Mathematics | Generating Functions – Set 2

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 [Tex]h_0, h_1, h_2, ………, h_n, …… [/Tex]e a sequence. Then its exponential generating function, denoted by [Tex]g^e(x) [/Tex]is given by,

[Tex]g^e(x) =\sum_{n=0}^{+\infty} \frac{x^n}{n!} h_n [/Tex]

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

[Tex]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 [/Tex]

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 [Tex]h_n=5h_{n-1}+6h_{n-2} [/Tex].
Given [Tex]h_0 [/Tex]=1 and [Tex]h_1=-2 [/Tex].

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

-5xg(x)= [Tex]-h_0x+h_1 x^2 + h_2 x^3 +……..+ h_n x^n+1+…. [/Tex]

[Tex]6x^2g(x) [/Tex]=[Tex]h_0 x^2+h_1 x^3 + h_2 x^4 +……..+ h_n x^n+2+…. [/Tex]

Adding these 3 quantities we obtain
[Tex](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+….. [/Tex]

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

[Tex](1+5x-6x^2)g(x)=h_0 + (h_1-5h_0)x = (1-7x) [/Tex]

Or g(x)=[Tex]\frac{(1-7x)}{(1+5-6x^2)} [/Tex]

Now [Tex](1+5x-6x^2) [/Tex]=(1-2x)(1-3x)

So, g(x)=[Tex]\frac{(1-7x)}{(1-2x)(1-3x)} [/Tex]

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

Now [Tex]\frac{1}{(1-2x)}=1 + 2x+2^2 x^2 +2^3 x^3+…. +2^n x^n+…… [/Tex]
And [Tex]\frac{1}{(1-3x)}=1 + 3x+3^2 x^2 +3^3 x^3+…. +3^n x^n+…… [/Tex]

So g(x)=[Tex]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+……) [/Tex]

Since this is the generating function for the sequence [Tex]h_0, h_1, ……h_n [/Tex]We observe that [Tex]h_n=5*2^n-4*3^n [/Tex]

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

Links of Various examples are given below regarding generating functions.

Mathematics | Sequence, Series and Summations

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Mathematics | Generating Functions – Set 2

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