathbf{C}_{(0)}$ be such that $\\rho(f, f^*), |t - t^*|, \\rho(g, g^*) < \\delta$, and set \\begin{aligned} \\phi & := \\Phi(f,t,g) \\\\ \\phi^* & := \\Phi(f^*, t^*, g^*) \\\\ m & := \\min(t, t^*) \\\\ M & := \\max(t, t^*) \\end{aligned} Note that $M \\leq k$.\n\nWe have $$\\sup_{0 \\leq s \\leq k}|\\phi(s) - \\phi^*(s)| = \\\\ \\max(\\sup_{0 \\leq s \\leq m}|\\phi(s) - \\phi^*(s)|, \\sup_{m \\leq s \\leq M}|\\phi(s) - \\phi^*(s)|, \\sup_{M \\leq s \\leq k)} |\\phi(s) - \\phi^*(s)|)$$\n\nNow, by lemma 6a (\"Equivalence of $\\rho$ and $\\sup$\"), $$\\sup_{0 \\leq s \\leq m}|\\phi(s) - \\phi^*(s)| = \\sup_{0 \\leq s \\leq m}|f(s) - f^*(s)| < 2^{- (n + 2)} < 2^{-k}$$\n\nSecondly, we have (by lemma 6a and by the assumptions on $k$ and on $n$), $$\\sup_{m \\leq s \\leq M} |\\phi(s) - \\phi^*(s)| \\leq \\sup_{m \\leq s \\leq M} |f(s) - f^*(s)| + \\sup_{0 \\leq s \\leq 2^{- (n + 2)}} |g^*(s)| + \\sup_{0 \\leq s \\leq 2^{- (n + 2)}} |g(s) - g^*(s)| \\\\ < \\frac{2^{-n}}{4} + \\frac{2^{-k}}{4} + \\frac{2^{-n}}{4} < 2^{-k}$$\n\nThirdly, we have (again, by lemma 6a and by the assumptions on $k$ and on $n$), $$\\sup_{M \\leq s \\leq k} |\\phi(s) - \\phi^*(s)| \\leq \\sup_{M \\leq s \\leq k} |f(s) - f^*(s)| + \\sup_{0 \\leq s \\leq k - M} |g(s) - g^*(s)| \\\\ + \\sup_{0 \\leq s \\leq k - m}|g^*(s + 2^{- (n + 2)}) - g^*(s)| < \\frac{2^{-n}}{4} + \\frac{2^{-n}}{4} + \\frac{2^{-k}}{4} < 2^{-k}$$\n\nIn conclusion, $$\\sup_{0 \\leq s \\leq k} |\\phi(s) - \\phi^*(s)| < 2^{-k}$$\n\nHence, by lemma 6b, $\\rho(\\phi, \\phi^*) < (k + 1)2^{-k} \\leq \\varepsilon$.\n\nQ.E.D.\n\nAUXILIARY FACTS AND DEFINITIONS\n\nDefinition 1 ($\\pi_t$, the canonical projection onto the $t$th coordinate). ([S], section 4.1, p. 40) Setting $I := [0,\\infty)$, denote by $\\pi_t : (\\mathbb{R}^d)\\rightarrow \\mathbb{R}^d$, $\\omega \\mapsto \\omega(t)$ the canonical projection onto the $t$th coordinate.\n\nDefinition 2 (The product $\\sigma$-algebra $\\mathcal{B}^I(\\mathbb{R}^d)$). ([S], section 4.1, p. 40) The natural $\\sigma$-algebra on the infinite product $(\\mathbb{R}^d)^I$ is the product $\\sigma$-algebra $$\\mathcal{B}^I(\\mathbb{R}^d) := \\sigma \\{\\pi_t^{-1}(B) : B \\in \\mathcal{B}(\\mathbb{R}^d), t \\in I\\} = \\sigma \\{\\pi_t : t \\in I\\}$$\n\nDefinition & Proposition 3 (The trace $\\sigma$-algebra). (a: [S2] Example 3.3 (vi), p. 16; b: [H] Theorem E, p. 25) Let $\\Omega$ be some non-empty set, let $\\mathcal{F}$ be a $\\sigma$-algebra over $\\Omega$ and let $S \\subseteq \\Omega$.\n\na) $S \\cap \\mathcal{F} := \\{S \\cap T : T \\in \\mathcal{F}\\}$ (called the trace $\\sigma$-algebra of $S$ w.r.t. $\\mathcal{F}$) is a $\\sigma$-algebra over $S$.\n\nb) If $\\mathcal{E} \\subseteq \\mathcal{F}$ is a generator of $\\mathcal{F}$ (i.e. if $\\mathcal{F} = \\sigma(\\mathcal{E})$), then $S \\cap \\mathcal{F} = \\sigma(S \\cap \\mathcal{E})$.\n\nCorollary 4 ($\\mathbf{C}_{(0)} \\cap \\mathcal{B}^I(\\mathbb{R}^d) = \\sigma(\\pi_t\\mid_{\\mathbb{C}_{(0)}} : t \\in I)$). ([S] section 4.1, p. 41) $$\\mathbf{C}_{(0)} \\cap \\mathcal{B}^I(\\mathbb{R}^d) = \\sigma(\\pi_t\\mid_{\\mathbb{C}_{(0)}} : t \\in I)$$\n\nDefinition & Proposition 5 ($\\rho$, the locally uniform convergence metric). (Based on, but not identical to, [S] section 4.1, p. 41) The function $\\rho: \\mathbf{C}_{(0)} \\rightarrow [0,\\infty)$, $$\\rho(w,v) := \\sum_{n = 1}^\\infty \\left(2^{-n} \\wedge \\sup_{0 \\leq t \\leq n}|w(t) - v(t)|\\right)$$ is a metric on $\\mathbf{C}_{(0)}$.\n\nLemma 6 (Equivalence of $\\rho$ and $\\sup$). Let $n \\in \\mathbb{N}_1$ and let $w,v \\in \\mathbf{C}_{(0)}$.\n\na. If $\\rho(w,v) < 2^{-n}$, $\\sup_{0 \\leq t \\leq n}|w(t) - v(t)| < 2^{-n}$.\n\nb. If $\\sup_{0 \\leq t \\leq n}|w(t) - v(t)| < 2^{-n}$, $\\rho(w,v) < (n + 1) 2^{-n}$.\n\nProof of lemma 6\n\na. Assume that $\\rho(w,v) < 2^{-n}$. Suppose, by way of contradiction, that $\\sup_{0 \\leq t \\leq n}|w(t) - v(t)| \\geq 2^{-n}$. Then $$\\rho(w,v) \\geq 2^{-n} \\wedge \\sup_{0 \\leq t \\leq n} |w(t) - v(t)| = 2^{-n}$$ contrary to the assumption.\n\nb. Assume that $\\sup_{0 \\leq t \\leq n}|w(t) - v(t)| < 2^{-n}$. Then $$\\rho(w, v) \\leq \\sum_{m = 1}^n 2^{-n} + \\sum_{m = n+1}^\\infty 2^{-m} = n2^{-n} + 2^{-n}$$\n\nQ.E.D.\n\nDefinition 7 ($\\mathcal{O}_\\rho$, $\\mathcal{B}(\\mathbf{C}_{(0)})$). ([S] section 4.1, p. 41) Denote by $\\mathcal{O}_\\rho$ the topology induced by $\\rho$ and consider the Borel $\\sigma$-algebra $\\mathcal{B}(\\mathbf{C}_{(0)}) := \\sigma(\\mathcal{O}_\\rho)$ on $\\mathbf{C}_{(0)}$.\n\nProposition 8 ($\\mathbf{C}_{(0)}$ is a Polish space). ([S] section 4.1, p. 41) Equipped with the metric $\\rho$ defined above, $\\mathbf{C}_{(0)}$ becomes a Polish space, i.e. a complete, separable metric space.\n\nLemma 9 ($\\mathbf{C}_{(0)}\\cap\\mathcal{B}^I(\\mathbb{R}^d) = \\mathcal{B}(\\mathbf{C}_{(0)})$). ([S] Lemma 4.1, p. 41) We have $$\\mathbf{C}_{(0)}\\cap\\mathcal{B}^I(\\mathbb{R}^d) = \\mathcal{B}(\\mathbf{C}_{(0)})$$\n\nProposition 10 (A Brownian motion $\\omega \\mapsto (t \\mapsto B(\\omega, t))$ is $\\mathcal{F}/\\mathbf{C}_{(0)}$-measurable). ([S] top paragraph on p. 42) A standard, $d$-dimensional Brownian motion $(B(t))_{t \\geq 0}$ over the probability space $(\\Omega, \\mathcal{F}, P)$, when considered as a function $\\in \\Omega \\mapsto \\mathbf{C}_{(0)}$, is $\\mathcal{F}/\\mathcal{B}(\\mathbf{C}_{(0)})$-measurable.\n\nDefinition 11 (Box topology). ([M] p. 114) Let $\\{X_\\alpha\\_{\\alpha \\in J}$ be an indexed family of topological spaces. Let us take as a bases for a topology on the product space $\\prod_{\\alpha \\in J}X_\\alpha$ the collection of all sets of the form $\\prod_{\\alpha \\in J}U_\\alpha$ where $U_\\alpha$ is open in $X_\\alpha$, for each $\\alpha \\in J$. The topology generated by this basis is called the box topology.\n\nDefinition 12 (Product topology, product space). ([M] p. 114) Let $\\mathcal{S}_\\beta$ denote the collection $$\\mathcal{S}_\\beta := \\{\\pi_\\beta^{-1}(U_\\beta) \\mid U_\\beta \\textrm{ open in } X_\\beta \\}$$ and let $\\mathcal{S}$ denote the union of these collections, $\\mathcal{S} := \\bigcup_{\\beta \\in J}\\mathcal{S}_\\beta$. The topology generated by the sub basis $\\mathcal{S}$ is called the product topology. In this topology, $\\prod_{\\alpha \\in J} X_\\alpha$ is called a product space.\n\nTheorem 13 (Comparison of the box and product topologies). ([M] Theorem 19.1, p. 115) The box topology on $\\prod X_\\alpha$ has as basis all sets of the form $\\prod U_\\alpha$, where $U_\\alpha$ is open in $X_\\alpha$ for each $\\alpha$. The product topology on $\\prod X_\\alpha$ has as basis all sets of the form $\\prod U_\\alpha$, where $U_\\alpha$ is open in $X_\\alpha$ for each $\\alpha$ and $U_\\alpha$ equals $X_\\alpha$ except for finitely many values of $\\alpha$.\n\nCorollary 14 (Equality of the box and product topologies for finite products). ([M] p. 115) For finite products $\\prod_{\\alpha = 1}^n X_\\alpha$, the two topologies are precisely the same.\n\nLemma 15 (Product and Borel $\\sigma$-fields). ([K] Lemma 1.2, p. 3) If $S_1, S_2, \\dots$ are separable metric spaces with induced topologies $\\mathcal{T}_1, \\mathcal{T}_2, \\dots$, respectively, and if $\\mathcal{T}$ is the product topology on $S_1 \\times S_2 \\times \\cdots$, then $$\\sigma(\\mathcal{T}) = \\sigma(\\mathcal{T}_1) \\otimes \\sigma(\\mathcal{T}_2) \\otimes \\cdots$$\n\nLemma 16 (Continuity and measurability). ([K] Lemma 1.5, p. 4) Let $f$ be a continuous mapping between two topological spaces $S$ and $T$ with Borel $\\sigma$-fields $\\mathcal{S}$ and $\\mathcal{T}$. Then $f$ is $\\mathcal{S}/\\mathcal{T}$-measurable.\n\nWORKS CITED\n\n[H] Halmos, Paul R. Measure Theory. Springer-Verlag, 1974\n\n[K] Kallenberg, Olav. Foundations of Modern Probability. 2nd edition. Springer, 2002\n\n[M] Munkres, James R. Topology. 2nd edition. Prentice Hall, 2000\n\n[S] Schilling, René L. and Partzsch, Lothar. Brownian motion - an introduction to stochastic processes. De Gruyter, 2012\n\n[S2] Schilling, René L. Measures, integrals and martingales. Cambridge University Press, 2011" ]

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[ "A Graphical Dice Simulator\n\n31May - by Alan - 0 -\n\nThis PyGame program simulates the roll of a die (or dice if you prefer). It’s a fairly simple, straightforward thing to do, though it may be worth noting that the spots display is oh-so-slightly clever. Some spots appear in more than one number, e.g. the middle spot is in all the odd numbers, hence the test if n % 2 == 1: serves to display the middle spot for 3 numbers. By the time we get down to n==6 we only have 2 more spots to display. It’s an example of DRY 🙂\n\nOf course, this isn’t the simplest possible program that could simulate the roll of a dice; this is:\n\nimport random\nprint(random.randint(1,6))\n\nHowever, this is a bit dull, hence we spruce it up with a graphical interface, and just to add a bit more realism, we let the dice ‘roll’ a little while. It displays a dozen faces with a small delay between each, and stops on the last. Without the delay you won’t see the faces before the last.\n\n'''\nDiceSimulator.py\nI don't know if I should say die or dice :()\nAuthor: Alan Richmond, Python3.codes\n'''\nimport pygame, random, time\n\nsize = 256 # Size of window/dice\nspsz = size//10 # size of spots\nm = int(size/2) # mid-point of dice (or die?)\nl=t=int(size/4) # location of left and top spots\nr=b=size-l # location of right and bottom spots\nrolling = 12 # times that dice rolls before stopping\ndiecol = (255,255,127) # die colour\nspotcol = (0,127,127) # spot colour\n\nd = pygame.display.set_mode((size, size))\nd.fill(diecol)\npygame.display.set_caption(\"Dice Simulator\")\n\nfor i in range(rolling): # roll the die...\nn=random.randint(1,6) # random number between 1 &amp;amp; 6\nd.fill(diecol) # clear previous spots\nif n % 2 == 1:\npygame.draw.circle(d,spotcol,(m,m),spsz) # middle spot\nif n==2 or n==3 or n==4 or n==5 or n==6:\npygame.draw.circle(d,spotcol,(l,b),spsz) # left bottom\npygame.draw.circle(d,spotcol,(r,t),spsz) # right top\nif n==4 or n==5 or n==6:\npygame.draw.circle(d,spotcol,(l,t),spsz) # left top\npygame.draw.circle(d,spotcol,(r,b),spsz) # right bottom\nif n==6:\npygame.draw.circle(d,spotcol,(m,b),spsz) # middle bottom\npygame.draw.circle(d,spotcol,(m,t),spsz) # middle top\n\npygame.display.flip()\ntime.sleep(0.2)" ]

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[ "# Sine of an Angle\n\n## Sine of an Angle\n\nThe sine of an angle is the y-coordinate of the corresponding point P on the unit circle (which has a radius of 1).", null, "When any real number is associated with the sine of an angle whose measure in the unit circle is that number, a circular function called a sine function is defined.", null, "Graphical representation of the function $$f(x)= sin(x)$$, which has an amplitude of 1 and a period of 2π.\n\n### Educational note\n\nThe argument of sine is a number (that is, the measure of an angle or an angle of rotation), not the geometric figure itself (angle). Therefore, the expression “sine of an angle” is a shortened version of “the sine of the measure of an angle”.\n\n### Etymologycal note\n\nThe term “sine” is derived from the Latin sinus meaning “bending”, “curve”, or “hollow”. In human anatomy, the term sinus refers to various cavities such as the carotid sinus, facial sinuses, venous sinuses and pilonidal sinuses (wounds). The term “sine” was introduced into mathematics through transcriptions from one language into another. Although the study of trigonometric ratios in right triangles or circles goes back to ancient Greece (around 400 BCE), the most important work on the subject was done by the Indian mathematician Ăryabhata who used the term jya, meaning “chord”, to refer to the half-chord that subtends an arc. The term was adopted and translated by Arab mathematicians as jb (pronounced ja-ib) which means “chord”, but also “breast”. European mathematicians, in turn, translated the Arabic texts into Latin, and the word jb became sinus, which means “curve” or “hollow”." ]

[ null, "https://media.scolab.com/Illustrations/Lexique/sinus_angle.svg", null, "https://media.scolab.com/Illustrations/Lexique/sinusoid(en).svg", null ]

[ "Cody\n\n# Problem 2016. Area of an equilateral triangle\n\nSolution 450222\n\nSubmitted on 5 Jun 2014 by Mohapi Mohlamonyane\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\n%% x = 1; y_correct = sqrt(3)/4; tolerance = 1e-12; assert(abs(equilateral_area(x)-y_correct)<tolerance)\n\n2   Pass\n%% x = 2; y_correct = sqrt(3); tolerance = 1e-12; assert(abs(equilateral_area(x)-y_correct)<tolerance)\n\n3   Pass\n%% x = 3; y_correct = sqrt(3)*9/4; tolerance =1e-12; assert(abs(equilateral_area(x)-y_correct)<tolerance)" ]

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[ "# How to show sum of two elements is decomposable.\n\nLet $V$ be a vector space over the field $F.$ Let $\\{v_1,\\ldots,v_4\\}$ be linearly dependent set. Then how do I show that $w=v_1 \\wedge v_2 + v_3 \\wedge v_4$ is decomposable ?\n\nNote that to show $w$ is decomposable we have to show that $w=x \\wedge y$ for some $x,y \\in V.$ What I tried: Atleast one of the $v_i$ can be written as linear combination of the others, so wlog let $v_1=c_2v_3 + c_3 v_3 + c_4 v_4$ then $w= v_3\\wedge (c_3v_2+v_4)+v_4\\wedge c_4 v_2.$ I think I need some other trick. Please help. Thanks.\n\nBy your work, we have $$w = c_3 v_2 \\wedge v_3 + c_2 v_4 \\wedge v_2 + 1 \\ v_3 \\wedge v_4$$ Now, it suffices to find two vectors $u_1,u_2 \\in \\Bbb F^3$ whose \"cross-product\" is $$u_1 \\times u_2 = (c_3,c_2,1)$$ for instance, we can take $u_1 = (1,0,-c_3)$ and $u_2 = (0,1,-c_2).$ With that in mind, try computing $$(v_4 - c_3 v_2) \\wedge (v_3 - c_2 v_2)$$" ]

