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THE HALF-MULTIPLIER OPERATOR
[A]Tij o [B]jk = [C]JK
AND
THE UNIFIED FIELD EQUATION
c([A]Tij o [B]jk)= [C]JK
([A]Tij o [B]jk)= [C]jk Inventories/Accounting
1/N([1]1j [A]Tij o [B]jk)2= [C]jk Between Subjects Statistic
1/N([[A]Tij o [B]jk [1]k1)2= [C]jk Within Subjects Statistic
m([B]Tij o [a]jk)= [F]jk Newtons Second Law
m/N([1]1j[B]Tij o [a]jk)2= [F]jk connection of statistics and Physics
m[a]jk= [F]jk B=I
8((([B]Tij o [T]jk)= [G]jk Einsteins First Law of Gravity
8((/N([1]1j[B]Tij o [T]jk)2= [G]jk connection of statistics and Astro-Physics
8(([T]jk= [G]jk B=I
BY
CLINTON L. HOLT
THE UGLY DUCKLING
CLASSICAL PHYSICS
C([A]Tij o [B]jk)= [C]jk
Let c = m, and [B]jk = [a]jk (acceleration) and [C]jk = Fjk, then
[C]jk = m([A]Tij o [B]jk ) but [A]ij = [I]jj and [B]jk = [a]jk and [C]jk = Fjk, so
Fjk = m([I]jj o [a]jk) for m = to a constant.
Fjk = m([a]jk)
If [A]ij ( [I]jj , then we have
Fjk = m([A]Tij o [a]jk )
EINSTEIN’S FIELD EQUATION FOR GRAVITATION
C([A]Tij o [B]jk)= [C]jk
Let [C]jk = [G]jk , [B]jk = [T]ij and [A]ij = [I]jj and c = 8((.
Then we have:
8((([A]Tij o[T]jk) = [G]jk but [A]ij = [I]jj so we have
8(([T]jk = [G]jk which is Einstein’s First Field Equation. This is the equation as we now understand it, but the true equation is 8((([A]Tij o[T]jk) = [G]jk.
GRAVITATIONAL WAVE EQUATION:
[G]jk = 8(( (T[T]jk( 8(( (T([A]Tij o [T]jk)(
or [G]jk =
(T ( (T (
Gjk is now solvable.
STATISTICS
There was, until now, no field equation describing the field of statistics.
Going back to the Unified Field Equation:
C([A]Tij o [B]jk)= [C]jk and for the simple case, letting [B]jk = [I]jk and c = 1/N, the equation becomes:
1/N([A]Tij o [B]jk )= [C]jk =1/N([A]Tij o [I]jk)= 1/N([A]Tij) = [C]ij
Since [A][I] is a straight multiplication problem, we do not need to transpose it. But we do need to sum the columns of [A], so the basic statistical equation becomes:
1/N([1]1,I[A]ij) = [C]1j
But to make this statistical, we must square the above expression so that it becomes:
1/N([1]1,i[A]ij)2 = [C]21j Where [C]21j = CCT. (this gives us a one by one matrix as a solution).
Suppose [B]jk is not = to [I]jk. Then we have our basic complex statistical field equation, the simplest form of which is the Analysis of Variance.
1/N([1]1,i[A]ij[B]jk)2 = [C]21,k where [C]2 = CCT
But we also need to subtract the correction factor(s), so the total statistical field equation becomes:
1/N([1]1,i[A]ij[B]jk)2 - correction factor(s) = C21,k - correction factor(s).
There are only 3, perhaps 4 operators from which statistics (perhaps all of statistics) may be computed:
1/2(([A]Tij o [B]jk = i Cjk matrices (Half-Multiplier mode)
( where i = #rows in un-transposed matrix)
C( 1/N([A]Tij o [B]jk ) = 1/N[C]ik regular matrix multiplication.
R( 1/N([A]Tij o [B]jk ) = 1/N([B]jk[1]k,1o[A]Tij)
M( 1/N([A]Tij o [B]jk ) = 1/N([A]Tij[1]i,1o[B]jk)
QUANTUM STATISTICS:
[A]TMN[A]MN = [A]NM[A]MN = [A]2NN
Then we do the following:
1/N[DB1]iN [A]2NN[DB1]TiN 1/N[DB1]iN [A]2NN[DB1]Ni 1/N[DB1]iN [A]2NN[DB1]Ni
= = =
[DB1]iN[DB1]TiN [DB1]iN[DB1]Ni [DB1]2ii
1/N[C]2ii
[DB1]2ii
QUANTUM MECHANICS FROM EINSTEIN’S EQUATION:
8(( (T[T]jk(
[G]jk =
(T (
c (Tjk[H]jk(jk c(T([A]Tij o [H]jk)(
Ejk or [E]jk =
(T ( (T (
INVENTORY/ACCOUNTING SYSTEM:
[A]Tij o [B]jk = [C]jk =
[INV]ij[DB]jk = [SOL]ik This is the sum total of everything bought, sold, manufactured, etc.
i[A]Tij o [B]jk = i[C]jk =
i[INV]Tij o [DB]jk = i[AP]jk This keeps and individual item accounting of everything bought, sold, manufactured, etc. The superscript i just tells us which row in the un-transposed matrix we have hollow-dotted onto the [DB] matrix.
j[B]jk o [A]Tij = j[C]jk =
j[DB]jk o [INV]Tij = j[IP]jk This takes an individual item from the database matrix (a column) and multiplies it across the inventory(or accounting page if we wish) giving us a slice of the total pie, so to speak. It tells us how much of that item was used, by whom or what machine or smokestack, and shows how it is distributed throughout the total inventory.
[A]Tij o [B]jk = i, [C]jk sub-matrices .
[INV]Tij o [DB]jk = i, [SOL]jk sub-matrices
This is the pure half-multiplier operation giving all the accountpage matrices, i of them, in a single operation.
TABLE OF CONTENTS
OPEN LETTER Pg. 1
INTRODUCTION Pg. 2
THE THEORY OF INFORMATION AND THE UNIFIED FIELD EQUATION Pg. 11
PROOF FOR HALF-MULTIPLIER OPERATOR Pg. 11
PROOF FOR REGULAR MATRIX MULTIPLICATION Pg. 12
ROW PRODUCT OF A MATRIX Pg. 17
MATRIC PRODUCT OF A MATRIX Pg. 22
TRANSPOSE COMMUTIVITY OF THE HALF-MULTIPLIER OPERATOR Pg. 26
NESTED ARRAYS Pg. 27
ONTO MULTIPLICATION OF MATRICES Pg. 30
THE UNIFIED FIELD EQUATION Pg. 33
PHILOSOPHY Pg. 36
GENERALIZED INVENTORY/ACCOUNTING SYSTEM Pg. 40
CHEMICAL USAGE INVENTORY Pg. 40
MULTIPLE CHEMICAL USAGE INVENTORY Pg. 53
CALCULATION OF A WATER BILL Pg. 60
TO BILL THE AREA, STATE, USA, WORLD Pg. 65
TWO FACTOR MIXED DESIGN: REPEATED MEASURES ON ONE FACTOR FOR
STATISTICS AND ACCOUNTING/INVENTORY SYSTEMS (EXAMPLE) Pg. 73
STATISTICAL ANALYSIS OF PROBLEM Pg. 89
MATHCAD PROGRAM FOR INVENTORY STATISTICS Pg. 94
PHILOSOPHICAL MEANING WHEN ([A]ijT o [B]jk = ( Pg. 98
EXAMPLE: COMPANY THAT MAKES SAUSAGE AND BOLOGNA Pg. 99
ARGONIA, KS STORE RECIPIES Pg. 119
LOVE BOX CO. INVENTORY Pg. 127
MATRIX SOLUTION FOR GAUSSIAN REDUCTION Pg. 133
SIMPLIFYIN’ Pg. 144
INVERSES Pg. 145
SYMMETRY Pg. 146
NESTED ARRAYS AND GAUSSIAN REDUCTION Pg. 147
GENERAL MATRIX SOLUTION SETS FOR ½ H ( AND ( Pg. 152
ELEMENTARY MOLECULAR ORBITAL METHODS Pg. 154
ETHYLENE Pg. 156
ACCOUNTING FOR OVERLAP INTERGAL Pg. 157
BUTADIENE Pg. 159
BOND ORDER, MOBILE BOND ORDER, TOTAL BOND ORDER AND FREE VALENCE INDEX Pg. 162
EXAMPLE: TRIMETHYLENE METHANE Pg. 162
MO’S AND NESTED ARRAYS Pg. 166
CYCLOBUTADIENE Pg. 167
OZONE Pg. 169
UTILIZING SYMMETRY IN BUTADIENE Pg. 173
COEFFICIENTS BY PROPERTIES OF NON-BONDING
MOLECULAR ORBITALS (NBMO’S) Pg. 176
HETEROCYCLIC COMPOUNDS Pg. 178
PERTUTBATION THEORY (SIMPLE) Pg. 181
PART 2. PERTURBATION THEORY (COMPLETE) Pg. 184
WAVE FUNCTION OF ACROLEIN FROM BUTADIENE Pg. 185
SAME COMPUTATION USING THE HALF-MULTIPLIER OPERATOR Pg. 188
COMPUTING (E Pg. 193
THE PICTURE FUNCTION Pg. 196
BOND ORDERS, FREE VALENCE INDEX AND CHARGE DENSITY FOR ACROLEIN Pg. 197
BOND ORDER AND CHARGE DENSITY USING THE HALF-MULTIPLIER OPERATOR Pg. 199
SECOND ORDER ENERGY CORRECTION Pg. 201
HIGHER ORDER ENERGY CORRECTIONS Pg. 202
MULTIPLE PERTURBATIONS Pg. 205
BIBLIOGRAPHY Pg. 211
STATISTICS Pg. 211
THE MEAN Pg. 212
STANDARD DEVIATION Pg. 212
t-TEST FOR A DIFFERENCE BETWEEN A SAMPLE MEAN AND A
POPULATION MEAN Pg. 213
t-TEST FOR THE DIFFERENCE BETWEEN TWO INDEPENDENT MEANS Pg. 215
t-TEST FOR RELATED MEASURES Pg. 219
SANDLERS A TEST Pg. 221
ANALYSIS OF VARIANCE Pg. 222
BASIC TRANSLATIONS FOR STATISTICS Pg. 225
COMPLETELY RANDOMIZED DESIGN Pg. 227
FACTORIAL DESIGN: TWO FACTORS Pg. 234
FACTORIAL DESIGN: THREE FACTORS (COMPLETE ANALYSIS OF STATISTICAL METHOD Pg. 240
WITH NESTED ARRAYS)
TREATMENT BY LEVELS DESIGN (WITH NESTED ARRAYS) Pg. 269
TREATMENT BY SUBJECTS: REPEATED MEASURES DESIGN Pg. 275
TREATMENT BY TREATMENT BY SUBJECTS, OR REPEATED MEASURES:
TWO FACTOR DESIGN Pg. 275
TWO FACTOR MIXED DESIGN: REPEATED MEASURES ON ONE FACTOR Pg. 284
THREE FACTOR MIXED DESIGN: REPEATED MEASURES ON ONE FACTOR Pg. 292
THREE FACTOR MIXED DESIGN: REPEATED MEASURES ON TWO FACTORS Pg. 305
LATIN SQUARE DESIGN: SIMPLE Pg. 321
LATIN SQUARE DESIGN: COMPLEX Pg. 330
TEST FOR DIFFERENCE BETWEEN VARIANCES OF TWO INDEPENDENT SAMPLES
(test for homogeneity of independent variances) Pg. 346
TEST FOR DIFFERENCE BETWEEN VARIANCES OF TWO RELATED SAMPLES
(test for homogeneity of related variances) Pg. 348
T-TEST FOR DIFFERENCES AMONG SEVERAL MEANS Pg. 354
DUNCANS MULTIPLE RANGE TEST Pg. 361
THE NEUMAN-KEUL’S MULTIPLE RANGE TEST Pg. 363
THE TUKEY TEST Pg. 365
SCHEFFES TEST Pg. 367
DUNNETT’S TEST Pg. 369
F-TESTS FOR SIMPLE EFFECTS Pg. 370
Question 1 Pg. 370
Question 2 Pg. 371
Question 3 Pg. 376
USE OF ORTHOGONAL COMPONENTS IN TESTS FOR TREND Pg. 382
EXAMPLE 2 (forgot example 1) Pg. 390
EXAMPLE 3 Pg. 400
CORRELATION AND RELATED TOPICS Pg. 400
PEARSON PRODUCT-MOMENT CORRELATION Pg. 402
SPEARMAN RANK-ORDER CORRELATION (RHO () Pg. 403
KENDALL RANK-ORDER CORRELATION ( (TAU) Pg. 406
POINT-BISERIAL CORRELATION Pg. 408
THE CORRELATION RATIO ( (ETA) Pg. 411
PARTIAL RANK-ORDER CORRELATION USING KENDALL’S TAU Pg. 413
MULTIPLE CORRELATION: THREE VARIABLES Pg. 414
SIMPLE ANALYSIS OF COVARIANCE: ONE TREATMENT VARIABLE Pg. 416
FACTORIAL ANALYSIS OF COVARIANCE: TWO TREATMENT VARIABLES Pg. 447
RELIABILITY OF MEASUREMENT: THE TEST AS A WHOLE Pg. 448
TEST FOR DIFFERENCE BETWEEN INDEPENDENT CORRELATIONS Pg. 449
TEST FOR DIFFERENCE BETWEEN DEPENDENT CORRELATIONS Pg. 445
NON-PARAMETRIC TESTS, MISCELLANEOUS TESTS OF SIGNIFICANCE
AND INDEXES OF RELATIONSHIP Pg. 453
TEST FOR SIGNIFICANCE OF A PROPORTION Pg. 454
TEST FOR SIGNIFICANCE OF THE DIFFERENCE BETWEEN TWO PROPORTIONS Pg. 454
THE MANN-WHITNEY U-TEST FOR DIFFERENCES BETWEEN INDEPENDENT SAMPLES Pg. 458
WILCOXON A SIGN TEST FOR DIFFERENCES BETWEEN RELATED SAMPLES Pg. 460
SIMPLE CHI SQUARE AND THE PHI COEFFICIENT Pg. 461
COMPLEX CHI-SQUARE AND THE CONTINGENCY COEFFICIENT C Pg. 463
QUANTUM STATISTICS Pg. 466
LINEAR REGRESSION Pg. 470
STRAIGHT LINE Pg. 470
SEMI-LOG Pg. 471
LOG-LOG Pg. 473
PARABOLIC FIT Pg. 474
CRYPTOGRAPHY Pg. 475
PHYSICS 101 Pg. 484
OPEN LETTER
Twenty years ago (in 1978) I received a double degree in Chemistry and Mathematics at Fort Hays State University in Hays, KS. I had one mediocre job in the next three years and few interviews. In 1981 I returned to Fort Hays and completed my degree in Physics. After receiving my third Bachelors degree, I got no further interviews, much less jobs over the next 16 years. I had hoped to complete my Masters and PhD while I was working as my family was poor and I couldn’t afford graduate school on my own. Well, right now, I have forgotten everything I ever knew, except for some simple algebra, and do not consider myself fit to enter the job market as a scientist anymore. I have given up.
Before reading the following book, I will say I did take a course in Linear Algebra and had two chapters of Matrix theory in an applied math class I took. The only other class work I took concerning matrices was Quantum Mechanics I & II. So if some of my indical notation is incorrect, I beg your pardons as I had to develop most of the math on my own.
A note about the proofs, just to keep them short and simple, I have shown an example for a small matrix, then for the proof, I proved the theorems for M,N and M+1, N+1 matrices both at the same time. I could prove them both separately, but this book is too long as it is.
When I took some classes in physics in college, I was impressed by some of the derivations because the mathematics was pretty. In most cases the math was boring or utilitarian, but in some cases it was just intellectually beautiful. In the case of the Half-Multiplier Operator, the math is quite pretty. It is also seductive. Just when I thought I was through with a problem, the operator would show me a simpler, easier way to solve the problem. But just the fact that we can solve all parts of a problem in one operation instead of many separate, similar calculations is only one aspect of this equations beauty (see Sect. 2.3 and 2.4 in Statistics). We can solve everything all at the same time if the math is set up correctly. Also, even though I derived Classical Physics, Einstein’s First Field Equation for Gravity and Quantum Mechanics, most of this book covers the solution for the field equations for accounting/inventory systems and the field equations for statistics. The main reason is that I have forgotten all my physics, although there is some at the end of this book and some quantum mechanics is covered in the section on quantum chemistry. Anyway, you folks have already covered these equations extremely well but no one has ever done statistics and inventories this way. I think that even the calculus will fall easily under this operator. Another part of the beauty of these equations is that the mathematics, once completed, is automatically it’s own computer program.
For the separate book on statistics, you will need a copy of THE COMPUTATIONAL HANDBOOK OF STATISTICS, 2nd Edition by James L Bruning and B. L. Kintz , 2nd Edition, Scott Foresman and Co., Glenview, Ill. to be able to understand my approach (unless you’re a professional statistician). This is an example of the beauty of this operator. I have never taken a statistics class above the most elementary, yet with the Statistical Field Equation, even I was able to re-write the entire field of Statistics very easily and simply. In fact, I was writing the programs after just looking at the way the problem was done without having to solve the problem in the first place or see the mathematical equations. You people who were good enough in math to get a job and are working daily with it should have no problem with this book. The most advanced techniques are in the 9th grade level (Logs and simple algebra). Most of the math is simple addition, subtraction, multiplication and division.
I’ve sent copies of this book when it was in a much shorter form to various journals, but they refused to read it. They say it works, nothing’s wrong with it, they’ve never seen it before, but it is impossible to do math this way. Perhaps their feelings are the very reasons the Unified Field Equation was never discovered before. About 200 years ago, some mathematician said there was only one way to multiply a matrix and that notion has been accepted as gospel ever since. No one has ever looked to see if this idea was true or not. In this book you will see the first effort to explore matrices as they are meant to be used.
At present, I am working in a machine shop gluing cardboard boxes together. I am only able to afford to print 33 copies of this book. Please, if you cannot read this book, return it to me at Clint Holt, 1751 S. Battin, Wichita KS 67218. I will pay the postage. If you send in examples of how you solved problems using this operator, I’ll include them in my next copy and name the equations after you.
Well, I’ll let you get on with the reading of this book.
Sincerely,
Clinton L. Holt
The “Ugly Duckling”
2-22-1998
INTRODUCTION
A little over a year ago, while I was on lay-off status and unemployed, I occupied my time by trying to write a simple program so that I could perform elementary Molecular Orbital computations on my HP-48SX calculator. While I was working with Gaussian reduction to solve for the roots of the secular equations of simple organic molecules, I made a singular discovery. If I took every step I computed in a Gaussian reduction sequence and wrote each solution step in a series of columns and made these columns into a matrix, then transposed and multiplied this new matrix to the secular determinant, this matrix multiplication automatically carried out the Gaussian reduction of computing zeros in each position under the principal diagonal. By all accounts, what I had done was impossible as defined by the mathematics of today. I mean, I had been told since I was a senior in high school and by my professors all through college that we can multiply matrices only one way, that all other multiplications any other way have no meaning. I had also been told that there never has been a proof for Gaussian reduction. It is used to solve for systems of linear equations because it works, but has no other mathematical validity or use. What I had done by accident was to discover a proof for Gaussian reduction. What I also discovered, but did not know until later, was that this is the connection between our regular mathematics that we all know (addition, subtraction, multiplication and division) and tensors and matrices. What this means is that if
1, 2, 3,. . . ( are numbers with all the properties mathematicians have discovered about numbers up to and including logarithms, calculus, trigonometry, differential equations, etc., that an mxn array of numbers can be considered to be a single number, not just an
array of separate numbers. In other words, a two by two matrix is not a matrix containing four separate numbers 1, 2, 3 and 4, but can be considered as a single number all by itself. To add, subtract, multiply and divide this number, all the other numbers we wish to operate on must be of the same size. To add 50 to 1, 50 can be considered as a one by one matrix and 1 is a one by one matrix, and we get [50] + [1] = [51]. We cannot add 1 to a number of a different size, i.e. [1] + [50 2] cannot be summed, but [1 -2] + [50 2] = [51 0] which can be summed. Let’s take two matrices with elements 1, 2, 3 and 4 and 5, 6, 7 and 8 respectively and see how their properties mimic our regular concept of mathematics. First we shall add the two numbers together and see that we add all four numbers at the same time to their counterparts in the second matrix in one operation.
[pic]
Here we add 1+5 = 6, 2+6=8, 3+7=10 and 4+8=12 all at the same time. They are commutative under addition, i.e if we reverse the order of addition, we get the same answer:
[pic]
Let’s now subtract the two numbers, and we will do it both ways:
[pic]
[pic]
Note that switching the order of subtraction changes the signs, so these two numbers are not in general commutative, but the absolute value of the solution would be commutative.
If we multiply the two matrices in the manner that is accepted by mathematics, we would proceed in the following manner:
[pic]
[pic]
To accomplish this, we proceed as follows: The top row in the first matrix is multiplied to the first column in the second matrix and the two products are added, i.e.
1x5 + 2x7 = 5+14 = 19. The first row is now multiplied to the second column in the second matrix and the sum of the products is places in the second top-hand position. 1x6 + 2x8 = 6+16=22. We have run out of columns in the second matrix to multiply [1 2] by, so now we go to the second row in the first matrix and multiply it to the first column in the second matrix: 3x5 + 4x7 = 15 + 28 = 43. This number is put in the first position in the second row. To finish up, we finally multiply the second row in the first matrix to the second column in the second matrix and sum their products. This solution is put in the second position in the bottom row of the solution matrix. 3x6 + 4x8 = 18 + 32 = 50. This sure doesn’t look like our regular multiplication that we are familiar with. If we multiply two numbers we should get a number we are familiar with. So now we get to the next step in our new definition of number. Mathematicians say this operation is illegal, that it doesn’t work and the results have no meaning, but we will look at it anyway.
I will define the operator ( as meaning we will multiply the first number in the top row of the first matrix by the first number in the top row of the second matrix, the second number in the top row of the first matrix by the second number in the top row of the second matrix, the first number in the second row of the first matrix by the first number in the bottom row of the second matrix and the second number in the bottom row of the first matrix by the second number in the bottom row of the second matrix, i.e.
1 2 ( 5 6 = 5 12
3 4 7 8 21 32
This actually makes more sense since it is logical that 5x1=5, 6x2=12, 7x3=21 and 8x4=32. Although this form of multiplication cannot work and is impossible, it’s use is found throughout statistics, but in the examples I give in this book, I solve for the grand sum of squares in a way more acceptable to mathematicians (the grand sum of squares means that we square each number in the matrix and then add all the numbers together to obtain a single number. Although I did not really write it this way in my book, we could write the sum of squares more simply and elegantly as
[1]1i([A]ij([A]ij)[1]j1
To obtain the sum of squares for the second matrix above, we would compute it as follows:
[pic] [pic]
[pic]
[pic]
Where [pic]
The arrow above the squared matrix is MathCad’s of saying we square each individual element in the matrix rather than squaring the matrix itself using regular matrix multiplication. So the grand sum is 25+36+49+64 = 174. (I use MathCad +6 as my computer’s math program, rather than Excel or Lotus or Quattro Pro spreadsheet because it can handle math better than these programs can.) This sum of squares is the correction factor in statistics. Note also that the ( operation is commutative, just like in our regular mathematics i.e.
[pic]and [pic]
Finally we get to division, there is no way to divide a number by a matrix, so we have to multiply the matrix by an inverse to accomplish division, but on the one-to-one system as shown just above, we can divide matrices. Let’s divide the second matrix by the first.
[pic]
or
[pic]
and
[pic]
The rightmost term under the arrow takes the inverse of each element in the 1, 2, 3, 4 matrix and multiplies it one-to-one to the 5,6,7,8 matrix (under the second, longer arrow). All computer programs are written using the inverse of matrices, so we must use their terminology to obtain the results we desire.
Let’s look at how the half-multiplier operation works using the two 4x4 matrices above. To half multiply the two matrices, we must first transpose the matrix containing the elements 1, 2, 3 and 4 and hollow dot multiply into the second matrix, i.e.
[pic]
[pic]o[pic]= [pic]o[pic] [pic]o[pic]
[pic] o[pic][pic][pic][pic] = [pic] [pic] = [pic][pic]
This is the multiplied matrix in half-multiplier mode. There are three ways to manipulate this data, actually there are four, we can leave the matrices as they are, or we can align the two matrices on top of each other and sum their rows:
[pic]
[pic]
Deleting the inner brackets we get
which is the solution for regular matrix multiplication.
[pic]
Instead of adding the rows, let's add the columns of each of the 2 sub-matrices:
Deleting the inner brackets and combining the two column matrices into a 2x2 matrix we get:
[pic][pic]=[pic]
To solve this mathematically according to the proof, we sum the columns in the
5, 6, 7, 8 matrix and half-multiply into the 1, 2, 3, 4 matrix, i.e.
[pic] o[pic]=[pic] o[pic]=[pic]
This is defined as the cross product of a matrix.
The next thing we can do is add the two sub-matrices together, i.e.
[pic]
To solve this mathematically according to the proof, we sum the columns in the
1, 2, 3, 4 matrix and half-multiply into the 5, 6, 7, 8 matrix, i.e.
[pic] o[pic]=[pic] o[pic]=[pic]=[pic]
This is defined as the matric product of a matrix.
Three of the above four computations, although derived by using Gaussian Reduction, are illegal, or at least unknown to modern mathematics. Examples of their use will follow in this book.
Now that our concept of number has been more completely defined, I wish that other mathematicians and physicists would check on this work and see how it applies or does not apply to other problems in mathematics, especially statistics and calculus. I have derived Newton’s Equation and Einstein’s First Field Equation of Gravity, so this equation should hold through all of Physics, unless the field equations are wrong. It would especially be neat to see if the completion of these equations hold promise for ease in computation. Please do not say yea or nay in words, since I have the mathematical proofs, please prove mathematically that this does or does not work.
Before we get into the book proper, I will present a further tutorial on how this math is used.
TUTORIAL
The first thing we will do is sum the columns of the following matrix. The row matrix ONE is the operator which will sum the columns.
[pic] [pic]
The problem looks like:
[pic]
[pic]
To sum the rows we post multiply by a column Matrix composed of ones equal to the number of columns in the
pre-multiplier Matrix. ie:
[pic] [pic]
The problem for summing the rows looks like:
[pic][pic] [pic]
To multiply each row in a Matrix by a constant, proceed as follows:
3 X 1 7 13 19 25 31
4 X 2 8 14 20 26 32
5 X 3 9 15 21 27 33 =
6 X 4 10 16 22 28 34
7 X 5 11 17 23 29 35
8 X 6 12 18 24 30 36
[pic]
Modern computer programs allow us to multiply in a one-to-one correspondence with two matrices, but none allow us to multiply in the manner of Gauss. This method has not been invented yet. Fortunately, there is an operation that can get us the same answer, but at the cost of more multiplication steps and computer memory. But we must do the math this way until programmers write more efficient programs. To do this, we must diagonalize the pre-multiplier matrix, in this example N:
[pic]
[pic] M = [pic]
[pic]
[pic]
To multiply the columns of the Matrix by the same constants, we post-multiply (instead of pre-multiplying) by the diagonal Matrix. ie:
[pic][pic]
[pic]
ELEMENTARY STATISTICAL CALCULATIONS:
Suppose we want to add the first two numbers, the third and fourth numbers and the fifth and sixth numbers together and keep the solutions separate:
[pic]
We can proceed as follows:
[pic]
[pic]
[pic]
[pic]
Or we can set up a database matrix that will do the same multiplications all at the same time, i.e.
[pic] or [pic]
[pic]
Suppose we want to add the first three numbers and the second three numbers and keep the solutions separate:
[pic]
[pic] [pic]
[pic]
[pic]
Suppose we want to add the first and third numbers, the second and fourth numbers and the fifth and sixth numbers together:
[pic] [pic]
[pic]
[pic]
An interesting problem is to take [1]8,8 and make a tic-tac-toe figure out of it. We may accomplish it in this manner: [DB1]8,8[1]8,8 + [1]8,8[DB1]8,8
[pic] [pic]
[pic]
If we do not wish to have 2’s at the point’s of intersection but wish them to be ones, we may proceed as follows: [DB1]8,8[1]8,8 + [1]8,8[DB1]8,8 - [DB1]8,8[1]8,8[DB1]8,8
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Suppose we wish to work only with the 4x4 array in the center of the [1]8,8 matrix. This may be accomplished as follows:
[pic] [pic]
[pic] [DB2]8,8[1]8,8[DB2]8,8
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Suppose we wish to make an X through the matrix, we can go through the following transformation:
[pic] [pic]
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The arrow above the multiplication means we multiply one-to-one rather than using regular matrix multiplication.
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Of course, we could just MAJORDIAG + MINORDIAG and get the same answer, but I’m trying to show how the math here works.
THE THEORY OF INFORMATION
AND
THE UNIFIED FIELD EQUATION
PROOF FOR THE HALF-MULTIPLIER OPERATOR
PROOF OF [A]T o [B]= diagonal(s)[A]T[B]
I must do this proof first, since I will need it to prove the rest of the math to follow.
Suppose we have the matrices:
A11 B11. . . B1K. . . B1M. . . B1,M+1
. . . . .
[A]T = A1J [B]= BJ1. . . BJK. . . BJM.. . .BJ,M+1
. . . . .
A1N BN1. . . BNK. . . BNM. . . BN,M+1
. . . . .
A1,N+1 BN+1,1. . BN+1,K . .BN+1,M . .BN+1,M+1
Changing the pre-multiplier into a diagonal matrix and multiplying:
A11. . . 0. . . 0. . .0 B11. . . B1K. . . B1M. . . B1,M+1
. . . . .. . . . . . . . . . . .
. . . . .. . . . . . . . . . . .
0. . . A1J. . . 0. . . 0 BJ1. . . BJK. . . BJM.. . .BJ,M+1
. . . . .. . . . . . . . . . . .
. . . . .. . . . . . . . . . . .
0. . . 0. . . A1M. . .0 BN1. . . BNK. . . BNM. . . BN,M+1
. . . . .. . . . . . . . . . . .
. . . . .. . . . . . . . . . . .
0. . . 0. . 0. . A1,M+1 BN+1,1. . BN+1,K . .BN+1,M . .BN+1,M+1
We get the solution:
A11B11 . . .+. A11B1K .+. . .A11B1M .+. . . .A11B1,M+1
. . . .
. . . .
A1JBJ1 . . .+. A1JBJK .+. . .A1JBJM .+. . . .A1JBJ,M+1
. . . .
. . . .
A1MBN1 . . .+. A1MBNK .+. . .A1MBNM .+. . . .A1MBN,M+1
. . . .
. . . .
A1,M+1BN+1,1 .+. A1,M+1BN+1,K .+. A1,M+1BN+1,M .+. A1,M+1BN+1,M+1)
This proof is in thousands of textbooks from High school to college to PhD so I won’t prove it here.
Taking the pre-multiplier and multiplying each element across it’s corresponding row we get:
A11 B11. . . B1K. . . B1M
. . . .
. o . . .
A1J BJ1. . . BJK. . . BJM
. . . . =
A1M BN1. . . BNK. . . BNM.
A11(B11 . . .B1K . . .B1M) A11B11 .+. A11B1K .+. A11B1M
. . . . . . .
. . . . . . .
A1J(BJ1 . . .BJK . . .BJM) = A1JBJ1 .+. A1JBJK .+. A1JBJM
. . . . . . .
. . . . . . .
A1M(BN1 . . .BNK . . . BNM) A1MBN1 .+. A1MBNK .+. A1MBNM
This is for all mxn Matrices, now to prove it works for all matrices m+1, n+1 :
A11 B11. . . B1K. . . B1M. . . B1,M+1
. . . . .
. . . . .
A1J BJ1. . . BJK. . . BJM.. . .BJ,M+1
. o . . . . =
. . . . .
A1M BN1. . . BNK. . . BNM. . . BN,M+1
. . . . .
. . . . .
A1,M+1 BN+1,1. . BN+1,K . .BN+1,M . .BN+1,M+1
A11(B11 . . .B1K . . .B1M . . . B1,M+1 ) A11B11 .+. A11B1K .+. A11B1M .+. A11B1,M+1
. . . . . . . . .
. . . . . . . . .
A1J(BJ1 . . .BJK . . .BJM. . . . BJ,M+1) A1JBJ1 .+. A1JBJK .+. A1JBJM .+. A1JBJ,M+1
. . . . . . . . .
. . . . . = . . . .
A1M(BN1 . . .BNK . . . BNM . . . BN,M+1) A1MBN1 .+. A1MBNK .+. A1MBNM .+. A1MBN,M+1
. . . . . . . . .
. . . . . . . . .
A1,M+1(BN+1,1. . .BN+1,K. . .BN+1,M. . . BN+1,M+1) A1,M+1BN+1,1 .+. A1,M+1BN+1,K .+. A1,M+1BN+1,M .+. A1,M+1BN+1,M+1)
QED
PROOF #1: Regular matrix multiplication: [A]ij[B]jk = [C]ik .
[A]Tij o [B]jk = Cjk
First I will do a micro-proof for simplicity, then a regular proof.
A11 A21 A31 B11 B12 B13
A12 A22 A32 o B21 B22 B23 =
A13 A23 A33 B31 B32 B33
A14 A24 A34 B41 B42 B43
Separating each column in the [A]T matrix into a column matrix, we cross multiply:
A11 B11 B12 B13 A21 B11 B12 B13 A31 B11 B12 B13
A12 o B21 B22 B23 A22 o B21 B22 B23 A32 o B21 B22 B23
A13 B31 B32 B33 A23 B31 B32 B33 A33 B31 B32 B33
A14 B41 B42 B43 A24 B41 B42 B43 A34 B41 B42 B43
We take the first column in [A]T and multiply it straight across the [B]jk matrix. This operation is equivalent to diagonalizing the first column of [A]ij and multiplying across [B]. i.e.
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This is equal to:
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But this is a nested array, and I will show later that when we transpose a
nested array, only the matrices are transposed and not their elements. Since in Half-Multiplying we originally transposed the spreadsheet Matrix, we must now re-transpose the nested array back in it’s un-transposed state. Transposing the nested array also rids us of the three inner matrix brackets and makes the array into a regular array of dimensions 12x3. (of course, I have not proved this yet). i.e.
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Now we will sum the columns for each individual matrix: ( we must remember that we are working with nested arrays, but computers are not programmed to handle these yet, so we must set up the pre-multiplier so that it operates on the individual sub-matrices)
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The answer to this is quite long on mathcad, but since we are working with nested arrays, we may also look at the multiplication in this manner:
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But this is equal to:
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micro-QED
Generalized Proof
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I am going to have to type this in by hand, Word 7 doesn’t have the memory to handle the math here. The Solution for a regular MxN matrix multiplication is given by:
A11B11. .+ .A1JBJ1+. . . A1NBN1 A11B1K+. . . .A1JBjK. .+ . A1NBNK.
. . . . .
. . . . .
Ai1B11. .+ .AiJBJ1+. . . AiNBN1 Ai1B1K+. . . .AiJBJK. + A1NBNK
. . . . .
. . . . .
AM1B11. .+ .AMJBJ1+. . . AMNBN1 AM1B1K+. . . .AMJBJK. + AiNBNK
A11B1M . + .A1JBJM . .+. A1NBNM A11B1M+1. .+.A1jBjM . . +.A1NBNM
. . . . . . . . . . . .
Ai1B1M . +. AiJBJM . .+..AiNBNM Ai1B1M+1. .+.AijBjM . . +.AiNBNM
. . . . . .
. . . . . .
AM1B1M . + .AMJBJM+. .+. AMNBNM AM1B1M. .+.AMjBjM. . .+.AMNBNM
Now we shall achieve the same answer by the properties of the half-multiplier operator. We shall first accomplish this by multiplying by the diagonal form for all matrices size MxN as this is the most familiar to mathematicians.
Diagonalizing the first transposed column (remember, for this to work, AiN must be transposed and we multiply across the Bnm matrix (Remember.to get our multiplied product o
A11 B11. . .B1K. . .B1M
. . . .
. . . .
A1J BJ1. . .BJK. . .BJM
. . . . =
. . . .
A1N BN1. . .BNK. . .BNM
A11(B11. . .B1K. . .B1M)
. . .
. . .
A1J (BJ1. . .BJK. . .BJM)
. . . =
. . .
A1N (BN1. . .BNK. . .BNM)
A11B11. . .A11B1K. . .A11B1M
. . .
. . .
A1J BJ1. . .A1JBJK. . .A1JBJM
. . .
. . .
A1N BN1. . .A1NBNK. . .A1NBNM
Now for the i th column:
Ai1 B11. . .B1K. . .B1M Ai1(B11. . .B1K. . .B1M)
. . . . . . .
. . . . . . .
AiJ BJ1. . .BJK. . .BJM AiJ (BJ1. . .BJK. . .BJM) =
. . . . = . . .
. . . . . . .
AiN BN1. . .BNK. . .BNM AiN (BN1. . .BNK. . .BNM)
Ai1B11. . .Ai1B1K. . .Ai1B1M
. . .
. . .
AiJ BJ1. . .AiJBJK. . .AiJBJM
. . .
. . .
AiN BN1. . .AiNBNK. . .AiNBNM
And now for the Mth column:
AM1 B11. . .B1K. . .B1M AM1(B11. . .B1K. . .B1M)
. . . . . . .
. . . . . . .
AMJ BJ1. . .BJK. . .BJM AMJ (BJ1. . .BJK. . .BJM) =
. . . . = . . .
. . . . . . .
AMN BN1. . .BNK. . .BNM AMN (BN1. . .BNK. . .BNM)
AM1B11. . .AM1B1K. . .AM1B1M
. . .
. . .
AMJ BJ1. . .AMJBJK. . .AMJBJM
. . .
. . .
AMN BN1. . .AMNBNK. . .AMNBNM
Now let’s make these into a Nested array and sum each column of the individual sub-matrice
A11B11. . .A11B1K. . .A11B1M
. . .
. . .
A1J BJ1. . .A1JBJK. . .A1JBJM
. . .
. . .
A1N BN1. . .A1NBNK. . .A1NBNM
Ai1B11. . .Ai1B1K. . .Ai1B1M
. . .
. . .
AiJ BJ1. . .AiJBJK. . .AiJBJM
. . .
. . .
AiN BN1. . .AiNBNK. . .AiNBNM
AM1B11. . .AM1B1K. . .AM1B1M
. . .
. . .
AMJ BJ1. . .AMJBJK. . .AMJBJM
. . .
. . .
AMN BN1. . .AMNBNK. . .AMNBNM
A11B11..+.. A1J BJ1..+.. A1N BN1 A11 B1K..+..A1JBJK..+.. A1NBNK A11B1M..+.. A1JBJM..+..A1NBNM
Ai1B11..+.. AiJ BJ1..+.. AiN BN1 Ai1B1K..+..AiJBJK..+.. AiNBNK Ai1B1M..+..AiJBJM..+.. AiNBNM
AM1B11..+.. AMJ BJ1 ..+..AMN BN1 AM1B1K..+.. AMJBJK..+.. AMNBNK AM1B1M..+.. AMJBJM..+..AMNBNM
Now let’s do the same for all M+1, N+1 Matrices:
A11 B11. . .B1K. . .B1M. . .B1,M+1
. . . . .
. . . . .
Ai1 BJ1. . .BJK. . .BJM. . .BJ,M+1
. . . . . =
. . . . .
A1N BN1. . .BNK. . .BNM. . .BN,M+1
. . . . .
. . . . .
A1,M+1 BN+1,1 . .BN+1,K . .BN+1,M. .BN+1,M+1
A11 (B11. . .B1K. . .B1M. . .B1,M+1) A11 B11. . . A11B1K. . . A11B1M. . . A11B1,M+1
. . . . . . . . . .
. . . . . . . . . .
AiJ (BJ1. . .BJK. . .BJM. . .BJ,M+1) AiJ BJ1. . . AiJ BJK. . . AiJ BJM. . . AiJ BJ,M+1
. . . . . = . . . . .
. . . . . . . . . .
A1N (BN1. . .BNK. . .BNM. . .BN,M+1) A1N BN1. . . A1N BNK. . . A1N BNM. . . A1N BN,M+1
. . . . . . . . . .
. . . . . . . . . .
A1,N+1 (BN+1,1. .BN+1,K. .BN+1,M. .BN+1,M+1) A1,N+1BN+1,1. . A1,N+1BN+1,K. . A1,N+1BN+1,M. . A1,N+1BN+1,M+1
The summed columns are equal to:
[A11B11. .+ .AiJBJ1+. . . A1NBN1 + A1,N+1BN+1,1 ] [A11B1K+. . . .AiJBJK. .+ . A1NBNK + A1,N+1BN+1,K]
A11B1M . + .AiJBJM+. .+. A1NBNM + . A1,N+1BN+1,M] [A11B1,M+1. .+.AijBj,M+1. .+.A1NBN,M+1 + A1,N+1BN+1,M+1]
Now we multiply by the Jth column:
Ai1 B11. . .B1K. . .B1M. . .B1,M+1
. . . . .
. . . . .
AiJ BJ1. . .BJK. . .BJM. . .BJ,M+1
. . . . . =
. . . . .
A1N BN1. . .BNK. . .BNM. . .BN,M+1
. . . . .
. . . . .
A1,N+1 BN+1,1 . .BN+1,K . .BN+1,M. .BN+1,M+1
Ai1 (B11. . .B1K. . .B1M. . .B1,M+1) Ai1B11. . . Ai1B1K. . . Ai1B1M. . . Ai1B1,M+1
. . . . . . . . . .
. . . . . . . . . .
AiJ (BJ1. . .BJK. . .BJM. . .BJ,M+1) AiJ BJ1. . . AiJBJK. . . AiJBJM. . . AiJBJ,M+1
. . . . . = . . . . .
. . . . . . . . . .
AiN (BN1. . .BNK. . .BNM. . .BN,M+1) AiN BN1. . . AiN BNK. . .AiN BNM. . . AiN BN,M+1
. . . . . . . . . .
. . . . . . . . . .
Ai,N+1 (BN+1,1. .BN+1,K. .BN+1,M. .BN+1,M+1) Ai,N+1 BN+1,1. . Ai,N+1 BN+1,K. . Ai,N+1 BN+1,M. . Ai,N+1 BN+1,M+1
Summing the columns we get:
[Ai1B11. .+ .AiJBJ1+. . . AiNBN1 + Ai,N+1BN+1,1] [ Ai1B1K+. . .AiJBJK. .+ . AiNBNK + Ai,N+1BN+1,K ]
[Ai1B1M . +. AiJBJM . .+..AiNBNM + . Ai,N+1BN+1,M ] [Ai1B1M+1. .+.AijBjM+1 . . +.AiNBNM+1 + Ai,N+1BN+1,M+1]
Now we multiply by the Mth column:
AM1 B11. . .B1K. . .B1M. . .B1,M+1
. . . . .
. . . . .
AMJ BJ1. . .BJK. . .BJM. . .BJ,M+1
. . . . . =
. . . . .
AMN BN1. . .BNK. . .BNM. . .BN,M+1
. . . . .
. . . . .
AM,N+1 BN+1,1 . .BN+1,K . .BN+1,M. .BN+1,M+1
AM1 (B11. . .B1K. . .B1M. . .B1,M+1) AM1B11. . . AM1B1K. . . AM1B1M. . . AM1B1,M+1
. . . . . . . . . .
. . . . . . . . . .
AMJ (BJ1. . .BJK. . .BJM. . .BJ,M+1) AMJBJ1. . . AMJBJK. . . AMJBJM. . . AMJBJ,M+1
. . . . . = . . . . .
. . . . . . . . . .
AMN (BN1. . .BNK. . .BNM. . .BN,M+1) AMNBN1. . . AMNBNK. . AMNBNM. . . AMNBN,M+1
. . . . . . . . . .
. . . . . . . . . .
AM,N+1 (BN+1,1. .BN+1,K. .BN+1,M. .BN+1,M+1) AM,N+1BN+1,1. . AM,N+1BN+1,K. . AM,N+1BN+1,M. . AM,N+1BN+1,M+1
Summing the columns we get:
[AM1B11. .+..AMJBJ1+. . . AMNBN1..+..AM,N+1BN+1,1 ] [AM1B1K..+. .AMJBJK. .+. . AMNBNK..+.. AM,N+1BN+1,K]
[AM1B1M..+..AMJBJM+. .+. AMNBNM..+..AM,N+1BN+1,M] [AM1B1,M+1. .+..AMJBJ,M+1..+..AMNBN,M+1 +..AM,N+1BN+1,M+1]
And finally we multiply by the M+1 th column:
AM+1,1 B11. . .B1K. . .B1M. . .B1,M+1
. . . . .
. . . . .
AM+1,J BJ1. . .BJK. . .BJM. . .BJ,M+1
. . . . . =
. . . . .
AM+1,N BN1. . .BNK. . .BNM. . .BN,M+1
. . . . .
. . . . .
AM+1,N+1 BN+1,1 . .BN+1,K . .BN+1,M. .BN+1,M+1
AM+1,1 (B11. . .B1K. . .B1M. . .B1,M+1) AM+1,1B11. AM+1,1B1K. . AM+1,1B1M. . AM+1,1B1,M+1
. . . . . . . . . .
. . . . . . . . . .
AM+1,J (BJ1. . .BJK. . .BJM. . .BJ,M+1) AM+1,JBJ1 .AM+1,J BJK. AM+1,J BJM. . AM+1,J BJ,M+1
. . . . . = . . . . .
. . . . . . . . . .
AM+1,N (BN1. . .BNK. . .BNM. . .BN,M+1) AM+1,NBN1 .AM+1,NBNK. . AM+1,NBNM. . AM+1,N BN,M+1
. . . . . . . . . .
. . . . . . . . . .
AM+1,N+1 (BN+1,1. .BN+1,K. .BN+1,M. .BN+1,M+1) AM+1,N+1BN+1,1. AM+1,N+1BN+1,K. AM+1,N+1BN+1,M. AM+1,N+1BN+1,M+1
[AM+1,1B11.+ .AM+1,JBJ1+. . . AM+1,NBN,1 + AM+1,N+1BN+1,1] [AM+1,1B1K+..AM+1,JBJK.+ . AM+1,NBNK + AM+1,N+1BN+1,K]
AM+1,1B1M + .AM+1,JBJM. +. AM+1,NBNM + AM+1,,N+1BN+1,M] [AM+1,1B1,M+1.+.AM+1,JBJ,M+1.+..AM+1,NBN,M+1 + AM+1,N+1BN+1,M+1]
Transposing them (Stacking the nested arrays on top of each other we get):
Note: When we first half-multiply, we transpose the Pre-multiplier matrix. In order to get our regular matrix multiplication results we have to re-transpose the nested arrays, then sum their columns. Wo could sum the columns then transpose also.
A11 B11. . . A11B1K. . . A11B1M. . . A11B1,M+1
. . . . .
. . . . .
AiJ BJ1. . . AiJ BJK. . . AiJ BJM. . . AiJ BJ,M+1
. . . . .
. . . . .
A1N BN1. . . A1N BNK. . . A1N BNM. . . A1N BN,M+1
. . . . .
. . . . .
A1,N+1BN+1,1. . A1,N+1BN+1,K. . A1,N+1BN+1,M. . A1,N+1BN+1,M+1
Ai1B11. . . Ai1B1K. . . Ai1B1M. . . Ai1B1,M+1
. . . . .
. . . . .
AiJ BJ1. . . AiJBJK. . . AiJBJM. . . AiJBJ,M+1
. . . . .
. . . . .
AiN BN1. . . AiN BNK. . .AiN BNM. . . AiN BN,M+1
. . . . .
. . . . .
Ai,N+1 BN+1,1. . Ai,N+1 BN+1,K. . Ai,N+1 BN+1,M. . Ai,N+1 BN+1,M+1
AM1B11. . . AM1B1K. . . AM1B1M. . . AM1B1,M+1
. . . . .
. . . . .
AMJBJ1. . . AMJBJK. . . AMJBJM. . . AMJBJ,M+1
. . . . .
. . . . .
AMNBN1. . . AMNBNK. . AMNBNM. . . AMNBN,M+1
. . . . .
. . . . .
AM,N+1BN+1,1. . AM,N+1BN+1,K. . AM,N+1BN+1,M. . AM,N+1BN+1,M+1
AM+1,1B11. AM+1,1B1K. . AM+1,1B1M. . AM+1,1B1,M+1
. . . . .
. . . . .
AM+1,JBJ1 .AM+1,J BJK. AM+1,J BJM. . AM+1,J BJ,M+1
. . . . .
. . . . .
AM+1,NBN1 .AM+1,NBNK. . AM+1,NBNM. . AM+1,N BN,M+1
. . . . .
. . . . .
AM+1,N+1BN+1,1. AM+1,N+1BN+1,K. AM+1,N+1BN+1,M. AM+1,N+1BN+1,M+1
Summing the columns in each sub-matrix we get:
[A11B11. .+ .AiJBJ1+. . . A1NBN1 + A1,N+1BN+1,1 ] [A11B1K+. . . .AiJBJK. .+ . A1NBNK + A1,N+1BN+1,K]
A11B1M . + .AiJBJM+. .+. A1NBNM + . A1,N+1BN+1,M] [A11B1,M+1. .+.AijBj,M+1. .+.A1NBN,M+1 + A1,N+1BN+1,M+1]
[Ai1B11. .+ .AiJBJ1+. . . AiNBN1 + Ai,N+1BN+1,1] [ Ai1B1K+. . .AiJBJK. .+ . AiNBNK + Ai,N+1BN+1,K ]
[Ai1B1M . +. AiJBJM . .+..AiNBNM + . Ai,N+1BN+1,M ] [Ai1B1M+1. .+.AijBjM+1 . . +.AiNBNM+1 + Ai,N+1BN+1,M+1]
[AM1B11. .+..AMJBJ1+. . . AMNBN1..+..AM,N+1BN+1,1 ] [AM1B1K..+. .AMJBJK. .+. . AMNBNK..+.. AM,N+1BN+1,K]
[AM1B1M..+..AMJBJM+. .+. AMNBNM..+..AM,N+1BN+1,M] [AM1B1,M+1. .+..AMJBJ,M+1..+..AMNBN,M+1 +..AM,N+1BN+1,M+1]
[AM+1,1B11.+ .AM+1,JBJ1+. . . AM+1,NBN,1 + AM+1,N+1BN+1,1] [AM+1,1B1K+..AM+1,JBJK.+ . AM+1,NBNK + AM+1,N+1BN+1,K]
AM+1,1B1M + .AM+1,JBJM. +. AM+1,NBNM + AM+1,,N+1BN+1,M] [AM+1,1B1,M+1.+.AM+1,JBJ,M+1.+..AM+1,NBN,M+1 + AM+1,N+1BN+1,M+1]
Separating them from each other we get:
A11B11. .+ .A1JBJ1+. . . A1NBN1 + A1,N+1BN+1,1 A11B1K+. . . .A1JBjK. .+ . A1NBNK + A1,N+1BN+1,K . .
. . . . . . . . .
. . . . . . . . .
Ai1B11. .+ .AiJBJ1+. . . AiNBN1 + Ai,N+1BN+1,1 Ai1B1K+. . . .AiJBJK. .+ . AiNBNK + Ai,N+1BN+1,K
. . . . . . . . .
. . . . . . . . .
AM1B11. .+ .AMJBJ1+. . . AMNBN1 + AM,N+1BN+1,1 AM1B1K+. . . .AMJBJK. .+ . AMNBNK + AM,N+1BN+1,K
. . . . . . .
. . . . . . .
AM+1,1B11.+ .AM+1,JBJ1+. . . AM+1,NBN+1 + AM+1,N+1BN+1,1 AM+1,1B1K+. . .AM+1,JBJK.+ . AM+1,NBNK + AM+1,N+1BN+1,K
A11B1M . + .A1JBJM . .+. A1NBNM + . A1,N+1BN+1,M A11B1M+1. .+.A1jBjM+1 . . +.A1NBNM+1 + A1,N+1BN+1,M+1
. . . . . . . .
. . . . . . . .
Ai1B1M . +. AiJBJM . .+..AiNBNM + . Ai,N+1BN+1,M Ai1B1M+1. .+.AijBjM+1 . . +.AiNBNM+1 + Ai,N+1BN+1,M+1
. . . . . . . .
. . . . . . . .
AM1B1M . + .AMJBJM+. .+. AMNBNM + . AM,N+1BN+1,M AM1B1M+1. .+.AMjBjM+1. . .+.AMNBNM+1 + AM,N+1BN+1,M+1
. . . . . . . .
. . . . . . . .
AM+1,1B1M + .AM+1,JBJM. +. AM+1,NBNM + AM+1,,N+1BN+1,M AM+1,1B1M+1.+.AM+1,JBJ,M+1.+..AM+1,NBN,M+1 + AM+1,N+1BN+1,M+1
Which is the solution for regular Matrix Multiplication.
QED
Now we will achieve the same results using the Half-Multiplier Operator (we proved the top portion with diagonals, here we will use the Half-Multiplier Operator:
First we will prove for all matrices sized MxN:
A11 B11. . .B1K. . .B1M
. . . .
. . . .
A1J o BJ1. . .BJK. . .BJM =
. . . .
. . . .
A1N BN1. . .BNK. . .BNM
A11(B11. . .B1K. . .B1M)
. . .
. . .
A1J (BJ1. . .BJK. . .BJM)
. . . =
. . .
A1N (BN1. . .BNK. . .BNM)
A11B11. . .A11B1K. . .A11B1M
. . .
. . .
A1J BJ1. . .A1JBJK. . .A1JBJM
. . .
. . .
A1N BN1. . .A1NBNK. . .A1NBNM
Now for the i th column:
Ai1 B11. . .B1K. . .B1M Ai1(B11. . .B1K. . .B1M)
. . . . . . .
. o . . . . . .
AiJ BJ1. . .BJK. . .BJM = AiJ (BJ1. . .BJK. . .BJM) =
. . . . . . .
. . . . . . .
AiN BN1. . .BNK. . .BNM AiN (BN1. . .BNK. . .BNM)
Ai1B11. . .Ai1B1K. . .Ai1B1M
. . .
. . .
AiJ BJ1. . .AiJBJK. . .AiJBJM
. . .
. . .
AiN BN1. . .AiNBNK. . .AiNBNM
And now for the Mth column:
AM1 B11. . .B1K. . .B1M AM1(B11. . .B1K. . .B1M)
. . . . . . .
. . . . . . .
AMJ o BJ1. . .BJK. . .BJM = AMJ (BJ1. . .BJK. . .BJM) =
. . . . . . .
. . . . . . .
AMN BN1. . .BNK. . .BNM AMN (BN1. . .BNK. . .BNM)
AM1B11. . .AM1B1K. . .AM1B1M
. . .
. . .
AMJ BJ1. . .AMJBJK. . .AMJBJM
. . .
. . .
AMN BN1. . .AMNBNK. . .AMNBNM
Now let’s make these into a Nested array and sum each column of the individual sub-matrices:
A11B11. . .A11B1K. . .A11B1M
. . .
. . .
A1J BJ1. . .A1JBJK. . .A1JBJM
. . .
. . .
A1N BN1. . .A1NBNK. . .A1NBNM
Ai1B11. . .Ai1B1K. . .Ai1B1M
. . .
. . .
AiJ BJ1. . .AiJBJK. . .AiJBJM
. . .
. . .
AiN BN1. . .AiNBNK. . .AiNBNM
AM1B11. . .AM1B1K. . .AM1B1M
. . .
. . .
AMJ BJ1. . .AMJBJK. . .AMJBJM
. . .
. . .
AMN BN1. . .AMNBNK. . .AMNBNM
Summing the columns we get:
A11B11..+.. A1J BJ1..+.. A1N BN1 A11 B1K..+..A1JBJK..+.. A1NBNK A11B1M..+.. A1JBJM..+..A1NBNM
Ai1B11..+.. AiJ BJ1..+.. AiN BN1 Ai1B1K..+..AiJBJK..+.. AiNBNK Ai1B1M..+..AiJBJM..+.. AiNBNM
AM1B11..+.. AMJ BJ1 ..+..AMN BN1 AM1B1K..+.. AMJBJK..+.. AMNBNK AM1B1M..+.. AMJBJM..+..AMNBNM
Which is true for all MxN Matrices, now to prove it is true for all Matrices M+1, N+1
A11 B11. . .B1K. . .B1M. . .B1,M+1
. . . . .
. . . . .
Aij BJ1. . .BJK. . .BJM. . .BJ,M+1
. o . . . . =
. . . . .
A1N BN1. . .BNK. . .BNM. . .BN,M+1
. . . . .
. . . . .
A1,N+1 BN+1,1. .BN+1,K. .BN+1,M. .BN+1,M+1
A11 (B11. . .B1K. . .B1M. . .B1,M+1) A11 B11. . . A11B1K. . . A11B1M. . . A11B1,M+1
. . . . . . . . . .
. . . . . . . . . .
AiJ (BJ1. . .BJK. . .BJM. . .BJ,M+1) AiJ BJ1. . . AiJ BJK. . . AiJ BJM. . . AiJ BJ,M+1
. . . . . = . . . . .
. . . . . . . . . .
A1N (BN1. . .BNK. . .BNM. . .BN,M+1) A1N BN1. . . A1N BNK. . . A1N BNM. . . A1N BN,M+1
. . . . . . . . . .
. . . . . . . . . .
A1,N+1 (BN+1,1. .BN+1,K. .BN+1,M. .BN+1,M+1) A1,N+1BN+1,1. . A1,N+1BN+1,K. . A1,N+1BN+1,M. . A1,N+1BN+1,M+1
The summed columns are equal to:
[A11B11. .+ .AiJBJ1+. . . A1NBN1 + A1,N+1BN+1,1 ] [A11B1K+. . . .AiJBJK. .+ . A1NBNK + A1,N+1BN+1,K]
A11B1M . + .AiJBJM+. .+. A1NBNM + . A1,N+1BN+1,M] [A11B1,M+1. .+.AijBj,M+1. .+.A1NBN,M+1 + A1,N+1BN+1,M+1]
Ai1 B11. . .B1K. . .B1M. . .B1,M+1
. . . . .
. . . . .
AiJ BJ1. . .BJK. . .BJM. . .BJ,M+1
. o . . . . =
. . . . .
AiN BN1. . .BNK. . .BNM. . .BN,M+1
. . . . .
. . . . .
Ai,N+1 BN+1,1 . .BN+1,K . .BN+1,M. .BN+1,M+1+
Ai1 (B11. . .B1K. . .B1M. . .B1,M+1) Ai1B11. . . Ai1B1K. . . Ai1B1M. . . Ai1B1,M+1
. . . . . . . . . .
. . . . . . . . . .
AiJ (BJ1. . .BJK. . .BJM. . .BJ,M+1) AiJ BJ1. . . AiJBJK. . . AiJBJM. . . AiJBJ,M+1
. . . . . = . . . . .
. . . . . . . . . .
AiN (BN1. . .BNK. . .BNM. . .BN,M+1) AiN BN1. . . AiN BNK. . .AiN BNM. . . AiN BN,M+1
. . . . . . . . . .
. . . . . . . . . .
Ai,N+1 (BN+1,1. .BN+1,K. .BN+1,M. .BN+1,M+1) Ai,N+1 BN+1,1. . Ai,N+1 BN+1,K. . Ai,N+1 BN+1,M. . Ai,N+1 BN+1,M+1
Summing the columns we get:
[Ai1B11. .+ .AiJBJ1+. . . AiNBN1 + Ai,N+1BN+1,1] [ Ai1B1K+. . .AiJBJK. .+ . AiNBNK + Ai,N+1BN+1,K ]
[Ai1B1M . +. AiJBJM . .+..AiNBNM + . Ai,N+1BN+1,M ] [Ai1B1,M+1. .+.AijBj,M+1 . +.AiNBN,M+1 + Ai,N+1BN+1,M+1]
AM1 B11. . .B1K. . .B1M. . .B1,M+1
. . . . .
. . . . .
AMJ BJ1. . .BJK. . .BJM. . .BJ,M+1
. o . . . . =
. . . . .
AMN BN1. . .BNK. . .BNM. . .BN,M+1
. . . . .
. . . . .
AM,N+1 BN+1,1 . .BN+1,K . .BN+1,M. .BN+1,M+1
AM1 (B11. . .B1K. . .B1M. . .B1,M+1) AM1B11. . . AM1B1K. . . AM1B1M. . . AM1B1,M+1
. . . . . . . . . .
. . . . . . . . . .
AMJ (BJ1. . .BJK. . .BJM. . .BJ,M+1) AMJBJ1. . . AMJBJK. . . AMJBJM. . . AMJBJ,M+1
. . . . . = . . . . .
. . . . . . . . . .
AMN (BN1. . .BNK. . .BNM. . .BN,M+1) AMNBN1. . . AMNBNK. . AMNBNM. . . AMNBN,M+1
. . . . . . . . . .
. . . . . . . . . .
AM,N+1 (BN+1,1. .BN+1,K. .BN+1,M. .BN+1,M+1) AM,N+1BN+1,1. . AM,N+1BN+1,K. . AM,N+1BN+1,M. . AM,N+1BN+1,M+1
Summing the columns we get:
[AM1B11. .+..AMJBJ1+. . . AMNBN1..+..AM,N+1BN+1,1 ] [AM1B1K..+. .AMJBJK. .+. . AMNBNK..+.. AM,N+1BN+1,K]
[AM1B1M..+..AMJBJM+. .+. AMNBNM..+..AM,N+1BN+1,M] [AM1B1,M+1. .+..AMJBJ,M+1..+..AMNBN,M+1 +..AM,N+1BN+1,M+1]
For the M+1th column:
AM+1,1 B11. . .B1K. . .B1M. . .B1,M+1
. . . . .
. . . . .
AM+1,J BJ1. . .BJK. . .BJM. . .BJ,M+1
. . . . . =
. . . . .
AM+1,N BN1. . .BNK. . .BNM. . .BN,M+1
. . . . .
. . . . .
AM+1,N+1 BN+1,1 . .BN+1,K . .BN+1,M. .BN+1,M+1
AM+1,1 (B11. . .B1K. . .B1M. . .B1,M+1) AM+1,1B11. AM+1,1B1K. . AM+1,1B1M. . AM+1,1B1,M+1
. . . . . . . . . .
. . . . . . . . . .
AM+1,J (BJ1. . .BJK. . .BJM. . .BJ,M+1) AM+1,JBJ1 .AM+1,J BJK. AM+1,J BJM. . AM+1,J BJ,M+1
. . . . . = . . . . .
. . . . . . . . . .
AM+1,N (BN1. . .BNK. . .BNM. . .BN,M+1) AM+1,NBN1 .AM+1,NBNK. . AM+1,NBNM. . AM+1,N BN,M+1
. . . . . . . . . .
. . . . . . . . . .
AM+1,N+1 (BN+1,1. .BN+1,K. .BN+1,M. .BN+1,M+1) AM+1,N+1BN+1,1. AM+1,N+1BN+1,K. AM+1,N+1BN+1,M. AM+1,N+1BN+1,M+1
[AM+1,1B11.+ .AM+1,JBJ1+. . . AM+1,NBN,1 + AM+1,N+1BN+1,1] [AM+1,1B1K+..AM+1,JBJK.+ . AM+1,NBNK + AM+1,N+1BN+1,K]
AM+1,1B1M + .AM+1,JBJM. +. AM+1,NBNM + AM+1,,N+1BN+1,M] [AM+1,1B1,M+1.+.AM+1,JBJ,M+1.+..AM+1,NBN,M+1 + AM+1,N+1BN+1,M+1]
Transposing the Nested Array (Stacking the sub-matrices on top of each other) we get
A11 B11. . . A11B1K. . . A11B1M. . . A11B1,M+1
. . . . .
. . . . .
AiJ BJ1. . . AiJ BJK. . . AiJ BJM. . . AiJ BJ,M+1
. . . . .
. . . . .
A1N BN1. . . A1N BNK. . . A1N BNM. . . A1N BN,M+1
. . . . .
. . . . .
A1,N+1BN+1,1. . A1,N+1BN+1,K. . A1,N+1BN+1,M. . A1,N+1BN+1,M+1
Ai1B11. . . Ai1B1K. . . Ai1B1M. . . Ai1B1,M+1
. . . . .
. . . . .
AiJ BJ1. . . AiJBJK. . . AiJBJM. . . AiJBJ,M+1
. . . . .
. . . . .
AiN BN1. . . AiN BNK. . .AiN BNM. . . AiN BN,M+1
. . . . .
. . . . .
Ai,N+1 BN+1,1. . Ai,N+1 BN+1,K. . Ai,N+1 BN+1,M. . Ai,N+1 BN+1,M+1
AM1B11. . . AM1B1K. . . AM1B1M. . . AM1B1,M+1
. . . . .
. . . . .
AMJBJ1. . . AMJBJK. . . AMJBJM. . . AMJBJ,M+1
. . . . .
. . . . .
AMNBN1. . . AMNBNK. . AMNBNM. . . AMNBN,M+1
. . . . .
. . . . .
AM,N+1BN+1,1. . AM,N+1BN+1,K. . AM,N+1BN+1,M. . AM,N+1BN+1,M+1
AM+1,1B11. AM+1,1B1K. . AM+1,1B1M. . AM+1,1B1,M+1
. . . . .
. . . . .
AM+1,JBJ1 .AM+1,J BJK. AM+1,J BJM. . AM+1,J BJ,M+1
. . . . .
. . . . .
AM+1,NBN1 .AM+1,NBNK. . AM+1,NBNM. . AM+1,N BN,M+1
. . . . .
. . . . .
AM+1,N+1BN+1,1. AM+1,N+1BN+1,K. AM+1,N+1BN+1,M. AM+1,N+1BN+1,M+1
Summing the columns in each sub-matrix we get:
[A11B11. .+ .AiJBJ1+. . . A1NBN1 + A1,N+1BN+1,1 ] [A11B1K+. . . .AiJBJK. .+ . A1NBNK + A1,N+1BN+1,K]
A11B1M . + .AiJBJM+. .+. A1NBNM + . A1,N+1BN+1,M] [A11B1,M+1. .+.AijBj,M+1. .+.A1NBN,M+1 + A1,N+1BN+1,M+1]
[Ai1B11. .+ .AiJBJ1+. . . AiNBN1 + Ai,N+1BN+1,1] [ Ai1B1K+. . .AiJBJK. .+ . AiNBNK + Ai,N+1BN+1,K ]
[Ai1B1M . +. AiJBJM . .+..AiNBNM + . Ai,N+1BN+1,M ] [Ai1B1M+1. .+.AijBjM+1 . . +.AiNBNM+1 + Ai,N+1BN+1,M+1]
[AM1B11. .+..AMJBJ1+. . . AMNBN1..+..AM,N+1BN+1,1 ] [AM1B1K..+. .AMJBJK. .+. . AMNBNK..+.. AM,N+1BN+1,K]
[AM1B1M..+..AMJBJM+. .+. AMNBNM..+..AM,N+1BN+1,M] [AM1B1,M+1. .+..AMJBJ,M+1..+..AMNBN,M+1 +..AM,N+1BN+1,M+1]
[AM+1,1B11.+ .AM+1,JBJ1+. . . AM+1,NBN,1 + AM+1,N+1BN+1,1] [AM+1,1B1K+..AM+1,JBJK.+ . AM+1,NBNK + AM+1,N+1BN+1,K]
AM+1,1B1M + .AM+1,JBJM. +. AM+1,NBNM + AM+1,,N+1BN+1,M] [AM+1,1B1,M+1.+.AM+1,JBJ,M+1.+..AM+1,NBN,M+1 + AM+1,N+1BN+1,M+1]
Separating each new column we get:
A11B11. .+ .A1JBJ1+. . . A1NBN1 + A1,N+1BN+1,1 A11B1K+. . . .A1JBjK. .+ . A1NBNK + A1,N+1BN+1,K . .
. . . . . . . . .
. . . . . . . . .
Ai1B11. .+ .AiJBJ1+. . . AiNBN1 + Ai,N+1BN+1,1 Ai1B1K+. . . .AiJBJK. .+ . AiNBNK + Ai,N+1BN+1,K
. . . . . . . . .
. . . . . . . . .
AM1B11. .+ .AMJBJ1+. . . AMNBN1 + AM,N+1BN+1,1 AM1B1K+. . . .AMJBJK. .+ . AMNBNK + AM,N+1BN+1,K
. . . . . . .
. . . . . . .
AM+1,1B11.+ .AM+1,JBJ1+. . . AM+1,NBN+1 + AM+1,N+1BN+1,1 AM+1,1B1K+. . .AM+1,JBJK.+ . AM+1,NBNK + AM+1,N+1BN+1,K
A11B1M . + .A1JBJM . .+. A1NBNM + . A1,N+1BN+1,M A11B1M+1. .+.A1jBjM+1 . . +.A1NBNM+1 + A1,N+1BN+1,M+1
. . . . . . . .
. . . . . . . .
Ai1B1M . +. AiJBJM . .+..AiNBNM + . Ai,N+1BN+1,M Ai1B1M+1. .+.AijBjM+1 . . +.AiNBNM+1 + Ai,N+1BN+1,M+1
. . . . . . . .
. . . . . . . .
AM1B1M . + .AMJBJM+. .+. AMNBNM + . AM,N+1BN+1,M AM1B1M+1. .+.AMjBjM+1. . .+.AMNBNM+1 + AM,N+1BN+1,M+1
. . . . . . . .
. . . . . . . .
AM+1,1B1M + .AM+1,JBJM. +. AM+1,NBNM + AM+1,,N+1BN+1,M AM+1,1B1M+1.+.AM+1,JBJ,M+1.+..AM+1,NBN,M+1 + AM+1,N+1BN+1,M+1
QED
This is the same answer we get as when we multiply the matrices the regular way.
We have now proved that the Half-Multiplier Operator when half multiplied into sub-matrices, transposing to a column matrix, summing the columns gives the same solution as regular Matrix Multiplication AijBjk = Cik.
THE ROW PRODUCT OF A MATRIX
OR
THE CROSS PRODUCT OF A MATRIX
When we took the half-multiplied sub-matrices and summed the columns, the solution was the same as matrix multiplication. Let’s look at what happens if we take the half-multiplied sub-matrices and sum their rows instead of their columns.
R([A]Tij o [B]JK = ([B]JK[1]k1) o [A]TiJ
A11 A21 A31 B11 B12 B13
A12 A22 A32 o B21 B22 B23 =
A13 A23 A33 B31 B32 B33
A14 A24 A34 B41 B42 B43
A11 B11 B12 B13 A21 B11 B12 B13 A31 B11 B12 B13
A12 o B21 B22 B23 A22 o B21 B22 B23 A32 o B21 B22 B23
A13 B31 B32 B33 A23 B31 B32 B33 A33 B31 B32 B33
A14 B41 B42 B43 A24 B41 B42 B43 A34 B41 B42 B43
[pic]
[pic]
[pic]
This is equal to:
[pic]
Here we are back to a nested array. We do not need to transpose this array, since after the row sums are done, the matrix is already in the form that is needed for the proper solution. I think that this the cross product of a matrix, (which is up until now undefined) and wait for better mathematicians than I to confirm or deny this hypothesis.
Let’s go ahead and sum the columns and see what we get.
