This paper review the method of determining the coefficients of a generating function. Generating functions are a convenient tool for handling special constraints in selection and arrangement problems. It can be used in recurrence relations, inclusion exclusion events study, and polya’s enumeration formula. It may also help to solve some other combinatorial problems. Generating functions are a kind of abstract problem-solving technique once we understand it may easy to model a broad spectrum of combinatorial problems. In this paper, we will usesome vivid examples to demonstrate both the theoretical and applicable results of generating function

Chen, William W.S.

English

This paper reviews the method of determining the coefficients of a generating function. Generating functions are a convenient tool for handling special constraints in selection and arrangement problems. It can be used in recurrence relations, inclusion exclusion events study, and polya’s enumeration formula. It may also help to solve some other combinatorial problems. Generating functions are a kind of abstract problem-solving technique once we understand it may easy to model a broad spectrum of combinatorial problems. In this paper, we will use some vivid examples to demonstrate both the theoretical and applicable results of generating function

William W.S. Chen

NEUROSURGERY ENTHUSIASTIC WOMEN SOCIETY

On Finding Coefficient Of Generating Function

Scitech Research Journals

Journal of Progressive Research in Mathematics(JPRM) ISSN: 2395-0218     Volume 16, Issue 1  available at www.scitecresearch.com/journals/index.php/jprm                 2837|   SCITECH                                                                         Volume 16, Issue 1 RESEARCH ORGANISATION                    Published online: February 12, 2020| Journal of Progressive Research in Mathematics www.scitecresearch.com/journals    On Finding Coefficient Of Generating Function            William W.S. Chen           Department of Statistics             The George Washington University            Washington D.C. 20013            E-mail: williamwschen@gmail.com   Abstract This paper reviews the method of determining the coefficients of a generating function. Generating functions are a convenient tool for handling special constraints in selection and arrangement problems. It can be used in recurrence relations, inclusion exclusion events study, and polya’s enumeration formula. It may also help to solve some other combinatorial problems. Generating functions are a kind of abstract problem-solving technique once we understand it may easy to model a broad spectrum of combinatorial problems. In this paper, we will use some vivid examples to demonstrate both the theoretical and applicable results of generating function.  Keywords and Phases: A sum of 25 when 10 dice are rolled, Generating functions, Inclusion exclusion events study, Polya’s enumeration formula, Recurrence relation, Selection and arrangement problems, Two equal unordered piles.    1. Introduction We consider three objects labeled , ,  1 2 3x x and x . From the algebraic Product (1 )(1 )(1  )1 2 3x t x t x t    multiplied out and arranged in powers of t, that is 2 31 ( +  ) ( +  ) ( )1 2 3 1 2 1 3 2 3 1 2 3x x x t x x x x x x t x x x t      or, in the notation of symmetric function Journal of Progressive Research in Mathematics(JPRM) ISSN: 2395-0218     Volume 16, Issue 1  available at www.scitecresearch.com/journals/index.php/jprm                 2838|  2 311 2 3a t a t a t    where , ,  1 2 3a a and a  are the elementary symmetric functions of the three variables , ,  1 2 3x x and x . These symmetric functions are identified by the equation ahead, and it will be noticed that ,  r= 1 ,2 ,3 ,ra  contains one term for each combination of the three things taken r at a time. Hence, the number of such combinations is obtained by setting each ix  to unity, that is, 33(1 ) (1,1,1) .0rt a trr  In the case of n distinct things labeled 1 n to  x ,x  it is clear that (1 )(1 )(1  ) . . . . . . . .