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[ "", null, "# Faster R-CNN: Towards Real-Time Object Detection with Region Proposal Networks\n\nState-of-the-art object detection networks depend on region proposal algorithms to hypothesize object locations. Advances like SPPnet and Fast R-CNN have reduced the running time of these detection networks, exposing region proposal computation as a bottleneck. In this work, we introduce a Region Proposal Network (RPN) that shares full-image convolutional features with the detection network, thus enabling nearly cost-free region proposals. An RPN is a fully convolutional network that simultaneously predicts object bounds and objectness scores at each position. The RPN is trained end-to-end to generate high-quality region proposals, which are used by Fast R-CNN for detection. We further merge RPN and Fast R-CNN into a single network by sharing their convolutional features---using the recently popular terminology of neural networks with 'attention' mechanisms, the RPN component tells the unified network where to look. For the very deep VGG-16 model, our detection system has a frame rate of 5fps (including all steps) on a GPU, while achieving state-of-the-art object detection accuracy on PASCAL VOC 2007, 2012, and MS COCO datasets with only 300 proposals per image. In ILSVRC and COCO 2015 competitions, Faster R-CNN and RPN are the foundations of the 1st-place winning entries in several tracks. Code has been made publicly available.\n\n## Code Repositories\n\nFaster R-CNN\n\n### glossary\n\nhttps://machinelearning.wtf/ - An online glossary of machine learning terms.\n\n### awesome_segmentation\n\nMethods of semantic segmentation with links to implementations and papers\n\n### faster_rcnn\n\nFaster RCNN pytorch\n\n### Summer-2017\n\nImage classification by Keras/CNN for bird species and bird positions detection (By FRCNN) in the given Images.\n\n##### This week in AI\n\nGet the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.\n\n## 1 Introduction\n\nRecent advances in object detection are driven by the success of region proposal methods (e.g., ) and region-based convolutional neural networks (R-CNNs) . Although region-based CNNs were computationally expensive as originally developed in , their cost has been drastically reduced thanks to sharing convolutions across proposals [1, 2]. The latest incarnation, Fast R-CNN , achieves near real-time rates using very deep networks , when ignoring the time spent on region proposals. Now, proposals are the test-time computational bottleneck in state-of-the-art detection systems.\n\nRegion proposal methods typically rely on inexpensive features and economical inference schemes. Selective Search , one of the most popular methods, greedily merges superpixels based on engineered low-level features. Yet when compared to efficient detection networks , Selective Search is an order of magnitude slower, at 2 seconds per image in a CPU implementation. EdgeBoxes currently provides the best tradeoff between proposal quality and speed, at 0.2 seconds per image. Nevertheless, the region proposal step still consumes as much running time as the detection network.\n\nOne may note that fast region-based CNNs take advantage of GPUs, while the region proposal methods used in research are implemented on the CPU, making such runtime comparisons inequitable. An obvious way to accelerate proposal computation is to re-implement it for the GPU. This may be an effective engineering solution, but re-implementation ignores the down-stream detection network and therefore misses important opportunities for sharing computation.", null, "Figure 1: Different schemes for addressing multiple scales and sizes. (a) Pyramids of images and feature maps are built, and the classifier is run at all scales. (b) Pyramids of filters with multiple scales/sizes are run on the feature map. (c) We use pyramids of reference boxes in the regression functions.\n\nIn this paper, we show that an algorithmic change—computing proposals with a deep convolutional neural network—leads to an elegant and effective solution where proposal computation is nearly cost-free given the detection network’s computation. To this end, we introduce novel Region Proposal Networks (RPNs) that share convolutional layers with state-of-the-art object detection networks [1, 2]. By sharing convolutions at test-time, the marginal cost for computing proposals is small (e.g., 10ms per image).\n\nOur observation is that the convolutional feature maps used by region-based detectors, like Fast R-CNN, can also be used for generating region proposals. On top of these convolutional features, we construct an RPN by adding a few additional convolutional layers that simultaneously regress region bounds and objectness scores at each location on a regular grid. The RPN is thus a kind of fully convolutional network (FCN) and can be trained end-to-end specifically for the task for generating detection proposals.\n\nRPNs are designed to efficiently predict region proposals with a wide range of scales and aspect ratios. In contrast to prevalent methods [8, 9, 1, 2] that use pyramids of images (Figure 1, a) or pyramids of filters (Figure 1, b), we introduce novel “anchor” boxes that serve as references at multiple scales and aspect ratios. Our scheme can be thought of as a pyramid of regression references (Figure 1, c), which avoids enumerating images or filters of multiple scales or aspect ratios. This model performs well when trained and tested using single-scale images and thus benefits running speed.\n\nTo unify RPNs with Fast R-CNN object detection networks, we propose a training scheme that alternates between fine-tuning for the region proposal task and then fine-tuning for object detection, while keeping the proposals fixed. This scheme converges quickly and produces a unified network with convolutional features that are shared between both tasks.111Since the publication of the conference version of this paper , we have also found that RPNs can be trained jointly with Fast R-CNN networks leading to less training time.\n\nWe comprehensively evaluate our method on the PASCAL VOC detection benchmarks where RPNs with Fast R-CNNs produce detection accuracy better than the strong baseline of Selective Search with Fast R-CNNs. Meanwhile, our method waives nearly all computational burdens of Selective Search at test-time—the effective running time for proposals is just 10 milliseconds. Using the expensive very deep models of , our detection method still has a frame rate of 5fps (including all steps) on a GPU, and thus is a practical object detection system in terms of both speed and accuracy. We also report results on the MS COCO dataset and investigate the improvements on PASCAL VOC using the COCO data. Code has been made publicly available at https://github.com/shaoqingren/faster_rcnn (in MATLAB) and https://github.com/rbgirshick/py-faster-rcnn (in Python).\n\nA preliminary version of this manuscript was published previously . Since then, the frameworks of RPN and Faster R-CNN have been adopted and generalized to other methods, such as 3D object detection , part-based detection , instance segmentation , and image captioning . Our fast and effective object detection system has also been built in commercial systems such as at Pinterests , with user engagement improvements reported.\n\nIn ILSVRC and COCO 2015 competitions, Faster R-CNN and RPN are the basis of several 1st-place entries \n\nin the tracks of ImageNet detection, ImageNet localization, COCO detection, and COCO segmentation. RPNs completely learn to propose regions from data, and thus can easily benefit from deeper and more expressive features (such as the 101-layer residual nets adopted in\n\n). Faster R-CNN and RPN are also used by several other leading entries in these competitions. These results suggest that our method is not only a cost-efficient solution for practical usage, but also an effective way of improving object detection accuracy.\n\n## 2 Related Work\n\nObject Proposals. There is a large literature on object proposal methods. Comprehensive surveys and comparisons of object proposal methods can be found in [19, 20, 21]. Widely used object proposal methods include those based on grouping super-pixels (e.g., Selective Search , CPMC , MCG ) and those based on sliding windows (e.g., objectness in windows , EdgeBoxes ). Object proposal methods were adopted as external modules independent of the detectors (e.g., Selective Search object detectors, R-CNN , and Fast R-CNN ).\n\nDeep Networks for Object Detection. The R-CNN method trains CNNs end-to-end to classify the proposal regions into object categories or background. R-CNN mainly plays as a classifier, and it does not predict object bounds (except for refining by bounding box regression). Its accuracy depends on the performance of the region proposal module (see comparisons in ). Several papers have proposed ways of using deep networks for predicting object bounding boxes [25, 9, 26, 27]. In the OverFeat method , a fully-connected layer is trained to predict the box coordinates for the localization task that assumes a single object. The fully-connected layer is then turned into a convolutional layer for detecting multiple class-specific objects. The MultiBox methods [26, 27] generate region proposals from a network whose last fully-connected layer simultaneously predicts multiple class-agnostic boxes, generalizing the “single-box” fashion of OverFeat. These class-agnostic boxes are used as proposals for R-CNN . The MultiBox proposal network is applied on a single image crop or multiple large image crops (e.g., 224\n\n224), in contrast to our fully convolutional scheme. MultiBox does not share features between the proposal and detection networks. We discuss OverFeat and MultiBox in more depth later in context with our method. Concurrent with our work, the DeepMask method\n\n is developed for learning segmentation proposals.\n\nShared computation of convolutions [9, 1, 29, 7, 2] has been attracting increasing attention for efficient, yet accurate, visual recognition. The OverFeat paper computes convolutional features from an image pyramid for classification, localization, and detection. Adaptively-sized pooling (SPP) on shared convolutional feature maps is developed for efficient region-based object detection [1, 30] and semantic segmentation . Fast R-CNN enables end-to-end detector training on shared convolutional features and shows compelling accuracy and speed.", null, "Figure 2: Faster R-CNN is a single, unified network for object detection. The RPN module serves as the ‘attention’ of this unified network.", null, "Figure 3: Left: Region Proposal Network (RPN). Right: Example detections using RPN proposals on PASCAL VOC 2007 test. Our method detects objects in a wide range of scales and aspect ratios.\n\n## 3 Faster R-CNN\n\nOur object detection system, called Faster R-CNN, is composed of two modules. The first module is a deep fully convolutional network that proposes regions, and the second module is the Fast R-CNN detector that uses the proposed regions. The entire system is a single, unified network for object detection (Figure 2). Using the recently popular terminology of neural networks with ‘attention’ mechanisms, the RPN module tells the Fast R-CNN module where to look. In Section 3.1 we introduce the designs and properties of the network for region proposal. In Section 3.2 we develop algorithms for training both modules with features shared.\n\n### 3.1 Region Proposal Networks\n\nA Region Proposal Network (RPN) takes an image (of any size) as input and outputs a set of rectangular object proposals, each with an objectness score.333“Region” is a generic term and in this paper we only consider rectangular regions, as is common for many methods (e.g., [27, 4, 6]). “Objectness” measures membership to a set of object classes vs. background. We model this process with a fully convolutional network , which we describe in this section. Because our ultimate goal is to share computation with a Fast R-CNN object detection network , we assume that both nets share a common set of convolutional layers. In our experiments, we investigate the Zeiler and Fergus model (ZF), which has 5 shareable convolutional layers and the Simonyan and Zisserman model (VGG-16), which has 13 shareable convolutional layers.\n\nTo generate region proposals, we slide a small network over the convolutional feature map output by the last shared convolutional layer. This small network takes as input an\n\nspatial window of the input convolutional feature map. Each sliding window is mapped to a lower-dimensional feature (256-d for ZF and 512-d for VGG, with ReLU\n\n following). This feature is fed into two sibling fully-connected layers—a box-regression layer (reg) and a box-classification layer (cls). We use in this paper, noting that the effective receptive field on the input image is large (171 and 228 pixels for ZF and VGG, respectively). This mini-network is illustrated at a single position in Figure 3 (left). Note that because the mini-network operates in a sliding-window fashion, the fully-connected layers are shared across all spatial locations. This architecture is naturally implemented with an convolutional layer followed by two sibling convolutional layers (for reg and cls, respectively).\n\n#### 3.1.1 Anchors\n\nAt each sliding-window location, we simultaneously predict multiple region proposals, where the number of maximum possible proposals for each location is denoted as . So the reg layer has outputs encoding the coordinates of boxes, and the cls layer outputs\n\nscores that estimate probability of object or not object for each proposal\n\n444For simplicity we implement the cls\n\nlayer as a two-class softmax layer. Alternatively, one may use logistic regression to produce\n\nscores.\n. The proposals are parameterized relative to reference boxes, which we call anchors. An anchor is centered at the sliding window in question, and is associated with a scale and aspect ratio (Figure 3, left). By default we use 3 scales and 3 aspect ratios, yielding anchors at each sliding position. For a convolutional feature map of a size (typically 2,400), there are anchors in total.\n\nTranslation-Invariant Anchors\n\nAn important property of our approach is that it is translation invariant, both in terms of the anchors and the functions that compute proposals relative to the anchors. If one translates an object in an image, the proposal should translate and the same function should be able to predict the proposal in either location. This translation-invariant property is guaranteed by our method555As is the case of FCNs \n\n, our network is translation invariant up to the network’s total stride.\n\n. As a comparison, the MultiBox method \n\nuses k-means to generate 800 anchors, which are\n\nnot translation invariant. So MultiBox does not guarantee that the same proposal is generated if an object is translated.\n\nThe translation-invariant property also reduces the model size. MultiBox has a -dimensional fully-connected output layer, whereas our method has a -dimensional convolutional output layer in the case of anchors. As a result, our output layer has parameters ( for VGG-16), two orders of magnitude fewer than MultiBox’s output layer that has parameters ( for GoogleNet in MultiBox ). If considering the feature projection layers, our proposal layers still have an order of magnitude fewer parameters than MultiBox666Considering the feature projection layers, our proposal layers’ parameter count is ; MultiBox’s proposal layers’ parameter count is .. We expect our method to have less risk of overfitting on small datasets, like PASCAL VOC.\n\nMulti-Scale Anchors as Regression References\n\nOur design of anchors presents a novel scheme for addressing multiple scales (and aspect ratios). As shown in Figure 1, there have been two popular ways for multi-scale predictions. The first way is based on image/feature pyramids, e.g., in DPM and CNN-based methods [9, 1, 2]. The images are resized at multiple scales, and feature maps (HOG or deep convolutional features [9, 1, 2]) are computed for each scale (Figure 1(a)). This way is often useful but is time-consuming. The second way is to use sliding windows of multiple scales (and/or aspect ratios) on the feature maps. For example, in DPM , models of different aspect ratios are trained separately using different filter sizes (such as 57 and 75). If this way is used to address multiple scales, it can be thought of as a “pyramid of filters” (Figure 1(b)). The second way is usually adopted jointly with the first way .\n\nAs a comparison, our anchor-based method is built on a pyramid of anchors, which is more cost-efficient. Our method classifies and regresses bounding boxes with reference to anchor boxes of multiple scales and aspect ratios. It only relies on images and feature maps of a single scale, and uses filters (sliding windows on the feature map) of a single size. We show by experiments the effects of this scheme for addressing multiple scales and sizes (Table VIII).\n\nBecause of this multi-scale design based on anchors, we can simply use the convolutional features computed on a single-scale image, as is also done by the Fast R-CNN detector . The design of multi-scale anchors is a key component for sharing features without extra cost for addressing scales.\n\n#### 3.1.2 Loss Function\n\nFor training RPNs, we assign a binary class label (of being an object or not) to each anchor. We assign a positive label to two kinds of anchors: (i) the anchor/anchors with the highest Intersection-over-Union (IoU) overlap with a ground-truth box, or (ii) an anchor that has an IoU overlap higher than 0.7 with any ground-truth box. Note that a single ground-truth box may assign positive labels to multiple anchors. Usually the second condition is sufficient to determine the positive samples; but we still adopt the first condition for the reason that in some rare cases the second condition may find no positive sample. We assign a negative label to a non-positive anchor if its IoU ratio is lower than 0.3 for all ground-truth boxes. Anchors that are neither positive nor negative do not contribute to the training objective.\n\nWith these definitions, we minimize an objective function following the multi-task loss in Fast R-CNN \n\n. Our loss function for an image is defined as:\n\n L({pi},{ti})=1Ncls∑iLcls(pi,p∗i) (1) +λ1Nreg∑ip∗iLreg(ti,t∗i).\n\nHere, is the index of an anchor in a mini-batch and is the predicted probability of anchor being an object. The ground-truth label is 1 if the anchor is positive, and is 0 if the anchor is negative.\n\nis a vector representing the 4 parameterized coordinates of the predicted bounding box, and\n\nis that of the ground-truth box associated with a positive anchor. The classification loss is log loss over two classes (object vs. not object). For the regression loss, we use where is the robust loss function (smooth L) defined in . The term means the regression loss is activated only for positive anchors () and is disabled otherwise (). The outputs of the cls and reg layers consist of and respectively.\n\nThe two terms are normalized by and and weighted by a balancing parameter . In our current implementation (as in the released code), the term in Eqn.(1) is normalized by the mini-batch size (i.e., ) and the term is normalized by the number of anchor locations (i.e., ). By default we set , and thus both cls and reg terms are roughly equally weighted. We show by experiments that the results are insensitive to the values of in a wide range (Table IX). We also note that the normalization as above is not required and could be simplified.\n\nFor bounding box regression, we adopt the parameterizations of the 4 coordinates following :\n\n tx =(x−xa)/wa,ty=(y−ya)/ha, (2) tw =log(w/wa),th=log(h/ha), t∗x =(x∗−xa)/wa,t∗y=(y∗−ya)/ha, t∗w =log(w∗/wa),t∗h=log(h∗/ha),\n\nwhere , , , and denote the box’s center coordinates and its width and height. Variables , , and are for the predicted box, anchor box, and ground-truth box respectively (likewise for ). This can be thought of as bounding-box regression from an anchor box to a nearby ground-truth box.\n\nNevertheless, our method achieves bounding-box regression by a different manner from previous RoI-based (Region of Interest) methods [1, 2]. In [1, 2], bounding-box regression is performed on features pooled from arbitrarily sized RoIs, and the regression weights are shared by all region sizes. In our formulation, the features used for regression are of the same spatial size () on the feature maps. To account for varying sizes, a set of bounding-box regressors are learned. Each regressor is responsible for one scale and one aspect ratio, and the regressors do not share weights. As such, it is still possible to predict boxes of various sizes even though the features are of a fixed size/scale, thanks to the design of anchors.\n\n#### 3.1.3 Training RPNs\n\nThe RPN can be trained end-to-end by back-propagation and stochastic gradient descent (SGD)\n\n. We follow the “image-centric” sampling strategy from to train this network. Each mini-batch arises from a single image that contains many positive and negative example anchors. It is possible to optimize for the loss functions of all anchors, but this will bias towards negative samples as they are dominate. Instead, we randomly sample 256 anchors in an image to compute the loss function of a mini-batch, where the sampled positive and negative anchors have a ratio of up to\n\n1:1. If there are fewer than 128 positive samples in an image, we pad the mini-batch with negative ones.\n\nWe randomly initialize all new layers by drawing weights from a zero-mean Gaussian distribution with standard deviation 0.01. All other layers (\n\ni.e., the shared convolutional layers) are initialized by pre-training a model for ImageNet classification , as is standard practice . We tune all layers of the ZF net, and conv3 and up for the VGG net to conserve memory . We use a learning rate of 0.001 for 60k mini-batches, and 0.0001 for the next 20k mini-batches on the PASCAL VOC dataset. We use a momentum of 0.9 and a weight decay of 0.0005 \n\n. Our implementation uses Caffe\n\n.\n\n### 3.2 Sharing Features for RPN and Fast R-CNN\n\nThus far we have described how to train a network for region proposal generation, without considering the region-based object detection CNN that will utilize these proposals. For the detection network, we adopt Fast R-CNN . Next we describe algorithms that learn a unified network composed of RPN and Fast R-CNN with shared convolutional layers (Figure 2).\n\nBoth RPN and Fast R-CNN, trained independently, will modify their convolutional layers in different ways. We therefore need to develop a technique that allows for sharing convolutional layers between the two networks, rather than learning two separate networks. We discuss three ways for training networks with features shared:\n\n(i) Alternating training. In this solution, we first train RPN, and use the proposals to train Fast R-CNN. The network tuned by Fast R-CNN is then used to initialize RPN, and this process is iterated. This is the solution that is used in all experiments in this paper.\n\n(ii) Approximate joint training. In this solution, the RPN and Fast R-CNN networks are merged into one network during training as in Figure 2. In each SGD iteration, the forward pass generates region proposals which are treated just like fixed, pre-computed proposals when training a Fast R-CNN detector. The backward propagation takes place as usual, where for the shared layers the backward propagated signals from both the RPN loss and the Fast R-CNN loss are combined. This solution is easy to implement. But this solution ignores the derivative w.r.t. the proposal boxes’ coordinates that are also network responses, so is approximate. In our experiments, we have empirically found this solver produces close results, yet reduces the training time by about 25-50% comparing with alternating training. This solver is included in our released Python code.\n\n(iii) Non-approximate joint training. As discussed above, the bounding boxes predicted by RPN are also functions of the input. The RoI pooling layer \n\nin Fast R-CNN accepts the convolutional features and also the predicted bounding boxes as input, so a theoretically valid backpropagation solver should also involve gradients w.r.t. the box coordinates. These gradients are ignored in the above approximate joint training. In a non-approximate joint training solution, we need an RoI pooling layer that is differentiable w.r.t. the box coordinates. This is a nontrivial problem and a solution can be given by an “RoI warping” layer as developed in\n\n, which is beyond the scope of this paper.\n\n4-Step Alternating Training. In this paper, we adopt a pragmatic 4-step training algorithm to learn shared features via alternating optimization. In the first step, we train the RPN as described in Section 3.1.3. This network is initialized with an ImageNet-pre-trained model and fine-tuned end-to-end for the region proposal task. In the second step, we train a separate detection network by Fast R-CNN using the proposals generated by the step-1 RPN. This detection network is also initialized by the ImageNet-pre-trained model. At this point the two networks do not share convolutional layers. In the third step, we use the detector network to initialize RPN training, but we fix the shared convolutional layers and only fine-tune the layers unique to RPN. Now the two networks share convolutional layers. Finally, keeping the shared convolutional layers fixed, we fine-tune the unique layers of Fast R-CNN. As such, both networks share the same convolutional layers and form a unified network. A similar alternating training can be run for more iterations, but we have observed negligible improvements.\n\n### 3.3 Implementation Details\n\nWe train and test both region proposal and object detection networks on images of a single scale [1, 2]. We re-scale the images such that their shorter side is pixels \n\n. Multi-scale feature extraction (using an image pyramid) may improve accuracy but does not exhibit a good speed-accuracy trade-off\n\n. On the re-scaled images, the total stride for both ZF and VGG nets on the last convolutional layer is 16 pixels, and thus is 10 pixels on a typical PASCAL image before resizing (500375). Even such a large stride provides good results, though accuracy may be further improved with a smaller stride.\n\nFor anchors, we use 3 scales with box areas of , , and pixels, and 3 aspect ratios of 1:1, 1:2, and 2:1. These hyper-parameters are not carefully chosen for a particular dataset, and we provide ablation experiments on their effects in the next section. As discussed, our solution does not need an image pyramid or filter pyramid to predict regions of multiple scales, saving considerable running time. Figure 3 (right) shows the capability of our method for a wide range of scales and aspect ratios. Table I shows the learned average proposal size for each anchor using the ZF net. We note that our algorithm allows predictions that are larger than the underlying receptive field. Such predictions are not impossible—one may still roughly infer the extent of an object if only the middle of the object is visible.\n\nThe anchor boxes that cross image boundaries need to be handled with care. During training, we ignore all cross-boundary anchors so they do not contribute to the loss. For a typical image, there will be roughly 20000 (\n\n) anchors in total. With the cross-boundary anchors ignored, there are about 6000 anchors per image for training. If the boundary-crossing outliers are not ignored in training, they introduce large, difficult to correct error terms in the objective, and training does not converge. During testing, however, we still apply the fully convolutional RPN to the entire image. This may generate cross-boundary proposal boxes, which we clip to the image boundary.\n\nSome RPN proposals highly overlap with each other. To reduce redundancy, we adopt non-maximum suppression (NMS) on the proposal regions based on their cls scores. We fix the IoU threshold for NMS at 0.7, which leaves us about 2000 proposal regions per image. As we will show, NMS does not harm the ultimate detection accuracy, but substantially reduces the number of proposals. After NMS, we use the top- ranked proposal regions for detection. In the following, we train Fast R-CNN using 2000 RPN proposals, but evaluate different numbers of proposals at test-time.\n\n## 4 Experiments\n\n### 4.1 Experiments on PASCAL VOC\n\nWe comprehensively evaluate our method on the PASCAL VOC 2007 detection benchmark . This dataset consists of about 5k trainval images and 5k test images over 20 object categories. We also provide results on the PASCAL VOC 2012 benchmark for a few models. For the ImageNet pre-trained network, we use the “fast” version of ZF net that has 5 convolutional layers and 3 fully-connected layers, and the public VGG-16 model that has 13 convolutional layers and 3 fully-connected layers. We primarily evaluate detection mean Average Precision (mAP), because this is the actual metric for object detection (rather than focusing on object proposal proxy metrics).\n\nTable II (top) shows Fast R-CNN results when trained and tested using various region proposal methods. These results use the ZF net. For Selective Search (SS) , we generate about 2000 proposals by the “fast” mode. For EdgeBoxes (EB) , we generate the proposals by the default EB setting tuned for 0.7 IoU. SS has an mAP of 58.7% and EB has an mAP of 58.6% under the Fast R-CNN framework. RPN with Fast R-CNN achieves competitive results, with an mAP of 59.9% while using up to 300 proposals888For RPN, the number of proposals (e.g., 300) is the maximum number for an image. RPN may produce fewer proposals after NMS, and thus the average number of proposals is smaller.. Using RPN yields a much faster detection system than using either SS or EB because of shared convolutional computations; the fewer proposals also reduce the region-wise fully-connected layers’ cost (Table V).\n\nAblation Experiments on RPN. To investigate the behavior of RPNs as a proposal method, we conducted several ablation studies. First, we show the effect of sharing convolutional layers between the RPN and Fast R-CNN detection network. To do this, we stop after the second step in the 4-step training process. Using separate networks reduces the result slightly to 58.7% (RPN+ZF, unshared, Table II). We observe that this is because in the third step when the detector-tuned features are used to fine-tune the RPN, the proposal quality is improved.\n\nNext, we disentangle the RPN’s influence on training the Fast R-CNN detection network. For this purpose, we train a Fast R-CNN model by using the 2000 SS proposals and ZF net. We fix this detector and evaluate the detection mAP by changing the proposal regions used at test-time. In these ablation experiments, the RPN does not share features with the detector.\n\nReplacing SS with 300 RPN proposals at test-time leads to an mAP of 56.8%. The loss in mAP is because of the inconsistency between the training/testing proposals. This result serves as the baseline for the following comparisons.\n\nSomewhat surprisingly, the RPN still leads to a competitive result (55.1%) when using the top-ranked 100 proposals at test-time, indicating that the top-ranked RPN proposals are accurate. On the other extreme, using the top-ranked 6000 RPN proposals (without NMS) has a comparable mAP (55.2%), suggesting NMS does not harm the detection mAP and may reduce false alarms.\n\nNext, we separately investigate the roles of RPN’s cls and reg outputs by turning off either of them at test-time. When the cls layer is removed at test-time (thus no NMS/ranking is used), we randomly sample proposals from the unscored regions. The mAP is nearly unchanged with (55.8%), but degrades considerably to 44.6% when . This shows that the cls scores account for the accuracy of the highest ranked proposals.\n\nOn the other hand, when the reg layer is removed at test-time (so the proposals become anchor boxes), the mAP drops to 52.1%. This suggests that the high-quality proposals are mainly due to the regressed box bounds. The anchor boxes, though having multiple scales and aspect ratios, are not sufficient for accurate detection.\n\nWe also evaluate the effects of more powerful networks on the proposal quality of RPN alone. We use VGG-16 to train the RPN, and still use the above detector of SS+ZF. The mAP improves from 56.8% (using RPN+ZF) to 59.2% (using RPN+VGG). This is a promising result, because it suggests that the proposal quality of RPN+VGG is better than that of RPN+ZF. Because proposals of RPN+ZF are competitive with SS (both are 58.7% when consistently used for training and testing), we may expect RPN+VGG to be better than SS. The following experiments justify this hypothesis.\n\nPerformance of VGG-16. Table III shows the results of VGG-16 for both proposal and detection. Using RPN+VGG, the result is 68.5% for unshared features, slightly higher than the SS baseline. As shown above, this is because the proposals generated by RPN+VGG are more accurate than SS. Unlike SS that is pre-defined, the RPN is actively trained and benefits from better networks. For the feature-shared variant, the result is 69.9%—better than the strong SS baseline, yet with nearly cost-free proposals. We further train the RPN and detection network on the union set of PASCAL VOC 2007 trainval and 2012 trainval. The mAP is 73.2%. Figure 5 shows some results on the PASCAL VOC 2007 test set. On the PASCAL VOC 2012 test set (Table IV), our method has an mAP of 70.4% trained on the union set of VOC 2007 trainval+test and VOC 2012 trainval. Table VI and Table VII show the detailed numbers.\n\nIn Table V we summarize the running time of the entire object detection system. SS takes 1-2 seconds depending on content (on average about 1.5s), and Fast R-CNN with VGG-16 takes 320ms on 2000 SS proposals (or 223ms if using SVD on fully-connected layers ). Our system with VGG-16 takes in total 198ms for both proposal and detection. With the convolutional features shared, the RPN alone only takes 10ms computing the additional layers. Our region-wise computation is also lower, thanks to fewer proposals (300 per image). Our system has a frame-rate of 17 fps with the ZF net.", null, "Figure 4: Recall vs. IoU overlap ratio on the PASCAL VOC 2007 test set.\n\nSensitivities to Hyper-parameters. In Table VIII we investigate the settings of anchors. By default we use 3 scales and 3 aspect ratios (69.9% mAP in Table VIII). If using just one anchor at each position, the mAP drops by a considerable margin of 3-4%. The mAP is higher if using 3 scales (with 1 aspect ratio) or 3 aspect ratios (with 1 scale), demonstrating that using anchors of multiple sizes as the regression references is an effective solution. Using just 3 scales with 1 aspect ratio (69.8%) is as good as using 3 scales with 3 aspect ratios on this dataset, suggesting that scales and aspect ratios are not disentangled dimensions for the detection accuracy. But we still adopt these two dimensions in our designs to keep our system flexible.\n\nIn Table IX we compare different values of in Equation (1). By default we use which makes the two terms in Equation (1) roughly equally weighted after normalization. Table IX shows that our result is impacted just marginally (by ) when is within a scale of about two orders of magnitude (1 to 100). This demonstrates that the result is insensitive to in a wide range.\n\nAnalysis of Recall-to-IoU. Next we compute the recall of proposals at different IoU ratios with ground-truth boxes. It is noteworthy that the Recall-to-IoU metric is just loosely [19, 20, 21] related to the ultimate detection accuracy. It is more appropriate to use this metric to diagnose the proposal method than to evaluate it.\n\nIn Figure 4, we show the results of using 300, 1000, and 2000 proposals. We compare with SS and EB, and the proposals are the top- ranked ones based on the confidence generated by these methods. The plots show that the RPN method behaves gracefully when the number of proposals drops from 2000 to 300. This explains why the RPN has a good ultimate detection mAP when using as few as 300 proposals. As we analyzed before, this property is mainly attributed to the cls term of the RPN. The recall of SS and EB drops more quickly than RPN when the proposals are fewer.\n\nOne-Stage Detection vs. Two-Stage Proposal + Detection. The OverFeat paper proposes a detection method that uses regressors and classifiers on sliding windows over convolutional feature maps. OverFeat is a one-stage, class-specific detection pipeline, and ours is a two-stage cascade consisting of class-agnostic proposals and class-specific detections. In OverFeat, the region-wise features come from a sliding window of one aspect ratio over a scale pyramid. These features are used to simultaneously determine the location and category of objects. In RPN, the features are from square (33) sliding windows and predict proposals relative to anchors with different scales and aspect ratios. Though both methods use sliding windows, the region proposal task is only the first stage of Faster R-CNN—the downstream Fast R-CNN detector attends to the proposals to refine them. In the second stage of our cascade, the region-wise features are adaptively pooled [1, 2] from proposal boxes that more faithfully cover the features of the regions. We believe these features lead to more accurate detections.\n\nTo compare the one-stage and two-stage systems, we emulate the OverFeat system (and thus also circumvent other differences of implementation details) by one-stage Fast R-CNN. In this system, the “proposals” are dense sliding windows of 3 scales (128, 256, 512) and 3 aspect ratios (1:1, 1:2, 2:1). Fast R-CNN is trained to predict class-specific scores and regress box locations from these sliding windows. Because the OverFeat system adopts an image pyramid, we also evaluate using convolutional features extracted from 5 scales. We use those 5 scales as in [1, 2].\n\nTable X compares the two-stage system and two variants of the one-stage system. Using the ZF model, the one-stage system has an mAP of 53.9%. This is lower than the two-stage system (58.7%) by 4.8%. This experiment justifies the effectiveness of cascaded region proposals and object detection. Similar observations are reported in [2, 39], where replacing SS region proposals with sliding windows leads to 6% degradation in both papers. We also note that the one-stage system is slower as it has considerably more proposals to process.\n\n### 4.2 Experiments on MS COCO\n\nWe present more results on the Microsoft COCO object detection dataset . This dataset involves 80 object categories. We experiment with the 80k images on the training set, 40k images on the validation set, and 20k images on the test-dev set. We evaluate the mAP averaged for IoU (COCO’s standard metric, simply denoted as mAP@[.5, .95]) and [email protected] (PASCAL VOC’s metric).\n\nThere are a few minor changes of our system made for this dataset. We train our models on an 8-GPU implementation, and the effective mini-batch size becomes 8 for RPN (1 per GPU) and 16 for Fast R-CNN (2 per GPU). The RPN step and Fast R-CNN step are both trained for 240k iterations with a learning rate of 0.003 and then for 80k iterations with 0.0003. We modify the learning rates (starting with 0.003 instead of 0.001) because the mini-batch size is changed. For the anchors, we use 3 aspect ratios and 4 scales (adding ), mainly motivated by handling small objects on this dataset. In addition, in our Fast R-CNN step, the negative samples are defined as those with a maximum IoU with ground truth in the interval of , instead of used in [1, 2]. We note that in the SPPnet system , the negative samples in are used for network fine-tuning, but the negative samples in are still visited in the SVM step with hard-negative mining. But the Fast R-CNN system abandons the SVM step, so the negative samples in are never visited. Including these samples improves [email protected] on the COCO dataset for both Fast R-CNN and Faster R-CNN systems (but the impact is negligible on PASCAL VOC).\n\nThe rest of the implementation details are the same as on PASCAL VOC. In particular, we keep using 300 proposals and single-scale () testing. The testing time is still about 200ms per image on the COCO dataset.\n\nIn Table XI we first report the results of the Fast R-CNN system using the implementation in this paper. Our Fast R-CNN baseline has 39.3% [email protected] on the test-dev set, higher than that reported in . We conjecture that the reason for this gap is mainly due to the definition of the negative samples and also the changes of the mini-batch sizes. We also note that the mAP@[.5, .95] is just comparable.\n\nNext we evaluate our Faster R-CNN system. Using the COCO training set to train, Faster R-CNN has 42.1% [email protected] and 21.5% mAP@[.5, .95] on the COCO test-dev set. This is 2.8% higher for [email protected] and 2.2% higher for mAP@[.5, .95] than the Fast R-CNN counterpart under the same protocol (Table XI). This indicates that RPN performs excellent for improving the localization accuracy at higher IoU thresholds. Using the COCO trainval set to train, Faster R-CNN has 42.7% [email protected] and 21.9% mAP@[.5, .95] on the COCO test-dev set. Figure 6 shows some results on the MS COCO test-dev set.\n\nFaster R-CNN in ILSVRC & COCO 2015 competitions We have demonstrated that Faster R-CNN benefits more from better features, thanks to the fact that the RPN completely learns to propose regions by neural networks. This observation is still valid even when one increases the depth substantially to over 100 layers . Only by replacing VGG-16 with a 101-layer residual net (ResNet-101) , the Faster R-CNN system increases the mAP from 41.5%/21.2% (VGG-16) to 48.4%/27.2% (ResNet-101) on the COCO val set. With other improvements orthogonal to Faster R-CNN, He et al. obtained a single-model result of 55.7%/34.9% and an ensemble result of 59.0%/37.4% on the COCO test-dev set, which won the 1st place in the COCO 2015 object detection competition. The same system also won the 1st place in the ILSVRC 2015 object detection competition, surpassing the second place by absolute 8.5%. RPN is also a building block of the 1st-place winning entries in ILSVRC 2015 localization and COCO 2015 segmentation competitions, for which the details are available in and respectively.", null, "Figure 5: Selected examples of object detection results on the PASCAL VOC 2007 test set using the Faster R-CNN system. The model is VGG-16 and the training data is 07+12 trainval (73.2% mAP on the 2007 test set). Our method detects objects of a wide range of scales and aspect ratios. Each output box is associated with a category label and a softmax score in [0,1]. A score threshold of 0.6 is used to display these images. The running time for obtaining these results is 198ms per image, including all steps.", null, "Figure 6: Selected examples of object detection results on the MS COCO test-dev set using the Faster R-CNN system. The model is VGG-16 and the training data is COCO trainval (42.7% [email protected] on the test-dev set). Each output box is associated with a category label and a softmax score in [0,1]. A score threshold of 0.6 is used to display these images. For each image, one color represents one object category in that image.\n\n### 4.3 From MS COCO to PASCAL VOC\n\nLarge-scale data is of crucial importance for improving deep neural networks. Next, we investigate how the MS COCO dataset can help with the detection performance on PASCAL VOC.\n\nAs a simple baseline, we directly evaluate the COCO detection model on the PASCAL VOC dataset, without fine-tuning on any PASCAL VOC data. This evaluation is possible because the categories on COCO are a superset of those on PASCAL VOC. The categories that are exclusive on COCO are ignored in this experiment, and the softmax layer is performed only on the 20 categories plus background. The mAP under this setting is 76.1% on the PASCAL VOC 2007 test set (Table XII). This result is better than that trained on VOC07+12 (73.2%) by a good margin, even though the PASCAL VOC data are not exploited.\n\nThen we fine-tune the COCO detection model on the VOC dataset. In this experiment, the COCO model is in place of the ImageNet-pre-trained model (that is used to initialize the network weights), and the Faster R-CNN system is fine-tuned as described in Section 3.2. Doing so leads to 78.8% mAP on the PASCAL VOC 2007 test set. The extra data from the COCO set increases the mAP by 5.6%. Table VI shows that the model trained on COCO+VOC has the best AP for every individual category on PASCAL VOC 2007. Similar improvements are observed on the PASCAL VOC 2012 test set (Table XII and Table VII). We note that the test-time speed of obtaining these strong results is still about 200ms per image.\n\n## 5 Conclusion\n\nWe have presented RPNs for efficient and accurate region proposal generation. By sharing convolutional features with the down-stream detection network, the region proposal step is nearly cost-free. Our method enables a unified, deep-learning-based object detection system to run at near real-time frame rates. The learned RPN also improves region proposal quality and thus the overall object detection accuracy." ]

[ null, "https://deepai.org/static/images/logo.png", null, "https://deepai.org/publication/None", null, "https://deepai.org/publication/None", null, "https://deepai.org/publication/None", null, "https://deepai.org/publication/None", null, "https://deepai.org/publication/None", null, "https://deepai.org/publication/None", null ]

[ "# How to find the median distance between points within an alpha shape?\n\n조회 수: 8 (최근 30일)\nBin Qi . 2021년 2월 20일\n편집: Gaurav Garg . 2021년 2월 26일\nI want to know the median and average distance between the points within the created alpha shape. How can I do that?\nI can use pdist([x,y]) to find all posible distances and remove the values larger than the alpha, then find the median but I think it is very ineficient.\n\n댓글을 달려면 로그인하십시오.\n\n### 채택된 답변\n\nGaurav Garg 2021년 2월 26일\n편집: Gaurav Garg 님. 2021년 2월 26일\nHi Bin,\nYou can find distance between a query point and multiple other points in the following 2 ways -\ntic;d = x-Y\na = abs(d(:,1))+abs(d(:,2));toc\ntic;\nfor i=1:242\nfor j=1:2\nd(i,j) = x(i,j) - Y(j)\nend\nb(i) = abs(d(i,1)) + abs(d(i,2))\nend\ntoc;\nHere, x is the array of 2-D points and Y is the 1-by-2 query point and you have 242 points in x.\nAlso, you would observe that the first way is almost 50x more efficient than the second way. This is because the first way uses vectorization, unline the second way which uses loops.\nYou can then apply median function to find median, or compare each distance with alpha. Note that this can also be done through vectorization.\n\n댓글을 달려면 로그인하십시오.\n\n### 카테고리\n\nHelp CenterFile Exchange에서 Bounding Regions에 대해 자세히 알아보기\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]

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[ "# Subspaces of finite dimensional Hilbert spaces\n\nThis might be a trivial question but please point out exactly where my reasoning is incorrect. Is every subspace of $\\mathbb{R}^n$ closed since $\\mathbb{R}^n$ with the dot product is a finite dimensional Hilbert space and all subspaces of finite dimensional Hilbert spaces are closed?\n\n• It's not incorrect, but you need much less than Hilbert. Every finite-dimensional subspace of a Hausdorff topological vector space is closed. – Daniel Fischer Feb 16 '14 at 16:21\n• Do you need the TVS to be Hausdorff? Does it not follow for all topological vector spaces? – Alex Feb 16 '14 at 16:44\n• @Alex Hausdorff is necessary. Otherwise $\\{0\\}$ is a finite-dimensional but not closed subspace. – Daniel Fischer Feb 16 '14 at 17:56\n• I don't understand your '$\\{0\\}$ example' above, could you expand on that? Why would $\\{0\\}$ not be closed if assuming that TVS is the only requirement and why is that a problem? – Alex Feb 18 '14 at 8:43\n• @Alex Let $X$ be the real line with the trivial topology: the only open sets are $\\varnothing$ and $X$. This is a TVS. The set $\\{0\\}$ is a zero-dimensional subspace of $X$ which is not closed. If you don't like zero-dimensional example, consider $\\mathbb R$ as a subspace of $\\mathbb R^2$, again with the trivial topology. – user127096 Feb 19 '14 at 2:23\n\n## 1 Answer\n\nThe statement in the comment is correct.\n\nThe reasoning in the question is also correct on the technical level, but I can't really tell how adequate it is in the context where you'll put it. Generally, by the time people learn what a Hilbert space is, they already know how to prove that a linear subspace of $\\mathbb R^n$ is closed (because they know enough linear algebra to represent such a subspace by a system of linear equations, i.e., as the kernel of a linear operator).\n\nWhat I'm saying is: a person who actually needs an explanation of why a linear subspace of $\\mathbb R^n$ is closed is unlikely to be satisfied with yours.\n\n• Okay I have not followed traditional course of studying functional analysis but am instead trying to learn independently. Okay but just to be clear my question was referring to any subset of $\\mathbb{R}^{n}$ which is endowed with the subspace topology, not just linear subspaces of $\\mathbb{R}^{n}$. Does the result I stated in my initial question only apply to linear subspaces? – user103184 Feb 23 '14 at 10:55\n• @Moses So you're asking if every subset of $\\mathbb R^n$ is closed? Does the interval $(0,1)$ look like a closed subset of $\\mathbb R$ to you? – user127096 Feb 23 '14 at 19:12" ]

