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.)

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1 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 b k x k, Exercise 37. Recall that the generating function for a sequence {a 0,a,a 2,...,} is X a k x k (Considered formally, i.e. without. consideration of convergence.) X! kx a i b k i x k. ( ) Give the generating functions, in terms of their series and in their simplified (closed) form, for each of the following sequences. In many cases, you ll use your answers from Exercise 36 (from last time). (a) a n = n + for n =0,, 2,... (b) a n =3 n, for n =0,,... ( n =3k for some k 2 Z, (c) a n = for n =0,,... 0, n 6= 3k for all k 2 Z, (i.e. a n is if n is a multiple of 3, and is 0 otherwise). (d) a n =/n for n =, 2,... i=0 Exercise 38. For each of the following recursively defined sequences, (i) solve using generating functions; (ii) solve using the methods of Section 8.2 and compare to part (i); (iii) check your answers by comparing the values you get by using your formula and by computing recursively for the first three terms of the sequence. (a) a n =2a n + 3, a 0 =. (b) a n =3a n +4 n.

2 Exercise 39. (a) Consider the integer partitions of 4. (i) Write a corresponding generating function, wherein the coe integer partitions of 4. cient of x 4 is the number of (ii) Make explicit the correspondence between the monomials in the expansion of your generating function and the integer partitions of 4 (like we did for the partitions of 5 in class). (b) Describe a combinatorial problem that is solved by calculating the coe following generating functions. (i) ( + x + x 2 + x 3 ) 4 (ii) ( + x + x 2 + +)( + x 2 + x 4 + x 6 )( + x 3 + x 6 ) (iii) (x + x 2 + x 3 ) 5 cient of x 6 for the (c) For each of the following questions, give the corresponding generating function and determine k such that the coe cient of x k is the answer the question. (i) How many di erent ways can 2 identical action figures can be given to five children so that each child receives at most three action figures? (ii) How many di erent ways 0 identical balloons can be given to four children if each child receives at least two balloons? (iii) How many di erent ways are there to choose a dozen bagels from three varieties plain, onion, and raisin if at least two bagels of each kind but no more than three plain bagels are chosen? (iv) How many ways can you make change for $00 using $ bills, $5 bills, $0 bills, and $20 bills? Exercise 40. (Generating functions for partitions) (a) Write the generating function for partitions with even-sized parts, i.e. or but not or. (b) Write the generating function for partitions with no more than two parts of each size, i.e. or but not or. (c) Write the generating function for partitions parts all of prime size, i.e. or but not or. (The prime numbers are the integers p greater than that are divisible by and p but no other positive integers.)

3 Taylor series to know and love: nÿ ˆn p ` xq n x k ` k ˆn 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! ˆn x 2 ` `x n 2 (finite) (finite) (infinite) ` (infinite) The lefthand side of each is called the closed form for the series. New Series from old: Let fpxq a k x k and gpxq b k x k. Then f pxq`gpxq pa k ` b k qx k and fpxqgpxq kÿ a i b k i x k. You can also di erentiate and integrate series to get new series. Section 8.4: Generating functions. 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.

4 Section 8.4: Generating functions. 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. Not every generating function has a nice closed form, butthat shouldn t stop you from writing it down. For example, the generating function for the sequence, 0, 2, 0, 3, 0,... is ` 0 x ` 2x 2 ` 0 x 3 ` 3x 4 ` pk ` q x 2k. And the generating function for the sequence0, 0, 2 2, 3 2, 0, 5 2, 0, 7 2,..., i.e. a n n 2 if n is prime and a n 0 otherwise is 2 2 x 2 ` 3 2 x 3 ` 5 2 x 5 ` 7 2 x 2 ` ÿ p 2 x p. p prime 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 a k x k. ˆn x k px`q n. You try Exercise 37 k

5 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!)

6 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 Gpxq a 0 ` a x ` a 2 x 2 ` a 3 x 3 `... a 0 ` xpa ` a 2 x ` q Set aside d terms, a 0 ` x a 0 ` x a 0 ` 3x a n` x n 3a n x n a n x n (where d degree of recurrence) and factor out x d from the rest. Plug in the recurrence relation. Simplify. a 0 ` 3xGpxq. Return to closed form. Now solve for Gpxq: Gpxq a 0 ` 3xGpxq. 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 Now solve for Gpxq: and so for x {3, Gpxq a 0 ` 3xGpxq. a 0 Gpxq 3xGpxq p 3xqGpxq; Gpxq a 0 3x a 0 ÿ 8 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).

7 Ex 2: suppose I have a sequence satisfying a n 9a n 2 ` 0 n 2 with a 0 3 and a 2. Let Gpxq 8 a nx n. Then Gpxq a 0 ` a x ` a 2 x 2 ` a 3 x 3 `... a 0 ` a x ` x 2 pa 2 ` a 3 x ` q Set aside d terms, a 0 ` a x ` x 2 a n`2 x n a 0 ` a x ` x 2 p9a n ` 0 n qx n a 0 ` a x ` 9x 2 8 ÿ a n x n ` x 2 (where d degree of recurrence) 8 ÿ and factor out x d from the rest. Plug in the recurrence relation. p0xq n Expand and simplify. ˆ a 0 ` a x ` 9x 2 Gpxq`x 2. Return to closed forms. 0x Ex 2: suppose I have a sequence satisfying a n 9a n 2 ` 0 n 2 with a 0 3 and a 2. Let Gpxq 8 a nx n. Then ˆ Gpxq a 0 ` a x ` 9x 2 Gpxq`x 2 0x Now solve for Gpxq: ˆ a 0 ` a x ` x 2 Gpxq 9x 2 Gpxq p 9x 2 qgpxq; 0x So Gpxq pa 0 ` a xqp 0xq`x 2 p 0xqp 9x 2 a 0 `pa 0a 0 qx `p 0a qx 2 q p 0xqp ` 3xqp 3xq 3 28x 9x 2 p 0xqp ` 3xqp 3xq ˆ ˆ46 0x ` 39 p 3xq ` 9 0x Review partial fractions decomposition!

8 Ex 2: suppose I have a sequence satisfying a n 9a n 2 ` 0 n 2 with a 0 3 and a 2. Let Gpxq 8 a nx n.then ˆ Gpxq a 0 ` a x ` 9x 2 Gpxq`x 2 0x Now solve for Gpxq: ˆ ˆ46 Gpxq 0x ` 39 p 3xq ` 9 0x Review partial fractions decomposition! Putting back into series form, we get ˆ ˆ46 Gpxq 0 n x n ` p 3q n x n ` 3 n x n 39 9 ˆ ˆ ˆ46 n 0 n ` p 3q n ` 3 x n So a n 0 n ` ˆ ˆ46 p 3q n ` 3 n Try Ex 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?

9 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? The answer is the same as the coe cient of x 0 in... This is something we can plug into a calculator like WolframAlpha:

10 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.

11 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

12 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