(1 )1 2 3x t x t x t x tn     21 ( , ,....... ) ....... ( , ,....... ) . ....... . ( , ,....... )1 1 2 2 1 2 1 2na x x x t a x x x t a x x x tn n n n       and  (1 ) (1,1,.....1) ( , ) .0 0n nn r rt a t C n r trr r     where ( , )C n r  is binomial coefficient. The number of ways to select an r-subsets from an n set. The expression (1 )nt  is called the enumerating generating function or simply the enumerator, of combinations of n distinct things. The problem of determining the coefficient of rt when we multiply a group of polynomial factors together can be stated in terms of exponents. For example, what is the coefficient of 4t  in the expansion of 3 2 2(1 ) (1 )t t t   . It is the number of different formal products whose sum of exponents is 4. Determining this coefficient can be modeled as an integer solution to an equation problem. The coefficient of 4t  in that form is the number of integer solution to 4  1 2 3 4 5e e e e e      w h e re   0 e ,e ,e 1 , 1 2 3  n d   0 e ,e 24 5a   . The variable ei stands for the value of exponents of the ith term in the formal product of the two polynomials. The integer solution to an equation problem is equivalent to the problem of selecting four objects from a collection of five types of objects, where there is one object of the first type, one of the second type, one of the third type, and two of each of fourth and fifth types. It is also equivalent to the problem of distributing four identical objects into five distinct boxes with at most one object in each of the first three boxes and at most two objects in the last two boxes. More generally, the coefficient of rt in 3 2 2(1 ) (1 )t t t    will be the number of integer solutions to  1 2 3 4 5e e e e e r      w h e re   0 e ,e ,e 1 , 1 2 3  n d   0 e ,e 24 5a   . We could repeat all the above statements but replace 4 by r. Thus 3 2 2(1 ) (1 )t t t   is the generating function for ra , the number of Journal of Progressive Research in Mathematics(JPRM) ISSN: 2395-0218     Volume 16, Issue 1  available at www.scitecresearch.com/journals/index.php/jprm                 2839|  ways to select r objects from the given collection of five types or to perform the equivalent distribution. In this paper, our objection is break down distinct generating function and finding the ra number. In each different case, we demonstrate by some examples.  2. Generating Functions We are developing algebraic techniques for calculating the coefficients of generating functions. All these methods seek to reduce a given generating function to a either simple binomial type or negative binomial type generating function. It might also be the product of these two types generating functions. The example (2.1), (2.2) and (2.3) demonstrate the standard of these three applications.    Example 2.1: Find the coefficient of 1 2x in 321 3xx x .     32 3 5 3 51 32 2x A Bx xx x        3 5 3 53 ( ) ( )2 2x A x B x       To determine constant A and B, we let 3 5 3 5  o r    2 2x x    Solve A and B, we get 1 9 1 9  a n d  B =2 22 5 2 5A     Therefore,  1 9 1 9 1 9 1 9 3 1 1 1 12 2 2 22 5 2 5 2 5 2 5,    a  c =   a  c1 1 2 22 3 5 3 5 3 5 3 5 3 5 3 51 3 1 13 5 3 52 2 2 2 2 22 2xx xx x                    3 1 1(1 ) (1 )1 1 2 221 3xa c x a c xx x       1 1 1 1 1{1 ( ) .. . . . . . ( 1) ( ) .. . .}1 1 111 1 1 1 1  {1 ( ) .. . . . . . ( 1) ( ) .. . .}2 2 21rr ra c x c xrrr ra c x c xr                                    , we are seeking coefficient of 1 2x  is given by,  1 9 1 91 2 1 22 22 21 2 1 2 1 2 1 22 5 2 5( 1) ( ) ( 1) ( )1 2 1 23 5 3 5 3 5 3 52 21 2 1 2( 0 .9 5 9 6 7 5 ) * 0 .3 8 1 9 6 6 ) ( 3 .9 5 9 6 7 ) * ( 2 .6 1 8 0 3 3 9 ) 0 3 .9 5 9 6 7 5 * 1 0 3 6 8 2 4 1 0 5 4 7                       Alternative way, we can use the recurrence relation,  Journal of Progressive Research in Mathematics(JPRM) ISSN: 2395-0218     Volume 16, Issue 1  available at www.scitecresearch.com/journals/index.php/jprm                 2840|  3 2.........0 1 221 3xa a x a xx x     2 23 (1 3 )( .........)0 1 22 3( 3 ) ( 3 ) ( 3 ) ....... .0 1 0 2 1 0 3 2 1.x x x a a x a xa a a x a a a x a a a x                 3 ,  a -3 a = 1 ,  a 1 0 ,   a 3 0 ,  a 3 * 1 0 3 2 70 1 0 1 2 1 0 2a a a          a 3 3 * 2 7 1 0 7 1,  a 3 3 * 7 1 2 7 1 8 6 ,  3 2 1 4 3 2a a a a            a 3 3 * 1 8 6 7 1 4 8 7 , a 3 3 * 4 8 7 1 8 6 1 2 7 5 ,5 4 3 6 5 4a a a a           a 3 3 * 1 2 7 5 4 8 7 3 3 3 8 ,   a 3 3 * 3 3 3 8 1 2 7 5 8 7 3 9 ,  7 6 5 8 7 6a a a a           a 3 3 * 8 7 3 9 3 3 3 8 2 2 8 7 9 ,   a 3 3 * 2 2 8 7 9 8 7 3 9 5 9 8 9 8 ,  9 8 7 1 0 9 8a a a a           a 3 3 * 5 9 8 9 8 2 2 8 7 9 1 5 6 8 1 5 ,   a 3 3 * 1 5 6 8 1 5 5 9 8 9 8 4 1 0 5 4 7 ,1 1 1 0 9 1 2 1 1 1 0a a a a           We often need to go through some computations of the generating function to find the required coefficients. The most common situation is multiplication. If 2 2h (x )= f(x )*g (x ),  w h ere  f(x )= a + a x + a x ..... ; g (x )= b + b x + b x .....0 1 2 0 1 2     2 rth en   h  (x )= a b + (a b + a b )x + (a b + a b + a b )x ..... (a b + a b + a b + ....+ a b )x ...0 0 1 0 0 1 2 0 1 1 0 2 r 0 r-1 1 r-2 2 0 r    We give some examples to demonstrate the usefulness of this product formula. Example 2.2. How many ways are there to get a sum of 25 when 10 dice are rolled? The generating function is giving by, 1 02 3 4 5 6 1 0 6 1 0 1 0 1 0 6 1 0( ) ( ) (1 ) (1 ) (1 )1xg x x x x x x x x x x xx               We need coefficient of 1 5x in 1 0 6 1 0(1 ) (1 )x x   1 1 0 1 2 1 0 1 1 0 11 0 2 f(x )= (1 ) 1 ..... ....1 2r rle t x x x xr                           1 0 1 0 1 06 1 0 6 6 2 6 6 1 0 f(x )= (1 ) 1 ( ) ..... ( 1) ( ) ....( 1) ( )1 2nr r nle t x x x x xr n                              coefficients of 1 5x  are 1 5 1 0 1 1 0 9 1 0 1 1 0 3 1 0 18 3 1 2 0 41 5 1 9 2 3                             Example 2.3. How many ways are there to divide five pears, five apples, five doughnuts, five lollipops, five chocolate cats, and five candy rocks into two equal unordered piles? The generating function of six piles is given by  ,  a 3 0n 1 2in g en e ra l a an n   Journal of Progressive Research in Mathematics(JPRM) ISSN: 2395-0218     Volume 16, Issue 1  available at www.scitecresearch.com/journals/index.php/jprm                 2841|  612 3 4 5 6 6 6 6 6( ) (1 ) ( ) (1 ) (1 )1xg x x x x x x x xx            1 6 1 2 6 1 6 1 6 6 6 62 6 1 2 6 6 3 6{1 ..... .....} {1 .....( 1) .. ( 1) }1 2 1 2 6r r r rx x x x x x xr r                                                        coefficient of 1 5x are 1 5 6 1 6 9 6 1 6 3 6 14 3 3 21 5 0 9 1 3 2 1 5 1 9 2 3a b a b a b                                      so left or right pile is 12 of coefficient 1 5x . We may sometimes meet the situation there are more than two product factors. The next example will give a technique how to handle the Problem. Example 2.4. Find the coefficient of rx  in 2( ) (1 ) (1 )n nf x x x x     This generating function can be rewritten as 31 3( ) (1 ) (1 ) (1 ) (1 )1nx n n n nf x x x x xx          , a product of three functions. It is very messy to try to apply the product rule to a multiple product repeatedly. Instead we view the two products as 2n-type factors: 2 2 2(1 )(1 )......(1 )(1 )1 )....1 )    (1 )               (2 )                     (n )                (1 )     (2 )       (n )x x x x x x x x x        , we can select k of 2x -type among the first n factors in    nk ways of exponent 2k. Since total exponents are assume r therefore there are r-2k remains for type x and there is 2n-k factor remain to be selected. There is 22n kr k    way for type-x factor to be selected. The coefficient of rx  will be 22m a x ( 2 ,0 )n n n ka rk r kk r n            A partition of a group of r identical objects divides the group into a collection of unordered subsets of various sizes. Analogously we define a partition of the integer r to be a collection of positive integers whose sum is r.  