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[ "# Definition:Many-to-One Relation\n\n(Redirected from Definition:Partial Mapping)\n\n## Definition\n\nA relation $\\RR \\subseteq S \\times T$ is many-to-one if and only if:\n\n$\\forall x \\in \\Dom \\RR: \\forall y_1, y_2 \\in \\Cdm \\RR: \\tuple {x, y_1} \\in \\RR \\land \\tuple {x, y_2} \\in \\RR \\implies y_1 = y_2$\n\nThat is, every element of the domain of $\\RR$ relates to no more than one element of its codomain.\n\n### Defined\n\nLet $f \\subseteq S \\times T$ be a many-to-one relation.\n\n#### Defined at Element\n\nLet $s \\in S$.\n\nThen $f$ is defined at $s$ if and only if $s \\in \\Dom f$, the domain of $f$.\n\n#### Defined on Subclass\n\nLet $R \\subseteq S$.\n\nThen $f$ is defined on $R$ if and only if it is defined at all $r \\in R$.\n\nEquivalently, if and only if $R \\subseteq \\Dom f$, the domain of $f$.\n\n## Also known as\n\nA many-to-one relation is also referred to as:\n\na rule of assignment\na functional relation\na right-definite relation\na right-unique relation\na partial mapping.\n\nSome sources break with mathematical convention and call this a (partial) function.\n\nThese sources subsequently define a total function to be what on $\\mathsf{Pr} \\infty \\mathsf{fWiki}$ is called a mapping.\n\nNone of these names is as intuitively obvious as many-to-one relation, so the latter is the preferred term on $\\mathsf{Pr} \\infty \\mathsf{fWiki}$.\n\nHowever, it must be noted that a one-to-one relation is an example of a many-to-one relation, which may confuse.\n\nThe important part is the to-one part of the definition, which is as opposed to the to-many characteristic of a one-to-many relation and a many-to-many relation.\n\nSome approaches, for example 1999: András Hajnal and Peter Hamburger: Set Theory, use this as the definition for a mapping from $S$ to $T$, and then separately specify the requisite left-total nature of the conventional definition by restricting $S$ to the domain.\n\nHowever, this approach is sufficiently different from the mainstream approach that it will not be used on $\\mathsf{Pr} \\infty \\mathsf{fWiki}$ and limited to this mention.\n\n## Also see\n\nIf in addition, every element of the domain relates to exactly one element in the codomain, the many-to-one relation is known as a mapping (or function).\n\n• Results about many-to-one relations can be found here." ]

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[ "Excellent Grade 10 Math Geometry Worksheets", null, "Worksheet. August 08th , 2020.\n\nHere you find our selection of geometry worksheets based on the singapore math curriculum. You will then have two choices.", null, "printablementalmathsheets4thgrade10.gif (1000×1294\n\n10th grade math worksheets with answer key feature are widely used in public schools because they give students a chance to make mistakes while working on a problem without consequence.", null, "Grade 10 math geometry worksheets. Printable math worksheets for grade 10. Geometry and measurement word problems grade 4. Click on the free geometry worksheet you would like to print or download.\n\nEasily download and print our 10th grade math worksheets. This resource was designed to keep math concepts fresh all year and to help you easily track student progress. 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Free geometry worksheets created with infinite geometry.\n\nAll math worksheets by grade: We have worksheets with basic shapes, compound shapes and polygons for grade levels 1 to 6. For the lower math grade levels we have.\n\nEasily download and print our geometry worksheets. Geometry worksheets are great learning tools for kids in preschool through high school! Some of the worksheets for this concept are mcas mathematics grade 10 practice.\n\nUse our printable 10th grade math worksheets written by expert math specialists! Math worksheets and lesson plans are aligned to the engageny/eureka math common core core curriculum, examples and step by step solutions, homework, lesson plans, worksheets, assessments that are suitable for common core math. 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[ "# Auto-improve parallel edges in tikz\n\nI have to draw a lot of graphs with a lot of edges. Some of the nodes are connected via edges in one direction, some in both. Edges that come from specific nodes are painted red, while the edges going into these nodes are painted black. Now I have two thinks I'd like to change to make the graphs more easy to understand:\n\n• If two nodes are connected in both ways change the two arrows <-- and --> into one <--> (the differences are small but they bug me).\n• If two nodes are connected in both ways, and one is one with the outgoing nodes in red automaticaly draw two arrows, either slightly bend or shifted, one in black one in red.\n\nIts important, that this is done automaticaly, so that the results look always the same, independent from the slope of the connection. Here is my current code with the result:\n\n\\documentclass[preview]{standalone}\n\n\\usepackage{tikz}\n\\usepackage{pgfplots}\n\\usetikzlibrary {positioning}\n\\usetikzlibrary{arrows,automata,shapes}\n\\usetikzlibrary{trees,fit,decorations.pathreplacing}\n\\usetikzlibrary{calc}\n\\usetikzlibrary{graphs}\n\\tikzset{near start abs/.style={xshift=1cm}}\n\n\\begin{document}\n\\pgfdeclarelayer{background}\n\\pgfdeclarelayer{foreground}\n\\pgfsetlayers{background,main,foreground}\n\nGraph type one:\n\\begin{figure}[H]\n\\centering\n\\begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=3cm,semithick]\n\\tikzstyle{every state}=[scale =1,fill=blue,draw=none,text=white]\n\n\\def \\n {12}\n\\def \\margin {8} % margin in angles, depends on the radius\n\n\\foreach \\s in {0,...,11}\n{\n\\node[draw, state] (\\s) at ({-360/\\n * (\\s +4.5)}:\\radius) {$\\s$};\n% \\draw[->, >=latex] ({360/\\n * (\\s - 1)+\\margin}:\\radius)\n% arc ({360/\\n * (\\s - 1)+\\margin}:{360/\\n * (\\s)-\\margin}:\\radius);\n}\n\\node[fit=(0)(1)(2)(3)](input){};\n\\node[fit=(4)(5)](output){};\n\\node[fit=(6)(7)(8)(9)](intern){};\n\\node[fit=(10)(11)](inhib){};\n\n% edges\n\\path\n(0) edge (9)\nedge (11)\n(1) edge (8)\nedge (9)\nedge (10)\nedge (11)\n(2) edge (6)\nedge (7)\nedge (8)\nedge (9)\n(3) edge (9)\nedge (11)\n(4) edge (7)\nedge (9)\nedge (11)\n(5) edge (6)\nedge (9)\nedge (10)\n(6) edge (4)\nedge (8)\nedge (9)\nedge (11)\n(7) edge (4)\nedge (5)\nedge [red,dashed,bend left =50] node {} (9)\nedge (11)\n(8) edge (4)\nedge (6)\nedge (11)\n(9) edge (4)\nedge (7)\nedge (10)\n(10) edge[red] (5)\nedge[red] (9)\n(11) edge[red] (4)\nedge[red] (5)\nedge[red] (6)\n;\n\n% Input area\n\\begin{pgfonlayer}{background}\n\\filldraw [line width=4mm,join=round,black!10]\n(input.north west) rectangle (input.south east);\n\\end{pgfonlayer}\n\n% Input area\n\\begin{pgfonlayer}{background}\n\\filldraw [line width=4mm,join=round,blue!10]\n(output.north west) rectangle (output.south east);\n\\end{pgfonlayer}\n\n% Intern area\n\\begin{pgfonlayer}{background}\n\\filldraw [line width=4mm,join=round,green!10]\n(intern.north west |- intern.north) rectangle (intern.south east |- intern.south);\n\\end{pgfonlayer}\n\n% Inhib area\n\\begin{pgfonlayer}{background}\n\\filldraw [line width=4mm,join=round,red!10]\n(inhib.north west |- inhib.north) rectangle (inhib.south east |- inhib.south);\n\\end{pgfonlayer}\n\n% captions\n% \\node[below=0.3cm] at (input.south) {Input};\n% \\node[above=0.3cm] at(intern.north) {Intern};\n% \\node[below=0.3cm] at (inhib.south) {Inhibitorisch};\n\n% Labels\n\\foreach \\c [count=\\x from 0] in {0,2,2,0,0,0,0,1,2,1,1,2} %{{a,f},{b,o},{c,o},{d,b},{e,a},{f,r}}\n{\n\\node (\\x) at ({-360/\\n * (\\x +4.5)}:\\labelrad) {\\c};\n}\n\\end{tikzpicture}\n\\end{figure}\n\nGraph type two:\n\\begin{figure}[H]\n\\centering\n\\begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=2cm,semithick]\n\\tikzstyle{every state}=[scale =1,fill=blue,draw=none,text=white]\n\n\\node[state] (4) {4};\n\\node[state] (7) [above right of=4] {7};\n\\node[state] (9) [below right of=7] {9};\n\\node[state] (10)[below left of=9] {10};\n\n% edges\n\\path\n(4) edge (7)\nedge (9)\n(7) edge (4)\nedge [red,dashed,bend left =50] node {} (9)\n(9) edge (4)\nedge (7)\nedge (10)\n(10) edge[red] (9)\n;\n\n% Labels\n\\node (l4) at ($(4)+(-.5,.7)$) {$(0,0\\rightarrow2)$};\n\\node (l7) at ($(7)+(-0,.7)$) {$(2,1\\rightarrow2)$};\n\\node (l9) at ($(9)+(.5,-.8)$) {$(4,1\\rightarrow2)$};\n\\node (l10) at ($(10)+(0,-.7)$) {$(2,1\\rightarrow2)$};\n\\node[red] (gew) at ($(7)!0.5!(9)+(1.3,.7)$) {$(0\\rightarrow1)$};\n\\end{tikzpicture}\n\\end{figure}\n\\end{document}\n\n\nWith the result:", null, "EDIT: Just ignore the dashed edge from node 7 to node 9 that is a different thing.\n\n• You can't do this with this approach, I don't think. (I'm pretty sure.) The problem is that you aren't drawing them as graphs. You're just drawing a bunch of nodes and edges. So the relational properties you want to take as the basis for altering the output are simply not there.\n– cfr\nJul 16, 2016 at 21:37\n• I looked around, and did not find any package or method that was better made for this. Did i miss something? Can you point me to a better suited method for doing this kind of graphs? Maybe one that can do what I asked for in the original question? Jul 16, 2016 at 22:00\n• I don't know the graph drawing facilities of TikZ well, but that is where I would be looking, provided LuaTeX is an option, of course. I don't know whether pstricks has anything like this, but the TikZ graphing stuff seems the best bet to me.\n– cfr\nJul 16, 2016 at 22:07\n• i tought i was using the tikz method for drawing graphs, am i not? Jul 16, 2016 at 22:11\n• You're using TikZ to draw graphs, but you are not using the graph syntax or the automatic layout facilities. (The latter requires LuaTeX.)\n– cfr\nJul 16, 2016 at 22:13\n\nThis is not an answer. It is an illustration of TikZ's graphing syntax and of its support for graph-drawing algorithms.\n\nThe first 2 graphs just illustrate the syntax. The third uses the algorithmic graph layout facilities and, therefore, requires LuaTeX.\n\nIf the following line gives an error, comment it out. It is needed to fix a compatibility issue with standalone and current LuaTeX.\n\n\\RequirePackage{luatex85}\n\\documentclass[border=10pt,multi,tikz]{standalone}\n\n\nLoad some TikZ libraries. graphs supports the syntax. graphdrawing supports the algorithmic graph layouts provided for use with LuaTeX.\n\n\\usetikzlibrary{graphs,graphdrawing,arrows.meta}\n\n\nThe graphdrawing algorithms support includes a number of libraries for various out-of-the-box layout algorithms. Here's one of them.\n\n\\usegdlibrary{circular}\n\\begin{document}\n\n\nThis example doesn't use an algorithm but it does illustrate the syntax. It uses a circular placement strategy which lays out groups of 12 nodes in circular fashion.\n\n\\begin{tikzpicture}\n[\n\n\nmy node is a style which sets incoming edges black and outgoing edges red.\n\n my node/.style={<red, >black},\n]\n\n\nHere's the graph specification itself. We turn the nodes through an angle and increase the default radius a bit to lay things out a bit differently from the default.\n\n \\graph [clockwise=12, radius=25mm, phase=-120, /tikz/>=Stealth, nodes={fill=blue, circle, node sep=12.5mm, minimum size=5mm, inner sep=1pt, text=white}]\n\n\nNow for the nodes in the graph.\n\n {\n\n\nLayout our 12 nodes.\n\n {\\foreach \\i in {0,...,11} \\i,},\n\n\nSpecify the various connections between them.\n\n {1, 9} -> 10 [my node] -> 9,\n{0, 1, 3, 4, 5, 6, 7, 8} -> 11 [my node] -> {4, 5, 6},\n{0, 1, 2, 3, 4, 5, 6} -> 9 -> {4, 7},\n{1, 2, 6} -> 8 -> {4, 6},\n{2, 4} -> 7 -> 4,\n{2, 4, 5} -> 6 -> 4,\n};\n\\end{tikzpicture}", null, "Here's a slightly different layout which explicitly uses circular placement.\n\n\\begin{tikzpicture}\n[\nmy node/.style={<red, >black},\n]\n\\graph [circular placement, group polar shift=(-30:0), /tikz/>=Stealth, nodes={fill=blue, circle, minimum size=5mm, inner sep=1pt, text=white}]\n{\na/ [fill=none] -!- \\foreach \\i in {0,...,11} { a -!- \\i },\n{1, 9} -> 10 [my node] -> 9,\n{0, 1, 3, 4, 5, 6, 7, 8} -> 11 [my node] -> {4, 5, 6},\n{0, 1, 2, 3, 4, 5, 6} -> 9 -> {4, 7},\n{1, 2, 6} -> 8 -> {4, 6},\n{2, 4} -> 7 -> 4,\n{2, 4, 5} -> 6 -> 4,\n};\n\\end{tikzpicture}", null, "The third example uses one of the algorithmic layouts we loaded with the circular library, called simple necklace layout.\n\n\\begin{tikzpicture}\n[\nmy node/.style={<red, >black},\n]\n\n\nNote the use of node sep here to increase the spacing between nodes, as opposed to the more manual approach above. Setting radius, for example, obviously applies only to a circular layout. In contrast, node sep is neutral between layouts.\n\n \\graph [simple necklace layout, grow'=60, /tikz/>=Stealth, nodes={fill=blue, circle, node sep=12.5mm, minimum size=5mm, inner sep=1pt, text=white}]\n{\n{\\foreach \\i in {0,...,11} \\i,},\n{1, 9} -> 10 [my node] -> 9,\n{0, 1, 3, 4, 5, 6, 7, 8} -> 11 [my node] -> {4, 5, 6},\n{0, 1, 2, 3, 4, 5, 6} -> 9 -> {4, 7},\n{1, 2, 6} -> 8 -> {4, 6},\n{2, 4} -> 7 -> 4,\n{2, 4, 5} -> 6 -> 4,\n};\n\\end{tikzpicture}", null, "\\end{document}\n\n\nIt is possible to design your own algorithms to control the layout and appearance of the graph. Since this way of proceeding specifies the structural relationships between the nodes, this information is available when drawing the graph.\n\nIn the original code, nothing specifies these structural relationships, so there is no information which might be used to automate the graph's appearance in the way desired.\n\nBy using \\graph, you can make the structural information available.\n\nWhether it is worth pursuing this approach probably depends on the number and sophistication of the graphs you want to draw in this way.\n\nComplete code:\n\n\\RequirePackage{luatex85}\n\\documentclass[border=10pt,multi,tikz]{standalone}\n\\usetikzlibrary{graphs,graphdrawing,arrows.meta}\n\\usegdlibrary{circular}\n\\begin{document}\n\\begin{tikzpicture}\n[\nmy node/.style={<red, >black},\n]\n\\graph [clockwise=12, radius=25mm, phase=-120, /tikz/>=Stealth, nodes={fill=blue, circle, node sep=12.5mm, minimum size=5mm, inner sep=1pt, text=white}]\n{\n{\\foreach \\i in {0,...,11} \\i,},\n{1, 9} -> 10 [my node] -> 9,\n{0, 1, 3, 4, 5, 6, 7, 8} -> 11 [my node] -> {4, 5, 6},\n{0, 1, 2, 3, 4, 5, 6} -> 9 -> {4, 7},\n{1, 2, 6} -> 8 -> {4, 6},\n{2, 4} -> 7 -> 4,\n{2, 4, 5} -> 6 -> 4,\n};\n\\end{tikzpicture}\n\\begin{tikzpicture}\n[\nmy node/.style={<red, >black},\n]\n\\graph [circular placement, group polar shift=(-30:0), /tikz/>=Stealth, nodes={fill=blue, circle, minimum size=5mm, inner sep=1pt, text=white}]\n{\na/ [fill=none] -!- \\foreach \\i in {0,...,11} { a -!- \\i },\n{1, 9} -> 10 [my node] -> 9,\n{0, 1, 3, 4, 5, 6, 7, 8} -> 11 [my node] -> {4, 5, 6},\n{0, 1, 2, 3, 4, 5, 6} -> 9 -> {4, 7},\n{1, 2, 6} -> 8 -> {4, 6},\n{2, 4} -> 7 -> 4,\n{2, 4, 5} -> 6 -> 4,\n};\n\\end{tikzpicture}\n\\begin{tikzpicture}\n[\nmy node/.style={<red, >black},\n]\n\\graph [simple necklace layout, grow'=60, /tikz/>=Stealth, nodes={fill=blue, circle, node sep=12.5mm, minimum size=5mm, inner sep=1pt, text=white}]\n{\n% 0 -!- \\foreach \\i in {1,...,11} { 0 -!- \\i },\n{\\foreach \\i in {0,...,11} \\i,},\n{1, 9} -> 10 [my node] -> 9,\n{0, 1, 3, 4, 5, 6, 7, 8} -> 11 [my node] -> {4, 5, 6},\n{0, 1, 2, 3, 4, 5, 6} -> 9 -> {4, 7},\n{1, 2, 6} -> 8 -> {4, 6},\n{2, 4} -> 7 -> 4,\n{2, 4, 5} -> 6 -> 4,\n};\n\\end{tikzpicture}\n\\end{document}" ]

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[ "## Algebraic Functions & Graphs", null, "New Topic: Algebraic Functions and Graphs\n\nBy the end of this topic you will be able to:\n\n• complete the square for quadratic functions\n\n• recognise the probable form of a function from its graph\n\n• draw graphs of related functions given f(x), f(x) being a simple polynomial or trigonometric function\n\n• know the general features of the graphs of y=ax and y=logax\n\nHomework: Algebraic Functions Set1\n\nDue: Wednesday 11th September\n\n## Algebraic Functions and Graphs", null, "New Topic: Algebraic Functions and Graphs\n\nBy the end of this topic you will be able to:\n\n• complete the square for quadratic functions\n\n• recognise the probable form of a function from its graph\n\n• draw graphs of related functions given f(x), f(x) being a simple polynomial or trigonometric function\n\n• know the general features of the graphs of y=ax and y=logax\n\n## Composite & Inverse Functions", null, "By the end of this topic you will:\n\n• know domain, range, inverse, composite functions\n• be able to find f(g(x)) given f(x) and g(x)\n\n## Trigonometric Functions & Graphs\n\nNew Topic: Trigonometric Functions & Graphs\n\nBy the end of this topic you will be able to:\n• convert Degrees to Radian measure and vice versa\n• recall exact values in and calculate using radian measure\n• translate graphs of trig functions\n• sketch trig functions given their equation\n• solve trig equations\n\n## Composite & Inverse Functions", null, "By the end of this topic you will:\n\n• know domain, range, inverse, composite functions\n• be able to find f(g(x)) given f(x) and g(x)\n\n## Algebraic Functions and Graphs", null, "New Topic: Algebraic Functions and Graphs\n\nBy the end of this topic you will be able to:\n\n• complete the square for quadratic functions\n\n• recognise the probable form of a function from its graph\n\n• draw graphs of related functions given f(x), f(x) being a simple polynomial or trigonometric function\n\n• know the general features of the graphs of y=ax and y=logax" ]

[ null, "https://htends2infinity.files.wordpress.com/2010/06/parabola-graph.png", null, "https://htends2infinity.files.wordpress.com/2010/06/parabola-graph.png", null, "https://htends2infinity.files.wordpress.com/2008/09/compositefunction-copy.png", null, "https://htends2infinity.files.wordpress.com/2008/09/compositefunction-copy.png", null, "https://htends2infinity.files.wordpress.com/2010/06/parabola-graph.png", null ]

[ "", null, "SPSS TUTORIALS BASICS DATA ANALYSIS REGRESSION ANOVA T-TESTS\n\n# Creating Dummy Variables in SPSS\n\nYou can't readily use categorical variables as predictors in linear regression: you need to break them up into dichotomous variables known as dummy variables.\n\nThe ideal way to create these is our dummy variables tool. If you don't want to use this tool, then this tutorial shows the right way to do it manually.\n\n## Example Data File\n\nThis tutorial uses staff.sav throughout. Part of this data file is shown below.", null, "## Example I - Any Numeric Variable\n\nLet's first create dummy variables for marit, short for marital status. Our first step is to run a basic FREQUENCIES table with frequencies marit. The table below shows the resulting table.", null, "So how to break up marital status into dummy variables? First off, we always omit one category, the reference category. You may choose any category as the reference category.\n\nSo for this example, we choose 5 (Widowed). This implies that we'll create 3 dummy variables representing categories 1, 2 and 4 (note that 3 does not occur in this variable).\n\nThe syntax below shows how to create and label our 3 dummy variables. Let's run it.\n\n*Create dummy variables for categories 1, 2 and 4.\n\ncompute marit_1 = (marit = 1).\ncompute marit_2 = (marit = 2).\ncompute marit_4 = (marit = 4).\n\n*Apply variable labels to dummy variables.\n\nvariable labels\nmarit_1 'Marital Status = Never Married'\nmarit_2 'Marital Status = Currently Married'\nmarit_4 'Marital Status = Divorced'.\n\n*Quick check first dummy variable\n\nfrequencies marit_1.\n\n## Results\n\nFirst off, note that we created 3 nicely labelled dummy variables in our active dataset.", null, "The table below shows the frequency distribution for our first dummy variable.", null, "Note that our dummy variable holds 3 distinct values:\n\n• respondents whose marital status is not “never married” score 0;\n• respondents whose marital status is “never married” score 1;\n• respondents whose marital status is a missing value (and therefore unknown) have a system missing value.\n\nWe may now check the results more thoroughly by running crosstabs marit by marit_1 to marit_4. Doing so creates 3 contingency tables, the first of which is shown below.", null, "On our dummy variable,", null, "respondents having other marital statuses than “never married” all score 0;", null, "respondents who “never married” all score 1;", null, "we've a sample size of N = 170 (this table only includes respondents without missing values on either variable).\n\nOptionally, a final -very thorough- check is to compare ANOVA results for the original variable to regression results using our dummy variables. The syntax below does just that, using monthly salary as the dependent variable.\n\n*Minimal regression using dummy variables.\n\nregression\n/dependent salary\n/method enter marit_1 to marit_4.\n\n*Minimal ANOVA using original variable.\n\noneway salary by marit.\n\nNote that both analyses result in identical ANOVA tables. We'll discuss ANOVA versus dummy variable regression more thoroughly in a future tutorial.\n\n## Example II - Numeric Variable with Adjacent Integers\n\nWe'll now create dummy variables for region. Again, we start off by inspecting a minimal frequency table which we'll create by running frequencies region. This results in the table below.", null, "We'll choose 1 (“North”) as our reference category. We'll therefore create dummy variables for categories 2 through 5. Since these are adjacent integers, we can speed things up by using DO REPEAT as shown below.\n\n*Create dummy variables for region categories 2 through 5.\n\ndo repeat #vals = 2 to 5 / #vars = region_2 to region_5.\nrecode region (#vals = 1)(lo thru hi = 0) into #vars.\nend repeat print.\n\n*Apply variable labels to new variables.\n\nvariable labels\nregion_2 'Region = East'\nregion_3 'Region = South'\nregion_4 'Region = West'\nregion_5 'Region = Top 4 City'.\n\n*Quick check.\n\ncrosstabs region by region_2 to region_5.\n\nA careful inspection of the resulting tables confirms that all results are correct.\n\n## Example III - String Variable with Conversion\n\nSadly, our first 2 methods don't work for string variables such as jtype -short for “job type”). The easiest solution is to convert it into a numeric variable as discussed in SPSS Convert String to Numeric Variable. The syntax below uses AUTORECODE to get the job done.\n\n*Convert jtype into numeric variable.\n\nautorecode jtype\n/into njtype.\n\n*Check result.\n\nfrequencies njtype.\n\n*Set missing values.\n\nmissing values njtype (1,2).\n\n*Recheck result.\n\nfrequencies njtype.\n\n## Result", null, "Since njtype -short for “numeric job type”- is a numeric variable, we can now use method I or method II for breaking it up into dummy variables.\n\n## Example IV - String Variable without Conversion\n\nConverting string variables into numeric ones is the easy to create dummy variables for them. Without this conversion, the process is cumbersome because SPSS doesn't handle missing values for string variables properly. However, syntax below gets the job done correctly.\n\n*Inspect frequencies.\n\nfrequencies jtype.\n\n*Chance '(Unknown)' into 'NA'.\n\nrecode jtype ('(Unknown)' = 'NA').\n\n*Set user missing values.\n\nmissing values jtype ('','NA').\n\n*Reinspect frequencies.\n\nfrequencies jtype.\n\n*Create dummy variables for string variable.\n\nif(not missing(jtype)) jtype_1 = (jtype = 'IT').\nif(not missing(jtype)) jtype_2 = (jtype = 'Management').\nif(not missing(jtype)) jtype_3 = (jtype = 'Sales').\nif(not missing(jtype)) jtype_4 = (jtype = 'Staff').\n\n*Apply variable labels to dummy variables.\n\nvariable labels\njtype_1 'Job type = IT'\njtype_2 'Job type = Management'\njtype_3 'Job type = Sales'\njtype_4 'Job type = Staff'.\n\n*Check results.\n\ncrosstabs jtype by jtype_1 to jtype_4.\n\n## Final Notes\n\nCreating dummy variables for numeric variables can be done fast and easily. Setting proper variable labels, however, always takes a bit of work. String variables require some extra step(s) but are pretty doable as well.\n\nNevertheless, the easiest option is our SPSS Create Dummy Variables Tool as it takes perfect care of everything.\n\nHope you found this tutorial helpful! Let us know by throwing a comment below.\n\nThanks for reading!\n\n# Tell us what you think!\n\n*Required field. Your comment will show up after approval from a moderator." ]

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[ "## Precalculus (6th Edition) Blitzer\n\na. $574$ million. b. $1148$ million. c. $2295$ million. d. $4590$ million. e. doubles every 27 years.\na. Given the function $f(x)=574(1.026)^x$, let $x=0$; we have $f(0)=574(1.026)^0=574$ millions. b. Let $x=27$; we have $f(27)=574(1.026)^{27}\\approx1148$ million. c. For year 2029, $x=2028-1974=54$; we have $f(54)=574(1.026)^{54}\\approx2295$ million. d. For year 2055, $x=2055-1974=81$; we have $f(81)=574(1.026)^{81}\\approx4590$ million. e. Based on the above results, it appears that India's population will double every 27 years." ]

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[ "# Profit percentage formula explained with examples\n\nLearn and know the profit percentage formula in math. In commercial maths, this is one of the important formulas we have to learn and remember.", null, "As we are learning about the formula for profit percentage, so first we have to know what is the meaning of profit. Suppose, if you want to know about what is profit? Then you should have known what is cost price and what is mean by selling price. Hope that you know these two definitions, so we will know about the meaning of profit and later profit percentage formula.\n\n## What is mean by profit in math?\n\nIn math, profit is also called as gain, both gives the same meaning. If we are selling an item for more than the cost price of the item then we say that we will get profit or gain. In simple words, if selling price > cost price, we get profit which is also called as gain.\n\n## What is the Profit percentage formula in math?\n\nWe know what is profit, so now we will know what is profit percentage? Then we will know the formula for profit percentage. Whatever the profit we got, that profit representing in terms of percentage is called as profit or gain percentage. To determine this gain percentage, we have a simple formula.\n\nThe formula for Gain or profit percentage = $\\frac { gain\\quad or\\quad profit }{ cost\\quad price }$ × 100\n\nTo find out the gain or profit use this formula.\n\nThe Gain or profit = selling price (S.P) – cost price (C.P)\n\nExamples:\n\nIt is given that the cost price (C.P) of notepad is equal to Rs.60 and the selling price (S.P) of notepad is equal to Rs. 80. Find out what is profit percentage.\n\nSolution:\n\nGiven that,\n\nCost price (C.P) of notepad = Rs. 60\n\nSelling price (S.P) of notepad = Rs. 80\n\nTherefore, gain or profit = S.P – C.P = 80 – 60 = 20\n\nWe know that the formula for profit percentage = $\\frac { gain\\quad or\\quad profit }{ cost\\quad price }$ × 100 = $\\frac { 20 }{ 60 }$× 100 = $\\frac { 1 }{ 3 }$ × 100 = 33.33%.\n\nIf cost price is Rs.40 and selling price is Rs.80 then find gain percentage.\n\nSolution:\n\nGiven\n\nCost price (C.P) = Rs.40\n\nSelling price (S.P)= Rs.80\n\nGain or profit = selling price (S.P) – Cost price (C.P)\n\n= 80 – 40\n\n= 40\n\nTherefore, we know that the formula for gain percentage = $\\frac { gain\\quad or\\quad profit }{ cost\\quad price }$ × 100 = $\\frac { 40 }{ 40 }$ × 100 = 100%", null, "" ]