A11(B11+B12+B13) A21(B11+B12+B13) A31(B11+B12+B13)
A12(B21+B22+B23) A22(B21+B22+B23) A32(B21+B22+B23) =
A13(B31+B32+B33) A23(B31+B32+B33) A33(B31+B32+B33)
A14(B41+B42+B43) A24(B41+B42+B43) A34(B41+B42+B43)
A11(B1 A21(B1 A31(B1
A12(B2 A22(B2 A32(B2
A13(B3 A23(B3 A33(B3
A14(B4 A24(B4 A34(B4
But this is equal to:
(B1 A11 A21 A31
(B2 o A12 A22 A32
(B3 A13 A23 A33
(B4 A14 A24 A34
So this is equivalent to the Itempage matrix in the inventory/accounting system. The neat thing about this operator is that it has it’s counterparts in regular mathematics. That is, the above expression can be calculated by the matric equation:
([B]JK[1]k1) o [A]TiJ
B11 B12 B13 1 A11 A21 A31
B21 B22 B23 1 o A12 A22 A32 =
B31 B32 B33 1 A13 A23 A33
B41 B42 B43 1 A14 A24 A34
This is an important property. It says that if you transpose [A] then [B] becomes commutative with matrix [A]. Let’s look at a simple example for illustration. Bill has just come into some money. He wants to buy a boat for $10,000, two cars, one for $20,000 and the other for $15,000 , a TV/VCR for $500 and some new clothes, two suits for $250 each. Lets make a matrix out of what Bill wants to buy, and how much of each.
The one boat, one car, one TV/VCR and two suits become: [1 1 1 1 2]
And the one for price is: 10,000
20,000
15,000
500
250
To find the total, we multiply the two together in the regular way:
[1 1 1 1 2] 10,000
20,000
15,000 = $46,000
500
250
Now let’s transpose [A] and see what happens:
1 10,000 10,000 1
1 20,000 20,000 1
1 o 15,000 = 15,000 o 1
1 500 500 1
2 250 250 2
One boat equals $10,000 no matter how you look at it, the first car is still $20,000 multiplied both ways, etc.
This works for [A] and [B] if [A] is a row or column matrix, whether [A] is a N x M matrix or just a row or column matrix. But if [B] and [A] both become a matrix of more than just a row or column, then multiplying [B]o[A]T gives the transpose of [A][B]. (This is proved somewhat a little later on in this section of my paper). i.e. Suppose:
D E F
[B] = G H I and [A]= [A B C]
J K L
[A B C] D E F
G H I = [AD+BG+CJ AE+BH+CK AF+BI+CL]
J K L
D E F
[B] = G H I
J K L
D E F 1 A A(D+E+F) AD AE AF
R([B]o[A] = G H I 1 o B = B(G+H+I) = BG + BH + BI
J K L 1 C C(J+K+L) CJ CK CL
Sum the columns and we have the same answer as in [A][B] above. Remove the inner brackets from the nested array above and the solutions are equal. But this is also a valid solution all in it’s own. Even each individual column is a valid solution (or valid sub-solution) also. A good example of this you will see in the section on Itempage accounting.
An interesting effect occurs when
B11 + B12 + B13 = 1
B21 + B22 + B23 = 1
B31 + B32 + B33 = 1
B41 + B42 + B43 = 1
The equation reduces to:
R([A]iJ[B]JK = ([B]JK[1]K1) o [A]TiJ = [A]TiJ
I am going to call the above inventory where [B]JK[1]K1 = [1]J1 a closed system (or closed inventory) as opposed to an open inventory where [B]JK[1]K1 = [IP]J1.
NOW LET’S PROVE THIS FOR ALL NxM MATRICES:
B11. . .B1K. . .B1M 1
. . . .
. . . .
BJ1. . .BJK. . .BJM 1 =
+. . . .
. . . .
BN1. . .BNK. . .BNM 1
NM N,1
B11. .+. .B1K. .+. .B1M
. . . . .. . . . . .
. . . . .. . . . . .
BJ1. .+. .BJK. .+. .BJM.
. . . . .. . . . . .
. . . . .. . . . . .
BN1. .+. .BNK. .+. .BNM. NM
Then ([B]JK[1]K1) o [A]TiJ = (this will give us i sub-matrices)
B11. .+. .B1K. .+. .B1M. A11 . . .Ai1 . . .AM1
. . . . .. . . . . .
. . . . .. . . . . .
BJ1. .+. .BJK. .+. .BJM. o A1J . . .AiJ . . .AMJ.
. . . . .. . . . . .
. . . . .. . . . . .
BN1. .+. .BNK. .+. .BNM. A1N . . .AiN . . . AMN
A11 . . .A1J . . .A1N
. . .
[A]MN = Ai1 . . .AiJ . . .AiN
. . .
AM1 . . .AMJ . . .AMN
LET (B1 = B11. .+. .B1K. .+. .B1M.
LET (B2 = BJ1. .+. .BJK. .+. .BJM.
LET (B3 = BN1. .+. .BNK. .+. .BNM.
And R([B] o [A]T= ((B1 o [A]T )+ ((B2 o [A]T) + ((B3 o[A]T)
((B1 o [A]T )=
B11. . . .0. . . .0 A11 . . .Ai1 . . .AM1
. . . . .. . . . .
. . . . .. . . . .
0. . . .B1K. . . 0 o A1J . . .AiJ . . .AMJ.
. . . . .. . . . .
. . . . .. . . . .
0. . . .0. . . .B1M. A1N . . .AiN . . . AMN
B11. . . .0. . . .0 A11 . . .Ai1 . . .AM1
. . . . .. . . .
. . . . .. . . .
0. . . .B1J. . . 0 o A1J . . .AiJ . . .AMJ.
. . . . .. . . .
. . . . .. . . .
0. . . .0. . . .BN1 A1N . . .AiN . . . AMN
B11(A11 . . .Ai1 . . .AM1) B11A11 .. B11Ai1 .. B11AM1
. . . . . . .
. . . . . . .
BJ1(A1J . . .AiJ . . .AMJ) = BJ1A1J .. BJ1AiJ .. BJ1AMJ
. . . . . . .
. . . . . . .
BN1(A1N . . .AiN . . . AMN) BN1A1N .. BN1AiN .. BN1AMN
((B2 o [A]T)= (MAKE COLUMN 2 OF [A] INTO A DIAGONAL MATRIX AND MULTIPLY x [A], WE GET:
B1K(A11 . . .Ai1 . . .AM1) B1KA11 .. B1KAi1 .. B1KAM1
. . . . . . .
. . . . . . .
BJK(A1J . . .AiJ . . .AMJ) = BJKA1J .. BJKAiJ .. BJKAMJ
. . . . . . .
. . . . . . .
BMK(A1N . . .AiN . . . AMN) BNKA1N .. BNKAiN .. BNKAMN
((B3 o [A]T)= (MAKE COLUMN 3 OF [B] INTO A DIAGONAL MATRIX AND MULTIPLY x [A], WE GET:
B1M(A11 . . .Ai1 . . .AM1) B1MA11 .. B1MAi1 .. B1MAM1
. . . . . . .
. . . . . . .
BJM(A1J . . .AiJ . . .AMJ) = BJMA1J .. BJMAiJ .. BJMAMJ
. . . . . . .
. . . . . . .
BNM(A1N . . .AiN . . . AMN) BNMA1N .. BNMAiN .. BNMAMN
Now that we’ve computed the three sub-matrices, we must add them together. I don’t have a lot of room on this computer to add the whole thing and keep the computations short and clear. Let’s first add the first three sums that occupy A11:
B11A11 + B1KA11 + B1MA11 = A11(B11 + B1K + B1M ).
These check, so let’s add them all together. Do you see anything that happens when we add the sub-matrices? That is the subject of my next proof.
A11B11.+.A11B1K.+.A11B1M. Ai1B11.+.Ai1B1K.+.Ai1B1M AM1B11.+. AM1B1K.+. AM1B1M
A1JBJ1.+.A1JBJK.+.A1JBJM. AiJBJ1.+.AiJBJK.+.AiJBJM AMJBJ1.+. AMJBJK.+. AMJBJM
A1NBN1.+.A1NBNK.+.A1NBNM AiNBN1.+.AiNBNK.+.AiNBNM AMNBN1.+. AMNBNK.+. AMNBNM
Which is equal to:
A11(B11.+.B1K.+.B1M) Ai1(B11.+.B1K.+.B1M) AM1(B11.+.B1K.+.B1M)
A1J(BJ1..+..BJK..+..BJM) AiJ(BJ1..+..BJK..+..BJM) AMJ(BJ1..+..BJK..+..BJM)
A1N(BN1..+..BNK..+..BNM) AiN(BN1..+..BNK..+..BNM) AMN(BN1..+..BNK..+..BNM)
QED
This is the proof for all NxM Matrices. Now to prove it works for all Matrices N+1,M+1.
Now let’s prove this:
B11. . .B1K. . .B1M. . .B1,M+1 1
. . . . .
. . . . .
BJ1. . .BJK. . .BJM. . .BJ,M1 1
+. . . . . =
. . . . .
BN1. . .BNK. . .BNM. . .BN,M+1 1
. . . . .
. . . . .
BN+1,1 . .BN+1,K . .BN+1,M. .BN+1,M+1 1
B11. .+. .B1K. .+. .B1M. . +. .B1,M+1
. . . . .. . . . . . . . .
. . . . .. . . . . . . . .
BJ1. .+. .BJK. .+. .BJM. . +. .BJ,M1
. . . . .. . . . . . . . .
. . . . .. . . . . . . . .
BN1. .+. .BNK. .+. .BNM. . +. .BN,M+1
. . . . .. . . . . . . . .
. . . . .. . . . . . . . .
BN+1,1.+. .BN+1,K .+. .BN+1,M. +. .BN+1,M+1
Then ([B]JK[1]K1) o [A]TiJ =
B11. .+. .B1K. .+. .B1M. . +. .B1,M+1 A11 . . .Ai1 . . .AM1 . . . AM+1,1
. . . . .. . . . . . . . .
. . . . .. . . . . . . . .
BJ1. .+. .BJK. .+. .BJM. . +. .BJ,M1 A1J . . .AiJ . . .AMJ. . . . AM+1,J
. . . . .. . . . . . . . . o
. . . . .. . . . . . . . .
BN1. .+. .BNK. .+. .BNM. . +. .BN,M+1 A1N . . .AiN . . . AMN . . . AM+1,N
. . . . .. . . . . . . . .
. . . . .. . . . . . . . .
BN+1,1.+. .BN+1,K .+. .BN+1,M. +. .BN+1,M+1 A1,N+1. . .Ai,N+1. . .AM,N+1. . . AM+1,N+1
A11 . . .A1J . . .A1N . . .A1,N+1
. . . .
Ai1 . . .AiJ . . .AiN . . .Ai,N+1
[A]MN = . . . .
AM1 . . .AMJ . . .AMN . . .AM,N+1
. . . .
AM+1,1. . .AM+1,J. . .AM+1,N. . .AM+1,N+1
LET (B1 = B11. .+. .B1K. .+. .B1M. . +. .B1,M+1
LET (B2 = BJ1. .+. .BJK. .+. .BJM. . +. .BJ,M1
LET (B3 = BN1. .+. .BNK. .+. .BNM. . +. .BN,M+1
LET (B4 = BN+1,1.+. .BN+1,K .+. .BN+1,M. +. .BN+1,M+1
And R([B] o [A]T= ((B1 o [A]T )+ ((B2 o [A]T) + ((B3 o[A]T) + ((B4 o [A]T)
((B1 o [A]T )=
B11. . . .0. . . .0. . . .0 A11 . . .Ai1 . . .AM1 . . . AM+1,1
. . . . .. . . . . . . . .
. . . . .. . . . . . . . .
0. . . .B1K. . . 0. . . 0 A1J . . .AiJ . . .AMJ. . . . AM+1,J
. . . . .. . . . . . . . . o
. . . . .. . . . . . . . .
0. . . .0. . . .B1M. . . .0 A1N . . .AiN . . . AMN . . . AM+1,N
. . . . .. . . . . . . . .
. . . . .. . . . . . . . .
0. . .0 . . . 0. . .B1,M+1 A1,N+1. . .Ai,N+1. . .AM,N+1. . . AM+1,N+1
N+1,M+1 N+1,M+1
B11. . . .0. . . .0. . . .0 A11 . . .Ai1 . . .AM1 . . . AM+1,1
. . . . .. . . . . . . . .
. . . . .. . . . . . . . .
0. . . .B1J. . . 0. . . 0 A1J . . .AiJ . . .AMJ. . . . AM+1,J
. . . . .. . . . . . . . . o
. . . . .. . . . . . . . .
0. . . .0. . . .BN1. . . .0 A1N . . .AiN . . . AMN . . . AM+1,N
. . . . .. . . . . . . . .
. . . . .. . . . . . . . .
0. . .0 . . . 0. . .BN+1,1 A1,N+1. . .Ai,N+1. . .AM,N+1. . . AM+1,N+1
N+1,M+1 N+1,M+1
B11(A11 . . .Ai1 . . .AM1 . . . AM+1,1 ) B11A11 .. B11Ai1 .. B11AM1 .. B11AM+1,1
. . . . . . . . .
. . . . . . . . .
BJ1(A1J . . .AiJ . . .AMJ. . . . AM+1,J) BJ1A1J .. BJ1AiJ .. BJ1AMJ .. BJ1AM+1,J
. . . . . . . . .
. . . . . = . . . .
BN1(A1N . . .AiN . . . AMN . . . AM+1,N) BN1A1N .. BN1AiN .. BN1AMN .. BN1AM+1,N
. . . . . . . . .
. . . . . . . . .
BN+1,1(A1,N+1. . .Ai,N+1. . .AM,N+1. . . AM+1,N+1) BN+1,1A1,N+1 .. BN+1,1Ai,N+1 .. BN+1,1AM,N+1 .. BN+1,1 AM+1,N+1)
((B2 o [A]T)= (MAKE COLUMN 2 OF [A] INTO A DIAGONAL MATRIX AND MULTIPLY x [A], WE GET:
B1K(A11 . . .Ai1 . . .AM1 . . . AM+1,1 ) B1KA11 .. B1KAi1 .. B1KAM1 .. B1KAM+1,1
. . . . . . . . .
. . . . . . . . .
BJK(A1J . . .AiJ . . .AMJ. . . . AM+1,J) BJKA1J .. BJKAiJ .. BJKAMJ .. BJKAM+1,J
. . . . . . . . .
. . . . . = . . . .
BMK(A1N . . .AiN . . . AMN . . . AM+1,N) BNKA1N .. BNKAiN .. BNKAMN .. BNKAM+1,N
. . . . . . . . .
. . . . . . . . .
BN+1,K(A1,N+1. . .Ai,N+1. . .AM,N+1. . . AM+1,N+1) BN+1,KA1,N+1 .. BN+1,KAi,N+1 .. BN+1,KAM,N+1 .. BN+1,K AM+1,N+1)
((B3 o [A]T)= (MAKE COLUMN 3 OF [B] INTO A DIAGONAL MATRIX AND MULTIPLY x [A], WE GET:
B1M(A11 . . .Ai1 . . .AM1 . . . AM+1,1 ) B1MA11 .. B1MAi1 .. B1MAM1 .. B1MAM+1,1
. . . . . . . . .
. . . . . . . . .
BJM(A1J . . .AiJ . . .AMJ. . . . AM+1,J) BJMA1J .. BJMAiJ .. BJMAMJ .. BJMAM+1,J
. . . . . . . . .
. . . . . = . . . .
BNM(A1N . . .AiN . . . AMN . . . AM+1,N) BNMA1N .. BNMAiN .. BNMAMN .. BNMAM+1,N
. . . . . . . . .
. . . . . . . . .
BN+1,M(A1,N+1. . .Ai,N+1. . .AM,N+1. . . AM+1,N+1) BN+1,MA1,N+1 .. BN+1,MAi,N+1 .. BN+1,MAM,N+1 .. BN+1,M AM+1,N+1)
((B4 o [A]T)= (MAKE COLUMN 4 OF [B] INTO A DIAGONAL MATRIX AND MULTIPLY x [A], WE GET
B1,M+1(A11 . . .Ai1 . . .AM1 . . . AM+1,1 ) B1,M+1A11 .. B1,M+1Ai1 .. B1,M+1AM1 .. B1,M+1AM+1,1
. . . . . . . . .
. . . . . . . . .
BJ,M+1(A1J . . .AiJ . . .AMJ. . . . AM+1,J) BJ,M+1A1J .. BJ,M+1AiJ .. BJ,M+1AMJ .. BJ,M+1AM+1,J
. . . . . . . . .
. . . . . = . . . .
BN,M+1(A1N . . .AiN . . . AMN . . . AM+1,N) BN,M+1A1N .. BN,M+1AiN .. BN,M+1AMN .. BN,M+1AM+1,N
. . . . . . . . .
. . . . . . . . .
BN+1,M+1(A1,N+1. . .Ai,N+1. . .AM,N+1. . . AM+1,N+1) BN+1,M+1A1,N+1 .. BN+1,M+1Ai,N+1 .. BN+1,M+1AM,N+1 .. BN+1,M+1AM+1,N+1)
Now that we’ve computed the four sub-matrices, we must add them together. I don’t have a lot of room on this computer to add the whole thing and keep the computations short and clear. Let’s first add the first four sums that occupy A11:
B11A11 + B1KA11 + B1MA11 +, B1,M+1 A11 = A11(B11 + B1K + B1M + B1,M+1).
These check, so let’s add them all together. Do you see anything that happens when we add the sub-matrices? That is the subject of my next proof.
A11B11.+.A11B1K.+.A11B1M.+.A11B1,M+ Ai1B11.+.Ai1B1K.+.Ai1B1M.+.Ai1B1,M+
A1JBJ1.+.A1JBJK.+.A1JBJM.+.A1JBJ,M1 AiJBJ1.+.AiJBJK.+.AiJBJM.+.AiJBJ,M1
A1NBN1.+.A1NBNK.+.A1NBNM.+.A1NBN,M+1 AiNBN1.+.AiNBNK.+.AiNBNM.+.AiNBN,M+1
A1,N+1BN+1,1.+.A1,N+1BN+1,K.+.A1,N+1BN+1,M.+.A1,N+1BN+1,M+1 Ai,N+1BN+1,1.+.Ai,N+1BN+1,K.+. Ai,N+1BN+1,M.+.Ai,N+1BN+1,M+1
AM1B11.+. AM1B1K.+. AM1B1M.+. AM1B1,M+ AM+1,1B11.+.AM+1,1B1K.+.AM+1,1B1M.+.AM+1,1B1,M+
AMJBJ1.+. AMJBJK.+. AMJBJM.+. AMJBJ,M1 AM+1,JBJ1.+.AM+1,JBJK.+.AM+1,JBJM.+.AM+1,JBJ,M1
AMNBN1.+. AMNBNK.+. AMNBNM.+. AMNBN,M+1 AM+1,NBN1.+.AM+1,NBNK.+.AM+1,NBNM.+.AM+1,NBN,M+1
AM,N+1BN+1,1.+.AM,N+1BN+1,K.+.AM,N+1BN+1,M.+.AM,N+1BN+1,M+1 AM+1,N+1BN+1,1.+.AM+1,N+1BN+1,K.+.AM+1,N+1BN+1,M.+.AM+1,N+1BN+1,M+1
Which is equal to:
A11(B11.+.B1K.+.B1M.+.B1,M+) Ai1(B11.+.B1K.+.B1M.+.B1,M+)
A1J(BJ1..+..BJK..+..BJM..+..BJ,M1 ) AiJ(BJ1..+..BJK..+..BJM..+..BJ,M1)
A1N(BN1..+..BNK..+..BNM..+..BN,M+1) AiN(BN1..+..BNK..+..BNM..+..BN,M+1)
A1,N+1(BN+1,1.+..BN+1,K.+..BN+1,M.+..BN+1,M+1) Ai,N+1(BN+1,1.+..BN+1,K.+..BN+1,M.+..BN+1,M+1)
AM1(B11.+.B1K.+.B1M.+.B1,M+)) AM+1,1(B11.+.B1K.+.B1M.+.B1,M+) )
AMJ(BJ1..+..BJK..+..BJM..+..BJ,M1) AM+1,J(BJ1..+..BJK..+..BJM..+..BJ,M1)
AMN(BN1..+..BNK..+..BNM..+..BN,M+1) AM+1,N(BN1..+..BNK..+..BNM..+..BN,M+1)
AM,N+1(BN+1,1.+..BN+1,K.+..BN+1,M.+..BN+1,M+1) AM+1,N+1(BN+1,1.+..BN+1,K.+..BN+1,M.+..BN+1,M+1)
QED
SUMMARY OF 3 EQUATIONS DESCRIBED
R([A]TiJ o [B]JK = ([B]JK[1]K1) o [A]TiJ FOR OPEN SYSTEMS
R([A]iJ[B]JK = ([B]JK[1]K1) o [A]TiJ = [I]jj[A]TiJ = [1]J1o[A]TiJ = [A]TiJ FOR CLOSED SYSTEMS
AND
R([B] o [A]T= ((B1 o [A]T )+ ((B2 o [A]T) + ((B3 o[A]T) + ((B4 o [A]T =
[IP1]+[IP2]+[IP3]+[IP4]
FOR EACH ITEM IN THE DATABASE MATRIX [B].
THE MATRIC PRODUCT OF A MATRIX:
M([A]TiJ o [B]JK = ([A]TiJ[1]J1) o [B]JK
A11 A21 A31 B11 B12 B13
A12 A22 A32 o B21 B22 B23 =
A13 A23 A33 B31 B32 B33
A14 A24 A34 B41 B42 B43
A11 B11 B12 B13 A21 B11 B12 B13 A31 B11 B12 B13
A12 o B21 B22 B23 A22 o B21 B22 B23 A32 o B21 B22 B23
A13 B31 B32 B33 A23 B31 B32 B33 A33 B31 B32 B33
A14 B41 B42 B43 A24 B41 B42 B43 A34 B41 B42 B43
[pic]
[pic]
[pic]
This is equal to:
[pic]
Here we are back to a nested array. We do not need to transpose this array, since after the row sums are done, the matrix is already in the form that is needed for the proper solution. To conserve on space, I am going to omit the first step here and just go ahead and collect terms, the complex stuff will be gotten to later.
Let’s go ahead and sum the matrices and see what we get:
B11(A11+A21+A31) B12(A11+A21+A31) B13(A11+A21+A31)
B21(A12+A22+A32) B22(A12+A22+A32) B23(A12+A22+A32) = M([A]T o [B]=
B31(A13+A23+A33) B32(A13+A23+A33) B33(A13+A23+A33)
B41(A14+A24+A34) B42(A14+A24+A34) B43(A14+A24+A34)
B11(A1 B12(A1 B13(A1
B21(A2 B22(A2 B23(A2
B31(A3 B32(A3 B33(A3
B41(A4 B42(A4 B43(A4
But this is equal to:
(A1 B11 B21 B31
(A2 o B12 B22 B32
(A3 B13 B23 B33
(A4 B14 B24 B34
So this is equivalent to the Accountpage matrix in the inventory/accounting system. The neat thing about this operator is that it also has it’s counterparts in regular mathematics. That is, the above expression can be calculated by the matric equation:
diag([A]TiJ[1]J1)[B]JK
This is also an important property. It says that if you transpose [A] and sum the rows of [A] and o multiply across [B], the solution is the same as if we took the half-multiplied matrices and added them all together. [A] is now commutative with [B].
For an open system, suppose:
(A11+A21+A31)= 1
(A12+A22+A32)= 1 , Then M([A]TiJ o [B]JK= [I]JJ o [B]JK = [B]JK
(A13+A23+A33)= 1
(A14+A24+A34)= 1
PROOF FOR ALL MxN MATRICES
A11. .. . Ai1. .. . AM1 1
. . . . .. . . . . .
. . . . .. . . . . .
A1J. .. . AiJ. .. . AMJ 1
. . . . .. . . . . . =
. . . . .. . . . . .
A1N. .. . AiN. .. . AMN 1
A11. .+. . Ai1. .+. . AM1
. . . . .. . . . . .
. . . . .. . . . . .
A1J. .+. . AiJ. .+. . AMJ
. . . . .. . . . . .
. . . . .. . . . . .
A1N. .+. . AiN. .+. . AMN
Then [A]TiJ[1]i1 o [B]JK =
A11. . . Ai1. . . AM1 1 B11. . . B1K. . . B1M
. . . . .. . . . . . . . .
. . . . .. . . . . . . . .
A1J. . . AiJ. . . AMJ 1 o BJ1. . . BJK. . . BJM
. . . . .. . . . . . . . .
. . . . .. . . . . . . . .
A1N. . . AiN. . . AMN 1 BN1. . . BNK. . . BNM
LET (A1T = A11. .+. . Ai1. .+. . AM1
LET (A2T = A1J. .+. . AiJ. .+. . AMJ
LET (A3T = A1N. .+. . AiN. .+. . AMN
(A1T. . . 0. . . 0 B11. . . B1K. . . B1M
. . . . .. . . . . . . .
. . . . .. . . . . . . .
0. . . (A2T. . . 0. BJ1. . . BJK. . . BJM =
. . . . .. . . . . . . .
. . . . .. . . . . . . .
0. . . 0. . . (A3T BN1. . . BNK. . . BNM
B11(A1T. . . B1K(A1T. . . B1M(A1T
. . . .
. . . .
BJ1(A2T. . . BJK(A2T. . . BJM(A2T =
. . . .
. . . .
BN1(A3T. . . BNK(A3T. . . BNM(A3T
This matrix is also equal to the sum of each individual half-multiplied sub-matrices.
M([A]To[B] = (([A1]T o [B]) + (([A2]T o [B]) + (([A3]T o [B])
(([A1]T o [B])
A11. . . 0. . . 0 B11. . . B1K. . . B1M. . . B1,M+1
. . . . .. . . . . . . . .
. . . . .. . . . . . . . .
0. . . A1J. . . 0 BJ1. . . BJK. . . BJM.. . .BJ,M+1
. . . . .. . . . . . . . . =
. . . . .. . . . . . . . .
0. . . 0. . . A1M. BN1. . . BNK. . . BNM. . . BN,M+1
A11(B11 . . .B1K . . .B1M) A11B11 .+. A11B1K .+. A11B1M
. . . . . . .
. . . . . . .
A1J(BJ1 . . .BJK . . .BJM) = A1JBJ1 .+. A1JBJK .+. A1JBJM
. . . . . . .
. . . . . . .
A1M(BN1 . . .BNK . . . BNM) A1MBN1 .+. A1MBNK .+. A1MBNM
(([A2]T o [B])= (MAKE COLUMN 2 OF [A] INTO A DIAGONAL MATRIX AND MULTIPLY x [B], WE GET:
Ai1(B11 . . .B1K . . .B1M) Ai1B11 .+. Ai1B1K .+. Ai1B1M
. . . . . . .
. . . . . . .
AiJ(BJ1 . . .BJK . . .BJM) = AiJBJ1 .+. AiJBJK .+. AiJBJM
. . . . . . .
. . . . . . .
AiN(BN1 . . .BNK . . . BNM) AiNBN1 .+. AiNBNK .+. AiNBNM
(([A3]T o [B])= (MAKE COLUMN 3 OF [A] INTO A DIAGONAL MATRIX AND MULTIPLY x [B], WE GET:
AM1(B11 . . .B1K . . .B1M) AM1B11 .+. AM1B1K .+. AM1B1M
. . . . . . .
. . . . . . .
AMJ(BJ1 . . .BJK . . .BJM) = AMJBJ1 .+. AMJBJK .+. AMJBJM
. . . . . . .
. . . . . . .
AMN(BN1 . . .BNK . . . BNM AMNBN1 .+. AMNBNi .+. AMNBNM
Now that we’ve computed the four sub-matrices, we must add them together. I don’t have a lot of room on this computer to add the whole thing and keep the computations short and clear. Let’s first add the first four sums that occupy A11:
A11B11 + B11Ai1 + AM1B11 = B11(A11 + Ai1 + AM1 ). = B11(A1T
These check, so let’s add them all together.
B11(A1T. . . B1K(A1T. . . B1M(A1T
. . . .
. . . .
BJ1(A2T. . . BJK(A2T. . . BJM(A2T =
. . . .
. . . .
BN1(A3T. . . BNK(A3T. . . BNM(A3T
B11(A11.+.Ai1.+.AM1) B1K(A11.+.Ai1.+.AM1) B1M(A11.+.Ai1.+.AM1)
BJ1(A1J.+.AiJ.+.AMJ) BJK(A1J.+.AiJ.+.AMJ) BJM(A1J.+.AiJ.+.AMJ)
BN1(A1N.+.AiN.+.AMN) BNK(A1N.+.AiN.+.AMN) BNM(A1N.+.AiN.+.AMN)
QED
This proves the operator works for all NxM matrices. Now to prove, by induction, that it works for all M+1, N+1 matrices. We will solve for a Nx1,Mx1 Matrix:
A11. .. . Ai1. .. . AM1. .. .AM+1,1 1
. . . . .. . . . . . . . . .
. . . . .. . . . . . . . . .
A1J. .. . AiJ. .. . AMJ. . . AM+1,J. 1
. . . . .. . . . . . . . . . =
. . . . .. . . . . . . . . .
A1N. .. . AiN. .. . AMN. . . AM+1,N 1
. . . . .. . . . . . . . . .
. . . . .. . . . . . . . . .
A1,N+1 . Ai,N+1.. . AM,N+1. . AM+1,N+1 1
A11. .+. . Ai1. .+. . AM1. .+. .AM+1,1
. . . . .. . . . . . . . . .
. . . . .. . . . . . . . .
A1J. .+. . AiJ. .+. . AMJ. . +. AM+1,J.
. . . . .. . . . . . . . .
. . . . .. . . . . . . . .
A1N. .+. . AiN. .+. . AMN. . +. AM+1,N
. . . . .. . . . . . . . .
. . . . .. . . . . . . . .
A1,N+1 . Ai,N+1.+. . AM,N+1. +. .AM+1,N+1
Then ([A]TiJ[1]i1) o [B]JK =
A11. . . Ai1. . . AM1. . .AM+1,1 1 B11. . . B1K. . . B1M. . . B1,M+1
. . . . .. . . . . . . . . . . . . .
. . . . .. . . . . . . . . . . . . .
A1J. . . AiJ. . . AMJ. . . AM+1,J. 1 BJ1. . . BJK. . . BJM.. . .BJ,M+1
. . . . .. . . . . . . . . . o . . . .
. . . . .. . . . . . . . . . . . . .
A1N. . . AiN. . . AMN. . . AM+1,N 1 BN1. . . BNK. . . BNM. . . BN,M+1
. . . . .. . . . . . . . . . . . . .
. . . . .. . . . . . . . . . . . . .
A1,N+1. . Ai,N+1. . AM,N+1. .AM+1,N+1 1 BN+1,1. . BN+1,K . .BN+1,M . .BN+1,M+1
N+1,M+1
LET (A1T = A11. .+. . Ai1. .+. . AM1. .+. .AM+1,1
LET (A2T = A1J. .+. . AiJ. .+. . AMJ. . +. AM+1,J.
LET (A3T = A1N. .+. . AiN. .+. . AMN. . +. AM+1,N
LET (A4T = A1,N+1 .+ Ai,N+1.+. . AM,N+1. +. .AM+1,N+1
(A1T. . . 0. . . 0. . .0 B11. . . B1K. . . B1M. . . B1,M+1
. . . . .. . . . . . . . . . . .
. . . . .. . . . . . . . . . . .
0. . . (A2T. . . 0. . . 0. BJ1. . . BJK. . . BJM.. . .BJ,M+1
. . . . .. . . . . . . . . . . . =
. . . . .. . . . . . . . . . . .
0. . . 0. . . (A3T. . . 0 BN1. . . BNK. . . BNM. . . BN,M+1
. . . . .. . . . . . . . . . . .
. . . . .. . . . . . . . . . . .
0. . . 0. . 0. . (A4T BN+1,1. . BN+1,K . .BN+1,M . .BN+1,M+1
B11(A1T. . . B1K(A1T. . . B1M(A1T. . . B1,M+1(A1T
. . . .
. . . .
BJ1(A2T. . . BJK(A2T. . . BJM(A2T.. . .BJ,M+1(A2T
. . . . =
. . . .
BN1(A3T. . . BNK(A3T. . . BNM(A3T. . . BN,M+1(A3T
. . . .
. . . .
BN+1,1(A4T. . BN+1,K(A4T . .BN+1,M(A4T . .BN+1,M+1(A4T
This matrix is also equal to the sum of each individual half-multiplied sub-matrix
M([A]To[B] = (([A1]T o [B]) + (([A2]T o [B]) + (([A3]T o [B]) + (([A4]T o [B])
(([A1]T o [B])
A11. . . 0. . . 0. . .0 B11. . . B1K. . . B1M. . . B1,M+1
. . . . .. . . . . . . . . . . .
. . . . .. . . . . . . . . . . .
0. . . A1J. . . 0. . . 0. BJ1. . . BJK. . . BJM.. . .BJ,M+1
. . . . .. . . . . . . . . . . . =
. . . . .. . . . . . . . . . . .
0. . . 0. . . A1M. . . 0 BN1. . . BNK. . . BNM. . . BN,M+1
. . . . .. . . . . . . . . . . .
. . . . .. . . . . . . . . . . .
0. . . 0. . 0. . A1,M+1 BN+1,1. . BN+1,K . .BN+1,M . .BN+1,M+1
A11(B11 . . .B1K . . .B1M . . . B1,M+1 ) A11B11 .+. A11B1K .+. A11B1M .+. A11B1,M+1
. . . . . . . . .
. . . . . . . . .
A1J(BJ1 . . .BJK . . .BJM. . . . BJ,M+1) A1JBJ1 .+. A1JBJK .+. A1JBJM .+. A1JBJ,M+1
. . . . . . . . .
. . . . . = . . . .
A1M(BN1 . . .BNK . . . BNM . . . BN,M+1) A1MBN1 .+. A1MBNK .+. A1MBNM .+. A1MBN,M+1
. . . . . . . . .
. . . . . . . . .
A1,M+1(BN+1,1. . .BN+1,K. . .BN+1,M. . . BN+1,M+1) A1,M+1BN+1,1 .+. A1,M+1BN+1,K .+. A1,M+1BN+1,M .+. A1,M+1BN+1,M+1)
(([A2]T o [B])= (MAKE COLUMN 2 OF [A] INTO A DIAGONAL MATRIX AND MULTIPLY x [B], WE GET:
Ai1(B11 . . .B1K . . .B1M . . . B1,M+1 ) Ai1B11 .+. Ai1B1K .+. Ai1B1M .+. Ai1B1,M+1
. . . . . . . . .
. . . . . . . . .
AiJ(BJ1 . . .BJK . . .BJM. . . . BJ,M+1) AiJBJ1 .+. AiJBJK .+. AiJBJM .+. AiJBJ,M+1
. . . . . . . . .
. . . . . = . . . .
AiN(BN1 . . .BNK . . . BNM . . . BN,M+1) AiNBN1 .+. AiNBNK .+. AiNBNM .+. AiNBN,M+1
. . . . . . . . .
. . . . . . . . .
Ai,N+1(BN+1,1. . .BN+1,K. . .BN+1,M. . . BN+1,M+1) Ai,N+1BN+1,1 .+. Ai,N+1BN+1,K .+. Ai,N+1BN+1,M .+. Ai,N+1BN+1,M+1)
(([A3]T o [B])= (MAKE COLUMN 3 OF [A] INTO A DIAGONAL MATRIX AND MULTIPLY x [B], WE GET:
AM1(B11 . . .B1K . . .B1M . . . B1,M+1 ) AM1B11 .+. AM1B1K .+. AM1B1M .+. AM1B1,M+1
. . . . . . . . .
. . . . . . . . .
AMJ(BJ1 . . .BJK . . .BJM. . . . BJ,M+1) AMJBJ1 .+. AMJBJK .+. AMJBJM .+. AMJBJ,M+1
. . . . . . . . .
. . . . . = . . . .
AMN(BN1 . . .BNK . . . BNM . . . BN,M+1) AMNBN1 .+. AMNBNi .+. AMNBNM .+. AMNBN,M+1
. . . . . . . . .
. . . . . . . . .
AM,N+1(BN+1,1. . .BN+1,i. . .BN+1,M. . . BN+1,M+1) AM,N+1BN+1,1 .+. AM,N+1BN+1,K .+. AM,N+1BN+1,M .+. AM,N+1BN+1,M+1)
(([A4]T o [B])= (MAKE COLUMN 4 OF [A] INTO A DIAGONAL MATRIX AND MULTIPLY x [B], WE GET
AM+1,1(B11 . . .B1K . . .B1M . . . B1,M+1 ) AM+1,1B11 .+. AM+1,1B1K .+. AM+1,1B1M .+. AM+1,1B1,M+1
. . . . . . . . .
. . . . . . . . .
AM+1,J(BJ1 . . .BJK . . .BJM. . . . BJ,M+1) AM+1,JBJ1 .+. AM+1,JBJK .+. AM+1,JBJM .+. AM+1,JBJ,M+1
. . . . . . . . .
. . . . . = . . . .
AM+1,N(BN1 . . .BNK . . . BNM . . . BN,M+1) AM+1,NBN1 .+. AM+1,NBNK .+. AM+1,NBNM .+. AM+1,NBN,M+1
. . . . . . . . .
. . . . . . . . .
AM+1,N+1(BN+1,1. . .BN+1,K. . .BN+1,M. . .BN+1,M+1) AM+1,N+1BN+1,1 .+. AM+1,N+1BN+1,K .+. AM+1,N+1BN+1,M .+. AM+1,N+1BN+1,M+1)
Now that we’ve computed the four sub-matrices, we must add them together. I don’t have a lot of room on this computer to add the whole thing and keep the computations short and clear. Let’s first add the first four sums that occupy A11:
A11B11 + B11Ai1 + AM1B11 +, AM+1,1 B11 = B11(A11 + Ai1 + AM1 + AM+1,1). = B11(A1T
These check, so let’s add them all together.
B11(A1T. . . B1K(A1T. . . B1M(A1T. . . B1,M+1(A1T
. . . .
. . . .
BJ1(A2T. . . BJK(A2T. . . BJM(A2T.. . .BJ,M+1(A2T
. . . . =
. . . .
BN1(A3T. . . BNK(A3T. . . BNM(A3T. . . BN,M+1(A3T
. . . .
. . . .
BN+1,1(A4T. . BN+1,K(A4T . .BN+1,M(A4T . .BN+1,M+1(A4T
B11(A11.+.Ai1.+.AM1.+.AM+1,1) B1K(A11.+.Ai1.+.AM1.+.AM+1,1)
BJ1(A1J.+.AiJ.+.AMJ.+.AM+1,J.) BJK(A1J.+.AiJ.+.AMJ.+.AM+1,J.)
BN1(A1N.+.AiN.+.AMN.+.AM+1,N) BNK(A1N.+.AiN.+.AMN.+.AM+1,N)
BN+1,1(A1,N+1.+.Ai,N+1.+.AM,N+1.+.AM+1,N+1) BN+1,K(A1,N+1.+.Ai,N+1.+.AM,N+1.+.AM+1,N+1)
B1M(A11.+.Ai1.+.AM1.+.AM+1,1) B1,M+1(A11.+.Ai1.+.AM1.+.AM+1,1 )
BJM(A1J.+.AiJ.+.AMJ.+.AM+1,J.) BJ,M+1(A1J.+.AiJ.+.AMJ.+.AM+1,J.)
BNM(A1N.+.AiN.+.AMN.+.AM+1,N) BN,M+1(A1N.+.AiN.+.AMN.+.AM+1,N)
BN+1,M(A1,N+1.+.Ai,N+1.+.AM,N+1.+.AM+1,N+1) BN+1,M+1(A1,N+1.+.Ai,N+1.+.AM,N+1.+.AM+1,N+1)
QED
We have now proved by induction that this operator works for Matrices of any size or dimension.
SUMMARY OF 3 EQUATIONS DESCRIBED
M([A]TiJ o [B]JK = ([A]TiJ[1]i1) o [B]JK FOR OPEN SYSTEMS
M([A]TiJ o [B]JK = ([A]TiJ[1]i1) o [B]JK = [I]JJ[B]JK = [1]J1 o [B]JK = [B]JK FOR CLOSED SYSTEMS
AND
M([A]To[B] = (([A1]T o [B]) + (([A2]T o [B]) + (([A3]T o [B]) + (([A4]T o [B])
[AP1]+[AP2]+[AP3]+[AP4]
FOR EACH ITEM IN THE SPREADSHEET MATRIX [A].
THE TRANSPOSE COMMUTIVITY OF THE HALF-MULTIPLIER OPERATOR
The proofs are long and confusing. I’m not even sure what I’m proving since we have no solutions in mathematics to compare them too, so I just tried to prove that the whole matrix is a sum of it’s sub-matrices. In view of this, I think I’ll just do a micro-proof here. If the professional mathematicians holler, they can substitute I, J, N and M for matrix [A] and J, K, M and N for matrix [B], or they can prove it for themselves.
(
([A]TiJ o [B]JK =
B11 B12 B13 A11 A21 A31
B21 B22 B23 o A12 A22 A32 =
B31 B32 B33 A13 A23 A33
B41 B42 B43 A14 A24 A34
B11A11 B11A21 B11A31 B12A11 B12A21 B12A31 B13A11 B13A21 B13A31
B21A12 B21A22 B21A32 B22A12 B22A22 B22A32 B23A12 B23A22 B23A32
B31A13 B31A23 B31A33 B32A13 B32A23 B32A33 B33A13 B33A23 B33A33
B41A14 B41A24 B41A34 B42A14 B42A24 B42A34 B43A14 B43A24 B43A34
Now we will transpose the nested array to put the array into a form we can mathematically use. Unlike the proof for the Half-multiplier Operator at the beginning of this section, when I transpose this time, we will have a single stacked matrix to contend with. We will sum using a database matrix. i.e. (I have separated the 3 sub-matrices just to make it easier to see what I’m doing, we do not need to do it when we are programming this on a computer. The [DB] matrix takes care of this).
1 1 1 1 0 0 0 0 0 0 0 0 B11A11 B11A21 B11A31
0 0 0 0 1 1 1 1 0 0 0 0 B21A12 B21A22 B21A32
0 0 0 0 0 0 0 0 1 1 1 1 B31A13 B31A23 B31A33
B41A14 B41A24 B41A34
B12A11 B12A21 B12A31
B22A12 B22A22 B22A32 =
B32A13 B32A23 B32A33
B42A14 B42A24 B42A34
B13A11 B13A21 B13A31
B23A12 B23A22 B23A32
B33A13 B33A23 B33A33
B43A14 B43A24 B43A34
B11A11+B21A12+B31A13+B41A14 B11A21+B21A22+B31A23+B41A24 B11A31+B21A32+B31A33+B41A34
B12A11+B22A12+B32A13+B42A14 B12A21+B22A22+B32A23+B42A24 B12A31+B22A32+B32A33+B42A34
B13A11+B23A12+B33A13+B43A14 B13A21+B23A22+B33A23+B43A24 B13A31+B23A32+B33A33+B43A34
There is no need to re-transpose back into the nested array because this is the final form we want. When we re-transpose, we can take two steps of action. The outer brackets can be discarded and we are left with the sub-matrices as a solution or as working operators. Or, we may leave it in the matrix-matrix form, manipulate the inner brackets to choose the size (columns only) of the solution matrices, then remove the outer bracket for the engineered solution. Let’s multiply [A]x[B] in the normal way:
[pic]
[pic]
[pic]
[pic]
Which is the transpose of the value computed above. It is also equal to [B]T[A]T i.e.
[pic]=
[pic]
micro-QED
NESTED ARRAYS
The process of half-multiplying two matrices seems to naturally produce a nested array.
These nested arrays have interesting properties, but properties I am unable to prove. They are free wheeling and we can do almost anything we want with them, as long as we do not break the rules of mathematics. When we half-multiply and obtain the i sub-matrices, if we transpose the nested array, the elements do not change their order, just the matrices. Also, the inner set of brackets around the sub-matrices disappear and we can treat this transposed nested array as a regular matrix for purposes of multiplication, addition and subtraction. We can then re-transpose the final matrix if we wish, or leave it in the same transposed form. Another interesting and perhaps a very important property is that after half-multiplication, we can adjust the brackets around the sub-matrices to any size we wish, though normally they would all be of the same dimensions, this need not always be the case (see the examples on statistics, Section 2.3 and 2.4). Let’s look at a simple example where we take two sets of data, X and Y.
We will wish to multiply each X by it’s corresponding Y value.
First though, I must show how to transpose a nested array.
TRANSPOSING NESTED ARRAYS
I do not know how to prove this yet, the transpose of a Nested Array comes from the work done in Statistics and Quantum Chemistry. Suppose we have 4 arrays:
[pic]
The Nested Array becomes:
[pic]
Or we can write it in terms of the sub-matrices:
[pic]
When we transpose this we get:
[pic]
[pic]
Which becomes:
[pic]
The arrays transpose, but the individual elements inside the arrays do not.
Now let’s look at a simple example where we take two sets of data, X and Y.
We will wish to multiply each X by it’s corresponding Y value.
[pic]
Half-multiply by [1]4,1 to create the nested array
[pic] o[pic]=
[pic]
Now we remove the inner bracket and create three 2x4 sub-matrices:
[pic]
Now we transpose this nested array:
T
[pic] =
[pic]
Note the inner brackets are gone, the elements themselves are not transposed. To multiply by adding, we need to take the log of the elements of the array:
[pic]
To take the log in MathCad, we must name the above matrix. I can't quite do it here because if you name a symbolic array, all the elements turn black. But if these were numbers and I called the transposed array A, then
( (
LOGA = log(A)vectorize (vectorize is the button on the matrix palette with the X * Y with an arrow over it. Now let's multiply the X's and Y's together, take the anti-log and obtain our solution.
[pic]
[pic]
(
To convert this to regular math, we do the following: XY= 10LOGAxTWO1 vectorize, where TWO1 represents a matrix with two rows and one column. Generally, the way I use these variables throughout this paper, every ONE2 (one row, 2 columns) are filled with ones only. Thus NINE1 has 9 rows and one column all filled with ones. Suppose one of the X or Y values is zero, how do we take the log of zero which equals minus infinity? All we need is a number that is very close to zero, not zero itself. For purposes of this paper, I define zero by default as 10EEX-100. This gives a log of -100, close enough to zero for the problems presented here and for most problems involving the universe. So I can show the program, I will call A=[1 1]. This program works if all the elements are numbers rather than symbols.
[pic]
[pic]
[pic]
[pic]
The solution we get back is:
[pic]
Now we re-transpose the matrix:
[pic]
And this is our statistical matrix with the corresponding X's and Y's multiplied.
ONTO MULTIPLICATION
Let’s now check on a micro-proof concerning onto multiplication of matrices (element by corresponding element, rather than the sum of row x column). This is illegal according to modern mathematics. I claim 3 x 2 is always equal to 6, no matter how you multiply the two numbers together. This is part of the connection between our numbers we are familiar with and matrices. Whatever we can do with single numbers, we can do thousands of times in one operation with matrices. I wish a professional mathematician would prove to me that 3 x 2 ( 6 when done in the following manner. Let’s take two general matrices:
[pic] [pic]
and multiply them such that we get JxA, MxD, PxG, etc. We first need to take the logs of the two matrices, add them to multiply the numbers, and take the anti-log to return the numbers in familiar form:
[pic] [pic]
[pic]
[pic] (here we take the anti-log 10x . We must do this by hand, MathCad can’t take the anti-log of a symbolic.)
=[pic]
Or putting in numbers:
[pic] [pic]
[pic]
[pic]
[pic]
Suppose one or more of the elements in M1 and M2 are zero's? We proceed as follows, letting zero, by default, equal 10EEX-100:
[pic] [pic]
[pic]
[pic]
[pic]
MathCad cannot remove a diagonal from a matrix, it can take a column matrix and make a diagonal, but not vice versa. In many applications, especially since the half-multiplier’s matrix equivalent is multiplication by a diagonal matrix, we need to be able to remove the diagonal from a matrix and make a separate matrix out of it. The following is how I do it with MathCad. The HP-48G will remove the diagonal from a matrix as a column matrix and allows us to re-make it into a diagonal matrix.
MATHCAD +6:
Suppose we have a matrix and we wish to square every element in the matrix and add them to get the grand sum of squares. We take the matrix, transpose it and multiply it by itself. i.e. [A]2 = [A]T[A]. The sum of the squares in each column lie in the diagonal. We need to separate the diagonal from the rest of the matrix, and then add the elements together to get a single sum. We proceed as follows:
[pic] [pic]
[pic]
[pic]
CHECK:
[pic]
[pic]
[pic]
[pic]
We need to remove the diagonal, MathCad will not do this for us. This is the best way: Write a template 4x4 identity matrix, but write it as the log (this helps take care of the problem of the log of zero). Take the log of ASQ, add the two and take the anti-log. This will return the diagonal of ASQ.
[pic] Here I take the log of each individual element in [A]2 .
[pic] This is the log of the [I]4,4 matrix.
[pic]
[pic] Here I take the anti-log of each element to obtain the final solution.
[pic]
Now we need to sum the values. Define:
[pic]
[pic]
[pic]
[pic]
HP-48G PROGRAM: MATRIX, ENTER VALUES, ((,(A STO (this stores the matrix in A)
RCL A
(
(( TRN
(
SWAP
x
MTH,MATR,NXT,(DIAG [30 174 446 846]
4
(
DIAG(
RCL ONE4 MUST ENTER ONE4 AS A COLUMN MATRIX FIRST, THEN TRANSPOSE TO GET ROW MATRIX, OR
SWAP MATH WILL NOT WORK.
x
RCL ONE4
(( TRN
(
x 1496
We can also accomplish the same thing by multiplying element by element then compute the grand sum of the matrix.
[pic] [pic]
[pic]
[pic]
[pic]
I should have done the math this way with statistics, bur did not really see how simple it was until now (8-12-97). But the way I did it with statistics follows the more acceptable rules of math. Feel free to find the grand sum of squares in this manner rather than the ways I solved them later in this book if you so wish.
Also, when we transpose a nested array, we may put it in the form of a diagonal matrix
instead of a column or row matrix. I’ll not get into that here, but will field some examples in
quantum chemistry, Gaussian reduction and statistics.
Below are some simple computations of Nested arrays from a request from Dr. Monroy in Juarez, Mexico.
Dear Dr. Monroy:
These are some of the properties of nested arrays as apply to the operator, since your problem uses square matrices I used examples as square matrices although they may be MxN just as easily. Hope they might be of help to you. The three arrays are defined as C, D and E and their nested form is defined as A. Imagine they are stacked on top of each other (like crackers) in 3-D but the only way to display them is in 2-D form. MathCad cannot perform computations on nested arrays, so in order to utilize them we ignore the brackets and just remember that they are there.
[pic] [pic] [pic]
First we combine the separate arrays into one single array A. They are still nested, but since computers can’t handle the math we must “remember” that there are brackets around them and work on them from this premise. This is the premise behind the mathematics in the section on Statistics later on in this book.
[pic] [pic]
TO ADD EACH THREE MATRICES SEPARATELY:
[pic]
CHECK: [pic]
LET'S ADD MATRIX 1 + MATRIX THREE AND IGNORE MATRIX 2:
[pic] [pic]
[pic] CHECK: [pic]
TO ONTO MULTIPLY EACH SEPARATE MATRIX BY ANOTHER:
[pic]
[pic]
[pic]
[pic]
TO MULTIPLY THE THREE MATRICES THE REGULAR WAY:
[pic]
[pic]
[pic]
CHECK:
[pic][pic][pic]
TO HALF MULTIPLY THE THREE MATRICES AND GET ALL THE SUB-MATRICES:
[pic] [pic] [pic]
[pic][pic][pic]
[pic][pic][pic]
[pic] [pic][pic]
[pic][pic][pic]
[pic][pic][pic]
[pic] [pic]
[pic]
Suppose the matrices in the nested array are not all the same dimensions, this is how we can multiply them and halfmultiply them.
[pic] [pic] [pic]
We make the nested array. Remember, MathCad cannot compute with nested arrays, so we just remove the brackets and “remember” that they are there. To make the nested array conformable, we must add zero’s where necessary to complete the array and legalize it mathematically.
[pic]
Or diagonalizing L we get:
[pic]
[pic]
L=12,4 M=12,12
WE CAN MULTIPLY THIS IN A CONDENSED MANNER ONLY ONE WAY, BECAUSE KTxK WOULD EQUAL A 5x5 MATRIX, WHICH IS OUT OF BOUNDS (INDICES DON'T CONFORM).
[pic]
[pic] [pic] [pic]
[pic]
[pic][pic]
[pic][pic]
[pic][pic]
BUT WE CAN MULTIPLY BOTH WAYS BY COMPUTING WITH THE DIAGONALIZED NESTED ARRAYS IN THIS MANNER:
[pic]
[pic]
Which gives us the product multiplying both ways.
Now let’s half multiply the nested array. Remember, computers cannot o multiply yet, this is a new discovery, so we must multiply by it’s equivalent diagonal matrix operator to get the solution.
[pic] [pic]
[pic]
[pic]
This multiplication gives us the transpose of the half-multiplication. Note the nested arrays are separated better than in the other multiplication.
[pic]
[pic]
CHECK:
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
THE THEORY OF INFORMATION
AND
THE UNIFIED FIELD EQUATION
FOR EVERY NUMERICAL OPERATION THAT EXISTS, THERE EXISTS A MATRIX OPERATION THAT WILL REPEAT THAT OPERATION ixj TIMES FOR ADDITION AND SUBTRACTION, OR (ixj)(jxk) TIMES FOR MULTIPLICATION.
Suppose we have a thousand numbers that we have to add to another thousand numbers.
The form is A+B=C. There exists a matrix [A]ij + [B]ij = [C]ij that will add A + B ixj times, all at the same time under one mathematical operation.
Suppose we have A x B ± C = 0. This has four matrix solutions, two additive and two multiplicative.
OPERATION TAKEN ONE AT A TIME MATRIX EQUIVALENT
A x B = C [A]ij[B]jk = [C]ik MULTIPLICATIVE (indices change)
A x B = -C [A]ij[B]jk = -[C]ik MULTIPLICATIVE (indices change)
A x B + C = 0 [A]Tij o [B]jk = -[C]jk ADDITIVE (indices don’t change)
A x B - C = 0 [A]Tij o [B]jk = [C]jk ADDITIVE (indices don’t change)
There are actually an infinite number of combinations. Suppose we have A/B = C, the matrix equivalents would be [A]ij[B]-1jk = [C]ik and/or [A]Tij o [B]-1jk = [C]jk; for A x B x C = D
the matric equivalents are [A]ij[B]jk[C]kL = [D]IL or [A]Tij o [B]jk [C]kL = [D]iL or
([A]Tij o [B]jk)T o [C]kL = [D]iL . Whatever the problem, there is a matrix solution that will solve everything all in one operation instead of perhaps thousands of separate operations conducted one at a time.
There is one equation above that seems to be really important, so I am going to concentrate mainly on this one. It is the equation [A]Tij o [B]jk = [C]jk . This looked so similar to Einstein’s First Field Equation in his theory of gravitation that I just had to explore it further. It was this exploration that leads me to believe that this is the Unified Field Equation.
[A]Tij o [B]jk = [C]jk describes a universal Inventory/Accounting system. If we multiply this equation through by a constant c, the equation is converted to a physics or statistics field equation. All we need to do is define c, [B] and [A]. Let’s look at some examples.
CLASSICAL PHYSICS
C([A]Tij o [B]jk )= [C]jk
Let c = m, and [B]jk = [a]jk (acceleration) and [C]jk = Fjk, then
[C]jk = m([A]Tij o [B]jk ) but [A]ij = [I]jj and [B]jk = [a]jk and [C]jk = Fjk, so
Fjk = m([I]jj o [a]jk) for m = to a constant.
Fjk = m([a]jk)
If [A]ij ( [I]jj , then we have
Fjk = m([A]Tij o [a]jk )
NOTE: We may be able to express this backwards, I don’t know. I mean, if [A]ij = [a]ij, [B]jk =[I]jk and [C]ij = Fij.
Then Fij = m([a]ij) and Fij = m([a]Tij o [B]jk )
Since there is a connection between Accounting/Inventories and Statistics..there ought to be a connection between Physics and Statistics…lets check out the Between association:
Fjk = m([A]Tij o [a]jk )
m([1]([A]Tij o [a]jk)2 = [C]2jk
If [A]ij = [I] this equation reduces to:
m([1]1,j[a]jk)2 = [C]2jk
But since this is statistics, we might have to divide by N, not sure..then the equation connecting statistics to physics becomes:
m/N([1]1,j[a]jk)2 = [C]2jk
But this is the equation of statistics for regular matrix multiplication. Can find for Astro-physics in the same way.
EINSTEIN’S FIELD EQUATION FOR GRAVITATION
C([A]Tij o [B]jk) = [C]jk
Let [C]jk = [G]jk , [B]jk = [T]ij and [A]ij = [I]jj and c = 8((.
Then we have:
8((([A]Tij o[T]jk) = [G]jk but [A]ij = [I]jj so we have
8(([T]jk = [G]jk which is Einstein’s First Field Equation. This is the equation as we now understand it, but the true equation might be 8((([A]Tij o[T]jk) = [G]jk.
Let’s see if we can derive a wave equation for gravity from this equation.
Let’s multiply Einstein’s equation through by the wave function (T(.
(T8(([T]jk( = (T[G]jk( but G is an energy term and , 8(( is a constant. The equation becomes:
8(( (T[T]jk( = [G]jk (T (
and therefore
[G]jk = 8(( (T[T]jk( 8(( (T([A]Tij o [T]jk)(
or [G]jk =
(T ( (T (
Gjk is now solvable.
Astrophysicists say Einstein’s equation is not complete. Knowing nothing about Astrophysics, I assume the mass term is included in the stress-metric tensor Tjk. Since Einstein’s equation resembles Newton’s equation, I suggest that the mass term may need to be removed from Tjk and hollow dotted into it. i.e.
[G]jk = 8((([m]Tij o [T]jk) or
[G]jk = 8(( [m]Tij o ([A]Tij o [T]jk)
or maybe: [G]jk = 8((m([A]Tij o [T]jk) I’m not sure which is correct.
Which has the form of Newton’s Second Equation.
This is all I can do with the math for this kind of mathematics. I never could get a handle on the Geometrodynamics equations of Space-time. I have no idea how to solve it or use it. Are there any astrophysicists out there who can make any sense out of it?
Einsteins Equation and Statistics:
Gjk = 8(( ([A]Tij o [T]jk )
8(( ([1]([A]Tij o [T]jk)2 = [C]2jk
If [A]ij = [I] this equation reduces to:
8(( ([1]1,j[T]jk)2 = [C]2jk
But since this is statistics, we might have to divide by N, not sure..then the equation connecting statistics to physics becomes:
8((/N([1]1,j[a]jk)2 = [C]2jk
But this is the equation of statistics for regular matrix multiplication.
STATISTICS
There was, until now, no field equation describing the field of statistics.
Going back to the Unified Field Equation:
C([A]Tij o [B]jk )= [C]jk and for the simple case, letting [B]jk = [I]jk and c = 1/N, the equation becomes: (this means there is no database matrix to multiply to, we manipulate only the data directly on matrix [A].)
1/N([A]Tij o [B]jk )= [C]jk =1/N([A]Tij o [I]jk)= 1/N([A]Tij) = [C]ij
Since [A][I] is a straight multiplication problem, we do not need to transpose it. But we do need to sum the columns of [A], so the basic statistical equation becomes:
1/N([1]1,I[A]ij) = [C]1j
But to make this statistical, we must square the above expression so that it becomes:
1/N([1]1,i[A]ij)2 = [C]21j Where [C]21j = CCT. (this gives us a one by one matrix as a solution).
Suppose [B]jk is not = to [I]jk. Then we have our basic complex statistical field equation, the simplest form of which is the Analysis of Variance. This is a straight multiplication so [A] does not need to be transposed. (We could transpose [A] first, and then sum the columns in a second operation). Therefore the standard complex statistical equation becomes:
1/N([1]1,i[A]ij[B]jk)2 = [C]21,k where [C]2 = CCT
But we also need to subtract the correction factor(s), so the total statistical field equation becomes:
1/N([1]1,i[A]ij[B]jk)2 - correction factor(s) = C21,k - correction factor(s).
Note, there are no (’s anywhere, since this approach to statistics is derived from Accounting and Inventories rather than random variables.
So there are only 3, perhaps 4 operators from which statistics (perhaps all of statistics) may be computed:
1/2(([A]Tij o [B]jk = i Cjk matrices (Half-Multiplier mode)
( where i = #rows in un-transposed matrix)
C( 1/N([A]Tij o [B]jk ) = 1/N[C]ik regular matrix multiplication.
R( 1/N([A]Tij o [B]jk ) = 1/N([B]jk[1]k,1o[A]Tij)
M( 1/N([A]Tij o [B]jk ) = 1/N([A]Tij[1]i,1o[B]jk)
Note: These equations also hold for Physics, Astrophysics and Inventory/Accounting systems also.
The mathematics also suggests that we may be able to create a quantum statistics (or perhaps quantum accounting/inventories?). Or more fundamentally, a quantum probability. What I am going to derive I will give an example of at the end of the section of statistics. I do not know how to interpret the solution, that will be left to the professional statisticians.
Suppose we have completed the experimental data on an experiment and put into the matrix form [A]MN. Now [A]MN is not a square matrix, so we must make a square out of it before we can do any sort of quantum statistical analysis on it. Squaring the data changes the matrix [A] from a statistical matrix to a probability matrix. To keep track of each of the sums of the squares for each column in the matrix, we must square it such that we end up with an NxN matrix.
[A]TMN[A]MN = [A]NM[A]MN = [A]2NN
Then we do the following:
1/N[DB1]iN [A]2NN[DB1]TiN 1/N[DB1]iN [A]2NN[DB1]Ni 1/N[DB1]iN [A]2NN[DB1]Ni
= = =
[DB1]iN[DB1]TiN [DB1]iN[DB1]Ni [DB1]2ii
1/N[C]2ii
[DB1]2ii
Now statistics and probability are directly associated with each other. How, for instance, can we compute 5! ? That is 5x4x3x2x1=120? Easy!
Anti-log(Log ([5 4 3 2 1]) 1 )
1
1 = 120
1
1
Of course, we don’t have to just use [DB1], we can use any of the database matrices that are pertinent to the analysis under scrutiny.
DERIVATION OF QUANTUM MECHANICS FROM EINSTEIN’S EQUATION
Since we have already derived Einstein’s equation above, I’ll start with it here.
8(( (T[T]jk(
[G]jk =
(T (
Since we are not dealing with stars, but with atoms, the stress-metric tensor Tjk may be simplified by calling it the Hamiltonian of the system. We are also not interested in the gravitational attraction of an atom, but with the total energy of its electro-magnetic field, or the total energy E. The constant at this time is unknown and will revert to c. Substituting the new values into Einstein’s equation (if it is a true field equation, it should be true for all of physics) we get:
c (Tjk[H]jk(jk c(T([A]Tij o [H]jk)(
Ejk or [E]jk =
(T ( (T (
Physics and astrophysics field equations are already known, so it should be simple for people proficient in these fields to check the correctness of these derivations. The field equations for statistics and Accounting/Inventory systems are brand new and have not been discovered before.
INVENTORY/ACCOUNTING SYSTEM:
[A]Tij o [B]jk = [C]jk =
[INV]ij[DB]jk = [SOL]ik This is the sum total of everything bought, sold, manufactured, etc.
i[A]Tij o [B]jk = i[C]jk =
i[INV]Tij o [DB]jk = i[AP]jk This keeps and individual item accounting of everything bought, sold, manufactured, etc. The superscript i just tells us which row in the un-transposed matrix we have hollow-dotted onto the [DB] matrix.
j[B]jk o [A]Tij = j[C]jk =
j[DB]jk o [INV]Tij = j[IP]jk This takes an individual item from the database matrix (a column) and multiplies it across the inventory(or accounting page if we wish) giving us a slice of the total pie, so to speak. It tells us how much of that item was used, by whom or what machine or smokestack, and shows how it is distributed throughout the total inventory. This one is best used taken one item at a time. If we want two items at a time, we have to add them together and compute a total for both. It would be like saying 35 apples and oranges instead of 15 apples and 20 oranges.
[A]Tij o [B]jk = i, [C]jk sub-matrices .
[INV]Tij o [DB]jk = i, [SOL]jk sub-matrices
This is the pure half-multiplier operation giving all the accountpage matrices, i of them, in a single operation.
PHILOSOPHY
There are more things under these heavens and earth than are dreamed of in your philosophy, Horatio. W. Shakespeare
In the history of mathematics, the origins of math follow two seemingly separate paths. The first was when man kept track of the rising of the sun to predict the seasons and later to predict eclipses. Careful records were kept, even of the positions of the planets. This system of accounting was discarded with the advent of Newton’s three equations of motion when modern physics was developed. The second system of mathematics was developed to keep track of what merchants and kings owned, what was owed them, and how much they owed to others. This system developed into modern accounting systems. Two seemingly different mathematical systems. With the half-multiplier operator, the two divergent systems are again brought together under the dominion of a single equation. Mathematics has now returned to it’s roots. A chapter in the history of math is now closed and a new chapter is open for exploration. The half-multiplier operator seems to say that the universe prefers matrices rather than partial differential equations and the calculus professional mathematicians and physicists have developed to explain the nature of matter and energy. This is the missing operator, the missing link in math, so to speak, that has been right under the noses of everyone but has escaped notice until now. They use the math all the time but have never consolidated it into one comprehensive set of proofs that I try to develop in this book. My proofs may be sloppy, it has been 19 years since I solved a math problem more difficult than a simple proportion, but they seem to hold for the whole of mathematics. Mathematicians say that 3x2=6 in most cases, but not in all cases. I prove here that 3x2=6 in all cases, even though mathematicians some 200 to 300 years ago said that what I am about to show you is impossible to do. Because everyone believed this, they missed out on discovering one of the most beautiful mathematical operator’s of all. This is basically all that this book is about, that the operation of multiplication holds in all cases i.e., 5x2=10 no matter how you multiply it. With this simple premise, we can do all our mathematical operations at the same time instead of doing just one calculation at a time.
There seems to be a direct relationship between ordinary addition, subtraction, multiplication, division and matrices. Suppose we want to add the numbers
24 We can re-write this as
-11
9
44
-6
[pic][pic][pic] = [pic]
Transposing and multiplying sums the numbers for us.
Suppose we wish to subtract 99, 22, 111 and 55 from 333. Ordinary subtraction looks like:
[pic] o[pic] = [pic] o[pic] = [pic]
So here we can convert ordinary addition and subtraction to matrix multiplication. This simple conversion is the missing link of mathematics, it is the matrix/tensor connection to our real number system operations of addition and subtraction. That is, if the pre-multiplier is a matrix all of one’s, we are adding/subtracting the numbers in the columns of the post-multiplier matrix.
The operations of multiplication and division we know about intuitively (because no one has proven it except for me) from it’s use in Gaussian reduction. i.e. 3x6=18, by using the half-multiplier we get the equal, but trivial equation [3]T o [6] = 18. Suppose we need to multiply 3x6, 4x5, 9x3, 22x11 and 7x11. I want to subtract the second product from the first, ignore the third product, double the fourth product and subtract and triple the fifth product and add. We have:
[pic] o[pic] o[pic] = [pic]
Or it is equal to:
[pic]
Or in terms of Mathcad programs:
[pic]
The three hollow dot multiplication’s on the left are illegal according to modern math, diagonalizing the first column and multiplying to the second column matrix is legal (I do it this way because it is the only way a computer or calculator can come up with the same solution, no one has ever thought about multiplying matrices in this manner before). And yet, what the hollow dot multiplication on the far left is saying is
1x3x6 = 18
-1x4x5 = -20
0x9x3 = 0
-2x22x11 = -484
3x7x11 = 231
Summing the separate answers we get 18 - 20 + 0 - 484 + 231 = -255.
Division is carried out the same as multiplication except we multiply by the inverse of the divisor.
This is all there is to this operator. In every case, there is an operation we already know and use that can be used to compute the solution. But look at the problem above. If we totally half-multiply all the numbers, we go through 15 separate multiplication’s and 5 sums for a total of 20 operations. When we have to diagonalize, the diagonal times the second column needs 25 multiplication’s and 25 sums. We then transpose this first product and multiply again to the third column matrix, needing 5 multiplication’s and 5 sums. This is a total of 25+25+5+5 = 60 separate mathematical operations to compute the same answer as 20 steps using the half-multiplier. That’s three times as many. If the problem concerned column matrices a million rows long, diagonalizing would produce a square matrix of dimensions 1x106 x 1x106 with 1x1012 elements, that would be a trillion elements, all zero’s except the diagonal. So we would need to perform a trillion multiplication’s and a trillion sums instead of a million multiplication’s and a million sums. Quite a savings on computer memory. Actually, MathCad +6 can perform this half-multiplication using the vectorize operation, but it cannot perform the following multiplication’s. Suppose we have the following 3x3 matrix, and we wish to multiply the first row by 3, the second row by 4 and the third row by 5. This is also illegal according to mathematicians, but they do it all the time when they use Gaussian reduction. i.e.
3 1 2 3 3 0 0 1 2 3 3 6 9
4 o 4 5 6 = 0 4 0 4 5 6 = 16 20 24
5 7 8 9 0 0 5 7 8 9 35 40 45
Suppose, on the other hand, we wish to multiply the first column in the 3x3 matrix by 3, the second column by 4 and the third column by 5. We proceed as follows:
1 2 3 3 1 2 3 3 0 0 3 8 15
4 5 6 4 = 4 5 6 0 4 0 = 12 20 30
7 8 9 5 7 8 9 0 0 5 21 32 45
These are important, because suppose we wish to deal only with the second row in the matrix and get rid of the first and third rows. We proceed as follows:
0 0 0 1 2 3 0 0 0
0 1 0 4 5 6 = 4 5 6
0 0 0 7 8 9 0 0 0 .
Or suppose we want only the first and third columns of the matrix, and ignore the middle column. We proceed as follows:
1 2 3 1 0 0 1 0 3
4 5 6 0 0 0 = 4 0 6
7 8 9 0 0 1 7 0 9
These are very important properties in accounting and inventories and statistics, especially when we have a large amount of data, but wish to compare only two or so columns or rows with each other in statistics, or to find the individual accountpage and itempage matrices in the generalized accounting/inventory system. These you will see more of in the examples to follow.
But, you might say, by the definition I give, that
1 2 3 3 3 6 9 3 6 9
4 5 6 o 4 = 16 20 24 and is also equivalent to 16 20 24
7 8 9 5 35 40 45 35 40 45
Depending upon which way we multiply. Right to left for the nested array and left to right for the larger single 3x3 matrix. Our choice, depending on the information we wish to obtain from the system. The nested array is ready for a transpose and further manipulation as a larger column matrix (in this case, a 9x1 column matrix), or just to separate the answers into separate arrays (for printing purposes say) or the matrix can be used any way we can use a regular 3x3 matrix.
I will show later (I can’t prove it) that when we are done with the math, we can drop the inner or the outer brackets of the nested array to obtain either three 3x1 matrices or one 3x3 matrix, whatever form we desire the solution to be. Matrices are freewheeling and we can do almost anything with them that we can do with single numbers, just as long as we keep to the rules of regular mathematics.
There is one further thing to notice. When we transpose the pre-multiplier matrix and half-multiply across the post-multiplier matrix, the transposed matrix becomes like a real set of numbers and is not like a matrix anymore. So just as we can take the square root of a single number, or the log of a single number, we can now take the square root of every number individually or the log of each number individually, all at the same time. If we then re-transpose the pre-multiplier matrix, it is in tensor or matrix form and we can multiply the matrices as they are now defined.
