Section 8.4: Generating functions.

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Pearl O’Neal’
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1 Section 8.4: Generating functions. Taylor series to know and love: nÿ ˆn ˆn p ` xq n x k ` k x ` x n n x ÿ x k ` x ` x 2 ` `x n x e x x k ` x ` x 2 ` x k {k! ` x ` x2 2 ` x3 3! Combining series: Let f pxq f pxq`gpxq ˆn x 2 ` `x n 2 a k x k and gpxq pa k`b k qx k and fpxqgpxq (finite) (finite) (infinite) ` (infinite) b k x k.then kÿ a i b k i x k. Example: Compute the series for p xq 2 using x 8 xk. Approach: Use derivatives, noting that d dx Thus, x d dx p xq p qp xq 2 p q p xq 2 d dx p xq d x k dx d dx xk kx k 0 ` ` 2x ` 3x 2 ` p xq 2. pj ` qx j (reindexing by j k ). j 0

2 A generating function for a sequence ta k u,,... is the series a k x k. ( Formal : forget about convergence!) When possible, we rewrite the generating function in terms of a simple expression of elementary functions, which we call closed solutions. For example, the generating function for the sequence,,, tu,,... is x k x. The generating function for the sequence, 2, 6, t{n!u is x n {n! e x. A generating function for a sequence ta k u,,... is the series a k x k. ( Formal : forget about convergence!) When possible, we rewrite the generating function in terms of a simple expression of elementary functions, which we call closed solutions. Note that a finite sequence a 0,a,...,a n is the same as the infinite sequence a 0,a,...,a n, 0, 0,...; similarly, the generating function for a finite sequence will be a finite degree polynomial: a k x k a 0 ` a x ` a 2 x 2 ` a n x n ` 0 ` 0 ` For example, for a fixed n, the generating function for the sequence t`n k u,,...,n is nÿ ˆn x k px`q n. k a k x k.

3 First application: solving recurrence relations Take a generating function for some sequence ta n u: Notice that Gpxq a n x n a 0 ` a x ` a 2 x 2 `. xgpxq a 0 x ` a x 2 ` a 2 x 3 ` x 2 Gpxq a 0 x 2 ` a x 3 ` a 2 x 4 `. n n 2 x d Gpxq a 0 x d ` a x d` ` a 2 x d`2 ` a n x n a n 2 x n n d a n d x n. (Rewrite sums so that the power of x matches the index, tomake it easier to collect like terms when adding series!) First application: solving recurrence relations So say I have a sequence ta n u that satisfies the recurrence relation a n 3a n. (Sanity check: we already know the general solution should look like a n a 0 3 n.) Let Gpxq 8 Gpxq 3xGpxq So a n x n 3 a 0 ` a 0 ` n n n a n x n a n x n 3 pa loooooomoooooon n 3a n q 0 a nx n. Then n x n a n x n (add and collect like terms) a 0. (use recursion relation to simplify) a 0 Gpxq 3xGpxq p 3xqGpxq; (line up indices)

4 First application: solving recurrence relations So say I have a sequence ta n u that satisfies the recurrence relation a n 3a n. (Sanity check: we already know the general solution should look like a n a 0 3 n.) Let Gpxq 8 a nx n. Then So and so for x {3, Gpxq 3xGpxq a 0 a 0 Gpxq 3xGpxq p 3xqGpxq; Gpxq a 0 3x a 0 a 0 p3xq n ˆ y ˇˇˇˇy 3x pa 0 3 n qx n. Now compare to the original formula for Gpxq! This shows that a n a 0 3 n as expected.

5 Last time, we looked at the recursion relation 0 a n 3a n and then started by simplifying Gpxq 3xGpxq. Eventually, we found that all but one term looked like a n 3a n and were able to cancel. You can also start with Gpxq, plug in the recursion relation and expand (instead of starting with an expression and simplifying). For example, suppose I have a sequence satisfying a n 8a n ` 0 n with a 0. Let Gpxq 8 a nx n (same as before). Then Gpxq a n x n a 0 ` a n x n a 0 ` ` 8 n n n p8a n ` 0 n qx n a n x n ` n (plug in recursion rel.) 0 n x n (separate series) For example, suppose I have a sequence satisfying a n 8a n ` 0 n with a 0. Let Gpxq 8 a nx n (same as before). Then Gpxq a n x n a 0 ` a n x n a 0 ` ` 8 ` 8x n n n p8a n ` 0 n qx n a n x n ` a n x n ` x n (plug in recursion rel.) 0 n x n (separate series) 0 n x n (turn into something we know) ` 8xGpxq`x. (put into closed form) 0x So, collecting Gpxq s to one side, we get Gpxqp 8xq Gpxq 8xGpxq ` x 0x 0x ` x 0x 9x 0x.

6 For example, suppose I have a sequence satisfying a n 8a n ` 0 n with a 0. Let Gpxq 8 a nx n (same as before). Then Gpxq `8xGpxq`x 0x So, collecting Gpxq s to one side, we get Gpxqp 8xq Gpxq 8xGpxq ` x 0x 0x ` x 0x So 9x Gpxq p 0xqp 8xq ˆ 2 0x ` 8x 9x 0x. (Review partial fractions decompositions over spring break!) n 0 n x n ` 8 n x 2 2 p0n ` 8 n qx n. Therefore a n 2 p0n ` 8 n q.