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[ "# Ryo Sakai's scientific contributions\n\n## Publications (11)\n\nPreprint\nFull-text available\nWe discuss the use of quantum simulation to study an $N$ flavor theory of interacting relativistic fermions in(1+1) dimensions on NISQ era machines. The case of two flavors is particularly interesting as it can be mapped to the Hubbard model. We derive the appropriate qubit Hamiltonians and associated quantum circuits. We compare classical simulati...\nPreprint\nFull-text available\nTensor network methods are becoming increasingly important for high-energy physics, condensed matter physics and quantum information science (QIS). We discuss the impact of tensor network methods on lattice field theory, quantum gravity and QIS in the context of High Energy Physics (HEP). These tools will target calculations for strongly interactin...\nPreprint\nFull-text available\nWe discuss recent progress in Tensor Lattice Field Theory and economical, symmetry preserving, truncations suitable for quantum computations or simulations. We focus on spin and gauge models with continuous Abelian symmetries such as the Abelian Higgs model and emphasize noise-robust implementations of Gauss's law. We discuss recent progress concer...\nPreprint\nFull-text available\nThe $q$-state clock model is a classical spin model that corresponds to the Ising model when $q=2$ and to the $XY$ model when $q\\to\\infty$. The integer-$q$ clock model has been studied extensively and has been shown to have a single phase transition when $q=2$,$3$,$4$ and two phase transitions when $q>4$.We define an extended $q$-state clock model...\nArticle\nFull-text available\nMotivated by recent attempts to quantum simulate lattice models with continuous Abelian symmetries using discrete approximations, we define an extended-O(2) model by adding a γcos(qφ) term to the ordinary O(2) model with angular values restricted to a 2π interval. In the γ→∞ limit, the model becomes an extended q-state clock model that reduces to t...\nPreprint\nFull-text available\nWe define an extended-O(2) model by adding a $\\gamma \\cos(q\\varphi)$ term to the ordinary O(2) model with angular values restricted to a $2\\pi$ interval. In the $\\gamma \\rightarrow \\infty$ limit, the model becomes an extended $q$-state clock model that reduces to the ordinary $q$-state clock model when $q$ is an integer and otherwise is a continuat...\nPreprint\nFull-text available\nWe discuss the successes and limitations of statistical sampling for a sequence of models studied in the context of lattice QCD and emphasize the need for new methods to deal with finite-density and real-time evolution. We show that these lattice models can be reformulated using tensorial methods where the field integrations in the path-integral fo...\nArticle\nFull-text available\nWe construct a tensor network representation of the partition function for the massless Schwinger model on a two-dimensional lattice using staggered fermions. The tensor network representation allows us to include a topological term. Using a particular implementation of the tensor renormalization group we calculate the average plaquette and topolog...\n\n## Citations\n\n... In recent years, Hamiltonian-simulation methods based on tensor networks have significantly advanced , targeting generally low-dimensional theories and systems without volume-law entanglement. The progress in the applications of tensor networks to lattice gauge theories in both the Hamiltonian and path-integral formulations will continue over the next decade, as discussed in a Snowmass whitepaper and recent reviews [356,357]. Nonetheless, more general Hamiltonian-simulation methods are needed, particularly pertinent to QCD and for real-time situations that exhibit an entanglement growth. ...\n... While the perturbation considered here had an explicit Z N symmetry, which may have had a dominant effect on the resulting phase structure, it would also be interesting to explore the effect of perturbations with less symmetry to see how the results are modified. For example, N could be extended away from integer values as was considered in the limit of a large perturbation here [47,48]. Determining the effect of small perturbations with varying degrees of symmetry could be important to understanding the errors inherent in simulations on resource limited and noisy near-term quantum simulators. ...\n... The Schwinger model, that is, 2D QED, shares some features with QCD [33,34] and is often used as a testbed for new algorithms for lattice gauge theories [21,. In the context of flow-based sampling and related approaches, this theory has already been investigated in Refs. ..." ]

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[ "# #0138a9 Color Information\n\nIn a RGB color space, hex #0138a9 is composed of 0.4% red, 22% green and 66.3% blue. Whereas in a CMYK color space, it is composed of 99.4% cyan, 66.9% magenta, 0% yellow and 33.7% black. It has a hue angle of 220.4 degrees, a saturation of 98.8% and a lightness of 33.3%. #0138a9 color hex could be obtained by blending #0270ff with #000053. Closest websafe color is: #003399.\n\n• R 0\n• G 22\n• B 66\nRGB color chart\n• C 99\n• M 67\n• Y 0\n• K 34\nCMYK color chart\n\n#0138a9 color description : Dark blue.\n\n# #0138a9 Color Conversion\n\nThe hexadecimal color #0138a9 has RGB values of R:1, G:56, B:169 and CMYK values of C:0.99, M:0.67, Y:0, K:0.34. Its decimal value is 80041.\n\nHex triplet RGB Decimal 0138a9 `#0138a9` 1, 56, 169 `rgb(1,56,169)` 0.4, 22, 66.3 `rgb(0.4%,22%,66.3%)` 99, 67, 0, 34 220.4°, 98.8, 33.3 `hsl(220.4,98.8%,33.3%)` 220.4°, 99.4, 66.3 003399 `#003399`\nCIE-LAB 28.639, 31.943, -64.071 8.587, 5.699, 38.181 0.164, 0.109, 5.699 28.639, 71.592, 296.499 28.639, -12.359, -82.832 23.872, 22.43, -78.12 00000001, 00111000, 10101001\n\n# Color Schemes with #0138a9\n\n• #0138a9\n``#0138a9` `rgb(1,56,169)``\n• #a97201\n``#a97201` `rgb(169,114,1)``\nComplementary Color\n• #018ca9\n``#018ca9` `rgb(1,140,169)``\n• #0138a9\n``#0138a9` `rgb(1,56,169)``\n• #1e01a9\n``#1e01a9` `rgb(30,1,169)``\nAnalogous Color\n• #8ca901\n``#8ca901` `rgb(140,169,1)``\n• #0138a9\n``#0138a9` `rgb(1,56,169)``\n• #a91e01\n``#a91e01` `rgb(169,30,1)``\nSplit Complementary Color\n• #38a901\n``#38a901` `rgb(56,169,1)``\n• #0138a9\n``#0138a9` `rgb(1,56,169)``\n• #a90138\n``#a90138` `rgb(169,1,56)``\n• #01a972\n``#01a972` `rgb(1,169,114)``\n• #0138a9\n``#0138a9` `rgb(1,56,169)``\n• #a90138\n``#a90138` `rgb(169,1,56)``\n• #a97201\n``#a97201` `rgb(169,114,1)``\n• #011f5d\n``#011f5d` `rgb(1,31,93)``\n• #012776\n``#012776` `rgb(1,39,118)``\n• #013090\n``#013090` `rgb(1,48,144)``\n• #0138a9\n``#0138a9` `rgb(1,56,169)``\n• #0140c2\n``#0140c2` `rgb(1,64,194)``\n• #0149dc\n``#0149dc` `rgb(1,73,220)``\n• #0151f5\n``#0151f5` `rgb(1,81,245)``\nMonochromatic Color\n\n# Alternatives to #0138a9\n\nBelow, you can see some colors close to #0138a9. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0162a9\n``#0162a9` `rgb(1,98,169)``\n• #0154a9\n``#0154a9` `rgb(1,84,169)``\n• #0146a9\n``#0146a9` `rgb(1,70,169)``\n• #0138a9\n``#0138a9` `rgb(1,56,169)``\n• #012aa9\n``#012aa9` `rgb(1,42,169)``\n• #011ca9\n``#011ca9` `rgb(1,28,169)``\n• #010ea9\n``#010ea9` `rgb(1,14,169)``\nSimilar Colors\n\n# #0138a9 Preview\n\nThis text has a font color of #0138a9.\n\n``<span style=\"color:#0138a9;\">Text here</span>``\n#0138a9 background color\n\nThis paragraph has a background color of #0138a9.\n\n``<p style=\"background-color:#0138a9;\">Content here</p>``\n#0138a9 border color\n\nThis element has a border color of #0138a9.\n\n``<div style=\"border:1px solid #0138a9;\">Content here</div>``\nCSS codes\n``.text {color:#0138a9;}``\n``.background {background-color:#0138a9;}``\n``.border {border:1px solid #0138a9;}``\n\n# Shades and Tints of #0138a9\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, #00040d is the darkest color, while #f9fbff is the lightest one.\n\n• #00040d\n``#00040d` `rgb(0,4,13)``\n• #000b21\n``#000b21` `rgb(0,11,33)``\n• #001134\n``#001134` `rgb(0,17,52)``\n• #001848\n``#001848` `rgb(0,24,72)``\n• #011e5b\n``#011e5b` `rgb(1,30,91)``\n• #01256f\n``#01256f` `rgb(1,37,111)``\n• #012b82\n``#012b82` `rgb(1,43,130)``\n• #013296\n``#013296` `rgb(1,50,150)``\n• #0138a9\n``#0138a9` `rgb(1,56,169)``\n• #013ebd\n``#013ebd` `rgb(1,62,189)``\n• #0145d0\n``#0145d0` `rgb(1,69,208)``\n• #014be4\n``#014be4` `rgb(1,75,228)``\n• #0152f7\n``#0152f7` `rgb(1,82,247)``\n• #0f5dfe\n``#0f5dfe` `rgb(15,93,254)``\n• #226afe\n``#226afe` `rgb(34,106,254)``\n• #3677fe\n``#3677fe` `rgb(54,119,254)``\n• #4984fe\n``#4984fe` `rgb(73,132,254)``\n• #5d91fe\n``#5d91fe` `rgb(93,145,254)``\n• #709ffe\n``#709ffe` `rgb(112,159,254)``\n• #84acfe\n``#84acfe` `rgb(132,172,254)``\n• #97b9fe\n``#97b9fe` `rgb(151,185,254)``\n• #abc6ff\n``#abc6ff` `rgb(171,198,255)``\n• #bed3ff\n``#bed3ff` `rgb(190,211,255)``\n• #d2e0ff\n``#d2e0ff` `rgb(210,224,255)``\n• #e5edff\n``#e5edff` `rgb(229,237,255)``\n• #f9fbff\n``#f9fbff` `rgb(249,251,255)``\nTint Color Variation\n\n# Tones of #0138a9\n\nA tone is produced by adding gray to any pure hue. In this case, #4f535b is the less saturated color, while #0138a9 is the most saturated one.\n\n• #4f535b\n``#4f535b` `rgb(79,83,91)``\n• #495161\n``#495161` `rgb(73,81,97)``\n• #424f68\n``#424f68` `rgb(66,79,104)``\n• #3c4c6e\n``#3c4c6e` `rgb(60,76,110)``\n• #354a75\n``#354a75` `rgb(53,74,117)``\n• #2f487b\n``#2f487b` `rgb(47,72,123)``\n• #284682\n``#284682` `rgb(40,70,130)``\n• #224388\n``#224388` `rgb(34,67,136)``\n• #1b418f\n``#1b418f` `rgb(27,65,143)``\n• #153f95\n``#153f95` `rgb(21,63,149)``\n• #0e3d9c\n``#0e3d9c` `rgb(14,61,156)``\n• #083aa2\n``#083aa2` `rgb(8,58,162)``\n• #0138a9\n``#0138a9` `rgb(1,56,169)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0138a9 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 to get current time in python and break up into year, month, day, hour, minute?\n\nI would like to get the current time in Python and assign them into variables like `year`, `month`, `day`, `hour`, `minute`. How can this be done in Python 2.7?\n\n## 11 Answers\n\nThe `datetime` module is your friend:\n\n``````import datetime\nnow = datetime.datetime.now()\nprint(now.year, now.month, now.day, now.hour, now.minute, now.second)\n# 2015 5 6 8 53 40\n``````\n\nYou don't need separate variables, the attributes on the returned `datetime` object have all you need.\n\n• just a suplyment: `import time \\n now=time.localtime() \\n print now.tm_year, now.tm_mon, now.tm_mday, now.tm_hour, now.tm_hour, now.tm_min, now.tm_sec, now.tm_wday, now.tm_yday, now.tm_isdst` – Sean.H May 24 '19 at 2:34\n\nHere's a one-liner that comes in just under the 80 char line max.\n\n``````import time\nyear, month, day, hour, min = map(int, time.strftime(\"%Y %m %d %H %M\").split())\n``````\n\nThe `datetime` answer by tzaman is much cleaner, but you can do it with the original python `time` module:\n\n``````import time\nstrings = time.strftime(\"%Y,%m,%d,%H,%M,%S\")\nt = strings.split(',')\nnumbers = [ int(x) for x in t ]\nprint numbers\n``````\n\nOutput:\n\n``````[2016, 3, 11, 8, 29, 47]\n``````\n\nBy unpacking `timetuple` of datetime object, you should get what you want:\n\n``````from datetime import datetime\n\nn = datetime.now()\nt = n.timetuple()\ny, m, d, h, min, sec, wd, yd, i = t\n``````\n• Heads up: you are shadowing the builtin function \"min\" – Peppa Aug 11 '20 at 23:30\n\nLet's see how to get and print day,month,year in python from current time:\n\n``````import datetime\n\nnow = datetime.datetime.now()\nyear = '{:02d}'.format(now.year)\nmonth = '{:02d}'.format(now.month)\nday = '{:02d}'.format(now.day)\nhour = '{:02d}'.format(now.hour)\nminute = '{:02d}'.format(now.minute)\nday_month_year = '{}-{}-{}'.format(year, month, day)\n\nprint('day_month_year: ' + day_month_year)\n``````\n\nresult:\n\n``````day_month_year: 2019-03-26\n``````\n\nFor python 3\n\n``````import datetime\nnow = datetime.datetime.now()\nprint(now.year, now.month, now.day, now.hour, now.minute, now.second)\n``````\n``````import time\nyear = time.strftime(\"%Y\") # or \"%y\"\n``````\n\nYou can use gmtime\n\n``````from time import gmtime\n\ndetailed_time = gmtime()\n#returns a struct_time object for current time\n\nyear = detailed_time.tm_year\nmonth = detailed_time.tm_mon\nday = detailed_time.tm_mday\nhour = detailed_time.tm_hour\nminute = detailed_time.tm_min\n``````\n\nNote: A time stamp can be passed to gmtime, default is current time as returned by time()\n\n``````eg.\ngmtime(1521174681)\n``````\n\nSee struct_time\n\nThree libraries for accessing and manipulating dates and times, namely datetime, arrow and pendulum, all make these items available in namedtuples whose elements are accessible either by name or index. Moreover, the items are accessible in precisely the same way. (I suppose if I were more intelligent I wouldn't be surprised.)\n\n``````>>> YEARS, MONTHS, DAYS, HOURS, MINUTES = range(5)\n>>> import datetime\n>>> import arrow\n>>> import pendulum\n>>> [datetime.datetime.now().timetuple()[i] for i in [YEARS, MONTHS, DAYS, HOURS, MINUTES]]\n[2017, 6, 16, 19, 15]\n>>> [arrow.now().timetuple()[i] for i in [YEARS, MONTHS, DAYS, HOURS, MINUTES]]\n[2017, 6, 16, 19, 15]\n>>> [pendulum.now().timetuple()[i] for i in [YEARS, MONTHS, DAYS, HOURS, MINUTES]]\n[2017, 6, 16, 19, 16]\n``````\n\nThis is an older question, but I came up with a solution I thought others might like.\n\n``````def get_current_datetime_as_dict():\nn = datetime.now()\nt = n.timetuple()\nfield_names = [\"year\",\n\"month\",\n\"day\",\n\"hour\",\n\"min\",\n\"sec\",\n\"weekday\",\n\"md\",\n\"yd\"]\nreturn dict(zip(field_names, t))\n``````\n\ntimetuple() can be zipped with another array, which creates labeled tuples. Cast that to a dictionary and the resultant product can be consumed with `get_current_datetime_as_dict()['year']`.\n\nThis has a little more overhead than some of the other solutions on here, but I've found it's so nice to be able to access named values for clartiy's sake in the code.\n\nyou can use datetime module to get current Date and Time in Python 2.7\n\n``````import datetime\nprint datetime.datetime.now()\n``````\n\nOutput :\n\n``````2015-05-06 14:44:14.369392\n``````\n• This doesn't answer the question. – Colin 't Hart May 31 '17 at 19:12" ]

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[ "# N&O Lecture Jesper Nederlof (TU/e): Faster Space Efficient Algorithms for Subset Sum and Knapsack\n\nEveryone is welcome to attend the public lecture of Jesper Nederlof. Abstract: Faster Space Efficient Algorithms for Subset Sum and KnapsackWe present a randomized Monte Carlo algorithm that solves a given instance of Subset Sum on n integers using O*(2^{0.86n}) time and O*(1) space, where O*() suppresses factors polynomial in the input size.\n\nEveryone is welcome to attend the public lecture of Jesper Nederlof. Abstract:\n\nFaster Space Efficient Algorithms for Subset Sum and Knapsack\nWe present a randomized Monte Carlo algorithm that solves a given instance of Subset Sum on n integers using O*(2^{0.86n}) time and O*(1) space, where O*() suppresses factors polynomial in the input size. The algorithm assumes random access to the random bits used. The same result can be obtained for Knapsack on n items, and the same methods also have consequences for the k-Sum problem. Joint work with Nikhil Bansal, Shashwat Garg and Nikhil Vyas." ]

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[ "# Probability\n\nEnikő Dinnyés\nStatus:\nConfirmed\nYear/Term:\n2017-2018 Summer\nLevel:\nImmersion 1\nDivision:\nNumerical Sciences\nCredit:\n8\n\nThe module covers major topics of probability theory and statistics. We will introduce theoretical concepts through simple experiments and games, before moving on to some important probability distributions: what kind of distribution shall we use for the lifetime of light bulbs, or for the number of mistyping errors on the pages of a newspaper, etc; how can we describe a probability distribution: a density function versus characteristic function; the law of large numbers in a mathematical formulation; the central limit theorem 7; or why the Gaussian (Normal) distribution is special. These concepts are used in financial mathematics, and they form the basis of statistics, thus providing tools for appraising the world around us, in the form of estimates, hypothesis testing, correlations, and regressions. These techniques are used in the social sciences as well. We are going to use calculus on a basic level only, but mathematical thinking is important. The course will show the theoretical background as well as the practice of statistics. It is an introductory course to a branch of applied mathematics. (In Cambridge, it is called “applicable maths”, and it falls under the group of pure maths subjects.)" ]

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[ "## Saturday, 23 December 2017\n\n### Python - Function\n\nMost applications weather they written in python or some other programming languages broke up in smaller block of codes known as functions. So in python there is no difference in functions.\n\n### Default / Simple Function\n\n```#!/usr/bin/python3\n\n# working with functions\n\n#define a function\ndef fun1():\nprint(\"This is a function\")\n```\n\nIn python we define a function with def keyword and given a name fun1. This function does not take any argument or parameter. Here colon indicates that you are starting the scope of that function. And in the scope of above function we have written print statement. Here you will notice that we have written the print statement within the scope of the function. You will also notice that we have intended the print statement as compare to the function scope, its mean this print statement is the part of the function.\n\nNow it is the time to call that function. To call the above function we use below code snippet.\n\n```def main():\nfun1()\nprint(fun1())\nprint(fun1)\nif __name__ == \"__main__\": main()` `\n```\n\nIn line no 11 you can see that when we call fun1() it prints out \"This is a function\". In line no 12 when this statement runs \"print(fun1())\" actually first fun1() will be called and it prints out, and after that print(fun1()) will be executed and this time our function return nothing (fun1 have no return value) that's why it prints out \"None\". In line no 13 we have print only fun1 with our parenthesis. Here you will notice that their print out is some hexadecimal numbers this is happening because everything in python is an object and function is also an object and have some value. In this case it is an object, a function object. Here print command is printing out the value of fun1 object.\n\nNow you know that how python treats object.\n\n### Parameterized Function\n\nParameterized functions take arguments. Below code snippet shows how to declare these types of functions in python.\n\n```#!/usr/bin/python3\n\n# working with functions\n\n#define a parameterize function\ndef parameter_fun(arg1,arg2):\nprint(arg1, \" \", arg2)` `\n```\n\nIn above code snippet we define a parameterize function that take two arguments (arg1 and arg2). And inside the function scope we have one print statement that will print the above arguments with a space separator.\n\nThe function call is same as we have been discussed in case of normal or default function. But the little difference is that in function call we pass two values or parameter. As shown below.\n\n```def main():\nparameter_fun(3,5)\nprint(parameter_fun(3,5))\nif __name__ == \"__main__\": main()```\n```\n```\n\n### Function with return value\n\nIn this example we will define a function that will return some value. For example here we are going to define a function that will return cubic value of a given number.\n\n```#!/usr/bin/python3\n\n# working with functions\n\n#define a function that will return some value\ndef cube(given_value):\nreturn given_value*given_value*given_value\n\ndef main():\nprint(cube(5))\nif __name__ == \"__main__\": main() ```\n```\n```\n\nIn above code snippet we define a function named cube(), this function take one parameter or argument of type number or integer. And inside the function scope we return three times of that number. So this function return cube of the given number.\n\nI hope you will understand functions in python. Functions can be used to code reusability. For example you have a function that will calculate some complex problems, algorithms and get values from the database and return the value. Now you can use this function inside your program multiple times whenever you need. You don't need to write code multiple times just define function and call that function.\n\n#### 1 comment:\n\n1.", null, "Very nice blog. keep it up" ]

[ null, "http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35", null ]

[ "# Outputs\n\nCompute the signals that this MATLAB S-function block emits\n\nYes\n\nMATLAB\n\n## Syntax\n\n```Outputs(s)\n\n```\n\n## Arguments\n\n`s`\n\nInstance of `Simulink.MSFcnRunTimeBlock` class representing the Level-2 MATLAB S-Function block.\n\n## Description\n\nThe Simulink® engine invokes this required method at each simulation time step. In a Level-2 MATLAB S-function, the `Outputs` method calculates the S-function's outputs at the current time step and store the results in the run-time object's `OutputPort(n).Data` property. In addition, for S-functions with a variable sample time, the `Outputs` method computes the next sample time hit.\n\nUse the run-time object method `IsSampleHit` to determine if the current simulation time is one at which a task handled by this block is active. For port-based sample times, use the `IsSampleHit` property of the run-time object's `InputPort` or `OutputPort` methods to determine if the port produces outputs or accepts inputs at the current simulation time step.\n\nSet the run-time object's `NextTimeHit` property to specify the time of the next sample hit for variable sample-time S-functions." ]