So now the concept of number has been re-defined. Just as 1, 2, 3, . . . are considered individual numbers, now
1 2 3 can be considered as a single number with all the properties of a single number.
4 5 6
7 8 9
But just as 1 + 1 = 2, the dimensions must be the same or conform with the rules of matrix multiplication. i.e.
Suppose we wish to multiply the two following numbers as we would two regular numbers:
1 2 3 9 8 7 9 16 21
4 5 6 ⊗ 6 5 4 = 24 25 24
7 8 9 3 2 1 21 16 9
Or we may add all the numbers at the same time, just as long as the matrices are the same size:
1 2 3 9 8 7 10 10 10
4 5 6 + 6 5 4 = 10 10 10
7 8 9 3 2 1 10 10 10
The former multiplication is very illegal in mathematics, but can anyone out there prove to me that 8x2 ( 16? Or 9x1 ( 9? This I would like to see. Not only is this multiplication legal, but it works! We cannot multiply all numbers in all problems at the same time without it. Statistics as a branch of mathematics would fall apart without it. But because mathematicians say it is illegal, statisticians must solve statistical problems the hard way, one multiplication and sum at a time, rather than all at once in one simplified operation.
GENERALIZED INVENTORY/ACCOUNTING SYSTEM
Let [A]ij = Inventory matrix
Let [B]jk = Database matrix
Then
[A]ij [B]jk = Cik = [SOL]ik
[INV]ij[DB]jk = [SOL]ik
where Cik = solution matrix for the inventory.
And
[A]Tij o [B]jk = Djk = [AP]jk
[INV]Tij o [DB]jk = [AP]jk
Where Djk = [AP]jk = the accounting matrix.
And
(
[A]Tij o [B]jk = Ijk = [IP]jk
(
[INV]Tij o [DB]jk = [IP]jk
Where Ijk = [IP]jk = The Itempage matrix.
The connection between the accounting/inventory field equation and statistics is given by:
[INV]ij[DB]jk = [SOL]ik
1/N([1]1,i[INV]ij[DBi]jk)2 = [SOL]1,k[SOL]k,1T - CORRECTION FACTOR
The best way to show how the inventory system works is to work an example. The first example will be a chemical usage inventory (CUI). For the following example:
1. the number of chemicals with unknown fractions = 9 = j.
1. the number of reportable chemical fractions = 15 = k.
1. the number of machines using smokestacks + one total inventory = 8 machines + 1 total inventory = 8 + 1 = 9 = i.
Therefore, the dimensional analysis of this example inventory =
1. For inventory: [A]ij [B]jk = Cik =[A]9,9 [B]9,15 = C9,15
2. For Accounting: [A]Tij o [B]jk = Djk = [A]T9,9 o [B]9,15 = D9,15
( (
For Itempage: [A]Tij o [B]jk = Ijk = [A]T9,9 o [B]9,15 = I9,15
THE INVENTORY MATRIX
We will write the inventory matrix first. We enter the inventory weights for the total yearly inventory (in pounds) in the first row of the matrix, for the second through 9th rows we will write the pounds of chemicals used in machines 1 through 8. Note: the sum of the columns 2 - 9 should equal the amount in the first row. i.e.
AP CSS DGP FC-37 KFR-18 NW-3A RW-41 TGC ZON
TOTAL INVENTORY 5590 55366 725 8610 49547 2750 21649 11000 880
MACHINE #1 5590 41605 0 0 0 0 0 0 0
MACHINE #2 0 12863 0 0 0 0 0 0 0
MACHINE #3 0 898.1 0 0 0 0 0 0 0
MACHINE #4 0 0 0 0 0 0 0 0 0
MACHINE #5 0 0 362.5 638 0 476.7 10824.5 5500 826.2
MACHINE #6 0 0 362.5 4023 0 325.6 10824.5 5500 0
MACHINE #7 0 0 0 3949.42 4952.7 1947.83 0 0 53.8
MACHINE #8 0 0 0 0 44574.3 0 0 0 0
Note: For the column under CSS, summing rows 2 through 8, we get 41605+12863+898.1 = 55366.1 which is the total yearly use for all machines.
This finished matrix I will call [INV]9,9 . For HP-48G program, STO INV99.
THE DATABASE MATRIX
The matrix [B]jk is written directly from the MSDS sheets for each chemical listed. It is permanent and will not change unless new chemicals are added to the inventory, old chemicals are not used anymore, or the manufacturer changes the proportions of the ingredients. First we make a vertical list of the 9 chemicals we use. In a row across the top, make a column for each of the 15 chemical fractions. Using the MSDS sheets and doing one chemical at a time, go along the row to the column where the chemical fraction is listed and enter the percent of that chemical as a decimal. Where there are no chemicals present, enter a zero. Approximately 5% of the DB matrix will be chemical fractions, and 95% will be zero’s. I will call this matrix [DB]9,15 .
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AP .02 .06 0 0 0 0 0 0 0 0 0 0 0 0 0
CSS 0 0 .12 .04 .10 .03 0 0 0 0 0 0 0 0 0
DGP 0 0 0 .02 0 0 0 0 0 0 0 0 0 0 0
FC-37 0 0 0 0 0 0 .0001 .001 0 0 0 0 0 0 0
KFR-18 0 0 0 0 0 0 0 0 .01 0 0 0 0 0 0
NW-3A 0 0 0 0 0 0 0 0 0 .006 0 0 0 0 0
RW-41 0 0 0 0 0 0 0 0 0 .001 .0002 .0002 .0001 0 0
TGC 0 0 0 0 0 0 0 0 0 .005 0 0 0 .02 0
ZON 0 0 0 0 0 0 0 .04 0 0 0 0 0 0 .075
Where:
APE: ALKYL PHENOLIC ETHER
DPE: DIPHENYL ETHER
NAP: NAPTHALENE
ISOP: ISOPROPANOL
2-EH:2-ETHYLHEXANE
oDCB: ortho-DICHLOROBENZENE
2EE: 2 ETHOXYETHANOL
EG: ETHYLENE GLYCOL
AMM: AMMONIA
FORM: FORMALDEHYDE
2BA: 2-BUTYLALCOHOL
DIOX: DIOXANE
ACRYL: ACRYLALDEHYDE
DEG: DIETHYLENEGLYCOL
ACET: ACETONE
We will store this matrix in DB915.
PROGRAM FOR HP-48G
ENTER INV99
STO INV99
ENTER DB915
STO DB915
RCL DB915
RCL INV99
x
MATHCAD +6 PROGRAM
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INV9,9*DB9,15 =
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[SOL]9,15 = INV9,9*DB9,15
INTERPRETATION: In the total inventory (row 1) we used 6643.92 lbs naphthalene in 1995. Of this 6643.92 lbs, 4492.6 lbs were used in machine 1, 1543.56 lbs were used in machine 2 and 107.772 lbs were used in machine 3 and none was used in any other machine. Or we used 43.81 lbs ethylene glycol total for 1995, of which 33.686 lbs were used in machine 5, 4.023 lbs were used in machine 6 and 6.1014 lbs was used in machine 7, etc.
Suppose we are not a synthetic minor or title V user and only need the total inventory, but not any breakdown by machine. We just take the first row in the inventory matrix and multiply it to the DB9,15 matrix. Rather than write out a new row matrix for the inventory, we will compute it from the INV9,9 matrix.
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Suppose we want the total inventory just for machine 7, then
FOR THE INVENTORY FOR MACHINE 7:
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Where R1 is the reduction matrix that returns only row one of INV9,9 upon multiplication, and R8 is the reduction matrix that returns only row 8 upon multiplication.
Suppose the EPA or the company managers wish to know only the throughput of machines 5,6 and 7, since they are the major users of the chemicals. Then
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ACCOUNTPAGE MATRIX FOR TOTAL INVENTORY: ROW 1
1[INV]T9,9 o [DB]9,15 = 1[AP]9,15
In the accountpage matrix, we can keep track of all the chemicals used in each machine, or keep track of the total inventory. The solution is the un-summed form of the inventory solution matrix. This solution remains a 9x15 matrix. If summed, it is the total inventory matrix for that particular machine. The 1 in the upper left hand corner of 1[INV] and 1[AP] means we are working with the first row transposed in the inventory matrix. For instance, for the total inventory, 1[AP]9,15 =
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5590 .02 .06 0 0 0 0 0 0 0 0 0 0 0 0 0
55371 0 0 .12 .04 .10 .03 0 0 0 0 0 0 0 0 0
725 0 0 0 .02 0 0 0 0 0 0 0 0 0 0 0
8630 0 0 0 0 0 0 .0001 .001 0 0 0 0 0 0 0
49547 o 0 0 0 0 0 0 0 0 .01 0 0 0 0 0 0
2750 0 0 0 0 0 0 0 0 0 .006 0 0 0 0 0
21649 0 0 0 0 0 0 0 0 0 .001 .0002 .0002 .0001 0 0
11000 0 0 0 0 0 0 0 0 0 .005 0 0 0 .02 0
880 0 0 0 0 0 0 0 .04 0 0 0 0 0 0 .075
But since the mathematics doesn’t exist in which this multiplication can take place, we must convert it to an equivalent form to multiply. i.e. We will take the column for the total inventory and make a diagonal matrix out of it and then multiply.
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[pic] Here I transpose [1]1,9 to [1]9,1 so I can post-multiply to sum the rows
AP = INV9,9[R1]9,9[1]9,1 This multiplication returns only column 1 of the transposed inventory matrix
[pic] Here I diagonalize the above column matrix
AP1 = [DIAGAP]9,9[DB]9,15 The accountpage matrix for the total inventory
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INTERPRETATION: In 1995, we used a total of 111.8 lbs of alkyl phenolic ether in chemical AP. The chemical CSS (row 2), of which the company used 55,366 lbs in 1995, used 6643.92 lbs naphthalene, 2214.64 lbs of isopropanol, 5536.6 lbs of 2-ethylhexane and 1660.98 lbs of o-dichlorobenzene. For the chemical RW-41 (row 7), the company used 129.894 lbs of formaldehyde, 4.3298 lbs 2-butoxyalcohol, 4.3298 lbs dioxane and 2.1694 lbs acrylaldehyde, etc.
Now let’s look at the total accountpage matrix for machine 7:
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INTERPRETATION: In the chemical FC-37, there was used .39494 lbs
2-ethoxyethylene and 3.94942 lbs ethylene glycol. For the chemical ZON, there is 2.152 lbs ethylene glycol and 4.035 lbs acetone, etc.
There will be 9 of these accountpage matrices, one for the total inventory, and 8 more, 1 for each of the eight machines.
Suppose we are Title V or synthetic minor and again the EPA or the state is mainly interested in machines 5, 6 and 7 because the bulk of the organic volatiles are used in these machines and expelled in their stacks which the EPA monitors. Then we have:
7
( [INV][DB] =
5
0 0 362.5 638 0 476.7 10824.5 5500 826
0 0 362.5 4023 0 325.6 10824.5 5500 0 [DB]9,15
0 0 0 3949.42 4952.7 1947.83 0 0 53.8
Or we can transpose and hollow dot the three of them separately to obtain three 9x15 sub-matrices that account for just these three particular stacks. Note also that we can take the three accountpage solutions and add each of the separate matrices together to get a single accountpage matrix that is the total accounting of the chemical usage of machines 5, 6 and 7.
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[pic] [pic] (6[INV]T9,9 [R6]9,9[1]9,1 )o [DB]9,15
[pic] [pic] (7[INV]T9,9 [R7]9,9[1]9,1 )o [DB]9,15
[pic] [pic] (8[INV]T9,9 [R8]9,9[1]9,1 )o [DB]9,15
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Or, we can transpose INV9,9 for 5, 6 and 7, sum the rows, diagonalize the column and multiply by DB9,15. i.e. (Note: R567 = R678 above).
[pic] (567[INV]T9,9 [R567]9,9[1]9,1 )o [DB]9,15
[pic] ( MathCad will not take matrices in parenthesis, so we must do the problem in steps).
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Because this may look complicated, let’s look at the last problem step by step.
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Let’s look at this more generally. In this example, I included the total inventory as row one in the inventory matrix. Then I broke this down to chemical usage per machine as an individual row each. Including the whole matrix except row 1, summing the columns equals the value of row 1. Therefore, we don’t even need row 1 for our computations. I just put it in for illustration and convenience. If we don’t include the total inventory in the inventory matrix, just the machine breakdowns, we get:
1[AP]9,15 = ( [AP]9,15 = 2( [AP]9,15 +3( [AP]9,15+. . . 9( [AP]9,15
Or if only the fractional parts are included but not the total inventory:
9
T[AP]9,15 = ([AP]9,15
1
Let’s do a few computations by hand and see if they check.
We had 5590 lbs AP total, of which 2% was alkylphenolic ether. So 5590 x .020 = 111.8 lbs APE. In machine 7, we used 3949.42 lbs FC-37 of which .01% is 2-ethoxyethylene.
3949.42 x .0001 = .3949 lbs 2-EE. Also, .1% of FC-37 is composed of ethylene glycol, so we have 3949.42 x .001 = 3.94942 lbs EG. To check on the total inventory, let’s do the formaldehyde total, since there is more in this column than any other. Of 2750 lbs NW-3A, .6% is formaldehyde: 2750 x .006 = 16.5 lbs. Of 21,649 lbs RW-41, .6% is also formaldehyde, so we have 21,649 x .006 = 129.894bs formaldehyde. The chemical mix called TGC contains .5% formaldehyde, so we have 11,000 x .005 = 55 lbs formaldehyde. The total formaldehyde is then 16.5+129.849+55 = 201.394 lbs formaldehyde used in 1995, which matches the matrix calculation.
THE ITEMPAGE MATRIX
c[DB]9,15 o [INV]T9,9
By all accounts, the solution to the itempage matrix is impossible, for we multiply the inventory matrix by the database matrix, multiplying from right to left, rather than left to right. (See proof in the theory section at the beginning of this paper.) What I’m doing is taking the individual chemical (item) and hollow dotting it into the transposed inventory matrix, receiving a sub-matrix that documents only that chemicals distribution through the inventory It’s main power is realized when we hollow dot multiply by one column in the database at a time. The lowercase c to the upper left of c[DB] stands for the number of the column we are multiplying by. There will be c = j = 15 itempage matrices if we wish to compute them all. Again, I will work with formaldehyde, since there are 3 entries in it’s column. In this case, c = 10 since formaldehyde is listed in the 10th column. We remove this column from [DB]9,15 and hollow dot multiply into the [INV]T9,9 matrix. i.e. Note also when we are done, that the solution is defined in four variables, rather than three. That is, in machine 5, the chemical RW-41 contains a total of 129.894 lbs of formaldehyde. We pick up an extra descriptive variable using this reverse multiplication.
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But a computer cannot multiply across, so we must diagonalize the column matrix first.
[pic] and 10[DB]9,15 =[pic]
10[DB]9,15 = FORM
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=
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Tot #1 #2 #3 #4 #5 #6 #7 #8
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INTERPRETATION: Looking at the Itempage matrix for formaldehyde, we can instantly see that three chemicals contain formaldehyde and their use is distributed among machines 5, 6 and 7 in the poundage’s listed. I really wish I could place the chemical names by the matrix, but MathCad won’t let me and I do not wish to type out the matrix again, so we just have to remember that in order, the chemicals are:
AP
CSS
DGP
FC-37
KFR-18
NW-3A
RW-41
TGC
ZON
Let me theorize here for a moment. I don’t really think that putting the c[DB]
in front of the [INV]T matrix is mathematically correct. I think it’s fundamentally simpler than this. I mean it works, but if you transpose the [DB] matrix and multiply by [INV]T there is no solution (the indices do not match), but according to the proof of the half-multiplier, if you can hollow dot multiply the matrices and then sum the columns, you can also transpose and multiply the regular way and get the same answer. i.e. [DB]T[INV]T=[SOL]T ; but [INV]T[DB]T([SOL]T .
What this seems to suggest is that the hollow dot operation is transpose commutative. That is [SOL]T = [IP].
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Let’s see if this makes sense. Let’s transpose the [DB] matrix, sum the new rows, and o multiply across the [INV]T matrix and see if we get the same answer.
ONE15 = (FIFTEEN1)T
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[pic] Here I sum the rows of the database matrix
[pic] Here I diagonalize the above solution
[pic] This is the total itempage solution
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INTERPRETATION: The chemical CSS (second row) has a total of 16,056.14 lbs of all organic’s, of which 12,065.45 lbs were used in machine 1, 3730.27 lbs were used in machine 2 and 260.449 lbs were used in machine 3. In the first row, the chemical AP contains a total of 447.2 lbs organic’s used in machine 1 only. Since this is a simple computation, let’s check and see if it is right. The chemical AP contains two organic’s, alkylphenolic
ether (2%) and diphenyl ether (6%). We used a total of 5590 lbs in 1995, so we have
5590x.02 + 5590x.06 = 111.8 + 335.4 = 447.2 lbs APE and DPE combined.
If we compute all the itempage matrices and add these sub-matrices together, we will come up with the above matrix. The itempage matrix does not keep track of the individual components, we must itempage for each separate chemical to find this out. It just gives us the total of volatiles used. Note also, that the post-multiplier matrix operates on the pre-multiplier matrix, giving a solution of chemicals per machine rather than chemical fraction per bulk chemical.
We can also take Log[IPTOT]9,9 - Log[INV]9,9, then take the anti-log which will give us the total % volatiles per chemical per machine.
We can also multiply the [APTOT]jk matrix by [1]1,j to sum the columns. i.e.
[1]1,j = [1]1,9 ; [1]1,9 [APTOT]9,15 = [111.8 335.4 6642 2229.3 5537 1661 .863 43.83 4955 84.5 4.33 4.33 2.16 220 66]
= [SOL]1,15.
Which, looking at the total inventory (row 1) in the [SOL]9,15 matrix are equal. Multiplying [SOL]1,15 by [1]15,1 we get [SOL]1,15[1]15,1 = [LBS]1,1 (a single number) which gives the total poundage for volatile organic’s and/or criteria pollutants in the inventory. i.e.
[SOL]1,15[1]15,1 = 21,897.513 lbs total in the inventory.
MULTIPLE CHEMICAL USAGE INVENTORY
Here I’m going to make simple computations comparing inventories over two years, comparing chemical usage in 1995 and 1996. This could also be for day after day comparisons, weekly, monthly, semi-annually, etc.
CHEMICALS USED (IN POUNDS)(
AP CSS DGP FC-37 KFR-18 NW-3A RW-41 TGC ZON
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MathCad cannot create a matrix larger than 100 elements at a time. To create
larger matrices, we must augment them (join two matrices side by side) or stack them,
(join two matrices on top of each other). Examples are as follows.
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APE DPE NAPTH ISOP 2-EH oDCB 2EtOH EtGly AMM FORM 2-BtOH DIOX ACRYL DEG ACET
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APE DPE NAPTH ISOP 2-EtHEX o DCB 2-EtOH EG AMMON FORM 2-BtOH DIOX ACRYL DEG ACET
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INTERPRETATION: In 1995, there was a total of 111.8 lbs alkylphenolic ether used, 111.8 lbs in machine #1. There was a total of 6644.52 lbs naphthalene used, 4992.6 lbs in machine #1, 1543.56 lbs in machine #2 and 107.772 lbs in machine #3. In 1996, there was 6777.72 lbs used, 5092.68 lbs in machine #1, 1571.196 lbs in machine #2 and 109.932 lbs used in machine #3. In 1996, a total of 4.55 lbs dioxane was used, 2.275 lbs in machine #5 and 2.275 lbs in machine #6 etc.
Now, to compare how much more or less of chemicals was used in 1996 than 1995, we proceed as follows
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INTERPRETATION: We used .20 lbs more alkylphenolic ether in 1996 than in 1995, 133.2 lbs more naphthalene in 1996 than 1995 and -.01 lbs less of 2-ethylhexane in 1996 than in 1995.
Since we are using a stacked matrix for the above calculation, we need to use the statistical database DB9.
We can also use the un-stacked Inventory matrices to calculate the differences in chemical usage from 1995 to 1996. The stacked matrix is used in the HP-48G only, but we can use the un-stacked matrices as follows: (this method gives the same answer as above, but it is computed differently).
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Therefore:
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Now we will look at some Accountpage Matrices. We can only extract a column at a time, not a row, so we have to transpose the Inventory Matrix to obtain whatever row we wish to account for. For this example, we shall extract the total inventory for 1995 and show its account page, and we shall extract the column for Machine 7 and show its account page. We proceed as follows:
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To extract column 0:
Press Matrix palette, type in INV1995, CLICK M TRANSPOSE,
CLICK M , TYPE IN 0, PRESS =.
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APE DPE NAPTH ISOP 2-ETHEX o-DCB 2-EtOH EtGly AMM FORMAL 2- BtOH DIOX ACRYL DEG ACET
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ANALYSIS:5500 LBS OF BURCO AP CONTAINS 111.8 LBS ALKYLPHENOLIC ETHER AND 335.4 LBS DIPHENYL ETHER
55,371 LBS SUPERSOL CSS CONTAINS 6644.52 LBS NAPTHALENE, 2214.84 LBS ISOPROPANOL, 5537.1 LBS
2-ETHYLHEXANE, AND 1661.13 LBS O-DICHLOROBENZENE
725 LBS DGP CONTAINS 14.5 LBS ISOPROPANOL
8630 LBS FC-37 CONTAINS .863 LBS 2-ETHOXYETHOXY ALCOHOL AND 8.63 LBS ETHYLENE GLYCOL
49,547 LBS KFR-18 CONTAINS 4954.7 LBS AMMONIA
2750 LBS NW-3A CONTAINS 16.5 LBS FORMALDEHYDE
21,649 LBS RW-41 CONTAINS 12.989 LBS FORMALDEHYDE, 4.33 LBS 2-BUTYL ALCOHOL, 4.33 LBS DIOXANE,
AND 2.165 LBS ACRYLALDEHYDE
11,000 LBS TGC CONTAINS 220 LBS DIETHYLENE GLYCOL
880 LBS ZONYL Y CONTAINS 66 LBS ACETONE
Now lets look at the account page Matrix for Machine 6 (it uses the most chemicals):
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APE DPE NAP ISOP 2-EH oDCB 2EE EG AMM FORM 2BA DIOX ACRYL DEG ACET [pic]
ANALYSIS: MACHINE 6 USED 362.5 LBS DGP CONTAINING 7.25 LBS ISOPROPANOL
4023 LBS FC-37 CONTAINING .402 LBS 2-ETHOXYETHOXY ALCOHOL
325.6 LBS NW-3A 1.954 LBS FORMALDEHYDE
10,824.5 LBS RW-41 CONTAINING 6.495 LBS FORMALDEHYDE, 2.165 LBS 2-BUTYL ALCOHOL,
2.165 LBS DIOXANE AND 1.082 LBS ACRYLALDEHYDE
5500 LBS TGC CONTAINING 27.5 LBS FORMALDEHYDE AND 110 LBS DIETHYLENE GLYCOL
AND FINALLY, FOR THIS SET OF EXAMPLES, WE WILL CALCULATE THE ITEMPAGE MATRIX FOR FORMALDEHYDE AND ISOPROPANOL.
FIRST WE MUST TRANSPOSE THE 1995 INVENTORY MATRIX, THEN FROM THE FORMALDEHYDE COLUMN ON THE DATABASE MATRIX WE MUST REMOVE THE COLUMNS REPRESENTING FORMALDEHYDE AND ISOPROPANOL. i.e.:
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TOTAL #1 #2 #3#4#5 #6 #7 #8
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ANALYSIS: NW-3A CONTAINED 16.5 TOTAL LBS OF FORMALDEHYDE, OF WHICH 2.86 LBS WERE USED IN
MACHINE 5, 1.954 LBS WERE USED IN MACHINE 6 AND 11.687 LBS WERE USED IN MACHINE 7.
RW-41 CONTAINED A TOTAL OF 12.989 LBS FORMALDEHYDE, 6.4955 LBS WERE USED IN MACHINE 5 AND 6.495 LBS WERE USED IN MACHINE 6.
TGC CONTAINED 55 LBS FORMALDEHYDE, 27.5 LBS WERE USED IN MACHINE 5 AND 27.5 LBS WERE USED IN MACHINE 6.
TO CALCULATE FORMALDEHYDE USAGE FOR BOTH 1995 AND 1996, WE PROCEED AS FOLLOWS:
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1995 1996
tot95 #1 #2 #3 #4 #5 #6 #7 #8 Tot96 #1#2 #3#4 #5 #6 #7 #8
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TO CALCULATE THE ITEMPAGE MATRIX FOR ISOPROPANOL:
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#1 #2 #3 #4 #5 #6 #7 #8 TOT96 #1 #2 #3 #4 #5#6 #7 #8
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ANALYSIS: FOR 1996, CSS CONTAINED A TOTAL OF 2259.24 LBS ISOPROPANOL, 1697.56 LBS USED IN
MACHINE 1, 523.732 LBS USED IN MACHINE 2 AND 36.644 LBS USED IN MACHINE 3.
DGP CONTAINED 16 LBS ISOPROPANOL, 8 LBS USED IN MACHINE 5 AND USED 8 LBS IN MACHINE6.
Normally, here is where I would do a statistical analysis on the chemical usage between 1995 and 1996, but I am told on the inner planes not to give everything away, so I think I’ll forego that part of my analysis for the time being and get on with other aspects of this inventory.
LABELING THE MATRICES
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Suppose we are a company who wishes to keep our inventory secret in case hackers break into our computer. Also suppose we are paranoid of any commercial encryption software out on the market (anyone who can understand the program can possibly decode our information). Lets encrypt the information ourselves. We must remember to save the key on a disk and delete all info as to the identity of the key from mathcad. (OF course, we can just keep the CUI on disk and not in the computer if we wish too) As you will see, if we wish to encrypt our data, we really should do it with our unlabled computations and add the labels in after de-encryption.
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First we will scramble just the rows, then we will scramble the columns. This simple CUI has 10 rows and 16 columns, we must make a 10x10 matris to interchange the rows. Note that there is only a single 1 in any row or column, you can scramble the ones as you wish.
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Now premultiplying on the data matrix we get:
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Now we will scramble the columns, since this is a 10x16 matrix, we need a 16x16 matrix, but Mathcad can only form single matrices of 100 elements. so we must create the matrices of less than 100 elements and stack them together.
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This is the scrambler matrix for the column scramble
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This matrix has it's rows and columns interchanged.
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To decode this and get our information back again we just multiply the coded matrix by the transpose of the encoding matrices i.e.:
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Which is the unscrambled inventory.
CALCULATION OF A WATER BILL
WATER BILL
METER READING THIS MONTH - METER READING LAST MONTH = GALLONS WATER USED
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TO COMPUTE THE CHARGE, SUPPOSE THE COST OF WATER IS 3¢ PER GALLON, THEN
[GWU]4,1 o c[1]4,1 , WHERE C = CHARGE PER GALLON.
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THE HOLLOW DOT MEANS TO MULTIPLY STRAIGHT ACROSS.
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NOTE: THESE VALUES ARE IN DOLLARS.
OR WE CAN COMPUTE THIS BY:
[pic][pic][pic][pic])=
PROGRAM:
[pic] Here we take the log of each individual element and add them together
[pic](THIS REPRESENTS THE SUMMED LOGS OF THE GALLONS + PRICE MATRICES)
[pic] (THIS TAKES THE ANTI-LOG OF THE ABOVE COMPUTATION.)
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THE TOTAL AMOUNT OWED BY ALL CUSTOMERS TO THE UTILITY IS:
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THE TOTAL OF MONTHLY BILLS PER CUSTOMER TOTALED FOR THE ENTIRE YEAR TO DATE =
LAST MONTHS BILL(JAN. + THIS MONTHS BILL(FEB) = TWO MONTHS TOTAL PER CUSTOMER
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BUT SUPPOSE THE CHARGES FOR THE FIRST 500 GALLONS = 3¢/GAL AND THE CHARGE FOR THE NEXT 1000 GALLONS RISES TO 5¢/GALLON, WE HAVE (WE IGNORE ZERO SIGNS):
[pic](CHANGE THE NEGATIVE NUMBERS TO ZERO)
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NOW WE NEED TO ADD THE CHARGES TOGETHER.
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THE FIRST CUSTOMER USED $40 WORTH OF WATER,THE SECOND CUSTOMER USED $15 WORTH OF WATER, THE THIRD CUSTOMER USED $25 WORTH OF WATER AND THE FOURTH CUSTOMER ONLY USED $6 WORTH OF WATER.
OR WE CAN USE LOGARITHMS:
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( NOTE: MUST CHANGE ZERO’S TO DEFAULT 0 ( 10-100)
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[pic]
[pic]
With MathCad, we don’t have to take the logs of the elements of the matrices, we are just multiplying the first element in matrix one by the first element in matrix two; the second element in matrix one by its corresponding element in matrix two, etc. But computers cannot do the matrix multiplication’s this way (we can with MathCad through the vectorize operation) so we have to use logarithms to get our solutions.
To find out who paid their bill, and who is behind in their payments, we proceed as follows:
We need an accounts paid matrix where each element represents the amount paid to the utility by the water user.
Owed paid
$40 $40 0 paid up through February
$15 - 0 = $15 owes $15 from February
$25 $35 -$10 credit $10 for February
$6 $6 0 paid up through February
Suppose the utility charges 1% per month penalty for the unpaid portion of the bill. We will ignore the $10 credit:
0 .01 0 0 0 0 MARCH’S
15 o .01 = .15 Then .15 + $15 = 15.15 + NEW
0 .01 0 0 -$10 -10 BILL
0 .01 0 0 0 0
Or we could compute the taxes, say at 6%, for the charges in the same manner:
40 .06 40 2.40 40 42.40
TAX = 15 o .06 + 15 = .90 + 15 = 15.90
25 .06 25 1.50 25 26.50
6 .06 6 .36 6 6.36
If a business gets a tax break, we just put in the value charged in place of the .06 and proceed as above.
One more example to show how to extend the calculations even further. Suppose after the first 1500 gallons, the charge increases to 10¢/gallon. Let’s add home 5 who used 1599 gallons of water last month. The first row will always be 500 gallons unless less water is consumed. Our matrix now becomes:
1000
500
700
200
1599
And the problem becomes:
[pic]
SINCE THE MAXIMUM FOR THE SECOND COLUMN IS 1000 GALLONS, ANYTHING OVER BELONGS IN THE THIRD COLUMN, SO LET'S SEE IF THERE ARE ANY VALUES ABOVE 1000 IN THE SECOND COLUMN. (HERE WE CAN SEE IT, BUT IF THIS WERE THE WATER BILL FOR NYC OR LA THERE WOULD BE MILLIONS OF LISTINGS, TOO MANY TO LOOK FOR BY HAND). IF ANY THREE VALUES IN ANY ROW ARE ALL POSITIVE, OVER 1500 GALLONS OF WATER WERE USED. A CORRECTION MATRIX CAN BE DERIVED AS FOLLOWS: REPLACE EVERY NEGATIVE VALUE WITH ZERO.
[pic]
[pic]
MATHCAD WILL LET US DELETE THE MIDDLE COLUMN, BUT IT WON'T LET US INSERT THE CORRECTED MATRIX IN IT'S PLACE. WE MUST SELECT THE 500 AT THE TOP OF THE MIDDLE COLUMN. CLICK THE MATRIX PALLET AND FOR ROWS INSERT A ZERO, AND IN COLUMNS INSERT A ONE. THEN CLICK THE DELETE BUTTON. THEN CLICK NEXT TO THE 500 IN THE FIRST COLUMN, BUT DON'T PUT IT IN A SELECTION BOX, JUST PUT THE CURSOR TO THE RIGHT OF IT AND PRESS INSERT. WE MUST RE-TYPE IN THE NEW VALUES. WE CAM EXTRACT A SINGLE COLUMN FROM A MATRIX, BUT WE CAN’T PUT ONE IN.
THE PROBLEM NOW BECOMES, AFTER SETTING UP THE COST MATRIX:
[pic] [pic]
AND THE TOTAL BILL IS:
[pic] [pic]
[pic](THE VECTORIZED INV x COST MULTIPLIES THE MATRICES ONE ON ONE.)
HONEOWNER ONE'S BILL IS $40, TWO'S BILL IS $15, THREE'S BILL IS $25, FOUR'S BILL IS $6 AND FIVE'S BILL IS $74.90.
PROGRAM: ([THIS MO. READING]5,1-[LAST MO. READING]5,1)o[.03]5,1= [BILL FOR SEPARATE HOMES]5,1
COMPLEX BILL: ANTI-LOG(LOG[SEP. INV.]5,3)+LOG([COSTS]5,3))[1]3,1 = [COMPLEX BILL FOR SEPARATE HOMES]5,1
TO KEEP TRACK OF WHOLE AREA, STATE, USA OR WORLD ALL AT THE SAME TIME
I am not going to solve this in detail like I did for the problem above, I’m just going to set this and solve for a simple system.
Suppose Butner charges 2¢/first 500 gal, 5¢/next 1000gal 11¢/all use above
Suppose Cozart charges 3¢/first 500 gal, 5¢/next 1000gal 10¢/all use above
Suppose Creedmore charges 4¢/first 500 gal, 6¢/next 1000gal 12¢/all use above
After reading the meters and calculating gallons used:
[pic] [pic] [pic]
TO COMPUTE THE TOTAL BILL:
[pic]
BUTNER: PRINT 1 - 4
COZART: PRINT 5 - 9
CREEDMORE: PRINT 10 - 13
OR: [pic]
[pic]
[pic] [pic]
THIS IS HOW TO ACHIEVE THE SOLUTION USING LOGS.
Now for one final example on how this method works. Suppose the government wishes to keep track of all the public utilities: water, gas and electricity. In the following example, this is the order that I will put in each value. Suppose the meter readings have been taken and subtracted. The cost for water is as above, the cost for gas is .001¢/cubic foot and the cost for electricity is 7¢/kilowatthour. The inventory and cost matrices are now:
[pic] [pic]
[pic]
[pic][pic]
THEN:
[pic]
[pic]
[pic]
[pic]
[pic]
THEN THE TOTAL BILL =:
[pic]
[pic]
[pic]
TOTAL ACCOUNTPAGE FOR TAXES TOTAL BILL SENT TO CUSTOMERS
[pic] [pic]
Household #1 received a bill of $26.50 for water, $5.83 for gas and $9.27 for electric with taxes of $1.50 for water, $0.33 tax on gas and $0.53 tax on electricity. Household #4 is a business, and paid $1229.6 for water, $8.48 for gas and $148.40 for electricity, etc.
TWO FACTOR MIXED DESIGN: REPEATED MEASURES ON
ONE FACTOR FOR STATISTICS AND ACCOUNTING/INVENTORY SYSTEMS
COMPUTATIONAL HANDBOOK OF STATISTICS, 2ND EDITION; SECTION 2.7, PG. 55-61.
EXAMPLE: A man has three hot-dog stands which he operates downtown. He sells only three products, two hot-dogs for $1, a burrito for $1.25 and a polish sausage for $1.50.These prices include the tax. He operates the stands 5½ hours and records the total sales of each item every 1/2 hour. Each employee makes $5/hr or $27.50 each day. We will find the total sales, the sales per cart, the sales by item, the taxes on the sales, the employees wages and federal, state and Social Security taxes. We shall then do a two factor mixed design analysis of variance on the sales to see if there is any statistical correlation between location of the carts, sales as the day goes by and if there is any statistical significance as to whether any item sold more than any other.
Note: This is problem 2.7 Analysis of Variance in Statistics section of book)
CART1 CART2 CART3
P B H P B H P B H
[pic] [pic] [pic] [pic]
[pic]NOTE: TAXDB with the rows summed equals MONEYDB
PROGRAM: [A]11,9 [MONEY]9,1 = MONEY MADE EVERY HALF HOUR
[pic]
[pic]
In the first half-hour he brought in $32, in the second half-hour he brought in $34, etc.
TOTAL SALES SEPARATED PER CART
[pic]: [pic]
[pic]
[pic]
[pic]
[pic]
[pic] [pic]
In the first half-hour, cart 1 had a total of $10.25 in sales, cart 2 sold $9.75 and cart 3 sold a total of $12 worth of polish sausage, burritos and hotdogs, etc.
To find the total sales for each cart, we proceed as follows:
[pic]
[pic]
[pic]
Cart 1 had a total of $108 in sales, cart 2 had a total of $118.25 and cart 3 had a total of $153.75 for the day.
The totals for polish sausage, burritos and hotdogs separated are: [pic]
[PBINV]11,9([MONEYDB]9,1 o[DB3]9,3)
Since we can’t compute matrices in parenthesis, we solve the equation as follows (using diagonal matrices):
[pic]
[pic]
[pic]
[pic]
Here we are just adding the cost of each polish sausage, each burrito and each hot-dog.
[pic]
[pic]
In the first half-hour, he sold $10.50 in polish sausages, $12.50 in burritos and $9 in hotdogs, in the 6th half-hour, he sold $13.50 in Polish sausages, $17.50 in burritos and $6.50 worth of hot-dogs, etc.
TOTAL SALES OF POLISH SAUSAGE, BURRITOS AND HOTDOGS
[pic]
[pic]
For the day, he sold $109.50 in polish sausage, $182.50 in burritos and $88 in hotdogs.
THE TAX DATABASE
Suppose taxes are 6%. There are two ways of computing the taxes.
METHOD #1:[PBHINV]([MONEYDB]o[TAX]) (This computes the taxes directly from the money taken
in that day)
[pic] [pic]
[pic]
[pic] [pic]
THIS IS THE TAX ON EACH HALF-HOUR OF SALES. THE TOTAL TAX WILL BE [1]1,11 x [TAX]11,1
[pic] [pic]THIS IS THE AMOUNT MADE AFTER TAXES
Summing the profits up to a single total, we find he made $357.20 after taxes.
[pic]TOTAL AMOUNT HE MADE AFTER TAXES
The government makes $22.80 for the total day, he made $357.20.
Or we can compute it with the TAXDB matrix: (This is computed directly from the inventory with no accounting necessary)
[pic]
[pic]
[pic] [pic]
Out of the $32 made in the first half-hour, we keep $30.08 and Uncle Sam gets 1.92. Net income = [pic]
[pic]
[pic]
For the day, we made $357.20, and paid the government $22.80. This is out of a total of $380 total brought in.
To check TAXES2 for error, multiply TAXES2 by TWO1. This sums the rows and they should equal to MONEYEACHHALFHOUR.
[pic] [pic]
[pic]= 0
EMPLOYEE WAGES AND TAXES
[pic] [pic]
[pic] E1 = employee # 1, etc. The -1 represents that that employee worked that half-hour period and that it is computed as a cost rather than income..
P B H P B H P B H E1 E2 E3
[pic] [pic]
[pic]
[pic]
CHECK:
[pic]
[pic]
[pic]
After paying his employees, he made $24.50 the first half-hour, $30 the 6th half-hour and $37.75 the last half-hour.
Suppose the first employee was 1/2 hour late, the second employee had to take an hour off for lunch and the third employee stayed an extra 1/2 hour for cleanup. The inventory matrix now looks like:
[pic]
[pic]
[pic]
[pic]
With some employees missing some time and another working an extra half-hour without overtime, in the first half-hour he made $27, in the last half-hour he made $32.25.
CHECK:
27+5=32
26.5+7.50=34, etc.
TOTAL WAGES PER EMPLOYEE
[pic]
[pic]
[pic]
[pic]
THE NEGATIVE SIGNS SHOW THAT THIS IS MONEY PAID OUT RATHER THAN INCOME RECEIVED.
[pic]
[pic]
[pic]
The first employee made $25, the second employee made $22.50 and the third employee made $30.
And finally, suppose the Social Security administration takes out 5%, Uncle Sam takes out 10% and the state takes out 6% of the employees wages:
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
To check the solution:
[pic]
[pic][pic][pic][pic]
[pic][pic][pic][pic]
[pic][pic][pic][pic]
Suppose now he wishes to check out how much money he made in the 6th half-hour and the first half-hour:
[pic] [pic]
[pic]
[pic] Removes the 6th column from the transposed inventory matrix
[pic] Removes the 1st column from the transposed inventory matrix
[pic] [pic]
[pic] [pic]
[pic] [pic]
[pic] [pic]
In the sixth half-hour period, he sold $6 worth of polish in the first cart, 6.25 worth of burritos in the second cart and $1 worth of hot-dogs in the third cart, etc. [AP1] says $4.50 worth of polish was sold in the first half-hour in the first cart, $3.75 worth of burritos in the second cart and $4 worth of hot-dogs in the third cart, etc. Let’s find the taxes for the first and sixth half-hours:
[pic]
[pic]
[pic]
[pic]
FOR THE AP MATRICES, FOR THE [SOL] MATRIX:
[pic]
[pic]
[pic]
USING THE PROPERTIES OF THE HALF-MULTIPLIER OPERATOR
[pic]
([INV]9,11[1]11,1) o [MONEYDB]9,1
[pic]
Here we sum the rows of the transposed inventory matrix.
[pic]
[pic]
Here we hollow dot multiply the summed rows by MONEYDB.
He sold 24 polish sausages in cart one in 5½ hours for a total of $36, he sold 43 burritos in cart 2 for a total of $53.75 and in cart 3, he sold 70 hot-dogs for a total of $35.
[pic]
CHECK: The sum total still sums to $380.
Now lets reverse multiply the transposed inventory matrix by the MONEYDB matrix.
[MONEYDB]o[PBHINVTRN]
[pic]
[pic]
[pic]
[pic]=
[pic]
We can see at a glance that in the second half-hour, he sold #$4.50 worth of polish sausage in cart one, $5 worth of burritos and $2 worth of hotdogs. In the eighth half-hour in cart 3, he sold $3 worth of polish, $6.25 worth of burritos and $3 worth of hot-dogs, etc.
Lets see how much tax he paid for each sum of items each half-hour. We can do this two ways, there is a tax of 9¢ on every polish sausage, .075¢ on each burrito and 3¢ on every hot-dog. We can multiply this by the inventory, or we can take the accountpage which has the sales in terms of money and multiply each value by .06 for 6%. i.e.
[pic]
[pic] [pic]
[ITEMTAX]o[PBHINVTRN] and [TAXA]o[ITEMPAGEMONEY]
[pic] [pic]
[pic] [pic]
[pic]
[pic]
27¢ of the $4.50 in polish sausage sales in the first half hour went to taxes. The highest taxes are in the 3rd, 4th, 5th and 11th half-hours in cart 3 for burritos, which tie at 52.5¢ and a high of 67.5¢ in the 11th half-hour..
The accounting is the same, it all depends upon whether it is easier to tax the inventory or tax the accounting.
CHECK: 4 polish sausage were sold in cart one in the first half-hour, there is a tax of 9¢
on each sausage. So we have 4x.09= 27¢. In the 11th half-hour in cart 3, 9 burritos were sold for a total tax of 9x.075 = 67.5¢.
$4.50 worth polish sausage were sold in cart one in the first half-hour, there is a tax of 6¢ on each sausage. So we have 4.50x.06= 27¢. In the 11th half-hour in cart 3, $11.25 worth of burritos were sold for a total tax of 11.25x.06= 67.5¢.
STATISTICAL ANALYSIS OF SALES
There is a lot more I can do to improve this inventory, but I can’t give everything away, besides, this paper is too long anyway. So I am going to go into the analysis of variance for the sales between the three carts. Note that the Database matrices I use in the statistical analysis are the same which are used to separate the sales by cart (or by store) and to separate the sales to each individual item (like only the sale of hot dogs, only the total of polish sausages sold and only the sales total of burritos (these could also represent every item sold in a store or in stores controlled by chains). This is the exact same statistical analysis you’ll find in Section 2.7 in the chapters on statistics. If you do not understand what I’ve done here, please read the part of this book dealing with statistics for a basic background.
STEP 1: Table the data:
P B H P B H P B H
3 3 4 2 3 6 2 4 8
3 4 4 1 4 5 3 6 3
1 3 4 3 6 7 4 7 7
1 2 3 2 4 6 1 7 4
2 3 5 1 2 3 3 7 12
[A]11,9 = 4 5 6 4 5 5 1 4 2
4 5 6 1 3 4 1 5 3
1 4 5 1 2 4 2 5 6
1 4 5 2 3 4 3 6 6
2 3 3 3 5 6 1 5 12
2 2 4 4 6 7 4 9 7
VARIABLES:
[pic] [pic]
[pic] [pic]
[pic] [pic] [pic]
[pic][pic][pic]
STEP 2: Add the scores in each column for each cart:
SUM1CARTS = [1]1,11[A]11,9
[pic]
[pic]
[pic]
HP-48G PROGRAM:
RCL ONE11
RCL A
x
STO SUM1CARTS
STEP 3: Obtain the sum of each cart by adding the sums of the individual sales.
PROGRAM: GROUPSUMSCARTS = [1]1,11[A]11,9[DB1]9,3
[pic]
[pic]
[pic]
HP-48G PROGRAM:
RCL SUM1
RCL DB1
x
STO GROUPSUMSCARTS [111 124 160]
STEP 4: Add the scores for each item in each Cart:
PROGRAM: ITEMSUMS = [A]11,9[DB1]9,3
[pic]
[pic] [pic]
HP-48G PROGRAM:
RCL A
RCL DB1
x
STO ITEMSUMS
STEP 5: Square each term in the table and sum:
PROGRAM: GRANDSUMSQ = [1]1,9 (diagonal[A]T[A])[1]9,1 or [1]1,11([A]11,9([A]11,9)[1]9,1 =
[pic]
[pic]
[pic]
[1 1 1 1 1 1 1 1 1] 66 1
142 1
229 1
66 0 1
189 1 = 2043
313 1
0 71 1
407 1
560 1
HP-48G PROGRAM:
RCL ONE9
(
TRN
SWAP
RCL A
(
TRN
SWAP
x
MTH,MATR,NXT (DIAG
9
(
DIAG(
x
SWAP
x
STO GNDSUMSQ
STEP 6: Sum each score in the table for the grand sum.
PROGRAM: GRANDSUM = [1]1,11[A]11,9[1]9,1
[pic]
[pic]
[pic]
HP-48G PROGRAM:
RCL ONE9TRN
RCL ONE11
RCL A
x
SWAP
x
STO GRNDSUM 395
STEP 7: Computation of the correction factor CORRFACT: Square the grand sum and divide by the total # of measures in the table. (NT = 99)
PROGRAM: CORRFACT = 1/NT ([1]1,11[A]11,9[1]9,1)2
[pic]
[pic]
[pic]
[pic]
(1/99)(395)(395) = 1576.0101 = CORRFACT
HP-48G PROGRAM: or:
RCL GRNDSUM 395 RCL ONE11
SQ RCL A
RCL NT = 99 x
/ RCL ONE9TRN
STO CORRFACT 1576.0101 x
SQ
RCL NT = 99
/
STO CORRFACT
STEP 8: Computation of the total sum of squares SST. Subtract the correction factor from the grand sum of squares (GRNDSUMSQ)
SST = GRNDSUMSQ - CORRFACT = 2043 - 1576 = 467
[pic]
[pic]
[pic]
BETWEEN SUBJECTS EFFECTS
STEP 9: Computation of between subjects sum of squares SSb . NB =# Items in each cart: Nb = 3.
PROGRAM: SSb = (1/Nb) [1]1,3diagonal([A]11,9[DB1]9,3 )2[1]3,1 - CORRFACT = SSb.
[pic]
[pic]
Again we must separate the problem into parts.
[pic]
[pic]
[pic]
[pic]
[pic]
(Tr1 (Tr2 (tR3
3 3 4 2 3 6 2 4 8 1 0 0 10 11 14
3 4 4 1 4 5 3 6 3 1 0 0 11 10 12
1 3 4 3 6 7 4 7 7 1 0 0 8 16 18
1 2 3 2 4 6 1 7 4 0 1 0 = 6 12 22
2 3 5 1 2 3 3 7 12 0 1 0 10 6 22
4 5 6 4 5 5 1 4 2 0 1 0 15 14 7
4 5 6 1 3 4 1 5 3 0 0 1 15 8 9
1 4 5 1 2 4 2 5 6 0 0 1 10 7 13
1 4 5 2 3 4 3 6 6 0 0 1 10 9 15
2 3 3 3 5 6 1 5 12 8 14 18
2 2 4 4 6 7 4 9 7 8 17 20
C2 = CTC (to get smallest possible square matrix, in this case a 3 x 3)
(1/NB) 10 11 8 6 10 15 15 10 10 8 8 10 11 14
11 10 16 12 6 14 8 7 9 14 17 11 10 12
14 12 18 22 22 7 9 13 15 18 20 8 16 18
6 12 22
10 6 22 1199 1218 1532
15 14 7 = 1218 1532 1826
15 8 9 1532 1826 2540
10 7 13
10 9 15
8 14 18
8 17 20
SSb = (1/3) [1 1 1] 1199 1
1532 1 - CORRFACT = 5271/3 - 1576 = 181
2540 1
HP-48G PROGRAM:
RCL ONE3 [1 1 1]
(
TRN
SWAP
RCL A
RCL DB1
x
(
TRN
SWAP
x
MTH,MATR,NEXT (DIAG
3
(
DIAG(
x
SWAP
x
RCL Nb = 3
/
RCL CORRFACT
-
STO SSb 181
STEP 10: Computation of the total sales between the three carts. (Performance of each cart in sales) (The meaningfulness effects). SSCARTS. NCARTS = NT/# columns in DB1 = 3.
NCARTS = 99/3 = 33.
PROGRAM: SSCARTS = 1/NC([1]1,11 [A]11,9 [DB1]9,3)2 - CORRFACT
= 1/33(SUM1CARTS x DB1)2 – CORRFACT
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
HP –48 PROGRAM
RCL ONE11
RCLA
x
RCL DB1
x
(
TRN
x
RCL NCARTS
/
RCL CORRFACT
-
STO SSCARTS
STEP 11: Computation of between subjects error term, SSErrb.
SSErrb = SSb - SSC = 181 - 39 = 142
[pic]
[pic]
[pic]
WITHIN SUBJECTS EFFECTS
STEP 12: Computation of within subjects sum of squares, SSW.
SSW = SST - SSb = 467 - 181 = 286
STEP 13: Computation for sum of squares for Polish, Burritos and Hot-dogs, SSPBH NPBH =
NT/# columns in DB3 = 3.
NPBH = 99/3 = 33.
PROGRAM: SSPBH = (1/NTr) ([1]1,11 [A]11,9 [DB3]9,3+)2 - CORRFACT
= (1/33) (SUM1 x DB3)2 - CORRFACT
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
HP-48G PROGRAM
RCL SUM1CARTS
RCL DB3
x
(
TRN
x
RCL NPBH
/
RCL CORRFACT
-
STO SSPBH
STEP 14: Computation of the sum of squares for the squaring of SUM1: SSTrxC. NTrxC = NT/# elements in SUM1 = 9. NTrxC = 99/9 = 11.
PROGRAM: SSPBHxCARTS = (1/NPBHxCARTS) ([1]1,11 [A]11,9)2 - CORRFACT - SSCARTS - SSPBH
=(1/11)([SUM1CARTS]1,9 [SUM1CARTS]1,9T) - CORRFACT - SSCARTS – SSPBH
[pic]
[pic]
[pic]
HP-48G PROGRAM
RCL SUM1CARTS
(
TRN
x
RCL NPBHxCARTS
/
RCL CORRFACT
-
RCL SSC ARTS
-
RCL SSPBH
-
STO SSPBHxCARTS
STEP 15: Computation of the within subjects error term, SSErrw.
PROGRAM AND SOLUTION: SSErrw = SSW - SSTr - SSTrxC = 286 - 170 - 19 = 97.
STEP 16: Tabling of data and calculation of degrees of freedom df.
SST = 467 df = # measures - 1 = 99 - 1 = 98
SSb = 181 df = # subjects - 1 = 33 - 1 = 32
SSCARTS = 39 df = # groups - 1 = 3 - 1 = 2
SSErrb = 142 df = dfSSb - dfSSC = 32 - 2 = 30
SSW = 286 df = dfSST - dfSSb = 98 - 32 = 66
SSPBH = 170 df = # trials - 1 = 3 - 1 = 2
SSPBHxCARTS = 19 df = dfSSTr x dfSSC = 2 x 2 = 4
SSErrw = 97 df = dfSSW - dfSSTr - dfSSTrxC = 66 - 2 - 4 = 60
STEP 17: Computation of F.
F = SS o df-1 o ERROR-1
[pic]
[pic][pic][pic]
[pic] [pic]
467 1/98 .2113 [1]8,1
181 1/32 .2113
39 1/2 .2113
142 1/30 .2113 =
286 1/66 .6186
170 1/2 .6186 19 1/4 .6186
97 1/60 .6186
F p
1.0068 - Hence it is concluded:
1.1950 - 1. The location of the carts significantly
4.1197 (.05 affected the overall amount sold
1.0000 ; - 2. Sales improved as the day went by
2.6806 -
52.5773 (.001 3. The items sold at different rates.
2.9381 (.05
1.0000 -
HP-48G PROGRAM
RCL SS
8
(
MATH,MATR,NXT,DIAG(
RCL df
8
(
DIAG(
INV
x
RCL ERROR
((TEACH
EXAM,PRG,APLY
x
STO F
POSTSCRIPT
Suppose the cost of each hot-dog is 15¢, the cost for each burrito is 35¢ and the cost for each polish sausage is 50¢. Then the total profits can be computed as:
[pic]
[pic]
But since the values in the second column are costs, it might be more illustrative to write the problem as:
[pic]
[pic]
CHECK: In the first half-hour, he brought in $32, there were $5 in wages paid out, and the total cost is:
[pic]
So his total profit is 32-5-9.7= $17.30.
One more check, in the fifth half-hour, he brought in
[pic]
Which also checks out.
THE PHILOSOPHICAL MEANING WHEN (AIJToBJK = ( 3-26/27-97
Suppose we take every atom in existence in all the galaxies of the Universe and we give each of them a place in the Grand Matrix. The zeros are empty space, and everything in creation has its unique matrix element or position in the Grand Matrix, even every Soul has to be somewhere. This is the Matrix Aij, the Datastream Matrix. it is the positioning and accounting Matrix for everything in creation. When the columns of the Datastream Matrix AijT are half-multiplied, we get the sum of All of creation at the moment of multiplication. We can identify position and account for everything in all the galaxies of all the stars of all the worlds and planets, their proper motions and directions through space, even to the vagaries that live on and between worlds (if we had a way to keep track of them, that is). So you see, with this equation we can keep track of Everything, but only at the moment of half-multiplication. We could continuously re-half-multiply by [1]1i and get a ‘picture ‘ or more likely a series of pictures ( like motion picture film) of the changes in motion for any given particle. Even waves in space are positioned somewhere! Even electrons in their orbits are accounted for, for they have their place in the Grand Matrix. Suppose the speed of the half-multiplier operator approaches the speed at which reality is perceived, we can keep up to date track of EVERYTHING in Existence! As it happens! (at least up to the last half- multiplication). And we don’t need calculus, nor algebra, or differential equations, all we need to know is how to add, subtract, multiply & divide! (and maybe a logarithm or two, just to satisfy the mathematics). And Accounting/Inventories is directly related to Statistics! Let the grand Sum of all creation be:
Grand Sum of everything in creation. [1]1,i[A]ij [1]j,1
[1]1,i[A]ij = The sum of the columns of Aij, here we keep track of each individual element.
[A]ij [1]j,1 = Sum of the rows of the Matrix These are the coordinates for each individual item in the Grand Matrix.
Lets limit the size of the Matrix and for every row in the Matrix we will put the name of every person in the world. In each column we put everything that is up for sale in the world. One’s. twos, threes, whatever quantity anyone buys or sells, the amounts exchanged, from whom to whom (buyer/seller) and for how much can easily be calculated for each person. The only practical restraint is memory capacity of the computer. For each new item added to the matrix we must add 3 billion rows, one for each person in the world. That’s 3 Gigabytes of memory for each item for sale to everyone in the world. This Grand Matrix of all things is [A]Tij o [ B]jk = ( [C]jk
Then: [1]J1 o [CJK] Accounts for everything bought and sold, keeping account of each individual row.
And: [1]1i o [A]IJ gives total sum of everything bought & sold and whatever data you may wish to include. Sums every column.
[1]J1 o [A]Tij Sums a total inventory of parts to a piece by piece amount for every single item bought/sold in the world.
[1]1i x [A]iJ Total summing of the columns, total sum of everything bought/sold in the world.
[1]iJ x [A]ij x [DB1]jk (Elementary statistics) Sums which groups you wish to add together.
1/N ([1]iJ x [A]ij[DB]jk)2 Analysis of variance of the chosen groups under study. We may be limited to the statistics we can do here, i.e., we can compare sales of brown shoes in comparison to sales of black shoes (even to a particular brand verses a similar brand item) and compare the sales. Or we can do this comparing sales of all brands and comparable items and their volume (inventory) and sales of each compared to the other. But comparing sales of apples and oranges or shoes Vs refrigerators isn’t practical unless we are looking for statistical relationships between the sales.
EXAMPLE: FARMLAND INDUSTRIES RECIPIE/EMPLOYEE HOUR/WAGE INVENTORY
Suppose we need to run a daily inventory for a company that makes sausages and bologna. The recipes are given below. All we wish to know is how much meat and spices are used daily. Note: This example is full of matrices that run off the page. When this is the case, I will try to place the complete matrix underneath the incomplete display.
THE RECIPIES
The basic equation for equating all ingredients in the above recipes to lbs is :
Nx1 lb = (1/N)oz 1 lb/16 oz. I assume 3 t =1/2 oz or 2 T = 1 oz
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BEEF LOG SUMMER SAUSAGE
1 LBS HAMB = 1 LB HMB
2.5t PEP = 2.5/5X6X16=.0052 lbs 1.25 t garlic = .0026 lbs
2.5 t TENDERIZER = .0052lb = oz/(5x6x16) = 480 3.75 t salt = .0078
3.75 t salt = .0078 2.5 t must seed = .0052
1.25 t garlic = .0052/2 = .0026 2.5 t pepper = .0052
2.5 t mustard seed = .0052 2.5 tsp liqsmk = .0052
2 t celery seed = 2/5x6x16 = .00417 2.5 t tenderizer = .0052
1 t dill = 1/480 = .00208
2.5 t liqsmk = .0052
BEEF BOLOGNA HERTERS SAUSAGE
lb = oz/(16x25) = oz/400 1lb = 1/(34x16) = 1/544
2.5 LBS BEEF = 1 lb water 4 oz = .00735 lbs
tenderizer = 1/25 = .04 lbs pepper 1.5 oz = 1.5/544 = .00276
1 oz pep. = .0625/5 = .0025 ginger 1 oz = 1/544 = .00184
1/2 cup sugar = 4/16 = .25/25 = .0100 nutmeg 1.5 oz = .00276 lbs
3 T liq smk = 1.5oz/25 = = .00375 allspice 1/2 oz = .000919
2 t saltpeter = .333/400 = .0008333 paprika 1/2 oz = .000919
garlic 1/3 oz = .000613
onion 1/3 oz = .000613 lbs
salt 12 oz = .02205
dry milk 8 oz = .0147
liqsmk 1.5 oz = .00276 lbs
THE DATABASE MATRIX
All ingredients are listed in pounds
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HMB PRK LQSMK PEP GIN NTMG ALSP PAP GAR ONION SALT DMLK TNDR MSTD SUG CLSD WT DIL SP
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The first row represents Herters sausage, the second row represents the beef log, the third row represents the summer sausage and the fourth row represents the beef bologna.
We now need to sum each row of the matrix to find the correction factors. We need a column matrix of 19 rows and one column:
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[pic]To compute the corrected matrix for final product, diagonalize this matrix, take it’s inverse and multiply to R:
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THIS PROCESS CHECKS TO MAKE SURE THE VALUES ARE CORRECTED SO THAT ONE LB OF FINISHED SAUSAGE OR BOLOGNA EQUALS ONE POUND OF INGREDIENTS. The values of the sums of the rows should equal one.
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TO FIND OUT HOW MUCH INGREDIENTS WERE USED IN ONE DAY: Suppose the final account for the day gives
50,000 LBS HERTERS SAUSAGE
45,000 LBS BEEF LOG
62,000 LBS SUMMER SAUSAGE
24,000 LBS BEEF BOLOGNA
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THE SOLUTION SAYS THAT FOR A TOTAL OF 171,000 POUNDS OF FINISHED PRODUCT, WE USED 140,174 LBS OF HAMBURGER, 23,650 LBS OF PORK, 731 LBS OF LIQUID SMOKE, 683 LBS OF PEPPER, 100 LBS OF GINGER, ETC.
LET'S DO A ROUGH CHECK ON THE AMOUNTS OF HAMBURGER AND PORK TO CHECK TO SEE IF THE SOLUTION IS CORRECT:
50,000x.5 = 25,000 LBS HAMBURGER.
45,000 x 1 = 45,000 LBS HAMBURGER
52,000X1=52,000 LBS HAMBURGER
24,000x1=24,000 LBS HAMBURGER
25,000+45,000+52,000 +24,000 = 146,000 LBS HAMBURGER
NOW LET'S CHECK THE ROUGH AMOUNT OF PORK:
.5x50,000 = 25,000 LBS PORK.
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THE CORRECT ANSWER FOR PORK IS 23,643 LBS WHICH IS WITHIN ROUNDING ERROR OF THE SPREADSHEET I'M USING.
Of course we can include the cost of each of the ingredients, federal and state taxes, and employee wages if we wish.
.PROGRAM CONDENSED
DAILY TOTAL = [50,000 45,000 52,000 24,000] changes daily THESE TOP THREE ARE THE DATA
R = [DATABASE] stays the same unless recipes change
NINETEEN1 = [1]19,1 stays the same unless recipes change
SUMR = R x NINETEEN1 THESE 5 STEPS ARE THE PROGRAM
T = diag(SUMR)-1
CORRECTEDTOTAL = T x R
FINALINVENTORY = DAILYTOTAL x CORRECTEDTOTAL
PRINT
Suppose that instead of just one department, the sausages and bologna are made in four different departments. We wish to keep track of not only the total inventory, but the amounts of ingredients used by each Department. The total usage for the day is:
H.S B.L S.S. BOL
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Suppose Dept. 1 made 20,000 lbs Herter’s sausage, 10,000 lbs beef log, 12,000 lbs summer sausage and no bologna.
Dept. 2 made 5,000 lbs Herter’s sausage, 0 lbs beef logs, 25,000 lbs summer sausage and 6,000 lbs bologna.
Dept. 3 made 25,000 lbs Herter’s sausage, 15,000 lbs beef logs, 0 lbs summer sausage and 12,000 lbs bologna.
Dept. 4 made 0 lbs Herter’s sausage, 20,000 lbs beef logs, 15,000 lbs summer sausage and 6,000 lbs bologna.
Then the inventory matrix becomes:
50,000 45,000 52,000 24,000
20,000 10,000 12,000 0
5,000 0 25,000 6,000
25,000 15,000 0 12,000
0 20,000 15,000 6,000
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HAMBURGER PORK LQ.SMK PEP GING NTMG ALSP PAP GAR ON SALT DRYMLK TNDR MSTD SUG CELERY WTR DILL SP.
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DEPT.1 used 30,740 lbs hamburger, DEPT.2 used 32,291 lbs hamburger, DEPT. 3 used 11,825 lbs pork and 645 lbs salt. DEPT. 4 used 39,506 lbs hamburger, 199 lbs liq. smoke, 175 lbs mustard and 6 lbs salt peter.
Just for ease in the computations, suppose that out of the 50,000 lbs of Herter's sausage produced, 1,000 lbs was waste, of the 45,000 lbs beef logs, 800 lbs was waste, of the 52,000 lbs summer sausage, 1,500 lbs was waste and of the 24,000 lbs bologna, 500 lbs was discarded. What is the new total inventory? To see what I am doing, I will use just the DAILY TOTAL matrix, not the DEPARTMENT TOTALS matrix.
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To find the total used that was good we add the two rows together:
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For the day, we passed 137,001.8 lbs hamburger, 23,177 lbs of pork, 1366.5 lbs of salt and 23.5 lbs salt peter, etc.
Let’s look at this problem in another light. Suppose that instead of computing the inventory at the end of the day, we wish to manufacture 50,000 lbs of Herter’s sausage, 45,000 lbs of beef logs, 52,000 lbs of summer sausage and 24,000 lb.s of bologna. We want to calculate the recipes. We will compute the recipes for all four departments as well as the total amount of ingredients. We will use the corrected database and DEPARTMENTTOTALS matrix, but with a twist, we will use t.he account page matrix to obtain the desired proportions. i.e.
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THE INGREDIENTS FOR BOLOGNA (TO MAKE 24,000 LBS FINAL PRODUCT) ARE 22,704 LBS HAMBURGER, 96 LBS LIQUID SMOKE, 48 LBS PEPPER, 912 LBS TENDERIZER, 216 LBS SUGAR AND 24 LBS SALT PETER. IN THIS RECIPIE, THE TENDERIZER CONTAINS THE SALT, SO I DO NOT INCLUDE SALT IN THE RECIPE. THIS IS THE TOTAL FOR ALL DEPARTMENTS. LET'S SEE WHAT INGREDIENTS ARE NEEDED FOR DEPARTMENT ONE.
MATHCAD CANNOT REMOVE A ROW OF A MATRIX, BUT IT CAN REMOVE A COLUMN. SINCE DEPARTMENT ONE IS REPRESENTED IN ROW 2, WE MUST TRANSPOSE THE DEPARTMENTTOTALS MATRIX AND REMOVE THE SECOND COLUMN, DIAGONALIZE IT AND MULTIPLY TO THE CORRECTED TOTAL MATRIX. WE MUST CHANGE THE ORIGIN OF THE MATRIX FROM 0 TO 1. i.e.
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FOR THE INGREDIENTS NEEDED BY DEPT. 1 TO MAKE 12,000 LBS OF SUMMER SAUSAGE, WE NEED 11,640 LBS HAMBURGER, 60 LBS LIQ. SMOKE, 60 LBS PEPPER, 36 LBS GARLIC, 96 LBS SALT, 60 LBS TENDERIZER AND 60 LBS MUSTARD.
LET'S NOW LOOK AT THE COST TO MAKE THE FINISHED PRODUCT. THE PRICES OF THE INGREDIENTS MUST BE IN DOLLARS PER POUND. SUPPOSE
HAMBURGER = .50/LB
PORK = .55/LB
LIQUID SMOKE = 1.00/LB
PEPPER = 2.50/LB
GINGER = .75/LB
NUTMEG = 2.75/LB
ALLSPICE = 2.35/LB
PAPRIKA = .35/LB
GARLIC = 1.00/LB
ONION = 1.00/LB
SALT = .35/LB
DRIED MILK = 2.50/LB
TENDERIZER = 1.25/LB
MUSTARD = .75/LB
SUGAR = .40/LB
CELERY SEED = 1.00/LB
WATER = .01/LB
DILL = 1.00/LB
SALT PETER = 1.50/LB
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IF WE WISH AN ACCOUNTING OF EVERY INDIVIDUAL COST, WE CAN COMPUTE THE ACCOUNT PAGE MATRIX.
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LET'S CHECK THE SOLUTIONS OF THE PRODUCTS THAT COST $1.00/LB. THESE ARE LIQUID SMOKE, GARLIC, ONION, CELERY SEED AND DILL. THE COST SHOULD EQUAL THE NUMBER OF POUNDS USED.
LIQUID SMOKE = 731 LBS = $731
GARLIC = 341 LBS = $341
ONION = 50 LBS = $50
CELERY SEED = 180 LBS = $180
DILL = 90 LBS = $90
THE ANSWERS ALL CHECK
TO CHECK THE TOTAL, WE SUM THE ACCOUNTPAGECOSTS COLUMN:
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BUT SOME OF THE PRODUCT WAS QA'd OUT, HOW MUCH MONEY WAS LOST?
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PROGRAMS CONDENSED
TO FIND POUNDS USED
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TO FIND POUNDAGE PER DEPARTMENT
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TO FIND FINAL POUNDAGE PLUS WASTE POUNDAGE
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TO FIND THE TOTAL USED THAT WAS GOOD MINUS THE WASTE
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TO FIND INGREDIENTS NEEDED
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COST TO MAKE FINISHED PRODUCT
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ACCOUNTPAGE FOR COSTS
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TO FIND MONEY LOST
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TO FIND INGREDIENTS NEEDED
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COST TO MAKE FINISHED PRODUCT
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ACCOUNTPAGE FOR COSTS
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TO FIND MONEY LOST
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LET'S LOOK AT OUR EMPLOYEES AND THEIR WAGES. SUPPOSE DEPT. 1 THROUGH DEPT. 4 HAVE TWO EMPLOYEES EACH (THIS IS JUST TO KEEP THE EXAMPLE TO A MANAGABLE SIZE). THE EMPLOYEES AND THEIR WAGES ARE:
DEPARTMENT 1: CHRIS $10/HR; $15/HR OT
FRAN $ 9/HR; $13.50 OT
DEPARTMENT 2: TINA $11/HR; $16.50 OT
BRUCE $8.50/HR; $12.75 OT
DEPARTMENT 3: JEANIE $10.50/HR; $15.75 OT
RAY $9.50/HR; $14.25 OT
DEPARTMENT 4: PENNY $12.00/HR; $18.00 OT
HELEN $11.50/HR; $17.25 OT
THE COMPANY WORKS 8 HOUR SHIFTS 6 DAYS A WEEK. WE WILL COMPUTE THE PAY MATRIX FIRST AS A SEPARATE ENTITY, THEN TRY TO CONSOLIDATE IT INTO THE COSTS MATRIX.
OR
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I'LL JUST GO AHEAD AND COMPUTE WAGES PLUS OVERTIME; TOTAL PAID OUT AND TOTAL TO EACH EMPLOYEE.
TOTAL PAID OUT BY COMPANY:
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I'LL JUST GO AHEAD AND COMPUTE WAGES PLUS OVERTIME; TOTAL PAID OUT AND TOTAL TO EACH EMPLOYEE.
TOTAL PAID OUT BY COMPANY:
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IF THE EMPLOYEES WORKED A REGULAR WORKDAY, THEIR WAGESWOULD BE $656, IF THEY WORKED OVERTIME, THEIR WAGES FOR THE DAY WOULD BE $984.
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WHEN I ADD THIS TO THE COSTS MATRIX, I WILL NOT DEDUCT FOR THE EMPLOYEES TAXES. THE EMPLOYEES WAGES WILL GO TOWARD THE COST OF MANUFACTURING THE SAUSAGES AND BOLOGNA. BUT HERE WE CAN COMPUTE THE TAXES AND WITHHOLDINGS. I DO NOT KNOW THE FEDERAL INCOME TAX WITHHOLDING FORMULAS, SO I WILL MAKE THE PERCENT DEDUCTION THE SAME PERCENTAGE AS THEIR OVERTIME WAGES. SOCIAL SECURITY WILL BE 10% AND STATE TAXES WILL BE 6%.
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TO ITEMIZE FOR FEDERAL, STATE AND SOCIAL SECURITY, LEAVE OUT THE MULTIPLICATION BY THREE1.
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FOR THE PURPOSES OF COMPUTING THE TOTAL COSTS OF PRODUCING THE SAUSAGES AND BOLOGNA, WE WILL ASSUME THE EMPLOYEES ARE PART OF THE INGREDIENTS IN MAKING THE PRODUCE.
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NOW LET'S THROW AWAY EVERYTHING WE HAVE DONE ABOVE THIS POINT AND COMPUTE EVERYTHING ALL AT THE SAME TIME, INCLUDING LOSSES. BECAUSE WE DO NOT WISH TO INCLUDE EMPLOYEE WAGES IN THE FIRST MULTIPLICATION, WE WILL INTRODUCE WHAT I WILL CALL TEMPLATE MATRICES.
