Generating Function October 21, 2014

Generating Functions Given an infinite sequence, its ordinary generating function is the power series: ‏G(x) = g 0 + g 1 x + g 2 x 2 + g 3 x 3 + … We will use a double-sided arrow to denote the correspondence between a sequence and its generating function. ‏  g 0 + g 1 x + g 2 x 2 + g 3 x 3 + …

Examples  0 + 0x + 0x 2 + … = 0  0 + 0x + 1x 2 + … = x 2  1 + 2x + 1x 2 + 0x 3 … = (1+x) 2  1 + x + x 2 + x 3 … = 1/(1-x)  1 - x + x 2 - x 3 … = 1/(1+x)  1 + ax + a 2 x 2 + a 3 x 3 … = 1/(1-ax)  1 + x 2 + x 4 … = 1/(1-x 2 )

Operations on Generation Functions Scaling: ‏Given  G(x) ‏Then,  c  G(x) Addition: ‏Given  G(x) ‏and  F(x) ‏Then,  G(x)+F(x)

Operations on Generation Functions Right Shift: ‏Given  G(x) ‏Then,  x k G(x) Differentiation: ‏Given  G(x) ‏Then,  G’(x) k zeroes

Class Discussion Give a closed form generating function for the sequence 1 + x + x 2 + x 3 + … = 1/(1-x) = f(x) 1 + 2x + 3x 2 + … = f’(x) x + 2x 2 + 3x 3 + … = xf’(x) = g(x) x x 2 + … = g’(x) x x x 3 + … = g’(x) But you need to work out g’(x).

Fibonacci Sequence Fibonacci numbers: ‏f 0 = 0 ‏f 1 = 1 ‏f n = f n-1 + f n-2 (n  2) What is the generating function of the Fibonacci sequence? Suppose: ‏  F(x)

Fibonacci Sequence  x  xF(x)  x 2 F(x) Then, ‏  x + xF(x) + x 2 F(x) Therefore, ‏F(x) = x + xF(x) + x 2 F(x) ‏F(x) = x / (1 - x - x 2 )

Fibonacci Numbers  x / (1 – x – x 2 ) We can get a closed form for the n th Fibonacci number by finding the coefficient of x n on the right hand side. We can show that: ‏F(x) = 1/5 0.5 (1/(1-  1 x) – 1/(1-  2 x)) ‏where  1 = 0.5( ) ‏and  2 = 0.5( )

Fibonacci Numbers The coefficient of x n in F(x) is: ‏1/5 0.5  1 n - 1/5 0.5  2 n Therefore,

Counting We can use generating function to count. Consider the following problem: ‏Suppose there are 4 types of fruits: apples, bananas, oranges and pears. In how many ways can we fill a bag with n fruits subject to the following constraints: The number of apples must be even. The number of bananas must be a multiple of 5. There can be at most 4 oranges. There can be at most 1 pear. Sources from Prof. Albert R. Meyer

Counting No. of apples must be even:  1/(1-x 2 ) No. of bananas must be a multiple of 5:  1/(1-x 5 ) At most four oranges:  (1-x 5 )/(1-x) At most one pear:  1 + x

Counting No. of ways to pick n fruits satisfying the constraints can be represented by the generating function: ‏1/(1-x 2 )  1/(1-x 5 )  (1-x 5 )/(1-x)  (1+x) ‏= 1/(1-x) 2 ‏= 1 + 2x + 3x 2 + 4x 3 + … ‏Therefore, no. of ways to pick n fruits satisfying all the constraints is n+1.

Counting Convolution Rule: ‏Let A(x) be the generating function for selecting items from set A, and let B(x) be the generating function for selecting items from set B. If A and B are disjoint, then the generating function for selecting items from the union A  B is the product A(x)  B(x). ‏In the previous example, A can be the set of apples and B the set of bananas.

Class Discussion A woman has a large supply of 1-, 2- and 3- cents stamps. Let a r denote the number of ways to make up a postage of r cents. Give a generating function for. How many ways can she pick the stamps so as to make up a postage of r cents? (1+x+x 2 +x 3 +…..) (1+x 2 +x 4 +…..) (1+x 3 +x 6 +…..) = (1-x) -1 (1-x 2 ) -1 (1-x 3 ) -1

Class Discussion The lifespan of a rabbit is 10 years. Suppose there are 2 newborn rabbits this year and the number of newborns each year is two times that in the previous year. Find the number of rabbits there are in the r th year.