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[ "", null, "Lesson: Heron’s Formula\nExercise 12.1\nQuestion: 1\nAn isosceles right triangle has area 8 cm\n2\n. The length\nof its hypotenuse is\n(A)\ncm32\n(B)\ncm16\n(C)\ncm48\n(D)\ncm24\nSolution\nA\nArea of an isosceles triangle\n1\n2\n(Base × Height)\n1\n8\n2\n(Base × Height) … (i)\nThe triangle is an isosceles triangle.", null, "Base Height\n2\nBase 16 Base 4 cm\n(From equation (i))\nNow, in ∆\nABC,\nusing Pythagoras theorem,\n2\n32 32 cmAC AC\nHence, the length of its hypotenuse is\n32 cm\n.\nQuestion: 2\nThe perimeter of an equilateral triangle is 60 m. The\narea is\n(A)\nm\n2\n10 3\n(B)\nm\n2\n15 3\n(C)\nm\n2\n20 3\n(D)\n2\nm100 3\nSolution\nD\nLet each side of the equilateral triangle be\nx\n.\nThe perimeter of the equilateral triangle\n60 m\n(Given)\n60x x x\n3 60x", null, "m\n60\n20\n3\nx\nArea of an equilateral triangle\nsid s e\n2\n3\n4\n( )\n3\n20 20\n4\nm\n2\n100 3\nThus, the area of the triangle is\nm\n2\n100 3 ,\nQuestion: 3\nThe sides of a triangle are 56 cm, 60 cm and 52 cm\nlong. Then the area of the triangle is\n2\nA 1322 cm\n2\nB 1311 cm\n2\nC 1344 cm\n2\nD 1392 cm\nSolution\nC\nThe sides of the triangle are\ncm 56a\n,\ncm60b\nand\ncm52 .c\n(Given)\nThen, the semi perimeter of the triangle is given by:", null, "cm\n56 60 52 168\n84\n2 2 2\na b c\ns\nNow, the area of the triangle\n( )( )( )s s a s b s c\n(By Heron’s formula)\n84(84 56)(84 60)(84 52)\n84 28 24 32\n4 7 3 4 7 4 2 3 4 4 2\n6 2 2\n(4) (7) (3)\ncm\n3 2\n(4) 7 3 1344\nHence, the area of triangle is 1344 cm\n2\n.\nQuestion: 4\nThe area of an equilateral triangle with side\ncm2 3\nis\n2\nA 5.196 cm\n2\nB 0.866 cm\n2\nC 3.496 cm\n2\nD 1.732 cm", null, "Solution\nA\nThe side of the equilateral triangle is\ncm2 3\n.\nHence, the area of the equilateral triangle\nside\n2\n3\n( )\n4\n2\n3\n(2 3)\n4\n3\n4 3\n4\ncm\n2\n3 3 3 1.732 5.196\nHence, the area of the equilateral triangle is 5.196 cm\n2\n.\nQuestion: 5\nThe length of each side of an equilateral triangle\nhaving an area of\n2\n9 3 cm\nis\nA 8 cm\nB 36 cm\nC 4 cm\nD 6 cm\nSolution\nD", null, "The area of the equilateral triangle is\ncm\n2\n9 3\nWe know that area of an equilateral triangle is\n(side)\n2\n3\n4\n(side)\n2\n3\n9 3\n4\nside\n2\n( ) 36\nSide = 6 cm\nHence, the length of the equilateral triangle is 6 cm.\nQuestion: 6\nIf the area of an equilateral triangle is\n16 3\ncm\n2\n,\nthen the perimeter of the triangle is\nA 48 cm\nB 24 cm\nC 12 cm\nD 36 cm\nSolution\nB", null, "The area of the equilateral triangle is\n16 3\ncm\n2\n.\nArea of the equilateral triangle is\n2\n3\n( )\n4\nside\nside\n2\n3\n( ) 16 3\n4\nSide\n2\n( ) 64\nSide = 8 cm\nPerimeter of the equilateral triangle\nside cm3 3 8 24\nHence, the perimeter of the equilateral triangle is 24\ncm.", null, "Question: 7\nThe sides of a triangle are 35 cm, 54 cm and 61 cm,\nrespectively. The length of its longest altitude is\n(A) 16 5 cm\n(B) 10 5 cm\n(C) 24 5 cm\n(D) 28 cm\nSolution\nC\nLet\nABC\nbe the triangle having sides\nAB\n= 35 cm,\nBC\n= 54 cm and\nCA\n= 61 cm.\nNow, the semi perimeter of the triangle,\ncm\n35 54 61 150\n75\n2 2 2\na b c\ns", null, "Area of the triangle\nABC\n( )( )( )s s a s b s c\n(By Heron’s formula)\n75(75 35)(75 54)(75 61)\n75 40 21 14\n25 3 4 2 5 7 3 7 2\n5 2 2 3 7 5\ncm\n2\n420 5 .\nNow, the area of\nAltitude.\n1\n2\nABC AB\nLet the altitude be CD.\n1\n35 420 5\n2\nCD\n420 2 5\n35\nCD\n24 5CD\ncm\nHence, the length of the altitude is\ncm24 5 .", null, "Question: 8\nThe area of an isosceles triangle having base 2 cm and\nthe length of one of the equal sides 4 cm, is\n2\n(A) 15 cm\n2\n15\n(B)\n2\ncm\n2\n(C) 2 15 cm\n2\n(D) 4 15 cm\nSolution\nA", null, "Let\nABC\nbe an isosceles triangle such that\nAB\n=\nAC\n=\n4 cm and BC = 2 cm.\nLet\nbe the perpendicular bisector from A on BC.\nNow, in the right angled\n,ABC\nAB\n2\n=\n2\n+\nBD\n2\n(By Pythagoras\ntheorem)\n(4)\n2 2\n2\n2\n1\n( ) 16 1\n15\n15\nArea of\n1\n2", null, "( Area of a triangle\n1\n2\n(Base× Height))\ncm\n2\n1\n2 15 15\n2\nQuestion: 9\nThe edges of a triangular board are 6 cm, 8 cm and 10\ncm. The cost of painting it at the rate of 9 paise per\ncm\n2\nis:\nA 2.00 Rs\nB 2.16 Rs\nC 2.48 Rs\nD 3.00 Rs\nSolution\nB", null, "Let the edges of the triangular board be\na\n= 6 cm,\nb\n=\n8 cm and\nc\n= 10 cm.\nNow, the semi perimeter of the triangular board is\ncm.\n6 8 10 24\n12\n2 2 2\na b c\ns\nArea of the triangle\n( )( )( )s s a s b s c\n(By Heron’s formula)\n12(12 6)(12 8)(12 10)\n12 6 4 2\n2 2\n(12) (2)\n= 12 × 2 = 24 cm\n2", null, "The cost of painting an area of 1 cm\n2\n= Rs 0.09.\nThe cost of painting the area of 24 cm\n2\n= 0.09 × 24\n= Rs 2.16.\nExercise 12.2\nQuestion: 1", null, "The area of a triangle with base 4 cm and height 6 cm\nis 24 cm\n2\n.\nSolution\nFalse.\nArea of a triangle =\nbase height\n1\n( )\n2\nBase = 4 cm and height = 6 cm\nArea of the triangle =\ncm\n2\n1\n4 6 12\n2\nQuestion: 2\nThe area of\nABC\nis 8 cm\n2\nin which\ncm4AB AC\nand\n90A\n.", null, "Solution\nTrue.\nLet\nABC\nbe right angled at\nA.\nSides of the triangle are\ncm4AB AC\nArea of a triangle\nABC\nbase height\n1\n2\ncm\n2\n1\n2\n1\n4 4\n2\n8\nAC AB\nQuestion: 3\nThe area of the isosceles triangle is\n5\n11\n4\ncm\n2\n, if the\nperimeter is 11 cm and the base is 5 cm.\nSolution\nTrue.\nLet the equal sides of the isosceles triangle be\na\n.", null, "Perimeter of the triangle\n5a a\n11 2 5a\n2 11 5\n2 6\na\na\n6\n2\na\ncm3\nHence, the semi perimeter of the isosceles triangle =\ncm\n3 3 5 11\n2 2 2\na b c\ns\nNow, the area of the triangle\n( )( )( )s s a s b s c\n(By Heron’s formula)\nThis implies,\n11 11 11 11\n( 3)( 3)( 5)\n2 2 2 2\n11 5 5 1\n2 2 2 2\n5\n11\n4\ncm\n2", null, "Question: 4\nThe area of an equilateral triangle is\n20 3\ncm\n2\nwhose\neach side is 8 cm.\nSolution\nFalse\nThe side of an equilateral triangle is 8 cm.\nArea of the equilateral triangle\n2\n2\n2\n3\n( )\n4\n3\n(8)\n4\n16 3\nside\ncm\nQuestion: 5\nIf the side of a rhombus is 10 cm and one diagonal is\n16 cm, the area of the rhombus is 96 cm\n2\n.\nSolution\nTrue\nThe side of the rhombus\nPQ\nRS is 10 cm and one of the\ndiagonal is 16 cm.", null, "PQ\n=\nQR\n= RS =\nSP\n= 10 cm and one of the\ndiagonal is 16 cm, say PR = 16 cm.\nSince, the diagonals of rhombus bisect each other at\nright angle (say at\nO\n).\nHence, in\n2 2 2\n,POQ PQ OP OQ\n(By Pythagoras theorem)\n2 2 2 2 2\n10 8OQ PQ OP\n2\n100 64 36OQ\n36 OQ\ncm6OQ\n2 2 6 12SQ OQ cm\nArea of a rhombus =\n1\n2\n(Product of diagonals)", null, "1\n2\n(\nQS\n×\nPR\n) =\n1\n2\n× 12 × 16 = 96 cm\n2\n.\nQuestion: 6\nThe base and the corresponding altitude of a\nparallelogram are 10 cm and 3.5 cm, respectively. The\narea of the parallelogram is 30 cm\n2\n.\nSolution\nFalse.\nBase and the corresponding altitude of a parallelogram\nare 10 cm and 3.5 cm.\nArea of a parallelogram", null, "= Base × Altitude\n10 × 3.5\n= 35 cm\n2\nQuestion: 7\nThe area of a regular hexagon of side ‘\na\n’ is the sum of\nthe areas of the five equilateral triangles with side\na\n.\nSolution\nFalse.\nSince, a regular hexagon is divided into six equilateral\ntriangles", null, "Area of a regular hexagon of side a\n= Sum of the area of the six equilateral triangles of\nside\na\n.\nQuestion: 8\nThe cost of levelling the ground in the form of a\ntriangle having the sides 51 m, 37 m and 20 m at the\nrate of Rs 3 per m\n2\nis Rs 918.\nSolution\nTrue.", null, "Let the sides of the triangle be\na\n= 51 m,\nb\n= 37 m and\nc\n= 20 m.\nSemi perimeter of the triangle,\na b c\ns\n2\n51 37 20 108\n2 2\n54 m\nArea of the triangle", null, "( )(s )( )a bs s cs\n(By Heron’s formula)\n2\n54 54 51 54 37 54 20\n54 3 17 34\n9 3 2 3 17 17 2\n3 3 2 17\n306 m\nCost of levelling per m\n2\n= Rs 3\nThe cost of levelling 306 m\n2\n= 3 × 306 = Rs 918\nQuestion: 9\nIn a triangle, the sides are given as 11 cm, 12 cm and\n13 cm. The length of the altitude is 10.25 cm\ncorresponding to the side having length 12 cm. The\narea of the triangle is 61.5 cm\n2\n.", null, "Solution\nTrue.\nThe sides of a triangle are\na\n=11 cm,\nb\n= 12 cm and\nc\n=13 cm.\nThe semi perimeter of the triangle is\na b c\ns\n2\n11 12 13 36\n18\n2 2\ncm", null, "Area of a triangle\ns(s )(s )(s )a b c\n(By Heron’s formula)\n18 18 11 18 12 18 13\n=\n18 7 6 5\n3 6 7 6 5\n6 3 7 5\n56 10\n2\n6 10.25\n61.5 cm\nWe can also find the area using:", null, "Area of a triangle\n1\n2\n(base × height)\ncm\n2\n12 10.25 6 10.25 61\n1\n5\n2\n.\nExercise 12.3\nQuestion: 1\nFind the cost of laying grass in a triangular field of\nsides 50 m, 65 m and 65 m at the rate of Rs 7 per m\n2\n.\nSolution\nLet ABC be a triangular field of sides\nAB\n=\na\n= 50 m,\nBC\n=\nb\n= 65 m and CA = c = 65 m.\nThe semi perimeter of the triangular field is,", null, "a b c 50 65 65\n2 2\n180\n90\n2\nm\ns\nArea of a triangle\ns(s ) (s ) (sb )a c\n(By Heron’s formula)\nArea of the triangular field =\n90 90 50 90 65 90 65\n90 40 25 25\n3 2 10 25\nm\n2\n1500\nCost of laying grass per 1 m\n2\n= Rs 7", null, "The cost of laying grass per 1500 m\n2\n= 7 × 1500 =\nRs 10,500\nQuestion: 2\nThe triangular side walls of a flyover have been used\nRs 2000 per m\n2\na year. A company hired one of its\nwalls for 6 months. How much rent did it pay?\nSolution\nThe sides of a triangular walls are\na\n=13 m,\nb\n= 14 m\nand\nc\n= 15 m.\nThe semi perimeter of the triangle is:", null, "m\n13 14 15 42\n2 2\n21\na b c\ns\n2\nArea of the triangular side wall\ns(s ) (s ) (sb )a c\n(By Heron’s formula)\n21 21 13 21 14 21 15\n21 8 7 6\n21 4 2 7 3 2\n2 2\n(21) (4)\nm\n2\n21 4 84\nm Rs\n2\n1 2000\n2", null, "2000 84 168000. Rs\nAs the company hired one of its walls for 6 months,\ntherefore the company pays the rent\n1\n168000 84000\n2\nRs\nHence, the company paid rent of Rs 84,000.\nQuestion: 3\nFrom a point in the interior of an equilateral triangle,\nperpendiculaRs are drawn on the three sides. The\nlengths of the perpendiculaRs are 14 cm; 10 cm and 6\ncm. Find the area of the triangle.\nSolution\nLet\nABC\nbe an equilateral triangle and the length of\neach side be\na\n.\nO\nis the interior point of the triangle\nand\nOQ\n,\nOR\nand\nOP\nare the perpendicular lines drawn\nfrom point\nO\n.\nArea of ∆\nOAB,\ncm\n2\n1\n2\n1\n14 7 (i)\n2\nAB OP\na a", null, "Area of\n1\n2\nΔO C OQB BC\ncm\n2\n1\n10 5 (ii)\n2\na a\nArea of\n1\n2\nΔ C RO OA AC\ncm\n2\n2\n6 3 iii\n1\na a\nArea of an equilateral ∆\nABC\n= Area of (∆OAB +\nOBC\n+\nOAC\n)\ncm\n2\n7 5 3 15 iva a a a\n[From (i), (ii) and (iii)]\nNow, the semi perimeter of the triangle", null, "s\n=\n+ +\n2\na a a\ncm\n3\n=\n2\na\ns\nArea of the equilateral\n( ) ( ) ( )s s sABC a b cs\n(By Heron’s formula)\n3\n2\na a a a\na a a\n3 3 3\n2 2 2\n2\n3 3\n(v)\n2 2 2 2 4\na a a a\na\nFrom equations (iv) and (v),\ncm\n2\n3\n15\n4\n15 4 3\n20 3\n3 3\na a\na", null, "On putting\n20 3\nin equation (v), we get\nArea of a\n2\n3\n(20 3)\n4\nABC\ncm\n2\n3\n400 3\n4\n300 3\nHence, the area of the equilateral triangle is\n2\n300 3 cm\n.\nQuestion: 4\nThe perimeter of an isosceles triangle is 32 cm. The\nratio of the equal side to its base is 3:2. Find the area\nof the triangle.\nSolution\nLet\nABC\nbe an isosceles triangle having perimeter 32\ncm.\nThe ratio of the equal side to its base is 3:2.\nLet the sides of the triangle be\nand 3 2AB AC x BC x.\nThe perimeter of the triangle =\n3 3 2 32x x x\ncm\n8\nx\n= 32\nx\n= 4 cm\nAB = AC = 3 × 4 = 12 cm\nand\ncm2 2 4 8BC x\n.\nThe sides of the triangle are\ncm cm and cm.12 , 12 8a b c", null, "cm.\n+ + 12 12 8 32\n16\n2 2 2\na b c\ns\nArea of the isosceles\n( ) Δ ( ) ( )s s sC csAB a b\n(By Heron’s formula)\n16 16 12 16 12 16 8\n16 4 4 8\ncm cm\n2 2\n4 4 2 2 32 2\nHence, the area of the isosceles triangle is\ncm\n2\n32 2\n.\nQuestion: 5\nFind the area of a parallelogram given in Fig. 12.2.\nAlso find the length of the altitude from vertex\nA\non\nthe side\nDC\n.", null, "Fig. 12.2\nSolution\nArea of the parallelogram\n2ABCD\n(Area\nof\nBCD\n) … (i)\nNow, the sides of\nBCD\nare\ncm cm12 , 17a b\nand\ncm25c\n.\nThe semi perimeter of\nBCD\n,", null, "cm\n2\n12 17 25 54\n2 2\n27\na b c\ns\nArea of\nBCD\n( ) ( ) ( )s s s b sa c\n(By Heron’s formula)\n27 27 12 27 17 27 5 2\n27 15 10 2\n9 3 3 5 5 2 2\ncm\n2\n3 3 5 2 90\nFrom equation (i), we get, the area of the\nparallelogram\ncm\n2\n2 90 180ABCD\n.\nNow, let the altitude of the parallelogram be\nh\n.", null, "Also, the area of a parallelogram\nABCD\n= Base ×\nAltitude\n180 DC h\n180 12 h\n180\n12\nh\n= 15 cm\nHence, the area of the parallelogram is\ncm\n2\n180\nand\nthe length of the altitude is 15 cm.\nQuestion: 6\nA field in the form of a parallelogram has sides 60 m\nand 40 m and one of its diagonals is 80 m long. Find\nthe area of the parallelogram.", null, "Solution\nLet\nABCD\nbe the field in the shape of a parallelogram\nwith following:\n(i) Sides\nm60AB CD\n, and\nm40BC DA\n(ii) Diagonal\nm80BD\nArea of the parallelogram\n2ABCD\n(Area of\nABD\n) …………… (i)\nNow, in ∆\nABD\n,\nThe semi perimeter of the triangle,\nm\n2\n60 80 40 180\n2 2\n90\na b c\ns\nArea of\n( ) Δ ( ) ( )s s sD csAB a b", null, "(By Heron’s\nformula)\n90 90 60 90 80 90 40\n90 30 10 50\n5100 13\nm\n2\n15300\nFrom equation (i), we get,\nThe area of the parallelogram\nABCD\n= 2 × 300\n15\n=\n600\n15\nm\n2.\nHence, the area of the parallelogram is 600\nm\n2\n15\n.\nQuestion: 7\nThe perimeter of a triangular field is 420 m and its\nsides are in the ratio\n6: 7: 8\n. Find the area of the\ntriangular field.", null, "Solution\nThe perimeter of the triangular field is 420 m and its\nsides are in the ratio\n6: 7: 8\n. (Given)\nLet the sides of the triangular field be\n6 , 7a x b x\nand\n8 .c x\nThe perimeter of the triangular field\na b c\n420 6 7 8 420 21x x x x\nm.\n420\n20\n21\nx\nThe sides of the triangular field are\nm,6 20 120a\nm7 20 140b\nand\nm8 20 160c\n.\nNow, the semi perimeter (s)\n420\n210\n2", null, "The area of the triangular field\n( ) ( ) ( )s s s b sa c\n(By Heron’s formula)\n210 210 120 210 140 210 160\n210 90 70 50\nm\n2\n100 7 3 15 2100 15\nHence, the area of the triangular field is\nm\n2\n2100 15 .\nQuestion: 8\nABCD\nare 6 cm, 8 cm, 12\ncm and 14 cm (taken in order) respectively, and the\nangle between the first two sides is a right angle. Find\nits area.\nSolution", null, "ABCD\nAB\n= 6 cm,\nBC\n= 8 cm,\nCD\n= 12 cm and\nDA\n= 14 cm.\nNow, we join\nA\nand\nC\n.\nABC\nis a right angled triangle at\nB\n.\nHence, by Pythagoras theorem,\n2 2 2\nAC AB BC\n2 2 2\n2\n6 8\n100\n100\n10\nAC\nAC\nAC\nAC\ncm10AC", null, "ABCD\n= Area of ∆\nABC\n+\nArea of ∆\nACD\nNow, area of\n1\n2\nBC BA CB A\ncm\n2\n1\n6 8\n2\n3 8 24\nNow, in\ncm cm, 10 , 12 ,ACD AC a CD b\nand\ncm.14DA c\nSemi perimeter of\nACD\ncm.\na b c 10 12 14\n2 2\n36\n18\n2\ns\nNow, area of\n( )(Δ )( )s s s sACD a b c", null, "(By Heron’s\nformula)\n18 18 10 18 12 18 14\n18 8 6 4\n2\n2\ncm\n(3) 2 4 2 3 2 4\n3 4 2 3 2\n24 6\nTherefore, the area of the quadrilateral\nABCD\n= 24 +\n24 cm\n2\nQuestion: 9\nA rhombus shaped sheet with perimeter 40 cm and one\ndiagonal 12 cm, is painted on both sides at the rate of\nRs 5 per m\n2\n. Find the cost of painting.\nSolution", null, "Let\nABCD\nbe a rhombus having each side equal to\nx\ncm.\ncmAB BC CD DA x\nThe perimeter of the rhombus = 40 cm.\n40AB BC CD DA\n40x x x x\n4 40x\n40\n4\nx\ncm 10x\nLet AC be the given diagonal.\nNow, in ∆\nABC\n,\nLet\ncm cm10 , 10a AB b BC\nand\ncm12c AC", null, "The semi perimeter of ∆\nABC,\na b c\ns\n10 10 12\n2 2\n32\n16\n2\ncm\nNow, by Heron’s formula,\nArea of\n( )(Δ )( )s s sBC csA a b\n16 16 10 16 10 16 12\n16 6 6 4\n4 6 2\n= 48 cm\n2\nArea of the rhombus\n2\n(Area of ∆\nABC\n)\n2 48\n= 96 cm\n2\nCost of painting of sheet of 1 m\n2\n= Rs 5\nThe cost of painting of the sheet of 96 cm\n2\n=\n96×5/10000 = Rs 0.048", null, "Hence, the cost of the painting of the sheet for both\nsides\n2 0.048 . Rs 0.096\nQuestion: 10\nFind the area of the trapezium\nPQ\nRS with height\nPQ\ngiven in Fig. 12.3\nFig. 12.3\nSolution", null, "In the trapezium\nPQ\nRS, draw a line\nRT\nperpendicular\nto\nPS\n.\nHence, the side,\nm12 7 5 .ST PS TP\nNow, in ∆\nSTR\n,\n2 2 2\nSR ST TR\n(By using Pythagoras theorem)\n2 2 2\n13 5 TR\nm\n2\n2\n169 25\n144\n12\nTR\nTR\nTR\nNow, the area of\n1\n2\nSTR TR TS.", null, "( Area of triangle =\n1\n2\n(Base × Height))\nm\n2\n1\n12 5 30\n2\nNow, area of the rectangle\nPQRT\n=\nPQ\n×\nRQ\n=12×7\n2\n84 m\nTherefore, the area of\nPQ\nRS =\nar(STR)\n+\nar(PQRT)\n= 30 + 84\n= 114 m\n2\nThe area of the trapezium is 114 m\n2\n.", null, "Exercise 12.4\nQuestion: 1\nHow much paper of each shade is needed to make a\nkite given in Fig. 12.4, in which\nABCD\nis a square\nwith diagonal 44 cm?", null, "Fig. 12.4\nSolution\nAll the sides of a square are always equal.", null, "AB BC CD DA\nNow, in ∆\nACD\n,\ncm 44 , 90AC D\n(\nABCD\nis a square)\nNow, in ∆\nACD\n, by Pythagoras theorem, we get\n2 2 2\n2 2 2\n2\n2\n44\n2 44 44\n22 44\n22 44\nHence,\nAB\n= BC = CD =\nDA\n= 22\n2\ncm.", null, "Area of a square\nABCD\n= Side × Side\n= 22\n2\n× 22\n2\n= 968 cm\n2\n.\nArea of the red portion =\n968\n4\n.\n= 242 cm\n2\n.\n(Since, the area of the square is divided into four\nparts)\nNow, area of the green portion\n968\n4", null, "= 242 cm\n2\n.\nArea of the yellow portion\n968\n2\n= 484 cm\n2\n.\nNow, in the triangle,\ncm cm20 , 20a b\nand\ncm14c\n.\nThe s ( )\n2\nemi perimeter\ncm\n20 20 1 54\n27\n2 2\n4\na b c\ns\nArea of the triangle =\n( )( )( )s s a s b s c\n(By Heron’s formula)", null, "27 27 20 27 20 27 14\n27 7 7 13\n3 3 3 7 7 13\n921 3\n21 6.24\n= 131.04 cm\n2\nTotal area of the green portion = 242 +131.04 =\n373.04 cm\n2\n.\nHence, the papers required for each shade to make a\nkite are:\nRed paper= 242 cm\n2\n, yellow paper= 484 cm\n2\nand green\npaper= 373.04 cm\n2\n.", null, "Question: 2\nThe perimeter of a triangle is 50 cm. One side of a\ntriangle is 4 cm longer than the smaller side and the\nthird side is 6 cm less than twice the smaller side. Find\nthe area of the triangle.\nSolution\nLet the smaller side of a triangle be\nx\ncm.\nAccording to the question, the second side is 4 cm\nlonger than the smaller side\ni.e.,\ncm,4x\nthe third side = 6 cm less than twice the smaller side\nc.e., mi .2 6x", null, "Given that the perimeter of the triangle = 50 cm.\n4 2 6 50x x x\n4 2 50x\n4 52x\n52\n4\nx\ncm13x\nThe smaller side, .\ncm.13a\nThe second side,\ncm.13 4 17b\nThe third side,\ncm.2 13 6 26 6 20c\nNow, the semi perimeter is given by\ncm\n13 17 20 50\n25\n2 2 2\na b c\ns\nArea of the triangle\ns s a s b s c\n(By Heron’s formula)", null, "25 25 13 25 17 25 20\n25 12 8 5\n5 5 2 2 3 2 2 2 5\n5 2 2 30\n20 30\n2\ncm\nHence, the area of the triangle is\ncm\n2\n20 30\n.\nQuestion: 3\nThe area of a trapezium is 475 cm\n2\nand the height is 19\ncm. Find the lengths of its two parallel sides if one\nside is 4 cm greater than the other.\nSolution\nLet one of the parallel sides of trapezium be\nx\ncm.\nHence, the length of the other parallel side\ncm4x\n(given in the question)\nArea of trapezium", null, "1\n2\n(Sum of parallel Sides) ×distance between the\nparallel sides\n475 =\n1\n2\n(\nx\n+\nx\n+ 4) × 19\n4\n475\n2 19\n2\nx\ncm\n2 4 475\n2 19\n2 2\n475\n2 19\n475\n2\n19\n2 25\n23\nx\nx\nx\nx\nx\nOther Side\ncm4 23 4 27x", null, "Hence, the parallel sides are 23 cm and 27 cm.\nQuestion: 4\nA rectangular plot is given for constructing a house,\nhaving a measurement of 40 m long and 15 m in the\nfront. According to the laws, a minimum of 3 m, wide\nspace should be left in the front and back each and 2 m\nwide space on each of other sides. Find the largest\narea where house can be constructed.\nSolution\nLet\nABCD\nbe a rectangular plot which is 40 m long\nand 15 m in the front.\nAs per the laws, a minimum of 3 m, wide space\nshould be left in the front and back each.\nThe length of the inner-rectangle\nm.40 3 3 34\nSimilarly, the breadth of the innerrectangle,\nm.15 2 2 11\nLet the rectangular plot be\nABCD\nand the other\nrectangle\nEFGH\nwill be formed inside the rectangle\nABCD.\nArea of inner rectangle\nEFGH", null, "(Area of a rectangle= length × breadth)\n=34 ×11\n= 374 m\n2\nHence, the largest area where the house can be\nconstructed in 374 m\n2\nQuestion: 5\nA field is in the shape of a trapezium having parallel\nsides 90 m and 30 m. These sides meet the third side at\nright angles. The length of the fourth side is 100 m. If", null, "it costs Rs 4 to plough 1 m\n2\nof the field, find the total\ncost of ploughing the field.\nSolution\nLet the trapezium be\nABCD\n.\nNow, let us draw a perpendicular line\nCE\nto the line\nAB.\nWe have,\nm.30DC AE\nNow,\nm.90 30 60BE AB AE\nIn the right angled Δ\nBEC,\n2 2 2\nBC BE EC\n(Using Pythagoras theorem)", null, "2 2 2\n2\n2\n100 60\n10000 3600\n6400\nEC\nEC\nEC\nm EC 6400 80\nArea of a trapezium\n1\n2\n(Sum of parallel Sides) ×distance between\nparallel sides\n1\n2\n(\nAB\n+\nCD\n)×\nEC\n=\n1\n2\n(90+30) × 80\nm\n2\n120 80 48\n1\n0\n2\n0\nCost of ploughing the field of 1 m\n2\n= Rs 4\nThe cost of ploughing the field of 4800 m\n2\n= 4800 ×\n4 = Rs 19200\nHence, the total cost of ploughing the field is Rs\n19200.", null, "Question: 6\nIn Fig. 12.5,\nABC\nhas\nsides\ncm cm7.5 , 6.5AB AC\nand\ncm.7BC\nOn\nthe base\nBC\na parallelogram\nDBCE\nof same area as\nthat of\nABC\nis constructed. Find the height\nDF\nof\nthe parallelogram.\nFig. 12.5", null, "Solution\nThe sides of the triangle are\ncm cm7.5 , 7AB a BC b\nand\ncm.6.5CA c\nThe semi perimeter of the triangle\nABC\n,\n10\n.\ncm\nArea of\n+ + 7.5+7+6.5 21\n.5\n2 2 2\nthe ΔABC\na b c\ns\ns s a s b s c\n(By Heron’s formula)\n2\n10.5 10.5 7.5 10.5 7 10.5 6.5\n10.5 3 3.5 4\n441 21 ..................... icm\nNow, the area of the parallelogram\nBCED\n= Base ×\nHeight\n7 iiBC DF DF\nAccording to the question,", null, "Area of a Δ\nABC\n=Area of parallelogram\nBCED\n21 7 DF\n(From Equations (i) and (ii))\ncm\n1\n7\n3\n2\nDF\nHence, the height of the parallelogram is 3 cm.\nQuestion: 7\nThe dimensions of a rectangle\nABCD\nare 51 cm × 25\ncm. A trapezium\nPQCD\nwith its parallel sides\nQC\nand\nPD\nin the ratio 9:8 is cut off from the rectangle as\nshown in the Fig. 12.6. If the area of the trapezium", null, "PQCD\nis\nth\n5\n6\npart of the area of the rectangle, find\nthe lengths\nQC\nand\nPD\n. Fig. 12.6\nFig. 12.6\nSolution\nGiven, the dimensions of the rectangle\nABCD\nas 51\ncm × 25 cm.\nAlso, in trapezium\nPQCD\n, the parallel sides\nQC\nand\nPD\nare in the ratio 9:8.", null, ": 9: 8 QC PD\nLet, the length of\nbe 9xQC\nand\n8 .PD x\nThe area of the trapezium\nPQCD\n5\n6\narea of the\nrectangle\nABCD,\n1\n2\n(Sum of the parallel Sides × distance between\nthe parallel sides )\n5\n6\nBC CD\n1\n2\n(8\nx\n+ 9\nx\n) × 25 =\n5\n6\n× 51 × 25", null, "1\n2\n× 17\nx\n× 25 =\n5\n6\n× 51 × 25\nx\n=\n5 51 25 2\n25 17 6\n5x\ncm\nSo,\ncm9 5 45QC\ncm8 5 40PD\nQuestion: 8\nA design is made on a rectangular tile of dimensions\n50 cm × 70 cm as shown in Fig.12.7. The design", null, "shows 8 triangles, each of sides 26 cm, 17 cm and 25\ncm.\nFind the total area of the design and the remaining\narea of the tile.\nFig.12.7\nSolution\nThe dimension of rectangular tile is 50 cm × 70 cm.", null, "The area of the rectangular title = 50 ×70 = 3500\ncm\n2\n.\nThe sides of the triangle are\na\n= 25 cm,\nb\n= 17 cm and\nc\n= 26 cm.\nHence, the semi perimeter of the triangle is\n25 1 26\n34\n.\ncm.\nArea of the triangle\n7 68\n2 2 2\na b c\ns\ns s a s b s c\n(By Heron’s formula)\n(34 25) (34 17) (34 26)\n2\n34\n34 9 17 8\n17 2 3 3 17 2 2 2\n17 3 2 2 204 cm\nThe design is made of eight triangles.\nThe area of the design\ncm\n2\n204 8 1632\n.\nAlso, the remaining area of the tile = Area of the\nrectangular tileArea of the design.\n= 3500 1632 = 1868 cm\n2\n.\nHence, the total area of the design is 1632 cm\n2\nand the\nremaining area of the tile is 1868 cm\n2\n." ]

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[ "Q\n\n#### Solving Mixed Problems (× ÷) - Quiz 2\n\n(15 Questions)\n0 viewed last edited 3 years ago\nSolving Mixed Problems (× ÷)\n0\n2\nFind the missing value in the problem ? = 60 ÷ 10\n0\n2\nFind the missing value in the problem ? × 7 = 21\n0\n2\nFind the missing value in the problem 16 ÷ 2 = ?\n0\n2\nFind the missing value in the problem ? = 7 × 10\n0\n2\nFind the missing value in the problem 2 = ? ÷ 4\n0\n2\nFind the missing value in the problem 14 ÷ ? = 7\n0\n2\nFind the missing value in the problem ? × 6 = 24\n0\n2\nFind the missing value in the problem ? ÷ 6 = 8\n0\n2\nFind the missing value in the problem 16 ÷ 2 = ?\n0\n2\nFind the missing value in the problem 60 ÷ 6 = ?\n0\n2\nFind the missing value in the problem ? = 4 × 3\n0\n2\nFind the missing value in the problem 16 ÷ ? = 8\n0\n2\nFind the missing value in the problem 4 × ? = 24\n0\n2\nFind the missing value in the problem 28 ÷ 7 = ?\n0\n2\nFind the missing value in the problem 14 ÷ ? = 7", null, "" ]

[ null, "https://d3648m43e37g8z.cloudfront.net/ailzx08rta8xrwh2d7zjhx0ht62xjp27bnltrvz", null ]

[ "", null, "", null, "## Encyclopedia > Semigroup\n\nArticle Content\n\n# Semigroup\n\nIn mathematics, a semigroup is a set with an associative binary operation on it.\n\nA semigroup with an identity element is called a monoid. Any semigroup S may be turned into a monoid simply by adjoining an element e not in S and defining ee = e and es = s = se for all sS.\n\nSome examples of semigroups:\n\n• The positive integers with addition.\n• Any monoid, and therefore any group.\n• Any ideal of a ring, with the operation of multiplication.\n• Any subset of a semigroup which is closed under the semigroup operation.\n• The set of all finite strings over some fixed alphabet Σ, with string concatenation as operation. If the empty string is included, then this is actually a monoid, called the \"free monoid over Σ\"; if it is excluded, then we have a semigroup, called the \"free semigroup over Σ\".\n\nTwo semigroups S and T are said to be isomorphic if there is a bijection f : ST with the property that, for any elements a, b in S, f(ab) = f(a)f(b). In this case, T and S are also isomorphic, and for the purposes of semigroup theory, the two semigroups are identical.\n\nA number of concepts are useful for understanding the structure of semigroups, some of which will be described below. For brevity, the semigroup operation will be shown simply by juxtaposition, that is, xy denotes the result of applying the semigroup operation to the ordered pair (xy). If A and B are subsets of some semigroup, then AB denotes the set { ab | a in A and b in B }.\n\nA subset A of a semigroup S is called a subsemigroup if it is closed under the semigroup operation, that is, AA is a subset of A. If A is nonempty then A is called a right ideal if AS is a subset of A, and a left ideal if SA is a subset of A. If A is both a left ideal and a right ideal then it is called an ideal (or a two-sided ideal). The intersection of two ideals is also an ideal, so a semigroup can have at most one minimal ideal. All nonempty finite semigroups have a minimal ideal. An example of semigroup with no minimal ideal is the set of positive integers under addition. The minimal ideal of a commutative semigroup, when it exists, is a group.\n\nIf S is a semigroup, then the intersection of any collection of subsemigroups of S is also a subsemigroup of S. So the subsemigroups of S form a complete lattice. For any subset A of S there is a smallest subsemigroup T of S which contains A, and we say that A generates T. A single element x of S generates the subsemigroup { xn | n is a positive integer }. If this is finite, then x is said to be of finite order, otherwise it is of infinite order. A semigroup is said to be periodic if all of its elements are of finite order. Finite semigroups are clearly periodic. A semigroup generated by a single element is said to be monogenic (or cyclic). If a monogenic semigroup is infinite then it is isomorphic to the semigroup of positive integers with the operation of addition. If it is finite, then it must contain an idempotent, in fact, exactly one. It follows that every nonempty periodic semigroup has at least one idempotent.\n\nA subsemigroup which is also a group is called a subgroup. There is a close relationship between the subgroups of a semigroup and its idempotents. Each subgroup contains exactly one idempotent, namely the identity element of the subgroup. For each idempotent e of the semigroup there is a unique maximal subgroup containing e. Each maximal subgroup arises in this way, so there is a one-to-one correspondence between idempotents and maximal subgroups. (It should be noted that the use of the term maximal subgroup is different here than it is in group theory. In group theory, a so-called \"maximal subgroup\" is really a maximal proper subgroup. When considered as a semigroup, a group has only one maximal subgroup, namely itself.)\n\nAll Wikipedia text is available under the terms of the GNU Free Documentation License\n\nSearch Encyclopedia\n Search over one million articles, find something about almost anything!\n\nFeatured Article\n U.S. presidential election, 1804 ... 1804, 1808, 1812, 1816 Source: U.S. Office of the Federal Register (http://www.archives.gov/federal_register/electoral_college/scores.html#1804) (Larg ...", null, "", null, "" ]