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NOTE: THE HOURTEMP MATRIX WILL ALWAYS BE THE SAME UNLESS WE HIRE MORE EMPLOYEES; THE CORRTEMP MATRIX WILL ALWAYS BE THE SAME UNLESS WE CHANGE THE RECIPES OR ADD NEW ONES OR DELETE OLD ONES. THE [pic] MATRIX WILL CHANGE DAILY. THE COSTS MATRIX WILL ALWAYS BE THE SAME UNLESS THERE IS A CHANGE IN PRICE OF THE INGREDIENTS OR AN EMPLOYEE GETS A RAISE. WE JUST CHANGE THOSE ITEMS AND PROCEED AS FOLLOWS. IN OTHER WORDS, ONCE THE DATABASES ARE ENTERED, WE NEED ONLY TO KEEP TRACK OF WHAT IS PRODUCED AND WASTED EACH DAY.
NOW THE ABOVE MATRIX IS FINE, BUT I AM NOT SATISFIED WITH IT. WE CANNOT DO SIMPLE STATISTICS AS IT STANDS, SO I AM GOING TO PUT THE WASTE PER DEPARTMENT IN THE [pic] MATRIX. WE'LL LET THE COMPUTER DO ALL THE EXTRA MATH FOR US.
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BEFORE WE ADD IN THE EMPLOYEES HOURS, I'LL COMPUTE BOTH THE WASTE AND THE TOTAL POUNDAGE THAT IS GOOD. FOR SIMPLICITY, ALTHOUGH WE DO NOT HAVE TO DO THIS, I'LL GET RID OF THE TOTAL POUNDS AND LEAVE ONLY THE WASTE POUNDAGE.
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THE COST IN DOLLARS FOR THE WASTEIS GIVEN BY:
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THE COSTS OF THE DEPARTMENT WASTES SHOULD EQUAL THE TOTAL WASTE ($2025.81):
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LET'S NOW LOOK AT THE TOTAL POUNDAGE THAT PASSED. DB1 IS A STATISTICAL DATABASE MATRIX THAT IS USED FOR ACCOUNTING PURPOSES:
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LET'S LOOK AT THE % LOSSES. % ERROR =100x (AMOUNT PASSED - TOTAL AMOUNT)/(TOTAL AMOUNT)
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NOTE: MATHCAD CANNOT TAKE THE INDIVIDUAL INVERSES OF THE ABOVE MATRIX BECAUSE IT CANNOT
DIVIDE BY ZERO, SO I HAD TO COMPUTE THE INVERSES BY HAND. THE EASIEST WAY WOULD NOT BE TO ITEMIZE THE % ERROR BUT TO COMPUTE THE TOTAL ERROR.
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DEPT. 2 HAD THE HIGHEST PERCENT WASTE, IT WASTED 6% OF THE PORK. DEPT. 3 HAD THE BEST SHOWING.
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THE TOTAL % WASTE = 2.22%; THE % WASTE FOR DEPT.1 = 2.86%; THE % WASTE FOR DEPT. 2 = 2.22%, ETC.
NOW LET'S GO AHEAD AND ADD IN THE EMPLOYEE HOURS. NOTE THAT THIS IS THE EASIEST WAY TO COMPUTE WAGES, BUT IT IS NOT THE BEST WAY. I MEAN, SUPPOSE ONE EMPLOYEE WORKED 8 HOURS A DAY IN THE REGULAR WORKWEEK, ANOTHER WORKED 7 HOURS THAT DAY, BUT BOTH ONLY WORKED FOUR HOURS SATURDAY? THIS EXAMPLE IS FOR DEMONSTRATION. THE OVERTIME PROBLEM IS SIMPLE BUT I WON'T GO INTO THAT HERE YET. I’LL JUST DEMONSTRATE THE PROBLEM AS DESCRIBED ABOVE.
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THIS IS GOOD,BUT I DON'T LIKE IT, WHAT IF WE WERE WORKING SATURDAY? LET'S RE-DEFINE THE COSTS MATRIX AS:
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IF WE RAN THIS ON A WEEKDAY, THE PRODUCTION COSTS WOULD BE:
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THE TOTAL COSTS FOR THE WORKDAY WOULD BE $92,097.50 FOR THE DAILY TOTAL, $22,865 FOR DEPT. 1; 19,141.20 FOR DEPT. 2; $28,495.20 FOR DEPT. 3 AND $21,596.10 FOR DEPT. 4.
IF THIS PRODUCTION WERE RUN ON SATURDAY, THE COSTS WOULD BE:
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THE TOTAL COSTS FOR THE SATURDAY WITH OVERTIME WOULD BE $92,425.50 FOR THE DAILY PRODUCTION TOTAL; $22,941 FOR DEPT. 1; $19,219.20 FOR DEPT. 2; $28,575.20 FOR DEPT. 3 AND $21,690.10 FOR DEPT. 4.
NOW THIS WOULDN'T BE A PROPER INVENTORY SYSTEM IF WE COULD NOT KEEP TRACK OF OUR TOTAL INVENTORY ALSO. WITH DECIMALS, THE MATRICES ARE TOO LARGE TO FIT ON THIS PAGE, SO I AM GOING TO ROUND OFF TO THE NEAREST POUND.
HMB PRK LQSMK PEP GIN NTMG ALSP PAP GAR ON SLT DMLK TNDR MSTD SUG CLSD WTR DIL SP
SUPPOSE IN-HOUSE WE HAVE:
1,000,000 LBS HAMBURGER
500,000 LBS PORK
100,000 LBS LIQUID SMOKE
50,000 LBS PEPPER
25,000 LBS GINGER
30,000 LBS NUTMEG
20,000 LBS ALLSPICE
75,000 LBS PAPRIKA
80,000 LBS GARLIC POWDER
100,000 LBS ONION POWDER
150,000 LBS SALT
55,000 LBS DRY MILK
125,000 LBS TENDERIZER
100,000 LBS MUSTARD
20,000 LBS SUGAR
40,000 LBS CELERY SEED
400,000 LBS WATER
15,000 LBS DILL SEED
99,000 LBS SALT PETER
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HMB PRK LQSMK PEP GIN NTMG ALSP PAP GAR ON SLT DMLK TNDR MSTD SUG CLSD WTR DIL SP
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EXAMPLE: ARGONIA STORE INVENTORY
SUPPOSE WE HAVE A STORE THAT SELLS 9 ITEMS. THE STORE IS IN A TOWN THAT HAS 10 RESIDENTS. THE 9 ITEMS, THEIR PRICES, THE SHOPKEEPERS COST TO BUY THE ITEMS, AND THE SALES TAX ARE LISTED BELOW. The shopkeeper wishes to keep a list of what is bought and who buys it, how much money she brought in, how much tax was collected, employee wages and total profits for the day. She also wishes to know how much of yesterdays inventory is left after today’s sales.
LET TOM CEL BEEF CHICK PORK MILK EGGS BEER
LARRY 2 3 1 0 0 0 2 1 0
HELEN 0 1 0 2 4 0 1 1 0
TANYA 1 1 0 1 2 0 1 0 0
BECKY 0 2 0 2 3 1 0 1 1
DON 0 0 0 4 2 2 1 0 0
SHANNA 2 1 1 1.5 1 0 2 1 0
JEANIE 1 0 1 2 2 0 1 1 1
LORI 1 0 0 3 3 2 1 0 1
BETTY 0 1 0 2 0 2 2 1 0
PAULA 1 0 1 3 1 0 1 1 0
PRICE COST TAX
L = LETTUCE (HEADS) $.99 $.75 .06
T = TOMATO'S (POUNDS) $1.09 $.50 .06
C = CELERY (STALKS) $.69 $.35 .06
B = BEEF (POUNDS) $2.29 $1.45 .06
C = CHICKEN (POUNDS) $1.59 $.85 .06
P = PORK (POUNDS) $2.25 $1.65 .06
M = MILK (HALF-GALLONS) $1.59 $1.00 .06
E = EGGS (DOZEN) $1.09 $.55 .06
B = BEER (SIX-PACK) $2.50 $1.25 .06
WE PUT THE SALES INVENTORY INTO MATRIX FORM AND NAME THE MATRIX SALES. WE THEN COMPUTE THE 6% TAX ON THE SALE PRICE AND ENTER IT INTO THE THIRD COLUMN. THE COST TO BUY THE ITEMS ARE ENTERED AS NEGATIVE VALUES IN THE SECOND COLUMN. WE NAME THIS MATRIX DB FOR DATABASE.
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THE TOTAL SALE TO LARRY IS $10.21, SALES TAX IS 61CENTS, YOUR COST IS $5.90 FOR A PROFIT OF 10.21 - 5.90 = $4.31. THE TOTAL SALE TO BOB IS $21.22, HE PAID 1.27 IN TAXES AND YOU MADE A PROFIT OF 21.22 - 13.20 = $8.02. LET'S FIND YOUR TOTAL PROFITS PER CUSTOMER:
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THESE VALUES SHOULD MATCH THE CASHREGISTER RECEIPTS.
BESIDES YOU, YOU HAVE TWO EMPLOYEES, ONE WORKS PART-TIME, THE OTHER WORKS FULL-TIME. THE PART-TIME WORKER MAKES $6.00/HR AND WORKED 20 HOURS THIS WEEK, THE FULL-TIME WORKER MAKES $7.50/HR AND WORKED 40 HOURS. BUT SUPPOSE THE ABOVE SALES INVENTORY WAS A ONE DAY TOTAL, AND THE FULL-TIME WORKER WORKED 8 HOURS WHILE YOUR PART-TIME HELP WORKED 3 HOURS.
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THE MATRIX IS TOO BIG FOR MATHCAD TO ALLOW US TO ADD NEW ROWS AND COLUMNS, SO WE MUST USE THE AUGMENT AND STACK FUNCTIONS TO INCREASE THE SIZE OF OUR INVENTORY MATRIX:
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WE PLACE THE EMPLOYEES HOURS IN THIS MANNER BECAUSE WE DO NOT WISH TO INTERFERE WITH CUSTOMER SALES INVENTORY OR SALES ACCOUNTING.
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THE FULL-TIME EMPLOYEE EARNED $60, AND THE PART-TIME EMPLOYEE EARNED $18. SO NOW, WHAT IS YOUR TOTAL PROFIT FOR THE DAY?
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TODAY WE MADE NO PROFIT, WE DID NOT LOSE ANY MONEY BECAUSE WE BROUGHT IN (MINUS WAGES):
[pic]
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WE HAVE $68 LEFT IN THE TILL, BUT MADE A PROFIT OF -$20.06 FOR THE DAY.
SUPPOSE WE ARE HAVING A SALE ON BEER AND TOMATOES, AND TO SAVE ON COSTS, WE WISH TO MAIL FLYERSOUT ONLY TO THOSE CUSTOMERS WHO BUY BEER AND TOMATOES. WE PROCEED TO HALF-MULTIPLY AS FOLLOWS:
[pic] [pic] [pic]
HERE WE REMOVE THE SECOND ROW (TOMATOES) AND THE NINETH COLUMN (BEER) FROM THE SALES MATRIX:
[pic] [pic]
[pic]
THE PROOFS SAY WE CAN ADD BOTH COLUMNS TOGETHER AND OBTAIN THE SUM OF BOTH INSTEAD OF FINDING THE SOLUTIONS FOR EACH COLUMN SEPARATELY AND ADDING THE TWO MATRICES. WE GET THE SAME ANSWER.
[pic]
NOW WE MUST HALF-MULTIPLY, BUT SINCE NO ONE HAS EVER DISCOVERED THIS TYPE OF MULTIPLICATION BEFORE, IT CAN NOT BE DONE ON ANY COMPUTER ON EARTH TODAY, SO WE MUST DIAGONALIZE THE COLUMN MATRIX AND MULTIPLY TO THE SALES MATRIX TO HALF-MULTIPLY.
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WE MAIL ADS FOR TOMATOES AND BEER TO THOSE PEOPLE WHO DO NOT HAVE A ROW OF ZEROS. SUPPOSE WE JUST HAVE A SPECIAL ON BEER?
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WE SEND ADS ONLY TO BECKY, JEANIE AND BOB.
FOR THE INVENTORY LEFT IN THE STORE, WE PROCEED AS FOLLOWS:
[pic] INVENTORY AS OF YESTERDAY
[pic]
[pic] SUMS THE COLUMNS IN THE SALES MATRIX
[pic] HERE WE FIND THE NEW IN-STORE INVENTORY
[pic]NEW INVENTORY
WE HAVE ONLY 8 HALF-GALLONS OF MILK LEFT, IT IS TIME TO ORDER MORE. WE MIGHT AS WELL ORDER SOME MORE BEER SINCE WE ARE RUNNING LOW ON IT TOO. WE HAVE ENOUGH OF THE REST TO LAST FOR AWHILE. i.e. 17 HEADS OF LETTUCE LEFT, 91 tomatoes left, 28 stalks of celery, 54.5 lbs of beef, 17 pounds of chicken, 35 pounds of pork and 18 dozen eggs.
LOVE BOX COMPANY BASIC INVENTORY PROGRAM
The following is an inventory system that keeps track of the different parts that go into making-up a single finished product. It is sort of simple, sort of complex, but if we do not have to keep track of everything entered here, just the inventory, the system would become much simpler to compute. The matrix SHIFT123 gives the total products completed by each shift. The first row is first shifts production, the second row is 2nd shifts daily production and the third row the 3rd shifts production. The matrix SHIFT123T gives the maximum total of items that should have been produces with no losses. The DATABASE matrix gives each product (the rows) and the columns give how many parts of each are needed to complete each item. The last column gives the total area in square feet for the finished item. We want to know the total square feet produced, square feet lost, the number of items produced per hour and the cost as applied to worker hours to produce each piece.
4 3 4 3
0 2 0 2
0 1 0 0
5 5 5 5
0 0 4 4
0 0 0 0
0 0 1 1
[pic] [pic]
4 4 3 3 4 4 3 3 S
0 0 2 2 0 0 2 2 Q
0 0 1 1 0 0 0 0 U
/ / / / / / / / A
5 5 5 5 5 5 5 5 R
0 0 0 0 4 4 4 4 E
0 0 0 0 0 0 0 0 F
0 0 0 0 1 1 1 1 E
A B A A2 A1 A E
T
[pic]
Total Produced 3000 1500 7500 1500 3000 1500 3000 1500 52687.65
400-5000 2 1 0 0 0 0 0 0 6.5078
321-5000 0 0 5 1 0 0 0 0 13.0156
400-5401 0 0 0 0 2 1 0 0 10.5023
320-5401 0 0 0 0 0 0 2 1 5.0994
The columns represent the number of pieces needed to make each item. i.e. item 321-5000 needs 5 pieces of item 321-5000A and 1 piece of item 321-5000. Of the items 321-5000A there were 7500 made and 1500 pieces of 321-5000.
[pic]
[pic] [pic]
[pic]
This gives the total number of pieces completed in one shift for all three shifts.
The following list gives the total number of pieces made by the corrugator for the three shifts.
[pic]
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This computation defines the total number of pieces for all three shifts.
[pic]
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This is the loss in pieces and square footage for all three shifts combined.
This is the theoretical inventory assuming no pieces are lost.
[pic]
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To compute the losses for each shift we proceed as follows:
[pic]
We must divide the totals of the corrugator by three to get the total for each shift.
[pic]
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50 pieces lost of 400-5000A, 24 pieces lost of 400-5401A2 were lost and first shift lost a total of 203.49 square feet. First shift lost 30 pieces of 400-5000A and second shift lost 20 pieces while third shift lost none.
To check if this correct, we sum each column and compare to the total losses for all three shifts.
[pic]
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Suppose we wish to keep track of the employees time per task:
[pic]
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First shift took 1.75 hours to complete the first project, second shift took 2.5 hours to complete the second project and third shift took 1 hour to complete the third project, etc.
[pic]
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Now lets compute the pieces per hour:
First let's add the total pieces for each product:
[pic] [pic]
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Here we sum the total number of separate pieces for each order.
[pic]
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Therefore for Kawasaki 400-5000, first shift did 831.43 pieces per hour, for Kawasaki 321-5000, second shift did 1200 pieces per hour, for Kawasaki 400-5401, third shift did 1470 pieces per hour and for kawasaki 320-5401, second shift did 848.57 pieces per hour.
Let's look at employee wages and the cost to produce each piece.
Now let's compute the cost per piece.
[pic]
Combined pay second shift = $18.06
combined pay first shift = $20.75
Combined pay third shift = $17.06
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Let's check to see if the above problem is correct. First shift produced 485 pieces in 1.75 hours with a combined pay of $20.75. The cost is 20.75 x 1.75 = $36.21. So cost per piece would equal 36.3125/485 = .074871.
MATRIX SOLUTION FOR GAUSSIAN REDUCTION
In this section, I will show how I discovered this operator. Mathematicians already know about this method, but I do not think anyone has ever derived it from a mathematical basis. Suppose we wish to reduce the following matrix:
a11 a12 a13 . . . a1n
a21 a22 a23 . . . a2n
Aij = a31 a32 a33 . . . a3n
. . . . . . .
. . . . . . .
an1 an2 an3 . . . ann
Now normally to perform the Gaussian Reduction, we write out the top row, then multiply the first row of the matrix by the element in another row that we are trying to reduce to zero. We multiply this second row by the first element and subtract so that the number in the row below row one equals zero under the diagonal. Instead of multiplying and subtracting by hand, one day I put zero’s in the places I wasn’t interested in and formed a column matrix from them. After playing around with the idea for awhile, this is what I came up with:
1 a11 a12 a13 . . . a1n a21 a11 a12 a13 . . . a1n
0 a21 a22 a23 . . . a2n -a11 a21 a22 a23 . . . a2n
0 a31 a32 a33 . . . a3n + 0 a31 a32 a33 . . . a3n + . . . +
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
0 an1 an2 an3 . . . ann 0 an1 an2 an3 . . . ann
ai1 a11 a12 a13 . . . a1n an1 a11 a12 a13 . . . a1n
0 a21 a22 a23 . . . a2n 0 a21 a22 a23 . . . a2n
0 o . . . . . . . +. . . + 0 o a31 a32 a33 . . . a3n
-a11 ai1 ai2 ai3 . . . ain . . . . . . . . . .
. . . . . . . . . . . . . . . . .
0 an1 an2 an3 . . . ann -a11 an1 an2 an3 . . . ann
Cross multiplying the matrices and adding we get:
a11 a12 a13 a1n
a21a11-a11a21 a21a12-a11a22 a21a13-a11a23 . . . a21a1n-a11a2n
a31a11-a11a31 a31a12-a11a32 a31a13-a11a33 . . . a31a1n-a11a3n
. . . . . . . . . . .
. . . . . . . . . . .
ai1a11-a11ai1 ai1a12-a11ai2 ai1a13-a11ai3 . . . ai1a1n-a11ain
. . . . . . . . . . .
an1a11-a11an1 an1a12-a11an2 an1a13-a11an3 . . . an1a1n-a11ann
But all the terms under a11 are equal to zero. To help keep things simple, I’m going to define b11 = a21a12-a11a22; b12 = a21a13-a11a23; bin = ai1a1n-a11ain; bnn = an1a1n-a11ann; etc.
But first, I’m going to collect all the pre-multipliers of [A]nn and form them into a matrix in the order I multiplied them.:
[pic]
And re-writing the new Matrix we get:
[pic]
Now we will get all the terms under b11 to equal zero: Note we are multiplying the same as in Gaussian Reduction except we multiply by zeros in those terms in which we are not currently interested.
1 a11 a12 a13 . . . a1n 0 a11 a12 a13 . . . a1n
0 0 b11 b12 . . . b1n-1 1 0 b11 b12 . . . b1n-1
0 0 b21 b22 . . . b2n-1 0 0 b21 b22 . . . b2n-1
. o . . . . . . . . + 0 o . . . . . . . . +
0 0 bi1 bi2 . . . bin-1 . 0 bi1 bi2 . . . bin-1
. . . . . . . . . . . . . . . . . .
0 0 bn-1,1 bn-1,2 . . bnn-1 0 0 bn-1,1 bn-1,2 . . bnn-1
0 a11 a12 a13 . . . a1n 0 a11 a12 a13 . . . a1n
b21 0 b11 b12 . . . b1n-1 bi1 0 b11 b12 . . . b1n-1
-b11 0 b21 b22 . . . b2n-1 0 0 b21 b22 . . . b2n-1 + . . . + 0 o . . . . . . . . + . o 0 . . . . . . .
0 0 bi1 bi2 . . . bin-1 -b11 0 bi1 bi2 . . . bin-1
. . . . . . . . . . . . . . . . . .
0 0 bn-1,1 bn-1,2 . . bnn-1 0 0 bn-1,1 bn-1,2 . . bnn-1
0 a11 a12 a13 . . . a1n
bn-1 0 b11 b12 . . . b1n-1
0 0 b21 b22 . . . b2n-1
0 o . . . . . . . . =
. 0 bi1 bi2 . . . bin-1
. . . . . . . . .
-b11 0 bn-1,1 bn-1,2 . . bnn-1
a11 a12 a13 a1n
0 b11 b12 b1,n-1
0 b21b11-b11b21 b21b12-b11b22 . . .b21b1n-1-b11b2,n-1 .
. . . . . . . . . . .
. . . . . . . . . . .
0 bi1b11-b11bi1 bi1b12-b11bi2 . . .bi1b1i-b11bi,n-1
. . . . . . . . . . .
0 bn-1,1b11-b11bn-1,1 bn-1,1b12-b11bn-1,2 bn-1,1b11-b11bn-1,n-1
And, again putting all the pre-multiplier column matrices into a single square matrix, and noting all the elements below b11 are equal to zero, we let c11 = b21b12-b11b22, etc., we get:
pre-multiplier for b
[pic]
[pic]
Note the inner Matrix is getting smaller by one row and column with each successive operation. The new c sub-matrix is 2 rows and two columns smaller than the size of the original a matrix.
Following the pattern, the pre-multiplier matrix for c is:
Pre-multiplier to reduce c:
1 0 0 0 0 . 0 . 0
0 1 0 0 0 . 0 . 0
0 0 1 0 0 . . . 0
0 0 0 c21 c31 . . . cn-3,1
0 0 0 -c11 0 . . . 0
0 0 0 0 -c11 . . . .
. . . . . . . .
0 0 0 0 0 -c11 0
0 0 0 0 0 0 -c11
This goes on in this pattern and we get to the I’th row:
a11 a12 a13 . . . a1n
0 b11 b12 . . . b1n-1
GI = 0 0 c11 . . . b1,n-2
. . . . . . . .
0 0 I11 . . . Ii,n-i
. . . . . . . .
0 0 In-i,1 . . In-i,n-i
1 a11 a12 a13 . . . a1n 0 a11 a12 a13 . . . a1n
0 0 b11 b12 . . . b1n-1 1 0 b11 b12 . . . b1n-1
0 0 0 c11 . . . b1,n-2 0 0 0 c11 . . . b1,n-2
. o . . . . . . . . + . o . . . . . . . . + . . . +
0 0 0 I11 . . . Ii,n-i 0 0 0 I11 . . . Ii,n-i
. . . . . . . . . . . . . . . . . .
0 0 0 In-i,1 . . In-i,n-I 0 0 0 In-i,1 . . In-i,n-I
0 a11 a12 a13 . . . a1n 0 a11 a12 a13 . . . a1n
0 0 b11 b12 . . . b1n-1 0 0 b11 b12 . . . b1n-1
0 0 0 c11 . . . b1,n-2 0 0 0 c11 . . . b1,n-2
. o . . . . . . . . + . o . . . . . . . . + . . . +
1 0 0 I11 . . . Ii,n-I I21 0 0 I11 . . . Ii,n-i
. . . . . . . . . -I11. . . . . . . . .
0 0 0 In-i,1 . . In-i,n-I 0 0 0 In-i,1 . . In-i,n-I
0 a11 a12 a13 . . . a1n
0 0 b11 b12 . . . b1n-1
0 0 0 c11 . . . b1,n-2
. o . . . . . . . . =
In-1 0 0 I11 . . . Ii,n-I
0 . . . . . . . .
-I11 0 0 In-i,1 . . In-i,n-I
The pre-multiplier matrix is:
1 0 0 . . 0 0
0 1 0 . . 0 0
0 0 1 . . 0 0
. . . . . . . . .
0 0 0 . . 1 . . .
0 0 0 . . I21 I31 In-i,1
0 0 0 . .-I11 0 . . 0
0 0 0 . . -I11 . . 0
0 0 0 . . . . . . .
0 0 0 . . 0 -I11
And the last row (which is now reduced to a 2x2 submatrix) is calculated by:
[pic]
The pre-multiplier matrix ix:
[pic]
And Gn =
[pic]
Before I go on, let me recap what I’ve done. I did the Gaussian reduction on an nxn square matrix (or an nxm augmented matrix, with the solutions = to zero). I multiplied two rows at a time and added such that the value below the diagonal under the top element would equal zero. I kept track of each hand calculated multiplication and addition process by putting them into a column matrix. When I was finished, I combined each column into a square nxn matrix. For an nxn matrix, there are n-1 cofactor matrices to set up. Rather than keep on going on an infinite nxn matrix, I will let n = 3 and we will solve for a general 3x4 augmented matrix. According to the proofs of the half-multiplier operator, we just take the pre-multiplier matrices developed by hand, transpose them and multiply. We will get the same answers.
[pic] The 3 x 4 augmented matrix.
[pic]
we must first find the values to put in for
the b matrix
[pic]
We use only the values under a12 of the first multiplication and ignore the rest.
b11 = a21 x a12 - a11 x a22
b21 = a31 x a12 - a11 x a23
[pic]
Note all the values under the principle diagonal are zero.
[pic]
For a 4x5 augmented matrix, evaluated symbolically by MathCad, we get:
[pic]
[pic]
[pic]
[pic]
[pic]= 0, MathCad doesn’t seem to want to do the subtraction.
SO OUR MATRIX HAS NOW BEEN REDUCED TO:
[pic]
WHICH IS WHERE WE WANT IT TO BE. THE VALUES FOR THE B'S, C'S AND D'S ARE: (THE EXPRESSIONS ARE WAY TOO LONG TO PUT INTO TERMS OF a) . The values for the B’s, C’s and D’s are:
B11 = A21 x A12 - A11 x A22 C11 = B21 x B12 - B11 x B22 D11 = C21 x C12 - C11 x C22
B21 = A31 x A12 - A11 x A32 C21 = B31 x B12 - B11 x B32 D11 = C21 x C2 - C11 x D2
B31 = A41 x A12 - A11 x A42 C12 = B21 x B13 - B11 x B23
B12 = A21 x A13 - A11 x A23 C22 = B31 x B13 - B11 x B33
B22 = A31 x A13 - A11 x A33 C2 = B21 x B1 - B11 x C1
B32 = A41 x A13 - A11 x A43 D2 = B31 x B1 - B11 x D1
B13 = A21 x A14 - A11 x A24
B23 = A31 x A14 - A11 x A34
B33 = A41 x A14 - A11 x A44
B1 = A21 x A - A11 x B
C1 = A31 x A - A11 x C
D1 = A41 x A - A11 x D
MathCad cannot define the above variables, so I’ve written them in Word 7.
The pre-multiplier matrices used to get zero’s below the diagonal will be denoted as ½H( . Now let’s try to complete the solution to Gaussian reduction by eliminating all off-diagonal elements going up. This set of matrices will be denoted as ½H( . I am not going to go through the problem like I did for the ½H( operator. I’m just going to derive the first Gaussian matrix and extrapolate from there following the pattern.
-D11 a11 a12 a13 a14 A 0 a11 a12 a13 a14 A 0 a11 a12 a13 a14 A
0 o 0 B11 B12 B13 B1 + -D11+ o 0 B11 B12 B13 B1 + 0 o 0 B11 B12 B13 B1 +
0 0 0 C11 C12 C2 0 0 0 C11 C12 C2 -D11 0 0 C11 C12 C2
a14 0 0 0 d11 D3 B13 0 0 0 d11 D3 C13 0 0 0 d11 D3
0 a11 a12 a13 a14 A
0 o 0 B11 B12 B13 B1 =
0 0 0 C11 C12 C2
1 0 0 0 d11 D3
Putting the pre-multipliers together and transposing we get:
[pic]
Likewise, we can just write out the other two Gaussian pre-multipliers, so that the total ½H( operator becomes:
[pic]
To find the values for the second pre-multiplier, we must first multiply the first pre-multiplier by the reduced Gaussian matrix:
[pic]
[pic]
The transformations are at the bottom of the problem, if I put them here, MathCad will put the values in the matrix and the algebra would be too long to put the solution on the same page much less the same line. To find the values for the third pre-multiplier matrix, we must now multiply by the second pre-multiplier to the fourth Gaussian matrix:
[pic]
We now fill in for the final pre-multiplier matrix times the fifth Gaussian matrix:
[pic]
And we now have the total solution to the system of equations.
The variable substitutions for the matrices are:
F11 = D11 x C11 I11= F11 x G11 D11 = D4
G11 = D11 x B11 J11= F11 x H11 F11 = C4
G12 = D11 x B12 J12= F11 x H12 I11 = -F11 x B3 + G12 x C4
H11 = D11 x a11 A2= F11 x A1 + H13 x C4
H12 = D11 x a12 B4= F11 x B3 + G12 x C4
H13 = D11 x a13
A1 = D11 x A + a14 x D4
B3 = D11 x B2 + B13 x D4
C4 = D11 x C3 + C12 x D4
-I11xJ11 = I11xF11xA1 - I11xH13xC4 - J12xF11xB3 + J12xG12xC4
Note that the solutions for the ½H matrixes, whether you are multiplying up or down, are pre-calculated for us by the solution of the previous Gaussian matrix. i.e. The first solution for the ½H( gives a column of zero’s under the position a11 . All the numbers needed for the solution of the next Gaussian reducer matrix now lie under the position a22. We must get all numbers below this to equal zero. The topmost number (it is on the next diagonal) is given a negative value and completes the rest of the diagonal.
The numbers just below it are placed in position under the new 1 (identity) matrix occupying the place (row) of the row just reduced. Their signs are unchanged. All other values in the Gaussian reduction matrix are made equal to zero. We continue in this manner until all elements under the principle diagonal are equal to zero. Then we repeat the process, but instead of listing all elements below the a11 column, we list them above the ann position. A one is placed in the ann position for the row one which is not to be changed. The negative of the value occupying the ann position is put along the rest of the Gaussian reduction matrix diagonal. Then we pre-multiply to the previous solution. In all instances the previous calculation calculates the next needed values for us. The matrix itself, in reduction, determines it’s own solution. This is why this method should be so powerful for computers, we don’t have to calculate the later values, just read the column under the diagonal of the Gaussian reduction matrix just computed, plug them into the next pre-multiplier matrix in the sequence and multiply on, repeating the process until the solution is complete.
Boy, that’s enough symbolic math, let’s try solving a real problem. This will be a 3x3 matrix both for brevity and illustration.
I will define the augmented matrix [A]34 as:
3 -1 -2 -1
[A]34 = -1 6 -3 0
-2 -3 6 -6
which represents the set of equations:
3x -y-2z = 1 Which is equivalent to: 3x -y-2z-1 = 0
-x+6y-3z = 0 -x+6y-3z = 0
-2x-3y+6z = 6 -2x-3y+6z-1= 0
Then ½H( is equal to: [pic]
Or if we wish to start with ½H( , we use the following Pre-multipliers:
[pic]
The values for the b's are computed during the first multiplication. So ½H( is equal to:
[pic]
The first multiplication is:
[pic]
The number -17 lies on the diagonal, and is represented by b11, and the number just below it is 11, which represents b21. We plug these values into the b pre-multiplier and multiply to the first Gaussian reduction matrix. Remember, we take the negative of
-17.
[pic]
We now have all zero's under the principle diagonal. It is now time to reduce the matrix going up: The number on the principal diagonal is -117. We take it's negative and place it on the principle diagonal. The two numbers above -117 are 11 and -2, they are placed above the 1 occupying the a33 position on the pre-multiplier matrix. i.e.
So ½H( is equal to:
[pic]
The first multiplication is:
[pic]
The number -1989 is the next number up on the principle diagonal with -117 above it. -b11 = -1989 and -117 = b12. Substituting and multiplying we get:
[pic]
Even though it is illegal (although mathematicians do it all the time) let's simplify by dividing each row by the value along it's diagonal. i.e.
[pic]
[pic]
[pic]
We can do this totally by matrix by using the following method (which is also legal, by the way):
[pic]
First we isolate the diagonal and then take the inverse of each of it's elements
[pic]
[pic]
[pic] I’ve isolated the diagonal
[pic] I’m taking the inverse of each element along the diagonal
[pic] I’ve taken the inverse of each element along the diagonal
[pic] I’m multiplying each element in A2 by the inverse of the diagonal
[pic]The solution of the augmented matrix
The equations are:
X-3=0 ; x = 3
y-2=0 ; y = 2
z-3=0 ; z = 3
Now that the four pre-multipliers are computed for ½H( and ½H(, we can multiply them together and get a single matrix for ½H( and a single matrix for ½H(. We can also multiply (½H()( ½H() or ( ½H()( ½H() to get a single matrix that will also solve the augmented matrix. Let’s see how this works.
The ½H(, =
Gaussian 2 up x Gaussian 1 up = ½H(,
[pic]
The ½H( matrix =
[pic]
Then ( ½H()( ½H() =
[pic]
And
( ½H()( ½H() x A =
[pic]
Note: These operators do not commute. The operator’s are computed starting with the down operation. Commuting them with these values of ½H( first gives invalid results. i.e.
[pic]
[pic]
But if we start reducing the augmented matrix by getting all zero's above the diagonal first, and then below the diagonal to finish the problem, we get the two Gaussian reduction matrices for the up first operation as:
Gaussian 2 up x Gaussian 1 up
[pic]
And Gaussian 2 down x Gaussian 1 down =
[pic]
And multiplying the two together we obtain:
: [pic]
Or:
[pic]
[pic]
Simplifying:
[pic]
[pic]
[pic]
[pic]These are not actually zero’s along the diagonal, the numbers are too small for MathCad to show.
[pic]
[pic]
[pic]
Note: I did this differently than from above because I just figured it out. Especially the part of reducing the 3 x 4 augmented matrix to its diagonal. (7-8-97).
This operator notation is confusing, even to me, so I am going to change the look of the operator a little to make it a little less confusing. If we start with a down operation, the down operators will be
( ½H()d( ½H()d
and the up operator that begins after the down operation is complete will be
( ½H()du( ½H()du
If we begin with the up operators, the notation will be
( ½H()u( ½H()u
and the down operator that begins after the up operation is complete will be
( ½H()ud( ½H()ud
And
H( = ( ½H()ud( ½H()ud( ½H()u( ½H()u
We can now define
H( = ( ½H()du( ½H()du( ½H()d( ½H()d
as the solution which begins with a down multiplication
And
H( = ( ½H()ud( ½H()ud( ½H()u( ½H()u
as the solution which begins with an up multiplication
Then
H( ( H(
but
(H()A = (H()A
SIMPLIFYING
Note that even with a 3x3 matrix, the elements in the matrix get very large for
H( & H(. By a rough calculation, a 9x9 matrix could give a value for H larger than the memory capacity of my calculator (10499). So like mathematicians have done inductively with Gaussian reduction for the past few hundred years or so, we will divide out the larger numbers as we come across them.
Lets look at the down operation again
( ½H()d( ½H()d = [pic]
[pic]
No simplification needed here, let’s go to the next step:
[pic]
Let’s get rid of the large numbers in the last row by dividing by the number on the diagonal:
1/117R3
1 3 -1 -2 -1 3 -1 -2 -1
1 o 0 -17 11 1 = 0 -17 11 1
1/117 0 0 -117 351 0 0 -1 3
So the new and different ( ½H()du operator matrix is denoted as
1/117R3
[pic] and the first up multiplication becomes
1/117R3
[pic]
Here in the second row we can divide by 17
1/17R2
1 3 -1 0 -7 3 -1 0 -7
1/17 o 0 -17 0 34 = 0 -1 0 2
1 0 0 -1 3 0 0 -1 3
So the final multiplication becomes:
[pic]
And our solution is complete.
Note: I write the divisor in the upper left hand corner of the matrix mainly for two reasons, the first so that it won’t be mistaken for an exponent, the second is to remind me of what steps I took to simplify the results and when and where, just in case I make a mistake and need to know where to look to correct it.
INVERSE OF A MATRIX
Suppose we extend A by adding on to it the identity matrix such that
a11 a12 a13 1 0 0 1 0 0 INVERSE
A = a21 a22 a23 0 1 0 ; Then ( ½H()d( ½H()d A = 0 1 0 OF
a31 a32 a33 0 0 1 0 0 1 A
We’ve already calculated H(, so let’s see how this works.
[pic]
[pic]
[pic]
R2 stands for reducer matrix 2. Reducer matrix 1 is here. This operator takes a larger matrix to legally makes a smaller matrix out of it. Multiplying by R2 will return the inverse
[pic]
[pic]
[pic]
In mathematics, we can’t legally multiply a single row by a single number. When we multiply a matrix by a number, we multiply every element in that matrix by that number. By separating the left diagonal, taking it’s inverse and multiplying, I am doing this operation, but I am doing it legally. The operation everyone has really been doing all these centuries and thinking it illegal is
1/698139 483,327 214,812 268,515 .692 .308 .385
-1/1989 o -612 -714 -561 = .308 .359 .282 =
-1/117 -45 -33 -51 .385 .282 .436
Or it is legally equal to:
[pic]
[pic]
[pic]
[pic]
To check this answer, let's multiply by the un-augmented matrix A
[pic]
[pic] Here I’m multiplying A by it’s inverse
[pic]
[pic] Here I’m multiplying in reverse order.
[pic]
The matrix A and it’s inverse are commutative.
SYMMETRY
This section is for that mathematician or physicist who finds it unnecessary to learn two solution sets when one will do. So let’s go ahead and solve the same augmented matrix using only the down operator:
[pic]
Now, since we are only using the ½H( operator, we must rotate the matrix along it’s central element a22 (in this case -17) 180(. In other words, we must rotate the first Gaussian solution matrix such that a11 and a33 change places, likewise a31 and a13 change places. We leave the solutions with their original rows, that part of the math does not change.
First I am going to divide row three by 117. The rotated matrix now becomes:
-1 0 0 3 -z +0y+0x = -3
11 -17 0 1 or 11z-17y+0x = -1
-2 -1 3 -1 -2z - y+3x = 1
So let’s calculate down again:
[pic]
Now we need to get the -1 under the -17 to be zero, we accomplish this as follows:
[pic]
Now we divide by the numbers along the principle diagonal to simplify:
[pic]
[pic]
Which is the solution we are looking for.
NESTED ARRAYS AND GAUSSIAN REDUCTION
Sometimes, especially in physics and engineering we have a matrix’s characteristic equation or determinant for which we have solved the roots or eigenvalues, but we need to plug each root back into the matrix and solve the system of equations for each root. Usually this is done one root at a time. I am going to use nested arrays to show how to solve all these matrix solutions at the same time. We will solve for the wave functions of butadiene all in one set of operations.
[pic] The butadiene secular matrix
[pic] The symbolic determinant of the butadiene matrix
[pic] The coefficients of the symbolic determinant in order from constant to X4
[pic] The roots of the symbolic determinant.
Normally, we would put -1.618 in the matrix for X and reduce the matrix to it’s simplest form, then put in -.618 in for X and reduce. We do this for each of the roots, and will end up with four solutions. As I do this all at the same time, although I could use row, column transformations, I will transform the transposed nested array into a diagonal. It takes up more memory, but shows what I am doing better.
[pic]
The first Gaussian reduction matrix is:
[pic]
In the next computation the matrix is too big to fit on the page, so I am redefining TOTALSOLMATRIX as TSM (for this one problem). Multiplying the nested array by the first Gaussian reduction matrix we get:
[pic]
[pic]
[pic]
Note that there are all zero’s under the diagonal in the 1st, 5th,9th and 13th columns. Now to get zero’s under the diagonal in the 2nd, 6th, 10th and 14th columns:
[pic]
Now we multiply the first product RM1 by the second Gaussian reduction matrix:
[pic]
[pic] (This is the computer program, believe it or not)
Now we want to get zero’s under the diagonal in the 3rd, 7th, 11th and 15th columns:
[pic]
Now we multiply the second reduced matrix by the third Gaussian reduction matrix:
[pic]
Woah! They’re all zero’s! This means we can’t reduce the matrix to simpler terms. We can play with it if we want, but no further reductions are really needed.
Let’s put all values in terms of the diagonal:
I16 = identity(16)
[pic]
[pic]
[pic]
MATHCAD CAN'T TAKE THE INVERSE BECAUSE OF THE ZERO'S ALONG THE DIAGONAL, SO I'LL HAVE TO DO IT BY HAND:
[pic]
I also clicked out the zero's on the diagonal and replaced them with real zero's. Due to rounding errors, the zero's in RM3 are very small numbers. HOW TO READ: For the first wave function a11=.618a2;a2=a3; a3=1.618a4. For the second wave function, a1=1.618a2; a2=-a3; a3=.618a4. For the third wave function, a1=-1.618a2; a2=a3; a3=-.618a4. For the fourth and last wave function, a1=-.618a2; a2=-a3; a3=-1.618a4
[pic]
GENERAL MATRIX SOLUTION SET FOR ½H(
1 0 0 . . . 0 0 1 0 0 . . . 0 0 0 1 0 0 . . . 0 0 0 0
0 1 0 . . . 0 0 0 1 0 . . . 0 0 0 0 1 0 . . . 0 0 0 0
0 0 1 . . . 0 0 0 0 1 . . . 0 0 0 0 0 1 . . . 0 0 0 0
. . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
0 0 0 . . . 1 0 0 0 0 . . . 1 0 0 0 0 0 . . . 1 0 0 0
0 0 0 . . . M21 -M11 0 0 0 . . . L21 -L11 0 0 0 0 . . . K21 -K11 0 0
0 0 0 . . . L31 0 -L11 0 0 0 . . . K31 0 -K11 0
0 0 0 . . . K41 0 0 -K11
1 0 0. . . 0 0 0 . . . 0 0 1. . . 0 0 0 . . . 0 0
0 1 0. . . 0 0 0 . . . 0 0 0. . . 1 0 0 . . . 0 0
0 0 1. . . 0 0 0 . . . 0 0 0. . . B21 -B11 0 0 . . . 0 0
. . .. . . . . . . . . . . 0. . . B31 0 -B11 0 0 . . . 0 0
. . .. . . . . . . . . . . 0. . . B41 0 -B11 0 . . . 0 0
0 0 0. . . 1 0 0 . . . 0 0 .. . . . . . . . . . .
0 0 0. . . I21 -I11 0 0 . . . 0 0 . . . .. . . . . . . . . . .
0 0 0. . . I31 0 -I11 0 0 . . . 0 0 0. . . Bi1 0 0 -B11. 0 0
0 0 0. . . I41 0 -I11 0 . . . 0 0 0. . . Bi1 0 0 . -B11 0
. . .. . . . . . . . . . . 0. . . B(n-I),1 0 0 . . . . -B11
. . .. . . . . . . . . . .
0 0 0. . . Ii1 0 0 -I11. 0 0
. . .. . . . . . . . . . .
. . .. . . . . . . . . . .
0 0 0. . . I(n-I),1 0 0 . . . . -I11
1 0 0 . . . 0 0
A21 -A11 0 0 . . . 0 0
A31 0 -A11 0 0 . . . 0 0
A41 0 -A11 0 . . . 0 0
. . . . . . . .
. . . . . . . .
Ai1 0 0 -A11. 0 0
Ai1 0 0 . -A11 0
A(n-I),1 0 0 . . . . -A11
GENERAL MATRIX SOLUTION SET FOR ½H(
A21 -A11 0. . .0 -B11 0 B13 0. . . 0 -C11 0 0 C14 . 0 . . . 0
0 1 0. . .0 021 -B11 B12 0. . . 0 0 -C11 0 C13 0 . . . 0
0 0 0. . .0 0 0 1 0. . . 0 0 0 -C11 C12 0 . . . 0
. . .. . .. . . . .. . . . 0 0 0 1 0 . . . 0
. . .. . .. . . . .. . . . 0 0 0 0 1 . . . 0
0 0 0 1 0 0 0 0 0. .1. 0 . . . . . . . . .
0 0 0 0 1 0 0 0 0. .0. 1 . . . . . . . . .
0 0 0 0 0 1 0
0 0 0 0 0 0 1
-I11 0 0 . . . . I(n-I),1 0 . . . . 0 -B11 0 0 . . . . B(n-2),1 0
0 -I11 0 . . . . 0 . . . . .. 0 -B11 0 . . . . 0
. . . . . . . . . . . . .. 0 . -B11 . B14 0
. . . . . . . . . . . . .. 0 . 0 -B11 . B13 0
0 . 0 -I11. . . I14 0 . . . . .0 0 . 0 -B11 B12 0
0 . 0 .-I11. I13 0 . . . . .0 0 . 0 -B11 B12 0
0 . 0 . . . .-I11 I12 0. . . . . 0. . . 0 . 0 1 0
0 . 0 . . . . 0 1 0. . . . . 0 0 . 0 0 1
. . . . . . . . . . . . ..
. . . . . . . . . . . . ..
0 . 0 . . . . 0 0 1. . . . . 0
0 . 0 . . . . 0 0 0. 1 0 0 0 0
0 . 0 . . . . 0 0 0. 0 1 0 0 0
0 . 0 . . . . 0 0 0. 0 0 1 0 0
0 . 0 . . . . 0 0 0. 0 0 0 1 0
0 . 0 . . . . 0 0 0. 0 0 0 0 1
-A11 0 0 . . . . A(n-1),1
0 -A11 0 . . . .
. . . . . . .
. . . . . . .
0 . -A11 . A14
0 . 0 -A11 . A13
0 . 0 -A11 A12
0 . 0 1
ELEMENTARY MOLECULAR ORBITAL METHODS
(PART ONE)
[1]HÜCKEL MO THEORY
In elementary Hückel Molecular calculations, there are three assumptions made:
The wave function can be written ((((((((( where ( is an anti-symmetric function of ( electrons, and ( is an anti-symmetric function of ( electrons.
( and ( are separately normalized to unity.
( and ( can each be expanded in terms of a set of orthonormal Slater determinants. In Hückel theory, ( is ignored and ( is crudely solved. The basis set is limited to 2p orbitals.1
It is also assumed that an eigenfunction can be used to represent the ( values of an electron in a molecule just as an electron in an atom can be described by a similar function. But in a molecule, one has to also worry about the electrostatic fields of the surrounding nuclei.2
Beyond the hydrogen atom, it is impossible to obtain an exact solution of the Schrodinger equation (hydrogen, with six vectors to describe the x, y and z positions of both the nucleus and electrons, forms a system of six equations in six unknowns, which we can solve. Helium, on the other hand, has the same six equations but nine unknowns, which we cannot solve exactly). Approximate methods of calculating these energies must be used. One method is the Theory of Variations.
THE THEORY OF VARIATIONS
From the Schrodinger Equation
H( = E(
Multiplying both sides by (* (the complex conjugate of () we get
(*H( = (*E(
But E (energy) is a constant, so the equation transforms to:
(*H( = E(2
or
E = ∫(*H(d(
∫(2 d(
Suppose that ( = (C1P1 + C2P2) is a solution to the above equation. We just plug it in and simplify the result.
E = ∫(C1P1 + C2P2)*H(C1P1 + C2P2) d(
∫(C1P1 + C2P2)2 d(
= ∫(C1*P1*HC1P1 + C1*P1*HC2P2 + C1*P2*HC1P1 + C2*P2*HC2P2)d(
∫ (C12 P12 + 2C1*C2P1*P2 + C22 P22)d(
Where ∫(P1*HP2)d( = ∫(P2HP1*)d( (The properties if the integrals are symmetric)
Simplifying:
H11 = ∫ (P1*HP1)d( ; H12 = H21 = ∫ (P1*HP1)d( = ∫(P1*HP2)d( = ∫(P2HP1*)d(
H22 = ∫(P2*HP2)d( ; S11 = ∫P12 d( ; S12 = S21 = ∫(P1*P2)d( = ∫(P2*P1)d( ; S22 = ∫P22 d(
Therefore
C12H11 + 2C1*C2H12 + C22H22
E = EQ. 1
C12S11 + 2C1*C2S12 + C22S22
Suppose ( = (ai(i where ai are real algebraic numbers, and ( is a set of arbitrary real well-behaved (conforms to physical reality) functions from which we seek to choose parameters for ai such that ( approximates most closely to (0.
Now we will substitute ( = (ai(I into the Schrodinger Equation:
E = ∫((ai(I)H((aj(j) d(
∫((ai(I)((aj(j) d(
Now simplify Hij = ∫((IH(j) d( ; Sij = ∫((I(j)d( and the equation becomes:
aiajHij
E = = f1/f2 EQUATION 2.
AijSij
where f1/f2 is a constant and the integrals Hij and Sij do not involve the parameters ai.
Now we must use the calculus of variations to solve this equation:
(E/(ai = 0
(E/(ai = (1/f2) (f1/(ai - (f1/f2) (f2/(ai = 0
and using Equation 2:
(E/(ai = (f1/(ai - E (f2/(ai (E=(f1/f2)
E=(f1/f2)= 2ajHij-E2ajSij
or
(2aj(Hij-ESij) = 0 3
To demonstrate, suppose ( = a1P1 + a2P2 : Lets solve for a2 first. ai = a2, then
(E/(a2 and Eq. 1 become:
(E/(a2= M(N/M2 - N(M/M2= 0
Where M = a12S11 + 2a1a2S12 + a22S22
and
N = a12H11 + 2a1a2H12 + a22H22
but (E/(a2=0, therefore
M(N/M2 = N(M/M2
(N=(N/M)(M,
but N/M = E (EQ.1)
therefore
(N=E(M
or
2a1H12 + 2a2H22 = E(2a1H12 + 2a2H22) the 2’s cancel leaving
a1H12 + a2H22 = E(a1H12 + a2H22),
a1(H12 -ES12) + a2(H22-ES22) = (E/(a2
by similar reasoning, we find
a1(H11 -ES12) + a2(H22-ES22) = (E/(a2 4
These are the Slater determinants that we will use in our calculations. Even if they are not understandable how they are derived, it is important to at least see where they come from.
BUTADIENE
The Hückel method chooses eigenfunctions which obey the rules of AO’s (Atomic Orbitals) and the Aufbau principle. With use of the above Slater determinant, energies of an electron in various possible MO’s (Molecular Orbitals) can be calculated where each AO accommodates electrons of opposite spins. Orbitals are filled starting from the lowest en ergies until the electrons are used up. When dealing with more than two electrons, two are placed in the most stable orbit, two in the next, then two more and so on until all the electrons are accounted for. In cases of degenerate orbitals, electrons with parallel spins go into each of the orbitals (with equal probability of occupation) with only one electron per orbital until each orbital is half-filled. Then, if there are enough electrons left over, we begin filling the half-filled orbitals until all electrons are used. i.e. in the case of butadiene, orbitals are filled as E1 = A-2B, E2 = A = E3 and
E4 = A+2B. And the orbitals are filled as indicated:
where represents electrons with
parallel + spins, & represents
E1 = A-2B E2 = A = E3 E4 = A+2B electrons with - spins.
One serious drawback to the Hückel MO method is that overlap is generally neglected (a simple method is included on pg. 110-111) and the energy associated with surrounding nuclei and orbitals is assumed to remain unchanged even in the presence of other electrons. To help visualize this a little better, suppose we have a molecule of ethylene before its ( bonds are formed:
CH3=CH3 ( C1 = C2 where the electron is in one of the p orbitals. Suppose we find it is in p1. Then Hp1 = Ep1. Multiplying both sides by p1 and integrating, we have:
E = ( p1Hp1d( = ( p1Ep1d(
where E is a constant, and
( p1p1d( = 1
Therefore
( p1Hp1d( = E
This is the coulomb integral representing the energy of an electron in a 2p orbital of carbon. It’s value must be negative since the electron is attracted to the nucleus. This particular integral is denoted as H11 or A. Similarly ( p2Hp2d( = H22 = A.
B represents the energy of an electron when it is between carbon atoms 1 and 2. It is also a negative quantity and is called the resonance integral. It is defined by:
( p1Hp2d( = H12 = B = ( p2Hp1d( = H21
To find the most probable location of an electron in a molecule, we describe a function ( such that ( = a1p1 + a2p2 (which represents the ( bonding of ethylene having the energy A + B). The more negative the energy, the more stable the bond.
ETHYLENE
We are now in a position to solve the energy states of ethylene. First we set up the secular determinant for ethylene with the conditions:
i=j = 1 coulomb computation and ( = a1p1 + a2p2.
Sij = i(j = 0 resonance with overlap neglected and E11 = E22
H11 -S11E11 H12 - S12E12 = H11 - E11 H12 = 0 ( A - E B
H21 -S21E21 H22 - S22E22 H21 H22 - E22 B A - E
We divide both rows by B and let (A-E)/B = X (EQ. 3), which gives us the simplified Slater determinant
X 1 = 0. The value of the determinant is (X2 - 1) = 0 The roots of this
1 X equation are X=1, X=-1. Putting each value of X into EQ. 3 and
solving for E1 (defined as the most negative or highest energy or bonding energy, and E2 as the least negative, lowest energy or non-bonding energy, we get:
(A-E)/B = 1; -E = B-A; E2 = A - B
(A-E)/B = -1; -E = -B-A; E1 = A + B
Or we can write the equation out as: (X+1)(X-1) = (A-E1+B)(A-E2-B) and solve for E.
(This proves our assumption above that the energy for ethylene is A + B.)
Since A & B are negative quantities, E1 is the most stable orbital and E2 is the least. E1 is called a bonding orbital (it is occupied) and E2 is called an anti-bonding orbital (it’s orbitals are unoccupied). There are two orbitals to fill (in the non-bonding orbital, p1 & p2), so the total ( energy (E() is 2A + 2B which is twice the energy of a single occupied orbital.
We have just computed E( = 2A + 2B as the total de-localized energy of ethylene. To determine the energy of localized ethylene (for resonance energy calculations), we [2]assume that the carbon atoms are separated in space (the electrons occupy orbitals on the carbon atoms and not in between). In this case, H12 = H21 = 0; and H11 = H22 = A. The secular determinant now becomes
A - E 0 = 0 = (A - E)2. E1 = A
0 A - E E2 = A
Which is degenerate (the roots are equal and of the same sign). This is interpreted to mean each of the carbon atoms contain one electron each in their p orbitals and E( = 2A.
The resonance energy DE( = E( (de-localized) - E( (localized) = 2A + 2B -2A = 2B. This difference is considered to be the energy difference between the most stable classical structure and the compound itself. It is the energy of the ( bond formation. (5)
A more exact energy diagram of ethylene may be constructed if overlap is included. (6)7
In this case, for Sij, i( j ( 0; H11 = H22 = A; S12 = S21 = S and H12 = H21 = (. The Slater determinant becomes:
H11 -S11E11 H12 - S12E12 = H11 - E11 (- S12E12 = 0 ( A - E (-SE
H21 -S21E21 H22 - S22E22 ( -S21E21 H22 - E22 (-SE A - E
This looks like a more complex determinant to solve, but if we let B = (-SE, the determinant changes to
A-E B Which is the same form as above, dividing by B, and letting
B A-E X = (A-E)/B or X = (A-E)/((-SE), we get (EQ.4)
X 1 = 0.
1 X
Solving the same way as outlined above, the solution to the determinant is (X2 - 1)
and X = ∀1. Now we must solve for E1 and E2:
(ACCOUNTING FOR OVERLAP INTERGAL)
Suppose X =(A-E)/((-SE) = -1; A - E1 = -( + SE1
A + ( = E1 + SE1
A + ( = E1(1 + S)
E1 = (A + ()/(1 + S)
Suppose X =(A-E2)/((-SE2) = 1; A - E2 = ( - SE2
then
E2 = (A - ()/(1 - S)
Where S is not a constant, but is a function of the distance separating the carbon atoms. For ethylene, experiment gives this distance as 1.33Ẵ or about a 25% overlap. Therefore,
S = .25. Plugging this value for S into both equations, we get:
E1 = (A + ()/(1 + .25) = (A + ()/(1.25) = .8(A + ()
E2 = (A - ()/(1 - .25) = (A - ()/(.75) = 1.33(A - ()
E( (de-localized) = 2E1 = 2x.8(A + () = 1.6(A + ()
E( (localized) = (A - E)2, E1 = E2 = A = 2A (A for one orbital, twice this for both orbitals).
DE( (resonance energy) = 1.6A + 1.6( -2A = -.4A + 1.6(.
To find the coefficients for ( = a1p1 + a2p2, we will reduce the secular determinant by use of Gaussian Reduction. Solving first for X=1, we plug this value for X into the determinant
X 1 = 0 = 1 1
1 X 1 1
This is simple, just subtract row 2 from row one, and we get (remembering the first column represents the values of a1 and the second column represents the values of a2)
a1 a2
1 1 a1 + a2 = 0 or a1 = -a2. We will now write a2 in terms of a1 and normalize,:
0 0 remembering a1 = 1 and a2 = 1:
a1 (normalized) = 1/(a12 +(-a12)) = 1/(12 + 12)1/2 = 1/(2)1/2 therefore a1 = 1/(2 , and
(2 = 1/(2 P1 - 1/(2 P2 = 1/(2 (P1 - P2)
Similarly, for X = -1, the determinant becomes
-1 1 = 0 = (adding row 2 to row 1) -1 1 or a1 = a2. Setting in terms of a1
1 -1 0 0 and normalizing:
a1 (normalized) = 1/(12 + 12 )1/2 = 1/(2 and (1 = 1/(2 (P1 + P2).
The total wave function describing ethylene is:
+ +
(1 = 1/(2 (P1 + P2) bonding
- -
+ -
(2 = 1/(2 (P1 - P2) anti-bonding
- +
Since only (1 is occupied, each electron contributes (1/(2)2 = 1/2 of the total electron density at each carbon, or 1/2 + 1/2 = 1 = (11 for the total molecule. But at some point on (2, the coefficient of a1 will equal a2 , and a1-a2 = 0, meaning that at some point the molecule will experience a state of zero electron density (a node). But (1 will always equal a negative quantity such that there is a non-vanishing electron density. Since in nature it is assumed the orbitals of an atom or molecule never vanish, (1 is chosen as the function to represent the bonding state of ethylene, (1 is therefore a well behaved function. 7
BOND ORDER
The bond order of a substance is an indication of the location of the ( bonds in a molecule. It is calculated from the occupied orbitals and is defined as
(occ
Pij = (Naiaj
Since we are calculating between two carbons (not the interaction between three or four),
N = 2 ( N will always be two as long as we are calculating for adjacent carbons). So for ethylene,
(occ
P12 = (2(1/(2)( 1/(2) = 2(1/2) = 1
It is defined that the bond order of ethylene is the maximum strength of a ( bond, with all other ( bonds in other, larger molecules either equal or less than this value.
Note here the above ( function can be represented as a matrix:
(1 (2 (2 and the bond order can be calculated by squaring the first row
(2 (2 -(2 of the matrix. We can do this as follows: (1occ(1Tocc
P12 = 2( 1 0 (2 (2 ) ( 1 0 (2 (2 )T = 2[(2 (2] (2 = 2(1/2) = 1
0 0 (2 -(2 0 0 (2 -(2 (2
Lets take a quantum leap in our mathematics and run a comparison of butadiene and cyclobutadiene and compare their relative stability’s using the Hückel method.
BUTADIENE
Butadiene has the structure CH2=CH-CH=CH2. To set up the Slater determinant, we number the carbons in order from left to right 1 2 3 4 , and set up the math such
C- C- C- C
that H11=H22=H33=H44 = A (each carbon atom has an orbital to fill). Also suppose the determinant of A is symmetric, then H12=H21=H23=H32=H34=H43 = B (the electron can exist equally in the regions between any two adjacent carbon atoms), and H13=H31=H14=H41=H24=H42 = 0 (there is assumed to be no inter-electronic interaction between non-adjacent carbon atoms). These assumptions are not really correct, for it is assumed here that B12=B21=B23=B32=B34=B43 = B and
S12=S21=S23=S32=S34=S43 = 0 which is not true as the bond lengths in butadiene are not the same. But since S=.25, overlap is usually ignored. 8
So the Slater determinant for butadiene becomes
A-E B 0 0 X 1 0 0
B A-E B 0 = 0 Letting X=(A-E)/B, the determinant becomes 1 X 1 0 = 0
0 B A-E B 0 1 X 1
0 0 B A-E 0 0 1 X
To solve the determinant and find it’s characteristic equation, we compute the following steps using Cramer’s rule:
(X X 1 0 - 1 1 1 0 ) = X(X X 1 - 1 1 1 ) -1(1 X 1 -1 0 1 )=
1 X 1 0 X 1 1 X 0 X 1 X 0 X
0 1 X 0 1 X
X( X(X2-1)-X)- ((X2-1)-X)= X(X3-X-X)-X2+1 = X4-2X2-X2+1= X4-3X2+1=0.
Note: This is one of the reasons I discovered the half-multiplier operator. I was trying to obtain the characteristic equation using Gaussian reduction (which I am still unable to do) when it occurred to me to prove it, as I had never been shown the proof in any math or physics class I had ever taken. The rest is mathematical history, as from it’s proof is derived the Unified Field Equation. It also puts the final touches on the definition of number and the operations of addition, subtraction, multiplication and division. Mathematicians and physicists have used Gaussian reduction for over 200 years, and not one of them ever set about to prove it, it’s usefulness was just intuitive, never proven, it was used because it worked. The missing link in mathematics was under their noses all this time, and none ever saw it.
The roots of this equation are ∀(3/2 ∀ 1/2(5) or ∀1.618 and ∀.618. The energy diagram is as follows:
E1 = A + 1.618B occupied bonding orbital
E2 = A + .618B occupied bonding orbital
E3 = A - .618B un-occupied anti-bonding orbital
E4 = A - 1.618B un-occupied anti-bonding orbital
We will now calculate the ai coefficients for the ( functions. First we write the secular determinant of butadiene:
a1 a2 a3 a4 Setting X=-1.618
a1 a2 a3 a4
[pic]= 0 [pic]= 0
Normally, in all the books, this determinant is solved using Cramer’s rule, but the math takes a quarter page per ai, or about an entire page to solve the wave function (i. Note also that in this method of solution, the above equation should be an augmented matrix of 4 rows and 5 columns, the fifth column being the negative of the matrix to the right of the equal sign. But since here this value is 0, I’m just going to ignore it, but remember that it’s there. i.e.
a1 a2 a3 a4 Setting X=-1.618
(1 X 1 0 0 -0 -1.618 1 0 0 -0
(2 1 X 1 0 -0 = 0 1 -1.618 1 0 -0 = 0
(3 0 1 X 1 -0 0 1 -1.618 1 -0
(4 0 0 1 X -0 0 0 1 -1.618 -0
To solve, we do the following steps:
1. Divide row 1 by 1.618 , and add to row two. 1/1.618R1+R2
Put solution in row 2, leaving row 1 unchanged.
-1 .618 0 0
0 -1 1 0
0 1 -1.618 1
0 0 1 -1.618
Add row 2 to row 3
Put solution in row 3, leaving rows 1&2 unchanged. R2+R3
-1 .618 0 0
0 -1 1 0
0 0 -.618 1
0 0 1 -1.618
3. Divide row 4 by 1.618 and add to row 3.
Put solution in row 4, leaving rows 1,2&3 unchanged. 1/1.618R4+R3
-1 .618 0 0
0 -1 1 0
0 0 -.618 1
0 0 0 0
This is as simple as we can get this reduction. Remembering that this is actually an augmented matrix, we will subtract the terms that are negative to get them on the other side of the equal sign. This gives us
a1 = .618a2
a2 = a3
.618a3 = a4
To make the calculations simpler, we will put all these values in terms of a1.
a1 = a1
a2 = 1.618a1
a3 = 1.618a1
a4 = a1
Therefore, (1 = P1 + 1.618P2 + 1.618P3 + P4 Now we must normalize (1:
(1(N) = (a12 + a22 + a32 + a42)1/2 = (12 + 1.6182 + 1.6182 + 12)1/2 = (1+2.618+2.168+1)1/2 =(7.75)1/2 = 2.69
This is also equal to ((1(1T)1/2 = ([1 1.618 1.618 1] 1 )1/2 = 2.69
1.618
1.618
1
So the equation in general is: (i/((i(iT)1/2
and for (1 = 1/2.69[1 1.618 1.618 1] = [.3717 .6015 .6015 .3717] or
(1 = .3717P1 + .6015P2 + .6015P3 + .3717P4
Lets look at another way to solve the secular determinant when X = -.618, also using Gaussian reduction, but in its pure matrix form (see section in paper about Gaussian reduction for details, theory and proofs). The matrix for (2 =
a1 a2 a3 a4 Setting X=-.618
(1 X 1 0 0 -.618 1 0 0
(2 1 X 1 0 = 0 1 -.618 1 0 = 0
(3 0 1 X 1 0 1 -.618 1
(4 0 0 1 X 0 0 1 -.618
To get zero’s under the -.618 in the first column:
G1 PSI PSI1
1 0 0 0 -.618 1 0 0 -.618 1 0 0
1 .618 0 0 1 -.618 1 0 = 0 .618 .618 0
0 0 .618 0 0 1 -.618 1 0 .618 -.382 .618
0 0 0 .618 0 0 1 -.618 0 0 .618 -.382
Using the newly calculated matrix, to get zero’s under the +.618, multiply by:
G1 PSI1 PSI2
1 0 0 0 -.618 1 0 0 -.618 1 0 0
0 1 0 0 0 .618 .618 0 = 0 .618 .618 0
0 .618 -.618 0 0 .618 -.382 .618 0 0 .618 -.382
0 0 0 -.618 0 0 .618 -.382 0 0 -.382 .236
Now we want a zero under the diagonal in the third column, so we multiply by:
G2 PSI2 PSI3
1 0 0 0 -.618 1 0 0 -.618 1 0 0
0 1 0 0 0 .618 .618 0 = 0 .618 .618 0
0 0 1 0 0 0 .618 -.382 0 0 .618 -.382
0 0 -.382 .618 0 0 -.382 .236 0 0 0 0
We are done with the Gaussian reduction, the coefficients ai are:
a1 = a1
.618a1 = a2
a2 = -a3
.618a3 = .382a4
Putting these all in terms of a1 we have:
a1 = a1
a2 = .618a1
a3 = -.618a1
a4 = (.618)(-.618a1)/.382 = -1a1
Before I finish, I am going to write out the MathCad +6 program for this Gaussian reduction. G1 is the first Gaussian multiplier, G2 is the second and G3 is the third. PSI is the Slater determinant, PSI1 is the determinant after being multiplied by G1. PSI2 is PSI1 after being multiplied by G2 and PSI3 is the solution, after multiplying PSI2 by G3.
MATHCAD +6 PROGRAM:
PSI1 = G1xPSI
psi2 = G2xPSI1
PSI3 = G3xPSI2 The math is done for (1
a1 = a1
.618a1 = a2
a2 = -a3
.618a3 = .382a4
Putting these all in terms of a1 we have:
a1 = a1
a2 = .618a1
a3 = -.618a1
a4 = (.618)(-.618a1)/.382 = -1a1
A2 = [1 1.618 -1.618 -1] Starting (2
A2SQ = A2 x A2T
ATWO = (A2SQ)1/2
ATWO = 1.662
PSITWO = ATWO-1 x A2
PSITWO = [.6015 .3717 -.3717 -.6015]
Note, in MathCad, the zero’s are mostly in decimals of 10-5 or 10-6. This is due to rounding error, just look at them as zero’s.
Using any method (even Cramer’s rule, if you prefer) let the remaining X’s = .618 and 1.618 respectively. The four wave functions for butadiene can be expressed as:
(1 = .3717P1 + .6015P2 + .6015P3 + .3717P4 bonding
(2 = .6015P1 + .3717P2 - .3717P3 - .6015P4 bonding
(3 = .6015P1 - .3717P2 - .3717P3 + .6015P4 anti-bonding
(4 = .3717P1 +-.6015P2 + .6015P3 - .3717P4 anti-bonding
or
.3717 .6015 .6015 .3717
( = .6015 .3717 -.3717 -.6015
.6015 -.3717 -.3717 .6015
.3717 -.6015 .6015 -.3717
Diagramatically, these functions look like:
+ + + -
+ + + -
(1 = (2 =
- - - +
- - - +
+ + - +
[3] - - + -
(3 = (4 =
+ + - +
- - + -
(1 is the most stable state where all the orbitals are reinforcing each other additively. In (2 the bonding between P2 and P3 cancels out at a node, reinforcing the bonding between P1-P2 and P3-P4. The relative strengths of these bonds can be found via the bond order. The bond order Pij is defined as:
Pij = (Naiaj
(Which is the same as for ethylene in the previous example).
P12 = (Na1a2)(1 + (Na1a2)(2
= 2(.6015)(.3717) + 2(.6015)(.3717) = .8942
P23 = (Na2a3)(1 + (Na2a3)(2
= = 2(.6015)(.6015) + 2(.3717)(-.3717) = .4473
P34 = P12 = .8942 (9)
Remember we are using only the occupied orbitals here.
To calculate these values using matrices, we proceed as follows: (TOCC(occ.
2( .3717 .6015 .3717 .6015 .6015 .3717 ) =
.6015 .3717 .6015 .3717 -.3717 -.6015
.6015 -.3717
.3717 -.6015
1 0.894 0 -0.447
0.894 1 0.447 0
0 0.447 1 0.894
-0.447 0 0.894 1
We ignore the values along the diagonal, and of course P14 and P41. Note, the in-between values are not necessarily all zero’s, it just happened to come out this way in this case.
These are defined as the mobile bond orders. To get the total bond order, the bond order of the ( bonds must be added. For simplicity, the hydrogen bonds will be defined as:
( = 1. Therefore, P12 = P21 = P34 = P43 = 1.8942; and P23 = P32 = 1.4473
The molecular diagram at this point is:
1.8942 1.4473 1.8942
CH2 - CH - CH - CH2
This shows us that there is more double bond character in the 1-2, 3-4 positions than in the 2-3 position. So in the butadiene molecule, the ( bonds should favor CH2=CH-CH=CH2 .
The Free-Valence Index (10) of a compound is the measure of the degree that the atoms in a molecule are bonded to adjacent atoms relative to their maximum bonding power. If particular atoms are not bonded much compared to this maximum, they are said to occupy especially reactive positions.
Free-Valence Index (F) = max. bonding energy - (Pij
EXERCISE: Using trimethylene methane, verify (PijMax = 4.732 . (or 1.732 ignoring ( bonds)
CH2
CH2 - C A = AT (Matrix is symmetric); H11=H22=H33=H44 = A; H12=H21=H23=H32=H34=H43 = B
CH2 and H13=H31=H14=H41=H24=H42 = 0;
1 2 3
The Secular determinant is: C - C - C
C 4
A-E B 0 0 X 1 0 0
B A-E B B = 0 Letting X=(A-E)/B, the determinant becomes 1 X 1 1 = 0
0 B A-E 0 0 1 X 0
0 B 0 A-E 0 1 0 X
The determinant is equal to:
X X 1 1 - 1 1 1 1 = X(1 1 1 + X X 1 ) - 1(1 X 0 ) =
1 X 0 0 X 0 X 0 1 X 0 X
1 0 X 0 0 X
X[(-X)+X(X2-1)]-X2) = X(-X+X3-X)-X2 = -X2+X4-X2-X2 = X4-3X2 = X2(X2-3)=0 .
