Generating Functions (Revised Edition)

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1 Math 700 Fall 06 Notes Generating Functions (Revised Edition What is a generating function? An ordinary generating function for a sequence (a n n 0 is the power series A(x = a nx n. The exponential generating function is defined to be a n n! x n. We define various arithmetical operations for power series: Let A(x = a nx n, B(x = b nx n. Then we define addition, subtraction and multiplication in the following way. (i Addition: (A + B(x = (a n + b n x n. (ii Subtraction: (A B(x = (a n b n x n. (iii Multiplication: (A B(x = ( n i=0 a ib n i x n. Example: Suppose A(x = xn, B(x = nxn. Then ( n (A B(x = (n i x n. Observing that ( n n(n+ i=0 (n i = n + (n + + =, we obtain (A B(x = i=0 n(n + x n. We shall use various power series. We mention a few useful examples below. x = + x + x + x 3 +. x n+ x = + x + x + + x n. + x = ( x = x + x x ( n x n +.

2 x = + x + x 4 + x 6 +. x k = + xk + x k + x 3k +. + x k = xk + x k x 3k + + ( n x nk +. ax = + (ax + (ax + (ax 3 + = + ax + a x + a 3 x a n x n +. ( = + x + 3x + 4x (n + x n + = d x dx e x x n = n!. log( + x = x x + x3 3 x4 4 + = n= n+ xn ( ( x. x We have various tricks and techniques that we use to manipulate power series that will be useful when determining generating functions. n. Finding power series by substitution The first trick involves substitution. For example to find the power series for x power series for = + x + x x +. Now for the variable x substitute x, then x = + (x + (x + = x n. We can use the same trick to find the power series for x. Observe that x = ( x get the power series for x So by substituting x for x in the power series for x. Thus x = + ( x + (x + = x = x n n. x n n. take the. We

3 3 Finding power series using derivatives Another trick involves differentiation of a power series. If f(x is a function having a power series f(x = a nx n, then f (x has a power series given by f (x = (n+a n+x n. Example: Find the power series for ( x. ( d We use the following observation: dx x = ( x = d ( dx x ( x. Thus = d ( + x + x + + x n + dx = + x + 3x + + nx n + = (n + x n. 4 Finding power series using partial fractions Example: Find the power series for 3x+x. To find the power series, we shall the partial fractions method. Observe that 3x + x = ( x( x = A x + B x for some choice of A and B. To find A, and B, cross multiply to get the equation A( x + B( x =. If we set x =, then we get A =. If we set x =, then we get B =. Thus ( 3x + x = x x. Given that = x xn, and = x n x, subtracting these series, we get n 3x + x = ( n+ xn. Example Find the coefficient of x n in G(x = ( x (+x. Solution: We shall use partial fractions. We need to find A, B, C, D such that The above can be rewritten as: ( x ( + x = A x + B ( x + C + x + D ( + x. = A( x( + x + B( + x + C( x ( + x + D( x. 3

4 When x =, we obtain = 4B and B = 4. When x =, we obtain = 4D and D = 4. When x = 0, we obtain = A + C. When x =, we obtain = 3A C. Solving the above two equations for A and C, we obtain that A = B = C = D = 4. Therefore, ( x ( + x = ( 4 x + ( x + + x +. ( + x Now given that the coefficient of x n in each of,,, and is, ( n+ x ( x +x (+x n, ( n, and ( n( n+ n, respectively, we see that the coefficient of x n in equals ( x (+x ( ( ( { n + n + 0, if n is odd; + + ( n + ( n = n+ 4 n n, if n is even. 5 Solving recurrence relations using generating functions We can solve recursions using generating functions, and this is illustrated in the following example. Example: Solve a n = a n +, a 0 = n. To solve this, let A(x = n 0 a nx n. Then using the recursion above we have x n. n a n x n = n (a n + x n = n a n x n + n The sum on the left side equals A(x. On the right side, the first sum is seen to be xa(x and the second sum equals (. Thus we have x Solving for A(x we obtain A(x = xa(x + x. A(x = ( x x = n 0(n + x n. Consequently, a n = n +, n = 0,,,.... 4

5 6 Partitions of an integer We shall examine how many ways one can express a positive integer n as a sum n = n + n + n k where n,..., n k are positive integers. There are two cases: the ordered case where we require that n n n 3 n k and the unordered where the numbers n, n,..., n k need not occur in descending order. 6. Compositions A composition of n is an unordered partition of n. We let c nk denote the number of compositions of n into k parts. For example, c 43 = 3: 4 = + + = + + = + +. Another way to describe c nk is that it is the number of solutions to x + x + + x k = n, x i, i =,..., k. Letting y i = x i, i =,..., k the number of compositions c nk also equals the number of solutions to y + y + + y k = n k, y i 0, i =,..., k. But as was shown before, the number of solutions to (see Example Text is the same as the number of ways of distributing n k balls into k boxes. Thus ( ( n k + k n c nk = =. k k What is the total number of compositions? Recall the Binomial Theorem (see Theorem 9.7. Text: 6. Theorem Binomial Theorem (a + b m = m i=0 ( m a m i b i. i As a consequence of the binomial theorem when a = b =, we obtain the identity m ( m m =. i i=0 5