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[ "MySQL Forums\n\nRunning python on wsl ubuntu while managing data with mysql workbench on windows\nPosted by: David Kaeser\nDate: March 04, 2022 06:57AM\n\nHello everyone\n\nI need to collect data out of a bunch of csv files (16 GB) and make with the usful data some new csv files. To achieve this I would like to acess the data with a python code in my ubuntu subsystem environment and then ananlyze the data with mysql workbench on windows. However as for now I do not understand how I can link mysql workbench with the gathered data in the subsystem. How can I achieve this?\n\nFor the gathering of the data I use the following python code:\n\n# -*- coding: utf-8 -*-\n\n#\n# The script will drop any existing table with the same name and schema then\n# set in the settings, then read all csv files that contain the key word\n# (string) and create a new table appending all data from the fitting csv files\n##############################################################################\nfrom sqlalchemy import create_engine\nimport import_csv_to_sql_functions as sqlf\n\ndata_dir = \"path\"\n#\n#\n#\n# filtering key word (only files containing this will be included)\nfiltering_kw = \"Unavailability\"\n#\n# name of mysql schema to add table\nmysql_schema = \"raw\"\n#\n# name of mysql table to add data in\nmysql_table = \"generator_data\"\n#\n##############################################################################\n\nif __name__ == '__main__':\n\nconnection = engine.connect()\nsqlf.insert_into_sql_table(engine, data_dir, mysql_schema, mysql_table,\nfiltering_kw)\nprint(\"Done!\")\n\nWith the function defined here:\n# -*- coding: utf-8 -*-\nimport os\nimport time\nimport pandas as pd\n\n# function used to insert data into sql database\n\ndef insert_into_sql_table(engine, data_dir, mysql_schema, mysql_table,\nfiltering_kw = None):\n# drop table if already existant\nconnection = engine.connect()\nconnection.execute(\"DROP TABLE IF EXISTS \"+mysql_schema+\".\"+mysql_table+\";\")\nconnection.close()\nfiles = os.listdir(data_dir)\nfor file in files:\n# checking if keyword is in filename, else file will not be imported\nif ((filtering_kw is not None and filtering_kw in file) or (filtering_kw is None)):\nprint(\"##############################################\")\nprint(file)\nstart_time = time.time()\n# load csv file as pandas dataframe\ndf = pd.read_csv(os.path.join(data_dir, file), sep = '\\t')\nintermediate_time = time.time()\n# insert dataframe into mysql database\ndf.to_sql(name = mysql_table, con = engine, schema = mysql_schema,\nif_exists = 'append', chunksize = 10000, method = 'multi', index = False)\nend_time = time.time()\nprint(\"csv load time: \", round(intermediate_time-start_time,0), \"s\")\nprint(\"Insert time: \", round(end_time - intermediate_time,0), \"s\")\nprint(\"Total Time: \", round((end_time - start_time)/60,1), \"m\")\nreturn True\n\nI would grealy appraciate any help or tipps.\nBest\nDavdi" ]

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[ "# How To Solve Decimal Ratio\n\nHow To Solve Decimal Ratio – Fraction to Decimal calculator is one of the best instruments utilized for fraction calculations. It’s hassle-free as well as simple to use. Fractional info are essential in scientific research and design job areas as they assist us fix issues. We could remedy for the beginnings of sophisticated difficulties using Fraction to Decimal Calculators.", null, "Fraction to Decimal Worksheets provide a number of valuable characteristics which you can use to perform a range of fractional data examination. The Fraction to Decimal Calculator worksheet is split into two key elements. Fraction to Decimal Fractionial and Worksheet Details Sheet. The Fractionial Data Sheet includes each of the Fraction ideals found in the Fraction to Decimal Worksheet. The Fraction to Decimal Worksheet can be used to perform a wide array of numerical computations.\n\nFraction to Decimal calculator is split into several major sections. Fractional Details Sheet, Fraction to Decimal Percentage calculator, Fraction to Decimal calculator with logarithm, and Fraction to Decimal calculator. Fraction to Decimal calculator with logarithm is popular in educational facilities to discover statistical principles.", null, "Fraction to Decimals Calculator with Logarithm offers a variety of options to solve for Fraction to Decimals proportions. Fraction to Decimals Proportion can be computed as long as you recognize how Fraction ideals are extracted. Fraction beliefs can even be produced from other products such as oz ., kgs and cereals and so forth. You may choose Fraction to Decimals rate work in Fraction to Decimal calculator with Logarithm alternative to select a desired model.", null, "" ]

[ null, "https://decimalworksheets.co/wp-content/uploads/2021/07/simplifying-ratios-decimals-youtube-5.jpg", null, "https://decimalworksheets.co/wp-content/uploads/2021/07/decimal-to-ratio-calculator-inch-calculator-36.png", null, "https://decimalworksheets.co/wp-content/uploads/2021/07/simplifying-ratios-involving-decimals-and-fractions-youtube-11.jpg", null ]

[ "", null, "SOLVED\n\n# Calculate difference, pos or neg between numbers\n\nI'm new to Excel and having a problem figuring this out. Here's a simplified explanation of what I'm trying to do:\n\nIn A2 I have 100\n\nIn A3 I have 150, an increase of 50\n\nIn B3 I want to show the increase or decrease between A2 and A3, in positive or negative numbers.\n\nIn B3 I used =A2-A3 and got -50 instead of 50\n\nI've not been able to find the correct formula to show positive or negative change.\n\n8 Replies\nbest response confirmed by Grahmfs13 (Microsoft)\nSolution\n\n# Re: Calculate difference, pos or neg between numbers\n\nI you want an increase to be positive and a decrease to be negative, use =A3-A2\n\nIf you always want a positive (or zero) result, use =ABS(A3-A2)\n\n# Re: Calculate difference, pos or neg between numbers\n\nThanks much. I see how that works now. Another question:\n\nIn the attached file, for blank row 4 cell C4 now shows -150. How can my formula be modified so this would remain blank until B4 is filled? I appreciate your assistance.\n\n# Re: Calculate difference, pos or neg between numbers\n\nIn C3:\n\n=IF(B3=\"\",\"\",B3-B2)\n\nFill down.\n\n# Re: Calculate difference, pos or neg between numbers\n\nThanks, Hans. That works beautifully, though I don't understand it. I inherited a workbook from the prior official handling it for our organization, and he didn't understand these things either! I will have to study and learn to read the formulas. I appreciate your time.\n\n# Re: Calculate difference, pos or neg between numbers\n\nThe formula says: if the value of the cell to the left (B3) is equal to an empty string \"\", i.e. if B3 is blank, then return an empty string \"\" so that C3 looks blank. Otherwise, return the difference of B2 and B3.\n\n# Re: Calculate difference, pos or neg between numbers\n\nA very succinct explanation of how the formula reads in a way even a caveman can understand. Would you know where I can find the same for other formulas?\n\n# Re: Calculate difference, pos or neg between numbers\n\nIf you'd like to learn about formulas and functions, see for example Formulas and Functions and the links provided there.\n\nThanks" ]

[ null, "https://www.facebook.com/tr", null ]

[ "", null, "## Total Pressure Boundary Condition\n\nTotal pressure is uniform total pressure applied everywhere in the inlet patch. Total pressure value is equal to p0.\n\np:\n\n``` boundaryField\n{\n\".*_inlet\"\n{\ntype totalPressure;\np0 uniform 97.5;\ngamma\t\t 0.0;\nvalue \\$internalField;\n}\n}\n```\n\nU:\n\n``` boundaryField\n{\n\".*_inlet\"\n{\ntype pressureInletVelocity;\nvalue \\$internalField;\n}\n}\n```\n\nor for velocity U:\n\n``` \".*_inlet\"\n{\ntype pressureDirectedInletVelocity;\ninletDirection uniform (0 0 1);\nvalue \\$internalField;\n}\n```\n\npressureDirectedInletVelocity prescribes velocity angles at the inlet, but the velocity magnitude is computed from total pressure – static pressure difference." ]

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[ "Algebra 1\n\na) They have opposite signs. b) $q$\na) Since $c=-30$, $p*q = -30$, so either $p$ or $q$ is negative. (The other is positive.) b) Since $13 > 0$, $q$ has the lower absolute value and is negative." ]

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[ "## Tuesday, December 10, 2019\n\n### Exploratory Data Analysis EDA Cheat Sheet\n\nPandas\n\n• pandas.DataFrame.shape -- > (row_count, col_count)\n• pandas.DataFrame.shape --> number of records, number of samples in the dataset\n• my_dataframe['my_series_name'].unique() --> returns a unique values of a column, \"radio button choices\"\n• len(my_dataframe['my_series_name'].unique()) --> number of unique values\n• import os os.listdir('name_of_directory_or_just_use_.') --> list the files in the current directory '.' os.listdir('.')  or a specific directory with a name\n• import os len(os.listdir('.') ) --> returns the number of files in the current directory\n• my_dataframe.groupby(['col_1', 'col_2']) --> groupby column 1 first then groupby column 2\n• Converting a Pandas GroupBy output from Series to DataFrame: .groupby() returns a groupby object with MultiIndex instead of a dataframe with a single index. it is also known as a hierarchical index. Will need to rename columns and reset index my_groupby.add_suffix('_Count').reset_index() or call the .size().reset_index() important to note that .size() is called on the groupby object not the usual dataframe. pandas.core.groupby.GroupBy.size calculates : Series Number of rows in each group\n• group = ['col_1', 'col_2']; my_df.groupby(group).size().reset_index(name=\"colum_name\")\n• df = df[(df.col_name < 1) & (df.col_name_2 < 1)] complex condition query / filter in dataframe\n• .value_count df.column.value_count()\n• pandas cheatsheet\n• .copy()\n• .unique()\n• df['col_name'].isnull().sum()\n• df['col_name'].min()\n• df['col_name'].max()\n• df.fillna(0) #fill the dataframe with zero the entire table" ]

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[ "# C++ Classes and Objects\n\nThis post will teach you everything you need to know about classes and objects in C++, an object-oriented feature. So, without further ado, let's get started with the \"class\" first.\n\n## C++ class\n\nAs you already know, a class represents a group of similar objects. A class is a way to bind the data describing an entity and its associated functions together. For instance, consider an account having characteristics such as number, type, and balance. Deposit and withdrawal are two of its associated operations.\n\nA class definition begins with the keyword \"class,\" then the class name and the class body, all surrounded by curly braces. A class definition must be followed either by a semicolon or a list of declarations.\n\nThe following is the general syntax for creating a class in C++.\n\n```class className\n{\n.\n.\n.\n};```\n\nThe keyword \"class\" is used to create a class, and \"className\" is the name of the class that will be created. We define attributes and methods within and outside of the access specifiers \"public,\" \"private,\" and \"protected\" within the class. As an example:\n\n```class CodesCracker\n{\nint tot;\npublic:\nvoid printMsg()\n{\ncout<<\"Sum = \";\n}\nvoid sum(int a, int b)\n{\ntot = a+b;\ncout<<tot;\n}\n};```\n\nIn the above code fragment, a class named \"CodesCracker\" is defined, in which an \"int\" variable named \"tot\" is defined. And then two methods, namely \"printMsg()\" and \"sum(),\" are defined in the \"public\" label.\n\n### Class Method's Definition: Outside the scope of the class definition\n\nThe functions \"printMsg()\" and \"sum()\" in the preceding code fragment are the two member functions of the class \"CodesCracker,\" which are defined in the class definition. However, we can also define the member function of a class later in the program instead of defining it inside the class.\n\nA member function definition that is not part of a class definition is similar to a function definition that you are already familiar with. The only difference here is that the function name is the full name of the function, also known as the qualified name of a function, which is written as:\n\n`class-name :: function-name`\n\nIn this case, class-name denotes that the function specified by function-name belongs to the class specified by class-name. And the symbol ::, also known as the scope resolution operator, specifies that the function's scope is limited to the class class-name. Outside of the class definition, the following is the general form of a member function definition:\n\n```return-type class-name :: function-name(parameter list)\n{\n// function body\n}```\n\n### Class Declaration in C++\n\nA class's declaration in C++ includes the declaration of its four associated attributes:\n\n• Data members are data type properties that describe a class's characteristics. A class can have zero or more data members of any type.\n• Members functions are the operations that can be performed on objects of that class. A class can have zero or more member functions.\n• Program access levels govern access to members within the program. These access levels are classified as private, protected, or public.\n• class tagname that serves as the class's type specifier, allowing objects of this class type to be created.\n\nWhen you create a new class, you're actually creating a template for a new data type. This doesn't specify what data will be used, but it does specify what a class object is and what operations can be applied to it.\n\n## C++ Object\n\nWhen you define a class, you are not defining or creating objects of that class; rather, you are specifying the types of information that the objects of this class type will contain.\n\n## Create C++ objects\n\nThe object of any class is declared in the same way that the basic type variables are declared. For example, the following statement declares an object of the class \"CodesCracker\" named \"obj.\"\n\n`CodesCracker obj;`\n\n## Accessing the data members of the class in C++\n\nUsing the \"dot (.)\" operator, the variable (object) \"obj\" can now access all of the attributes and methods of the class \"CodesCracker\" as follows:\n\n`obj.printMsg();`\n\nThe preceding statement called the method \"printMsg()\" from the class \"CodesCracker.\" Outside of the class, attributes and members of the \"PRIVATE\" label are inaccessible.\n\nNow is the time to look at an example to gain a better understanding of C++ classes and objects.\n\n```#include <iostream>\nusing namespace std;\n\nclass CodesCracker\n{\nint tot;\npublic:\nvoid printMsg()\n{\ncout<<\"Sum = \";\n}\nvoid sum(int a, int b)\n{\ntot = a+b;\ncout<<tot;\n}\n};\n\nint main()\n{\nint x = 10, y = 50;\n\nCodesCracker ob;\n\nob.printMsg();\nob.sum(x, y);\n\ncout<<endl;\nreturn 0;\n}```\n\nThis program's output should be exactly:\n\n`Sum = 60`\n\nLet me now walk you through the above program step by step:\n\n• I created a class named \"CodesCracker.\"\n• I declared a variable of type \"int\" called \"total\" within the class.\n• Then I made a \"public\" label and defined two methods within it.\n• That is, I created the first method, \"printMsg(),\" which does not take an argument and does not return a value. This method only prints \"Sum = \" to the output console when called.\n• The second method, \"sum(),\" takes two \"int\" arguments, adds them, and then prints the addition result on the output console.\n• Now inside the main(), I declared and initialized two variables of the \"int\" type, namely \"x\" and \"y.\"\n• Then I made a \"ob\" object of the class \"CodesCracker.\"\n• Now, using the object \"ob,\" I accessed the methods \"printMsg()\" and \"sum().\"\n• On the output console, the function \"printMsg()\" displays the text \"Sum = \".\n• Then I invoked the \"sum()\" method, passing \"x\" and \"y\" as parameters; thus, the values \"x\" and \"y\" will be copied to \"a\" and \"b.\" That is, 10 and 50 will be added and printed on the output console.\n\n## C++ Classes and Objects Example Program\n\nHere is a practical example program demonstrating the concept of classes and objects in C++.\n\n```#include <iostream>\nusing namespace std;\nclass ITEM\n{\nint itemcode;\nfloat itprice;\n\npublic:\nvoid initialize(void);\nfloat largest(void);\nfloat sum(void);\nvoid displayitems(void);\n};\nvoid ITEM::initialize(void)\n{\nfor (int i = 0; i < 5; i++)\n{\ncout << \"---Supply data for item no. \" << (i + 1) << \"---\";\ncout << \"\\nEnter the item code: \";\ncin >> itemcode[i];\ncout << \"Enter the item price: \";\ncin >> itprice[i];\ncout << \"\\n\";\n}\n}\nfloat ITEM::largest(void)\n{\nfloat larg = itprice;\nfor (int i = 1; i < 5; i++)\n{\nif (larg < itprice[i])\n{\nlarg = itprice[i];\n}\n}\nreturn larg;\n}\nfloat ITEM::sum(void)\n{\nfloat sum = 0;\nfor (int i = 0; i < 5; i++)\n{\nsum = sum + itprice[i];\n}\nreturn sum;\n}\nvoid ITEM::displayitems(void)\n{\ncout << \"\\nCode\\tPrice\\n\";\nfor (int i = 0; i < 5; i++)\n{\ncout << itemcode[i] << \"\\t\";\ncout << itprice[i] << \"\\n\";\n}\n}\n\nint main()\n{\nITEM order;\norder.initialize();\nfloat tot, big;\nint ch = 0;\n\ndo\n{\ncout << \"\\n1. Display the Largest Price\\n\";\ncout << \"2. Display the Sum of Prices\\n\";\ncout << \"3. Display Item Lists\\n\";\ncout << \"4. Exit\\n\";\ncout << \"Enter your choice(1-4): \";\ncin >> ch;\n\nswitch (ch)\n{\ncase 1:\nbig = order.largest();\ncout << \"Largest Price = \" << big;\nbreak;\ncase 2:\ntot = order.sum();\ncout << \"Sum of Prices = \" << tot;\nbreak;\ncase 3:\norder.displayitems();\nbreak;\ncase 4:\nexit(0);\ndefault:\ncout << \"\\nWrong choice.\";\nbreak;\n}\ncout << \"\\n\";\n} while (ch >= 1 && ch <= 4);\n\ncout<<endl;\nreturn 0;\n}```\n\nThe following snapshot depicts the above program's initial output:\n\nNow supply the inputs for five items; for example, type \"1001\" as an item code and hit the ENTER key, then type \"430\" as its price and hit the ENTER key again. Now supply the data for item number two: say \"1002\" as the item code and \"455\" as its price. In this way, supply item codes and their prices for five items. The following snapshot shows the sample run:\n\nYou can now enter \"1\" and press the ENTER key to display the highest price, or enter 2, 3, or 4 to display the sum of prices, item lists, or exit the program. Please take a look at the image below:\n\nYou can type \"4\" and hit the ENTER key to exit the program.\n\nIn the preceding program, I created an object \"order\" of the type \"ITEM\" (class) and used it to call the \"initialize()\" method, which receives five item codes and their prices, which are stored in the \"itemcode\" and \"itprice\" variables. Then I used the \"do-while\" loop to show the user the menu so they could select the appropriate option to get the desired result on the output console.\n\nThat is, when the user enters \"1\" as an option and presses the ENTER key, the class \"ITEM\"'s member function named \"largest()\" is invoked via its object \"order.\" And this function will find the highest price among all five items' prices. If the user selects \"2\" or \"3,\" I called the other two methods \"sum()\" and \"displayitems()\" in a similar manner.\n\nPlease note: Only the class's member functions have access to private members. These are not directly accessible through objects. The scope of public members, on the other hand, is determined by the referencing object. The scope of the public member is local if the referencing object is local. The scope of the public member is global if the referencing object is global.\n\nWhen you need to use the value of the global variable instead of the local one, then you can use the scope resolution (::) operator. For example:\n\n```#include <iostream>\nusing namespace std;\nint a = 100;\nint main()\n{\nint a = 150;\ncout << \"a = \" << a << \"\\n\";\ncout << \"::a = \" << ::a << \"\\n\\n\";\n\n{\nint a = 250;\ncout << \"Inner Block\\n\";\ncout << \"a = \" << a << \"\\n\";\ncout << \"::a = \" << ::a << \"\\n\\n\";\n}\n\nreturn 0;\n}```\n\nThe output should be:\n\n```a = 150\n::a = 100\n\nInner Block\na = 250\n::a = 100```\n\n### C++ classes and objects: created an object array\n\nIn C++, you can make an array of objects. As an example, consider the following program:\n\n```#include <iostream>\nusing namespace std;\nclass myClass\n{\npublic:\nint num;\n};\nint main()\n{\nmyClass ob;\nint i;\n\ncout<<\"Enter five numbers: \";\nfor(i=0; i<5; i++)\ncin>>ob[i].num;\n\ncout<<\"\\nYou entered: \";\nfor(i=0; i<5; i++)\ncout<<ob[i].num<<\" \";\n\ncout<<endl;\nreturn 0;\n}```\n\nThe following snapshot shows the sample run of the above program with user inputs of 10, 20, 40, 56, and 87 as five number inputs.\n\nC++ Quiz\n\n« Previous Tutorial Next Tutorial »" ]

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[ "# StrictMath min() Method in Java with Examples\n\nThe java.lang.StrictMath.min() method returns the smaller of two values. There are four variations of this method with different types of parameters passed.\nAll of them are discussed below :\n\n1. min(double num1, double num2) is the inbuilt method of StrictMath class which is used to get the least of given two double values arguments. It returns NaN when any of the arguments is NaN. It returns the same value when num1 and num2 have the same value. min() method assumes negative zero to be strictly smaller than positive zero.It returns negative zero when one argument is positive zero and the other is negative zero.\nSyntax :\n\n`public static double min(double num1, double num2)`\n\nParameters : The method accepts a two parameter :\n\n• num1 of double type representing one parameter\n• num2 of double type representing another parameter\n\nReturn Value : The method returns the least of num1 and num2.\n\nExamples :\n\n```Input:\nnum1 = 9\nnm2 = 99\nOutput: 9.0\n```\n\nBelow programs illustrate the Java.lang.StrictMath.min() Method.\nProgram 1:\n\n `// Java praogram to illustrate the ` `// Java.lang.StrictMath.min() Method ` `// with double values passed ` `// as parameters ` `import` `java.lang.*; ` ` `  `public` `class` `Geeks { ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` ` `  `        ``double` `num1 = ``10``, num2 = ``40``, num3 = -``25``, num4 = -``25``, ` `               ``num5 = -``17``; ` ` `  `        ``double` `min_Value = StrictMath.min(num1, num2); ` `        ``System.out.println(``\"min of the two num is \"` `+ min_Value); ` ` `  `        ``min_Value = StrictMath.min(num3, num4); ` `        ``System.out.println(``\"min of the two num is \"` `+ min_Value); ` ` `  `        ``min_Value = StrictMath.min(num4, num5); ` `        ``System.out.println(``\"min of the two num is \"` `+ min_Value); ` `    ``} ` `} `\n\nOutput:\n\n```min of the two num is 10.0\nmin of the two num is -25.0\nmin of the two num is -25.0\n```\n\nExample of Error Condition :\n\n `// Java praogram to illustrate the ` `// error condition in ` `// Java.lang.StrictMath.min() Method ` `import` `java.lang.*; ` ` `  `public` `class` `Geeks { ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` ` `  `        ``double` `num1 = ``51``, num2 = ``71``, num3 = ``3``, num4 = -``93``, ` `               ``num5 = -``93``; ` `        ``double` `a = ``0.0``; ` ` `  `        ``num1 = a / ``0.0``; ` `        ``double` `min_Value = StrictMath.min(num1, a); ` `        ``System.out.println(``\"min of the two num is \"` `+ min_Value); ` `    ``} ` `} `\n\nOutput:\n\n```min of the two num is NaN\n```\n2. The min(float num1, float num2) is the inbuilt method of StrictMath class which is used to get the least of given two float values arguments.It returns NaN when any of the arguments is NaN.It returns the same value when num1 and num2 have the same value. min() method assumes negative zero to be strictly smaller than positive zero.It returns negative zero when one argument is positive zero and the other is negative zero.\nSyntax :\n\n`public static float min(float num1, float num2)`\n\nParameters : The method accepts a two parameter :\n\n• num1 of float type representing one parameter\n• num2 of float type representing another parameter\n\nReturn Value : The method returns the least of num1 and num2.\n\nExamples :\n\n```Input:\nnum1 = 9\nnm2 = 5\nOutput: 5.0\n```\n\nBelow programs illustrate the Java.lang.StrictMath.min() Method.\nProgram 1:\n\n `// Java praogram to illustrate the ` `// Java.lang.StrictMath.min() Method ` `// with float values passed ` `// as parameters ` `import` `java.lang.*; ` ` `  `public` `class` `Geeks { ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` ` `  `        ``float` `num1 = ``28``, num2 = ``82``, num3 = -``23``, num4 = -``23``, ` `              ``num5 = -``11``; ` ` `  `        ``float` `min_Value = StrictMath.min(num1, num2); ` `        ``System.out.println(``\"min of the two num is \"` `+ min_Value); ` ` `  `        ``min_Value = StrictMath.min(num3, num4); ` `        ``System.out.println(``\"min of the two num is \"` `+ min_Value); ` ` `  `        ``min_Value = StrictMath.min(num4, num5); ` `        ``System.out.println(``\"min of the two num is \"` `+ min_Value); ` `    ``} ` `} `\n\nOutput:\n\n```min of the two num is 28.0\nmin of the two num is -23.0\nmin of the two num is -23.0\n```\n3. The min(int num1, int num2) is the inbuilt method of StrictMath class which is used to get the least of given two int values arguments.It returns NaN when any of the arguments is NaN.It returns the same value when num1 and num2 have the same value. min() method assumes negative zero to be strictly smaller than positive zero.Simply the argument closer to the value of Integer.MIN_VALUE is the result.\nSyntax :\n\n`public static int min(int num1, int num2)`\n\nParameters : The method accepts a two parameter :\n\n• num1 of int type representing one parameter\n• num2 of int type representing another parameter\n\nReturn Value : The method returns the least of num1 and num2.\n\nExamples :\n\n```Input:\nnum1 = 61\nnm2 = 18\nOutput: 5.0\n```\n\nBelow programs illustrate the Java.lang.StrictMath.min() Method.\nProgram 1:\n\n `// Java praogram to illustrate the ` `// Java.lang.StrictMath.min() Method ` `// with int values passed ` `// as parameters ` `import` `java.lang.*; ` ` `  `public` `class` `Geeks { ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` ` `  `        ``int` `num1 = ``51``, num2 = ``71``, num3 = -``74``, num4 = -``93``, ` `            ``num5 = -``93``; ` ` `  `        ``double` `min_Value = StrictMath.min(num1, num2); ` `        ``System.out.println(``\"min of the two num is \"` `+ min_Value); ` ` `  `        ``min_Value = StrictMath.min(num3, num4); ` `        ``System.out.println(``\"min of the two num is \"` `+ min_Value); ` ` `  `        ``min_Value = StrictMath.min(num4, num5); ` `        ``System.out.println(``\"min of the two num is \"` `+ min_Value); ` `    ``} ` `} `\n\nOutput:\n\n```min of the two num is 51.0\nmin of the two num is -93.0\nmin of the two num is -93.0\n```\n4. The min(long num1, long num2) is the inbuilt method of StrictMath class which is used to get the least of given two long values arguments.It returns NaN when any of the arguments is NaN.It returns the same value when num1 and num2 have the same value. min() method assumes negative zero to be strictly smaller than positive zero.Simply the argument closer to the value of Long.MIN_VALUE is the result.\nSyntax :\n\n`public static long min(long num1, long num2)`\n\nParameters : The method accepts a two parameter :\n\n• num1 of long type representing one parameter\n• num2 of long type representing another parameter\n\nReturn Value : The method returns the least of num1 and num2.\n\nExamples :\n\n```Input:\nnum1 = 51617\nnm2 = 1345\nOutput: 1345.0\n```\n\nBelow programs illustrate the Java.lang.StrictMath.min() Method.\nProgram 1:\n\n `// Java praogram to illustrate the ` `// Java.lang.StrictMath.min() Method ` `// with long values passed ` `// as parameters ` `import` `java.lang.*; ` ` `  `public` `class` `Geeks { ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` ` `  `        ``long` `num1 = ``265626``, num2 = ``66671``, num3 = -``776264``, num4 = -``9263``, ` `             ``num5 = -``97623``; ` ` `  `        ``double` `min_Value = StrictMath.min(num1, num2); ` `        ``System.out.println(``\"min of the two num is \"` `+ min_Value); ` ` `  `        ``min_Value = StrictMath.min(num3, num4); ` `        ``System.out.println(``\"min of the two num is \"` `+ min_Value); ` ` `  `        ``min_Value = StrictMath.min(num4, num5); ` `        ``System.out.println(``\"min of the two num is \"` `+ min_Value); ` `    ``} ` `} `\n\nOutput:\n\n```min of the two num is 66671.0\nmin of the two num is -776264.0\nmin of the two num is -97623.0\n```\n\nAttention reader! Don’t stop learning now. Get hold of all the important Java and Collections concepts with the Fundamentals of Java and Java Collections Course at a student-friendly price and become industry ready.\n\nMy Personal Notes arrow_drop_up", null, "lets make it a lil simple\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nArticle Tags :\nPractice Tags :\n\nBe the First to upvote.\n\nPlease write to us at [email protected] to report any issue with the above content." ]

[ null, "https://media.geeksforgeeks.org/auth/profile/lkhw2dg8v5u237flnw3o", null ]

[ "# The difference between == and === in JavaScript\n\n## \"==\" case\n\n1. Two different basic data types: if the two basic data types (string, number) are different, they will be converted into data types, and then compared. For example:\n\n``````0 == '' // true\n0 == '0' // true\n1 == '1' // true\n'0' == '' // false\n\nfalse == '0' // true\nfalse == 'false' // false\nfalse == null // false\nfalse == undefined // false\nnull == undefined // true\n\n0 === '' // false\n0 === '0' // false\n1 === '1' // false\n\nfalse === '0' // false\nnull === undefined // false``````\n\n2. Basic Data Type and Reference Data Type: Transform Reference Data Type (function, object, which includes array) into its original value (background automatically calls `Object.prototype.valueOf()) and compare them. For example:\n\n``````[1,2] == '0' // true\nfunction(){return ['1,2']}() == '1,2' // true\n\n[1,2] === '0' // false\nfunction(){return ['1,2']}() === '1,2' // false``````\n\n3. Two reference types: judging the reference path between them. For example:\n\n``````[1,2] == [1,2] // false\n[1,2] === [1,2] // false``````\n\n## \"===\" case\n\nAs can be seen from the above examples, without type conversion, different types must return to fasle.\n\nSummary: Recommended use \"===\", as long as the type is inconsistent, return to fasle directly; \"==\" will cause type conversion and comparison, not rigorous." ]