We don’t need any special math to determine the roots of this equation. From inspection, we can easily see X = 0, 0, (3, -(3 . Therefore:
E1 = A+(3B
E2 = A
E3 = A
E4 = A-(3B
To solve for the coefficients ai, let X = -(3 = -1.732, putting this value for X into the secular determinant, we get:
-1.732 1 0 0
1 -1.732 1 1 = 0
0 1 -1.732 0
0 1 0 -1.732
1/1.732R1+R2
-1 .577 0 0
0 -1.155 1 1 = 0
0 1 -1.732 0
0 1 0 -1.732
1/1.732(R3+R2)
-1 .577 0 0
0 -1.155 1 1 = 0
0 .577 -1 0
0 .577 0 -1
-R3 + R1
-1 0 1 0
0 -1.155 1 1 = 0
0 .577 -1 0
0 .577 0 -1
R3+R2
-1 0 1 0
0 -.578 0 1 = 0
0 .577 -1 0
0 .577 0 -1
R2+R4
-1 0 1 0
0 -.578 0 1 = 0
0 .577 -1 0
0 0 0 0
R3+R2
-1 0 1 0
0 0 -1 1 = 0
0 .577 -1 0
0 0 0 0
Note: Even though the problem is solved in the step above this one, it is necessary to reduce the determinant to its simplest terms to obtain the correct ( function.
a1 = a3
a3 = a4
.577a2 = a3
To be different, I am going to put these values in terms of a3 instead of a1, just to show how to do it and that we don’t have to do it for a1 all the time.
a3 = a1
a3 = a4
.577a2 = a3 ; 1.732a3 = a2
a3 = a3
Therefore, ((T()1/2 = (11 + (31/2)2 12 +12 )1/2= (6)1/2 and
( = 1/(6)1/2P1 + (3)1/2/(6)1/2P3 1/(6)1/2P1 + 1/(6)1/2P4
( = .4082P1 +.4082P2 +.7071P1 +.4082P4
Pij = (Naiaj =(TOcc(Occ = .4082 [.4082 .7071 .4082 .4082] =
.7071
.4082
.4082
Note: With this method ((TOcc(Occ), the value of N is automatically taken care of in the math, so we do not need to include it in our mathematics. To find the maximum energy, we must also include the values of the ( bonds for (12 = (23 = (24 = 1, or for the total ( energy, we have (12 + (23 + (24 = 3, and the total energy (PijMax = (T( + (ij .
The MathCad program and mathematics are as follows:
.4082 ( = [.4082 .7071 .4082 .4082] ONE4 = [1 1 1 1]
.7071
(TRN = .4082 1
.4082 1
FOUR1 = 1
1
-100 0 -100 -100 0 1 0 0
0 -100 0 0 Pij = 1 0 1 1
LOGPij = -100 0 -100 -100 0 1 0 0
-100 0 -100 -100 0 1 0 0
Where the one’s in Pij occupy the positions for P12=P21= P23=P32= P24=P42=1. All other values in the matrix are identical to zero. This way, in a one to one multiplication, we retrieve only the values in these positions and ignore the rest. Compare with the Slater determinant, these values all occupy the positions where the one’s are located, with X=0.
This is the easiest way to set up the Pij and LOGPij matrix templates.
Note: The log of 0 is undefined, so to be able to use it in a problem, I give 0 a default value of 10-100. MathCad can only go as low as 10-336, so the default value has to be at least half that value, because as we add 0 + 0, we will get 10-200 which the computer can handle. I am not including the HP-48G program here, mainly because I cannot multiply the matrices one on one like I can in MathCad.
[pic]
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Here I take the log of each individual element in the matrix (SQ.
[pic]
Here I add the two log matrices (in effect multiplying each element one on one) and take the anti-log of each element to return the numbers to our real number system.
[pic]
[pic] Here I sum all the values in the matrix.
[pic]
To complete the calculation: (PijMax = (T( + (ij = 1.732 + 3 = 4.732.
Because of having to now subtract the ( bonds during the calculations, I do not furthermore use the value of 4.732, I just use the 1.732 value (SUMPij) because it is much easier to do. The value 4.732 is for advanced quantum chemistry users.
So for butadiene, Fi = 1.732 - (Pij
F1 = F4 = 1.732 - .8942 = .8378
F2 = F3 = 1.732 - (.8942 + .4473) = .391
Or to do this totally by matrices, we proceed as follows:
Fi = [1.73]4,1 -( [P]4,4[1]1,4 )
(Remembering to use only those values that correspond to the positions of the one’s in the original Slater determinant).
1.732 0 0.894 0 0 1 1.732 .8942 .8378
1.732 - 0.894 0 0.447 0 1 = 1.732 - 1.341 = .3910
1.732 0 0.447 0 0.894 1 1.732 1.341 .3910
1.732 0 0 0.894 0 1 1.732 .894 .8378
Our molecular diagram now looks like:
.391
.8378
1.8942 1.4473 1.8942
CH2 - CH - CH - CH2
The charge distribution qi (the deviation from the ‘normal’ electron density at a given ( bonded atom) is obtained by summing the electron’s probabilities corresponding to the contribution of that particular AO (Atomic Orbital) to the other various occupied orbitals. Corrections have to be made concerning formal charges resulting from the ( bonds to obtain the overall charge. Therefore, if a carbon forms 3 ( bonds and is ( bonded, it will be neutral if there is an average of one electron in it’s 2p ( bonded orbitals.
If qi is defined as the deviation from neutrality of such a carbon, then
qi = 1.000 - (Nai2(I 11
For butadiene, q1 = q4 q = 1.000 - (Nai2(I -(Nai2(2 =
1.000 - 2(.3717)2 - 2(.6015)2 =
1.000 - 2(.1382) - 2(.3618) = 1.000 - .2763 - .7236 = 0
for q2 = q3 q = 1.000 - 2(.6015)2 - 2(.3717)2 = 0
Now this seems quite simple to do by hand, much more simple than the following MathCad programs. But suppose we were dealing with a protein that has a matrix of dimensions 100,000x100,000, then we would need a computer because it would be impossible to add these values by hand. I am going to show two ways, the first by calculating each value of q separately, then by using the method of nested arrays. Remember, I’m just going to use the (Occ matrix here, not the whole ( matrix.
MATHCAD +6 PROGRAM # 1
[pic] [pic] [pic] [pic]
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Solving for q1: (Am using only the values for the occupied orbitals.)
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Now to solve for q2:
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[pic] [pic]
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(See part 2, Bond order for Acrolein for simpler method)
NESTED ARRAYS
Here we will compute q1 and q2 at the same time using nested arrays:
Use the variable definitions above plus: (See below how we obtain NESTARRAYB)
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First we must compute the half-multiplication the hard way, mainly because there is no computer program that can accomplish this (programmers don’t even know this mathematical operator exists yet). The neat thing about this operator is that there is a regular matrix method that can accomplish the same results, but with a lot more mathematical steps, so instead of the elegant [b] o [1], we must use the following method:
B1 = [B][A1][TWO1]
B2 = [B][A2][TWO1]
Then
.3717 .6015 1 .3717 .6015
[B] o [1] = .6015 .3717 o 1 = .6015 .3717
.6015 -.6015 1 .6015 -.6015
.3717 -.3717 1 .3717 -.3717
Now to transform this so we can use it in our regular mathematics, we transpose the nested array, remembering that when we transpose, we transpose the matrices (not the elements) and drop the inner brackets.
.3717
NESTARRAYB = stack(B1,B2) .6015
.6015
The equation now to solve is: [DB1]([NESTARRAYB]o[I8])2 NESTARRAYB = .3717
([DB1] is used to keep the solutions q1 and q2 separate) .6015
.3717
PROGRAM: -.6015
DIAGONALNESTARRAYB = diag(NESTARRAYB) -.3717
DIAGONALNESTARRAYBSQ = [DIAGONALNESTARRAYB]2
q = 1.000 - [DB1][ DIAGONALNESTARRAYBSQ][EIGHT1] = 0
or due to rounding errors is equal to:
q = 7.372EEX-5 = 0
7.372EEX-5 0
We have completed our basic calculations. The complete molecular diagram for butadiene is:
.391
.8378
1.8942 1.4473 1.8942
CH2 - CH - CH - CH2
0 0 0 0
The bond order on the 1,4 carbons deviates the most from maximum, they are predicted to be the most reactive positions. Since qi = 0, butadiene is designated as a molecule with a self-consistent field.
CYCLOBUTADIENE
I shall now do the molecule cyclobutadiene. I am going to consolidate the math to save space, for explanation of the steps and computations, see the solution for butadiene above. We start by numbering the carbons, which we do in consecutive order, starting from left and counting to the right.
C1 C2 H11=H22=H33=H44 = A
H12=H21=H23=H32=H34=H43=H14=H41= B
H13=H31=H24=H42 = 0
C4 C3
The secular determinant is:
X 1 0 1 And its roots (determined by Cramer’s rule) are:
1 X 1 0 = 0
0 1 X 1 X4- 4X2 = X2(X2 - 4); and the roots are (0, 0, +2, -2)
1 0 1 X
X = -2; E1 = A+2B
X = 0; E2 = A
X = 0; E3 = A
X = 2; E4 = A - 2B
At X=2: -2 1 0 1 1 0 1 -2 1 0 1 -2 1 0 0 0 a1=a1
1 -2 1 0 0 1 -2 1 0 1 -2 1 0 1 0 0 a1=a2
0 1 -2 1 0 -2 0 2 0 -2 0 2 0 0 1 0 a1=a3
1 0 1 -2 1 0 2 -2 0 0 2 -3 0 0 0 1 a1=a4
Normalize: (12+12+12+12)1/2 = (4)1/2 = 2; a1 = 1/2 and (1 = 1/2(P1+P2+P3+P4)
At X=0: 0 1 0 1 1 0 1 0 a1=-a3 Putting in terms of a1: a1=a1
1 0 1 0 0 1 0 1 a2=-a4 a2=0
0 1 0 1 0 0 0 0 a1=-a3
1 0 1 0 0 0 0 0 a4=0
Normalize: (12+(-12))1/2 = (2)1/2 = 2; a2=a3 = 1/(2 and (2 and (3 = 1/(2(P1-P3)
At X=2: 2 1 0 0 1 0 1 2 1 0 0 0 a1=a1
1 2 1 0 0 1 2 1 0 -1 0 0 a1=-a2
0 1 2 1 2 1 0 0 0 0 1 0 a1=a3
1 0 1 2 1 2 1 0 0 0 0 -1 a1=-a4
Normalize: (12+(-1)2+12+(-1)2)1/2 = (4)1/2 = 2; a1 = 1/2 and (4 = 1/2(P1+P2+P3+P4)
The wave functions are:
(1 = 1/2(P1+P2+P3+P4)
(2 = 1/(2(P1-P3)
(3 = 1/(2(P2-P4)
(4 = 1/2(P1+P2+P3+P4)
And in matrix form:
.5 .5 .5 .5
(I = 1.4142 0 -1.4142 0
0 1.4142 0 -1.4142
.5 .5 .5 .5
Bond Order: P12=P23=P34=P14 = 2(1/2)(1/2)(1 + 2(1/(2)(0)(2 +2(1/(2)(0)(3 = 1/2
Fi : a12=a14 ; P12+P14 = (1/2)+(1/2)= 1, ( = 3; (Pij = 1+3=4. F = 4.7318 - 4 = .7318
or (Pij=1, F = 1.7318 - 1 = .7318 ignoring the ( bonds.
There is no need to compute qi since cyclobutadiene is a molecule which contains a self-consistent field, thus qi = 0.
The molecular diagram is:
.7318
C1 C2
.50
C4 C3
qi = 0
This molecule has never been isolated, presumably due to it’s instability in it’s resonance stabilization.12
E1 (BUTADIENE) = A + 1.618B
E1 (CYCLOBUTADIENE) = A + 2B
Because E1 (BUTADIENE) has a lower energy than E1 (CYCLOBUTADIENE), it would be predicted to be more stable than cyclobutadiene.
OZONE
Lets look at a non-carbon molecule ozone and see if we can mathematically determine if ozone is a linear molecule, or if all it’s bonds are connected forming a triangular structure. We’ll look at the linear model first.
1 2 3
0 O O H11=H22=H33 = A+2B
H12=H21=H23=H32 = (2B A+2B-E (2B 0
H13=H31 = 0 And (2B A+2B-E (2B = 0
Sij = i=j = 1 0 (2B A+2B-E
i(j = 0
Let X = (A+2B-E)/ (2B, then the Slater determinant becomes
X 1 0 The determinant is equal to: X( X 1 ) - 1( 1 1 ) =
1 X 1 = 0 1 X 0 X
0 1 X
X(X2-1)-X = X3-X-X = X3-2X = X(X2-2)
and the roots are (by inspection): (0, (2, -(2)
E1 = (A+2B-E)/(2B = -(2 = (A+2B-E)= -2B; E1 = A+4B
E2 = (A+2B-E)/ (2B = 0; E2 = A+2B
E3 = (A+2B-E)/ (2B = (2; =(A+2B-E)= 2B ; E3 = A+2B-2B = A
E1 = A+4B
E2 = A+2B
E3 = A
Now we’ll look at the triangular model for ozone:
1
O
3 O O 2
H11=H22=H33 = A+2B A+2B-E (2B (2B
H12=H21=H23=H32=H13=H31 = (2B and (2B A+2B-E (2B
H13=H31 = 0 (2B (2B A+2B-E
Sij = i=j = 1; i(j = 0
Let X = (A+2B-E)/ (2B, then the Slater determinant becomes:
X 1 1
1 X 1 = 0; The equation for this determinant is X3-3X+2; The roots are (1, 1, -2).
1 1 X
E1 = (A+2B-E)/(2B = -2 = (A+2B-E)= -2(2B; E1 = A+4.83B
E2 = (A+2B-E)/ (2B = 1; E2 = A+(2-(2)B = A+.586B
E1 = A+4.83B
E2 = A+.568B
E3 = A+.586B
Because E1 is smaller in the linear form of ozone, it is predicted to be the most stable. Just for the heck of it, let’s compute the ( function for ozone.
The Slater determinant is:
Normalizing: (12+0+(-1)2)1/2 = (2 and (3 = 1/(2(P1-P3) = [.7071 0 -.7071]
R1-(2R2 R2+R3
X 1 0 1.41 1 0 (2 1 0 (2 1 0
1 X 1 =0 ; At X=(2 we get 1 1.41 1 = 0 -1 -(2 0 -1 -(2
0 1 X 0 1 1.41 0 1 (2 0 0 0
(2a1 = -a2 ; a1 = -.707a2
a2 = -(2a3 (-(2a1) = -(2a3 ; a1=a3
a1 = a1 a1 = a1
Normalizing: (12+(-(2)2+12)1/2 = (4 = 2; a = 1/2 ; (2 = 1/2(P1-(2P2+P3) = [.5 -.7071 .5]
For X = -(2
1/(2R1+R2 (2R2+R1
X 1 0 1.41 1 0 -1 1/(2 0 -1 0 1
1 X 1 =0 ; At X=-(2 we get 1 1.41 1 = 1 -(2 1 0 -1 -(2
0 1 X 0 1 1.41 0 1 -(2 0 0 0
a1=a3
a2=(2a3
Putting in terms of a1 :
a1=a1
a1=1/(2a2
a2=(2a3
Normalizing: (12+((2)2+12)1/2 = (4 = 2; a = 1/2 ; (1 = 1/2(P1+(2P2+P3) = [.5 .7071 .5]
(1 = 1/2(P1+(2P2+P3)
(2 = 1/2(P1-(2P2+P3)
(3 = 1/(2(P1-P3)
or
.5 .7071 .5
( = .5 -.7071 .5
.7071 0 -.7071
Let’s use the matrix solution of Gaussian reduction to solve all three states of ozone at the same time. We will solve them in the order of X = -(2, 0 and (2. But since putting 0 in the diagonal for X = 0 gives us a diagonal of zero, I am going to switch rows two and one. This makes the calculations easier. Let B equal the matrix of matrices where the roots replace X, and let A be the first Gaussian reduction matrix needed to get zero’s under the values of the elements in the topmost corner of the diagonal.
[pic] [pic]
Then multiplying AxB will give zeros below each of the a11 elements. Note the matrix for X=0 already has all zero’s under it, so I just multiply by identity and leave it unchanged for this computation.
A x B = C
[pic]
Now we want to make the elements under the a22 positions of the matrices zero, so we multiply C by the matrix A2:
[pic]
Then A2 x C = D
[pic]
We are almost done, all we need to do is get rid of the 1 in the a12 position in the matrix where X = - (2. We do not wish to change anything else, so the rest of the sub-matrices equal identity.
[pic]
And A3 x D = E
[pic]
So lets see what we’ve got:
For the sub-matrix representing the roots of X, and putting in terms of a1:
X = - (2:
a1=a3
a2= (2a3 ; a2= (2a1; a1= 1/(2a2
a1=a1
X = 0:
a1=-a3
a2 = 0
X = (2:
a1=a1
(2a1=-a2 ; a1=-1/(2a2
a2=- (2a3 ; (2a1=- (2a3; a1=-a3
Which are the same values computed above.
UTILIZING SYMMETRY IN BUTADIENE 13
We can take advantage of the symmetry of butadiene to write:
C1=C2-C3=C4 C1=C4; C2=C3. The ( function can be written as:
( = a1/(2(P1 ± P4) + a2/(2(P2 ± P3)
Let’s choose the positive values to find those roots:
( = a1/(2(P1 + P4) + a2/(2(P2 + P3) =
½∫(P1 + P4)H(P1 + P4)d( - E ½∫(P1 + P4)H(P2 + P3)d(
= 0 =(Multiply & add terms)
½∫(P2 + P3)H(P1 + P4)d( ½∫(P2 + P3)H(P2 + P3)d( - E
½∫(P1HP1 +P1HP4 + P4HP1 + P4HP4)d( - E ½∫(P1HP2 +P1HP3 + P4HP2 + P4HP3)d(
=0
½∫(P2HP1 +P2HP4 + P3HP1 + P3HP4)d( ½∫(P2HP2 + P2HP3 + P3HP2 + P3HP3)d( - E
= 1/2(H11 + 2H14 + H44)-E 1/2(H12 + H13 + H42 + H43)
= 0
1/2(H21 + H31 + H24 + H34) 1/2(H22 + 2H23 + H33)-E
= 1/2(A+0+A)-E 1/2(B+0+0+B) A-E B X 1
= = = X2+X-1=0
1/2(B+0+0+B) 1/2(A+2B+A)-E B A+B-E 1 X+1
X = (-1 ± (5)/2 = -1.618; .618
Before we compute for the negative values of these roots, let’s see how we might solve this problem using nested arrays. For convenience, I’m going to ignore the integral sign
P2 H [P1 P4] = P2H [P1 P4] = P2HP1 P2HP4
P3 P3H P3HP1 P3HP4
P1 H [P1 P4] = P1H [P1 P4] = P1HP1 P1HP4
P4 P4H P4HP1 P4HP4
P2 H [P2 P3] = P2H [P2 P3] = P2HP2 P2HP3
P3 P3H P3HP2 P3HP3
P1 H [P2 P3] = P1H [P2 P3] = P1HP2 P1HP3
P4 P4H P4HP2 P4HP3
So the problem becomes:
P1H [P1 P4]-E P1H [P2 P3] P1HP1 P1HP4 -E P1HP2 P1HP3
P4H P4H = P4HP1 P4HP4 P4HP2 P4HP3
P2H [P1 P4] P2H [P2 P3]-E P2HP1 P2HP4 P2HP2 P2HP3 -E
P3H P3H P3HP1 P3HP4 P3HP2 P3HP3
Now E is single valued and I do not want to drag it along when I transpose, so I will separate it from the nested array and re-combine it later. i.e.
P1HP1 P1HP4 P1HP2 P1HP3
P4HP1 P4HP4 P4HP2 P4HP3 -E 0
+
P2HP1 P2HP4 P2HP2 P2HP3 0 -E
P3HP1 P3HP4 P3HP2 P3HP3
Now we will transpose the nested array so that we can use it mathematically. In this case, we wish to compute each sub-matrix separately. We do not want to operate on them all at the same time, so the transposed matrix will be a diagonal rather than a column matrix.
We wish to sum the four elements in each sub-matrix, so we will multiply by [DB1]4,16 .
T Let’s # these as
[DB1]4,16 P1HP1 P1HP4 P1HP2 P1HP3
P4HP1 P4HP4 P4HP2 P4HP3 1 2
+ [-E]2,2 3 4
P2HP1 P2HP4 P2HP2 P2HP3
P3HP1 P3HP4 P3HP2 P3HP3
1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 P1HP1
0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 P4HP1 0
0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 P1HP4
0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 P4HP4
P1HP2
P1HP3
P4HP2
P4HP3 + [-E]2,2 =
P2HP1
[DB1] 1 P2HP4
2 -E P3HP1
3 P3HP4
4 P2HP2
P2HP3
0 P3HP2
P3HP3
P1HP1+P4HP1+P1HP4+P4HP4 0
P1HP2+P1HP3+P4HP2+P4HP3
P2HP1+P2HP4+P3HP1+P3HP4
0 P2HP2+P2HP3+P3HP2+P3HP3
Now we re-transpose, returning the matrix to it’s original form, and add [-E] to complete the computation.
P1HP1+P4HP1+P1HP4+P4HP4 -E P1HP2+P1HP3+P4HP2+P4HP3 1-E 2
;
P2HP1+P2HP4+P3HP1+P3HP4 P2HP2+P2HP3+P3HP2+P3HP3 -E 3 4-E
Of course the numbers inside the brackets are matrices, but since they sum to a single number, it is superfluous whether I add them in or not.
Lets also solve this when we put the transposed nested array into a single column:
1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 P1HP1
0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 P4HP1 [DB1] 1
0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 P1HP4 2
0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 P4HP4 3
P1HP2 4
P1HP3
P4HP2
P4HP3 + [-E]2,2 =
P2HP1
P2HP4
P3HP1
P3HP4
P2HP2
P2HP3
P3HP2
P3HP3
1 2 3 4
[P1HP1+P4HP1+P1HP4+P4HP4 P1HP2+P1HP3+P4HP2+P4HP3 P2HP1+P2HP4+P3HP1+P3HP4 P2HP2+P2HP3+P3HP2+P3HP3]
Which gives us the same answer, but not in the matrix form we need. Of course, this is the first time I’ve encountered a nested array whose dimensions were greater than Nx1 or
1xN. But if we follow the rules, the resulting array re-formed from this row matrix would be the first two elements forms the first row, and the second two would form the second row. i.e.
1 2
3 4
Hey, it works! So I guess we can use nested arrays both as diagonals and columns and most probably as a row also, but I will not get into that here, maybe the next example. When I started this example, I did not know this column method would work, but I’m going to keep my doubts and accomplishments as they were written. The exploration of mathematics is a beautiful wilderness, we learn as we explore it’s beauty. Sometimes, even when I think something can’t work, the math proves me wrong. The above example is such a case.
Letting ( = a1/(2(P1 - P4) + a2/(2(P2 - P3) =
1/2(H11 -2H14 +H44)-E 1/2(H12 - H13 - H24 - H43)
= 0
1/2(H21 - H31 - H42 - H34) 1/2(H22 -2H23 +H33)-E
1/2(A-0+A)-E 1/2(B-0-0+B) A-E B X 1
= = = X2-X-1 = 0
1/2(B-0-0+B) 1/2(A-2B+A)-E B A-B-E 1 X-1
X = (1 ± (5)/2 = 1.618; -.618
COEFFICIENTS BY PROPERTIES OF NON-BONDING MOLECULAR ORBITALS
(NBMO’S)14
Let’s look at methyl hexane:
7 CH2
CH 1
6 CH CH 2
5 CH CH 3
CH
4
First we number the carbons, starting with the longest continuous chain. Then the positions on the molecule are alternately starred to get the greatest number of starred positions.
*
7 CH2
CH 1
6 CH* *CH 2
5 CH CH 3
*
CH
4
Then the sum of the coefficients of the AO’s of the starred atoms directly linked to a given un-starred atom is zero. Thus the sum of a2 + a4 attached to the un-starred atom 3 must equal zero, etc.. Thus
a2 + a4 = 0
a6 + a4 = 0
a2 + a6 + a7 = 0
If we set a4 = 1, then a2 = a6 = -1 and a7 = 2. These coefficients are not normalized. To normalize, we must first determine the energies of each carbon.
X 1 0 0 0 1 1 1 0 0 0 1 1 X 1 0 0 0 1
1 X 1 0 0 0 0 X 1 0 0 0 0 1 X 1 0 0 0
0 1 X 1 0 0 0 (1( 1 X 1 0 0 0 ) + (X( 0 1 X 1 0 0 ) =
0 0 1 X 1 0 0 = 0 1 X 1 0 0 0 0 1 X 1 0
0 0 0 1 X 1 0 0 0 1 X 1 0 0 0 0 1 X 0
1 0 0 0 1 X 0 0 0 0 1 X 0 1 0 0 0 1 X
1 0 0 0 0 0 X
1[-1( 1 0 0 1 1 + X( 1 0 0 0 1 ] + X[X( X 1 0 0 0 )- 1( 1 X 1 0 0 )]
X 1 0 0 0 X 1 0 0 0 1 X 1 0 0 0 1 X 1 0
1 X 1 0 0 1 X 1 0 0 0 1 X 1 0 0 0 1 X 1 =
0 1 X 0 0 0 1 X 1 0 0 0 1 X 0 0 0 0 1 X
0 0 1 1 0 0 0 1 X 0 0 0 0 1 X 1 0 0 0 1
1[1(1[ 1 0 0 1 + -1[1( 1 0 0 1 + X(1( 1 0 0 1 )] +X[-X([ 1 0 0 1
X 1 0 0 X 1 0 0 X 1 0 0 X 1 0 0
1 X 1 0 1 X 1 0 1 X 0 0 1 X 1 0
0 1 X 1 0 1 X 0 0 1 1 0 0 1 X 1
I am going to stop here, find the determinant is quite complex, as you can see. The MathCad +6 program can find it easily for us, and also the roots. Proceed as follows:
X 1 0 0 0 1 1
1 X 1 0 0 0 0
0 1 X 1 0 0 0
D = 0 0 1 X 1 0 0 = 0
0 0 0 1 X 1 0
1 0 0 0 1 X 0
1 0 0 0 0 0 X
click SYMBOLIC
choose MATRIX OPERATIONS
click DETERMINANT OF MATRIX. =
= [pic]
To find the roots, click matrix and set for 8 rows, 1 column. Put the coefficients, starting from the constant and going on up to the coefficient of X7. Type in polyroots(P) and press the = key.
0 -2.101
-7 -1.259
0 -1.000
P = 13 ; polyroots(P) = 0
0 1.000
-7 1.259
0 2.101
1
E1 = (A-E)/B = -2.101; A-E=-2.101B; E1 = A+2.101B
E2 = A+1.259B
E3 = A+B
E4 = A
E5 = A-B
E6 = A-1.259B
E7 = A-2.101B
E( = 7A + 8.72B
To normalize: (a22 + a42 + a62 + a72)1/2 = (12+12+12+22)1/2 = (7.
( = 1/(7(-P2+P4-P6+2P7 -1/(7
-
C .
+ 2/(7
1/(7 + C
-
C
-1/(7
This alternate carbon structure has a self-consistent field, qi = 0. Therefore, if an electron is removed from the NBMO to obtain a benzyl cation, the positive charge will be distributed solely over those atoms whose orbital coefficients are not zero for the NBMO.
The same is true if we add an electron to get a benzyl anion.
CYCLOBUTADIENE by NBMO’s:
First we must find the coefficients ai for cyclobutadiene.
C1 C2
*
C4 *C3
First we star the greatest number of non-adjacent points on the molecule (in this case I choose C1 and C3 ). Then for the point about C2, the starred atoms must equal 0.
C1 + C3 = 0. Since there are tow NBMO’s about C3 (C2 and C4), then C2 + C4 = 0 also. We arbitrarily choose C1 = 1 and C2 = -1 (C1 + C3 = 0; 1 + (-1) = 0). Then (2 (the point between C1 and C3) = 1/(2(P1 -P3) and rotating in the starred direction, (3 = 1/(2(P2 -P4).
And, of course, (1 is in general always positive, so (1 = 1/(4(P1 + P2 + P3 + P4).
(4 is determined by, starting at C1 = 1,(the first adjacent atom) then C2 = -1, C3 = 1 and C4 = -1 as (4 = 1/2(P1 - P2 + P3 - p4).
These correspond to the earlier equations. This works on molecules where qi = 0.
HETEROCYCLIC COMPOUNDS15
Wieland and Pauli adjusted values of AN (the A value for nitrogen) and AO for oxygen by expressing them in terms of A plus some multiple of B.
AN = A + (NB AN(1) = A + 1/2B AN(2) = A + 3/2B
and where and
AO = A + (OB AO(1) = A + 3/2B AO(2) = A + 5/2B
where the one in parenthesis denotes one electron has been donated, and the two in parenthesis denotes that 2 electrons have been donated. Contribution of two electrons by the heteroatom decreases shielding at the position. When ( bonding involves the P2 orbital, the electrons in the system are under the influence of a more attractive potential, and the energy of the electron in the 2Pz orbital decreases. Starting with a molecule of pyrrole:
4 C C 3
5 C C 2
N
1
(AN(2)-E)
The Slater determinant is: A+3/2B-E B 0 0 B X+3/2 1 0 0 0
B A-E B 0 0 1 X 1 0 0
0 B A-E B 0 0 1 X 1 0 = 0
0 0 B A-E B 0 0 1 X 1
B 0 0 B A-E 1 0 0 1 X
The determinant is equal to: X5+3/2X4+-5X3-9/2X2+5X+7/2=0
And has roots: X = (-2.55, -1.15, -.618, 1.20, 1.62)
Substituting each root for X in the Slater determinant, we determine the wave function to be:
(1 = .749P1 + .393P2 + .254P3 + .254P4 + .393P5
(2 = .503P1 - .089P2 - .605P3 - .605P4 - .089P5
(3 = .602P2 - .372P3 - .372P4 - .602P5
(4 = .430P1 - .580P2 + .267P3 + .267P4 - .580P5
(5 = .372P2 - .602P3 + .602P4 - .372P5
For this exercise, we will check the correctness of the coefficients. We proceed as follows:
Checking for (1.
H(1=E(1
Where E = A + 2.55B
Multiply both sides by (1 and integrate
((1H(1d( = ((1E(1d(
((1H(1d(
E =
((1(1d(
but ((1(1d( = 1
Therefore
E = ((1H(1d(
(1 = .749P1 + .393P2 + .254P3 + .254P4 + .393P5 and HAB=HBA.
H11 = A + 3/2B; H22=H33=H44=H55=A
H12=H21=H15=H51=H23=H32=H34=H43=H45=H54=B
H13=H31=H14=H41=H24=H42=H25=H52=H35=H53=0
I am going to do this problem using MathCad again, mainly to reduce the amount of typing needed to illustrate this problem. I can’t double up on lines imported from MathCad, so I’ll have to put in these items one line at a time which really takes up space, but first we’ll solve this by hand.
E = (.749)2H11 + 2(.749)(.393)H12 + (H13,31 & H14,41=0) + 2(.749)(.393)H15 + (.393)2H22 +
2(.393)(.254)H23 + (H24,42 & H25,52 =0) + (.254)2H33 + 2(.254)(.254)H34 +
(.254)2H44 + (H35,53=0) + 2(.254)(.393)H45 + (.393)2H55 =
H11 = A + 3/2B; H22=H33=H44=H55=A
H12=H21=H15=H51=H23=H32=H34=H43=H45=H54=B
E = (.749)2(A+3/2B) + (.393)2A + (.254)2A + (.254)2A + (.393)2A + 4(.294)B +
4(.1)B + 2(.065)B = .999A + .842B + 1.18B + .400B + .130B = .999A + 2.552B
MATHCAD +6 PROGRAM
.749
.393
(T = .254 ( = [.749 .393 .254 .254 .393] ONE5 = [1 1 1 1 1]
.254
.393
1
1
FIVE1 = 1 I5 = identity(5) (5x5 identity matrix)
1
1
(SQ = (T( = .561 .294 .190 .190 .294
.294 .154 .100 .100 .154
.190 .100 .065 .065 .100
.190 .100 .065 .065 .100
.294 .154 .100 .100 .154
We want only the real values (the values where the carbons are connected) so we must make a template matrix. Go to the Slater determinant and for every element that isn’t zero, put in a one in it’s place. Put zero’s in for zero’s. But we want the log of this matrix, and computers can’t take the log of zero, in fact it is undefined, so we must make it ourselves. I will define 0(10-100 so the log 0 ( -100 which is close enough to zero for the purposes of this problem. So to write the log of the template matrix, for every non-zero element in the Slater determinant we will put a zero, and for every zero, we will replace it with -100.
TEMPLATE1 = 0 0 -100 -100 0
0 0 0 -100 -100
-100 0 0 0 -100
-100 -100 0 0 0
0 -100 -100 0 0
We must take the log of each individual element in (SQ, in MathCad, this accomplished by:
LOG(SQ = log((SQ)
LOGENERGY = LOG(SQ + TEMPLATE1
ENERGY = 10LOGENERGY
This takes the anti-log of every element in LOGENERGY.
.561 .294 0 0 .294
.294 .154 .100 0 0
ENERGY = 0 .100 .065 .065 0
0 0 .065 .065 .100
.294 0 0 .100 .154
Now A is equal to the sum of the diagonal, and B is equal to the sum of the elements of the matrix minus the diagonal. Since this is a heteroatom, we must correct the B term in H11. MathCad can make a diagonal from a column matrix, but it cannot remove one, a big problem that they must correct, so we must construct the template to remove the diagonal.
[4]
The log if the 5x5 identity matrix is:
LOGI5 = log(I5)
0 -100 -100 -100 -100
-100 0 -100 -100 -100
LOGI5 = -100 -100 0 -100 -100
-100 -100 -100 0 -100
-100 -100 -100 -100 0
ENERGYA1 = LOGI5 + ENERGY
ENERGYA2 = 10ENERGYA1
.561
.154
ENERGYA2 = .065
.065
.154
ENERGYA = ONE5*ENERGYA2*FIVE1 = .999
To compute B, we must first correct for the Nitrogen term in a11:
3/2 0 0 0 0
0 0 0 0 0
ENERGYCORR = 0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
ENERGYB1 = ENERGYA2*ENERGYCORR
.842 0 0 0 0
0 0 0 0 0
ENERGYB1 = 0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
ENERGYBCORRECTED = ENERGY - ENERGYA2 + ENERGYB1
.561 .294 0 0 .294 .561 .842 0 0 0 0
.294 .154 .100 0 0 .154 0 0 0 0 0
0 .100 .065 .065 0 - .065 + 0 0 0 0 0 =
0 0 .065 .065 .100 .065 0 0 0 0 0
.294 0 0 .100 .154 .154 0 0 0 0 0
ENERGYB = ONE5*ENERGYBCORRECTED*FIVE1
ENERGYB = 2.547
E = .999A + 2.547B which matches.
PERTURBATION THEORY16
This will be the largest section in this part of my paper, mainly because of the importance of perturbation theory to the chemist. The majority of chemical problems are rarely concerned with the absolute energies of molecules, but more with the relative energies of pairs of molecules that are nearly identical. Since chemical reactions rarely involve the formation/breaking of more than one bond, the products and reactants in reactions are necessarily closely related in structure. Even in the examples involving localized and de-localized forms of the same molecule, the calculated resonance energy depends only upon a slight change in the classical structure of the molecule. Thus if one molecule of naphthalene (for instance) is resolved by the Hückel method, the relative energies of mono and di substituted naphthalene compounds can readily be calculated by setting up the un-perturbed matrix and the perturbation matrix for the perturbed state of naphthalene. Since naphthalene is known (we presumably have already calculated the un-perturbed wave function), the perturbed state can easily be found by multiplying the perturbation matrix by the wave function found for the un-perturbed molecule.
Approximate solutions to many problems may be calculated if it is assumed that it’s solution differs only slightly from one that is already known. Suppose H( = E( and that H is the sum of two symmetric matrices H = HO + (H1 , where ( is a scalar, (H1 is small, and Ho(o = Eo(o is known. So if (H1 is small, the eigenvalues E of the perturbed system should not differ much from (O and EO.
Suppose some EOK is non-degenerate with eigenvector (OK and a perturbation of (H1.
Then E = EOK + (EK1 + O((2) + . . . where O((2) means all subsequent terms are multiplied by powers of ( greater than or equal to 2. We shall determine EK1 and the coefficients of the higher powers of (. Similarly, ( = (OK + ((1K + O((2) + . . . where (1K and E are to be determined. Substituting these values of ( and E into H(=E( we get:
(HO + (H1)( (OK + ((1K + O((2) + . . .) = (EOK + (EK1 + O((2) + . . .)( (OK + ((1K + O((2) + . . .)
or
Ho(Ko + E[HO(K1 + H1(OK + . . .] = EOK(OK + ([ EOK(1K + EK1(OK + . . .]
and equating coefficients of like powers of (, we have for the zeroth order term
HO(OK = EoK(Ok
which is where we started. The first order correction (from the coefficients of () is:
HO(K1 + H1(Ok = EOK(1K + EK1(OK,
with EK1 and (1K to be determined. HO forms a basis in a vector space of n dimensions, so we can expand (1K in terms of (Oj.
(1K = (cjk(Oj, and H1(Ok = (bjk(Oj
j j
cjk is unknown, bjk is known: bjk = ((Oj)TH1(OK. Since the matrix is symmetric, bjk = bkj. These values can now be substituted into the above equation:
HO(K1 = HO(cjk(Oj = (cjkEOj(Oj
and the first order equation becomes:
(cjkEOj(Oj + (bjk(Oj = EOK(cjk(Oj + EK1(2K .
Again, (Oj is a set of n linearly independent vectors in n dimensions, so the above equation can be solved only if the coefficients of identical vectors on either side are separately equal. From the coefficients of (OK we have bKK = E1K. which determines the first order correction to the eigenvalue. From the coefficient of (Oj i(k, we get
cjk = bjk/(EOK - EOJ),
the values of which determine the first order correction to the eigenvector. So we now have the equation:
( = (Ok + (((bjk(Oj)/(EOK - EOJ) + O(()2
and it’s associated eigenvalue
E = EOK + (((Ok)TH1(OK + O(()2
EXAMPLE: FORMALDEHYDE
In this example, H2C=O is considered to be a perturbation of ethylene where a CH2 is replaced by an oxygen atom. Since oxygen is more electronegative than carbon, the effect of such a change will make the diagonal element corresponding to oxygen more negative. Assume H22 for oxygen = A + B, then the determinant corresponding to H2C=O is
A B = 0 ; H = HO + (H1 and ( = 1. (If H22 = A + .5B, then ( = .5)
B A+B
So in this case, HO = A B and H1 = 0 0
B A 0 B .
The solution for HO (the un-perturbed function for ethylene) is:
E1 = A + B and (1 = 1/(2(P1 + P2)
E2 = A - B and (2 = 1/(2(P1 - P2)
The coefficients bjk are calculated from the zero order eigenvectors
bjk = ((Oj)TH1(OK
(O1 (in matrix form) = [1/(2, 1/(2]
(O2 (in matrix form) = [1/(2, -1/(2]
b11 = [1/(2, 1/(2] 0 0 1/(2 = [0 1/(2B] 1/(2 = 1/2B
0 B 1/(2 1/(2
b22 = [1/(2, -1/(2] 0 0 1/(2 = [0 -1/(2B] 1/(2 = 1/2B
0 B -1/(2 -1/(2
b12 = [1/(2, -1/(2] 0 0 1/(2 = [0 -1/(2B] 1/(2 = -1/2B
0 B 1/(2 1/(2
[pic]
H1 = 0 0
0 B
. BE SURE THAT IN THE MATH MENU AUTOMATIC MODE AND LIVE SYMBOLICS ARE CHECKED. TYPE IN THE FUNCTION AND CLICK TO PUT EXPRESSION IN A SELECTION BOX, THEN PRESS CONTROL PERIOD (CTRL . ), THIS WILL GIVE THE SYMBOLIC EQUAL SIGN . THEN CLICK ANYWHERE OUTSIDE THE EXPRESSION TO GET THE SOLUTION.
[pic]
OR WE CAN TAKE THE ROW FROM THE PSI1 MATRIX THAT CORRESPONDS TO THE ROW THAT THE B IS IN AND SQUARE IT AS PSI1SQ = PSI[2]^TxPSI[2]. (IN THIS CASE, IT IS IN ROW 2) i.e.
[pic]
WHICH GIVES US THE SAME ANSWER (SEE PART TWO OF THIS PAPER).
The un-normalized perturbed ground state function is given by:
(1 = (O1 + b21/(EO1 - EO2) = (O1 - 1/2B/(A+B)-(A-B) = (O1 - 1/2B/2B (O2 = (O1 -1/4(O2
Normalizing (1
(1NORM = 1/[11 + (-1/4)2]1/2 = 1/1.033
Therefore (1NORM = 1/1.033((O1 -1/4(O2 ) = (1 = .968(O1 - .243(O2 =
(To see how to normalize using matrices, see example in part two of this paper.)
.968[ 1/(2 1/(2] - .243 [1/(2 -1/(2] ( [1 1] .968 0 1/(2 1/(2 =
0 -.243 1/(2 -1/(2
[pic]
(1NORM = [1 1] .68448 .68448 =
-.17183 .17183
The program for the rest of this problem is:
[pic]
[pic]
[pic]
And now we will solve for (2 .
E2 = EO2 +(b22 = A-B +1/2B = A-1/2B
(2 = (O2 + b12/(EO2 - EO1) = (O2 - 1/2B/(A-B)-(A+B) = (O1 - 1/2B/-2B (O1 = (O2 +1/4(O1
Normalizing (2
(2NORM = 1/[11 + (1/4)2]1/2 = 1/1.033
(2NORM = -.968(O1 - .243(O2 =
-.968[ 1/(2 1/(2] - .243 [1/(2 -1/(2] ( [1 1] -.968 0 1/(2 1/(2 =
0 -.243 1/(2 -1/(2
The rest I will solve using MathCad +6:
[pic]
[pic]
[pic]
or
( = .514 .858
-.858 -.514
The two electrons occupy the ground state (, so E1 = A + 1.5B and
E( = 2(A + 1.5B) = 2A + 3B. If the bond is broken, the electrons occupy the separate atoms, A for carbon, A + B for oxygen, or E = A + (A + B) = 2A + B. So
DE( = 2A = 3B - 2A - B = 2B, which is the same DE( of un-perturbed ethylene.
The charge distribution qi for carbon is 2(.514)2 = .528. The charge distribution q2 for oxygen is 2(.858)2 = 1.476. Subtracting the residual charge of 1 from each value we get the molecular charge distribution. C = .472; O = -.472.
d=1.21 Δ
.472 -.472
C O
The dipole moment ( is defined as the product of the charge concentration about the positive pole times the inter-atomic distance. .so here
( = (.472)e-(1.21x10-5cm) = (.472)(4.803x10-10esu)(1.21x10-5cm) = 2.74 Debye units.
Experimental values for ( for formaldehyde are found to be from 2.29 - 2.34 Debye units. But note, the perturbation for ethylene covers 50% of the molecule, which is a very large percentage (50%). If the molecule had 100 carbons, the perturbation of adding one oxygen would be 1%, so the larger the molecule, the closer to the actual approximated value the answers will be.
PART 2
PERTURBATION THEORY
This part of my paper on Molecular Orbital methods was sent to 9 different math and physics journals. It was rejected in total by all of them. Ms. Frobosi of the Indiana Journal of Mathematics said the method was a fluke, it works, but it is impossible for it to work.
Around October 1995, I discovered what seems to be a new matrix operator. This operator that cannot work solves problems in statistics, quantum chemistry, the solution of equations (Gaussian reduction) and can create a universal accounting/inventory system that can keep track of everything that is made and the parts it is made of, everything that is bought and sold, and who bought it and from whom and for how much, and also can keep track of taxes, all in four program steps. Not millions, like in Excel or Lotus databases, but 4 steps, six steps if we want specialized information from the system. I say this because I proved we are able to multiply matrices differently from the ordinary dot product and still come up with the correct answer. In the half-multiplier mode, though, we can compute the intermediate values of matrix multiplication before we sum, and many of these values are of interest and some are very important to understanding the system we are working on.
The half-multiplier operator works the same as the normal matrix multiplier, but it ‘freezes’ the operation of multiplication before summation is done. Information about a system not available through accepted methods of multiplication are easily accessible through utilization of this operator. The sub-matrices calculated each represent one summed row, or one summed column in the final matrix. Although not necessarily a linear combination, the sub-matrices seem more like nested arrays that await our use. We may add/subtract them as in a linear combination of matrices, cross multiply them, or re-define the dimensions of the matrix, transpose and by transposing, drop the inner brackets to produce one large grand matrix that follows the rules of our regular mathematics, and use this to solve many different problems all at the same time, keeping them all separate, or adding them all together in one lump. One of the really neat properties of this operator is if a computer is not set up to do the math, there is a one to one correspondence with regular matrices, but the normal way uses many more steps than by using the half-multiplier. For instance, [A]T31 o [B]33 needs nine multiplication’s to get the answer, but the equivalent matrix operation is to change [A] (a column matrix) into a square diagonal matrix of 9 elements. The first row needs 9 multiplication’s, as does the second and third row, which makes 27 multiplication’s, even though 6 of every 9 is a multiplication by zero. Plus we must add the 27 numbers three at a time to obtain the final solution. In other words, with the half multiplier, we get the answer in 32 or 9 steps, but with regular matrix multiplication, the answer needs 2(3)3 or 54 mathematical steps to get the same result. Or better yet, suppose we are working with the anti-AIDS medicine AZT which has about 100,000 carbons and various other atoms composing it. I will show that for a single perturbation, we can calculate bij in 100,0002 steps, or 1010 mathematical operations. The accepted way is bij = (TH(. (TH needs 2(105)3 or 2x1015 mathematical steps. Then ((TH)( also needs 2(105)3 steps for a total of 4(1015) mathematical operations, which is a factor of 40 thousand more than with the half-multiplier. Of course, with bij = (TH( we can calculate all the perturbations at once, with the half multiplier though, we must compute one perturbation at a time and add the solutions. So for two perturbations, rather than one, we need 2x1010 steps to multiply and add another 100,000 to add them, for a total of 3x1010 steps, still much fewer steps than by computing bij the accepted way.
Although the original derivation of the half-multiplier operator had nothing to do with science, one of the field equations I developed (c((AijT o Tjk) = Gjk (see derivation of Einstein’s First Field Equation for Gravity and the derivation for the wave equation for gravity at the beginning of this paper) reminded me so much of Einstein’s First Field Equation (8((Tjk = Gjk ) that I thought the operator might be applicable to physics as well as accounting/inventory systems and the solving of linear equations. If in the equation
(c((AijT o Tjk) = Gjk
Aij = I and c = 8((, the equation reduces to Einstein’s equation.
In Einstein’s equation, Tjk is a square matrix, if in general, Bjk is not a square matrix, [I]Tij o [B]jk = [B]jk. Therefore [I]Tij can also be defined as the identity matrix in the half-multiplier mode. i.e.
1 o a11 a12 a11 a12 1 a11 a12
1 o a21 a22 a21 a22 1 a21 a22
. o . . . . . . .
1 o ai1 ai2 = ai1 ai2 ( 1 ai1 ai2
. o . . . . . . .
. o . . . . . . .
1 o an1 an2 an1 an2 1 an1 an2
There is no new physics presented here, all the derivations were obtained from books on Quantum Chemistry and my notes from Quantum Mechanics I & II I took while in college at Fort Hays State in Hays, KS. What this paper covers is how to simplify the calculations so I could program the problem on my HP-48G calculator and MathCad +6 computer program.
So long as an exact or as close to exact solution can be obtained by hand or computer program first, no matter how much calculus or differential equations are used, the perturbation calculation can be used to further approximate solutions for tiny changes in the system using only regular addition, subtraction, multiplication and division, with some miscellaneous logarithms thrown in for good measure. I remember virtually nothing about matrix mechanics since, when I graduated 19 years ago and have yet to receive my first interview for a job, I have forgotten almost everything. When I wanted to see if the operator worked in physics, I had to find a simple problem that even I could understand and solve. Non-degenerate perturbation theory stood out from all other matrix mechanics problems for it’s simplicity and ease of comprehension. Everything else was beyond my mathematical abilities. This part of the paper is long, but that is because I cover every step with few or no shortcuts. To conserve on the length, therefore, there is more math than words of explanation. The math is so simple it should be understood with little trouble (as long as you understand how matrices are multiplied). Perturbation theory is not only applicable to chemistry, but also to electrical networks, mechanical vibrations, statistics and everything else where matrices are used to calculate linear systems.
EXAMPLE: COMPUTING THE WAVE FUNCTION OF ACROLEIN FROM BUTADIENE
In Quantum Mechanics, the perturbation calculation is given as
(i1PERTURBED = (Cij)NORM(i0
with
Cij = bij/(E0I - E0J); bij = ((0i)TH1(0i
Because there is no onto mapping of matrices for matric multiplication defined, only into mappings for multiplication, (onto mappings are defined only for addition and subtraction), I get around this difficulty by taking the log of the bij matrix and the log of the (Eij matrix defining any zero’s as 10-100. Since we are dividing the bij by (Eij, we just subtract the logs. It is now a simple subtraction problem to divide each element individually without manually placing the divisions by hand. The half-multiplier operator takes care of this automatically. I have also extended the theory of multiple perturbations later in this paper. In this example, the energy for oxygen = A + (B where ( = 1. ( is not necessarily = 1, it depends upon the value determined by experiment. For butadiene,
(1 (2 (3 (4
.3717 .6015 .6015 .3717
(01 = .6015 .3717 -.3717 -.6015
.6015 -.3717 -.3717 .6015
.3717 -.6015 .6015 -.3717
Suppose we wish now to solve for the ( function for acrolein (C=C-C=O). Instead of solving exactly, perturbation theory may be used. NOTE: For purposes of this paper,
(I is defined as the I’th row of the (ij matrix; (0i is defined as the whole ( matrix; (j is the j’th column of the (ij matrix. (I may also be denoted as (1j for the first row, (2j for the second row, etc.. (ij = one single element of the matrix (0i. i.e. (23 is the element in (0I corresponding to the third row, second column, in this case, it equals -.3717.
(1K = (Cjk(Oj and bjk = ((Oj)TH1(OK ; with E for oxygen = A + B. The off-diagonal elements are given by Cjk = bjk/(EOK - EOJ), and Ei = bii. The Slater determinant is given by:
H0 H1
A-E B 0 0 X 1 0 0 X 1 0 0 0 0 0 0
B A-E B 0 = 0 ; letting X = (A-E)/B; 1 X 1 0 = 1 X 1 0 + 0 0 0 0
0 B A-E B we get 0 1 X 1 0 1 X 1 0 0 0 0
0 0 B (A+B)-E 0 0 1 X+1 0 0 1 X 0 0 0 1
First we will calculate bjk:
bjk = ((Oj)TH1(OK
.3717 .6015 .6015 .3717 0 0 0 0 .3717 .6015 .6015 .3717
bjk = .6015 .3717 -.3717 -.6015 0 0 0 0 .6015 .3717 -.3717 -.6015 =
.6015 -.3717 -.3717 .6015 0 0 0 0 .6015 -.3717 -.3717 .6015
.3717 -.6015 .6015 -.3717 0 0 0 1 .3717 -.6015 .6015 -.3717
.1382 -.2236 .2236 -.1382
-.2236 .3618 -.3618 .2236
.2236 -.3618 .3618 -.2236
-.1382 .2236 -.2236 .1382
The energy correction lies along the principle diagonal bii.
HP-48G PROGRAM
RCL PSI01
TRN
RCL H1
x
RCL PSI01
x
STO bij
MATHCAD PROGRAM
bij: (01T*H1*(01
To calculate the energy for acrolein, we proceed as follows:
BY HAND:
E1 = E01 + E11 = (A + 1.618B) + .1382B = A + 1.7562B
E2 = E02 + E12 = (A + .618B) + .3618B = A + .9798B
E3 = E03 + E13 = (A - .618B) + .3618B = A - .2562B
E4 = E04 + E14 = (A - 1.618B) + .1382B = A - 1.4798B
Or in matrix form:
A+1.618B .1382B
A+.618B + .3618B =
A-.618B .3618B
A-1.618B .1382B
A+1.7562B
A+.9798B
A-.2562B
A-1.4798B
Or
(A + 1.618B) + .1382B = A + 1.7562B
(A + .618B) + .3618B = A + .9798B
(A - .618B) + .3618B = A - .2562B
(A - 1.618B) + .1382B = A - 1.4798B
HP-48G PROGRAM: (2 methods)
RCL bij RCL bij
(DIAG (DIAG
RCL E 4
+ (
STO EACROLIN DIAG(
RCL E
4
(
DIAG(
+
STO EACROLIN
To get MathCad to add symbolically, we must first divide by B:
[pic]
We cannot do the following computations symbolically, so we must just use the values of B to calculate the energy.
[pic] [pic]
[pic] creates a 4x4 identity matrix
[pic] multiplies onto bjk to give diagonal of bjk
[pic] diagonalizes the E matrix
[pic] adds the two diagonalized matrices giving us the B values.
[pic]
Another, more elegant method to compute the Energy of acrolein is to square by half- multiplier the row in (01 that corresponds to the row in which the perturbation occurs:
E1I = Ei + (B((4T)O2
Ei (4T (4T
1.618 .3717 .3717 1.618 .1382 1.7562
E1I = .618 + -.6015 o -.6015 = .618 +.3618 = .9798
-.618 .6015 .6015 -.618 .3618 -.2562
-1.618 -.3717 -.3717 -1.618 .1382 -1.4798
To calculate all the non-diagonal elements, for all i(j; Cjk = bjk/(E0i - E0j).
For a solution to the operator using the half-multiplier operator, we will calculate the bjk matrix as follows:
bjk = ((Oj)TH1(OK = (((((Oj)T o H1)T o (OK
((Oj)T o H1)T = For (1, take the first row of ((Oj) [.3717 .6015 .6015 .3717], transpose, and o multiply into H1 We will have four solutions to this first part. For (2, take the second row of ((Oj) [.6015 .3717 -.3717 -.6015], transpose, and o multiply into ((Oj). Repeat for (3 and (4.
.3717 .6015 .6015 .3717 0 0 0 0
bjk = .6015 .3717 -.3717 -.6015 o 0 0 0 0
.6015 -.3717 -.3717 .6015 0 0 0 0
.3717 -.6015 .6015 -.3717 0 0 0 1
= .3717 0 0 0 0 .6015 0 0 0 0 .6015 0 0 0 0 .3717 0 0 0 0
.6015 o 0 0 0 0 .3717 o 0 0 0 0 -.3717 o 0 0 0 0 -.6015 o 0 0 0 0
.6015 0 0 0 0 -.3717 0 0 0 0 -.3717 0 0 0 0 .6015 0 0 0 0
.3717 0 0 0 1 -.6015 0 0 0 1 .6015 0 0 0 1 -.3717 0 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 .3717 0 0 0 -.6015 0 0 0 .6015 0 0 0 -.3717
Then using the cross product property of the half-multiplier, we sum the rows to form a single matrix from the nested array.
0 0 0 0
0 0 0 0
0 0 0 0
.3717 -.6015 .6015 -.3717
Now we o multiply this value into (01: Note: the pre-multiplier matrix is already transposed, so we do not have to transpose it again.
0 0 0 0 .3717 .6015 .6015 .3717
0 0 0 0 o .6015 .3717 -.3717 -.6015 =
0 0 0 0 .6015 -.3717 -.3717 .6015
.3717 -.6015 .6015 -.3717 .3717 -.6015 .6015 -.3717
0 .3717 .6015 .6015 .3717 0 .3717 .6015 .6015 .3717
0 o .6015 .3717 -.3717 -.6015 0 o.6015 .3717 -.3717 -.6015
0 .6015 -.3717 -.3717 .6015 0 .6015 -.3717 -.3717 .6015
.3717 .3717 -.6015 .6015 -.3717 -.6015 .3717 -.6015 .6015 -.3717
=
0 .3717 .6015 .6015 .3717 0 .3717 .6015 .6015 .3717
0 o.6015 .3717 -.3717 -.6015 0 o.6015 .3717 -.3717 -.6015
0 .6015 -.3717 -.3717 .6015 0 .6015 -.3717 -.3717 .6015
.6015 .3717 -.6015 .6015 -.3717 -.3717 .3717 -.6015 .6015 -.3717
Note: these matrices should be in a single row, I just do not have the room on this page to place them so. Now we o multiply all the sub-matrices and keep them in a single row:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
.1382 -.2236 .2236 -.1382 -.2236 .3618 -.3618 .2236 .2236 -.3618 .3618 -.2236 -.1382 .2236 -.2236 .1382
Now we wish to sum the columns and create a new matrix of these summed columns, so lets transpose the nested array, keeping each sub-matrix as a 4x4 square of elements.
T
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 =
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
.1382 -.2236 .2236 -.1382 -.2236 .3618 -.3618 .2236 .2236 -.3618 .3618 -.2236 -.1382 .2236 -.2236 .1382
1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 .1382 -.2236 .2236 -.1382
0 0 0 0
0 0 0 0 .1382 -.2236 .2236 -.1382
0 0 0 0 -.2236 .3618 -.3618 .2236
-.2236 .3618 -.3618 .2236 = .2236 -.3618 .3618 -.2236
0 0 0 0 -.1382 .2236 -.2236 .1382
0 0 0 0
0 0 0 0
.2236 -.3618 .3618 -.2236
0 0 0 0
0 0 0 0
0 0 0 0
-.1382 .2236 -.2236 .1382
Which is the same solution we got for bjk = ((Oj)TH1(OK But what a lot of math to go through to get the same answer. There is a pattern, let’s see if it can take us anywhere. All values in the problem after the first half-multiplication are between the elements in row four and row four itself. i.e.
(41 o (4 .3717 .3717 -.6015 .6015 -.3717
bjk = (42 o (4 = -.6015 o.3717 -.6015 .6015 -.3717 =
(43 o (4 .6015 .3717 -.6015 .6015 -.3717
(44 o (4 -.3717 .3717 -.6015 .6015 -.3717
.1382 -.2236 .2236 -.1382 4
-.2236 .3618 -.3618 .2236 = bjk = (B((4k o (4
.2236 -.3618 .3618 -.2236 1
-.1382 .2236 -.2236 .1382
Hmmm, 16 mathematical steps, this is a step in the right direction. Since the matrix is symmetric, bjk = (B((j4 o (4 also works. The energy E is equal to the diagonal, or
E = (B(jj.
GENERAL PROOF
To generalize, suppose we have
011
022
033
.
H1 = . and (I = (jT (matrices are symmetric)
(Bjj
.
.
0nn
then (j1 o (j
(j2 o (j
(j3 o (j
.
bjk = . (B = (B((jk o (I = (B(((kj)T o (I
.
(jk o (j E = (B(B((jk o [(I]jj
.
. I put [(I]jj in brackets to show it is a square diagonal
(jn o (j matrix, not the jj’th element of the matrix.
But there is a simpler way to express this for a perturbation of the fourth row, we take the original multiplication and get rid of all the zero’s, leaving only the rows with numbers. i.e. For a perturbation at the fourth carbon, we can write
[.3717 -.6015 .6015 -.3717] o[.3717 -.6015 .6015 -.3717] = (4 o (4 = (4o2 which is different from ((4T)o2 which gives the energy correction term. I am not going to solve
(4 o (4 because I have already solved it above. Remember, the half-multiplier has a one-to-one relationship with regular matrix multiplication, all we need to do is transpose the first matrix and multiply, so lets try it.
(4o2 = (4 o (4 = (4T (4 = .3717 [.3717 -.6015 .6015 -.3717] =
-.6015
.6015
-.3717
.1382 -.2236 .2236 -.1382 4
-.2236 .3618 -.3618 .2236 = bjk = (B((4k o (4 = ((Oj)TH1(OK = (4o2 = (4 o (4 = (4T (4
.2236 -.3618 .3618 -.2236 1
-.1382 .2236 -.2236 .1382
For convenience, I’m going to call the row in which the perturbation occurs as (B, in this case, (B = (4 . I am now going to attempt to prove ((Oj)TH1(OK = (4o2 .
a11 a21. . . aj1 . . . an1 an+1,1
a12 a22. . . aj2 . . . an2 an+1,2
. . . . . . . . . . . .
. . . . . . . . . . . .
a1k a2k. . . ajk . . . ank an+1,k
. . . . . . . . . . . .
. . . . . . . . . . . .
a1n a2n. . . ajn . . . ann an+1,n
a1,n+1 a2,n+1. . aj,n+1 . . an,n+1 an+1,n+1
a11 a21. . . aj1 . . . an1 an+1,1 0
a12 a22. . . aj2 . . . an2 an+1,2 0
. . . . . . . . . . . . 0
. . . . . . . . . . . . .
a1k a2k. . . ajk . . . ank an+1,k .
. . . . . . . . . . . . (Bjj
. . . . . . . . . . . . .
a1n a2n. . . ajn . . . ann an+1,n .
a1,n+1 a2,n+1. . aj,n+1 . . an,n+1 an+1,n+1 0n+1,n+1
a11 a21. . . aj1 . . . an1 an+1,1 0 a11 a12. . . a1k . . . a1n a1,n+1
a12 a22. . . aj2 . . . an2 an+1,2 0 a21 a22. . . a2k . . . a2n a2,n+1
. . . . . . . . . . . . 0 . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
a1k a2k. . . ajk . . . ank an+1,k . aj1 aj2. . . ajk . . . ajn aj,n+1
. . . . . . . . . . . . (Bjj . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
a1n a2n. . . ajn . . . ann an+1,n . an1 an2. . . ank . . . ann an+1,n
a1,n+1 a2,n+1. . aj,n+1 . . an,n+1 an+1,n+1 0n+1,n+1 an+1,1 an+1,2. . an+1,k . . an+1,n an+1,n+1
STEP 1: Multiply ((Oj)TH1
a11 a21. . . aj1 . . . an1 an+1,1 0 0 0 0 . . aj1 . 0
a12 a22. . . aj2 . . . an2 an+1,2 0 0 0 0 . . aj2 . . 0
. . . . . . . . . . . . 0 . . . . . . 0
. . . . . . . . . . . . . . . . . . . . . . 0
a1k a2k. . . ajk . . . ank an+1,k . (B =(B . . . . . .ajk . . 0
. . . . . . . . . . . . 1 . . . . . . . . . 0
. . . . . . . . . . . . . . . . . . . . . . 0
a1n a2n. . . ajn . . . ann an+1,n . 0 0 0 . . .ajn . . 0
a1,n+1 a2,n+1. . aj,n+1 . . an,n+1 an+1,n+1 0n+1,n+1 0 0 0 . . .aj,n+1 . 0
STEP 2: Multiply (Oj by the product of ((Oj)TH1 :
0 0 0 . . aj1 . 0 a11 a12. . . a1k . . . a1n a1,n+1
0 0 0 . . aj2 . . 0 a21 a22. . . a2k . . . a2n a2,n+1
. . . . . . 0 . . . . . . . . . . . .
. . . . . . . . . 0 . . . . . . . . . . . .
. . . . . .ajk . . 0 (B aj1 aj2. . . ajk . . . ajn aj,n+1 =
. . . . . . . . . 0 . . . . . . . . . . . .
. . . . . . . . . 0 . . . . . . . . . . . .
0 0 0 . . .ajn . . 0 an1 an2. . . ank . . . ann an+1,n
0 0 0 . . .aj,n+1 . 0 an+1,1 an+1,2. . an+1,k . . an+1,n an+1,n+1
aj1aj1 aj1aj2. . . aj1ajk . . . aj1ajn aj1aj,n+1
aj1aj2 aj2aj2. . . aj2ajk . . . aj2ajn aj2aj,n+1
. . . . . . . . . . . .
. . . . . . . . . . . .
aj1ajk aj2ajk. . . ajkajk . . . ajkajn ajkaj,n+1 (B = ((Oj)TH1(OK
. . . . . . . . . . . .
. . . . . . . . . . . .
aj1ajn aj2ajn. . . ajkajn . . . ajnajn ajnaj,n+1
aj1aj,n+1 aj2aj,n+1. . ajkaj,n+1 . . ajnaj,n+1 aj,n+1aj,n+1
Now we’ll multiply (BT*(B but I’ll do this by MathCad symbolic:
[pic]
NOTE: In MathCad symbolic, the program won’t take commas in the subscripts, nor will it take the . . . notation, so to rewrite it in proper form we get:
aj1aj1 aj1aj2. . . aj1ajk . . . aj1ajn aj1aj,n+1
aj1aj2 aj2aj2. . . aj2ajk . . . aj2ajn aj2aj,n+1
. . . . . . . . . . . .
. . . . . . . . . . . .
aj1ajk aj2ajk. . . ajkajk . . . ajkajn ajkaj,n+1 (B = ((Oj)TH1(OK = (Bo2 = (BT*(B = bjk
. . . . . . . . . . . .
. . . . . . . . . . . .
aj1ajn aj2ajn. . . ajkajn . . . ajnajn ajnaj,n+1
aj1aj,n+1 aj2aj,n+1. . ajkaj,n+1 . . ajnaj,n+1 aj,n+1aj,n+1
QED
Now we will compute Cjk = bjk/(Ej - Ek)
First I am going to re-define the [1] operator. Up until now I have been using it as a row or a column matrix, but now I want to define it as an n x m matrix, or in the case of Quantum Chemistry, as an n x n matrix.
Define (jj[1]jj j=k=1; and -1(jj[1]jj j=k=0
j(k=0 j(k=1
such that (jj[1]jj + -1(jj[1]jj = [1]jj . To illustrate for j=k=4:
(jj[1]jj + -1(jj[1]jj = [1]jj
1 0 0 0 0 1 1 1 1 1 1 1
0 1 0 0 + 1 0 1 1 = 1 1 1 1
0 0 1 0 1 1 0 1 1 1 1 1
0 0 0 1 1 1 1 0 1 1 1 1
Define Bjk = (bjk - Ejj + (jj[1]jj ) = (bjk - bjj + (jj[1]jj)
Define (Ejk - (jj[1]jj = (E1)T o (-1(jj[1]jj)
Define Subtraction as (E-jk - (jj[1]jj = (E1)T o (-1(jj[1]jj) - ((E1)T o (-1(jj[1]jj))T
Define addition as (E+jk - (jj[1]jj = (E1)T o (-1(jj[1]jj) + ((E1)T o (-1(jj[1]jj))T
Note: This mess is not actually necessary if we only wish to perform subtraction, the subtraction process itself puts zero’s in the principle diagonal. But if for some reason we wish to add the energies, the above equation is the only way it will work.
Again
Cjk = bjk/(Ej - Ek)
Therefore, Log Bjk - Log (E-jk = Log Cjk
and thus
Anti-log (Cjk)NORM(01 = (i1PERTURBED
Then for the perturbation at (4:
b4 = (4o2 = (4T(4
b4 =(4o2 = (4 o (4 = (4T (4 =
[.3717 -.6015 .6015 -.3717] = .3717 .1382 -.2236 .2236 -.1382
-.6015 = -.2236 .3618 -.3618 .2236
.6015 .2236 -.3618 .3618 -.2236
-.3717 -.1382 .2236 -.2236 .1382
Bjk = (bjk - Ejj + (jj[1]jj ) = (bjk - bjj + (jj[1]jj)
.1382 -.2236 .2236 -.1382 .1382 1 0 0 0
B4 = -.2236 .3618 -.3618 .2236 - .3618 + 0 1 0 0 =
.2236 -.3618 .3618 -.2236 .3618 0 0 1 0
-.1382 .2236 -.2236 .1382 .1382 0 0 0 1
1 -.2236 .2236 -.1382
B4 = -.2236 1 -.3618 .2236
.2236 -.3618 1 -.2236
-.1382 .2236 -.2236 1
COMPUTING ΔE
Now we’ve computed bij and put ones in the diagonal. Now we must divide each element by it’s Ej - Ek value. We don’t have to do it this way, because when we subtract the E matrix and it’s transpose, the diagonal will be all zero’s anyway. But I’m going to show the other way, just in case we ever have to add the energies rather than subtract them.
The subtraction matrix is (in general)
E1 E2 E3 E4
-E1 -E2 -E3 -E4
Then (E-jk - (jj[1]jj = (E1)T o (-1(jj[1]jj) - ((E1)T o (-1(jj[1]jj))T
Let’s translate so we can understand what this equation means:
E1 -E1 0 1 1 1 E1 0 1 1 1 E1 0 1 1 1 T
E2 -E2 o1 0 1 1 = E2 o 1 0 1 1 - E2 o 1 0 1 1 + (jj[1]jj
E3 -E3 1 1 0 1 E3 1 1 0 1 E3 1 1 0 1
E4 -E4 1 1 1 0 E4 1 1 1 0 E4 1 1 1 0
0 E1 E1 E1 0 E2 E3 E4 1 0 0 0 1 E1-E2 E1-E3 E1-E4
(E-jk = E2 0 E2 E2 - E1 0 E3 E4 + 0 1 0 0 = E2-E1 1 E2-E3 E2-E4
E3 E3 0 E3 E1 E2 0 E4 0 0 1 0 E3-E1 E3-E2 1 E3-E4
E4 E4 E4 0 E1 E2 E3 0 0 0 0 1 E4-E1 E4-E2 E4-E3 1
Now (E-jk = 1.618 .618 -.618 -1.618
-1.618 -.618 .618 1.618
1.618 -1.618 0 1 1 1 1.618 0 1 1 1 -1.618 0 1 1 1 T
.618 -1.618 o 1 0 1 1 = .618 o 1 0 1 1 + -.618 o 1 0 1 1 + (jj[1]jj =
-.618 .618 1 1 0 1 -.618 1 1 0 1 .618 1 1 0 1
-1.618 1.618 1 1 1 0 -1.618 1 1 1 0 1.618 1 1 1 0
[pic]
[pic]
[pic] We use the vectorize operator to take the log of (E .
[pic]
[pic]
[pic] We use the vectorize operator to take the log of B4 .
[pic]
[pic] We subtract the logs, effectively dividing each element in B4 by it’s corresponding element in (E.
[pic]
[pic] We subtract the logs, effectively dividing each element in B4 by it’s corresponding element in (E.
[pic]
This is the un-normalized form of Cjk. We will use this when we compute the second and higher order energy corrections. But for the purpose of finding the wave function of acrolein, we must normalize this matrix. Let’s calculate them all at the same time. All we need to do is square CjkUN-NORM, the diagonal is the sum of the separate functions squared. We must then invert the values and take the square root of the values. Then we can o multiply the normalization matrix into CjkUN-NORM, and multiply by the original un-perturbed wave function to obtain the corrected wave function of acrolein.
To find the normalization factors: (I hope my notation here is correct)
(jj = diagonal(CjkTCjk); then
((jj-1)1/2Cjk(01 = (11
[pic]
MATHCAD CANNOT TAKE THE SQUARE ROOT OF THIS MATRIX, SO WE'LL HAVE TO MANIPULATE IT A LITTLE SO WE CAN TAKE THE SQUARE ROOT. FIRST WE MUST GET RID OF THE ZERO'S BY MAKING γJJ INTO A COLUMN MATRIX.