6 Using the above, the total number of compositions of n equals n c nk = k= n k= ( n k = n. 6. Ordered Partitions We define p k (n to be the number of ordered partitions of a positive integer n into k parts. For example, there are exactly four ordered partitions of 7 into 3 parts: 7 = = = = Thus p 3 (7 = 4. Another way of describing p k (n is that it is the number of solutions to x + x + + x k = n, where x x x k and x,..., x k are positive integers. If we let y i = x i, i =,..., k then p k (n is the number of solutions to y + y + + y k = n k, y y y k 0. The number of solutions to where i of the variables y,..., y k equal zero is equal to p k i (n k (according to. Since i = 0,,..., k, we get the recursion p k (n = p k (n k + p k (n k + + p (n k = k p k i+ (n k. i= The above recursion is useful when calculating p k (n. Observe also, (i p n (n = p (n =. (ii p (n = n. (iii p 3 = { n }. Here {x} denote the integer closest to x. We leave the proofs of (i, and (ii as exercises. The proof of (iii is somewhat more challenging. We define the partition function p(n to be the number of ordered partitions of the integer n; that is, n p(n = p j (n. For example, p(5 = 7 since there are exactly 7 ordered partitions of 5 j= 5 = 4 + = 3 + = = + + = =

7 Let P (x = n= p(nxn be the generating function for (p n. To find the generating function for P (x we do the following: for a partition of n, let α i be the number of parts of size i, i =,,..., n. To a partition of n, we associate the product x α x α x αn n = x n. Using this correspondence, one sees that p(n equals the coefficient of x n in ( + x + x ( + x + x 4 + ( + x 3 + x 6 + ( + x 4 + x 8 + = ( x k. Thus this is exactly the generating function P (x. However, for all practical purposes, it is hard to determine the exact coefficients of P (x. k= 6.3 The Power Series for ( k x What is ( k? We know x ( x k = ( + x + x + k = ( + x + x + ( + x + x + ( + x + x +. When we expand the above, every term x n x n x n k in the expansion where n + n + n n k = n, corresponds to an unordered partition of n into k parts where each part is nonnegative; that is, we allow n i = 0 for any i =,..., k. 6. Theorem ( see Example text The number of partitions of a positive integer n into k nonnegative parts( equals the number n + k of nonnegative integer solutions to x + x + + x k = n which equals. n From the above theorem it follows that ( k = x. ( n + k n x n Thus we have Example What is the coefficient of x n in ( x k? Solution: Let y = x. Then the coefficient of x n in ( k n x equals the coefficient of y in ( ( k n + k y which is n. 7

8 6.4 Restricted Partitions of an Integer In this section, we examine the problem of finding the number of ordered partitions an integer where certain restrictions apply, e.g. partitions only containing odd integers, or partitions only containing distinct integers. Example How many ways can we write the integer n as a sum of 5 positive odd integers. Solution: Let a n be the number of ways of doing this. Note that a n = 0 if n is even. The generating function for a n, n is G(x = (x + x 3 + x 5 + x = (x( + x + x 4 + x = x 5 ( + x + x 4 + x ( 5 = x 5. x Thus we see that a n is the coefficient of x n in x ( 5 5. x Therefore an equals the coefficient of x n 5 in ( ( n 5 ( n x. This coefficient equals n 5 = n 5. Example In how many ways can we fill a bag with n fruits if: the number of applies is even. the number of bananas is a multiple of 5. there can be at most 4 oranges. there can be at most pear. For example, there are 7 ways to fill a bag with 6 fruits as illustrated below. Applies Bananas Oranges Pears Let a n be the number of ways of filling the bag with n fruits. Then the generating function 8