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[ "•", null, "# Quantum computing: Reversible evolution\n\nHow do we design and execute quantum algorithms? Learn how quantum gates work, and why they all have to be reversible.\n\nIn quantum mechanics, every change except measurement (and noise that damages the state, known as decoherence, which we will discuss in the last week when we discuss hardware and quantum error correction) must be reversible. (In mathematical terms, this means that it can be represented by a unitary matrix.) This means that it must be possible for us to recreate the initial state using only the output state, without additional information. Here, we will introduce some of those changes as discrete operations which we call gates.\n\n## Basic Classical Logic Gates\n\nYou may already be familiar with the most basic logic gates we use in creating a classical computer, but just in case you’re not, let’s review them. Some of the gates have one input bit and one output bit, others have two input bits and only one output bit.\n\n### NOT\n\nThe NOT gate, as you might expect, flips a single bit.\n\ninput output\n0 1\n1 0\n\nThe NOT gate is reversible: if you apply a NOT gate twice to the same signal, you get out the same value you started with: NOT(NOT(X)) = X.\n\n### AND\n\nThe AND gate takes two input bits, and its output is one only if both input bits are one:\n\ninput (A) input (B) output (C)\n0 0 0\n0 1 0\n1 0 0\n1 1 1\n\nThe AND gate is not reversible: there are four different possible input states (00, 01, 10, 11) and only two possible output states (0 and 1), so there isn’t enough information in the output to know for sure what the inputs were. In the case that the output C is 1, you know that both inputs were 1, but in all three of the other cases, you can’t tell.\n\n### OR\n\nThe OR gate takes two input bits, and its output is zero only if both input bits are zero:\n\ninput (A) input (B) output C\n0 0 0\n0 1 1\n1 0 1\n1 1 1\n\nIts behavior in a lot of ways is similar to that of the AND gate. The OR gate is not reversible: there are four different possible input states (00, 01, 10, 11) and only two possible output states (0 and 1), so there isn’t enough information in the output to know for sure what the inputs were. In the case that the output C is 0, you know that both inputs were 0, but in all three of the other cases, you can’t tell.\n\nIn fact, with the AND or OR gate, even if I gave you the output and one of the inputs, you wouldn’t be able to tell unambiguously what the other input value was in all cases.\n\n### XOR\n\nThe XOR, or exclusive OR, gate takes two input bits, and its output is one only if exactly one input bit is one:\n\ninput (A) input (B) output (C)\n0 0 0\n0 1 1\n1 0 1\n1 1 0\n\nThe XOR gate is close to reversible: with two input bits and one output bit, there still isn’t enough information in the output to know for sure what the inputs were. But, if I give you one of the inputs, you can now tell unambiguously what the other input was! We will see below that one of the important reversible two-qubit gates is related to XOR.\n\n## Reversible Classical Gates\n\n### Entropy and Information\n\nIn the early 1970s, Charles Bennett of IBM, later joined by Richard Feynman of Caltech, and Tommaso Toffoli and Ed Fredkin of MIT, laid the groundwork for discrete reversible logic operations. Bennett was an acolyte of Rolf Landauer, who recognized that the destruction of information adds to the entropy of the Universe, and must generate waste heat.\n\nEntropy is the amount of disorder in something. If you take a physics class, or possibly chemistry, you will learn about entropy when you learn about thermodynamics. If you have a lot of energy confined in one place, you can use it to do work, such as turning a motor. This spreads energy around. Once the energy is spread out evenly, the entropy is maximized and it is no longer possible to do useful work with it.\n\nIn computer science we also talk about the entropy of information. If you have a certain amount of data, it might be all zeroes or all ones, in which case it’s very easy to tell a friend how to recreate the same state: just say, “I have seven hundred ones.” If your data is more complex, though, it gets harder to describe; it has high entropy.\n\nI (Van Meter) can recall taking Feynman’s class at Caltech, in which he explained the relationship between information and entropy in terms of thermodynamics. For quite a while, I believed that he was speaking allegorically, and couldn’t understand why he was bringing in physical constants. When I finally understood that he meant it quite literally, I could feel the hair on the back of my neck stand up. It remains one of the most startling ideas I have ever encountered.\n\n### Reversible Computing\n\nGenerally, entropy increases when we lose information, or when information is erased. The AND and OR gates we discussed above necessarily lose information: they start with two input bits, and end with only one, so there is no way to carry all of the information through.\n\nIf, instead, we require all of our gates to have the same number of input and output bits, it’s possible that we might be able to undo our computation. A second additional criteria is necessary to make it reversible: every possible output state comes from exactly one input state. Then we say that the function is “one to one”, or bijective.\n\nAny logic function can be computed using one, two, and three-bit reversible gates, so we will look at a few of them.\n\n### Identity\n\nThe identity gate does nothing to the state: its output is the same as its input. It is obviously reversible; since it does nothing, in order to get back to where we started, we only have to do nothing one more time!\n\ninput output\n0 0\n1 1\n\n### NOT\n\nWe have already seen the NOT gate above. Executing NOT twice in a row brings us back to where we started: NOT(NOT(X)) = X.\n\n### CNOT\n\nThe controlled NOT gate, or CNOT, takes two input bits and produces two output bits. If one of the bits (called the control bit) is one, the other bit (called the target bit) is flipped. If the control bit is zero, the target bit is passed through unchanged.\n\ninput (A) input (B) output (A’) output (B’)\n0 0 0 0\n0 1 0 1\n1 0 1 1\n1 1 1 0\n\nA quick examination of the table shows that (A’) is the same as (A), and (B’) is the XOR of (A) and (B), (B’ = A oplus B). If we apply the same gate twice, obviously (A) stays the same, and the new (B” = A oplus (A oplus B) = B), and we are back where we started.\n\n### CCNOT (Toffoli Gate)\n\nThe control-control-NOT gate, or CCNOT gate, is also called the Toffoli gate, named for Tommaso Toffoli. It uses two control bits instead of one.\n\ninput (A) input (B) input (C) output (A’) output (B’) output (C’)\n0 0 0 0 0 0\n0 0 1 0 0 1\n0 1 0 0 1 0\n0 1 1 0 1 1\n1 0 0 1 0 0\n1 0 1 1 0 1\n1 1 0 1 1 1\n1 1 1 1 1 0\n\n### CSWAP (Fredkin Gate)\n\nThe control-SWAP gate, or CSWAP gate, is also called the Fredkin gate, named for Ed Fredkin. If the control bit is zero, the other two bits are left alone. If the control bit is one, the other two bits are swapped.\n\ninput (A) input (B) input (C) output (A’) output (B’) output (C’)\n0 0 0 0 0 0\n0 0 1 0 0 1\n0 1 0 0 1 0\n0 1 1 0 1 1\n1 0 0 1 0 0\n1 0 1 1 1 0\n1 1 0 1 0 1\n1 1 1 1 1 1\n\nEither the CCNOT or the CSWAP gate is powerful enough to be able to create any classical logic circuit.\n\n## Single-Qubit Gates\n\nThe reversible gates above all have quantum equivalents, and we will use them in constructing quantum algorithms. But qubits are more complex objects than simple classical bits, allowing a broader range of operations that all qualify as unitary. Here, we will look at some single-qubit gates that go beyond just the Identity and NOT gates above.\n\n### Rotations on the Bloch Sphere\n\nWhen we introduced qubits, we introduced the notion of the Bloch sphere, and the idea of the state of a single qubit as a point on the sphere. The one-qubit gates can be understood as rotations about the X, Y, or Z axis of the Bloch sphere.\n\nRotation is a naturally reversible operation: just rotate back in the opposite direction by the same amount. Every point on the sphere has a starting position and an ending position, and no two points ever come together. No two different starting points come to the same ending position. This is a key factor maintaining reversibility. If two points did come together, when we attempted to reverse the operation, we would have no way of distinguishing them and returning them to separate starting positions.\n\n### The NOT Gate and Other X Gates\n\nA simple reversible gate, like the classical NOT gate, the (X) gate takes 0 to 1 and 1 to 0. Since our state can be a superposition, it does a little more than this, it swaps our 0 and 1 values, or the contents of our 0 and 1 dials. It is a rotation of 180 degrees, or (pi), about the (X) axis. It’s also possible to rotate by any angle, which swaps the 0 and 1 values less completely.\n\n### The Phase Gates\n\nAlthough rotating around the (X) axis is somewhat like the classical NOT gate, rotating around the (Z) axis has no classical equivalent. (Z) rotation modifies the phase of the state. We call a 180 degree rotation simply the Z gate. For other angles, we call it a Z rotation or a phase gate, and specify the angle.\n\n### The Hadamard Gate\n\nThe Hadamard gate is our most basic means of creating superposition. It takes our (vert0rangle) state to a 50/50 superposition of (vert0rangle) and (vert1rangle):\n\n[vert0rangle rightarrow frac{vert0rangle + vert1rangle}{sqrt{2}}]", null, "If it also took the (vert1rangle) state to the same 50/50 superposition, it wouldn’t be reversible. Instead, it takes it to a superposition with a phase twist on the (vert1rangle) state:\n\n[vert1rangle rightarrow frac{vert0rangle + (pi)vert1rangle}{sqrt{2}}]", null, "Applying the Hadamard twice to the same qubit returns it to its original state. In fact, this is our first use of interference! If we began with the (vert0rangle) state, we get constructive interference that builds us a new (vert0rangle) state, and destructive interference that eliminates the (vert1rangle) state:\n\n(frac{vert0rangle + vert1rangle}{sqrt{2}}rightarrow vert0rangle).\n\nHere we can see that the phase of the state has a critical impact on the interference; reversing the other state results in constructive interference on the (vert1rangle) state, and destructive interference on the (vert0rangle) state:\n\n(frac{vert0rangle + (pi)vert1rangle}{sqrt{2}}rightarrow vert1rangle).\n\n#### Understanding Quantum Computers", null, "", null, "## Our purpose is to transform access to education.\n\nWe offer a diverse selection of courses from leading universities and cultural institutions from around the world. These are delivered one step at a time, and are accessible on mobile, tablet and desktop, so you can fit learning around your life.\n\nWe believe learning should be an enjoyable, social experience, so our courses offer the opportunity to discuss what you’re learning with others as you go, helping you make fresh discoveries and form new ideas.\nYou can unlock new opportunities with unlimited access to hundreds of online short courses for a year by subscribing to our Unlimited package. Build your knowledge with top universities and organisations." ]

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[ "TITLE\n\n# Bifurcation of Limit Cycles for 3D Lotka-Volterra Competitive Systems\n\nAUTHOR(S)\nWang, Qinlong; Huang, Wentao; Wu, Haotao\nPUB. DATE\nJune 2011\nSOURCE\nActa Applicandae Mathematica;Jun2011, Vol. 114 Issue 3, p207\nSOURCE TYPE\nDOC. TYPE\nArticle\nABSTRACT\nBifurcation of limit cycles is discussed for three-dimensional Lotka-Volterra competitive systems. A recursion formula for computation of the singular point quantities is given for the corresponding Hopf bifurcation equation. Some new results are obtained for 6 classes 26-31 in Zeeman's classification, especially, an example with four limit cycles in class 29 is given for the first time. The algorithm applied here is effective for solving the above general cyclicity.\nACCESSION #\n60175670\n\n## Related Articles\n\n• EVALUATING CYCLICITY OF CUBIC SYSTEMS WITH ALGORITHMS OF COMPUTATIONAL ALGEBRA. Levandovskyy, Viktor; Pfister, Gerhard; Romanovski, Valery G. // Communications on Pure & Applied Analysis;Sep2012, Vol. 11 Issue 5, p2029\n\nWe describe an algorithmic approach to studying limit cycle bifurcations in a neighborhood of an elementary center or focus of a polynomial system. Using it we obtain an upper bound for cyclicity of a family of cubic systems. Then using a theorem by Christopher we study bifurcation of limit...\n\n• FUNCTIONS ON SPACE CURVES. GORYUNOV, V. V. // Journal of the London Mathematical Society;06/01/2000, Vol. 61 Issue 3, p807\n\nSimple singularities of functions on space curves are classified. It is shown that their bifurcation sets have properties very similar to those of functions on smooth manifolds and complete intersections: the k(À, 1)-theorem for the bifurcation diagram of functions is true, and both this...\n\n• The Center Conditions and Bifurcation of Limit Cycles at the Degenerate Singularity of a Three-Dimensional System. Shugang Song; Jingjing Feng; Qinlong Wang // Journal of Applied Mathematics;2014, p1\n\nWe investigate multiple limit cycles bifurcation and center-focus problem of the degenerate equilibrium for a three-dimensional system. By applying the method of symbolic computation, we obtain the first four quasi-Lyapunov constants. It is proved that the system can generate 3 small limit...\n\n• On the Limit Sets of Orbits of Systems of Vector Fields. Azamov, A.; Narmanov, A. Ya. // Differential Equations;Feb2004, Vol. 40 Issue 2, p271\n\nStudies the limit sets of orbits of systems of vector fields. One-dimensional manifold; Topology of an immersed submanifold; Foliation with singularities; Homeomorphism.\n\n• Moment Lyapunov exponent of three-dimensional system under bounded noise excitation. Fang, Ci-jun; Yang, Jian-hua; Liu, Xian-bin // Applied Mathematics & Mechanics;May2012, Vol. 33 Issue 5, p553\n\nIn the present paper, the moment Lyapunov exponent of a codimensional two-bifurcation system is evaluted, which is on a three-dimensional central manifold and subjected to a parametric excitation by the bounded noise. Based on the theory of random dynamics, the eigenvalue problem governing the...\n\n• Boundary-equilibrium bifurcations in piecewise-smooth slow-fast systems. Kowalczyk, P.; Glendinning, P. // Chaos;Jun2011, Vol. 21 Issue 2, p023126\n\nIn this paper we study the qualitative dynamics of piecewise-smooth slow-fast systems (singularly perturbed systems) which are everywhere continuous. We consider phase space topology of systems with one-dimensional slow dynamics and one-dimensional fast dynamics. The slow manifold of the reduced...\n\n• Bogdanov-Takens and Triple Zero Bifurcations of a Delayed Modified Leslie-Gower Predator Prey System. Xia Liu; Jinling Wang // Abstract & Applied Analysis;2013, p1\n\nA delayed modified Leslie-Gower predator prey system with nonlinear harvesting is considered. The existence conditions that an equilibriumis Bogdanov-Takens (BT) or triple zero singularity of the systemare given. By using the centermanifold reduction, the normal form theory, and the formulae...\n\n• ON SECOND ORDER BIFURCATIONS OF LIMIT CYCLES. ILIEV, I. D. // Journal of the London Mathematical Society;10/01/1998, Vol. 58 Issue 2, p353\n\nThe paper derives a formula for the second variation of the displacement function for polynomial perturbations of Hamiltonian systems with elliptic or hyperelliptic Hamiltonians H(x, y)=½y2-U(x) in terms of the coefficients of the perturbation. As an application, the conjecture stated by C....\n\n• On bifurcations of a system of cubic differential equations with an integrating multiplier singular along a second-order curve. Alekseev, Aleksandr A. // Electronic Journal of Qualitative Theory of Differential Equatio;2015, Issue 40-47, p1\n\nWe establish necessary and sufficient conditions for existence of an integrating multiplier of a special form for systems of two cubic differential equations of the first order. We further study bifurcations of such systems with the change of parameters of their integrating multipliers.\n\nShare" ]

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[ "# Find:10th term of the A.P. $1, 4, 7, 10, ŌĆ”ŌĆ”ŌĆ”$\n\nGiven:\n\nGiven A.P. is $1, 4, 7, 10, ………$\n\nTo do:\n\nWe have to find the 10th term of the given A.P.\n\nSolution:\n\nHere,\n\n$a_1=1, a_2=4, a_3=7, a_4=10$\n\nCommon difference $d=a_2-a_1=4-1=3$\n\nWe know that,\n\nnth term $a_n=a+(n-1)d$\n\nTherefore,\n\n10th term $a_{10}=1+(10-1)3$\n\n$=1+9\\times3$\n\n$=1+27$\n\n$=28$\n\nThe 10th term of the given A.P. is $28$.\n\nUpdated on: 10-Oct-2022\n\n237 Views", null, "" ]

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[ "kmmiles.com\n\n# 57.819 miles in km\n\n## Result\n\n57.819 miles equals 93.0308 km\n\nYou can also convert 57.819 mph to km.\n\n## Conversion formula\n\nMultiply the amount of miles by the conversion factor to get the result in km:\n\n57.819 mi × 1.609 = 93.0308 km\n\n## How to convert 57.819 miles to km?\n\nThe conversion factor from miles to km is 1.609, which means that 1 miles is equal to 1.609 km:\n\n1 mi = 1.609 km\n\nTo convert 57.819 miles into km we have to multiply 57.819 by the conversion factor in order to get the amount from miles to km. We can also form a proportion to calculate the result:\n\n1 mi → 1.609 km\n\n57.819 mi → L(km)\n\nSolve the above proportion to obtain the length L in km:\n\nL(km) = 57.819 mi × 1.609 km\n\nL(km) = 93.0308 km\n\nThe final result is:\n\n57.819 mi → 93.0308 km\n\nWe conclude that 57.819 miles is equivalent to 93.0308 km:\n\n57.819 miles = 93.0308 km\n\n## Result approximation\n\nFor practical purposes we can round our final result to an approximate numerical value. In this case fifty-seven point eight one nine miles is approximately ninety-three point zero three one km:\n\n57.819 miles ≅ 93.031 km\n\n## Conversion table\n\nFor quick reference purposes, below is the miles to kilometers conversion table:\n\nmiles (mi) kilometers (km)\n58.819 miles 94.639771 km\n59.819 miles 96.248771 km\n60.819 miles 97.857771 km\n61.819 miles 99.466771 km\n62.819 miles 101.075771 km\n63.819 miles 102.684771 km\n64.819 miles 104.293771 km\n65.819 miles 105.902771 km\n66.819 miles 107.511771 km\n67.819 miles 109.120771 km\n\n## Units definitions\n\nThe units involved in this conversion are miles and kilometers. This is how they are defined:\n\n### Miles\n\nA mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada.\n\n### Kilometers\n\nThe kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world." ]

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[ "Secant method is used for finding root of the function in given interval.\n\n## Algorithm:\n\n```IN:\nFunction f, which is continous function and interval [a,b]. Function must satisfy given equation: f(a) * f(b) < 0 - signs of that values are different, which means that given function in given interval has at least one root in interval [a,b].\nOUT:\nRoot in given interval.\n1. Make xn1 = a, xn = b, x = xn1.\n2. Check if abs(f(x)) < precision.\n3. If it is end algorithm and return x.\n4. Make x = xn - ((f(xn) * (xn - xn1)) / (f(xn) - f(xn1))).\n5. Make xn1 = xn and xn = x.\n6. Go to step 2.```\n\n## Sample Output:\n\n```f(x) = x * (x + 2) -1\n[-3, 0]\n\nroot: -2.414201183431953```\n\n## Chord Method Step By Step\n\nIteration Numberabxn1xnxf(x)\n1-3-1-3-1-32\n2-3-1-1-2-2-1\n3-3-1-2-3-32\n4-3-1-3-3-2.333-0.2222\n5-3-1-3-2.333-2.4-0.04\n6-3-1-2.333-2.4-2.410.00118\n7-3-1-2.4-2.41-2.414-6.007E-6\n\n## Cons:\n\n• has trouble with some functions.\n• initial approximations are not accurate enough.\n\nSecant Method Algorithm\nTagged on:" ]

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[ "# T-Test Essentials: Definition, Formula and Calculation", null, "## Course description\n\nThe t-test can be defined as a statistical test used to compare two means. This guide provide multiple tutorials describing the different types of t-test, including:\n\n• one-sample t-tests. Compares one-sample mean to a know standard mean.\n• independent samples t-tests: Student’s t-test and Welch’s t-test. Compares two independent groups\n• paired samples t-test. Compares two related samples.\n\nYou will learn the t-test formula and how to:\n\n• Compute the different t-tests in R\n• Check t-test assumptions\n• Calculate and report t-test effect size using Cohen’s d.\n\nContents:\n\n#### Related Book\n\nPractical Statistics in R II - Comparing Groups: Numerical Variables\n\n## Prerequisites\n\nMake sure you have installed the following R packages:\n\n• tidyverse for data manipulation and visualization\n• ggpubr for creating easily publication ready plots\n• rstatix provides pipe-friendly R functions for easy statistical analyses.\n• datarium: contains required data sets for this chapter.\n\nlibrary(tidyverse)\nlibrary(ggpubr)\nlibrary(rstatix)\n\n## Examples of R codes\n\nComparing two independent groups:\n\n# Data preparation\ndata(\"genderweight\", package = \"datarium\")\nhead(genderweight, 3)\n## # A tibble: 3 x 3\n## id group weight\n## <fct> <fct> <dbl>\n## 1 1 F 61.6\n## 2 2 F 64.6\n## 3 3 F 66.2\n# Statistical test\nstat.test <- genderweight %>%\nt_test(weight ~ group) %>%\nstat.test\n## # A tibble: 1 x 9\n## .y. group1 group2 n1 n2 statistic df p p.signif\n## <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>\n## 1 weight F M 20 20 -20.8 26.9 4.30e-18 ****\n# Visualization:\n# Create a box-plot\nbxp <- ggboxplot(\ngenderweight, x = \"group\", y = \"weight\",\nylab = \"Weight\", xlab = \"Groups\", add = \"jitter\"\n)\n# Add p-value and significance levels\nstat.test <- stat.test %>% add_xy_position(x = \"group\")\nbxp +\nstat_pvalue_manual(stat.test, tip.length = 0) +\nlabs(subtitle = get_test_label(stat.test, detailed = TRUE))", null, "1. ## Types of T-Test\n\nDescribes the different types of t-test for comparing the means of groups. These include: one-sample t-tests, unpaired t-test and paired t-test.\n1. ## One Sample T-Test\n\nDescribes the one-sample t-test, which is used to compare the mean of one sample to a known standard (or theoretical / hypothetical) mean. You will learn the formula, assumptions, calculation, visualization, effect size measure using the Cohen's d, interpretation and reporting in R.\n2. ## Unpaired T-Test\n\nDescribes the unpaired t-test, which is used to compare the mean of two independent groups. You will learn the formula, assumptions, calculation, visualization, effect size measure using the Cohen's d, interpretation and reporting in R. The Student's t-test and the Welch t-test are described.\n1. ## Student's T-Test\n\nDescribes the Student's t-test, which is used to compare the mean of two independent groups. You will learn the formula, assumptions, calculation, visualization, effect size measure using the Cohen's d, interpretation and reporting in R.\n2. ## Welch T-Test\n\nDescribes the Welch t-test, which is used to compare the mean of two independent groups. You will learn the formula, assumptions, calculation, visualization, effect size measure using the Cohen's d, interpretation and reporting in R.\n3. ## Paired T-Test\n\nDescribes the paired t-test, which is used to compare the mean of two related groups of samples. You will learn the formula, assumptions, calculation, visualization, effect size measure using the Cohen's d, interpretation and reporting in R.\n2. ## Pairwise T-Test\n\nDescribes how to compute the pairwise T-test in R between groups with corrections for multiple testing. The pairwise t-test consists of calculating multiple t-test between all possible combinations of groups. You will learn how to: 1) Calculate pairwise t-test for unpaired and paired groups; 2) Display the p-values on a boxplot.\n3. ## T-Test Formula\n\nDescribes the t-test formula for one-sample, two-independent samples and paired samples t-test.\n1. ## One Sample T-Test Formula\n\nDescribes the one sample t-test formula, which is used to compare the mean of one sample to a known standard mean.\n2. ## Independent T-Test Formula\n\nDescribes the independent t-test formula, which is used to compare the means of two independent groups. You will learn the Student t-test formula and the Weltch t-test formula.\n3. ## Paired T-Test Formula\n\nDescribes the paired t-test formula, which is used to compare the means of two related groups or samples.\n4. ## T-Test Assumptions\n\nDescribes the t-test assumptions and provides examples of R code to check whether the assumptions are met before calculating the t-test. You will learn the assumptions of the different types of t-test, including the one-sample t-test, independent t-test and paired t-test.\n1. ## One Sample T-Test Assumptions\n\nDescribes the one sample t-test assumptions and provides examples of R code to check whether the assumptions are met before calculating the t-test.\n2. ## Independent T-Test Assumptions\n\nDescribes the independent t-test assumptions and provides examples of R code to check whether the assumptions are met before calculating the t-test.\n3. ## Paired T-Test Assumptions\n\nDescribes the paired t-test assumptions and provides examples of R code to check whether the assumptions are met before calculating the t-test.\n5. ## How to Do a T-test in R: Calculation and Reporting\n\nDescribes how to do a t-test in R/Rstudio. You will learn how to 1) interpret and report the t-test; 2) add p-values and significance levels to a plot and 3) calculate and report the t-test effect size.\n1. ## How To Do a One-Sample T-test in R\n\nDescribes how to do a one-sample t-test in R/Rstudio. You will learn the calculation, visualization, effect size measure using the Cohen's d, interpretation and reporting.\n2. ## How To Do Two-Sample T-test in R\n\nDescribes how to do a two-sample t-test in R/Rstudio. You will learn the calculation, visualization, effect size measure using the Cohen's d, interpretation and reporting.\n3. ## How to Do Paired T-test in R\n\nDescribes how to do a paired t-test in R/Rstudio. You will learn the calculation, visualization, effect size measure using the Cohen's d, interpretation and reporting.\n6. ## T-test Effect Size using Cohen's d Measure\n\nDescribes the t-test effect size using the Cohen's d. You will learn Cohen's d formula, calculation in R, interpretation of small, medium and large effect.\n\nNo ratings yet." ]

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[ "# What is the y intercept of -2y=3x^2-3?\n\nApr 22, 2018\n\nThe y-intercept is $\\left(0 , \\frac{3}{2}\\right)$ or $\\left(0 , 1.5\\right)$.\n\n#### Explanation:\n\nGiven:\n\n$- 2 y = 3 {x}^{2} - 3$\n\nThe y-intercept is the value of $y$ when $x = 0$. Substitute $0$ for $x$ in the equation and solve for $y$.\n\n$- 2 y = 3 {\\left(0\\right)}^{2} - 3$\n\n$- 2 y = - 3$\n\nDivide both sides by $- 2$.\n\n$y = \\frac{- 3}{- 2}$\n\n$y = \\frac{3}{2}$\n\nThe y-intercept is $\\left(0 , \\frac{3}{2}\\right)$ or $\\left(0 , 1.5\\right)$.\n\nThis is also the vertex for this particular parabola.\n\ngraph{-2y=3x^2-3 [-10, 10, -5, 5]}" ]

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[ "", null, "", null, "", null, "", null, "0\n\n# How many vertices facesand edges has pentagonal pyramid?\n\nUpdated: 8/21/2019", null, "Wiki User\n\n8y ago\n\nIt has 6 vertices, 6 faces and 10 edges", null, "Wiki User\n\n8y ago", null, "", null, "Study guides\n\n14 cards\n\n## Which of the following is not a step in the inquiry process\n\n➡️\nSee all cards\n3.89\n148 Reviews", null, "Earn +20 pts\nQ: How many vertices facesand edges has pentagonal pyramid?\nSubmit\nStill have questions?", null, "", null, "Related questions\n\n### How many faces and vertices's and edges does a pentagonal pyramid have?\n\nA pentagonal pyramid has 6 faces, 6 vertices and 10 edges.\n\n### How many edges faces and vertices does a pentagonal pyramid?\n\nA pentagonal pyramid has 6 faces, 6 vertices and 10 edges.\n\n### How many faces edges vertices does a pentagonal based pyramid have?\n\nA pentagonal based pyramid has 6 faces, ten edges and 6 vertices.\n\n### How many faces edges and vertices does a pentagonal-based pyramid have?\n\nA pentagonal based pyramid has 6 faces, ten edges and 6 vertices.\n\n### How many faces edges and vertices does the pentagonal have?\n\nPentagonal what? Pentagonal Prism-&gt; Faces:7 Edges:15 Vertices:10 Pentagonal Pyramid-&gt; Faces:6 Edges:10 Vertices:6 There are others as well.\n\n### Pentagonal pyramid faces vertices edges?\n\nYes, they have them.\n\n10\n\n### How many faces edges and vertices does a pentagonal pyrmaid have?\n\nA pentagonal based pyramid has 6 faces, 10 edges and 6 vertices\n\n### How many faces edges and vertices does a pentagonal pyramid have?\n\nA pentagonal pyramid has 6 faces, 6 vertices and 10 edges.\n\n### What has 6 faces 10 edges 6 vertices and is a pyramid?\n\na pentagonal pyramid\n\n### What is a pyramid with 6 faces 10 edges and 6 vertices?\n\nA pentagonal pyramid.\n\n### How many vertices edges and faces on a pentagonal pyramid?\n\nA pentagonal pyramid has 6 faces. It has 10 edges and 6 vertices. It looks just like a pentagon on face and has a point at the top." ]

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[ "# CROCHET EASY CUTE BABY BOOTIES", null, "Hello everybody, there are many booties for baby, but it is easiest and cute baby booties to make. You can make them with the your preferred technique. Very comfortable and useful at home. You can increase and decrease what size you want.\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\nI hope you’ve got enough experience of knitting and crocheting one among the most effective stitches, as a result of our today’s project completely depends on it ability. So, let’s get right down to the business. Start crochet today and get a big pleasure.\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\nAs always, Before you go I want to ask you if you value my work, did you like the knitting and crochet course, share it with all your friends on social networks and share it in their crochet groups so all the little spiders in the world will benefit from art crochet!\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\nTutorial and photos by : Mrs. Oren\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\n​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​\n\nThanks so much for visiting and be sure to come back every day for more free tutorials.​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​", null, "" ]

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[ "", null, "# A tetrahedron has vertices $O(0,0,0),\\:A(1,2,1)\\:B(2,1,3)\\:C(-1,1,2)$. The angle between the faces $OAB$ and $ABC$ is ?\n$(a)\\:\\:90^{\\circ}\\:\\:\\:\\qquad\\:\\:(b)\\:\\:30^{\\circ}\\:\\:\\:\\qquad\\:\\:(c)\\:\\:cos^{-1}(19/35)\\:\\:\\:\\qquad\\:\\:(d)\\:\\:cos^{-1}(17/31)$" ]