[pic]
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Now we are ready to finish the calculation ((jj-1)1/2Cjk(01 = (11
.9705 1 -.2236 .1000 -.0427 .3717 .6015 .6015 .3717
(11 = .9343 .2236 1 -.2928 .1000 .6015 .3717 -.3717 -.6015 =
.9343 -.1000 .2928 1 -.2236 .6015 -.3717 -.3717 .6015
.9705 .0427 -.1000 -.2236 1 .3717 -.6015 .6015 -.3717
.2732 .4919 .6034 .5650
.5098 .5184 -.0637 -.6835 = (11
.6142 -.1761 -.6307 .4403
.1873 -.5142 .7254 -.4175
We are finished! The rest of the calculations we have already done, except for the second order and higher energy corrections. Let’s proceed:
.6034 .5650
.2732 .4919
(1 = bonding (1 = [.2732 .4919 .6034 .5650]
.5098 .5184
(2 = bonding (2 = [.5098 .5184 -.0637 -.6835]
-.0637 ..68
-.6835
.6142
.4403
.
(3 = anti-bonding (3 = [.6142 -.1761 -.6307 .4403]
-.1761
-.6307
.7254
.1873
(4 = anti-bonding (4 = [.1873 -.5142 .7254 -.4175]
-.4175
-.5142
Let’s look at this function as the half-multiplier acts as a ‘Picture Function”. We will freeze the computation of (11 just before summing. First we multiply Cjk by the normalization matrix:
[pic]
[pic]
Now we take the first row, diagonalize it, and multiply into (01:
[pic]
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Now we'll do the same for the second row:
[pic]
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Let’s graph the picture function for (1:
[pic]
.5838 .5838
.3607 .3607
.0807 .1305
.0583 .0593
.0249 .0154
(1 =
-.0154 -.0361 -.0361 -.0249
-.1305 -.0807
Each group of four represents the first column, the second column, etc.. When these
are summed, we get the functions calculated for on page 41. These represent the interactions of the individual Atomic Orbitals.
The picture function for (2 is represented by:
[pic]
.5620
.3473
.1257 .1257
.0776 .0776
.0347 .1017 .1017 .0562
(2 =
-.0562 -.0347
-.1645 -.1645
Now to finish the problem by computing the bond order, free-valence index and the charge density. As you remember, the bond order is given by:
Pij = (Naiaj =(TOcc(Occ =
.2732 .4919 .2732 .4919 .6034 .5650 = .6691 .7973 .2647 -.3882
.4919 .5184 .4919 .5184 -.0637 -.6835 .7973 1.0214 .5276 -.1528
.6034 -.0637 . 2647 .5276 .7363 .7689
.5650 -.6835 -.3882 -.1528 .7689 1.5728
and qi = 1.000 - (Nai2(i
but (Nai2(I = the diagonal of the product of = 2((TOcc(Occ) . We must subtract this value from 1.
q1 = 1 - .6691 = .3309 P12 = P21 = .7973
q2 = 1 - 1.0214 = -.0214 P23 = P32 = .5276
q3 = 1 - .7363 = .2637 P34 = P43 = .7689
q4 = 1 - 1.5728 = -.5728
Hey! That’s even easier to compute qi than in the section on butadiene! Let’s see if we can write a program for this.
[pic] This is the (OCC matrix
[pic] This is a 4x4 identity matrix
[pic] The value is 2 tines the square of (
[pic]
[pic] The second term removes the diagonal from (SQ, then it is subtracted from 1.000 .
[pic]
So the equation is qi = I44 - diag(2((24)T(24) = q44
Now to calculate the Free-valence Index. We make the template matrix by taking the secular determinant and replacing all the X’s with zero’s. We then take the log of this matrix, letting all zero’s equal 10-100 (this is so we can take the log, there is no definition for the log of zero.)
[pic] [pic] [pic] [pic]
[pic]
[pic] [pic]The molecular diagram thus becomes:
.9327 .4050 .4334 .9611
.7973 .5276 .7689
CH2 = CH - CH = O
q1 = .3309 q2 = -.0214 q3 = .2637 q4 = -.5728
The most reactive positions are at the CH-CH bond (free-radical attack), the C-O bond undergoes acid attack while the CH2 -CH undergoes alkali attacks most readily than the rest of the molecule. (Remember, it has been 20 years of scientific exile since I took any chemistry classes, so I do not really remember very well if the interpretations above for the reactivity of acrolein are correct).
Before I get to the section on multiple perturbations and the higher order energy corrections, let me show you how I came up with the bond order and Pij and charge density qi.
.2732 .4919 .6034 .5650
.5098 .5184 -.0637 -.6835 = (11
.6142 -.1761 -.6307 .4403
.1873 -.5142 .7254 -.4175
By hand
P12 = P21 = 2(.2731)(.4919) + 2(.5098)(.5185) = .7973
P23 = P32 = 2(.4919)(.6035) + 2(.5185)(-.0637) = .5277
P34 = P43 = 2(.6035)(.5649) + 2(-.0637)(-.6836) = .7689
Putting the pre-multipliers in a separate matrix and half-multiplying we get:
.2732 o .2732 = q1 = 2(.0746 + .2599) = .6691
.5098 .5098
.2732 o .2732 .4919 = 2( .0747 .1344 )
.5098 .5098 .5185 +.2599 .2634
q1 = .6691 P12 = .7974
.4919 o .4919 .6035 = 2( .2420 .2969 )
.5185 .5185 -.0637 + .2688 -.0330
q2 = 1.0216 P23 = .5278
.6035 o .6035 .5650 = 2( .3641 .3409 )
-.0637 -.0637 -.6835 + .0041 .0435
q3 = .7364 P34 = .7688
.5650 o .5650 = 2(.3192 + .4672) = q4 = 1.5728
-.6835 -.6835
All the other element values are don’t cares, but we cannot replace their values with zero. Now according to the properties of the half-multiplier operator, we take the pre-multipliers and form a matrix out of them. When this matrix is transposed, we multiply it to the post-multiplier to get the same solution. i.e.
.2732 .4919 .6034 .5650 o .2732 .4919 .6035 .5650 =
.5098 .5185 -.0637 -.6835 .5098 .5185 -.0637 -.6835
.2732 .4919 .6034 .5650 T .2732 .4919 .6035 .5650 =
.5098 .5185 -.0637 -.6835 .5098 .5185 -.0637 -.6835
.2732 .4919 .2732 .4919 .6034 .5650 = .6691 .7973 .2647 -.3882
.4919 .5184 .4919 .5184 -.0637 -.6835 .7973 1.0214 .5276 -.1528
.6034 -.0637 . 2647 .5276 .7363 .7689
.5650 -.6835 -.3882 -.1528 .7689 1.5728
Before I get into multiple perturbations, let’s check the energy of the perturbed
acrolein against the direct calculation from the Slater determinant.
TYPE IN THE MATRIX, SELECT IT (ENCLOSE IT IN A BOX), CLICK SYMBOLIC ON THE TOOL BAR, CLICK MATRIX OPERATIONS, CLICK SYMBOLIC DETERMINANT).
[pic]
[pic]
PUT THE COEFFICIENTS IN A COLUMN MATRIX, STARTING WITH THE CONSTANT, THEN THE COEFFICIENT OF X AND ON TO THE HIGHEST POWER OF X. tHE MATRIX MUST BE DEFINED WITH A VARIABLE NAME. THEN TYPE IN polyroots(ACROLEIN). NOTE: POLYROOTS MUST BE LOWER CASE. THEN PRESS THE = SIGN TO GET ROOTS.
[pic]
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TO COMPARE WITH THE PERTURBATION CALCULATIONS, WE SUBTRACT THE ENERGY WE GOT FROM THE PERTURBATION CALCULATIONS FROM THE ENERGY CALCULATED FROM THE SLATER DETERMINANT, THEN DIVIDE THE DIFFERENCE BY THE ENERGY FROM THE SLATER DETERMINANT x 100 TO PUT THE ANSWER IN PERCENTS. SINCE IT IS DEFINED bjj = E, WE TAKE THE DIAGONAL FROM THE bjk MATRIX AND ADD IT TO THE UNPERTURBED E OF BUTADIENE.
[pic]
[pic]This is how I got the energy for the perturbed acrolein.
[pic]
[pic]
[pic] Here I take the inverse of the ENERGYACROLEIN matrix.
[pic] Here I half-multiply the inverse to the difference in energy, effectively dividing
each element in the ENERGYDIFFERENCE matrix
by each element in the ENERGYACROLEIN matrix.
% error = 100x(correct answer-your answer)/correct answer
[pic]
Not too bad an approximation for a perturbation that perturbs 25% of the molecule, but the third value is quite high, perhaps we can correct it a little bit see if it makes a better fit.
SECOND ORDER ENERGY CORRECTION
In Quantum Mechanics, the term for the second order energy correction (for diagonal elements only) is given by:
E2N’,N = ( HN’,N 2
N’(N E0n’ - E0n
Note, the 2 in E2N’,N is not E2, but a notation stating that this is a second order perturbation. It is an index number only. This equation looks like gibberish, let’s see if we can simplify it a little bit. The energy computed for acrolein is the sum of the four energies of butadiene, the molecule we started out with, plus the values of the diagonal of the un-normalized bjk matrix, placed in order of the most negative energy first to the most positive energy last for E01. Then the energy correction term for (1 is
E21n = H12 2 (02 + H132 (03 + H142 (04 = (b12) 2 (02 + (b13)2 (03 + (b14)2 (04
E0n’ - E0n E0n’ - E0n E0n’ - E0n E0n’ - E0n E0n’ - E0n E0n’ - E0n
We’ve got to remember that b21 is not a matrix, but is the number that occupies the position of row 2, column 1 in the bjk matrix. Calculating likewise for the energy terms
E22n, E23n and E24n and putting in matrix form, we get:
(01 (02 (03 (04
= E21n 0 (b12) 2 (b13)2 (b14)2
E01 - E02 E01 - E03 E01 - E04
E22n (b21) 2 0 (b23)2 (b24)2
E02 - E01 E02 - E03 E02 - E04
E23n (b31) 2 (b32)2 0 (b34)2
E03 - E01 E03 - E02 E03 - E04
E24n (b41) 2 (b42)2 (b43)2 0
E04 - E01 E04 - E02 E04 - E03
The second order energy correction matrix is exactly the same in form as the un-normalized Cjk matrix except that the non-diagonal bjk elements are squared. So all we need to do to compute the second order correction is multiply the non-diagonal bjk matrix by the un-normalized matrix. i.e. for a perturbation at location (4 we have:
(B4 - Bjj)(Cjk)UN-NORM =
SECOND ORDER CORRECTION
[pic] [pic]
[pic]
So, taking the diagonal of this matrix and adding it to the first-order energy computation of acrolein, we have:
E21 1.618 + .1382 +.0783 = 1.8345 = 2.39% ERROR
E22 = .618 + .3618 +.0783 = 1.0581 = 5.8% ERROR
E23 = -.618 + .3618 -.0783 = -.3662 = 3.2% ERROR
E24 = -1.618 + .1382 +.0217 = -1.5015 = 1.99% ERROR
HIGHER ORDER ENERGY CORRECTIONS
The higher order energy corrections are given by:
H = H0 + bH1 + b2H2 + b3H3 + . . .
Where b2 and so on represent the square of the element of b, and H2,3,4,… are still indexes, not squares. So lets take bi, replace the diagonal with ones, take the log of the elements and store in location LOGb. Take Cjk un-normalized, replace the diagonal with zero’s, (10-100 in this case) take the log of it’s elements and store in location LOGC.
HP-48G PROGRAM:
RCL LOGb
RCL LOGC
+
(
((TEACH,EXAM,PRGS,APLY, ((TEACH
RCL FOUR1
x
STO E2
[pic] [pic] [pic] [pic]
[pic] [pic]
[pic](FIRST ORDER CORRECTION)
SECOND ORDER CORRECTION
METHOD 1 METHOD 2
[pic] [pic]
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[pic] PERCENT ERROR FOR SECOND ORDER ENERGY CORRECTION
[pic] [pic]
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[pic]
THIRD ORDER CORRECTION
METHOD1: MATRIC METHOD 2: LOGARITHMIC
[pic]
[pic] HERE I’M SQUARING EACH INDIVIDUAL ELEMENT IN b AND PRE-MULTIPLYING IT BY Cjk
[pic]
[pic] [pic]
[pic] HERE I’M REMOVING THE DIAGONAL OF b3 AND MAKING A COLUMN MATRIX OUT OF IT
[pic]
PERCENT ERROR FOR THIRD ORDER CORRECTION
[pic]
[pic]
[pic]
[pic] HERE I’M COMPUTING THE % ERROR FOR THE THIRD ORDER ENERGY CORRECTION
[pic]
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IT LOOKS LIKE THE SECOND ORDER CORRECTION WILL BE THE BEST, BUT LETS TRY A FOURTH ORDER CORRECTION JUST FOR THE HECK OF IT, AND ALSO THIS MAY HELP TO SEE THE PATTERN BETTER OF WHAT I’M TRYING TO DO.
FOURTH ORDER ENERGY CORRECTION
METHOD 1: MATRIC METHOD 2: LOGARITHMIC
[pic] [pic]
HERE I’M CUBING EACH INDIVIDUAL ELEMENT IN b AND PRE-MULTIPLYING BY Cjk
[pic] [pic]
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[pic]
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[pic] HERE I’M COMPUTING THE PERCENT ERROR FOR THE FOURTH ORDER CORRECTION
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SO IT LOOKS LIKE THE SECOND ORDER CORRECTION WAS THE BEST FOR THISPROBLEM.
MULTIPLE PERTURBATIONS
Finally, suppose we have perturbed two or more carbons on a molecule rather than just the one I’ve calculated for:
Suppose:
0 0 0 0
H1 = 0 0 0 0
0 0 1 0
0 0 0 1
.3717 .6015 .6015 .3717 0 0 0 0 .3717 .6015 .6015 .3717
bjk = .6015 .3717 -.3717 -.6015 0 0 0 0 .6015 .3717 -.3717 -.6015 =
.6015 -.3717 -.3717 .6015 0 0 1 0 .6015 -.3717 -.3717 .6015
.3717 -.6015 .6015 -.3717 0 0 0 1 .3717 -.6015 .6015 -.3717
.5000 -.4472 0 .2236
-.4472 .5000 -.2236 0
0 -.2236 .5000 -.4472
.2236 0 -.4472 .5000
but this is equal to (3T(3 + (4T(4 =
.6015 [.6015 -.3717 -.3717 .6015] + .3717 [.3717 -.6015 .6015 -.3717] =
-.3717 -.6015
-.3717 .6015
.6015 -.3717
[pic] [pic]
[pic] [pic]
[pic]
[pic]
[pic]
[pic]
[pic]
But b3 + b4 = b34
[pic]
[pic]
Therefore:
b34 = (TH1(3,4)( = (3(3T(3 + (4(4T(4
And Cjk = ((3b3 + (4b4)/(Ei - Ej) = (3(3o2 + (4(4o2
(Ei - Ej)
With (3 = (4 = 1 for this problem.
To generalize, suppose:
0
0
(3B33
0
(5B55
0
H1 = .
.
(jBjj
0
.
.
(nBnn
Then bjk = (3b3 + (5b5 + (jbj + (nbn = (3(3T(3 + (5(5T(5 + (j(jT(j + (n(nT(n =
n n n
= (j(bj = ((j(jo2 = (j((jT(j
j=1 j=1 j=1
And
n
Cjk = (TH1(3,5,j,n)(/(E0i - E0j) = (j((jT(j
j=1
((Ei-)nnT o [1]nn) + Inn
And the working computer program for this problem in perturbations is:
(11 = (((jj)-1)1/2[10EEX(LOG(((j(jT(j-Ejj)+[I] - LOG((Ejo[I] -(Ejo[I])T + [I])](01
Lets look at the perturbation of butadiene to the molecule C=N-C-O
EO = A + B (O = 1
EN = A + 1/2B (N = .5
Let’s use MathCad to determine the energies:
MULTIPLE PERTURBATIONS
[pic] [pic] [pic]ENERGIES FOR THE MO WAVE FUNCTION
[pic] Symbolic determinant of matrix for C=N-C=O .
To compute the bjk matrix:
b24 = (2(2T(2 + (4(4T(4 =
.5( .6015 [.6015 .3717 -.3717 -.6015]) + .3717 [.3717 -.6015 .6015 -.3717]
.3717 -.6015
-.3717 .6015
-.6015 -.3717
[pic] [pic]
[pic] [pic]
E1 = [pic]
E2 = [pic]
E3 = [pic]
E4 = [pic]
[pic] [pic]
[pic] [pic]
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[pic]
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(1 =
(2 =
PERCENT ERROR
[pic] [pic] [pic]
[pic]
[pic] [pic]
SECOND ORDER CORRECTION
[pic]
[pic] [pic]
[pic] [pic]
[pic]
[pic] [pic]
SOME OF THESE ROOTS ARE NOT VERY CLOSE, BUT REMEMBER, WE PERTURBED 50% OF THE BUTADIENE MOLECULE, WHICH IS A VERY LARGE CHANGE IN THE ENERGY. PERTURBATION THEORY MUST BE BASED UPON SMALL CHANGES FOR IT TO WORK THE BEST.
BIBLIOGRAPHY
Goodrich, “A Primer of Quantum Chemistry”, John Wiley & Sons, New York, 1972
Turner, “Methods in MO Theory”, Prentice-Hall, N.J. 1974
Roberts, “MO Calculations”, W.A. Benjamin Co., N.Y. 1962
Flurry, “MO Theories of Bonding in Organic Molecules”, Marcel Dekker Inc., N.Y. 1968
Dewar, “MO Theory of Organic Chemistry”, McGraw-Hill, N.Y. 1969
Liberles, “Introduction to MO Theory”, Holt, Rinehart and Winston, N.Y., 1966
STATISTICS
Yesterday (8-26-96) I received a response from the Indiana Journal of Mathematics. They rejected my paper as it was written up to this point without even reading it (the paper clip had never been removed from the manuscript). On the phone earlier with Elana Frobosi ( one of the editors at the Journal) she was at a loss to explain why it was rejected. She told me that everything worked, nothing was wrong with it, that they at the journal had never seen anything like it before, but it could only work for the examples included in my paper, but was impossible to work for anything else, the Half-multiplier Operator was a fluke. I did not have the heart to tell her that I had already begun working on Statistics. She made me so mad I decided that rather than solve a few problems, that I would rewrite the entire textbook, if only to find where this operator did not work in Statistics. That is why this section is so long. With every problem, I learned something new about the way this new operator worked. I did rewrite the whole book called the Computational Handbook of Statistics, 2nd edition, by James L. Bruning and B. L. Kintz. As a note, this book is a mish-mash of styles. I started to write it on October 22, 1996 (believe it or not, it was at about 11:53 PM) when I saw the connection between cross multiplying a matrix and matrix multiplication as we now know it.
Lets go back to the Unified Field equation: c( ([A]Tij o [B]jk) = c Cjk
Letting [B]jK = [I], the equation reduces to:
c(Aij = c Cij
where c is a constant that defines what the field equation will represent. In the case of Statistics, c = 1/N. So the Field equation for a simple Statistical system becomes:
(1/N) ((Aij) = (1/N) Cij
But to make it Statistical, we must square the above expression so that it becomes:
(1/N) (Aij)2 = (1/N) CCT
Which is our basic Statistical equation.
The ( operators are:
C( = [1]1,i Aij = Cj,1
R( = Aij [1]j,1 = Ci,1
M( = Aij
This is the Statistics for simple systems, for Analysis of Variance, [B] ( [I].
(1/N) (([A]Tij o Bjk) = (1/N)Cjk
And the Statistics Become:
(1/N) (([A]Tij o [B]jk)2 = (1/N) CCT
The ( operators are:
Where A represents the data taken and B = Database Matrix.
1/2(([A]Tij o [B]jk) = Cj,k (Half-multiplier mode)
C(([A]Tij o [B]jk) = Ci,k (regular matrix multiplication)
R(([A]Tij o [B]jk) = Bjk [1]j,1 o [A]Tij
M(([A]Tij o [B]jk) = [A]Tij [1]i,1 o Bjk
Note, these equations hold for Physics, Astrophysics and Accounting & Inventories also.
EXAMPLES
(The rewriting of the Statistics texts, with programs)
THE MEAN
The Statistics equation for the mean is:
0= (1/N)((X)
The matric equivalent is: (1/N)C(Aij =
(1/N) [1 1 1 . . . 1]1,j D
A
T
A j,1
EXAMPLE: We have 5 people whose height in inches has been recorded. What is the average height:
(1/5) [1 1 1 1 1] 60 = 280/5 = 56 Inches
53
57
52
58
STANDARD DEVIATION
Computational Handbook of Statistics, 2nd Edition, section 1.2, pg. 4-6
The Statistics equation for standard deviation is given by:
[pic]
The matric equivalent is given by:
[pic]
Where (x2 = [DATA] D = a
A
T
A We combine the two into one matrix operation.
And ((x) = [DATA] 1 = b When we’ve calculated b, we plug it in the
1 matrix, first dividing by N and then
1 multiply. We then take the square root of
1 the resulting single number.
EXAMPLE: Find the standard deviation (s.d.) for the following list of heights measured from twenty 12 year old students.
VARIABLES:
[pic] [pic] [pic][pic]
[pic]
[pic]
Mathcad cannot solve this as is, so we have to break the
problem up in two parts, first solve the problem, then take
the square root.
[pic]
[pic]
Putting the arrow over the second expression allows us to take the square root of a 1x1 Matrix, which Mathcad won't allow us to do.
[pic]
[pic]
s.d. = ((1/19)[64 48 55 68 72 59 57 61 63 60 60 43 67 70 65 55 56 64 61 60] 64 1 )2 1
48 1 -b/20
55 1
.
.
.
60 1 20,2
1/19 ([73,894 1208] 1 )1/2
-1208/20 = (1/19)[73,894 -72,963] = (931/19)1/2 = (49)1/2 = 7
STANDARD ERROR OF THE MEAN: s.d/(# of terms)1/2 = 7/(20)1/2 = 7/4.47 = 1.57.
To do this totally by computer, proceed as follows:
(1/(N-1) ([DATA]2 -(1/N)[DATA][1])2 )1/2
TWENTY1
HP-48G PROGRAM
((1/19) ([DATA] D - 1/20[DATA] 1 ))1/2 RCL DATA
A . (
T . TRN
A 20,1 1 20,1 SWAP
x
RCL DATA
RCL TWENTY1
x
RCL N = 20
/
-
RCL N-1 = 19
/
SQRT
STO SD 6.999
t-TEST FOR A DIFFERENCE BETWEEN A SAMPLE MEAN & POPULATION MEAN
Computational Handbook of Statistics, 2nd Edition, section 1.5, pg. 8-10
The Statistical equation is given by:
[pic]
The matric equation is given by:
[pic]
Using the data from the preceding problem, ( is obtained from tables as the average height of 12 year olds in the whole USA = 58”. We have already calculated
[pic]
Again Mathcad can't handle the math, so we must break it into simpler parts.
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
We multiply by t3 because we have already taken it's inverse, so we multiply to divide.
[pic]
To do this totally by computer or calculator, proceed as follows:
1/20 [DATA]1,20 [1]20,1 -( PROGRAM:
RCL DATA
RCL TWENTY1
x
( 1 [DATA] D -1/20([DATA] 1 2 )1/2 STO SUM
20(20-1) A . RCL N = 20
T . /
A 1 RCL ( = 58
-
RCL DATA
(
TRN
SWAP
x
RCL SUM
SQ
RCL N = 20
/
-
RCL N(N-1) = 380
/
SQRT
/
STO t-TEST2 1.53
t-TEST FOR THE DIFFERENCE BETWEEN TWO INDEPENDENT MEANS
(COMPUTATIONAL HANDBOOK OF STATISTICS, SECOND EDITION SEC. 1.6, PG 10 - 13)
The Statistical equation is given by:
01 - 02
( (X12 - 1/N1((X1)2 + (X22 - 1/N2((X2)2 (1/N1 + 1/N2))1/2
(N1 + N2) - 2
The matrix form of this equation is:
b1/N1 - b2/N2
(1/N1 - 1/N2)([a1 b1] 1 + [a2 b2] 1 1/2
-b1/N1 -b2/N2
N1 + N2 - 2
Where [DATA 1] D 1 And [DATA 2] D 1
A 1 = [a1 b1] A 1 = [a2 b2].
T . T .
A . A .
1 1 2 1
EXAMPLE:A principal wishes to determine whether there is a significant difference
between the IQ scores of boys and girls in a particular grade. The score on the Wechsler Intelligence Scale for Children is recorded for each student.
GROUP 1 GROUP 2
(BOYS) (GIRLS)
107 109
96 94
88 127
131 76
109 115
84 121
79 87
105 92
108 91
92 98
96 104
101 96
0 110
0 108
VARIABLES:
[pic] [pic] [pic]
[pic][pic][pic][pic]
[pic]
[pic]
[pic]
we are going to compute this completely from the Matric equation given below. To see how I derived this equation, please go to the next section (Sect. 1.7) for the derivation.
To calculate this problem totally on a computer and/or calculator:
[pic]
[pic]
[pic]
Of this big equation,
[pic]
We can’t take the square root of a matrix, so we use the arrow function of mathcad which evaluates matrices element by element, thus we can take the square root as follows:
[pic]
[pic]
[pic]
Now we compute the numerator, Mathcad can do this expression in one step, so we have:
[pic]
[pic]
[pic]
Whoops! Can't divide Matrices, we must multiply by the inverse:
[pic]
[pic]
Since the t-test is the absolute value, t = .426
This program is for using another method.
HP-48G PROGRAM:
RCL FOURTEEN1 RCL N1
RCL A INV
x [1196 1428] RCL N2
STO SUM1 INV
RCL A +
( x 30
TRN ((SQRT((
SWAP TEACH
x VAR,EXAM,PRGS, APLY =5.48
MTH,MATR,NXT,(DIAG STO SUM4
2 RCL SUM4
( RCL SUM1
DIAG( TRN
RCL TWO1 1 RCL NDIAG
1 SWAP
x x
STO SUM2 121,318 RCL MIN1 [1 -1]
RCL SUM2 148,202 SWAP
RCL SUM1 x use absolute value if
TRN SWAP negative.
( /
2 STO t 0.43
(
DIAG( Note: This is long and complex
SWAP compared to the Analysis of Variance
x 1/12 due to all the little nit-picking math
RCL N = 1/14 steps that need to be done.
2
(
MTH,MATR,NXT,DIAG(
STO NDIAG
RCL NDIAG
SWAP
119,201
145,656
x
-
STO SUM3 2116.667
2546.000
RCL TWO1
TRN
SWAP
x 4662.667
RCL N1 = 12
RCL N2 = 14
+
2
-
/ 194.277
STEP 17: To determine if the t value is significant, the degrees of freedom
df = (N1 + N2) - 2 = 24. The t value which is significant at the .05 level for df = 24 is 2.064. But we obtained t = .43, therefore it is concluded there is no significant difference between the IQ scores of boys and girls in this particular class.
MATHCAD +6 PROGRAM FOR t-TEST FOR DIFFERENCE BETWEEN TWO INDEPENDENT MEANS.
[pic]
[pic]
[pic] [pic] [pic]
[pic]
[pic]
[pic] The above equalities are all we need to solve this problem.
[pic]
[pic]
[pic] Sum the columns of A and multiply to the diagonal of N
[pic]
[pic] Add the two values obtained in t1
[pic]
[pic]
This is ( (X12 - 1/N1((X1)2 + (X22 - 1/N2((X2)2 (1/N1 + 1/N2))1/2 in matrix form.
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
ALL WE NEED FOR t IS THE ABSOLUTE VALUE, t = 0.426.
t-TEST FOR RELATED MEASURES
(COMPUTATIONAL HANDBOOK OF STATISTICS, 2ND EDITIONSectino 1.7, PG. 13-16
EXAMPLE: An experimenter is interested in determining the effects of a special education program on the intelligence test scores of underprivileged children. The first step is to match several pairs of students on the basis of their Wechsler IQ test scores. One student from each pair is randomly assigned to either the special training group or the control group that receives no special treatment. After 6 weeks of training, an alternate form of the Wechsler test is given to the students in both groups to determine the effects of the program. The score on this second test is recorded for each student.
STATISTICS FORMULA
X - Y
t=
ρ((D2 - 1/N((D)2)/N(N-1)
MATRIX METHOD
STEP 1: Table the data:
VARIABLES:
GROUP 1 GROUP 2
(NO SPECIAL (SPECIAL
TREATMENT) TREATMENT)
[pic] [pic] [pic]
[pic]
[pic]
EQUATION:
[pic]
STEP 2: Obtain the difference between each pair of scores:
[pic]
[pic] [pic]
HP-48G PROGRAM:
RCL A
RCL MINUS1
x
STO DIFF
STEP 3: Square the difference scores and sum:
[pic]
[pic]
[pic]
hp-48g program:
RCL DIFF
(
TRN
SWAP
x
STO GNDSUMSQ
STEP 4: Obtain the algebraic sum of the difference scores, square it and divide by N = 14.
PROGRAM: GRANDSUM = (1/N)([diff]1,14[1]14,1)2
[pic]
[pic]
[pic]
[pic]
HP-48G PROGRAM:
RCL ONE14
RCL DIFF
x
((SQ((
TEACH
VAR,EXAM,PRG,APLY
RCL N = 14
/
STO GRNDSUM 186
STEP 5: Subtract step 4 from step 3. ((x2 - ((x)2) = 293 - 186 = 107.
[pic]
[pic]
[pic]
STEP 6: Divide by N-1. 107/13 = 8.23.
[pic]
[pic]
[pic]
STEP 7: Take the square root:
θ8.23 = 2.87
[pic]
[pic]
STEP 8: Divide by (N)1/2 = (14)1/2 = 3.74.
[pic]
[pic]
[pic]
2.87/3.74 = .767
STEP 9: Obtain the mean score of each of the two groups:
PROGRAM: (1/N)[1]1,14[A]14,2
[pic]
[pic]
[pic]
[pic]
HP-48G PROGRAM:
RCL ONE14
RCL A
x
RCL N = 14
/
STO MEAN
STEP 10: Subtract the mean from group 1 from group 2:
PROGRAM: MEANDIFF = [MEAN]1,2[MINUS1]2,1.
[93.50 97.14] -1 = 3.64 (use absolute value if difference is negative.
1
[pic]
[pic]
[pic]
STEP 11: Divide value obtained in step 10 by step 8.
[pic]
[pic]
[pic]
t = 3.64/.767 = 4.74.
STEP 12: To determine significance:
Degrees of freedom df = N-1 = 13. From tables, the t value that is significant at the .05 level for df = 13 is 2.16. But we obtained t = 4.74, so it is concluded that the special training program improved the IQ test scores.
The entire equation in one step (one step, that is, if modern computers could handle the math) is given by:
[pic]
SANDLERS A TEST
An alternate and somewhat simpler technique for determining whether the difference between correlated samples is significant is by using the A test which is derived directly from the t-test.
STATISTICS FORMULA
A = (D2/ ((D)2 = 293/(-51)2 = 293/2601 = .113
df = 13 (N-1)
From tables, for an A value to be significant at the .05 level, it must be equal to or smaller than 0.270. Since A = .113, it’s concluded the difference is significant.
STEP 3: Square the difference scores and sum:
[-5 -8 -5 -5 -6 2 0 -3 -3-1 -5 -5 -3 -6] -5 hp-48g program:
-8 RCL DIFF
-5 (
-5 TRN
-6 SWAP
2 x
0 = 293 STO GNDSUMSQ
-3
-3
1
-5
-5
-3
-6
(1/14) ([-5 -8 -5 -5 -6 2 0 -3 -3-1 -5 -5 -3 -6] 1 )2
1
1
1
1
1
1 = -512/14 = 186
1
1
1
1
1
1
1
(1/N)([diff]21,14
PROGRAM: (1/N)([diff]1,14[1]14,1)2
THE ANALYSIS OF VARIANCE
INTRODUCTION
Before we begin with the analysis of variance and other advanced statistical testing, I am going to try to explain a little about what I will be doing, especially since the method is new and there is no information on it anywhere in the world except here in this book. (And in my head).
First about the structure of the examples. I wrote the book up to section 2.9, Simple Latin Square when my word processor got electrocuted. I had all the information on floppy disk, but when I got use of a computer the matrices were all garbled and it would be simpler to re-type the information rather than try to rearrange the numbers. I got use of a computer and started at section 2.10, Latin Square Design: Complex. At this point I did not have MathCad +6 nor a HP-48G calculator. All I had was my HP-48 SX calculator to write the programs with. When I got to trend analysis, somewhere in there I got my HP-48G calculator and the MathCad program. When I got to the end of trend analysis, I began retyping the paper from page one. The book is presented as a progression of my exploration of the properties of this operator, sort of like a journal, except as I rewrote the earlier portions, I used the most recent properties I discovered and left out most of my earlier archaic methods. Then we get to section 2.10, which I am going to leave as is, though at the end of each section I will update the method using MathCad +6 and HP-48G programs. So don’t be surprised if the math looks different after Latin Square: Simple ends. It will also help those who have the SX model to set up the statistical programs.
The math is basically simple. The basic function for the analysis of variance is
1/N(1 x A x B)2 = C2
In matrix form, A is equal to the data we have taken, I call this the datastream matrix.
B is the database matrix [DB]. This matrix is arbitrary, and with it you define how you manipulate the information in the datastream matrix [A]. C2 is the solution matrix.
Usually, but not in all cases, it will be a single number. So basically all advanced statistics is based on A x B = C. The [1] matrix when it is in a row in front of A just sums the columns of [A] when multiplied. The number of 1’s must equal the number of rows in [A]. If [1] is a column on the right side of the matrix [A], it sums the rows of [A] when multiplied. The number of 1’s must equal the number of columns in [A]. i.e.
[1 1 1] 1 2 3 1 2 3 1 6
4 5 6 = [12 15 18] and 4 5 6 1 = 15
7 8 9 7 8 9 1 24
SUMMING THE COLUMNS SUMMING THE ROWS
(WITHIN SUBJECTS) (BETWEEN SUBJECTS)
The third operator is new, no one knows about it, so there is no computer program that can calculate it, although mathematicians have been using it for centuries, no one has really understood what it was that they were really doing. It is found in Gaussian reduction. Had anyone tried to prove Gaussian Reduction, the half-multiplier operator would have been discovered over 200 years ago. It is simply this:
2 5 10 (5x2=10)
3 o 6 = 18 (6x3=18)
4 7 28 (4x7=28)
But mathematicians have been told throughout the years that this is impossible, therefore, illegal to use. I have proved it in this paper. The translation I use to accomplish this half-multiplication is to diagonalize the operator and matrix multiply:
2 0 0 5 10
0 3 0 6 = 18
0 0 4 7 28
And now we get to the mysterious looking, but beautifully symmetric statistical database matrix. This is the heart and soul of statistics, physics, accounting and inventories and whatever other math this half-multiplier operator covers.
So far, I have covered [1][A] and [A][1] and [A] o [B], but the equation for the analysis of variance is:
1/N([1][A][DB])2 = 1/NC2 where C2 = CCT
We multiply C2 as CCT to get a single number as an answer. CTC will give us a NxN matrix, though the sum of its diagonal is equal to CCT. The use and meaning of the statistical database is as simple as it is elegant. Once we have multiplied [1][A] we get a single row matrix. The statistical database tells us how to manipulate the data in this row matrix. For example, suppose we have the datastream matrix:
1 2 3 4 5 6 [1 1 1] 1 2 3 4 5 6 1+6+2=9
[A] = 6 5 4 3 2 1 then 6 5 4 3 2 1 = [9 11 13 8 10 12] 2+5+4=11
2 4 6 1 3 5 2 4 6 1 3 5 6+4+3=13
etc.
Now for accounting or statistical purposes, suppose we wish to add the first three numbers and ignore the last three, then ignore the first three numbers and add the last three numbers, and keep the sums separate. The database matrix that will accomplish this is:
1 0
1 0 I call this [DB1]. In fact, I define this form as database 1 in all cases.
1 0 Where the 1’s in a column are not separated by a zero. Another form of
0 1 database 1 is if we wish to add the first two numbers and ignore the rest,
0 1 add the third and fourth numbers only and ignore the rest, and add the
0 1 fifth and sixth numbers and ignore the rest. This database matrix is:
1 0 0 Note again the ones in the columns are next to each other. To be
1 0 0 consistent, since the above database matrix is called [DB1] I call
0 1 0 this second matrix [DB11} to show it is the second database
0 1 0 matrix of type one. i.e.
0 0 1
0 0 1
9+11+13=33 [9 11 13 8 10 12] 1 0 = [33 30] and
8+10+12=30 1 0
1 0
0 1
0 1
0 1
9+11=20 [9 11 13 8 10 12] 1 0 0 = [20 21 22]
13+8=21 1 0 0
10+12=22 0 1 0
0 1 0
0 0 1
0 0 1
Note: for all you experimenters out there, we can use the database matrices to help set up our experiment properly. Lets look at databases 1 and 11. For each [DB1] the number of columns (we can have as many rows as we want) must be evenly divisible by 3. That is, we can have 6 columns, 9 columns, 12 columns, etc. For the [DB11] matrix the number of columns in [A] must be divisible by two. If we want to calculate both [DB1] and [DB11], the number of columns in [A] must be divisible by both 2 and 3. We can have 6 columns, 12 columns, 18 columns etc. for the additions to work out, but we can’t have 9, 12, etc.
In the [DB2] type matrices, we add every other number, i.e. the first and third, the second and fourth, fifth and seventh, and the sixth and eighth numbers. We will end up with four sums with eight columns of data. The number of columns in [A] must therefore be divisible by four. So if we want to analyze with [DB1], [DB11] and [DB2], our common divisors are 2, 3 and 4. The only numbers that fit this category are 12, 24 etc. so in the experiment we must have 12 columns (again the number of rows (subjects) is arbitrary since we add them together, the number doesn’t really count. The [DB2] matrix looks like:
1 0 0 0
0 1 0 0
1 0 0 0
[DB2]= 0 1 0 0 Adding arbitrarily two more columns to [1][A] example above (so we can
0 0 1 0 do this problem) we get:
0 0 0 1
0 0 1 0
0 0 0 1 [9 11 13 8 10 12 6 14] 1 0 0 0 = [22 19 16 26]
0 1 0 0
1 0 0 0
0 1 0 0 9+13=22
0 0 1 0 11+8=19
0 0 0 1 10+6=16
0 0 1 0 12+14=26
0 0 0 1
In the [DB3] type matrices, we add every third number, i.e. first to the fourth, second to the fifth, etc. Note there are two zero’s underneath every 1.
1 0 0
0 1 0
[DB3]= 0 0 1 And [1][A][DB3] = [9 11 13 8 10 12] 1 0 0 =[17 21 25]
1 0 0 0 1 0
0 1 0 0 0 1 9+8=17
0 0 1 1 0 0 11+10=21
0 1 0 13+12=25
0 0 1
Finally, we get to the calculation of N for each of these databases. N for each database equals the number of elements used to calculate C. It may or may not be the same as other databases. With the exception of the calculation of (x2 and ((x)2, N is equal to the number of elements in [A] divided by the number of columns in the database matrix. That is, in the matrix [A], there are 18 elements, [DB1] has two columns, so N[DB1]=18/2=9.
N[DB11]=18/2=9. N[DB3] =18/3=6. It’s that simple.
Once this first part is complete, we square the resulting row matrix. To get a single number as our solution, we multiply as (1/N)CCT. We do not square the 1/N term. i.e. to get the variance for [DB1] we proceed as follows:
1/N [1][A][DB1] = (1/9)[33 30]. And 1/N ([1][A][DB1])2 = 1/9 [33 30] 33 =
30
(1/9) (332 + 302) = 1/9 (1089 + 900) = 1/9 (1989) = [221].
Note that if we multiplied as 1/N CTC we would get: 1/9 33 [33 30] =
30
(1/9) 1089 990 and we would have to sum the diagonal and divide by 1/N = 1/9
990 900 to get (1/N)C2. Which is a lot more math steps to go through.
A few more items to cover and we can get to the math. When we need to determine the grand sum of squares ((x2), the easiest way I’ve found is to square the [A] matrix and sum the diagonal. In general I multiply to get the smallest NxN matrix, but when we get to F-tests for simple effects and trend analysis, we have to multiply to get the largest matrix possible because we need to keep track of each of the squares separately. Using the HP-48G calculator to get the diagonal matrix we must pull out the diagonal and make a row matrix, then reassemble it to a diagonal. In MathCad +6 I create an identity matrix and vectorize an onto mapping rather than the usual into mapping. This multiplies one element by one element rather than the normal row x column and summing. I’ll not give an example here, you’ll see it in the math soon enough.
As I began to write this paper, I started out by using a [b] post-multiplier. But I discovered if all the rows are equal, all we need to do is square the C matrix. The [b] seems to be archaic, but in reality it is very important, especially if you are running an experiment and one or more of the subjects has to drop out early and you do not want to discard the data. The N’s then are not all the same and you must use the [b] form. I will introduce it in the first three problems so you can see how to set it up, then mostly ignore it for the rest of the paper.
Lets see how using the b method will work for the t-test for the difference between two independent means.
b1/N1 - b2/N2
(1/N1 + 1/N2)([a1 b1] 1 + [a2 b2] 1 )
-b1/N1 -b2/N2
N1 + N2 -2
Where [DATA X] D 1 = [a1 b1] and [DATA Y] D 1 1 = [a2 b2]
A . A . -b2/N2
T . T .
A . A .
X 1 Y 1
1196/12 - 1428/14 =.43
[((1/12 + 1/14)([121,318 1196] 1 + [148,202 1428] 1 ))/24]1/2
-1196/12 -1428/14
As a final note, throughout this paper, I am going to refer to the sum [1][A] = SUM1.
So whenever we encounter the term SUM1 we know we are talking about the datastream matrix
[A] whose columns have been added. Therefore, there will always be room for SUM1 in
this mathematics.
BASIC TRANSLATIONS FOR STATISTICS
N(X2 - ((X)2 = [DATA] D 1 NT = [DATA] D 1 1 = N[DATA][DATA]T -([DATA] 1 2)
A . -b A . -b/NT .
T . T . .
A 1 A 1 1
N(XY = NT[DATAX] D
A
T
A
Y
((X)((Y) = ([DATAX] 1 )([DATAY] 1 ) Or =Anti-Log([1 1](Log DATAX 1 ))
. . DATAY .
. . .
1 1 1
CROSS PRODUCT: (XY - (1/N)(X(Y = [DATAX] D -(1/N)([DATAX] 1 )([DATAY] 1 )
A . .
T . .
A 1 1
Y
PEARSON PRODUCT-MOMENT CORRELATION
r = NP(XY - ((X)((Y) Which can be translated to:
([N(X2 - ((X)2][ N(Y2 - ((Y)2])1/2
(NP[DATAX] D - ([DATAX] 1 )([DATAY] 1 )
A . .
T . .
A 1 1
Y
[([DATAX] D 1 1 )([DATAY] D 1 1 )]1/2
A . -bx/Np A . -by/NP
T . T .
A . A .
X 1 Y 1
Or calculating without the b’s:
NP[DATAX][DATAY]T - Anti-Log([1 1] LOG DATAX 1 )
DATAY .
.
1
[NP[DATAX][DATAX]T - ([DATAX] 1 )2 (NP[DATAY][DATAY]T - ([DATAY] 1 )2]
. .
. .
1 1
SECTION 2.1: COMPLETELY RANDOMIZED DESIGN
COMPUTATIONAL HANDBOOK OF STATISTICS, SECOND EDITION PG. 24-27.
For this first problem, I am going to go through it step by step, trying to leave nothing out, especially in using the calculator keystrokes, some of which are implied but not written down.
STATISTICAL EQUATIONS
SUM OF DEGREES MEAN SQUARE COMPUTING F
SQUARES OF FREEDOM FORMULA
k k k k
Among (Ni(xi - x)2 k-1 (Ni(xi - x)2 ((Ti2/Ni -T2/N) ((Ti2/Ni -T2/N)/k-1
Samples I=1 i=1 I=1 i=1
k-1 k-1 k Ni k
((xij2-(Ti2/Ni/(N-k)
i=1j=1 i=1
k Ni k Ni k Ni
Within (((xij-xi,)2 N-k (((xij-xi,)2 ((xij2-(Ti2/Ni
Samples I=1 j=1 I=1j=1 I=1j=1
N-k N-k
k Ni
Total (((xij-xi,)2 N-1
i=1j=1
You can see why you need a Masters or Ph.D. in Statistics to solve the above statistical formulas. Now we’ll see a way that high school seniors or freshmen in college can use to solve the same equations.
[1 1 1… 1] a11 b12 c13 d14
a21 b22 c23 d24
. . . . = [A B C D] and NT = total # of data elements
. . . .
an1 bn2 cn3 dn4
To sum the squares of each term and add them together, square the matrix, remove the diagonal and sum. i.e.
[A/N1 B/N2 C/N3 D/N4] A = 1/NT[A2/N1 B2/N2 C2/N3 D2/N4]
B
C
D
Let b = [A B C D] 1 = [A+B+C+D]
1
1
1
So now the total sum of squares, including subtraction of the correction factor is:
SSb = [A B C D] A 1 1
B 1 -b/NT
C 1
D 1
EXAMPLE: Assume an experimenter is interested in determining the effect of shock on the time required to solve a set of difficult problems. Subjects are randomly assigned to four experimental conditions: Subjects in group 1 receive no shock, group 2 receives very low intensity shocks, group 3 medium shocks and group 4 receives high intensity shocks. The total time required to solve all the problems is the measure recorded for each subject.
STEP 1: Table the data.
G1 G2 G3 G4
10 3 19 23
7 8 12 14
9 7 16 16
8 5 14 18
[A]12,4 = 15 6 7 12
3 10 8 13
8 12 13 16
9 4 10 17
11 7 19 19
9 6 9 14
5 5 15 16
17 15 14 17
STEP 2: Add the scores in each group. PROGRAM: [1]1,12[A]12,4
[pic]
[pic]
[pic]
HP-48G PROGRAM:
RCL ONE12
RCL A
x
STO SUM1
STEP 3: Square each number in the table and sum the values.
PROGRAM: GRANDSUMSQ = [1]1,12([A]4,12([A]4,12)[1]4,1
[pic]
[pic]
[pic]
Or METHOD 2: PROGRAM: [1]1,4diagonal(ATA)[1]4,1
ATA = 10 7 9 8 15 3 8 9 11 9 5 17 10 3 19 23
3 8 7 5 6 10 12 4 7 6 5 15 7 8 12 14
19 12 16 14 7 8 13 10 19 9 15 14 9 7 16 16 1189 852 1456 1820
23 14 16 18 12 13 16 17 19 14 16 17 8 5 14 18 852 778 1125 1397
15 6 7 12 = 1456 1125 2202 2644
3 10 8 13 1820 1397 2644 3265
8 12 13 16
9 4 10 17
11 7 19 19
9 6 9 14
5 5 15 16
17 15 14 17
And to sum the squares:
[1 1 1 1] 1189 0 1
778 1 = 7434
2202 1
0 3265 1
MICRO-PROOF: Log I = 0 -1000 -1000 -1000 Log A2 = 3.0752 2.93 3.163 3.26
-1000 0 -1000 -1000 2.93 2.891 3.051 3.145
-1000 -1000 0 -1000 3.163 3.051 3.343 3.4223
-1000 01000 -1000 0 3.26 3.145 3.4223 3.514
Anti-Log(Log I + Log A2) = 1189 0 HP-48G PROGRAM:
778 RCL A
2202 (
0 3265 ((TRN
(
x
MTH,MATR,NXT,(DIAG
4
(
DIAG(
RCL ONE4
x
RCL ONE4
((TRN
(
SWAP
x
STO GNDSUMSQ (OR SUM2)
STEP 4: Find grand sum of the table.
PROGRAM: GRANDSUM = [1]1,12[A]14,4[1]4,1
[pic]
[pic]
[pic]
HP-48G PROGRAM:
RCL ONE12
RCL A
x
RCL FOUR1
x
STO GRNDSUM (OR SUM3)
STEP 5: Square the grand sum (step 4) and divide by the total # of measures recorded,
NT = 48. 1/NT([1]1,12[A]12,4[1]4,1)2
[pic]
[pic]
[pic]
(1/48)([1 1 1 1 1 1 1 1 1 1 1 1] 10 3 19 23 1 )2 = (550)2/48 = 6302
7 8 12 14 1
9 7 16 16 1
8 5 14 18 1 HP-48G PROGRAM:
15 6 7 12 RCL ONE12
3 10 8 13 RCL A
8 12 13 16 x [111 88 156 195]
9 4 10 17 RCL FOUR1
11 7 19 19 x 550
9 6 9 14 (
5 5 15 16 x 5502
17 15 14 17 RCL N=48
/
STO CORRFACT 6302
Note: In steps 6,7 & 8, I wrote the program as: x(x when all I wanted to do was square 550. I just wanted to show this way of squaring 550.
STEP 6: Subtract the correction factor from SST from step 3.
SST = GRANDSUMSQ - CORRFACT = 7434 - 6302 = 1132.
STEP 7,8,9: Square the sum of each of the groups and divide by the number of measures in each group. NG = NT/4 columns in A=48/4=12.
PROGRAM: SSb = (1/NG)([SUM1]1,4[SUM1T]4,1-CORRFACT
HP-48G PROGRAM:
RCL SUM1 /
( RCL CORRFACT
TRN -
x STO SSb
RCL N=12
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
We can divide by 12 because all the [pic] terms are the same, that is, all the columns in a are filled and contain a value that is not zero. But suppose some of the subjects had to drop out of the experiment early but the results were valid and the experimenter did not wish to discard the results just to make the math easier. We would then multiply as follows:
[pic]
A neat shortcut may be calculated for SSb and we proceed as follows:
(1/NG)[111 88 156 195] 111 - [111 88 156 195] 111 1 1 =
88 88 1 -b/NT
156 156 1
195 195 1
[7434 550] 1 = 6869 - 6302 = 567 = SSb
-550/48
STEP 9: [pic]
[pic]
[pic]
[pic]
STEP 10: Finding the degrees of freedom. These are needed because the F-test is a ratio of the mean squares SS/df.
df SST = Total # of elements in the matrix minus 1= 12 x 4 = 48-1 = 47.
df SSb = # of experimental groups - 1 = 4 - 1 = 3.
df SSW = df SST - DFSSW = 47 - 3 = 44.
STEP 11: Calculation of F.
[pic] [pic] [pic]
F= SS o df-1 o ERROR-1
[pic] [pic]
[pic][pic][pic]
Row Matrix for F:
[pic]
column Matrix for f:
[pic]
What we do here is multiply straight across i.e. 1132 x 1/47 x .0779 = 1.8757
567 x 1/3 x .0779 = 14.7186
565 x 1/44 x .0779 = 1.0000
But a computer cannot do this, it is part of the math I discovered. So we must translate the math to a method a computer can use to get the same solution. Remember the o product is equivalent to multiplying by a diagonal matrix, so we will diagonalize one or more of the above matrices to get the solution. Since this is a small multiplication compared to some of those that will follow, I will try to solve this in as many different ways that I can. The easiest way is to just diagonalize all three and multiply by [1 1 1] if we want a row matrix solution, or post-multiply by if we want a column matrix
for a solution. 1
1
1
i.e.
[1 1 1] 1132 1/47 44/565 [1 1 1] 1.8757
567 1/3 .0779 = 14.7186 =
565 1/44 .0779 1.000
[1.8757 14.7186 1.000] or
i.e.
1132 1/47 44/565 1 1.8757 1
567 1/3 .0779 1 = 14.7186 1 =
565 1/44 .0779 1 1.000 1
1.8757
14.7186
1.0000
Or we can solve it in this manner for a row matrix:
[1132 567 565] ( .0213 .0779 )= [1.8757 14.7186 1.000]
.333 .0779
.0227 .0779
or
1132 .0213 .0779 1.8757
567 .333 .0779 = 14.7186
565 .0227 .0779 1.0000
Note: I put parenthesis in the preceding expression because here I am multiplying the two diagonal matrices first, out of the normal order of multiplication for matrices, but we don’t really need the parenthesis, the math will work out exactly the same if we multiply in the normal order of left to right. There are so many ways to do this multiplication that there is no one way to write the program. This will make it hard to patent, for as soon as someone comes up with a program, there are automatically three or four alternate ways of writing the same expression.
HP-48G PROGRAM #1:
RCL SS
3
mth,matr,nxt,diag( This takes the matrix of SS values (of which there are 3 elements,
and makes a diagonal matrix out of it.
RCL df
3
(
DIAG( This takes the df matrix and diagonalizes it.
INV This inverts the diagonal df matrix
x This multiplies diagSS x diagdf
RCL ERROR
x Gives a column matrix, the center value being the F value we are to STO F look at for significance.
HP-48G PROGRAM #2:
RCL SS
3
mth,matr,nxt,diag( This takes the matrix of SS values (of which there are 3 elements,
and makes a diagonal matrix out of it.
RCL df
3
(
DIAG( This takes the df matrix and diagonalizes it.
INV This inverts the diagonal df matrix
x This multiplies diagSS x diagdf
RCL ERROR
3
(
MTH,MATR,NXT,DIAG( Diagonalizes ERROR matrix
x Final multiplication of the 3 diagonalized matrices
RCL THREE1 Post-multiplies by [1 1 1]T
x
STO F
The probabilities are obtained from tables. The final tabling of the data is performed as follows:
SOURCE SS df ms F p
TOTAL 1132 47 - - -
BETWEEN GROUPS 567 3 189 14.71 (.001
WITHIN GROUPS 565 44 12.84 - -
The F value of 14.71 with a df of 3 and 44 would occur by chance less than once in a thousand times, it is concluded that the level of shock intensity does affect the time required to solve these problems.
MATHCAD +6 PROGRAM
WE START OUT BY DEFINING THE VARIABLES. TYPE IN VARIABLE NAME THEN PRESS SHIFT : (COLON). IF THE VARIABLE IS A MATRIX, THE EASIEST WAY I THINK TO DO IT IS TO CLICK THE MATRIX BUTTON IN THE TOOLBAR. ON THE DROPDOWN MENU, PRESS THE MATRIX IN THE UPPER LEFT-HAND CORNER AND FILL IN THE APPROPRIATE NUMBER OF ROWS AND COLUMNS. PRESS CREATE. IT IS IMPORTANT IF WE WISH TO CREATE A ROW MATRIX TO ENTER IT AS A COLUMN THEN TRANSPOSE IT RATHER THAN ENTERING IT DIRECTLY AS A ROW MATRIX. THIS IS BECAUSE MATHCAD ENTERS A SINGLE ROW MATRIX AS A COLUMN, AND WE MAY HAVE PROBLEMS DOWN THE ROAD TRYING TO TRANSPOSE IT OR ANY OF ITS PRODUCTS IF WE DON'T DO IT AT THE VERY BEGINNING.
VARIABLES:
[pic] [pic] [pic] [pic] [pic]
STEP 2: ADD THE SCORES IN EACH GROUP.
[pic]
[pic][pic]
STEP 3: SQUARE EACH ELEMENT IN THE MATRIX AND SUM THE VALUES.
[pic]
[pic]
[pic]
[pic][pic]
STEP 4: FIND THE GRAND SUM OF THE MATRIX.
[pic]
[pic]
[pic]
STEP 5: SQUARE THE GRANDSUM AND DIVIDE BY THE TOTAL # OF ELEMENTS IN MATRIX A.
[pic]
[pic]
STEP 6: SUBTRACT THE CORRECTION FACTOR FROM GRANDSUMASQUARE TO GET TOTAL SUM OF SQUARES..
[pic]
[pic]
STEPS 7,8: SQUARE THE SUMS OF EACH OF THE GROUPS AND DIVIDE BY THE # OF ELEMENTS IN EACH GROUP. N=48/4=12.SSb.
[pic]
[pic]
STEP 9: SUBTRACT SSb FROM SST TO GET THE WITHIN GROUPS SUM OF SQUARES SSW.
[pic]
[pic]
(SEE ABOVE FOR CALCULATION OF df VALUES.)
STEP 10,11 & 12: We’ve already calculated the df and ERROR matrices above, so using the vectorize operator, we’ll calculate f directly.
1131.917 1/47 44/565.167
SS = 566.750 df = 1/3 ERROR = 44/565.167
565.167 1/44 .0779
[pic] [pic]
INTERPRETATION: The Only significant value is 14.708, with a df of 3 and 44 would occur by chance once in a thousand times, it is concluded that the level of shock intensity does affect the time required to solve these problems.
STEP 2: PROGRAM: SUM1 = [1]1,12[A]12,4 [pic]
STEP 3: PROGRAM: [1]1,12([A]4,12([A]4,12)[1]4,1 [pic]
STEP 4: PROGRAM: GRANDSUM = [1]1,12[A]14,4[1]4,1 [pic]
STEP 5: PROGRAM: 1/NT([1]1,12[A]12,4[1]4,1)2 [pic]
STEP 6: PROGRAM: SST = GRANDSUMSQ – CORRFACT
STEP 7: PROGRAM: SSb = (1/NG)([SUM1]1,4[SUM1T]4,1-CORRFACT
[pic]
STEP 8: PROGRAM = [1]1,12([A]12,4 ([A]12,4)[1]4,1- CORRFACT) – (1/12)(([SUM1]1,4[SUM1T]4,1))-CORRFACT
[pic]
[pic]
STEP 10:
[pic] [pic] [pic]
STEP 11: PROGRAM: F= SS o df-1 o ERROR-1
[pic] [pic]
FACTORIAL DESIGN: TWO FACTORS
SECTION 2.2, COMPUTATIONAL HANDBOOK OF STATISTICS, SECOND EDITION. PG. 27-31.
STATISTICAL EQUATIONS
r r
ROWS cm((xi..- x)2 LR=r(Ri2-T2 r-1 MR MR/MW MR/MI MR/MI
i=1 i=1
c c
COLUMNS rm((x.j.- x)2 LC=c(Cj2-T2 c-1 MC MC/MW MC/MI MC/MW
j=1 j=1
r c r c
INTERACTION m(((xij.-xi..-x.j.+x)2 LI=rc((Tij2 (r-1)(c-1) MI MI/MW MI/MW MI/MW
i=1j=1 i=1 j=1
r c
-r(Ri2-c(Cj2+T2
i=1 j=1
r c m r c m
WITHIN ( ( ((xijk- xij)2 LW=n( ( (x2ijk rc(m-1) MW
CELLS i=1 j=1 m=1 i=1 j=1 m=1
r c
-rc((T2ij
i=1j=1
r c m r c m
TOTAL ( ( ((xijk- x)2 LT=n( ( (x2ijk-T2 n-1
i=1 j=1 m=1 i=1 j=1 m=1
SOURCE: STATISTICS WITH APPLICATIONS TO THE BIOLOGICAL AND HEALTH SCIENCES, REMINGTON & SCHORK,PRENTICE/HALL PG. 298.
MATRIX METHOD
EXAMPLE: An experimenter is interested in determining the effects of high Vs low intensity shock on the memorization of a hard Vs easy list of nonsense syllables. Subjects are randomly assigned to four experimental conditions. Group 1 receives periodic low intensity shocks and memorizes an easy list; group 2, high shock and an easy list; group 3 low shock and a hard list; group 4, high shock and a hard list. The total number of errors made by each subject is the measure recorded. (Note: If more than a 2x2 factorial is used i.e. a 2x3, 3x3, etc., simply table the additional treatment groups in their appropriate row or column.)
ARCHAIC METHOD: Used if # of elements in each column are not equal.
and NT = total # of data elements
[pic]=(A B C . D)
[pic]= For all N’s not being equal
[pic]= For all N’s being equal
FIRST FACTOR EFFECT:
[pic]
[pic]- CORRECTION FACTOR
UPGRADED:
[pic]
[pic]
[pic]
PROGRAM: 1/NS([SUM1]1,4[DB2]4,2)2-CORRFACT
SECOND FACTOR SHOCK: SSLIST=SSL.
[pic]
[pic]
[pic]
Upgraded:
[pic]
[pic]
PROGRAM: (1/NL)([SUM1]1,4[DB1]4,2)2 - C0RRFACT
INTERACTIVE SHOCK: SSSHOCK x LIST = SSSxL
[pic]
[pic]
[pic]
[A B C D] A/N1 - b2/NT -SSL - SSS = SSSxL
B/N2
C/N3
D/N4
UPGRADE: [SUM1][SUM1]T - CORRFACT - SSL - SSS
SSERROR = SST -SSL - SSS - SSSxL
Solution to example:
STEP 1: Table the data. AND THE VARIABLES
[pic] [pic]
[pic] [pic] [pic] [pic]
[pic] [pic] [pic] [pic] [pic]
STEP 2: Add the scores in each group. = SUM1.
PROGRAM: SUM1 = [1]1,6[A]6,4
[pic]
[pic]
[pic]
HP-48G PROGRAM:
RCL ONE6
RCL A
x
STO SUM1
STEP 3: Square each number in the matrix and sum.
PROGRAM: or [1]1,6([A]6,4([A]6,4)[1]4,1
[pic]
[pic]
[pic]
[1]1,4(diag([A]4,6T[A]6,4))[1]4,1
[1 1 1 1](diag( 9 16 14 11 12 8 9 15 10 19 )) 1
15 13 9 9 8 11 16 13 12 16 1 =
` 10 12 18 16 17 15 14 9 18 18 1
19 16 18 23 14 15 11 9 16 23 1
12 8 17 14
8 11 15 15
[1 1 1 1](diag 862 752 1034 1220 ) 1
752 741 913 1139 1 =
1034 913 1338 1537 1
1220 1139 1537 1891 1
[1 1 1 1] 862 1
741 1 = 4832 = (x2
1338 1
1891 1
HP-48G PROGRAM:
RCL A
(
TRN
SWAP
x
MTH,MATR,NXT,(DIAG EXTRACTS DIAGONAL AS COLUMN MATRIX
4
(
DIAG( MAKES COLUMN INTO A DIAGONAL MATRIX
RCL ONE4
SWAP
x
RCL FOUR1
x
STO SUM2 (OR GNDSUMSQ)
STEP 4: Add the scores in the entire table, sum, square, and divide by the total number
of elements in the table. NT = 6x4=24.
PROGRAM: 1/NT( [1]1,6[A]6,4[1]4,1)2
[pic]
[pic]
[pic]
[pic]
HP-48G PROGRAM:
RCL ONE6
RCL A
x
RCL FOUR1
x
SQ
RCL N=24
/
STO CORRFACT [4482.6667]
STEP 5: computation of total sum of squares SST.
[pic]
[pic]
[pic]
SST = GNDSUMSQ-CORRFACT
SST = 4832 - 4483 = 349
STEP 6: Computation of the effects of the first factor (the overall effects of low shock
vs. High shock). SSS. NS = NT/# columns in [DB2] 24/2 = 12.
PROGRAM: SSS = (1/NS)([SUM1]1,4 [DB2]4,2)2 – CORRFACT
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
(1/12)([70 65 88 105] 1 0 )2 - 4483 = (1582 + 1702)/12 - 4483 = 4489 - 4483 = 6
0 1
1 0
0 1
HP-48G PROGRAM:
RCL SUM1
RCL DB2
x
(
TRN
x
SQUARE
RCL N = 12
/
RCL CORRFACT
-
STO SSS 6
STEP 7: Computation of the effects of the second factor (the overall effects of the hard Vs easy lists) disregarding the shock dimension = SSLIST = SSL. NL = NT/# columns in DB1. NL = 24/2=12.
PROGRAM: (1/NL)([SUM1]1,4 [DB1]4,2)2 – CORRFACT C2 = CCT
(1/12)([SUM1]1,4[DB1]4,2)2 – CORRFACT
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
(1/12)([70 65 88 105] 1 0 )2 - 4483 = (1352 + 1932)/12 - 4263 = 4489 - 4483 = 140
1 0
0 1
0 1
HP-48G PROGRAM:
RCL SUM1
RCL DB1
x
( Note this program is exactly the same as step 6 above except we
TRN change DB2 to DB1 and SSS to SSL.
x
SQUARE
RCL N = 12
/
RCL CORRFACT
-
STO SSL 140
STEP 8: Computation of the interactive effects of shock and list difficulty. SSSxL.
NSxL = NT/# elements in SUM1 = 24/4 = 6.
PROGRAM: (1/NSxL)[SUM1]1,42 - CORRFACT - SSS - SSL
(1/6)([SUM1]1,4[SUM1]T4,1)- CORRFACT - SSS - SSL
[pic]
[pic]
[pic]
[pic]
(1/6) [70 65 88 105] 70 -4483 - 6 - 140 = 4649 - 4483 - 6 - 140 = 20 = SSSxL.
65
88
105
HP-48G PROGRAM:
RCL SUM1
(
TRN
x
RCL N = 6
/ 4649
RCL CORRFACT 4483
-
RCL SSS 6
-
RCL SSL 140
-
STO SSSxL 20
STEP 9: Computation of the error term sum of squares SSERROR.
SSERROR = SST - SSS - SSL - SSSxL
SSERROR = 349 - 6 - 140 - 20 = 183
STEP 10: All computations are done. Now we must compute the degrees of freedom df.
dfSST = TOTAL # ELEMENTS IN A - 1 = 24 - 1 = 23
dfSSS = TOTAL # SHOCK CONDITIONS - 1 = 2 - 1 = 1
dfSSL = TOTAL # LIST CONDITIONS - 1 = 2 - 1 = 1
dfSSSxL = dfSSS x dfSSL = 1 x 1 = 1
dfSSERROR = dfSST - dfSSS - dfSSL - dfSSSxL = 23 - 1 - 1 - 1 = 20
STEPS 11 & 12: Computation of F.
[pic][pic]
[pic]
[pic] [pic]
If we would rather use diagonalized Matrices rather than the Half-Multiplier Operator, we proceed as follows. To get F as a column Matrix:
[pic]
[pic]
To compute F as a Row Matrix, proceed as follows:
[pic]
[pic]
Here is another way to compute F:
[pic]
[pic]
Or we can compute F in the following manner:
349 23 20/183 183/20 -1
6 1 20/183 9.15
SS = 140 df = 1 ERROR = .1093 or 9.15
20 1 .1093 9.15
183 20 .1093 183/20
349 23 -1 9.15 -1 [349 6 140 20 183] 1/23 .1093
6 1 9.15 1 .1093
F = 140 o 1 o 9.15 = 1 .1093
20 1 9.15 1 .1093
183 20 9.15 1/20 .1093
F = [15.1739 6 140 20 9.15] .1093
.1093
.1093 = [1.6585 .6558 15.30 2.186 1]
.1093
.1093
Which gives us the values of F as a row matrix. In the HP-48G program, I’m going to calculate a column matrix for F.
HP-48G PROGRAM:
RCL SS
5
mth,matr,nxt,diag( This takes the matrix of SS values (of which there are 5 elements,
and makes a diagonal matrix out of it.
RCL df
5
(
DIAG( This takes the df matrix and diagonalizes it.
INV This inverts the diagonal df matrix
x This multiplies diagSS x diagdf
RCL ERROR
x Gives a column matrix, the center value being the F value we are to STO F look at for significance.
STEP 13: Table the final analysis as follows, the probabilities are obtained from tables.
SOURCE SS df ms F p
TOTAL 349 23 - - -
SHOCK 6 1 6 .66 n.s.
LIST 140 1 140 15.30 ± 2.23 is significant at the .05 level using a two-tailed test (see Appendix B).
THE CORRELATION RATIO ( (ETA)
COMPUTATIONAL HANDBOOK OF STATISTICS, SECT. 4.5, PG 185 - 187.
( is a measure of the relationship between two variables. It is useful in two different situations:
When the relationship is sufficiently non-linear that it makes you hesitant about using the Pearson correlation.
When the experiment involves independent and dependent measures and you wish to know the degree of their relationship.
STEPS 1 & 2: Eta is defined as:
( = (SSb/SST)1/2
For illustration, we will use SSb & SST from section 2.1. SSb = 567; SST = 1132
( = (567/1132)1/2 = (.50)1/2 = .71
STEP 3: ( can also be computed from ratios of the mean squares obtained in computing the F ratios. i.e.
F = msbetween/mswithin = 189/12.84 = 14.71 with a df = 3/44
SUPPLEMENT
It is sometimes important to know the ( for research reported in a publication. We will use data presented in sect. 2.7:
STEP 1: Obtain the F of interest; FTr = 52.46, df = 2/60
( = (2(52.46)/2(52.46 + 60/2))1/2 = (104.92/164.92)1/2 = (.64)1/2 = .80
EQUATION: (num(F)/num(F + denom/num))1/2
PARTIAL CORRELATION: THREE VARIABLES
COMPUTATIONAL HANDBOOK OF STATISTICS, SECT. 4.6, PG 187 - 188
Partial correlation is used when you have 3 sets of measures that are related, and you wish to find the relationship between any two when the effect of the third has been taken out of both variables.
EXAMPLE: Suppose we have measures of :
a. Incomes of a group of fathers
b. Incomes of their eldest sons
c. # years of formal education of the sons.
A correlation of +.60 between incomes of fathers and sons (rab) might indicate that money making characteristics run in families. However, one might argue the similarity in incomes is due to their level of education’s. A partial correlation between incomes of fathers & sons with education partialed out would help answer this question. The notation for partial correlation for this instance is rab,c being read as the partial correlation between variables a and b with the effects of c taken out. The formula for this partial correlation is:
rab,c = (rab - racrbc)/((1-rac2)1/2(1-rbc2)1/2)
The correlation’s are all Pearson product-moment correlation’s (Sect. 4.1). Suppose they are computed to be:
rab = .60; rac = .70; rbc = .80
Compute the partial correlation between a & b. The general matrix form is:
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Again Mathcad can't perform matrices in parenthesis in order, so we have to do it one
step at a time:
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The other values are calculated as follows:
rac,b = (rac - rabrbc)/((1-rab2)1/2(1-rbc2)1/2)
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rbc,a = (rbc - rabrac)/((1-rab2)1/2(1-rac2)1/2)
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MATHCAD +6 PROGRAM
PARTIAL CORRELATION: THREE VARIABLES
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AGAIN, THIS EQUATION DOES NOT WORK IN THIS FORM BECAUSE MATHCAD CANNOT PERFORM MATRICES IN PARENTHESIS FIRST, SO MUST USE SUBSTITUTION:
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The correlation rab,c = .09 when the mutual relationships of the variables with C are partialed out. The data suggests the hypothesis that money making abilities is hereditary is false & that the reason fathers & sons have similar incomes is due to similar levels of education.
PARTIAL RANK-ORDER CORRELATION (USING KENDALL’S TAU)
COMPUTATIONAL HANDBOOK OF STATISTICS, SECT. 4.7, PG 189 - 190.
The tau statistics of section 4.3 may be extended to the case of partial correlation:
When there are no tied ranks
When there are tied ranks, but the rank-order correlation’s are expected to give only rough indications of true values
EXAMPLE: Suppose a judge has rank-ordered 15 people on the basis of IQ, personality and poise. The data appears as follows:
Subject rank 1 rank 2 rank 3
(IQ) personality poise
S1 1 3 1
S2 2 5 4
. . . .
. . . .
. . . .
S15 15 13 12
STEP 1: Go to sect. 4.3 and compute the tau value for each of the possible pairs of
rankings. Comparison 1 & 2 = (12; comparison 1 & 3 = (13; comparison 2 & 3 = (23.
For this example, suppose (12 = .50; (13 = .60; (23 = .80
Compute the partial correlation of 1 with 2 partialing out all relationships with 3.