7 Counting problems and Generating functions Example: What is the coe cient on x 2 in px looooooooooomooooooooooon 2 ` x 3 ` x 4 ` x 5 q px loooomoooon 4 ` x 5 q px looooooomooooooon ` x 2 ` x 3 q? e, glazed e 2, choc. e 3, jelly This is equivalent to the question How many integer solutions are there to the equation e ` e 2 ` e 3 2 with 2 e 5, 4 e 2 5, e 3 3? 2 Which is the same as How many ways can you pick 2 doughnuts to bring to the o ce if you ve had requests for at least 2 glazed, 4 chocolate, and one jelly-filled, but when you get to the store, they only have 5 glazed, 5 chocolate, and 3 jelly-filled left? Example: Use a generating function to answer the question How many non-negative integer solutions are there to e ` e 2 ` e 3 0 where e 2 is a multiple of 2 and e 3 is a multiple of 3? The answer is the same as the coe cient of x 0 in p loooooooooomoooooooooon ` x ` x 2 ` qp looooooooooooooomooooooooooooooon ` x 2 ` x 4 ` x 6 ` qp looooooooooooooomooooooooooooooon ` x 3 ` x 6 ` x 9 ` q e e 2 e 3, which is the same as the coe cient of x 0 in p loooooooooooooomoooooooooooooon ` x ` x 2 ` `x 0 q p looooooooooooooomooooooooooooooon ` x 2 ` x 4 ` `x 0 q p loooooooooomoooooooooon ` x 3 ` x 6 ` x 9 q e e 2 e 3, since we would never use any terms that came from x a for a 0. This is something we can plug into a calculator like WolframAlpha.

Example: Use a generating function to answer the question How many non-negative integer solutions are there to e ` e 2 ` e 3 0 where e 2 is a multiple of 2 and e 3 is a multiple of 3?

8 Example: Use a generating function to answer the question How many non-negative integer solutions are there to e ` e 2 ` e 3 0 where e 2 is a multiple of 2 and e 3 is a multiple of 3? The answer is the same as the coe cient of x 0 in... This is something we can plug into a calculator like WolframAlpha: Integer partitions How many integer partitions are there of 5? This is the same as the coe cient of x 5 in p ` x ` x 2 ` x 3 ` x 4 ` x 5 qp ` x 2 ` x 4 qp ` x 3 qp ` x 4 qp ` x 5 q `px q 0 `px q `px q 2 `px q 3 `px q 4 `px q 5 (pts of length ) `px 2 q 0 `px 2 q `px 2 q 2 (pts of length 2) `px 3 q 0 `px 3 q (pts of length 3) `px 4 q 0 `px 4 q (pts of length 4) `px 5 q 0 `px 5 q (pts of length 5) Why? For example, consider the partition.

9 Integer partitions Counting integer partitions of 5 by looking at the coe. of x 5 in p ` x ` x 2 ` x 3 ` x 4 ` x 5 qp ` x 2 ` x 4 qp ` x 3 qp ` x 4 qp ` x 5 q... corresponds to px q 2 from first factor, since there are 2 parts of length, px 2 q 0 from second factor, since there are 0 parts of length 2, px 3 q from third factor, since there is part of length 3, px 4 q 0 from fourth factor, since there are 0 parts of length 4, and px 5 q 0 from fourth factor, since there are parts of length 5. Integer partitions Counting integer partitions of 5 by looking at the coe. of x 5 in p ` x ` x 2 ` x 3 ` x 4 ` x 5 qp ` x 2 ` x 4 qp ` x 3 qp ` x 4 qp ` x 5 q... corresponds to px q 2 px 3 q. Similarly, the correspondence between the other partitions of 5 and the monomials goes like px 5 q x px 4 q px 2 q px 3 q px q px 2 q 2 px q 3 px 2 q px q 5

10 Notice that the coe cient of x 5 in the polynomial from the previous slide is the same as the coe cient of x 5 in i 2i 3i 4i 5i x x x x x 5π k x ki 5π k Which is the same as the coe cient of x 5 in i 2i 3i x x x ˆ x k. x 4i 5i 6i 7i x x x loooomoooon loooomoooon must use the term must use the term Notice that the coe cient of x 5 in the polynomial from the previous slide is the same as the coe cient of x 5 in i 2i 3i 4i x x x x 5π ki 5π ˆ x x k. k k Which is the same as the coe cient of x 5 in i 2i 3i x x x 8π k x 5i x ki So in general, the number of integer partitions of n, denoted ppnq, is the coe cient of x n in 8π ˆ ppnqx n x k. k

Math 365 Monday 3/25/19. 1 x n n 1. 1 x = X. a k x k (Considered formally, i.e. without consideration of convergence.)

Math 365 Monday 3/25/19. 1 x n n 1. 1 x = X. a k x k (Considered formally, i.e. without consideration of convergence.) Recall: Also, for series we have f(x)+g(x) = Math 365 Monday 3/25/9 ( + x) n = e x = f(x) = nx X x k /k!, n x k, k and x n n x = X x k, x = X X a k x k and g(x) = X (a k + b k )x k and f(x)g(x) = x k X