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[ "Sale!\n\n# Project 04 – Graphs\n\n\\$30.00\n\nCategory:\n\nProject 04 – Graphs\n\nIntroduction\nYou are required to implement two solutions for problems that are embedded\nin a graph data structure: Prim’s algorithm in a undirected weighted graph,\nDijkstra’s algorithm in a directed weighted graph; you can only use adjacency\nlists to represent your graphs. You are also required to create UML diagrams\nfor each of the classes that you implement. Finally, you have to provide the\nmeans to test your system by developing a menu program that allows the user\nYou may use the queue, stack, and list algorithms defined in the C++ standard library (STL). You must use your own implementation for the hash table.\nAlso, any array must be dynamically allocated.\nUnderflow exceptions might be generated in some of the functions you implement. Make sure exceptions are thrown and caught where appropriate.\nDeliverables\n• A 1 page report that explains the design of your solutions. Please include\nwhat each teammate did and approximate hours spent on the project.\n• An implementation of Prim’s algorithm.\n• An implementation of Dijkstra’s algorithm.\n• A menu program to test the implemented data structures.\n1 Prim’s algorithm\nIn this part of the project, you need to implement three classes, a Graph Class,\na Vertex Class, and an Edge Class; create their respective UML diagrams. Calculate the running time of your functions and include them in your report.\n1\n1.1 Description Graphs\n1.1 Description\nA spanning tree of graph G=(V, E) is another graph T = (V, F) with the same\nvertices as G, and |V | − 1 edges of E that form a tree.\nA minimum spanning tree (MST) T of G is a spanning tree whose total\nweight (summed over all edges of T) is minimal.\nFor this part of the project you will implement Prim’s algorithm to find\nan MST in a undirected weighted graph; an adjacency list has to be\nused to represent the graph. Vertices are represented by their names in the\ngraph. To ease the execution of some of your operations, your graph has to have\na hash table to store all the vertices, where the key is the name of the vertex\nand the value is the vertex object. In the event you need information (data,\nedges, etc) about a particular vertex, you will have to hash the vertex’s name\nand find the vertex object in the table (you may use linear probing).\n1.2 Data Members\nSpecify all the data members your implementation needs.\n1.3 Member Functions\nConstructors\nDefines constructor. Maximum graph size is 20 vertices.\nDestructor\nDefines destructor. Clean up any allocated memory.\nAccessors\nbool empty() Returns true if the graph is empty, false otherwise.\nint degree(string v) Returns the degree of the vertex v. Throw an illegal\nargument exception if the argument does not correspond to an existing vertex.\nint edgeCount() Returns the number of edges in the graph\nbool isConnected() Determines if the graph is connected.\ndouble adjacent( string u, string v ) Returns the weight of the edge\nconnecting vertices u and v. If the vertices are the same, return 0. If the vertices\nare not adjacent, return -1 (our representation of infinity). Throw an illegal\nargument exception if the arguments do not correspond to existing vertices.\n2\n1.3 Member Functions Graphs\nDFS(string v) Performs DFS traversal starting on vertex v. Reset vertices after the traversal. Prints the order of vertices visited.\nBFS(string v) Performs BFS traversal starting on vertex v. Reset vertices after the traversal. Prints the order of vertices visited.\nMST( string v ) Returns the minimum spanning tree using Prim’s algorithm of those vertices which are connected to vertex v. Throw an illegal\nargument exception if the argument does not correspond to an existing vertex.\nMutators\nvoid buildGraph() Reads structure from a text file and builds a undirected weighted graph.\nclear() Removes all the elements in the undirected weighted graph\nreset() Iterates over all vertices in the graph and marks them as unvisited.\ninsert(string u, string v, double w) If the weight w < 0 or w = ∞,\nthrow an illegal argument exception. If the weight w is 0, remove any edge\nbetween u and v (if any). Otherwise, add an edge between vertices u and v\nwith weight w. If an edge already exists, replace the weight of the edge with the\nnew weight. If the vertices do not exist or are equal, throw an illegal argument\nexception.\nFriends\nDefines friends for this class.\n3\nGraphs\n2 Dijkstra’s algorithm\nIn this part of the project, you need to implement three classes, a DirGraph\nClass, a Vertex Class, and an Edge Class; create their respective UML diagrams.\nCalculate the running time of your functions and include them in your report.\n2.1 Description\nDijkstra’s algorithm is an algorithm to find the shortest path among vertices in\na graph.\nFor this part of the project you will implement Dijkstra’s algorithm to\nfind the shortest path between a source vertex and the rest of the\nvertices in a directed weighted graph; an adjacency list has to be\nused to represent the graph. Vertices are represented by their names in the\ngraph. To ease the execution of some of your operations, your graph has to have\na hash table to store all the vertices, where the key is the name of the vertex\nand the value is the vertex object. In the event you need information (data,\nedges, etc) about a particular vertex, you will have to hash the vertex’s name\nand find the vertex object in the table.\n2.2 Data Members\nSpecify all the data members your implementation needs.\n2.3 Member Functions\nConstructors\nDefines constructor. Maximum graph size is 20 vertices.\nDestructor\nDefines destructor. Clean up any allocated memory.\nAccessors\nbool empty() Returns true if the graph is empty, false otherwise.\nint inDegree(string v) Returns the indegree of the vertex v. Throw an\nillegal argument exception if the argument does not correspond to an existing\nvertex.\nint outDegree(string v) Returns the outdegree of the vertex v. Throw\nan illegal argument exception if the argument does not correspond to an existing\nvertex.\n4\n2.3 Member Functions Graphs\nint edgeCount() Returns the number of edges in the graph\ndouble adjacent( string u, string v ) Returns the weight of the edge\nconnecting vertices u and v. If the vertices are the same, return 0. If the vertices\nare not adjacent, return -1 (our representation of infinity). Throw an illegal\nargument exception if the arguments do not correspond to existing vertices.\nDFS(string v) Performs DFS traversal starting on vertex v. Reset vertices after the traversal. Prints the order of vertices visited.\nBFS(string v) Performs BFS traversal starting on vertex v. Reset vertices after the traversal. Prints the order of vertices visited.\nshortPath( string u, string v ) Returns the shortest path using Dijkstra’s between vertices u and v. Throw an illegal argument exception if the\narguments do not correspond to existing vertices.\ndouble distance( string u, string v ) Returns the shortest distance\nbetween vertices u and v. Throw an illegal argument exception if the arguments\ndo not correspond to existing vertices. The distance between a vertex and\nitself is 0.0. The distance between vertices that are not connected is -1 (our\nrepresentation of infinity).\nMutators\nvoid buildGraph() Reads structure from a text file and builds a directed\nweighted graph.\nclear() Removes all the elements in the undirected weighted graph\nreset() Iterates over all vertices and marks them as unvisited.\ninsert(string u, string v, double w) If the weight w ≤ 0, throw an\nillegal argument exception. If the weight is w 0, add an edge between vertices\nu and v. If an edge already exists, replace the weight of the edge with the\nnew weight. If the vertices do not exist or are equal, throw an illegal argument\nexception.\nFriends\nDefines friends for this class.\n5\nGraphs\n3 Graph Representation\nFormat for the text file\nThe example text file for the buildGraph() method is uploaded to Canvas. Up\nto twenty vertices (string) will be listed with white spaces between, followed by\na new line. Next, each edge will be listed: u (string), space, v (string), space,\nweight (double), new line. The same format is used for directed and undirected\ngraphs.\nFormat for printing Prim’s MST (edges and total weight)\nA B 4\nA C 6.4\nC F 3\nA D 9\nDistance: 22.4\nFormat for printing Dijkstra’s shortest path (distance to each node\nfrom A)\nA 0\nB 4\nC 3.1\nD 6\nF 7\nIn order to test your program, you are required to implement a menu program\nthat provides the means to run each of the functions in your classes (name your\nexecutable proj4). The TA will choose one group to demo the project.\nFirst, prompt the user for the type of graph (char ’d’ for directed or char ’u’\nfor undirected). Next, prompt for a file name (using path provided) for the .txt\nfile to load the graph. The format of an example .txt file is provided on Canvas\nfor you to be able to test your work. Next, give the following options for the\nspecific graph (please have them in this EXACT order for grading purposes):\nUndirected Graph (Prim’s)\n1. Empty?\n6\nGraphs\n2. Degree (v)\n3. Edge count\n4. Connected?\n6. DFS (v)\n7. BFS (v)\n8. Print MST (v)\n9. Clear\n10. Insert (u, v, w)\n11. Exit\nDirected Graph (Dijkstra’s)\n1. Empty?\n2. InDegree (v)\n3. OutDegree (v)\n4. Edge count\n6. DFS (v)\n7. BFS (v)\n8. Print Short path (v)\n9. Clear\n10. Insert (u, v, w)\n11. Exit\nSubmit the following files to Canvas (named EXACTLY as shown below\nwithout zipping for grading purposes: (you may also include implementation\nfiles with the header files if you would like)\n2. graph.h\n3. vertex.h\n4. edge.h\n5. dirGraph.h\n6. makefile (name your executable ’proj4’)\n7. project4.pdf\nThe rubric is as follows:\n1. A program that does not compile will result in a zero\n2. Runtime error and compilation warning 5%\n3. Commenting and style 15%\n4. Report 10%\n5. Functionality 70% (functions were declared and implemented as required)\n7\nGraphs\n5 The Project Report\nYou must include everything you consider relevant in your design, UML diagrams, and algorithm analysis.\n6 Acknowledgment\nThis project was created based on the work shared by the University of Waterloo.\n8", null, "Project 04 – Graphs\n\\$30.00\nHello\nCan we help?" ]

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[ "", null, "", null, "", null, "Algebra 2 Worksheets\n\nSystems of Equations and Inequalities Worksheets\n\nHere is a graphic preview for all of the Systems of Equations and Inequalities Worksheets. You can select different variables to customize these Systems of Equations and Inequalities Worksheets for your needs. The Systems of Equations and Inequalities Worksheets are randomly created and will never repeat so you have an endless supply of quality Systems of Equations and Inequalities Worksheets to use in the classroom or at home. We have handout for systems of equations, solving algebraically two variable systems of equations, solving graphically two variable systems of equations, systems of two linear inequalities, points in three dimensions, planes, systems of three equations, Cramer's rule and systems of two equations word problems for your use.\n\nOur Sequences and Series Worksheets are free to download, easy to use, and very flexible.\n\nThese Systems of Equations and Inequalities Worksheets are a good resource for students in the 8th Grade through the 12th Grade.\n\nClick here for a Detailed Description of all the Systems of Equations and Inequalities Worksheets.\n\nQuick Link for All Systems of Equations and Inequalities Worksheets\n\nClick the image to be taken to that Systems of Equations and Inequalities Worksheets.\n\nHandout forSystems of Equations", null, "Solving AlgebraicallyTwo VariableSystems of Equations", null, "Solving GraphicallyTwo VariableSystems of Equations", null, "Solving GraphicallyTwo VariableSystems of Inequalities", null, "Systems ofThree Equations", null, "Systems of Two EquationsWord Problems", null, "Systems of Inequalities Multiple Choice Problems", null, "Cramer's Rule 2x2Matrices Worksheets", null, "Cramer's Rule 3x3Matrices Worksheets", null, "Recommended Videos\n\nDetailed Description for All Systems of Equations and Inequalities Worksheets\n\nHandout for Systems of Equations Worksheets\nThis Systems of Equations worksheet will produce a handout for the students concerning Systems of Equations. You may select which methods will be produced for the handout. This Systems of Equations worksheet is a good resource for students in the 9th Grade, 10th Grade, 11th Grade, and 12th Grade.\n\nSolving Algebraically Two Variable Systems of Equations Worksheets\nThis Systems of Equations worksheet will produce problems for solving two variable systems of equations algebraically. You may select which type of method the student should use to solve the problems, as well as the types of solutions. This Systems of Equations worksheet will produce twelve problems per page. This Systems of Equations worksheet is a good resource for students in the 9th Grade, 10th Grade, 11th Grade, and 12th Grade.\n\nSolving Graphically Two Variable Systems of Equations Worksheets\nThis Systems of Equations worksheet will produce problems for solving two variable systems of equations algebraically. You may select which type of method the student should use to solve the problems, as well as the types of solutions. This Systems of Equations worksheet will produce eight problems per page. This Systems of Equations worksheet is a good resource for students in the 9th Grade, 10th Grade, 11th Grade, and 12th Grade.\n\nSolving Graphically Two Variable Systems of Inequalities Worksheets\nThis Systems of Equations worksheet will produce problems for solving two variable systems of inequalities graphically. You may select which type of method the student should use to solve the problems. This Systems of Equations worksheet will produce eight problems per page. This Systems of Equations worksheet is a good resource for students in the 9th Grade, 10th Grade, 11th Grade, and 12th Grade.\n\nSystems of Three Equations Worksheets\nThis Systems of Equations worksheet will produce problems for working with systems of three equations. You can select the method with which to solve the systems as well as the types of solutions. This Systems of Equations worksheet will produce twelve problems per page. This Systems of Equations worksheet is a good resource for students in the 9th Grade, 10th Grade, 11th Grade, and 12th Grade.\n\nSystems of 2 Equations Word Problems Worksheets\nThis Systems of Equations worksheet will produce problems for working with systems of 2 equations word problems. This Systems of Equations worksheet will produce four problems per page. This Systems of Equations worksheet is a good resource for students in the 9th Grade, 10th Grade, 11th Grade, and 12th Grade.\n\nSystems of Inequalities Multiple Choice Problems\nThis Systems of Equations worksheet will produce multiple choice problems for solving two variable systems of inequalities graphically. You can select the types of problems and number of problems. This Systems of Equations worksheet will produce two problems per page. This Systems of Equations worksheet is a good resource for students in the 9th Grade, 10th Grade, 11th Grade, and 12th Grade.\n\nCramer's Rule 2x2 Matrices Worksheets\nThis Systems of Equations worksheet will produce problems for using Cramer's rule with 2x2 matrices. You can select the types of solutions you want the problems to have. This Systems of Equations worksheet will produce twelve problems per page. This Systems of Equations worksheet is a good resource for students in the 9th Grade, 10th Grade, 11th Grade, and 12th Grade.\n\nCramer's Rule 3x3 Matrices Worksheets\nThis Systems of Equations worksheet will produce problems for using Cramer's rule with 3x3 matrices. You can select the types of solutions you want the problems to have. This Systems of Equations worksheet will produce twelve problems per page. This Systems of Equations worksheet is a good resource for students in the 9th Grade, 10th Grade, 11th Grade, and 12th Grade.", null, "", null, "", null, "" ]

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[ "These solutions are for reference only.\n\ntry to solve on your own\n\nbut if you get stuck in between than you can refer these solutions\n\nthere are different set of questions ,\n\nwe have provided the variations in particular question at the end.\n\n-----------------------------------------------------------------------------------------\n\n## Neural Networks: Representation\n\nTOTAL POINTS 5\n\nEXPLANATION:\nA two layer (one input layer, one output layer; no hidden layer) neural network can represent the XOR function. (False)\n=>We must compose multiple logical operations by using a hidden layer to represent the XOR function.\n\nAny logical function over binary-valued (0 or 1) inputs x1 and x2 can be (approximately) represented using some neural network.(True)\n=>Since we can build the basic AND, OR, and NOT functions with a two layer network, we can (approximately) represent any logical function by composing these basic functions over multiple layers.\n\nSuppose you have a multi-class classification problem with three classes, trained with a 3 layer network. Let a(3)1=(hΘ(x))1 be the activation of the first output unit, and similarly a(3)2=(hΘ(x))2 and a(3)3=(hΘ(x))3. Then for any input x, it must be the case that a(3)1+a(3)2+a(3)3=1. (False)\n=>The outputs of a neural network are not probabilities, so their sum need not be 1.\n\nThe activation values of the hidden units in a neural network, with the sigmoid activation function applied at every layer, are always in the range (0, 1).\n(True)\n=>fact\n\nQuestion 2\n\nConsider the following neural network which takes two binary-valued inputs $x_1, x_2 \\in \\{0, 1\\}$ and outputs $h_\\Theta(x)$. Which of the following logical functions does it (approximately) compute?", null, "VARIATION IN 2ND QUESTION:\n\nEXPLANATION:\nThis network outputs approximately 1 when atleast one input is 1.\n\nEXPLANATION:\n\n$a_1^{(3)} = g(\\Theta_{1,0}^{(2)}a_0^{(2)} + \\Theta_{1,1}^{(2)}a_1^{(2)} + \\Theta_{1,2}^{(2)}a_2^{(2)})$\nThis correctly uses the first row of Θ(2) and includes the \"+1\" term of a0(2)  This correctly uses the first\nrow of Θ(2) and includes the \"+1\" term of   a0(2)\n\nEXPLANATION:\na(2) = g(Θ(1)x)=g(z)\n\nEXPLANATION:\nSwapping Θ(1) swaps the hidden layers output a^{(2)}. But the swap of Θ(2) cancels out the change, so the output will remain unchanged.\n\nVARIATION IN 5 TH QUESTION:\n\nEXPLANATION:\nSwapping Θ(1) swaps the hidden layers output a^{(2)}. But the swap of Θ(2) cancels out the change, so the output will remain unchanged.\n\n---------------------------------------------------------------------------------\n\nreference : coursera\n\ndarkmode" ]

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[ "# Fun Standardizing Non Standard Evaluation\n\nUsing dplyr calls on the back-end of the rfishbase re-write means working around the non-standard evaluation (NSE), as described in the dplyr vignette.\n\nGrab the data I was using for this:\n\nlibrary(\"dplyr\")\nall_taxa <- read.csv(\"taxa.csv\")\n\nConsider a simple NSE dplyr call:\n\nx <- filter(all_taxa, Family == 'Scaridae')\n\nThe best SE version of this just needs to use the formula expression, ~, the _ SE version of the function and it’s .dots argument:\n\n.dots <- list(~Family == 'Scaridae')\nx1 <- filter_(all_taxa, .dots=.dots)\n\nidentical(x, x1)\n TRUE\n\nThis lets us treat the arguments (e.g. values of the factor on which we filter) as variables:\n\nfamily <- 'Scaridae'\n.dots <- list(~Family == family)\nx2 <- filter_(all_taxa, .dots=.dots)\nidentical(x, x2)\n TRUE\n\nIf we want both the key and value to vary, we need to get pretty fancy to subvert the non-standard evaluation:\n\nlibrary(lazyeval)\nfamily <- 'Scaridae'\nfield <- 'Family'\n.dots <- list(interp(~y == x,\n.values = list(y = as.name(field), x = family)))\nx3 <- filter_(all_taxa, .dots=.dots)\nidentical(x, x3)\n TRUE\n\nAt bit more fun to wrap this into a function where we take arbitrary number of arguments as name-value pairs:\n\nquery <- list(Family = 'Scaridae', SpecCode = 5537)\ndots <- lapply(names(query), function(level){\nvalue <- query[[level]]\ninterp(~y == x,\n.values = list(y = as.name(level), x = value))\n})\n\nx3 <- filter_(all_taxa, .dots = dots) \n\n## More fun standardizing NSE\n\nThe previous examples show only applications to filter_(). While the general idea is the same, this pattern doesn’t translate directly for other functions, such as mutate_. Here’s some common patterns I’ve adopted when using mutate_(). First consider the familiar NSE useage:\n\ndf <- mutate(mtcars, displ_l = disp / 61.0237)\nhead(df)\n mpg cyl disp hp drat wt qsec vs am gear carb displ_l\n1 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 2.621932\n2 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 2.621932\n3 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 1.769804\n4 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1 4.227866\n5 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2 5.899347\n6 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1 3.687092\n\nAgain we use list(interp( pattern, but note that we specify the name for our new column using setNames (naming the elements of the list).\n\ndots <- setNames(list(lazyeval::interp(~x / y, x = quote(disp), y=61.0237)), \"displ_l\")\ndf2 <- mutate_(mtcars, .dots = dots)\nidentical(df, df2)\n TRUE\n\nOf course the use y could be skipped for a more direct value if that was not a variable.\n\n## More dplyr patterns\n\nAlso thought I would scribble down some other common dplyr patterns I find myself re-using.\n\n• applying a function that returns a data.frame to each element of a list and coercing the combined output to a data.frame:\nmylist %>% lapply(myfun) %>% dplyr::bind_rows() \n\nTo place this deeper in the hadleyverse, purrr::map could be dropped in for lapply in the above example.\n\n• Another common pattern for me is expand.grid() %>% group_by() %>% do(), Here’s a recent example of mine\n\nAlso includes an example of how to define group_by_all() since that is usually the grouping I need from an expand.grid() call (that is, I want to apply over all combinations of some parameter settings, etc)\n\nSomething I hope is not a common pattern but one I struggled with for a bit: making recursive calls of the above pattern for nested lists. This code in RNeXML illustrates my solution, which required both function recursion and function closure." ]

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[ "# Mathematics of the Pure Charge Theory\n\nThis is one section from the Mathematics of the Pure Charge Theory. It is located on this separate page to speed downloading times.\n\n### Keywords\n\nmass, inertia, electric charge, charge, kinetic energy, field of motion (1), (2)\n\n## Accelerating a charge\n\nA charge q has a velocity", null, "and an electric field gives it an acceleration", null, ". If we assume that", null, "is not parallel to", null, ", then we can define Cartesian coordinates such that the X-axis is in the direction of", null, ". The point", null, "has polar coordinates", null, "such that its Cartesian coordinates are:", null, "The charge will be surrounded by a magnetic field, which we call its field of motion, given by", null, "The acceleration produces a change in the magnetic field given by", null, "The vector cross product is distributive across vector addition and subtraction while the process by which we prove that", null, "relies on vector addition and subtraction. We can therefore differentiate the cross product of", null, "times the constant vector", null, "giving", null, "This rate of change of the magnetic intensity changes the energy density of the magnetic field and this results in a flow of magnetic energy density flux both into and/or out of of the surface of the charge depending on the relative direction of", null, "compared with", null, "and on the position on the surface of the spherical charge. The energy density at", null, "in its scalar and vector forms is", null, "and we can expand these to get three more equations.", null, "", null, "The effect of acceleration on a body is twofold. The speed of the body may change and its direction of motion may change. There are two special cases: acceleration parallel to the velocity changes only the speed; acceleration perpendicular to the velocity changes only the direction. When we consider a single spherical charge, the effects on the field of motion are a change in its energy content caused by a change in speed and a rotation caused by a change in direction. The movement of magnetic energy density flux is constrained so that it can only move parallel to the electric field of the charge. Furthermore, the electric field is fixed in direction and cannot rotate. So rotation of the magnetic field is accomplished by the movement of energy density flux both into and out of the charge.\n\nA simplistic approach might be to differentiate the first equation for", null, "with respect to time to give", null, "", null, "", null, "", null, "This does not however take into account the fact that magnetic energy density flux has vector like properties. If in a small time", null, "the magnetic intensity changes from", null, "to", null, ", the energy density vector changes from", null, "to", null, ". However, the vector addition of", null, "must have the same geometry as that of", null, ". If we draw vector triangles to represent the addition, they must be similar.\nWe look to the vector triangle from which the limiting result for", null, "is derived and scale it by a factor of", null, ".", null, "This means that", null, "and this corresponds to the following result.", null, "", null, "At this point, it is best to enter into Cartesian coordinates and simplify the vector product because algebraic expansion would introduce extra sines and cosines of unknown angles. We have", null, "", null, "", null, "The magnitude", null, "is easy enough to work out as", null, "giving", null, "Which gives", null, "At this point, we are in danger of running into complications. We really need this vector in the form of a magnitude times a unit vector, but this expression is rather messy. So we will delay dealing with this problem by making a substitution.", null, "Then we can write", null, "This is a rate of change of the magnetic energy density vector. We need to divide the space surrounding the charge in such a way as to facilitate integration. We do this by dividing it into conic elements extending outwards from the sphere with rectangular cross section", null, "by", null, ". In a conic element, the rate of change of the magnetic energy density is inversely proportional to fourth power of the radius r. We need to integrate this over r to find the rate of flow of energy density flux into the conic element.", null, "", null, "To find the velocity with which magnetic energy density flux is flowing from the surface of the charge, we must divide by the energy density times the area of the surface of the charge within the conic element. Note that at this point, we need the magnitude and so we omit the direction which is given by the unit vector", null, ".", null, "", null, "We now know that magnetic energy density flux of density", null, "and in the direction of", null, "is flowing out of the surface at velocity", null, ". Since we have correctly calculated the velocity by considering changes in the magnetic energy density flux within the conic element, we may now use that velocity in a form of Faraday's law to calculate the electric field experienced by the surface of the charge.", null, "What we need to understand here is that we are generating (or adsorbing) magnetic energy density flux at the surface of the charge. Its energy density is the same as the energy density that is already there, but its directional property is different. The value of", null, "to use is given by", null, "Where the value of B is that surrounding the moving charge. We are more familiar with", null, "(where", null, "is the angle between", null, "and", null, ") than with", null, ", but the angle between", null, "and", null, "is not known, so we use the modulus of the cross product. This gives us the resulting electric field experienced by the surface of the charge.", null, "", null, "Now we find that", null, "and", null, "are reunited and we can substitute for", null, "", null, ".", null, "The force felt by the charge acting on the surface within the conic element is", null, "where the factor ½ is the penetration coefficient. This is needed because the surface of the charge has a finite thickness. Energy density flux is generated throughout the thickness of the surface. We have used the value of", null, "at the outer face of the surface, but this is its maximum value and we need to average it over the thickness of the surface.", null, "", null, "", null, "", null, "This integration is not as hard as it looks because a lot of the integrals of individual terms are zero. It simplifies to", null, "", null, "", null, "We find that effecting the change in the field of motion generates a force proportional to and opposing the acceleration. Thus a simple spherical charge with no other properties has an inherent electromagnetic reaction to acceleration. The charge has the property which we call inertial mass. We may write a familiar equation.", null, "where", null, "is now the force required to accelerate the charge and", null, "The inertial mass of a pure charge is thus dependent on its charge and its radius.", null, "", null, "", null, "So we find that we can account for the generation of a force resisting the acceleration, and of the correct magnitude to account for the changes in the energy stored in the magnetic field which surrounds a charge by virtue of its velocity. To do this, we have had to refine our understanding of the way magnetic flux behaves. Instead of measuring the bulk of a magnetic field as the integral over a surface of magnetic induction through that surface, we take the volume integral of the energy density of the field. When a field changes in size, it is the flow of energy into and out of the field which is calculated. We still need to take into account the directional properties of magnetic flux when considering it as an energy density field. We need also to specify that movement of energy density flux can only take place parallel to the electric field of the charge.\n\nWhen we consider real electrons and take the charge on an electron together with the Bohr radius of the electron and Einstein's E = mc2, we can calculate that the mass equivalent of the energy of the electric field of an electron is", null, "The similarity in the two suggests that we are close to explaining the inertial of electrons as a purely electromagnetic phenomena.\n\nGo to next section" ]

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[ "", null, "0\n\n# What is 35 in 10?\n\n10 miles equals approximately 16 km.\n\n35 miles equals approximately 56 km.\n\nYou can use a formula to figure out the conversion from miles to km.\n\nThe formula is: miles x 1.609 = kilometers\n\nIf you want to convert km to miles, the formula is:\n\nkilometers x 0.6214 = miles\n\n🙏\n0\n🤨\n0\n😮\n0\nStudy guides\n\n## 📓Create a Study Guide\n\nEarn +20 pts\nQ: What is 35 in 10?\nSubmit" ]

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[ "kmmiles.com\n\nSearch\n\n# 492.4 km in miles\n\n## Result\n\n492.4 km equals 305.7804 miles\n\nYou can also convert 492.4 km to mph.\n\n## Conversion formula\n\nMultiply the amount of km by the conversion factor to get the result in miles:\n\n492.4 km × 0.621 = 305.7804 mi\n\n## How to convert 492.4 km to miles?\n\nThe conversion factor from km to miles is 0.621, which means that 1 km is equal to 0.621 miles:\n\n1 km = 0.621 mi\n\nTo convert 492.4 km into miles we have to multiply 492.4 by the conversion factor in order to get the amount from km to miles. We can also form a proportion to calculate the result:\n\n1 km → 0.621 mi\n\n492.4 km → L(mi)\n\nSolve the above proportion to obtain the length L in miles:\n\nL(mi) = 492.4 km × 0.621 mi\n\nL(mi) = 305.7804 mi\n\nThe final result is:\n\n492.4 km → 305.7804 mi\n\nWe conclude that 492.4 km is equivalent to 305.7804 miles:\n\n492.4 km = 305.7804 miles\n\n## Result approximation\n\nFor practical purposes we can round our final result to an approximate numerical value. In this case four hundred ninety-two point four km is approximately three hundred five point seven eight miles:\n\n492.4 km ≅ 305.78 miles\n\n## Conversion table\n\nFor quick reference purposes, below is the kilometers to miles conversion table:\n\nkilometers (km) miles (mi)\n493.4 km 306.4014 miles\n494.4 km 307.0224 miles\n495.4 km 307.6434 miles\n496.4 km 308.2644 miles\n497.4 km 308.8854 miles\n498.4 km 309.5064 miles\n499.4 km 310.1274 miles\n500.4 km 310.7484 miles\n501.4 km 311.3694 miles\n502.4 km 311.9904 miles\n\n## Units definitions\n\nThe units involved in this conversion are kilometers and miles. This is how they are defined:\n\n### Kilometers\n\nThe kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world.\n\n### Miles\n\nA mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada." ]

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[ "It\"s really common as soon as learning around fractions to want to know how transform a fraction like 43/50 right into a percentage. In this step-by-step guide, we\"ll present you how to revolve any fraction into a percentage really easily. Let\"s take it a look!\n\nWant to conveniently learn or present students just how to transform 43/50 to a percentage? pat this an extremely quick and fun video clip now!\n\nBefore we acquire started in the portion to percentage conversion, let\"s go over some really quick portion basics. Remember the a numerator is the number over the portion line, and the denominator is the number listed below the fraction line. We\"ll use this later in the tutorial.\n\nYou are watching: What percent of 50 is 43?\n\nWhen we are using percentages, what we are really saying is the the percent is a fraction of 100. \"Percent\" means per hundred, and so 50% is the same as speak 50/100 or 5/10 in fraction form.\n\nSo, because our denominator in 43/50 is 50, we could adjust the fraction to make the denominator 100. To perform that, we divide 100 by the denominator:\n\n100 ÷ 50 = 2\n\nOnce we have that, we have the right to multiple both the numerator and also denominator through this multiple:\n\n43 x 2/50 x 2=86/100\n\nNow we deserve to see the our portion is 86/100, which means that 43/50 together a percentage is 86%.\n\nWe can also work this out in a simpler way by an initial converting the fraction 43/50 to a decimal. To do that, we just divide the molecule by the denominator:\n\n43/50 = 0.86\n\nOnce we have actually the prize to the division, we deserve to multiply the answer by 100 to do it a percentage:\n\n0.86 x 100 = 86%\n\nAnd there you have it! Two different ways to transform 43/50 to a percentage. Both are pretty straightforward and easy to do, yet I personally like the convert to decimal method as that takes much less steps.\n\nI\"ve viewed a most students get puzzled whenever a question comes up about converting a portion to a percentage, however if you follow the actions laid out right here it need to be simple. The said, you might still need a calculator for more complicated fractions (and you can constantly use our calculator in the kind below).\n\nIf you desire to practice, grab you yourself a pen, a pad, and a calculator and shot to transform a few fractions come a portion yourself.\n\nHopefully this tutorial has actually helped you to understand how to convert a fraction to a percentage. You deserve to now walk forth and convert fractions to percentages as lot as your small heart desires!\n\nIf you discovered this content advantageous in your research, please perform us a an excellent favor and use the tool listed below to make certain you correctly reference united state wherever you usage it. We really appreciate your support!\n\n\"What is 43/50 as a percentage?\". tasiilaq.net. Accessed on September 18, 2021. Https://tasiilaq.net/calculator/fraction-as-percentage/what-is-43-50-as-a-percentage/.\n\n\"What is 43/50 together a percentage?\". tasiilaq.net, https://tasiilaq.net/calculator/fraction-as-percentage/what-is-43-50-as-a-percentage/. Accessed 18 September, 2021.\n\nSee more: How Do I Make My Nipples Bigger ? How To Make Nipples Look Larger" ]

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