(12,3 = ((12 - (13(23)/((1 - (122)1/2(1 - (232)1/2)
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Poise is very important in the judges rankings over IQ and personality.
Computing the partial correlation of 2 with 3, partialing out 1:
(23,1 = ((23 - (12(31)/((1 - (122)1/2(1 - (132)1/2)
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Computing the partial correlation of 1 with 3 partialing out 2:
(13,2 = ((13 - (12(23)/((1 - (122)1/2(1 - (232)1/2)
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The MathCad + 6 program is the same as Sect. 4.6, so I’ll not cover it here.
MULTIPLE CORRELATION: THREE VARIABLES
COMPUTATIONAL HANDBOOK OF STATISTICS, SECT. 4.8, PG. 190 - 192.
Multiple correlation is used when we wish to determine the relationship of one set of numbers with two other sets of numbers.
EXAMPLE: First we set up a matrix containing the criterion variable (freshman GPA) and two other sets of numbers ( predictors), here they are high school grades and college entrance examination scores. The multiple correlation coefficient between the criterion variable and the two predictor variables gives an indication of the degree to which the predictors, taken together, actually predict.
STEP 1: If we have the matrix of scores and values, we can go to the Pearson product moment correlation in section 4.1 and compute the correlation coefficients as follows: lets abbreviate the 3 variables as:
the criterion variable: Freshman GPA
the first predictor: high-school grades
the second predictor: college entrance exam scores
The first correlation coefficient we have to find is r12, the correlation between GPA and high-school grades; the second correlation we must find is r13, the correlation between GPA and exam scores; the third correlation we must find is r23, the correlation between high-school grades and exam scores. I have no such data, but the illustration in Section 4.8 works fine. Suppose we completed all the computations and found:
r12 = .60 r13 = .50 r23 = .70
STEP 2: The multiple correlation of 1 with 2 & 3 is given by:
R1.23 = ((r122 + r132 -2r12r13r23)/(1 - r232))1/2
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HP-48G PROGRAM:
RCL R1 [.6 .5 -2(.6)(.5)] = [ .6 .5 -.6]
RCL R [.6 .5 .7]T
x
RCL ONER23 [1 .7]
(
TRN
RCL PLMN1 1
2 -1
(
DIAG(
SWAP
x
x
/ [.3725]
(
((TEACH,EXAM,PRGS,APLY,((TEACH
STO R1.23 .6104
The value of the multiple correlation is .61. The correlation between GPA and high-school grades was .6, ant the correlation between GPA and entrance exams was .5, but by combining high-school grades & entrance exams, the multiple correlation was raised to .61. Thus the combination of both predictors produced a higher correlation with the criterion variable than either predictor taken separately. Note: concerning the size of the correlation between the correlation variables and their correlation with the criterion variable: if the simple correlation’s between each of the predictors and the criterion remain unchanged, R will increase as the correlation between predictors become smaller than .65, and the value of R will decrease as the correlation between predictors increases above .65. i.e. the more independent of each other the predictors are, the greater their value in jointly predicting the criterion, providing the correlation’s between each predictor remain constant. If we have more than 2 predictor variables, see the Doolittle solution, Q. McNemar, Psychological Statistics, 4th Ed. New York: John Wiley & Sons, Inc., 1969, Chapter 11, pp. 199-202.
SIMPLE ANALYSIS OF COVARIANCE: ONE TREATMENT VARIABLE
COMPUTATIONAL HANDBOOK OF STATISTICS, SECT. 4.9, PG. 192 - 197.
When an experimental control of an important variable (such as age or IQ) is impossible or impractical, statistical control may be achieved by use of a covariance analysis. Below is the method when there is only one treatment variable.
EXAMPLE: In order to make a decision concerning future equipment needs, an elementary school principal designed the following experiment to evaluate 3 different methods of teaching spelling. Eighteen first-grade children were given a vocabulary test (the control) represented by the X column. Three different teaching methods were used for 3 months and then a standard spelling test was given. The Y columns represent these criterion spelling scores.
STEP 1: Table the scores as a matrix:
X Y1 X Y2 X Y3
32 62 21 72 38 95
46 66 24 85 45 88
[A]6,6 = 27 64 18 61 52 104
35 48 32 87 48 86
31 51 35 69 41 72
40 74 26 74 37 63
VARIABLES:
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STEP 2: Sum the columns:
PROGRAM: [1]1,6[A]6,6
[1 1 1 1 1 1] 32 62 21 72 38 95
46 66 24 85 45 88
27 64 18 61 52 104 = [211 365 156 448 261 508] STO SUM1
35 48 32 87 48 86
31 51 35 69 41 72
40 74 26 74 37 63
STEP 3: Square all the numbers in the X columns and sum. There are two ways I’ll show to solve this, the first is for the X2 values only. But since we have to square the Y columns also, I prefer to use the second method.
METHOD 1: PROGRAM: GRANDSUMX = [A][DBdiag2]
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METHOD 2:
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PROGRAM: GRANDSUMSQX = [1]1,6diag([X]T6,6[X]6,6)
HP-48G PROGRAM:
RCL ONE6
(
TRN
SWAP
RCL X
(
TRN
SWAP
x
(DIAG
6
(
DIAG(
x [7655 0 4266 0 11,527 0]
SWAP
x 23,448
STO GNDSMSQX We can do the same for Y, except the diagonal DB begins with 0 & alternates with 1’s & 0’s.
METHOD 2:STEPS 3 & 10: This is the way I prefer because we square both X & Y at the same
time.
PROGRAM: SS310 = [1]1,6(diag([A]6,6T[A]6,6)[DB2XY]6,2)
[1 1 1 1 1 1] diag( 32 46 27 35 31 40 32 62 21 72 38 95 ) 1 0
62 66 64 48 51 74 46 66 24 85 45 88 0 1
21 24 18 32 35 26 27 64 18 61 52 104 1 0 =
72 85 61 87 69 74 35 48 32 87 48 86 0 1
38 45 52 48 41 37 31 51 35 69 41 72 1 0
95 88 104 86 72 63 40 74 26 74 37 63 0 1
X Y
[1 1 1 1 1 1]( 7655 1 0 ) [1 1 1 1 1 1] 7655 0
22677 0 1 0 22677
4266 1 0 = 4266 0 =
33936 0 1 0 33936
11527 1 0 11527 0
44134 0 1 0 44134
( X2 ( Y2
[23,448 100,747] STO SS3-10
STEPS 4-11: Here we compute the X and Y values at the same time.
skip step 5, not needed; skip step 8, not needed; skip step 9, redundant
1/NX = 1/18; 1/NX1 = 1/6
STEP 4/11
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Or alternately:
PROGRAM: GRNDSUMSQX - 1/NX(([SUM1][DB2X])T)o2 = 1537.778 = SSTX
23448 21,910.222
( (1/6)([1]1,6(Diag([SUM1]6,6)2([DBXY]6,2)))T - 1/NX([SUM1][DB2X])2 = 919.445
22,829.667 21,910.222
For the sake of clarification, I present my reasoning why steps 5, 8 & 9 are not necessary. (Step 8 must be calculated for it is used later, but not here).
Let G = GRNDSUMSQX
H = ( (1/NX1)([1]1,6(Diag([SUM1]6,6)2([DBXY]6,2)))T
X = 1/NX(([SUM1][DB2])T)o2
THEN (G - X)-(H - X) = G - H
step 9 = GRNDSUMSQX - (1/NX1)([1]1,6(Diag([SUM1]6,6)2([DB2X]6,1)) = 618.333
23,448 22,829.667
(step 3) (step 7) (step 9)
(I’m not showing the math here, I’ll do all the math in one step with both X & Y in a bit)
STEP 5 & 12:
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STEP 10: Square the numbers in the Y columns and sum. Done in step 3 = 100,747.
STO GRNDSUMSQY.
STEP 11, 12 & 13: Add the sums of the Y columns, square and divide by 1/NY = 18, & subtract from
step 10: GRNDSUMSQY - 1/NY([SUM1][DB2Y])2 = 100,747 - 96,946.722 = 3800.278 = SSTY
STEP 14: Square the three sums of the Y columns & sum, ? by # of scores composing each
sum (1/NY1 = 6)
( (1/6)([1]1,6(Diag([SUM1]6,6)2([DB2Y]6,1)))T = 98,665.5
Final value Y: 100,747 - 98,665.5 = 2081.5
MATH EXAMPLE:
STEP 4, 5 & 8:
PROGRAM: [1]1,6(diag([A]6,6T[A]6,6)[DB2XY]6,2) - 1/NY1([SUM1][DB2Y])2 =
GRNDSUMSQXY - 1/NY1(([SUM1][DBXY])T)o2 =
[23,448 100,747] - (1/18)(([211 365 156 448 261 508] 1 0 )T)o2 =
0 1
(X (Y 1 0
[628 1321] STO SUMXY 0 1
1 0
0 1
(1/18) 628 628 = (1/18) 394,384 = 21,910.222 ((X)2
1321 1321 1,745,041 96,946.722 ((Y)2
STO SS4-5-8
STEP 6 & 13: Subtract CORRFACT from GRNDSUMSQXY:
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23,448 - 21,910.222 = 1537.778
100,747 96,946.722 3800.279 STO SS6-13
STEPS 7 & 14: Square the three sums of the X & Y columns (keeping them separate) and
divide by the # of scores composing each sum NXY1 = 6.
( (1/6)([1]1,6(Diag([SUM1]6,6)2([DBXY]6,2)))T
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(((1/6) [1 1 1 1 1 1] 211 2 1 0 ))T
365 0 1
156 1 0
448 0 1
261 1 0
508 0 1
((1/6) [1 1 1 1 1 1] 44,521 0 )T ((1/6) [136,978 591,993])T =
0 133,225
24,336 0 =
0 200,704
40,401 0
0 258,064
= [22,829.667 98,665.500] STO SS7-14
STEPS 8 & 15:
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OR
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STEPS 9 & 16:
[23,448 100,747] - [22,829.667 98,665.500] = [618.333 2081.500]
STO SS9-16
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OR
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STEP 17: Multiply each X score by its paired Y score and sum the products: Because of
the way I tabled the values, , we are going to have to unscramble the matrix. We will do this as follows:
PROGRAM: [1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1
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The arrow above log(A6,6) allows us to take the log of each element in A
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Or by using Logs:
32 62 21 72 38 95 1 0 0
46 66 24 85 45 88 1 0 0
[1]1,6 (ANTI-LOG(LOG 27 64 18 61 52 104 0 1 0 ))[1]3,1 =
35 48 32 87 48 86 0 1 0
31 51 35 69 41 72 0 0 1
40 74 26 74 37 63 0 0 1
[1]1,6 ANTI-LOG( 1.5051 1.7924 1.3222 1.8573 1.5798 1.0777 1 0 0 )[1]3,1
1.6628 1.8195 1.3802 1.9294 1.6532 1.9445 1 0 0
1.4314 1.8062 1.2553 1.7853 1.7160 2.0170 0 1 0
1.5441 1.6812 1.5051 1.9395 1.6812 1.9345 0 1 0
1.4914 1.7076 1.5441 1.8388 1.6128 1.8573 0 0 1
1.6021 1.8692 1.4150 1.8692 1.5682 1.7993 0 0 1
RCL LOG A
RCL DB1
x [1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1
(
((TEACH,EXAM,PRGS,APPLY,((TEACH
RCL ONE6
SWAP
x
RCL THREE1
x
STO XYPROD 47,131
[1 1 1 1 1 1] 1984 1512 3610 1 = [12,969 11,773 22,389] 1
3036 2040 3960 1 1 = [47,131]
1728 1098 5408 1 1
1680 2784 4128
1581 2415 2952
2960 1924 2331 STO SS17
STEP 18: Multiply the sum of column X and column Y together and divide by the # of XY pairs. NP = 18.
PROGRAM: SS18 = (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1)
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[pic]=
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(1/18)10EEX([2.7980 3.1209] 1 ) = (1/18)ANTI-LOG[5.9189] = (1/18)(829,588) =
1
[46,088.2222] STO SS18
STEP 19: Subtract step 18 from step 17.
PROGRAM: SS19 =[1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1)
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= 47,131 - 46,088.2222 = 1042.778 STO SS19
STEP 20: Multiply each X sum by its paired Y sum, sum them and divide by the # of pairs
used to obtain this sum. NPXY = 6.
PROGRAM: SS20 = (1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))[1]3,1)=)
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RCL SUM1
(
EXAM,PRGS,APLY,((TEACH,(
RCL DB1
x
(
EXAM,PRGS,APLY,((TEACH,( [77,015 69,888 132,588]
RCL THREE1
x [279,491]
RCL NPXY = 6
/
STO SSPXY [46,581.8333]
(1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))[1]3,1)=)
(1/6)((ANTI-LOG([2.3243 2.5623 2.1931 2.6513 2.4166 2.7059] 1 0 0 ) 1 )
1 0 0 1
0 1 0 1
0 1 0
0 0 1
0 0 1
= (1/6)(ANTI-LOG([4.8866 4.8444 5.1225]) 1 = (1/6)( [77,015 69,888 132,588] 1 =
1 1
1 1
(1/6)(279,491) = 46,581.833 STO SS20
STEP 21: Subtract result of step 18 from result of step 20.
PROGRAM: SS21 = (1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))[1]3,1)=) - (1/NP)ANTI-
LOG(LOG([SUMXY])[1]2,1)
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46,581.833 - 46,088.222 = 493.611
HP-48G PROGRAM:
RCL SSPXY
RCL SSP
-
STO SS21
STEP 22: Subtract the value of step 21 from step 19:
PROGRAM: [1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1) -
(1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))[1]3,1)=) - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1)
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= 1042.778 - 493.611 = 549.167 STO SS22
STEP 23: Square result of step 19 and divide by result of step 6:
PROGRAM: ([1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1))2
(GRNDSUMSQXY - 1/NY1(([SUM1][DBXY])T)o2 ) 1 =
0
STEP 23:
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1042.7782/1537.558 = 707.115 STO SS23
STEP 24: Subtract the result of step 23 from the result of step 13:
PROGRAM: (GRNDSUMSQXY - 1/NY1([SUM1][DBXY])o2) 0 - ([1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1))2
1
(GRNDSUMSQXY - 1/NY1([SUM1][DBXY])o2 ) 1
0
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= 3800.278 - 707.115 = 3093.163 STO SS24
STEP 25: Square the result of step 22 & divide by the result of step 9.
([1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1) -(1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))[1]3,1)=) - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1))2 =
GRNDSUMSQXY - 1/NXY1([SUM1][DBXY])o2 (1/6)([1]1,6(Diag([SUM1]6,6)2([DBXY]6,2)) 1
0
STEP 25 EQUATION IS TOO LONG TO PUT ON SINGLE PAGE:
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= 549.1672/618.333 = 487.738 STO SS25
STEP 26: Subtract the result of step 25 from the result of step 16:
GRNDSUMSQXY-1/NXY1([SUM1][DBXY])o2 - ([1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 0 -(1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1) -(1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))[1]3,1)=) - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1))
1
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= 2081.5 - 487.738 = 1593.76 STO SS26
GRNDSUMSQXY-1/NXY1([SUM1][DBXY])o2 - ([1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 0 -
1
(1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1) -(1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))[1]3,1)=) - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1))
(This is the above equation written in a point that can be more easily read, but it’s not all on the same line).
STEP 27: Divide the value of step 26 by the # of degrees of freedom for the adjusted within groups measures. The df’s will always be NPXY - a - 1 where NPXY is the total # of XY pairs, & a is the # of experimental groups (, corresponding to the three methods of teaching).
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1593.762/(NPXY - a - 1) = 1593.762/(18-3-1) = 1593.762/14 = 113.840 STO SS27
STEP 28: Subtract the value of step 26 from the value of step 24.
PROGRAM: (GRNDSUMSQXY - 1/NY1([SUM1][DBXY])o2) 0 - ([1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1))2
1- -
(GRNDSUMSQXY - 1/NY1([SUM1][DBXY])o2 ) 1
0
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GRNDSUMSQXY-1/NXY1([SUM1][DBXY])o2 - ([1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 0 -
1
(1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1) -(1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))[1]3,1)=) - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1))
= 3093.163 - 1593.762 = 1499.401 = SS28
STEP 29: Divide SS28 by the degrees of freedom for the adjusted between groups measure.
DF = a - 1 = 3-1 = 2
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SS28/14 = 1499.401/2 = 749.700 STO SS29
STEP 30: Compute the F ratio: F = SS29/SS27
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F = 749.7/113.840 = 6.59
This F has a df of (a-1)/(N-a-1) ( 2/14 for this example. From Appendix E, with a df =2/14, an F ratio larger than 6.51 would be expected by chance less than one time in 100. The F value is significant at the .01 level. We conclude, since the groups differed in vocabulary ability before training, the methods of teaching produced significantly different results on the spelling test.
MATHCAD +6 PROGRAM
32 62 21 72 38 95
46 66 24 85 45 88 ONE6 = [1 1 1 1 1 1]
A = 27 64 18 61 52 104
35 48 32 87 48 86
31 51 35 69 41 72 ZERO1 = 0
40 74 26 74 37 63 1
ONE0 = 1 1 0 1 0 0
0 0 1 1 0 0
DB2XY= 1 0 DB1 = 0 1 0
TWO1 = 1 0 1 0 1 0
1 1 0 0 0 1
0 1 0 0 1
THREE1 1
1
1
I6 = identity(6)
STEP 1: COMPUTE SUM1:
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STEP 2: SQUARE BOTH THE X & Y COLUMNS AND KEEP VALUES SEPARATE. REMEMBER, MATHCAD CANNOT HANDLE MATRICES IN PARENTHESIS, AND IT CANNOT DIVIDE MATRICES, EVEN WHEN THEY ARE 1X1'S, SO WE MUST MULTIPLY BY THEIR INVERSES:
PROGRAM: [1]1,6(diag[A]^T[A][DB2X,Y]
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STEPS458: STEP 4:ADD THE SUMS OF THE X & Y COLUMNS; STEP 5: SQUARE AND DIVIDE BY Np =18;STEP 8: SUBTRACT STEP 5 FROM STEP 8.
PROGRAM: SS310 - (1/Np)( [SUM1][DBXY] )^o2
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STEP 7&14: SQUARE THE THREE SUMS OF THE X&Y COLUMNS (KEEPING THEM SEPARATE) & DIVIDE BY THE # OF SCORES COMPOSING EACH SUM. Nxy1 = 6.
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STEPS 9&16:SUBTRACT SS310 FROM SS714
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STEP 17: MULTIPLY EACH X SCORE BY ITS PAIRED Y SCORE AND SUM:
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STEP 18: MULTIPLY THE ELEMENTS IN SS41 BY EACH OTHER & DIVIDE BY THE # OF XY PAIRS. Nxy = 18.
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STEP 19: SUBTRACT SS18 FROM SS17.
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STEP 20: MULTIPLY EACH X TERM IN SUM1 BY ITS PAIRED Y SUM & DIVIDE BY # OF PAIRS USED TO OBTAIN THIS SUM. Npxy = 6.
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STEP 21: SUBTRACT SS18 FROM SS20:
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STEP 22: SUBTRACT SS21 FROM SS19:
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STEP 23: SQUARE SS19 AND DIVIDE BY SS613x:
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OR
[pic]
[pic]
STEP 24: SUBTRACT SS23 FROM SS13y:
[pic]
[pic]
[pic]
[pic]
OR
[pic]
[pic]
STEP 25: SQUARE THE RESULT OF SS22 & DIVIDE BY SS916x
[pic]
[pic]
[pic]
[pic]
OR
[pic]
[pic]
STEP 26: SUBTRACT SS25 FROM SS16y
[pic]
[pic]
[pic]
[pic]
OR
[pic]
[pic]
STEP 27: DIVIDE VALUE OF SS26 BY # OF DEGREES OF FREEDOM FOR ADJUSTED WITHIN GROUPS MEASURES.
df = N-a-1 = 18 - 3 - 1 = 14.
[pic]
[pic]
OR
[pic]
[pic]
STEP 28: SUBTRACT SS26 FROM SS24
[pic]
[pic]
OR
[pic]
[pic]
STEP 29: DIVIDE SS28 BY THE df FOR THE ADJUSTED BETWEEN GROUPS MEASURE: df = a-1 = 3-1=2.
[pic]
[pic]
OR
[pic]
[pic]
STEP 30: COMPUTATION OF THE F RATIO:
[pic]
[pic]
OR (THIS EQUATION WITH ILLEGAL ARRAY OPERATION IS INCLUDED TO SHOW WE MUST MULTIPLY BY THE INVERSE
[pic]
[pic]
F2 =[pic] X
[pic]
[pic]
SUPPLEMENT
In order to determine whether the individual group regressions are homogeneous with the overall regression, first compute the sum of squares and the sum of the cross products (SP) for each group of data.
STEP 1: Using the data from the original analysis, compute SSx (sum of squares) for each group. The formula for each computation is:
(X2 - 1/NG((X)2
PROGRAM: SUM1X = [1]1,6[A]6,6[DB2X]6,6
[pic]
[pic]
[pic]
HP-48G PROGRAM:
RCL ONE6
RCL A
x
STO SUM1X
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
PROGRAM: SSX = [1]1,6diag([X]T6,6[X]6,6)
HP-48G PROGRAM:
RCL ONE6
(
TRN
SWAP
RCL X
(
TRN
SWAP
x
(DIAG
6
(
DIAG(
x [7655 0 4266 0 11,527 0]
STO SSX
Or using the Half-Multiplier Operator:
SSX = (X2 - 1/NG((X)2 = [SP]T - ([SUM1X]T)o2
7655 (1/6) 211 211 7655 (1/6) 44,521 234.8
0 0 0 0 0 0
4266 - 156 o 156 = 4266 - 24,336 = 210.0
0 0 0 0 0 0
11527 261 261 11,527 68,121 173.5
0 0 0 0 0 0
STO SSX
NOTE: We could have solved this easier if we re-tabled the data matrix to hold only the X values, but it really isn’t necessary, this math is so free-wheeling we can solve the math any way we want, just as long as we follow the rules.
STEP2: The SP values are computed using the formula: (XY - (1/NP) (X(Y
PROGRAM: XYPROD = [1]1,6(ANTI-LOG(LOG([A])[DB1]6,3
[pic]
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32 62 21 72 38 95 1 0 0
46 66 24 85 45 88 1 0 0
[1]1,6 (ANTI-LOG(LOG 27 64 18 61 52 104 0 1 0 ) =
35 48 32 87 48 86 0 1 0
31 51 35 69 41 72 0 0 1
40 74 26 74 37 63 0 0 1
[1]1,6 ANTI-LOG( 1.5051 1.7924 1.3222 1.8573 1.5798 1.0777 1 0 0 )
1.6628 1.8195 1.3802 1.9294 1.6532 1.9445 1 0 0
1.4314 1.8062 1.2553 1.7853 1.7160 2.0170 0 1 0
1.5441 1.6812 1.5051 1.9395 1.6812 1.9345 0 1 0
1.4914 1.7076 1.5441 1.8388 1.6128 1.8573 0 0 1
1.6021 1.8692 1.4150 1.8692 1.5682 1.7993 0 0 1
RCL LOG A
RCL DB1
x [1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)
(
((TEACH,EXAM,PRGS,APPLY,((TEACH
RCL ONE6
SWAP
x
STO XYPROD
[1 1 1 1 1 1] 1984 1512 3610 = [12,969 11,773 22,389]
3036 2040 3960
1728 1098 5408
1680 2784 4128
1581 2415 2952
2960 1924 2331
PROGRAM: SSPXY = (1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))
RCL SUM1
(
EXAM,PRGS,APLY,((TEACH,(
RCL DB1
x
(
EXAM,PRGS,APLY,((TEACH,( [77,015 69,888 132,588]
RCL NPXY = 6
/
STO SSPXY [12,835.8333 11,648 22,098]
(1/NPXY)(ANTI-LOG(LOG([SUM1])[DB1]))
(1/6)((ANTI-LOG([2.3243 2.5623 2.1931 2.6513 2.4166 2.7059] 1 0 0 )
1 0 0
0 1 0
0 1 0
0 0 1
0 0 1
= (1/6)(ANTI-LOG([4.8866 4.8444 5.1225]) = (1/6)( [77,015 69,888 132,588] =
[12,835.8333 11,648 22,098]
HP-48G PROGRAM:
RCL [XYPROD] 12,969 12,835 133.2
TRN 11,773 - 11,648 = 125.0
RCL SSPXY 22,389 22,098 291.0
TRN
-
STO SP
STEPS 3 & 4: Compute SP2/SSX and sum the products:
=
SP2/SSX = [pic]
[pic]
HERE WE ARE GOING TO SHORTEN SSX WHICH A 6 BY ONE MATRIX INTO A THREE BY ONE MATRIX WE WILL CALL SSX1:
[pic]
[pic]
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[pic]
[pic]
PROGRAM: SP2/SSX = [1]1,3(([SP]o2)o[SSX]-1)
[1 1 1] ( 133.2 133.2 ) 234.8 -1 [1 1 1] 17,742 .0043
125.0 o 125.0 o 210.0 = 15,625 o .0048 = 638
291.0 291.0 173.5 84,681 .00
STEP 5: Get the error sum of squares for Y by subtracting SS4 (sum of X columns) from
SS16. PROGRAM: ERRY = ([SS916][ZERO1]-[SUMXY][ONE0]
[SS916] 0 - [1 1][SS51] 1
0
[618.333 2081.5] 0 - [628 1321] 1 = [0 2081.5] - [628 0]
1 0
[-628 2081.5] 1 = 1453.5 NOTE: On Pg. 197, the value in step 5 638 is
1 incorrect, it should be 628.
Or by another method:
[pic]
[pic]
[pic]
HP-48G PROGRAM:
RCL SS916
RCL ZERO1
x
RCL TWO1
RCL SS51
x
RCL ONE0
x
-
RCL TWO1
x
STO ERRY
STEP 6: Divide ERRY by df = N - 2a = 18 - 6 = 12.
DENOM = ERRY – (N-2a)
HP-48 PROGRAM
RCL ERRY
RCL N-2a
/
STO DENOM 120.3
STEP 7: Subtract Step 5 error sum of squares for Y (SS26)
[pic]
[pic]
ERRYADJ = SS916 x ZERO1 - ((SS17 -SS18) - (SS20 -SS18))2 x (SS916 x ONE0)-1 - ([SS916][ZERO1]-[1]1,2[SS51][ONE0])[TWO1]
= SS26 - ERRY = 1593.762 - 1453.5 = 140.26 STO ERRYADJ
STEPS 8 & 9: Divide the result of step 7 by df = a - 1 = 3-1 = 2; Divide this by DENOM
in step 7:
[pic]
[pic]
[pic]
HP-48G PROGRAM:
RCL ERRYADJ
RCL a-1 = 2
/
RCL DENOM
/
STO F .5830
Since F is very small, we conclude the hypothesis that the group regressions are homogeneous.
PROGRAMS FOR EXAMPLE 4.9
STEP 2:
SUM1 = [1]1,6 [A]6,6
STEP 3 & 10:
METHOD 1:
GRANDSUMSQX = [1]1,6diag([X]T6,6[X]6,6)
PROGRAM: GRANDSUMX = [A][DBdiag2]
[pic]
[pic]
METHOD 2:
PROGRAM: SS310 = [1]1,6(diag([A]6,6T[A]6,6)[DB2XY]6,2)
STEP 4 & 11:
[pic]
STEPS 5 & 12:
[pic]
STEPS 6 & 13:
[pic]
STEPS 7 & 14:
[pic]
STEPS 8 & 15:
[pic]
STEPS 9 & 16:
[pic]
STEP 10:
STEP 17:
PROGRAM: [1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1
[pic]
STEP 18:
SS18 = (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1)
[pic]
STEP 19:
PROGRAM: SS19 =[1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1)
[pic]
STEP 20:
PROGRAM: SS20 = (1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))[1]3,1)=)
[pic]
STEP 21:
PROGRAM: SS21 = (1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))[1]3,1)=) - (1/NP)ANTI-
LOG(LOG([SUMXY])[1]2,1)
[pic]
[pic]
STEP 22:
PROGRAM: [1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1) -
(1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))[1]3,1)=) - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1)
[pic]
[pic]
STEP 23:
PROGRAM: ([1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1))2
(GRNDSUMSQXY - 1/NY1(([SUM1][DBXY])T)o2 ) 1 =
0
[pic]
[pic]
STEP 24:
PROGRAM: (GRNDSUMSQXY - 1/NY1([SUM1][DBXY])o2) 0 - ([1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1))2
1
(GRNDSUMSQXY - 1/NY1([SUM1][DBXY])o2 ) 1
0
[pic]
[pic]
STEP 25:
([1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1) -(1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))[1]3,1)=) - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1))2 =
GRNDSUMSQXY - 1/NXY1([SUM1][DBXY])o2 (1/6)([1]1,6(Diag([SUM1]6,6)2([DBXY]6,2)) 1
0
STEP 26:
GRNDSUMSQXY-1/NXY1([SUM1][DBXY])o2 - ([1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 0 -(1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1) -(1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))[1]3,1)=) - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1))
1
[pic]
STEP 27:
[pic]
STEP 28:
PROGRAM: (GRNDSUMSQXY - 1/NY1([SUM1][DBXY])o2) 0 - ([1]1,6(ANTI-LOG(LOG([A])[DB1]6,3)[1]3,1 - (1/NP)ANTI-LOG(LOG([SUMXY])[1]2,1))2
1- -
(GRNDSUMSQXY - 1/NY1([SUM1][DBXY])o2 ) 1
0
[pic]
STEP 29:
[pic]
STEP 30:
[pic]
SUPPLEMENT:
STEP 1:
PROGRAM: SUM1X = [1]1,6[A]6,6[DB2X]6,6
[pic]
PROGRAM: SSX = [1]1,6diag([X]T6,6[X]6,6)
SSX = (X2 - 1/NG((X)2 = [SP]T - ([SUM1X]T)o2
STEP 2:
PROGRAM: XYPROD = [1]1,6(ANTI-LOG(LOG([A])[DB1]6,3
[pic]
[pic]
[pic]
PROGRAM: SSPXY = (1/NPXY)((ANTI-LOG(LOG([SUM1])[DB1]))
STEPS 3 & 4:
PROGRAM: SP2/SSX = [1]1,3(([SP]o2)o[SSX]-1)
SP2/SSX = [pic]
STEP 5:
[pic]
ERRY = PROGRAM: ([SS916][ZERO1]-[SUMXY][ONE0]
STEP 6:
DENOM = ERRY – (N-2a)
STEP 7:
ERRYADJ = SS916 x ZERO1 - ((SS17 -SS18) - (SS20 -SS18))2 x (SS916 x ONE0)-1 - ([SS916][ZERO1]-[1]1,2[SS51][ONE0])[TWO1]
[pic]
STEP 8:
[pic]
FACTORIAL ANALYSIS OF COVARIANCE: TWO TREATMENT VARIABLES
COMPUTATIONAL HANDBOOK OF STATISTICS, 2ND EDITION, SECT. 4.10, PG. 198-209
EXAMPLE: A university was engaged in teaching Peace Corps volunteers the foreign languages they would need during their tours of duty. The following experiment was carried out to determine the effectiveness of the different teaching methods used:
A language aptitude test was given, this served as the control. (X column). The average aptitude scores differed considerably, so covariant analysis was needed to equate these scores, so that any differences found after the experiment could be interpreted as results of the experimental manipulations rather than the original differences in aptitude.
two different teaching methods were used, formal classroom meetings with lectures
no formal classes, only conversation periods held in a congenial atmosphere.
Half the students being taught by each teaching method spent 3 hours a day in the language lab using recording equipment and language tapes., the other half never entered a lab.
Two years later, when the volunteers returned from overseas, they were asked to evaluate the degree to which their language training prepared them. The ratings were on a 10 point scale ranging from one to 10. These ratings served as the criterion measures (Y columns).
STEP 1: Table the data: I am going to table this data differently from the simple
covariance problem in section 4.9 just to see if the other way I see is really simpler.
Now for the problem I am currently working on in Section 4.10 of the Computational Handbook of Statistics, 2nd Ed., PP. 199-200, steps 2-6,16-20. (5-26-97)
62 5 46 5 84 2 58 9 1 62 5 46 5 84 2 58 9
75 7 53 4 91 3 72 10 1 75 7 53 4 91 3 72 10
41 3 57 3 68 1 61 8 o 1 = 41 3 57 3 68 1 61 8
88 8 49 7 77 1 65 8 1 88 8 49 7 77 1 65 8
72 7 62 6 85 3 59 10 1 72 7 62 6 85 3 59 10
When I am doing the proofs, I use the symbol + between the sub-matrices, this does not necessarily mean to add, but to connect. If we transpose as is, we get a 1x40 matrix with alternating X & Y values in sets of 5. This I do not want, I want the X with its corresponding Y pair, so I do the connection [X]1 + [y]1 = [X Y]1 or [X]I + [Y]I = [X Y]I and then transpose the resulting matrix to get: Now the column matrix of
VARIABLES:
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[pic] [pic] [pic]
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[pic] [pic]
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[pic][pic][pic][pic][pic]
[pic][pic][pic][pic][pic][pic]
[pic][pic]
Now to begin the Factorial Analysis of Covariance: Two treatment variables.
STEP 2: Add the scores in each column:
[pic]
[pic]
OR
[pic]
[pic]
(X’s (Y’s
SUM1A = [DB1]4,20[A1]20,2 = (XG1 338 30 (YG1
(XG2 267 25 (YG2
(XG3 405 10 (YG3
(XG4 315 45 (YG4
or: SUM1 = [1]1,5[A]5,8 = [338 30 267 25 405 10 315 45]
THE CONTROL VARIABLE (steps 3-15)
STEPS 3-6: Find the sum of squares for the X column:
(3 & 4): Square all values in the X column and sum the values in the X column:
PROGRAM: [X]T20,1[X1]20,1
[pic]
[pic]
[pic]
[pic]
a = sum of squares, b = sum of elements
[pic]
a b
[pic]
(5) Square the sum of the X column and divide by 20:
(6) Subtract result of step 5 from step 3:
We will do this all in one operation. b = 1325
Using the b method:
[pic]
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[pic]
CORRECTION FACTOR, 3 METHODS:
Method 1:
[pic]
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[pic]
Method 2:
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Method 3:
[pic]
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[pic]
CORRFACT = (1325)(1325/20) = 87,781.250
CORRFACT = [b][-b/NT]
STEPS 7 & 8 : (7) Add the sums of the X column of the two classroom groups, square and
divide by # of cases on which the sum is based Nclass = 10. (8) Add the sums of the X column of the two conversation groups, square and divide by # of cases on which the sum is based Nconv = 10. (9) Add the two & subtract the correction factor.
PROGRAM: (1/N)(DB11] ([SUM1A][DBXCLXCON])T) -[CORRFACT]
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[pic] [pic]
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[pic][pic]
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Or alternatively:
(1/10)([338 30 267 25 405 10 315 45] 1 0 1 0 )2 =
0 0 0 0
0 1 1 0
0 0 0 0
1 0 0 1
0 0 0 0
0 1 0 1
0 0 0 0
= (1/10)[(338 + 405) (267 + 315) (338 + 267) (405 + 315)]2 =
(1/10) ([DB11]([743 582 605 720]T)o2) = (1/10)[DB11] ( 743 743 ) =
582 o 582
605 605
720 720
(1/10)[DB11C] 552,049 (1/10)( 1 1 0 0 552,049 ) = (1/10)[890,773 884,425] =
338,724 0 0 1 1 338,724
366,025 366,025 [89,077.3 88,442.5]
518,400 518,400
STEPS(9 & 12):
Here we have computed two values at the same time which need for the correction factor to be subtracted from each. First we compute the appropriate correction factor, then subtract
[pic]
[pic]
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[pic] [pic]
[pic][pic]
Method 2:
We need to add the first and third numbers, the second and fourth, the first and second and the third and fourth numbers. We do this with DB2:
[pic][pic]
[pic]=
[pic]
We now proceed as we did above.
Or alternatively:
[89,077.3 88,442.5] - [87,871.25 87,871.25] = [1296.05 661.25]
CORRFACT
(1/10) ([SUM1A][DBXCLXCON])T)o2
SUM1B = [DB1]4,20[A1]20,2 = (XG1 338 30 (YG1 ; SS7 (1/10) 552,049 =55,204.9
(XG2 267 25 (YG2 ; SS8 338,724 = 33,872.4
(XG3 405 10 (YG3 ; SS10 366,025 = 36,602.5
(XG4 315 45 (YG4 ; SS11 518,400 = 51,840.0
SUMX = (1/N)([DB11])([SUM1B][ONE0])o2)- [CORRFACT]
338 30 1 = 338 Finish as above.
267 25 0 267
405 10 405
315 45 315
HP-48G PROGRAM:
RCL B11
RCL SUM1B
RCL ONE0
x
(
TRN
SWAP
4
(
DIAG(
x
x
RCL N = 10
/
RCL [CORRFACT]
-
STO SS912 [1296.05 661.25] SS9 = [SS912][ONE0] = 1296.05; SS12 = [SS912][ZERO1] = 661.25
NOTE: I can’t check the program right now, my calculator seems to be on the fritz).
STEP 13-14: (13) Square each of the sums under the X column, divide by # of cases upon which each sum is based and sum (N = 5); (14) Subtract results from steps 5, 9 and 12 from SS13; (15) Subtract SS13 from SS3.
METHOD 1: PROGRAM: (1/NX) [1]1,8([SIM1A]1,8diag[DB2X]8,8)o2 -[SS9]1,1 - [SS12] - [SS3]1,1
(1/NX) [1]1,4([SUM1B]4,2[ONE0]2,1)o2- [SS9] - [SS12] - [CORRFACT]
[pic]
[pic]
[pic]
[pic]
We can also do the computation this way:
[pic]
[pic]
[pic]
Or alternatively:
[DBdiag2]
(1/5) (([338 30 267 25 405 10 315 45] 1 )T)o2 338 338
0 0 0
1 267 267
0 0 o 0 =
1 405 405
0 0 0
1 315 315
0 0 0
(1/5) [1 1 1 1 1 1 1 1] 114,244
0
71,289
0 = [89,756.6] - 87,781.25 - 1296.05 - 661.25 = 18.05
164,025
0
99,225
0
HP-48G PROGRAM:
RCL ONE8
RCL SUM1A
RCL DBdiag2
x
(
TRN
8
(
DIAG(
SWAP
x
x
RCL NX = 5
/ 89,756.6
RCL CORRFACT 87,781.25
-
RCL SS5 1296.05
-
RCL SS12 661.25
-
STO SS14 18.05
METHOD 2:
[pic]
[pic]
HP-48 PROGRAM:
RCL ONE5 (1/5)[1 1 1 1 1] ( 338 30 1 )o2
RCL SUM1B 267 25 0
RCL ONE0 405 10
x 315 45
(
4
(
DIAG(
SWAP
x
RCL NX = 5
/
RCL CORRFACT + SS9 + SS12
-
STO SS14 18.05
STEP 15: Subtract SS13 from SS3:
PROGRAM: [X]T20,1[X]20,1 - (1/NX) [1]1,5([SUM1B]4,2[ONE0]2,1)o2- SS9 - SS12 - CORRFACT =
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[pic]
SS15 = 91,587 - 89,756.6 = 1830.4
THE CRITERION VARIABLE (steps 16 - 28)
NOTE: The math is exactly the same as for the X column above, except we post-multiply by
ZERO1 instead of ONE0.
STEPS 16 - 19: Find the sum of squares for the Y column:
(16 & 17): Square all values in the Y column and sum the values in the Y column:
PROGRAM: [Y]T20,1[Y1]20,1
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a b
=[pic]
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(16) Square the sum of the Y column and divide by 20:
(19) Subtract result of step 18 from step 16:
CORRFACT = (110)(110/20) = 605
CORRFACT = [b][-b/NT]
STEPS 20, 21 & 22;23-25: (20) Add the sums of the Y column of the two classroom groups, square and
divide by # of cases on which the sum is based Nclass = 10. (21) Add the sums of the Y column of the two conversation groups, square and divide by # of cases on which the sum is based Nconv = 10. (22) Add the two & subtract the correction factor. ( Will actually do all steps in one set of operations from 20 - 25)
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Method 2:
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[pic]
We again sum the first and third numbers, the second and fourth, the first and second and the third and fourth numbers. This matrix is smaller because we are dealing with 4 elements rather than 8 in method 1.
[pic]
[pic]
We proceed from here as we did above in method 1.
Or alternatively:
PROGRAM: (1/N)(DB11]([SUM1][DBYCLYCON])T)o2-[CORRFACT]
(1/10)([338 30 267 25 405 10 315 45] 0 0 0 0 )o2 = 0 0 0 0
1 0 1 0 1 0 1 0
0 0 0 0 0 0 0 0
0 1 1 0 DBYCLYCON = 0 1 1 0
0 0 0 0 0 0 0 0
1 0 0 1 1 0 0 1
0 0 0 0 0 0 0 0
0 1 0 1 0 1 0 1
= (1/10)[(30 + 10) (25 + 45) (30 + 25) (10 + 45)]2 =
(1/10) ([DB11]([40 70 55 55]T)o2) = (1/10)[DB11] ( 40 40 ) =
70 o 70
55 55
55 55
(1/10)[DB11] 1600 (1/10)( 1 1 0 0 1600 ) = (1/10)[6500 6050] =
4900 0 0 1 1 4900
3025 3025 SS24 SS25
3025 3025 [650 605]
SS25 = SS23 + SS24 = 605
=(SS24 & SS25 ) [650 605] - [605 605] = [45 0] = SS2225
CORRFACT
METHOD 2:
SUM1B = [DB1]4,20[A1]20,2 = (XG1 338 30 (YG1
(XG2 267 25 (YG2
(XG3 405 10 (YG3
(XG4 315 45 (YG4
SUMY = (1/N)([DB11])([SUM1B][ZERO1])o2)- [CORRFACT]
338 30 0 = 30 Finish as above.
267 25 1 25
405 10 10
315 45 45
HP-48G PROGRAM:
RCL B11
RCL SUM1B
RCL ZERO1
x
(
TRN
SWAP
4
(
DIAG(
x
x
RCL N = 10
/
RCL [CORRFACT]
-
STO SS2225 [45 0] SS22 = [SS2225][ONE0] = 45; SS25 = [SS2225][ZERO1] = 0
STEP 26 - 27: (26) Square each of the sums under the Y column, divide by # of cases upon
which each sum is based and sum (N = 5); (27) Subtract results from steps SS18, SS22 and SS25 from SS26; (28) Subtract SS26 from SS16.
METHOD 1: PROGRAM: (1/NY) [1]1,8([SIM1A]1,8diag[DB2Y]8,8)o2-[SS22]1,1 - [SS25] - [SS18]1,1
(1/NY) [1]1,4([SUM1B]4,2[ZERO1]2,1)o2- SS22 - SS25 – CORRFACT
[pic]
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Or we can compute this in this manner:
[DBdiag2]
(1/5)(([338 30 267 25. 405 10 315 45] 0 )T)o2 0 0
1 30 30
0 0 0
1 25 o 25 =
0 0 0
1 10 10
0 0 0
1 45 45
(1/5) [1 1 1 1 1 1 1 1] 0
900
0
. 625 = [730] - 605 - 45 - 0 = 80
0
100
0
2025
HP-48G PROGRAM:
RCL ONE8
RCL SUM1A
RCL DBdiag2Y
x
(
TRN
8
(..
DIAG(
SWAP
x
x
RCL NY = 5
/ 730
RCL CORRFACT 605
-
RCL SS22 45
-
RCL SS25 0
-
STO SS27 80
METHOD 2: PROGRAM: (1/5)[1]1,4[SUM1B][ZERO1] - CORRFACT - SS22 - SS25
[pic]
[pic]
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Or we can compute this in the following manner:
HP-48 PROGRAM:
RCL ONE5 (1/5)[1 1 1 1 ] ( 338 30 0 )o2
RCL SUM1B 267 25 1
RCL ZERO1 405 10
x 315 45
(
4
(
DIAG(
SWAP
x
RCL NY = 5
/
RCL CORRFACT + SS22 + SS25
-
STO SS27 80
STEP 28: Subtract SS26 from SS16:
PROGRAM: [Y]T20,1[Y]20,1 - (1/NY) [1]1,5([SUM1B]4,2[ZERO1]2,1)o2- SS22 - SS25 - CORRFACT =
[pic]
[pic]
[pic]
[pic]
SS28 = 764 - 730 = 34
PRODUCTS OF THE TWO VARIABLES (steps 29 - 53)
STEP 29: Multiply each of the X scores with its paired Y score and add the products:
PROGRAM: SS29 = [X]T20,1 [Y]20,1
[pic]
[pic]
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STEP 30: Multiply SS4 by SS17 and divide by NXY # of XY pairs = 20).
PROGRAM: SS30 = (1/NXY)([X]T20,1[X1]20,2[ZERO1]2,1)( [Y]T20,1[Y1]20,2[ZERO1]2,1)
[pic]
[pic]
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STEP 31: Subtract SS30 from SS29
PROGRAM: SS31 = XTY - (1/NXY)([X]T20,1[X1]20,2[ZERO1]2,1)( [Y]T20,1[Y1]20,2[ZERO1]2,1)
[pic]
[pic]
[pic]
[pic]
([A1][ONE0])T[A1][ZERO1] -(1/NXY)([A1][ONE0])T[X1][ZERO1])([A1][ZERO1]T[Y1][ZERO1])
(The fully developed equation)
STEP 32: Obtain the X sum and the Y sum for the two classroom groups, multiply X by Y and
divide by the #of scores in each group. NCL = 10.
[pic]
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STEP 33: Obtain the X sum and the Y sum for the two conversation groups multiply X by Y and
divide by the #of scores in each group. Ncon = 10.
PROGRAM: [SS3233] = (1/NCLCON)(10EEX(LOG([DB2]2,4([DB1]4,20[A1]20,2 )))[[1]2,1)
[pic]
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OR ALTERNATIVELY:
DB2
HP-48G PROGRAM:
RCL DB2 1 0 1 0 338 30 = 743 40
RCL SUM1B 0 1 0 1 267 25 582 70
x 405 10
315 45
(
APLY
RCL TWO1 (1/10)(Anti-Log(Log( 743 40 ) 1 )=
x 582 70 1
( (1/10) 10EEX( 2.8710 1.6021 1 )(1/10)(10EEX 4.4730 ) =
APLY 2.7649 1.8451 1 4.6100
RCL N = 10
/ (1/10) 29720 = 2972 = [SS3233]
STO SS3233 40740 4074
STO SS3233 SS32 = [SS 3233][ONE0] = 2972;
SS33 = [SS3233][ZERO1] = 4074
STEP 34: Add SS32 + SS33 - SS30 NCLCON = 10; NXY = 20
PROGRAM:SS34 = (1/NCLCON)[1]1,2(10EEX(LOG([DB2]2,4([DB1]4,20[A1]20,2 )))[[1]2,1) -
(1/NXY)([X]T[X1][ZERO1])( [Y]T[Y1][ZERO1])
[SS34] = [1 1] 2972 - 7287.5 = [-241.5]
4074
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STEP 35 & 36: Obtain the X sum and the Y sum for the two LAB groups, multiply X by Y and
divide by the #of scores in each group. NLAB = 10.
STEP 36: Obtain the X sum and the Y sum for the two non-LAB groups multiply X by Y and
divide by the #of scores in each group. NNL = 10.
PROGRAM: [SS3536] = (1/NLAB/NL)(10EEX(LOG([DB2L/NL]2,4([DB1]4,20[A1]20,2 )))[[1]2,1)
[pic]
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Or solving in another way:
DB1L/NL
HP-48G PROGRAM:
RCL DB1L/NL 1 1 0 0 338 30 = 605 55
RCL SUM1B 0 0 1 1 267 25 720 55
x 405 10
315 45
(
APLY
RCL TWO1 (1/10)(Anti-Log(Log( 605 55 ) 1 )=
x 720 55 1
( (1/10) 10EEX( 2.7818 1.7404 1 )(1/10)(10EEX 4.5221 ) =
APLY 2.8573 1.7404 1 4.5977
RCL NLNL = 10
/ (1/10) 33275 = 3327.5 = [SS3536]
STO SS3536 39600 3960.0
STO SS3536 SS35 = [SS 3536][ONE0] = 3327.5;
SS36 = [SS3536][ZERO1] = 3960
STEP 37: Add SS35 + SS36 - SS30 NCLCON = 10; NXY = 20
[pic]
[pic]
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PROGRAM: SS37 = (1/NL/NL)[1]1,2(10EEX(LOG([DB1LNL]2,4([DB1]4,20[A1]20,2 )))[[1]2,1) -
(1/NXY)([X]T[X1][ZERO1])( [Y]T[Y1][ZERO1])
[SS37] = [1 1] 3327.5 - 7287.5 = [0]
3960.0
STEP 38: Multiply each X sum by its paired Y sum in SUM1B. Divide each product by the number of scores upon which each sum is based (NXYP = 5) and add the quotients.
SS38 = (1/Nxyp)[1]1,4(o(([SUM1B]4,2o[1]1,4)
[pic]
[pic][pic]
[pic]
[pic]
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Or alternatively:
SUM1B = [DB1]4,20[A1]20,2 = (XG1 338 30 (YG1
(XG2 267 25 (YG2
(XG3 405 10 (YG3
(XG4 315 45 (YG4
338 30 1 = 338 30 The connector between the nested arrays is the
267 25 o 1 267 o 25 hollow-dot multiplier, so we just multiply across.
405 10 1 405 10
315 45 1 315 45
= (1/5)[1]1,4 10,140 [1 1 1 1] 2028.0
6675 = 1335.0 = [7008]
4050 810.0
14,175 2835.0
HP-48G PROGRAM: SS38 = ([SUMB]4,2[ONE0]2,1)T([SUMB]4,2[ZERO1]2,1)
RCL SUMB
RCL ONE0 ( 338 30 1 )T( 338 30 0 )
x 267 25 0 267 25 1
TRN 405 10 405 10
RCL SUMB 315 45 315 45
RCL ZERO1
x
RCL NXYP = 5
/
STO SS38
METHOD 2:
NOTE: Here is a way to get the same solution using pure half-multipliers.
[pic]
[pic][pic]
Where Σ^O denotes we half multiply the results computed from the sub-matricies obtained from SUM1A o FOUR1. Can't think of a better way to denote the double half-multiplication yet.
[pic] o[pic]=[pic] o[pic]=[pic]
[pic]
STEP 39: Subtract SS30, SS34 and SS37 from SS38
SS39 = 7008 - 7287.5 - (-241.5) - 0 = -38
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NOTE: Both methods work, the first one I showed uses the properties of nested arrays, and is longer and more complex than the second method, which utilizes our regular mathematics.
The computations are over, we now have just a complex subtraction problem, starting with our previously calculated values. I’m not going to shrink any problem’s print size to less than 6 point(in section 4.9 I shrunk some down to four point size and could barely read the math) just to get everything on a single line. To help, I’m going to consolidate all the SS values obtained up to this point along with their programs.
(X’s (Y’s
SUM1B = [DB1]4,20[A1]20,2 = (XG1 338 30 (YG1
(XG2 267 25 (YG2
(XG3 405 10 (YG3
(XG4 315 45 (YG4
SUM1A = [1]1,5[A]5,8 = [338 30 267 25 405 10 315 45]
(X’s (Y’s
SUM1B = [DB1]4,20[A1]20,2 = (XG1 338 30 (YG1
(XG2 267 25 (YG2
(XG3 405 10 (YG3
(XG4 315 45 (YG4
STEP 40: Subtract SS38 from SS29:
SS40 = SS29-SS38
SS40 = XTY - ([SUMB]4,2[ONE0]2,1)T([SUMB]4,2[ZERO1]2,1)
SS40 = 7175 - 7008 = 167
STEP 41: SS28 - Square SS40 and divide by SS15:
[pic]
SS28 - (SS29 - SS38)2/SS29
(SS16 -SS26) - (SS29 - SS38)2/SS29
(764 - 730) - (7175 - 7008)2/7175 = 34 - 15.237 = 18.763
STEP 42: Add SS9 + SS15:
[pic]
SS42 = SS9 + SS15 = SS9 + (SS3 - SS13) = (1296.05 + 1830.4) = 3126.45
STEP 43: Add SS22 + SS28
[pic]
SS43 = (SS22 + SS28) = (SS22 + (SS16 - SS26) = (45 + (764 - 730)) = 45 + 34 = 79
STEP 44: Add SS40 + SS34:
[pic]
SS44 = (SS29 - SS38) + SS32 + SS33 - SS30 = -241.5 + 167 = -74.5
STEP 45: Square SS44 and divide by SS42, then subtract quotient and SS41 from SS43
[pic]
= SS43 - SS442/SS42 - SS41
= (SS22 + (SS16 - SS26) - ((SS29 - SS38) + SS32 + SS33 - SS30)2/ SS9 + (SS3 - SS13) -
(SS16 -SS28) - (SS29 - SS38)2/SS29 =
(SS22 + SS28) - (SS40 + SS34)2/(SS9 + SS15) - SS28 - (SS29 - SS38)2/SS29
SS45 = (-74.5)2/3126.5 = 5550.25/3126.5 = 1.775
STEP 46: Add SS12 + SS15:
[pic]
(SS11 + SS10 - SS5) + (SS3 - SS13) =
SS46 = (51,840 + 36,602.5 - 87,781.25) + (91,587 - 89,756.6) = (661.25 + 1830.4) = 2491.65
STEP 47: Add SS28 + SS25:
[pic]
SS47 = (SS16 - SS26 ) + SS23 + SS24 - SS18 = (764 - 730) + (302.5 + 302.5 - 605) = 0 + 34 = 34
STEP 48: Add SS37 + SS40:
[pic]
SS48 =(SS35 + SS36 - SS30) + (SS38 - SS29) = 0 + 167 = 167
STEP 49: Square SS48 and divide by SS46, then subtract the quotient and SS41 from SS46:
[pic]
SS49 =(SS16 - SS26 ) + SS23 + SS24 - SS18)- ((SS35 + SS36 - SS30) + (SS38 - SS29))2/((SS11 + SS10 - SS5) + (SS3 - SS13)) - ((SS16 -SS26) - (SS29 - SS38)2/SS29) = 34 - 11.193 - 18.763 = 4.044
STEP 50: Add SS14 + SS15:
[pic]
SS50= (SS13 - SS5 - SS9 - SS12) +(SS3 - SS13) = (91,587 - 89,756.6) = 18.05 + 1830.4 = 1848.450
STEP 51: Add SS27 + SS28
[pic]
SS51 = (SS26 - SS18 - SS22 - SS25) + (SS16 - SS26) = (730-605-45-0)+(764-730) = 80 + 34 = 114
STEP 52: Add SS39 + SS40
[pic]
= (SS38 - SS30 - SS34 - SS37) + (SS29 - SS38) =
SS52 = = -38 + 167 = 129
STEP 53: Square SS52 and divide by SS50, then add SS41 to the quotient and subtract the sum from SS51:
[pic]
((SS26 - SS18 - SS22 - SS25) + (SS16 - SS26) ) -
((SS38 - SS30 - SS34 - SS37) + (SS29 - SS38))2/ ((SS13 - SS5 - SS9 - SS12) +(SS3 - SS13)) -
((SS16 -SS26) - (SS29 - SS38)2/SS29 ) =
(SS14 + SS15) - (SS39 + SS40)2 /(SS15 + SS15) - SS28 - (SS29 - SS38)2/SS29
(80 + 34)- (-38 + 167)2/(91,587 - 89,756.6) - (764 - 730) - (7175 - 7008)2/7175 =
= 114 - 1292/1848.450 - 18.763 =
SS53 = 114 - 9.003 - 18.763 = 86.234
MEAN SQUARES AND F RATIOS (steps 54-60)
STEP 54: Divide SS41 (which is the adjusted error sum of squares) by the degrees of freedom (df) for
the adjusted error. Df for the adjusted error = N - CL -1 where N = total # of XY pairs (20 in this example), C is the # of experimental conditions on the first variable (2 in this example, classroom Vs conversation, and L = # of experimental conditions on the other variable (2 in this example, lab Vs no lab). So N-CL-1 =20 - (2x2)-1=15.
SSEMS = SS41/(N-CL-1) = 18.763/15 = 1.251
STEP 55: Divide SS45 (adjusted sum of squares for teaching methods) by df =C-1 where C = the
# of experimental conditions in this variable (2 in this example, classroom Vs conversation.
SSECOND = SS45/(C-1) = 58.462/(2-1) = 58.462
STEP 56: Divide SSECOND/SSEMS, the df = (C-1)/(N-CL-1) = 1/15
F = SSECOND/SSEMS = 58.462/1.251 = 46.73
From Appendix E, with a df = 1/15, an F ratio larger than 16.59 would be expected by chance alone one in a thousand times, therefore, F is significant at the .001 level. It is concluded that the different teaching methods were significantly different in their effectiveness.
STEP 57: Divide SS49 (the adjusted sum of squares for language-lab usage) by df = L-1 where L = the # of experimental conditions in this variable (2 for this example: lab Vs no lab).
msLL = SS49/(L-1) = 4.044/1 = 4.044
STEP 58: Divide SS57/SS54 to get the F for language/lab usage: df = (L-1)/(N-CL-1)=1/15
F = 4.044/1.251 = 3.23.
The relationship is non-significant (in 10% range). It is concluded there is no difference in language effectiveness depending upon whether one spends 3 hours a day in lab or not.
STEP 59: Divide SS53 (the adjusted sum of squares for the methods by lab interaction by
df = (C-1)(L-1)
msMxL = SS53/(C-1)(L-1) = 86.234/1 = 86.234
STEP 60: Divide SS59/msMxL to yield F for methods by lab interaction:
F = 86.234/1.251 = 68.93 with a df = (C-1)(L-1)/(N-CL-1) = 1/15
Appendix E shows this F is significant at the .001 level. Since the teaching methods affect (interact with) the lab experience, we need to seek out the source of the interaction. The procedures used are presented in section 3.10.
Just to show how complex these calculations are, I’ve translated the SS values back into their equations.
SS3 = [X]T[X1][ONE0] 91,587
SS4 =[X]T[X1][ZERO1] 1325
SS5 = CORRFACT = [b][-b/NT] = 87,781.25
SS6 = [X]T[X1][1 -b/NT]
PROGRAM: (1/N)(DB11]([SUM1A][DBXCLXCON])T)-[CORRFACT]
SS7 =(1/10) ([SUM1A][DBXCLXCON])T)o2 [1 0 0 0]T = 55,204.9
SS8 = (1/10) ([SUM1A][DBXCLXCON])T)o2 [0 1 0 0]T = 33,872.4
SS9 = (1/N)(DB11]([SUM1A][DBXCLXCON])T)[ONE0]-[CORRFACT] = 1296.05
SS10 = (1/10) ([SUM1A][DBXCLXCON])T)o2 [0 0 1 0]T = 36,602.5
SS11 = (1/10) ([SUM1A][DBXCLXCON])T)o2 [0 0 0 1]T = 51,840
SS12 = (1/N)(DB11]([SUM1A][DBXCLXCON])T)[ZERO1] - [CORRFACT] = 661.250
(SS11 + SS10 - SS5)
SS13 = (1/NX) [1]1,4([SUM1B]4,2[ONE0]2,1)o2 = 89,756.6
SS14 = ((1/NX) [1]1,4([SUM1B]4,2[ONE0]2,1)o2 )- ([b][-b/NT] - (1/N)(DB11]([SUM1A][DBXCLXCON])T)[ONE0]-
[CORRFACT] - (1/N)(DB11]([SUM1A][DBXCLXCON])T)[ZERO1] - [CORRFACT])
= (SS13 - SS5 - SS9 - SS12 = 18.05
SS15 = [X]T[X] - (1/NX) [1]1,5([SUM1B]4,2[ONE0]2,1)o2- SS9 - SS12 - CORRFACT
= (SS3 - SS13) = (91,587 - 89,756.6) = 1830.4
SS2225 = (1/N)(DB11]([SUM1A][DBYCLYCON])T)
SS22 = [SS2225][ONE0] = 45
(1/N)(DB11]([SUM1A][DBYCLYCON])T)[ONE0] - CORRFACT
SS25 = SS23 + SS24 - SS18 = 302.5 + 302.5 - 605 = 0
SS25 = [SS2225][ZERO1] = 0
(1/N)(DB11]([SUM1A][DBYCLYCON])T)[ZERO1] - CORRFACT
SS28 = [Y]T[Y] - (1/NY) [1]1,5([SUM1B]4,2[ZERO1]2,1)o2- SS22 - SS25 - CORRFACT
= SS16 - SS26 = 34
SS29 = XTY = 7175
SS30 =(1/NXY)([X]T[X1][ZERO1])( [Y]T[Y1][ZERO1]) = 7287.5
SS3233 =(1/NCLCON)[1]1,2(10EEX(LOG([DB2]2,4([DB1]4,20[A1]20,2 )))[[1]2,1) = [2972 4074]
SS34 = (1/NCLCON)[1]1,2(10EEX(LOG([DB2]2,4([DB1]4,20[A1]20,2 )))[[1]2,1) -
(1/NXY)([X]T[X1][ZERO1])( [Y]T[Y1][ZERO1])
SS32 + SS33 - SS30 = -241.5
SS37 =(1/NL/NL)[1]1,2(10EEX(LOG([DB1LNL]2,4([DB1]4,20[A1]20,2 )))[[1]2,1) -
(1/NXY)([X]T[X1][ZERO1])( [Y]T[Y1][ZERO1])
SS35 + SS36 - SS30= 0
SS38 =([SUMB]4,2[ONE0]2,1)T([SUMB]4,2[ZERO1]2,1) = 7008
SS39 = ([SUMB]4,2[ONE0]2,1)T([SUMB]4,2[ZERO1]2,1) (1/NXY)([X]T[X1][ZERO1])( [Y]T[Y1][ZERO1])-
( (1/NCLCON)[1]1,2(10EEX(LOG([DB2]2,4([DB1]4,20[A1]20,2 )))[[1]2,1) -
(1/NXY)([X]T[X1][ZERO1])( [Y]T[Y1][ZERO1]))-
((1/NL/NL)[1]1,2(10EEX(LOG([DB1LNL]2,4([DB1]4,20[A1]20,2 )))[[1]2,1) -
(1/NXY)([X]T[X1][ZERO1])( [Y]T[Y1][ZERO1]))
= SS38 - SS30 - SS34 - SS37 =-38
SS40 = XTY - ([SUMB]4,2[ONE0]2,1)T([SUMB]4,2[ZERO1]2,1)
= (SS29 - SS38) =
SS40 = 7175 - 7008 = 167
SS41 = ([Y]T[Y] - (1/NY) [1]1,5([SUM1B]4,2[ZERO1]2,1)o2- SS22 - SS25 - CORRFACT) - (XTY - ([SUMB]4,2[ONE0]2,1)T([SUMB]4,2[ZERO1]2,1) )2/([X]T[X] - (1/NX) [1]1,5([SUM1B]4,2[ONE0]2,1)o2-
SS9 - SS12 - CORRFACT)
SS28 - (SS29 - SS38)2/SS29
(SS16 -SS28) - (SS29 - SS38)2/SS29
(764 - 730) - (7175 - 7008)2/7175 = 34 - 15.237 = 18.763
SS42 =(1/N)(DB11]([SUM1A][DBXCLXCON])T)[ONE0]-[CORRFACT] + [X]T[X] -
(1/NX)[1]1,5([SUM1B]4,2[ONE0]2,1)o2- SS9 - SS12 - CORRFACT
SS42 = SS9 + SS15 = SS9 + (SS3 - SS13)
SS42 = 1296.05 + 1830.4 = 3126.45
SS43 = (SS22 + SS28) = (SS22 + (SS16 - SS26) = (45 + (764 - 730)) = 45 + 34 = 79
((1/N)(DB11]([SUM1A][DBYCLYCON])T)[ONE0] - CORRFACT) +
([Y]T[Y] -(1/NY) [1]1,5([SUM1B]4,2[ZERO1]2,1)o2- SS22 - SS25 - CORRFACT)
SS44 = (SS29 - SS38) + SS32 + SS33 - SS30 = -241.5 + 167 = -74.5
= (XTY - ([SUMB]4,2[ONE0]2,1)T([SUMB]4,2[ZERO1]2,1) ) +
((1/NCLCON)[1]1,2(10EEX(LOG([DB2]2,4([DB1]4,20[A1]20,2 )))[[1]2,1) -
(1/NXY)([X]T[X1][ZERO1])( [Y]T[Y1][ZERO1]))
SS45 = (((1/N)(DB11]([SUM1A][DBYCLYCON])T)[ONE0] - CORRFACT) +
([Y]T[Y] -(1/NY) [1]1,5([SUM1B]4,2[ZERO1]2,1)o2- SS22 - SS25 - CORRFACT)) -
((XTY - ([SUMB]4,2[ONE0]2,1)T([SUMB]4,2[ZERO1]2,1) ) +
((1/NCLCON)[1]1,2(10EEX(LOG([DB2]2,4([DB1]4,20[A1]20,2 )))[[1]2,1) -
(1/NXY)([X]T[X1][ZERO1])( [Y]T[Y1][ZERO1]))2)/
((1/N)(DB11]([SUM1A][DBXCLXCON])T)[ONE0]-[CORRFACT]) +
( [X]T[X] -(1/NX)[1]1,5([SUM1B]4,2[ONE0]2,1)o2- SS9 - SS12 - CORRFACT) -
([Y]T[Y] - (1/NY) [1]1,5([SUM1B]4,2[ZERO1]2,1)o2- SS22 - SS25 - CORRFACT) -
(XTY - ([SUMB]4,2[ONE0]2,1)T([SUMB]4,2[ZERO1]2,1) )2/([X]T[X] - (1/NX) [1]1,5([SUM1B]4,2[ONE0]2,1)o2-
SS9 - SS12 - CORRFACT)
= (SS22 + (SS16 - SS26) - ((SS29 - SS38) + SS32 + SS33 - SS30)2/ SS9 + (SS3 - SS13) -
(SS16 -SS28) - (SS29 - SS38)2/SS29 =
(SS22 + SS28) - (SS40 + SS34)2/(SS9 + SS15) - SS28 - (SS29 - SS38)2/SS29
SS45 = (-74.5)2/3126.5 = 5550.25/3126.5 = 1.775
SS46 = ((1/N)(DB11]([SUM1A][DBXCLXCON])T)[ZERO1] - [CORRFACT]) +
([X]T[X] - (1/NX)[1]1,5([SUM1B]4,2[ONE0]2,1)o2- SS9 - SS12 - CORRFACT)
(SS11 + SS10 - SS5) + (SS3 - SS13) =
SS46 = (51,840 + 36,602.5 - 87,781.25) + (91,587 - 89,756.6) = (661.25 + 1830.4) = 2491.65
SS47 = ([Y]T[Y] - (1/NY) [1]1,5([SUM1B]4,2[ZERO1]2,1)o2- SS22 - SS25 - CORRFACT) +
((1/N)(DB11]([SUM1A][DBYCLYCON])T)[ZERO1] - CORRFACT)
SS47 = (SS16 - SS26 ) + SS23 + SS24 - SS18 = (764 - 730) + (302.5 + 302.5 - 605) = 0 + 34 = 34
SS48 =
=((1/NL/NL)[1]1,2(10EEX(LOG([DB1LNL]2,4([DB1]4,20[A1]20,2 )))[[1]2,1)-(1/NXY)([X]T[X1][ZERO1])([Y]T[Y1][ZERO1]))+
(XTY - ([SUMB]4,2[ONE0]2,1)T([SUMB]4,2[ZERO1]2,1))
SS48 =(SS35 + SS36 - SS30) + (SS38 - SS29) = 0 + 167 = 167
SS49 = ((1/N)(DB11]([SUM1A][DBXCLXCON])T)[ZERO1] - [CORRFACT]) +
([X]T[X] - (1/NX) [1]1,5([SUM1B]4,2[ONE0]2,1)o2- SS9 - SS12 - CORRFACT)-
((1/NL/NL)[1]1,2(10EEX(LOG([DB1LNL]2,4([DB1]4,20[A1]20,2 )))[[1]2,1)-(1/NXY)([X]T[X1][ZERO1])([Y]T[Y1][ZERO1]))+
(XTY - ([SUMB]4,2[ONE0]2,1)T([SUMB]4,2[ZERO1]2,1))2/((1/N)(DB11]([SUM1A][DBXCLXCON])T)[ZERO1] - [CORRFACT]) + ([X]T[X] - (1/NX) [1]1,5([SUM1B]4,2[ONE0]2,1)o2-SS9 - SS12 - CORRFACT)-
([Y]T[Y] - (1/NY) [1]1,5([SUM1B]4,2[ZERO1]2,1)o2- SS22 - SS25 - CORRFACT) - (XTY - ([SUMB]4,2[ONE0]2,1)T([SUMB]4,2[ZERO1]2,1) )2/([X]T[X] - (1/NX) [1]1,5([SUM1B]4,2[ONE0]2,1)o2-
SS9 - SS12 - CORRFACT)
SS49 =(SS16 - SS26 ) + SS23 + SS24 - SS18)- ((SS35 + SS36 - SS30) + (SS38 - SS29))2/((SS11 + SS10 - SS5) + (SS3 - SS13)) - ((SS16 -SS26) - (SS29 - SS38)2/SS29) = 34 - 11.193 - 18.763 = 4.044
SS50 = ((1/NX) [1]1,4([SUM1B]4,2[ONE0]2,1)o2 )- ([b][-b/NT] - (1/N)(DB11]([SUM1A][DBXCLXCON])T)[ONE0]-
[CORRFACT] - (1/N)(DB11]([SUM1A][DBXCLXCON])T)[ZERO1] - [CORRFACT])
([X]T[X] - (1/NX) [1]1,5([SUM1B]4,2[ONE0]2,1)o2- SS9 - SS12 - CORRFACT)
SS50= (SS13 - SS5 - SS9 - SS12) +(SS3 - SS13) = (91,587 - 89,756.6) = 18.05 + 1830.4 = 1848.450
SS51 = SS27 + SS28 =
((1/5)[1]1,5[SUM1B][ZERO1] - CORRFACT - SS22 - SS25) + ( [Y]T[Y] - (1/NY) [1]1,5([SUM1B]4,2[ZERO1]2,1)o2- SS22 - SS25 - CORRFACT )
SS51 = (SS26 - SS18 - SS22 - SS25) + (SS16 - SS26) = (730-605-45-0)+(764-730) = 80 + 34 = 11
SUPPLEMENT
In order to determine whether the individual group regressions are homogeneous with the overall regression, first compute the sum of squares and the sum of the cross products (SP) for each group of data.
STEP 1: Using the data from the original analysis, compute SSx (sum of squares) for each group. The formula for each computation is:
(X2 - 1/NG((X)2
I will do this supplement with the MathCad +6 program
MATHCAD +6 PROGRAM FOR SUPPLEMENT
62 5 46 5 84 2 58 9
A = 75 7 53 4 91 3 72 10
41 3 57 3 68 1 61 8 ONE20=[1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
88 8 49 7 77 1 65 8
72 7 62 6 85 3 59 10 ONE8 = [1 1 1 1 1 1 1 1]
ONE5 = [1 1 1 1 1]
1 0 1 0 0 0
0 1 1 0 0 0
1 0 1 1 0 0 0
ONEXA = 0 ONEYA = 1 FOUR1 = 1 ONE0 = 1 ZERO1 = 0 1 0 0 0
1 0 1 0 1 1 0 0 0
0 1 1 0 1 0 0
1 0 0 1 0 0
0 1 0 1 0 0
1 0 0 0 0 1 0 0
1 0 0 0 0 1 0 0
I8 = identity(8) N = 20 0 1 0 0 DB11 = 0 0 1 0
ONEX = diag(ONEXA) a = 4 DB1 = 0 1 0 0 0 0 1 0
ONEY = diag(ONEYA) 0 0 1 0 0 0 1 0
0 0 1 0 0 0 1 0
0 0 0 1 0 0 1 0
0 0 0 1 0 0 0 1
0 0 0 1
0 0 0 1
0 0 0 1
0 0 0 1
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TO GET RID OF THE ZERO'S:
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PART 2, STEP 1: COMPUTE THE SUM OF THE CROSS PRODUCTS:
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STEP 2: COMPUTE: CROSSPRODUCT^2/SSX
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STEP 3 & 4: ADD THE RESULT OF STEP 2 AND SUBTRACT FROM THE ERROR SUM OF SQUARES FOR Y (SS28). THIS SS REPRESENTS WITHIN GROUPS REGRESSIONS.