Usually we write this collection as a sum and list the integers of the partition in increasing order. We will give an example to illustrate the application. Example 2.5. Find a generating function ( , , )g x y z  whose coefficient of r s tx y z  is the number of ways n people can each pick two different fruits from a bowl of apples, Journal of Progressive Research in Mathematics(JPRM) ISSN: 2395-0218     Volume 16, Issue 1  available at www.scitecresearch.com/journals/index.php/jprm                 2842|  oranges, and bananas for a total of r apples, s oranges, and t bananas. For each person, we consider x y y z z x  . Hence n people we have the generating function ( ) .nx y y z z x   Let us assume that the coefficient of r s tx y z  is na  and partition of a positive integer n has . . . .   w h e re  1 e .. . .1 2 1 2e e e n e ek k        . Except a recurrence relation we will prove in the appendix, there is no general rule to help us to count the number of ways to partitions a positive integer. We give a list of the number of ways to partition the first eight positive integer in the tables below.                           Table 2.1   N 1 2 3 4 5 6 7 8 R 1 2 3 5 7 11 15 22  For example, n=8, there are 22 ways to partition 8. We list Them as follow: 8,7+1,6+1+1,5+1+1+1,4+1+1+1+1,3+1+1+1+1+1,2+1+1+1+1+1+1, 1+1+1+1+1+1+1+1,6+2,5+2+1,4+2+1+1,3+2+1+1+1,2+2+1+1+1+1, 5+3,4+3+1,3+3+1+1,2+2+2+1+1,4+4,4+2+2,3+2+2+1,3+3+2, 2+2+2+2.  3. Concluding Remarks  The concept of a generating function is a convenient tool for handling special constraints in selection and arrangement problems. They are a simpleminded and a sophisticated mathematical model for counting problems; simpleminded because polynomial multiplication is a familiar, well-understood part of high school algebra and sophisticated since with standard algebraic manipulations on generating function, we can solve complicated counting problems that usually cannot be solved by the combinatorial arguments. These algebraic manipulations automatically perform the correct combinatorial reasoning for us. In this paper, we first use the regular hand calculator to compute data. Then we recomputed data by R-package.        Journal of Progressive Research in Mathematics(JPRM) ISSN: 2395-0218     Volume 16, Issue 1  available at www.scitecresearch.com/journals/index.php/jprm                 2843|  4. References [1] Feller, W. (1957) An Introduction to Probability Theory and its Applications. Volume 1. Second Edition, John Wiley & Sons, Inc. [2] Feller, W. (1965) An Introduction to Probability Theory And Its Applications. Volume II. John Wiley & Sons.  [3] Riordan, J. (1958) An Introduction to Combinatorial Analysis. John Wiley & Sons. New York.  [4] Tucker, A. (1980) Applied Combinatorics. John Wiley & Sons. [5] Crawley, M.J. (2007) The R Book. Imperial College London at Silwood Park, UK. John Wiley & Sons, Ltd.  5. Appendix Let ( , )R r k  denote the number of partitions of the integer r into k parts. Show that ( , ) ( 1, 1) ( , )R r k R r k R r k k           . . . .   w h e re  1 e .. . .1 2 1 2e e e r e ek k         Case 1. Assume 11e   and . . . . 1      w h e re  1 .. . .2 2e e r e ek k                  Hence there are ( 1, 1)R r k   ways. Case 2. / / /1,     let e 1,  e 1,  ......e 1,1 1 21 2 ke e e ek                 / / / / / /   . ...      w h ere  1 ....1 2 1 2th en e e e r k e e ek k                Hence there are ( , )R r k k  ways.    We further breakdown table 2.1, for each given number r       K: number of parts.          r: Given number         k: Number of parts.      Journal of Progressive Research in Mathematics(JPRM) ISSN: 2395-0218     Volume 16, Issue 1  available at www.scitecresearch.com/journals/index.php/jprm                 2844|                             k r\\k 1 2 3 4 5 6 7 8 Row sum 1 1        1 2 1 1       2 3 1 1 1      3 4 1 2 1 1     5 5 1 2 2 1 1    7 6 1 3 3 2 1 1   11 7 1 3 4 3 2 1 1  15 8 1 4 5 5 3 2 1 1 22  

ON FINDING COEFFICIENT OF GENERATING FUNCTION

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ON FINDING COEFFICIENT OF GENERATING FUNCTION

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