Mathematics | Generating Functions – Set 2

Difficulty Level : Hard
Last Updated : 29 May, 2021

Prerequisite – Generating Functions-Introduction and Prerequisites
In Set 1 we came to know basics about Generating Functions. Now we will discuss more details on Generating Functions and its applications.

Exponential Generating Functions –
Let $h_0, h_1, h_2, ........., h_n, ......$ e a sequence. Then its exponential generating function, denoted by $g^e(x)$ is given by,

$g^e(x) =\sum_{n=0}^{+\infty} \frac{x^n}{n!} h_n$

Example 1:- Let {1, 1, 1…….} be a sequence . The generating function of the sequence is
$g^e(x) = \sum_{n=0}^{+\infty} \frac{x^n}{n!}$ ( Here $h_n$ =1 for all n )
Example 2:- Let $perm{n}{k}$ be number of k permutation in an n- element set. Then the exponential generating function for the sequence $^nP_0, ^nP_1, ......., ^nP_n$ is

$g^e(x) =\sum_{k=0}^{n} \frac{x^n}{n!} ^nP_k = \sum_{k=0}^{n} \frac{x^k}{k!} \frac{n!}{(n-k)!} = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} x_k = \sum_{k=0}^{n} \^nC_k x_k =(1+x)^n$

Exponential Generating Function is used to determine number of n-permutation of a set containing repetitive elements. We will see examples later on.

Using Generating Functions to Solve Recurrence Relations –
Linear homogeneous recurrence relations can be solved using generating function .We will take an example here to illustrate .

Example :- Solve the linear homogeneous recurrence equation $h_n=5h_{n-1}+6h_{n-2}$ .
Given $h_0$ =1 and $h_1=-2$ .

We use generating function to solve this problem. Let g(x) be the generating function of the sequence $h_0, h_1, h_2, ......, h_n, ....$ .
Hence g(x)= $h_0+h_1 x + h_2 x^2 +........+ h_n x^n+....$
So we get the following equations.
g(x)= $h_0+h_1 x + h_2 x^2 +........+ h_n x^n+....$

-5xg(x)= $-h_0x+h_1 x^2 + h_2 x^3 +........+ h_n x^n+1+....$

$6x^2g(x)$ = $h_0 x^2+h_1 x^3 + h_2 x^4 +........+ h_n x^n+2+....$

Adding these 3 quantities we obtain
$(1+5-6x^2)g(x)=h_0 + (h_1-5h_0)x +(h_2-5h_1+6h_0)+....... +(h_n-5h_{n-1}+6h_{n-2})x^n+.....$

Now $h_n-5h_{n-1}+6h_{n-2}$ =0 for all n>1. So,

$(1+5x-6x^2)g(x)=h_0 + (h_1-5h_0)x = (1-7x)$

Or g(x)= $\frac{(1-7x)}{(1+5-6x^2)}$

Now $(1+5x-6x^2)$ =(1-2x)(1-3x)

So, g(x)= $\frac{(1-7x)}{(1-2x)(1-3x)}$

It is easy to see that $\frac{(1-7x)}{(1-2x)(1-3x)}=\frac{5}{(1-2x)}-\frac{4}{(1-3x)}$

Now $\frac{1}{(1-2x)}=1 + 2x+2^2 x^2 +2^3 x^3+.... +2^n x^n+......$
And $\frac{1}{(1-3x)}=1 + 3x+3^2 x^2 +3^3 x^3+.... +3^n x^n+......$

So g(x)= $5(1 + 2x+2^2 x^2 +2^3 x^3+.... +2^n x^n+......)-4(1 + 3x+3^2 x^2 +3^3 x^3+.... +3^n x^n+......)$

Since this is the generating function for the sequence $h_0, h_1, ......h_n$ We observe that $h_n=5*2^n-4*3^n$

Thus we can solve recurrence equations using generating functions.

Proving Identities via Generating Functions –
Various identities also can also be proved using generating functions.Here we illustrate one of them.

Example: Prove that : $^nC_r=^{(n-1)}C_r+^{(n-1)}C_{r-1}$
Here we use the generating function of the sequence $^nC_0, ^nC_1, ......^nC_r....$ i.e $(1+x)^n$ .
Now, $(1+x)^n=(1+x)^{n-1}(1+x)=(1+x)^{n-1}+x(1+x)^{n-1}$
For LHS the term containing $x^n$ is $^nC_r$ .For RHS the term containing $x^n$ is $^{(n-1)}C_r+^{(n-1)}C_{r-1}$ . So $^nC_r=^{(n-1)}C_r+^{(n-1)}C_{r-1}$ (proved)