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STEP 5: GET THE ADJUSTED ERROR SUM OF SQUARES FOR Y (SS41) - SSREGRESSIONS
THIS SS REPRESENTS BETWEEN GROUPS REGRESSION.
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STEP 6: DIVIDE THE RESULT OF STEP 4 BY ITS DEGREE OF FREEDOM:
# of paired groups
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teaching;conversation,lab,no lab = 4
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NOTE: VALUE OF THIS DIVISION IN BOOK ON PAGE 208 IS INCORRECT, THEY CALCULATE 12.4.
STEP 7: DIVIDE SSbGROUPREGRESSION BY ITS DEGREES OF FREEDOM:
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STEP 8: DETERMINE THE F RATIO BY DIVIDING SS7A BY SS6:
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SINCE THIS F IS SMALL, WE ACCEPT THE HYPOTHESIS THAT THE REGRESSIONS ARE HOMOGENEOUS.
METHOD 2: LETS LOOK AT THE FIRST PART OF THIS SOLUTION USING THE NESTED ARRAY B:
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SUM OF THE CROSS PRODUCT:
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ALL OTHER COMPUTATIONS ARE THE SAME AS ABOVE STARTING WITH THE COMPUTATION OF STEP 2: SP^2/SSx. NOTE: IN METHOD 2, WE DO NOT HAVE TO TAKE THE LOG OF A MATRIX, WE HAVE ONLY TO USE THE VECTORIZE OPERATION IN MATHCAD.
I TRIED TO COMPUTE THE COVARIANCE ON MATHCAD, BUT MATHCAD TAKES THE EQUATIONS ON A SINGLE LINE AND THEY KEPT RUNNING OVER THE PAGE. IT WAS JUST TOO MUCH OF A MESS TO TRY TO DO.
PROGRAMS FOR EXAMPLE 4.10
STEP 2:
SUM1A = [DB1]4,20[A1]20,2
SUM1 = [1]1,5[A]5,8
STEPS 3-6:
PROGRAM: [X]T20,1[X1]20,1
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CORRFACT = [b][-b/NT]
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CORRECTION FACTOR, 3 METHODS:
Method 1:
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Method 2:
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Method 3:
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STEPS 7 & 8 :
PROGRAM: (1/N)(DB11] ([SUM1A][DBXCLXCON])T) -[CORRFACT]
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STEPS 9 – 12:
SUMX = (1/N)([DB11])([SUM1B][ONE0])o2)- [CORRFACT]
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Step 13-14:
METHOD 1:
PROGRAM: SS14 = (1/NX) [1]1,8([SIM1A]1,8diag[DB2X]8,8)o2 -[SS9]1,1 - [SS12] - [SS3]1,1
(1/NX) [1]1,4([SUM1B]4,2[ONE0]2,1)o2- [SS9] - [SS12] - [CORRFACT]
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Method 2:
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STEP 15:
PROGRAM: [X]T20,1[X]20,1 - (1/NX) [1]1,5([SUM1B]4,2[ONE0]2,1)o2- SS9 - SS12 - CORRFACT
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STEP 16-19:
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STEPS 20, 21 & 22;23-25:
PROGRAM: (1/N)(DB11]([SUM1A][DBYCLYCON])T)o2-[CORRFACT]
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METHOD 2:
SUMY = (1/N)([DB11])([SUM1B][ZERO1])o2)- [CORRFACT]
STEP 26 - 27:
METHOD 1: PROGRAM: (1/NY) [1]1,8([SIM1A]1,8diag[DB2Y]8,8)o2-[SS22]1,1 - [SS25] - [SS18]1,1
(1/NY) [1]1,4([SUM1B]4,2[ZERO1]2,1)o2- SS22 - SS25 - CORRFACT
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METHOD 2:
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PROGRAM: (1/5)[1]1,5[SUM1B][ZERO1] - CORRFACT - SS22 - SS25
STEP 28:
PROGRAM: [Y]T20,1[Y]20,1 - (1/NY) [1]1,5([SUM1B]4,2[ZERO1]2,1)o2- SS22 - SS25 – CORRFACT
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STEP 29:
PROGRAM: SS29 = [X]T20,1 [Y]20,1
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STEP 30:
PROGRAM: SS30 = (1/NXY)([X]T20,1[X1]20,2[ZERO1]2,1)( [Y]T20,1[Y1]20,2[ZERO1]2,1)
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STEP 31:
PROGRAM: SS31 = XTY - (1/NXY)([X]T20,1[X1]20,2[ZERO1]2,1)( [Y]T20,1[Y1]20,2[ZERO1]2,1)
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STEP 32:
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STEP 33:
PROGRAM: [SS3233] = (1/NCLCON)(10EEX(LOG([DB2]2,4([DB1]4,20[A1]20,2 )))[[1]2,1)
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STEP 34:
PROGRAM:SS34 = (1/NCLCON)[1]1,2(10EEX(LOG([DB2]2,4([DB1]4,20[A1]20,2 )))[[1]2,1) -
(1/NXY)([X]T[X1][ZERO1])( [Y]T[Y1][ZERO1])
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STEP 35 & 36:
PROGRAM: [SS3536] = (1/NLAB/NL)(10EEX(LOG([DB2L/NL]2,4([DB1]4,20[A1]20,2 )))[[1]2,1)
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STEP 37:
PROGRAM: SS37 = (1/NL/NL)[1]1,2(10EEX(LOG([DB1LNL]2,4([DB1]4,20[A1]20,2 )))[[1]2,1) -
(1/NXY)([X]T[X1][ZERO1])( [Y]T[Y1][ZERO1])
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STEP 38:
SS38 = (1/Nxyp)[1]1,4(o(([SUM1B]4,2o[1]1,4)
SS38 = ([SUMB]4,2[ONE0]2,1)T([SUMB]4,2[ZERO1]2,1)
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METHOD 2:
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SS39:
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SS40:
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SS41:
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SS42:
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SS43:
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SS44:
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SS45:
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SS46:
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SS47:
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SS48:
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SS49:
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SS50:
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SS51:
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SS52:
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SS53:
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SS54:
SSEMS = SS41/(N-CL-1)
STEP 55:
SSECOND = SS45/(C-1)
STEP 56:
F = SSECOND/SSEMS
STEP 57:
msLL = SS49/(L-1)
STEP 58:
df = (L-1)/(N-CL-1)
STEP 59:
msMxL = SS53/(C-1)(L-1)
STEP 60:
F = SS59/msMxL
RELIABILITY OF MEASUREMENT: THE TEST AS A WHOLE
(TEST-RETEST, PARALLEL FORMS AND SPLIT HALVES)
COMPUTATIONAL HANDBOOK OF STATISTICS, 2ND EDITION SECT. 4.11 , PG. 209 - 210.
These three tests refer back to the Pearson product-moment correlation (Section 4.1). There is no math, just three different ways to use and interpret the different correlation’s.
Test-retest: This measure of reliability is computed when you give the same test to the same people at different times. Label the scores of the first test as X and the scores of the second test as Y. Do the Pearson correlation and the computed value will be the test-retest reliability coefficient.
Parallel-forms: This measure of reliability is when the same people take the same tests that are not the same, but are composed of equated items. Label the scores of the first test as X and the scores of the second test as Y. Do the Pearson correlation and the computed value will be the parallel forms reliability coefficient.
Split-halves reliability: This measure is computed when you have given a test to several people and then formed pairs of scores by dividing the test items into two equal groups. The usual method is to put even numbered questions in one group (X) and odd numbered questions into the other (Y). Compute the correlation and carry out the following steps to compute the split-halves reliability coefficient:
Compute the Pearson product-moment correlation.
We divided the test in half. Therefore the real test is twice as long as either half. When a test is increased in length, it’s reliability increases, so we must correct the value in step 1 so it pertains to the whole test rather than to just half of it.
Multiply the obtained correlation by 2.
Add 1 to the above correction.
Divide the result of step 1 by step 2. i.e. 2(correlation)/1+2(correlation).
The above ratio is the corrected split-halves reliability coefficient that applies to the whole test. A high reliability value (.70 or higher) shows that the test is accurately measuring the characteristic it was designed to measure.
RELIABILITY OF MEASUREMENT: THE INDIVIDUAL ITEMS
(KUDER-RICHARDSON AND HOYT)
COMPUTATIONAL HANDBOOK OF STATISTICS, 2ND EDITION SECT. 4-12, PG. 211 - 213.
We wish to determine whether the items in a test are homogeneous in terms of how the subjects responded to the questions:
STATISTICAL EQUATION
rtt = (Ve - Vr)/Ve
EXAMPLE: Suppose an experimenter wishes to test the reliability of a ten item test given to eight subjects. For purposes of determining reliability, the experimenter records for each subject on each test item whether the question was answered correctly (indicated by 1) or incorrectly (indicated by 0).
STEP 1: Table the data:
TEST ITEMS
SUB. 1 2 3 4 5 6 7 8 9 10
S1 1 1 1 1 1 1 1 1 1 1
S2 0 0 0 1 1 0 1 1 1 1
S3 0 0 1 1 1 0 0 0 0 0
S4 0 0 1 1 1 1 1 0 1 1
S5 0 1 1 1 1 0 1 0 1 1
S6 1 1 0 1 0 0 1 1 1 0
S7 0 0 1 1 0 0 1 0 0 1
S8 1 1 0 1 1 1 1 0 1 1
Make a matrix out of it.
1 1 1 1 1 1 1 1 1 1
0 0 0 1 1 0 1 1 1 1
0 0 1 1 1 0 0 0 0 0
[A]8,10 = 0 0 1 1 1 1 1 0 1 1
0 1 1 1 1 0 1 0 1 1
1 1 0 1 0 0 1 1 1 0
0 0 1 1 0 0 1 0 0 1
1 1 0 1 1 1 1 0 1 1
VARIABLES:
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STEP 2: Count the number of items each subject answered correctly:
PROGRAM: SUM2 = [A]8,10[1]10,1
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HP-48G PROGRAM:
RCLA
RCL TEN1
x
STO SUM2
STEP 3: Sum the number of correct scores in step 2.
PROGRAM: SS3 = [1]1,8([A]8,10[1]10,1)
HP-48G PROGRAM:
RCL ONE8
RCL SUM2
x
STO SS3
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STEP 4: Square each number of correct answers, sum and divide by the number of questions in the test. NQ = 10.
PROGRAM: SS4 = (1/NQ)( [A]8,10[1]10,1 )2 Where C2 = CTC OR (1/10)([SUM2]T[SUM2]).
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HP-48G PROGRAM:
RCL SUM2
(
TRN
SWAP
x
RCL NQ = 10
/
STO SS4
STEP 5: Square the result of step 3 and divide by the product of the number of subjects by the
number of items (or divide by number of elements in datastream matrix) NT = 80.
(1/80)(51)2 = 2601/80 = 32.512 STO SS5
STEP 6: Subtract SS3-SS5:
SS6 = SS3-SS5:
SS6 = 51 - 32.512 = 18.488
STEP 7: SUBTRACT SS4 - SS5
SS7 = SS4 - SS5
SS7 = 35.9 - 32.512 = 3.388
STEP 8: Count the number of subjects who correctly answered each question:
PROGRAM: SUM1 = [1]1,8[A]8,10
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HP-48G PROGRAM:
RCL ONE8
RCL A
x
STO SUM1
STEP 9: Square each element in SUM1 and divide by NS = 8 (# of people who took the test)
PROGRAM: SS9 = (1/NS)([SUM1][SUM1]T) =
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RCL SUM1
(
TRN
x
RCL NS = 8
/
STO SS9
STEP 10: Subtract SS9-SS5:
SS10 = SS9 – SS5
SS10 = 36.125 - 32.512 = 3.613
STEP 11: Subtract SS6-SS7-SS10
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SS11 = 18.488 - 3.388 - 3.613 = 11.487
STEP 12: Divide SS7 by N-1 where N = # of subjects who took the test; NS = 8.
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SS12 = 3.388/7 = .484
STEP 13: Divide SS11 by (N-1)(I-1) where N = 8 and I = # of questions = 10.
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SS13 = SS11/(N-1)(I-1) = 11.487/(7x9) = .182
STEP 14: Subtract SS12-SS13:
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SS14 = .484 - .182 = .302
STEP 15: Divide SS14/SS12:
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SS15 =.302/.484 = .62
A high reliability coefficient (.70 or higher) means the test accurately measured some characteristic of the people taking it. Further, it means that the individual items on the test were producing similar patterns of response in different people. A high value means that the test items were homogeneous and, therefore, valid.
SS14 = (SS12-SS13) = SS7/(N-1) - SS11/(N-1)(I-1)
SS12 = SS11/(N-1)(I-1) = (SS6-SS7-SS10)/(N-1)(I-1)
=(SS4-SS5))/ (N-1)
SS7 = SS4-SS5
SS14 = (SS4-SS5)/(N-1) - ((SS3-SS5) - (SS4-SS5) - (SS9-SS5))/ (N-1)(I-1)
SS12 = (SS4-SS5))/ (N-1)
SS14/SS12 =((SS4-SS5)/(N-1) - ((SS3-SS5) - (SS4-SS5) - (SS9-SS5))/ (N-1)(I-1))/ ((SS4-SS5)/ (N-1))
MATHCAD +6 PROGRAM
1 ONE8 = [1 1 1 1 1 1 1 1]
1
1 1 1 1 1 1 1 1 1 1 1 NT = 80
0 0 0 1 1 0 1 1 1 1 1
0 0 1 1 1 0 0 0 0 0 1 NS = 8
[A]8,10 = 0 0 1 1 1 1 1 0 1 1 TEN1 = 1
0 1 1 1 1 0 1 0 1 1 1 NQ = 10
1 1 0 1 0 0 1 1 1 0 1
0 0 1 1 0 0 1 0 0 1 1 I = 10
1 1 0 1 1 1 1 0 1 1 1
STEP 2 & 3: FIND SUM2 AND THE GRAND SUM OF MATRIX A:
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STEP 4: SQUARE SUM2 AND DIVIDE BY # OF QUESTIONS ON TEST:
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STEP 5: SQUARE SS3 AND DIVIDE BY # OF ELEMENTS IN A: NT=80.
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STEPS 8 & 9: FIND SUM1, SQUARE AND DIVIDE BY # OF SUBJECTS
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FINAL STEP: WE DO NOT NEED ANY OTHER VALUES, WE CAN COMPUTE THE SOLUTION FROM HERE:
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PROGRAMS FOR EXAMPLES 4.12-4.14
STEP 2:
SUM2 = [A]8,10[1]10,1
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STEP 3:
PROGRAM: SS3 = [1]1,8([A]8,10[1]10,1)
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STEP 4:
PROGRAM: SS4 = (1/NQ)( [A]8,10[1]10,1 )2 Where C2 = CTC OR (1/10)([SUM2]T[SUM2]).
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STEP 5:
STEP 6:
SS6 = SS3-SS5
STEP 7:
SS7 = SS4 - SS5
STEP 8:
PROGRAM: SUM1 = [1]1,8[A]8,10
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STEP 9:
PROGRAM: SS9 = (1/NS)([SUM1][SUM1]T)
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STEP 10:
SS10 = SS9 – SS5
STEP 11:
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STEP 12:
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STEP 13:
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STEP 14:
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STEP 15:
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TEST FOR DIFFERENCE BETWEEN INDEPENDENT CORRELATIONS
COMPUTATIONAL HANDBOOK OF STATISTICS, 2ND EDITION SECT. 4-13, PG. 214 - 215.
NOTE: The next 2 sections have nothing to do with the half-multiplier operator, I’m just adding them for completion on the section for correlation’s.
If we have two correlation’s computed from data gathered from two different groups of individuals, the correlation coefficients will be experimentally independent. If so, use the following procedure to test for significance of the difference between the two correlation’s.
EXAMPLE: Suppose we have a correlation coefficient of .68 computed between grades in an English class and IQ scores for 38 people. Also suppose we have a correlation coefficient of .36 between grades and IQ in a similar English class for a different group of 78 people. We wish to know if these correlation’s are different.
STEP 1: Change the two correlation’s into Fisher z scores (Appendix F):
Correlation of .68 = z of .829
Correlation of .36 = z of .377
STEP 2: Subtract either z score from each other: .829 - .377 = .452; STO SS2
STEP 3: Subtract 3 from the number of people in the group from which the first correlation was
computed. (38 in this example). 38 - 3 = 35. STO SS3
STEP 4: Take the reciprocal of SS3 (Carry the decimal to four places): 1/SS3 = 1/35 = .0286
STO SS4
STEP 5: Subtract 3 from the number of people in the group from which the second correlation was
computed. (73 in this example). 73 - 3 = 70. STO SS5
STEP 6: Compute the reciprocal of SS5: 1/SS5 = 1/70 = .0143 STO SS6.
STEP 7: Take the square root of the sum 1/SS4 + 1/SS6:
SS7 = (.0286 + .0143)1/2 = (.0429)1/2 = .207
STEP 8: Divide SS2/SS7 to get the z statistic:
z = .452/.207 = 2.18
A z larger than 1.96 is significant at the .05 level using a two-tailed test (see Appendix A). A significant z tells us that the two correlation values are very likely different.
TEST FOR DIFFERENCE BETWEEN DEPENDENT CORRELATIONS
COMPUTATIONAL HANDBOOK OF STATISTICS, 2ND Ed. SECT. 4.14 Pg. 215 - 217,
This procedure is used to determine the significance of the difference between experimentally dependent correlation’s. I.e. correlation’s based on data taken from the same group of people.
EXAMPLE: Suppose we’ve computed a correlation coefficient between grades in a statistics course and GPA for 63 students as .70. Suppose we also have the correlation between grades in a Psychology class and the same 63 students as .40. If we’ve also computed the correlation between statistics and psychology grades as .30, we can test for the significance of the difference of the first two correlation’, being aware that the scores are related.
Let statistics grades = 1, psychology grades = 2 and GPA = 3, then
r13 = .7
r23 = .40
r12 = .30
STEP 1: Compute the difference of the two correlation’s of interest (r13 - r23) = (.70-.40) = .30
STEP 2: Subtract 3 from the # of people involved in the correlation’s: 63-3 = 60. STO SS2
STEP 3: Add 1 to the third correlation = (r12 + 1) = .30 + 1 = 1.30 STO SS3
STEP 4: Multiply SS2 x SS3: (N-3)(r12 + 1) = 60(1.30) = 78 STO SS4
STEP 5: Take the square root of SS4 ((N-3)(r12 + 1))1/2 = (78)1/2 = 8.832 STO SS5
STEP 6: Square each of the three correlation’s and sum:
[r]T1,3[r]3,1 = [.70 .40 .30] .70
.40 = .74 STO SS6
.30
STEP 7: Multiply each of the three correlation values:
METHOD 1: (Anti-Log(Log ( r )T) [1]3,1)
10EEX([-.1549 -.3979 -.5229] 1 )
1 = 10EEX(-1.0757) = .0840 STO SS7
1
METHOD 2: USING NESTED ARRAYS:
1,o([r]t1,3o[1]1,1 = [.7 .4 .3]o[1]1,1 = [.7] [.4] [.3] with the connector being the o
operator: [.7] o [.4] o [.3] = [.0840]
STEP 8: Multiply SS7 x 2 and add 1 to the product (NOTE: 1 & 2 are always used).
SS8 = 2(SS7) + 1 = .168 + 1 = 1.168 STO SS8
STEP 9: Subtract SS8 - SS6:
(2(Anti-Log(Log ( r )T) [1]3,1) + 1) - [r]T1,3[r]3,1 = 1.168 - .74 = .428 STO SS9
STEP 10: Multiply SS9 by 2 and take the square root of the product:
PROGRAM: (2((2(Anti-Log(Log ( r )T) [1]3,1) + 1) - [r]T1,3[r]3,1))1/2 = (2x.428)1/2 = .925
2(SS8 - SS6)1/2 = (2(2(SS7)+1)-SS6)1/2
STEP 11: Divide SS5/SS10 to get the t statistic:
((N-3)(r12 + 1))1/2/(2((2(Anti-Log(Log ( r )T) [1]3,1) + 1) - [r]T1,3[r]3,1))1/2 =
t = 2.65/.925 = 2.6
CHAPTER 5
NONPARAMETRIC TESTS’ MISCELLANEOUS TESTS OF SIGNIFICANCE, AND INDEXES OF RELATIONSHIP
TEST FOR SIGNIFICANCE OF A PROPORTION
COMPUTATIONAL HANDBOOK OF STATISTICS, 2ND EDITION, SECT. 5.1, PG 220 - 221.
In a situation where a logically dichotomous variable, such as passing or failing a course is being considered, the use of proportions can sometimes help in the statistical analysis of significance. Before any test of significance can be computed, it is necessary to know the proportion to be expected. Therefore, we must know this proportion from:
the expected proportion from previous experience.
by some a priori proportion, such as an educated guess that 20% will fail
by some other method determine the expected proportion.
EXAMPLE: A teacher wants to determine whether a particular class of 15 students is under par in its performance on an arithmetic test. This can be accomplished by determining if the proportions of those who passed and failed correspond to those proportions in classes taught in previous years.
STEP 1: Record and table the data:
Student Score
A 64 64
B 35 35
C 81 81
D 77 77
E 65 65
F 48 48
G 56 [A] = 56
H 31 31
I 64 64
J 72 72
K 43 43
L 44 44
M 63 63
N 59 59
0 38 38
STEP 2: To compute the proportion of students who passed, the teacher must set a pass-fail cutoff point. Suppose a score of 50 or higher is passing, and 49 or less is passing. We can count by hand, or use the following method:
p = (1/N)(+)([A]15,1 - S[1]15,1)
This is elementary, but after the subtraction, we count all non-negative values. i.e.
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64 1 64 50 14
p = (1/15) 35 1 35 50 -15
81 1 81 50 31
77 1 77 50 27
65 1 65 50 15
48 1 48 50 -2
56 - 50 1 = 56 - 50 = 6
31 1 31 50 -19
64 1 64 50 14
72 1 72 50 22
43 1 43 50 -7
44 1 44 50 -6
63 1 63 50 13
59 1 59 50 9
38 1 38 50 -12
There are 9 non-negative numbers, therefore: p = 9/15 = .60. Here the number of students is small and we could count by hand, but if this test represented all the students in the US taking the test, it could run into the millions. We would just program the computer to count the non-negative elements (or to return a one if the value is positive, and a zero if the value is negative, then find the grand sum of the matrix).
STEP 3: Suppose the teachers past records indicate that, on average, 80% of the students got a passing grade on the test. This .80 is the P value. The significance test (z-test) for the obtained proportion is computed using the following formula:
z= (p-P)/(P(1-P)/N)1/2
z = (.6-.8)/(.8(1-.8)/15)1/2 = -.2/(.0107)1/2 = -1.93
To be significant at the .05 level, z would need to be at least ± 1.96. A significant value would mean that the p value is significantly different from the P value. In this case, it is concluded that the scores of the 15 students are not under par.
TEST FOR SIGNIFICANCE OF DIFFERENCE BETWEEN TWO PROPORTIONS
COMPUTATIONAL HANDBOOK OF STATISTICS, 2ND EDITION, SECT. 5.2, PG. 222 - 224.
EXAMPLE: (This problem is basically the same as in the previous section, but we are looking at two groups instead of one). In an experiment on short term memory for a single item, the following conditions were used: (1.The item was shown to all 40 subjects for two seconds. (2. One group of 20 people was tested to see if they recalled the item after 5 seconds. (3. Another group of 20 people was tested after 10 seconds.
If the item was correctly recalled, a 1 was recorded for that subject. If an incorrect response was made, the subject scored a 0.
STATISTICAL FORMULA
z = P1 - P2
( p(1-p)/N1 + p(1-p)/N2 )1/2
Where p = N1P1 + N2P2
N1 + N2
STEP 1: To compute the proportions P1 and P2, table the data as follows:
5 Sec. 10 Sec.
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STEPS 2 & 3: Sum the columns of A and divide by number of subjects in each group: N = 20 .
PROGRAM: SS23 = (1/NP)[1]1,20[A]20,2 = [P1 P2]
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(1/20)[1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1] 1 0
1 0
1 1
1 0
1 1
1 1
1 0
1 1
1 0 STO SUM1
0 0 = (1/20)[16 8] =
1 1
0 0 P1 P2
0 0 [.80 .40]
1 0
1 0
0 1 SS2 = .80
1 0 SS3 = .40
1 0
1 1
1 1
STEP 4:
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[.80 .40] 20
20 16 + 8 24/40 = .60 = p
= =
[20 20] 1 20 + 20
1
Or METHOD 2: (1/NT)[SUM1][1]2,1 = (1/40)[16 8] 1 = 24/40 = .60 = p
1
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STEP 6: Subtract p from 1 and multiply by p
1 - p = 1 - .60 = .40
p(1-p) = p(SS6) = (.60)(.40) = .24 STO SS6
STEP 7: Divide SS6 by N1: .24/20 = .012 STO SS7
STEP 8: Divide SS6 by N2: p(1-p) /N2 = .24/20 = .012 STO SS8
STEP 9: Add SS7 to SS8 and Take the square root of the sum:
p(1-p)/N1 + p(1-p)/N2 = .012 + .012 = .024
(p(1-p)/N1 + p(1-p)/N2)1/2 = (.024)1/2 = .155 STO SS9
STEP 10 Subtract SS3 from SS2: .80 - .40 = .40 STO SS10
STEP 11: Divide SS10/SS9:
(P1-P2)/(p(1-p)/N1 + p(1-p)/N2)1/2 = .40/.155 = 2.48 STO z
A z having a value greater than or equal to ± 1.96 or less is considered significant at the .05 level using a two-tailed test (see Appendix A). A significant z tells us that the two proportions are significantly different. Therefore there was a difference in short term memory retention from 5 to 10 seconds.
THE MANN-WHITNEY U-TEST FOR DIFFERENCES BETWEEN INDEPENDENT SAMPLES
COMPUTATIONAL HANDBOOK OF STATISTICS, 2ND EDITION, SECT. 5.3. PG. 224 - 227.
If the data obtained by an experimenter is badly skewed (one sample population is much larger than the other), the U-test might be used rather than the t-test for independent groups.
STATISTICAL FORMULA
U = N1N2 +(N1(N1+1))/2 - R1
or
U’ = N1N2 +(N2(N2+1))/2 - R2
Where N1 = size of the smaller sample
N2 = size of the larger sample
R1 = sum of the ranks of the smaller sample
R2 = sum of the ranks of the larger sample
When either U or U’ has been calculated, the other can quickly be found by the relations:
U = N1N2 - U’
or
U’ = N1N2 - U
We need compute only one, and the other is obtained by subtraction.
EXAMPLE: The following data has been collected representing weekly incomes for secretaries that are members of two ethnic groups. We wish to determine whether there is a significant difference between these incomes.
STEP 1: Table the data, then rank all the numbers in both groups taken together according to
the size of the number, beginning with the smallest. If any values are tied, the same rank is assigned to each, but the rank assigned is the mean value of the tied ranks. i.e. two incomes of $76 are tied for the 6th and 7th ranking, therefore, they both get the rank of 6.5.
Group 1 Rank Group 2 Rank
87 11.5 131 20
72 4 94 15
65 2 77 8.5
54 1 88 13
67 3 116 18
[A]12,4= 76 6.5 90 14
73 5 87 11.5
82 10 76 6.5
104 17 95 16
0 0 164 21
0 0 127 19
0 0 77 8.5
NOTE: Although there are only 9 measures in Group 1, I added the three zero’s to complete the matrix. They do not count as measures.
VARIABLES:
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STEP 2: Count the # of measures in each group and multiply them together.
G1: N1 = 9; G2: N2 = 12.
N1N2 = (9)(12) = 108 STO SS2
STEP 3: Add 1 to the number of measures in Group 1, multiply by the original measure and
divide by 2:
SS3 = (N1(N1+1))/2 = 9(9+1)/2 = 45 STO SS3
STEP 4: Add the measures for Group 1:
PROGRAM: [1]1,12 ([A]12,4 0 0 0 0 )[1]4,1 = R1
0 1 0 0
0 0 0 0
0 0 0 0
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[1 1 1 1 1 1 1 1 1 1 1 1] 0 11.5 0 0 1 = [60] = R1
0 4 0 0 1
0 2 0 0 1 STO SS4
0 1 0 0 1
0 3 0 0
0 6.5 0 0
0 5 0 0
0 10 0 0
0 17 0 0
0 0 0 0
0 0 0 0
0 0 0 0
NOTE: When we pre-multiply by a diagonal matrix, it is the same as multiplying the corresponding row by that number. When we post-multiply a matrix by a diagonal matrix, it is the same as multiplying the columns by the corresponding elements. In the example above, a11, a33 and a44 = 0 and a22 = 1, so the multiplication returns only the values in column 2. Then we just take the grand sum of the matrix to compute R1. If we wished to obtain the columns of both R1 and R2, we would post multiply by:
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
STEP 5: Add SS2+SS3-SS4 = U; 108+45-60 = 93 = U STO U
STEP 6: Perform the subtraction: U’ = SS2-U = 108 - 93 = 15 STO U’
If the number of measures in the larger group is greater than 8, and the # of measures in either group is smaller than 21, take whichever answer in steps 5 & 6 that is smaller and look in tables of U values (Appendix H) to check for significance. Since N1=9 and N2=12, both requirements are satisfied. For a .05 confidence level, we see that if the smaller value is 26 or less, the difference is significant.
Suppose one of the measures in either group is greater than 20, then we must revert to the z critical-ratio test:
STEP 7: Divide SS2 by 2 and subtract from U.
SS7 = U - N1N2/2 = 93-108/2 = 39
STEP 8: Add N1+N2+1: SS8 = 9+12+1=22
STEP 9: Multiply SS2 by SS8: SS9 =(N1N2)(N1+N2+1)= 108x22 = 2376
STEP 10: Divide SS9 by 12 (Note: 12 is always used), then take the square root of that
quotient:
((N1N2)(N1+N2+1)/12)1/2= (2376/12)1/2 = (198)1/2 = 14.071 STO SS10
STEP 11: Divide SS7 by SS10:
z =(U - N1N2/2)/((N1N2)(N1+N2+1)/12)1/2 = 39/14.071 = 2.77
Looking again in Appendix A, a value greater than ± 1.96 is significant at the .05 level. The Mann-Whitney analysis hypothesized the two medians are equal, a z value that is large enough so that the hypothesis is rejected tells us the chance of the medians being the same is very small. It is concluded that there is a significant difference between the median incomes of secretaries in the two ethnic groups.
WILCOXON A SIGN TEST FOR DIFFERENCES BETWEEN RELATED SAMPLES
COMPUTATIONAL HANDBOOK OF STATISTICS, 2ND EDITION, SECT. 5.4, PG. 228 - 229.
Unlike the U-test in the previous section, which is a non-parametric test for independent samples, the Wilcoxon signed-ranks test is applicable where the samples are related.
EXAMPLE: Suppose we have collected the reading rate scores for a group of 10 people before and after they enrolled in a course in speed reading. We want to find out if the course made any significant difference in reading rate. We can also look at this as training, did sales increase? Did production increase? Etc.
STEP 1: Table the data:
Rate Rate
before after
150 145
135 138
102 121
96 115
[A]10,2 = 127 134
118 132
132 138
124 145
115 126
103 94
STEP 2: Subtract the first column from the second column:
PROGRAM: [A]10,2[+ 1] =
150 145 -1 = -5
135 138 1 3
102 121 19
96 115 19
[A]10,2 = 127 134 7
118 132 14
132 138 6
124 145 21
115 126 11
103 94 -9
STEP 3: Rank the differences according to the absolute size of the number (ignore the signs), remembering to split any ties:
Diff Rank
-5 2
3 1
19 8.5
19 8.5
7 4
17 7
6 3
21 10
11 6
-9 5
STEP 4: Add the ranks of all the negative numbers: i.e. the two negative differences, -5 and
-9 are ranked 2 & 5. Note: Zero differences are not used in this test, in fact, if there is a zero difference, the N which is used in the consulting table (Appendix H) must be reduced by the # of zero difference cases.
R- = 2+5=7
STEP 5: Add the ranks of all the positive numbers:
R+ =1+8.5+8.5+4+7+3+10+6= 48
STEP 6: 7 and 48 are the sign-ranked values. Take the smallest result and look in a table of
Wilcoxon’s probabilities (Appendix M). When the # of pairs (N) = 10, a value of 8 or less is necessary for significance at the .05 level. We conclude that the reading rate after training was significantly faster than before training.
SIMPLE CHI-SQUARE AND THE PHI COEFFICIENT
COMPUTATIONAL HANDBOOK OF STATISTICS, 2ND EDITION, SECTION 5.5, PG. 230 - 232.
When we have frequency data comparing the effects of two variables and there are two groups associated with each variable, we can compute the phi coefficient to find the degree of relationship between the two variables, while the chi-square test will determine whether the variables are related. A significant chi-square is interpreted as showing a relationship between the two variables. The phi coefficient gives a numerical value (from 0 to 1) for that relationship.
EXAMPLE: Consider the data collected on the two variables: gum chewing and education. The experimenter asked two groups, one of high school dropouts and one of college graduates whether or not they chewed gum. Are educational backgrounds and gum chewing related?
STATISTICAL FORMULA
N(AD-BC)2
(2 = PHI (() = ( (2/N)1/2
(A+B)(C+D)(A+C)(B+D)
Where:
[A]2,2 = A B
C D
VARIABLES:
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STEP 1: Organize the data into a contingency table, the values in the table represent the number of persons in each category.
Chews Abstains
Dropout 90 50 = [A]2,2
Graduate 20 30
STEP 2: Add all the numbers in the table:
PROGRAM: [1]1,2[A]2,2[1]2,1 = N
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STEP 3: Find the row and column sums:
PROGRAMS: ROW SUMS = ([A]2,2[1]2,1 )T = [140 50]
COLUMN SUMS = [1]1,2[A]2,2 = [110 80]
We might as well do step 4 here also, since they are all related:
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STEP 4: Multiply together the four sums in step 3:
MATHCAD +6 PROGRAM: (10EEX(LOG(ROWSUMS)T+LOG(COLSUM))[TWO1])
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STEPS 5,6AND 7: This is just calculating the determinant of the matrix A:
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i.e. detA = 90x30-50x20 = 2700 - 1000 = 1700
STEP 8: Square the detA and multiply by the grand sum of the matrix:
PROGRAM: A 2 [1]1,2[A]2,2[1]2,1
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17002x190 = 549,100,000
STEP 9: Divide the result of step 8 by step 4 to get:
PROGRAM: ( A )2 [1]1,2[A]2,2[1]2,1 549,100,000/61,600,000 = 8.91
=
(10EEX(LOG(ROWSUMS)T+LOG(COLSUM))[TWO1])
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The # of degrees of freedom for a 2x2 chi-square will always be one. From Appendix D, we find a chi-square value greater than 3.8 is significant at the .05 level. It is concluded that educational background and gum chewing behavior are significantly related.
STEP 10: Computation of phi ():
( = ((2/N)1/2 = (8.91/190)1/2 = .217
( represents the degree of relationship between the two variables.
SUPPLEMENT
THE YATES CORRECTION FOR CONTINUITY
In a 2x2 contingency table, if the frequency of any cell is less than 10, it is recommended that the Yates correction for continuity be used to account for the fact that (2
computed with very small N’s will over-estimate the true (2 value. Yates correction is defined as:
N((AD-BC) - N/2)2
(2 = PHI (() = ( (2/N)1/2
(A+B)(C+D)(A+C)(B+D)
190((2700 - 1000)-(190/2)2)
(2 = = 489,444,750/61,600,000 = 7.94
140x50x110x80
COMPLEX CHI-SQUARE AND THE CONTINGENCY COEFFICIENT C
COMPUTATIONAL HANDBOOK OF STATISTICS, 2ND EDITION, SECTION 5.6, PG. 233 - 237.
Note: I’ve already worked this problem out in advance, and it’s solution is really neat, for it uses everything mathematicians say is illegal. This means either chi-square complex is a false statistical method, or that mathematicians have been wrong about the very basic tenets of mathematics for over 200 years. Chi-square complex is strikingly similar to perturbation theory in quantum chemistry, and I predict that this method is one of the connecting equations between statistics and physics.
When we have frequency data comparing the effects of two variables and there are more than two groups associated with either of the two variables, the complex chi-square can be used to test the hypothesis of no relationship between the variables. If we find a relationship, we can compute the contingency coefficient to determine the degree of relationship.
EXAMPLE: Suppose we gather information for three groups of people:
those < 26 years old
those > 26 and < 60 years old
those 61 and older
Their incomes are categorized in $3,000 ranges and a frequency count is made for each category. We wish to find if age and income are related.
STATISTICAL FORMULA
(2 = ((O-E)2/E
Where O = the observed frequency for a particular cell in the contingency table
E = the expected frequency for a cell, based upon marginal totals (steps 2-6).
And
C = ((2/((2 + N))1/2
Where N = grand sum of contingency table.
VARIABLES:
[pic] [pic] [pic]
STEP 1: Table the data into a matrix:
INCOME GROUPS
UNDER 3,000- 6,000- OVER
3,000 6,000 9,000 9,000
under 26 40 30 10 20
26-60 50 60 20 70
over 60 10 20 20 10
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STEP 2: Add the numbers in the matrix to obtain the grand sum:
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PROGRAM: SUM1 = [1]1,3[A]3,4[1]4,1
STEP 3: Add the numbers in each row:
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PROGRAM: ROWSUM = [A]3,4[1]4,1
STEP 4: Add the numbers in each column:
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PROGRAM: COLSUM = [1]1,4[A]3,4
STEPS 5 & 6: Make a new table, but unlike the book, we make the table by multiplying the sum times the column matrices. This is a more generalized form of the perturbation relation Ψi οΨi.
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Now we subtract the perturbed function from the contingency table:
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We now invert each individual element in the matrix e, we do not take the inverse of the matrix. This is an illegal mathematical transformation as mathematics views division today.
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We now square each individual element in the matrix F, we do not square the matrix. This is an illegal mathematical transformation as mathematics views matrix multiplication today.
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STEP 7:
We now multiply each individual element in the matrix F by its corresponding element in matrix g, we do not multiply the matrix column by row. This is an illegal mathematical transformation as mathematics views matrix multiplication today.
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STEP 8: Add all the elements in SS7 to compute (2:
PROGRAM: (2 = [1]1,3[A]3,4[1]4,1
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STEP 9: Compute the degrees of freedom: df = (# rows-1)(# columns-1) = (3-1)(4-1) = 2x3=6.
A chi-square value larger than 12.6 with 6 degrees of freedom is significant at the .05 level (Appendix D).
STEP 10 & 11: Find the contingency coefficient by computing the following equation:
C = ((2/((2 + N))1/2
C = (36.985/(36.985+360))1/2 = (.093)1/2 = .305
The contingency coefficient is significantly different from 0, it is concluded that age and income are significantly related.
All the math just done here in this book is impossible according to the knowledge of today’s mathematics; We cannot square the individual elements of a matrix; We cannot invert the individual elements of a matrix; We cannot multiply one on one with matrices and have them mean anything. Therefore the answers must be incorrect, even if they are right. But it works for statistics! So therefore, either statistics must be a fluke and really does not work, which hypothesis would invalidate 80 to 90% of our knowledge of chemistry and physics, or mathematicians have been wrong in their assumptions of the definition of addition, subtraction, multiplication and division for the past 200 to 2000 or so years. A number is a one by one matrix, but a row, column, square or rectangular matrix can itself be considered a single number that follows the rules of single numbers. . Each individual row or column can be added or subtracted from each other, element from corresponding element. Each individual row or column can be multiplied by a single number, the solution being the same as doing the multiplication’s one at a time on each element of the second matrix. But we can do these all at the same time using the half-multiplier operator. Each element can be multiplied by it’s corresponding element in another matrix on a one-to-one correspondence without summing, or we can transpose the pre-multiplier, matrix multiply and sum the products if we wish.
PROGRAMS FOR EXAMPLE 5.6
STEP 2:
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SUM1 = [1]1,3[A]3,4[1]4,1
STEP 3:
ROWSUM = [A]3,4[1]4,1
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STEP 4:
COLSUM = [1]1,4[A]3,4+
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STEPS 5 & 6:
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Or [pic]
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STEP 7:
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STEP 8:
PROGRAM: (2 = [1]1,3[A]3,4[1]4,1
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STEP 9:
df = (# rows-1)(# columns-1)
STEPS 10 & 11:
C = ((2/((2 + N))1/2
QUANTUM STATISTICS
Suppose we have completed the experimental data on an experiment and put into the matrix form [A]MN. Now [A]MN may not be a square matrix, so we must make a square out of it before we can do any sort of quantum statistical analysis on it. Squaring the data changes the matrix [A] from a statistical matrix to a probability matrix. To keep track of each of the sums of the squares for each column in the matrix, we must square it such that we end up with an NxN matrix.
[A]TMN[A]MN = [A]NM[A]MN = [A]2NN
Then we do the following:
1/N[DB1]iN [A]2NN[DB1]TiN 1/N[DB1]iN [A]2NN[DB1]Ni 1/N[DB1]iN [A]2NN[DB1]Ni
= = =
[DB1]iN[DB1]TiN [DB1]iN[DB1]Ni [DB1]2ii
1/N[C]2ii
[DB1]2ii
EXAMPLE: Since we’ve already done a series of different analysis on the problem in section 2.7, I will do the quantum statistics on that set of data. I will not try to interpret the results since I do not know what it means, but I can do the math. Professional statisticians will have to determine whether there is any significance in the answers or not. The data of the problem in section forms a 11x9 matrix, [A]11,9 .
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THIS IS THE MATHCAD +6 PROGRAM FOR THE QUANTUM STATISTIC FOR DB1
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THE FOLLOWING IS THE ABOVE PROGRAM ELUCIDATED STEP BY STEP:
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QUANTUM STATISTIC FOR DB1
THIS IS THE TERM IN THE DENOMINATOR:
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THE QUANTUM STATISTIC FOR DB3
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[DB3]T3,9 [DB3]9,3 = VALUE IN THE DENOMINATOR
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(DB3 =(1/99)([DB3]T[A]T[A][DB3])([DB3]T[DB3])-1
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I do not know if these values have any meaning. We may just have to look at the diagonal for the statistical significance, or we may have to take the grand sum of the solution and check it. Or we may find that both the diagonal and grand sum are useful, or that both of them have no meaning at all. I leave this to the professionals to determine.
Just for the heck of it, (and because they do it in quantum mechanics) lets normalize the solutions. Going back to the chapter on quantum chemistry (I do not have my book paged yet so I can’t tell you the page, it is all in separate files in my word processor waiting to be put together into final publishing form, so I have to say it is on the 40th page of that chapter), you’ll find that the equation for normalizing the solution is:
(jj = diag(CTjkCjk)
then
(NORM = ((jj-1)1/2(
I’ll solve the two problems side by side. The matrix solutions using DB1 and DB3 are:
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[pic] [pic] Here I am squaring the DB matrices
[pic] [pic] Here I am making a matrix only of the diagonal
[pic] [pic] Here I’m taking the inverse of the matrix
[pic] [pic] Here I take the square root of each element
[pic] [pic]These are the normalizing multipliers
[pic] [pic] Here I multiply the normalizing
factor times the solution matrix
[pic] [pic] Normalized solutions.
CHECK: FOR DB1 AND DB3, THE NORMALIZED SOLUTIONS SQUARED SHOULD EQUAL 1.
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
LINEAR REGRESSION
Note: I did not develop this, it is how linear regression is computed on computers and calculators. I thought I’d not only show how it works, but also show how close statisticians were to finding that statistics is truly a matrix mathematics rather than a random variable mathematics.
No book on statistics would be complete without a discussion of linear regression, or the best fit of a line or a curve through a set of experimental data points. In regular algebra, the equation for a line is given by Y = mX + b. The equation for a line in matrix form is given by Y = MX. Suppose we generate a set of data points in a laboratory experiment such that:
y1 = ax1 + b
y2 = ax2 + b
. . .
. . .
yn = axn + b
The matrix equation looks like:
M X = Y
x1 1 a y1
x2 1 b y2
x3 1 = y3
. . .
. . .
xn 1 yn
Now take a general vector MV such that V = to the perpendicular distance from the x axis. We then multiply the matrix equation for a line by the general vector MV such that:
(MV)T(MX-Y) = 0
expanding we get
VT(MTMX - MTY) = 0
VT(MTMX - MTY) = 0
We divide out the VT term, and get
MTMX - MTY = 0
MTMX = MTY
and solving for X we get
X = (MTM)-1MTY
First let’s look at regular linear regression to see how to apply this equation to our experimental data.
REGULAR LINEAR REGRESSION
THE FORCE CONSTANT OF A SPRING
The equation for a force acting on a spring is given by F = kX + b. Suppose the data recorded in our experiment was as follows:
Ypounds Xspring stretch
0 6.1”
2 7.6”
4 8.7”
6 10.4”
[pic] [pic] [pic]
Then
[pic]= [pic]
or
[pic]
The first value is the value of the slope m, and the lower value is the value of the y intercept b. The equation F = kX + b becomes F = 1.42X - 8.643 .
SEMI-LOG LINEAR REGRESSION
Suppose you are a doctor or pharmacist. Your patient needs to have at least a .50mg per 100 cc’s concentration of Dilantin in their blood to keep them from having epileptic seizures. If the concentration falls below this level the patient may experience seizures. The patient is injected with a .5 gram per 70 Kg body weight sample of the drug. After 30 minutes a sample of blood is drawn and analyzed. For the next hour a sample is drawn every 30 minutes. Thereafter a sample is drawn every two hours and analyzed. Calculate the half-life of Dilantin in your patients body.
The semi-log equation is given by Y = aebx .
Rewriting this as a logarithmic equation we get:
Log Y = Log A + bX
Let x = X; Log Y = Y and Log A = A
Substituting these changes of variables into the logarithmic equation, we get:
Y = bX + A
The logarithmic equation is now linear.
Suppose the experimental values are as follows:
time mg/100cc
0.5 1.12
1.0 .90
1.5 .80
2.0 .73
4.0 .50
6.0 .36
8.0 .20
The logarithmic equation looks like:
x1 1 b Log y1
x2 1 Log a Log y2
x3 1 Log y3
. . = .
. . .
xn 1 Log yn
[pic] [pic][pic] [pic]
[pic]
[pic]
Where -.09018 = b
Log .0624 Log a
But Y = aebx; Log Y = Log a + bx; a = .0624; anti-log a = 10.0624 = 1.1545 and the equation for the best fit becomes:
Y = 1.1545e-.0918x
To find the half-life, use the equation
Log .5000 = Log 1.1545 - .0918t
[pic]
[pic]Hours
8 * *
7
6 * *
5
4 * *
HALF-LIFE 3.9 hours
3
* *
2
* *
1 * *
* *
.2 .4 .6 .8 1.0 1.2 .2 .4 .6 .8 1.2
LOGARAMITHIC GRAPH LINEARIZED GRAPH
LOG-LOG LINEAR REGRESSION
The equation for the Log-Log relationship is given by:
Y = kXn
Changing this to a logarithmic equation we get:
Log Y = Log k + nLog X
Letting Log Y = Y; Log k = K; and Log X = X we get:
Y = nX + K
The equation is now linear.
Suppose through experiment we obtain the following set of data:
X 1 2 3 4
Y 2.5 8.0 19.0 50.0
The matrix equation becomes:
Log x1 1 n Log y1
Log x2 1 Log K Log y2
. . = .
. . .
Log xn 1 Log yn
[pic] [pic] [pic]
Then [pic]
Or expanding the above multiplication out we have:
[pic]
But
2.0953 n n = 2.0953
Log .3467 = Log K k = anti-log K = 10.3467 = 2.2218
But Y = kXn , so the best fit is Y = 2.2218X2.0953
Log-Log Graph Linearized Graph
* *
* *
*
*
* *
PARABOLIC FIT
Let’s see how linear regression fits a parabolic equation Y= a + bX + cX2 .
X = -3 -2 0 3 4
Y = 18 10 2 2 5
In the matrix X1, the first column consists of the values X2, the second column consists of the values of X, and the third column consists of the value 1. The Y matrix is a column matrix consisting of the values of Y.
[pic] [pic]
[pic]
Y = 1.8247 - 2.6455X + .8728X2
Let's look at this step by step:
[pic]=[pic]
[pic]=[pic]
[pic]=[pic]
[pic]=[pic]=[pic]
Y = .8718X2 - 2.6434X + 1.8224
CRYPTOGRAPHY
Just recently, I have found an example that follows the rules of quantum statistics. In the following example, though, the datastream matrix must be a square matrix. We cannot transpose and square because it is important and necessary that we do not add the matrix elements to each other. They must always keep their pristine identity. Since the matrix is square, we can multiply in two ways: This is not new code, but is well known in cryptographic circles.
[DB1][A][DB1] = [SOL1] or [DB1]T[A][DB1]= [SOL2]. We do not normalize the resulting solution matrix. In the middle to last part of the following example, if [V] is defined as the Vinegeir transform, and [DB11] is defined as the pre-scrambler matrix, the above equations become (computing the values inside the parenthesis first):
[DB1] (([A] + [V])[DB11]) [DB1] =[ SOL3] or [DB1]T(([A] + [V])[DB11])[DB1] = [SOL4].
To reverse the above transformation and return to the original matrix, we put the solution back in the place of [A], subtract [V] and multiply by the transpose of the database matrices. i.e.
[DB1]T[SOL1][DB1]T = [A] or [DB1][SOL2][DB1]T = [A].
Or for the more complex forms:
[[DB1]T[SOL3][DB1]T)[DB11]T = [SOL3B] = [SOL3A][DB11]T
or
([[DB1][SOL4][DB1]T)[DB11]T) - [V] = [A] = [SOL4B] - [V]
PLAINTEXT CIPHERCODE
[pic] [pic]
CHANGING LETTERS TO CORRESPONDING NUMBER IN ALPHABET
[pic]= [pic]
We change the letters in the text where H is the 8th letter in the alphabet, I is the 9th letter in the alphabet, etc.
COLUMNS MIXED UP [A][DB1]
[pic] [pic]
ROWS MIXED UP [DB1][A]
[pic] [pic]
ROWS AND COLUMNS MIXED UP. [DB1][A][DB1] = [SOL1]
[pic][pic]
ROWS UN-MIXED
[pic] [pic]
COLUMNS UN-MIXED
[pic] [pic]
ROWS AND COLUMNS UM-MIXED [DB1]T[SOL1][DB1]T = [A]
[pic][pic]
NOTE: These letters are not coded, they are scrambled.
To code the message first, we will use a Vinegeir transform letting C = 2 (A = C = 3;
B = D = 4; Or:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
C D E F G H I J K L M N O P Q R S T U V W X Y Z A B
=
[pic]=[pic]
Applying the Vinegeir transform:
[pic] [pic]
Note the number 27, it is greater than 26, the number of letters in our alphabet. What we can do is replace the 27 with the remainder of 27/26 = 1, but leaving the numbers unchanged is vastly simpler. If we wish to keep all numbers from 1 - 26, we divide by 26 in all cases and replace the number by it's remainder.
This is the pre-coded message before dis-assembly. [DB1] ([A] + [V]) [DB1] = [SOL3]
[pic]=[pic]
To re-assemble, we proceed as follows:
[pic]
[pic]
Suppose we are really security conscious and want to make the de-coding of this message as hard as possible, even for someone with sophisticated code breaking computers like governments (hackers couldn't break this code, a single typewritten page contains about 75 lines by 75 letters or spaces or 1,125 spaces/words of approximately 5 letters each, or 1,125 letters per page. If the pre-multiplier matrix if a 75x75 matrix, there will be 75 combinations to de-code. If we use a Vinegier distribution, they must also compute 26 variations for each guess in the 75x75 combination to see if their guess is right. The number of combinations are approximately 5,625 elements taken 75 at a time = 4.4604x10171, or a number about as large as all the atoms in all the stars and planets in all the galaxies in the universe put together. But we can only put one number in each row and one in each column, so we must divide this number by 75! And the number of combinations becomes [pic]
How big is this number? Suppose we had a super computer that was able to multiply one million of these pre and post multiplier matrices per second until a solution was found. Also suppose that the computer found the correct solution on the very last try.
[pic]
[pic] seconds.
60sec/min x 60min/hr x 24 hr/day x 365 days/year = 31,536,000 seconds per year.
If we divide 1.798x1056 by 31,536,000 we will get the number of years it will take to break this code.
[pic] years
If the age of the universe is five billion years since the big bang, (5x109 years), it would take a supercomputer
[pic] lives of the universe to decode a one page message.
Now suppose I am wrong and it is not the number of combinations. In a 75x75 character word-processor sheet, the first 1 in the pre-multiplier can occupy any of the 75 positions. In the second row, the 1 can occupy any of 74 positions since it cannot be in the same column as the 1 in the first row. We continue down the matrix until we reach the very last row where there is only one position the 1 can occupy. In this case, the number of combinations is 75 factorial, [pic]
which is a number a hell-of-a-lot bigger than [pic]
So we would have
[pic] lives of the universe to find the solution.
But suppose it's possible. We can confound any computer by employing the following method. Instead of a simple Vinegeir transform, we employ a complex transform, i.e. if we put a 2 in the transform matrix in position 1,1, H = J; if in position 1,2 we add 5, Y = D and C = H, etc. We add a different number from 1 to 26 instead of adding an entire matrix of 2's, for instance. Then before we transform to encode, we pre-encode the message with a different post-multiplier matrix. This way, even if they do stumble upon the proper combination to de-code your message, it is pre-scrambled and they will never know whether they’re close to a solution or not. Let me show you what I am talking about. We will use the same pre and post multiplier matrices as above, but will define the first post multiplier matrix = [DB11]as:
[pic]= [DB11].
And I will define the complex Vinegeir transform [V] as:
[pic]= [V]
First we change the letters one for the other: [A] + [V]
[pic]
Now we pre-scramble the text:
[pic] (([A] + [V])[DB11])
Now we scramble according to the way discussed above: [DB1] (([A] + [V])[DB11]) [DB1] = [SOL3]
[pic]
Again, to de-code the cipher, we do the problem backwards: [DB1]T[SOL3][DB1]T = [SOL3A]
[pic]= [SOL3A]
Even if they discovered the proper ciphercode above, the message is so garbled they would never know they found the key. Next we un-scramble the pre-scrambled message. [[DB1]T[SOL3][DB1]T)[DB11]T = [SOL3B] = [SOL3A][DB11]T
[pic]= [SOL3B]
Finally we revert from the Vinegeir transform back to our regular alphabet:
([[DB1]T[SOL3][DB1]T)[DB11]T) - [V] = [A] = [SOL3B] - [V]
[pic]
Now I will solve the encryption using the equation for quantum statistics. ([DB]T[A][DB])
[DB1]T(([A] + [V])[DB11])[DB1] = [SOL4].
[pic]
[pic]
For the simple case.
For the more complex case, I am going to create here a special Vinegeir transform, just to show how really difficult de-coding this message will be without the key.
[pic]= [V]
[A] + [V] =
[pic]= [pic]
We do not have to scramble the letters, multiplying this by different databases returns the same solution. But you can see how difficult to de-code the cipher is, where do we start, how many E’s are in the message? S’s? A’s? But let’s see what happens when we try to scramble it:
The column pre-scrambling operation is:
[pic]
and the final scramble becomes:
final ciphertext
[pic]= [pic]
If you intercepted this code, you wouldn't even know if it was a code, or if it were, how to go about de-coding it.. So let's de-scramble it.
[DB1][SOL4][DB1]T = [SOL4A]
[pic]= [SOL4A]
[[DB1][SOL4][DB1]T)[DB11]T = [SOL4B] = [SOL4A][DB11]T
[pic]= [SOL4B]
([[DB1][SOL4][DB1]T)[DB11]T) - [V] = [A] = [SOL4B] - [V]
[pic]
Adding the negative numbers to 26 we get:
[pic]
Which is where we started from.
Here I just wish to show that the eleven’s matrix (or whatever same number matrix) is independent of all pre and post multiplier operations.
.
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
In this special case, we do not have to go through all the de-scrambling steps, just take the K matrix and subtract the Vinegeir transform to get the plain text message back. So it is best to keep all the numbers different rather than the same if we wish to make the de-coding of the message more difficult for hackers.
Yeah, having a code with all the same letters is neat, but won't it be a hassle to transform many pages of plaintext into code this way? I mean, is there an easier way to encode your text than to do it letter by letter and by hand? Yes! And it is very simple to do. We go about it in this manner. In basic algebra, we have the equation ( for the letter S):
19 - x = 11; Where x is the number we wish to put in the key for our code. In this case
x = 19 - 11 = 8. In matrix form we will let A = whatever letter we wish to transform and X be the number in the key. Then we have A - X = 11; X = A - 11. Let's see how this works for the example above. Changing each letter into it's numerical equivalent we get:
[pic]=[pic]
Solving the equation A-11 = X we get:
[pic]
The matrix to the right of the equal sign is the key.
[pic] [pic]
[pic]
or
[pic]
Which is where we wish to be with this code.
Every message using this method will have it's unique key. A friend cannot have a copy in advance to use to decode your message, EXCEPT IN THE MANNER DESCRIBED BELOW. This is a code that is best used when you put information into a computer and do not wish for a hacker to download it. You transform the plaintext into ciphertext and copy the key onto a floppy disk or CD then erase it from your computer memory. If anyone hacks into your files, all they will get is a code of all the same letters, but the key is inaccessable to anyone but you because the key is not in computer memory where someone might discover it. For instance, any 25 letter message is a solution for the 5x5 eleven's matrix above. The solution "my dog had fleas and died" is a solution, but is incorrect. There are thousands of 25 letter messages that are solutions, but which one is your message? Without the key, deciphering the message is almost impossible. Recall the coded 5x5 eleven matrix message, recall the KEY, add them together and your message is returned.
[pic]
[pic] [pic] [pic]
Code1 means it needs KEY1 to decode it, CODE2 means it needs KEY2 to decode it, CODE3 means it needs KEY3 to decode it. In a 5x5 message, there are 120 possible KEYS (5!).
[pic] [pic] [pic]
[pic][pic][pic]
[pic]
[pic] [pic]
Suppose there are 3 agencies, one the headquarters and it has all three keys, the second agency has keys 1 and 3, the third agency has keys 2 and three. If the message is for agency one only, it is sent encoded in KEY1, the third agency can't decode it. If the third agency is the only recipient it is sent in KEY2 and the second agency can't decode it. If it is meant for both agencies it is sent in KEY3. It is the same if the two sub-agencies are writing to headquarters, they can send it so it can be read by HQ only by sending it in KEY1, if the third agency needs to read it, it is sent in KEY3, etc.
[pic][pic]
[pic][pic]
PHYSICS 101
EXAMPLE: A particle moving in one direction has a speed of v m/s at time t goven by the equation v = at2 + bt + c . We define (for this case) that a = 3, b = 4 and c = 5.
Find the initial speed.
V(0) = c = 5 m/s
[3 4 5] 0 = 5 m/s
0
1
Find the speed when 3 seconds have passed.
V(3) = 3(3)2 + 4(3) + 5 = 27 + 12 + 5 = 44m/s
[3 4 5] 9 = [27+12+5] = [44 m/s]
3
1
What is the acceleration when 4 seconds have passed?
a(t) = dv(t)/dt = 2at + b = 2(3)t + 4 = 6t + 4 .
At t = 4 sec. , a(t) = 6(4) + 4 = 28 m/s2
To do the calculus derivative using matrices, shrink the matrix by one element and multiply the first number in the new matrix by its position number (in this case it is the number 2) and the second number in the new matrix by it’s position number (in this case it is in the first position).
d[3 4 5]/dt = 2 o 3 = 6 = 6t + 4
1 4 4
or [3 4] ( [2 1] = [6 4] = 6t + 4 .
At t = 4s, we get
[6 4] 4 = [24 + 4] = 28 m/s .
1
Find an equation for the displacement x(t) using the equation v = 3t2 + 4t +5.
dx/dt = at2 + bt +c.
[pic] =[pic]
= x - x0 = a t3/3 + bt2/2 + ct
x = a t3/3 + bt2/2 + ct +x0 Substituting the given values for a, b and c we get
x = 3 t3/3 + 4t2/2 + 5t +x0 = t3 + 2t2 + 5t + x0 .
Using matrices, to integrate, we expand the matrix by one element and divide each element by it’s corresponding position in the new matrix (we ignore the position of the new constant). i.e.
[pic] = [3 4 5 x0 ] ( [1/3 1/2 1 1] = [3/3 4/2 5/1 x0/1] = [1 2 5 x0]
x = t3 + 2t2 + 5t + x0 .
Let’s look at the above problem using vectors.
Let V = 3t2I + 4tj +5k .
Using the displacement vector, find the displacement in 10 seconds.
x = t3i + 2t2j+ 5tk + x0 .
[3i 4j 6k 0] 103 = [3(103)i + 4(102j + 50k]
102
10
1
or
3i 1000 3000i
4j o 100 = 400j
5k 10 50k
0 1 0
Find the instantaneous and average velocity vectors.
t2 t 1
d[3i 4j 5k x0]/dt = [3i 4j 5k] ( [3 2 1] = [9i 8j 5k] evaluating at 10 seconds:
[9i 8j 5k] 102
10 = [900i +80j + 5k]
1
The average velocity is given by:
(1/10)[3i 4j 5k 0] 103
102 = (1/10)[3000i +400j + 50k + 0] = [300i +40j + 5k]
10
1
Find the instantaneous acceleration.
a = dv/dt = 18ti + 8j or
d[9i 8j 5k]/dt = [9i 8j] ( [2 1] = [18i 8j] and at t = 10s
[18i 8j] 10 = [180i 8j]
1
EXAMPLE: A particle moves along the x-axis. At t = 0, it’s position is -4 m. It’s instantaneous velocity in meters/sec is given by v = 6t2 - 5t + 4.
What is it’s velocity at t = 2 seconds?
[6 -5 4] 4 t2
2 t = [24 - 10 + 4] = 18 m/s
1 1
What is it’s instantaneous acceleration at t = 3 seconds?
d[6 -5 4]dt = [6 -5]( [2 1] = [12 -5]
a = [12 -5] 3 = [36 - 5] = [31m/s2]
1
What is it’s average acceleration for the time interval t = 2 to t = 4 seconds?
a = v(4) - v(2)
2
v(2) = 18 m/s; v(4) = 6(4)2 - 5(4) + 4 = 96 - 20 + 4 = 80m/s
a = (80 - 11)/2 = 34.5m/s2
or
(1/2)[1]1,3([v]1,3[(a]3,2[( 1]2,1 )
(1/2) [1 1 1] (6 -5 4] 16 4 1 ) =
4 2 -1
1 1
[pic]m/s2
Where is the particle at t = 3 s ?
[pic] =[6 -5 4 1] ( [1/3 1/2 1 1] = [2 -5/2 4 1] [pic]
Then
[pic]m
Where -4 is the initial displacement at -4 meters, 27 = 33; 9 = 32 and 3 = 3 seconds.
Or we can solve this in the normal manner:
x(3) - x(0) = [pic] = [pic] = x(3) - (-4)= [pic][pic] =
=
[pic]m
EXAMPLE: The position of a particle is described by r = (4t3 - 12t +9)I + (6t + 4)j meters with t in seconds.
Calculate the instantaneous velocity at t = 0, 1, 2, and 3 seconds.
r = [4 0 -12 9]i + [6 4]j
v = [4 0 -12]I ( [3 2 1] + [6]j ( [1] = [12 0 -12]i + [6]j = (12t2 -12)i +6j
v0 = [12 0 -12]i 0 + [0 0 6]j 0 = -12i + 6j
0 0
1 1
Or we can solve this using nested arrays, i.e.
i j
[ [12 0 -12] [0 0 6] ] 0
0
1
Transposing the nested array (so we can properly multiply them) we get
12 0 -12 0 = 0 = 0i +6j
0 0 6 0 6
1
Re-transposing the solution, we get [ [-12] [6] ], or removing the matrices and re-formatting into equation form we get -12i +6j.
v t
v1 = 12 0 -12 1 = 0
0 0 6 1 6
1
Or perhaps we don’t need to transpose the nested array, but multiply each sub-matrix as if it were a single element, i.e.
i j
[ [12 0 -12] [0 0 6] ] 1 [ 0 6]
1 =
1
v1 = 12 0 -12 4 = 36 = 36i + 6j .
0 0 6 2 6
1
v1 = 12 0 -12 9 = 96 = 96i + 6j .
0 0 6 3 6
1
Let’s solve them all at once:
[pic] [pic] [pic]
Let
[pic]
[pic][pic][pic][pic]
All these values are in meters/second (velocity).
b) Compute the average velocity between t = 1 and t = 2 seconds.
4 0 -12 9 8 1
0 0 6 4 4 1
2 1
1 1
[pic] [pic] [pic]
To subtract r(1) from r(2), define
[pic]
[pic]= 16i +6j
c) Find the speed at t = 1 second.
v1 = 12 0 -12 1 = 0
0 0 6 1 6
1
Speed = v(t) = 0i + 6j = 6 m/s
d) Find the instantaneous acceleration at t = 2 seconds.
a = dv/dt = d[12 0 -12]/dt +d[0 0 6]/dt = [12 0] ( [2 1] + [0 0] ( [2 1] =
[24i + 0j]. Evaluating at t=2 we get
[24 0] 2 = 48i m/s2
1
e) Find the average acceleration between t = 1 and 3 seconds.
aav = v(3) - v(1) (96i + 6j) - 6j 96i
= = = 48i m/s
3 - 1 2 2
[pic] [pic] [pic]
[pic] [pic] [pic]
[pic] [pic]
[pic] [pic]
[pic] [pic]
Let’s solve all possible acceleration states for this example. The first column represents 0-1 seconds, the second column represents 0-2 seconds, the third column represents 0-3 seconds, the last column represents 2-3 seconds. We divide by the number of seconds difference between the two time values.
[pic] [pic]
[pic]
[pic]
[pic]
[pic]
i.e., in 0-1 sec, the average acceleration is 12m/s2, 0-2 = 24m/s2, 0-3= 36m/s2,
1-2sec = 36m/s2, 1-3sec= 48m/s2 and 2-3sec = 60 m/s2.
EXAMPLE: Consider a weight W suspended from two ropes as shown. At what angle will the tensions in T1 and T2 be equal?
( (
T1 T2
W
Computing the forces along the x-axis, we get
Fx = -T1cos( + T2cos( = 0
T1 = T2
Computing the forces acting along the y-axis we get:
Fy = T1sin( + T2sin( - W = 0
Putting the two equations in matrix form, letting the top row represent the forces on x (cos() and the bottom row represent the forces along y (sin()
[pic]
Using Gaussian Reduction and adding row 1 to row 2, we get
[pic]
We want the tensions to be equal, so we will define W = T2; we now have
[pic]
Dividing out T2 in the second row, we get
[pic]
Remembering the second row is the sin row, we get
[pic]= 0
[pic]= 1/2
[pic]= 30o
When both the angles are equal to 30o, the tensions in the ropes will be equal.
b) Find the angle ( which makes T2 twice T1. Find the tension in T1.
60o (
T1 T2
W = mg
Fx = T2cos( - T1cos60o = 0
Fy = T2sin( o + T1sin60 = mg
Row one represents the cos terms, row 2 represents the sin terms
[pic] 60
[pic]
T2 = 2T1,
[pic]
Looking at row one we have
[pic]=[pic]
[pic]=[pic]
[pic]
[pic] RADIANS, OR 75.52 DEGREES
To find the tension in T1, go to the second row of the matrix
[pic]
[pic]=0
[pic]=[pic]
[pic]=[pic]
[pic]=[pic]=[pic]
EXAMPLE: A bar one meter long weighing 100 grams supports a 200g weight at one end and a 75 gram weight at the other end. It is balanced on a fulcrum 20 cm from the 200g weight. Find the center of gravity for this system.
X = point of center of gravity
0 20 100
X
X-20 100-20
200g 100g 75g
[200 75] X - 20 = [200][ 20]
100 - 20
200x-4000 = [200 20] 200X-4000+6000=4000
7500-1500
200X = 8000-6000;
X = 2000/200 = 10cm
EXAMPLE: a) Three particles have masses m1 = 3kg, m2 = 2kg and m3 = 5kg. They lie along a line such that x1 = -4m, x2 = 3m and x3 = 8m. Find the position of the center of mass.
m1 = 3kg m2 = 2kg m3 = 5kg
x1 = -4m x2 = 3m x3 = 8m
Then:
[m1 m2 m3] x1 [3 2 5] -4 [10]-1 = [34][10]-1 = 3.4kg m/kg = 3.4m
x2 3
x3 8
=
[m1 m2 m3] 1
1
1
b) Two particles have mass m1 = .3kg and m2 = .4kg with velocities v1 = (3i +4j)
and v2 = (5i -2j). Find the velocity of the center of mass of this two particle system.
[m1 m2 ] v1 [.3 .4] 3 4 [2.9 .04][.7]-1 =
v2 5 -2
= =
[m1 m2 ] 1 [.3 .4] 1
1 1
[pic]
vcm= (4.143i + .571j)m/s
EXAMPLE: Two forces F1 and F2 act on a mass of 1.5kg. F1 = 12N and is directed 30o below the horizontal, F2 = 5N and is directed 45o above the horizontal. Find the magnitude and direction of the resultant force and the magnitude and direction of it’s acceleration.
[pic] [pic]
[pic]=[pic]
[pic][pic]
FX = 13.927N, FY = -2.465N
F = [pic]
[pic]
[pic]
[pic][pic]
[pic] = [pic] = [pic]
[pic][pic] = [pic]
[pic] = -[pic] (10o below the horizontal)
b) a = F/m = 14N/1.5kg = 9.3m/s2
The acceleration is in the same direction as F.
-----------------------
1Turner, Methods in MO Theory, pp. 107
2Liberales, Introduction to MO Theory, pp. 82-85
3 Dewar, MO Theory of Organic Chemistry
4 Roberts, MO Calculations, pp. 27-28
5 Liberles, Intro. To MO Theory, pp. 119-121
6 Ibid, pp. 114-116
7 Goodrich, A Primer in Quantum Chemistry, pp. 35-36
8 Flurry, MO Theory of Bonding in Organic Molecules, pp. 54-55
9 Roberts, MO Calculations, pp. 53-54
10 Liberles, pp.57
11 Liberles, pp.59
12 Lederle, pp. 98-102 Note: I wrote this paper in a shorter form 20 years ago & didn’t write down the book or publisher.
13 Turner, Methods in MO Theory, pp. 108
14 Roberts, MO Calculations, pp. 106-107
15 Liberles, Introduction to MO Theory, pp. 170-174
16 Goodrich, A Primer of Quantum Chemistry, pp. 98-103
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