Skip to content
Related Articles

Related Articles

Improve Article
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 containingx^n is ^nC_r .For RHS the term containingx^n is ^{(n-1)}C_r+^{(n-1)}C_{r-1} . So ^nC_r=^{(n-1)}C_r+^{(n-1)}C_{r-1} (proved) 

Links of Various examples are given below regarding generating functions. 

  1. GATE CS 2018 | Question 18
  2. GATE-CS-2017 (Set 2) | Question 52


Attention reader! Don’t stop learning now. Learn all GATE CS concepts with Free Live Classes on our youtube channel.

My Personal Notes arrow_drop_up
Recommended Articles
Page :