9 for a n is given by G(x = ( + x + x 4 + x 6 + ( + x 5 + x 0 + ( + x + x + x 3 + x 4 ( + x = x x x5 ( + x 5 x = ( x. Therefore, the coefficient of x n in G(x = n + ways to fill a bag with n fruits. ( x is ( n+ n ( = n+ = n +. So there are n Note In the above calculations, we used the fact that + x + x + x 3 + x 4 = x5 x. Suppose we want to calculate the generating function for ordered partitions of n where each part is distinct. For example, there are five different such partitions of 7: 7 = = = 5 + = 6 +. To determine the generating function for these partitions, we observe that for each such partition of n, it holds that α i = 0, ; i =,,... n. Thus the generating function for these partitions is just ( + x( + x ( + x 3 = ( + x k. For example, the partition 7 = has the associated product of terms illustrated below ( + x( + x ( + x 3 ( + x 4. k= To find the generating function for ordered partitions with only odd parts we observe that α i = 0, if i is even. Thus the generating function is ( + x + x + ( + x 3 + x 6 + ( + x 5 + x 0 + = ( x k. k odd Example: In how many ways can a dollar be exchanged for quarters, dimes, and nickels? To solve this, let U(x be the generating function for the partitions which have parts of size 5, 0, and 5. Then U(x = ( + x 5 + x 0 + ( + x 0 + x 0 + ( + x 5 + x Letting y = x 5 we can write this series in terms of y, U(y = (+y +y + (+y +y 4 + (+y 5 +y 0 + = ( y ( y ( y 5. To solve the problem, we need to find the coefficient of y 0 in U(y. To do this, we shall determine the coefficients of the terms, y, y,..., y 0 in ( y ( y 5. We have ( y ( y 5 = ( + y + y y 0 + ( + y 5 + y 0 + y 5 + y

10 We calculate the coefficients of, y, y,, y 0 in the product by multiplying each term in (+y 5 +y 0 +y 5 +y 0 with (+y +y 4 + +y 0, throwing out any terms with y, y,.... Doing this we get + y + y 4 + y y 0 y 5 + y 7 + y y 9 y 0 + y + + y 0 y 5 + y 7 + y 9 y 0 We represent this in table form, each sum represented in a separate row in the table The numbers in the last line of the table represent the coefficients of, y, y,..., y 0 in the product ( y ( y 5. To get the coefficient of y 0 in ( y ( y ( y 5, we simply add the numbers in the last row, which gives us 9. Thus there are 9 different ways of changing one dollar into quarters, dimes, and nickels. 7 Ferrers diagrams Each ordered partition of an integer can be represented as rows of dots in a Ferrers diagram. For example, we represent = with If one transposes a Ferrers diagram, one obtains the Ferrers diagram of another partition, which is called the conjugate partition. For example, if we transpose the above diagram 0

11 we get So the conjugate partition is = Using Ferrers diagrams, the following theorem becomes apparent: 7. Theorem The number of ordered partitions of n into k parts equals the number of ordered partitions of n whose largest part equals k. Proof Using Ferrers diagrams, it is seen that an ordered partition has k parts if and only if its conjugate has largest part equal to k. Another theorem that one can prove using Ferrers diagrams is the following (proof left as exercise: 7. Theorem The number of ordered partitions of n into odd parts equals the number of ordered partitions of n into distinct parts. 8 Exercises. Find the generating function for the number of ways to have n cents in pennies, nickels, and dimes.. Find the generating function for the number of ways to select 0 candy bars from a large supply of 6 different kinds. 3. Find the generating function for the number of ways to select, with repetition allowed, r objects from a collection of n distinct objects. 4. Find the coefficient of x 7 in ( + x + x + x

12 5. Find the coefficient of x 50 in (x 7 + x 8 + x Find the coefficient of x 5 in each of the following: a x 3 ( x 0. b (x 3 5x( x 3. c (+x4 ( x In how many ways can a person select n marbles from a large supply of blue, red, and yellow marbles if the selection must include an even number of blue ones? 8. In how many ways can 3000 identical envelopes be divided in packages of 5, among 4 student groups so that each group gets at least 50, but not more than 000 of the envelopes? 9. Find the generating function F (x for the Fibonacci sequence {a n } where a 0 =, a = and a k = a k + a k, k. Using this function, determine a k. 0. In a certain game it is possible to score,, or 4 points each turn. Find the generating function for the number of ways to score n points in a game in which there are at least two turns where 4 points are scored.. Using generating functions, show that for every positive integer n, the number of partitions of n into parts congruent to ±(mod 6 equals the number of partitions of n into distinct parts congruent to ±(mod 3.. Show the following: (i p n (n = p (n =. (ii p (n = n. 3. Use the previous exercise and the recursion p k (n = k i= p k i+(n k to determine p 3 (6. 4. Show that the number of ordered partitions of n with no parts equal to is p(n p(n. 5. When the following expression is expanded and all like terms are gathered together, how many terms are there: (a + b + c + d + e + f 8? 6. Find the number of solutions to the inequality a + a + a 3 + a 4 68 where the a i s are nonnegative integers. 7. Prove that the number of partitions of n with no part greater than k is equal to the number of partitions of n + k with exactly k parts.

1 Answers to Chapter 8, Odd-numbered Exercises

1 Answers to Chapter 8, Odd-numbered Exercises Answers to Chapter 8, Odd-numbered Exercises ) (a) + 4x + 6x + 4x 3 + x 4. (b) + x + x + x 3 + x 4 + x 7. (c) x 3 + x 4 + 3x 5 + 4x 6 + 6x 7. (d) + x + x + x 3 + x 4 + x 5 + x 6. (e) 3 + x 3 4x 4 + 7x