Generating Functions | Engineering Mathematics - Civil Engineering (CE) PDF Download

Introduction

Generating function is a method to solve the recurrence relations.
Let us consider, the sequence a0, a1, a2....ar of real numbers. For some interval of real numbers containing zero values at t is given, the function G(t) is defined by the series
G(t)= a0, a1t + a2 t2+⋯+ ar tr+............equation (i)
This function G(t) is called the generating function of the sequence ar.
Now, for the constant sequence 1, 1, 1, 1.....the generating function is

Generating Functions | Engineering Mathematics - Civil Engineering (CE)

It can be expressed as
G(t) =(1-t)-1 = 1 + t + t2 + t3 + t+ ⋯[By binomial expansion]
Comparing, this with equation (i), we get
a0 = 1, a1 = 1, a2 = 1 and so on.
For, the constant sequence 1,2,3,4,5,..the generating function is
G(t) =  Generating Functions | Engineering Mathematics - Civil Engineering (CE)because it can be expressed as
G(t) =(1-t)-2 = 1 + 2t + 3t2  + 4t+⋯ +(r+1) tr Comparing, this with equation (i), we get

a0 = 1, a1 = 2, a2 = 3, a3 = 4 and so on.
The generating function of Zr,(Z≠0 and Z is a constant)is given by
G(t) = 1+Zt + Z2 t+ Z3 t3 +⋯+Zr tr
G(t) =  Generating Functions | Engineering Mathematics - Civil Engineering (CE)[Assume |Zt|<1]
So,  G(t) =  Generating Functions | Engineering Mathematics - Civil Engineering (CE) generates Zr, Z ≠ 0
Also, If a(1)r has the generating function G1(t) and a(2)r has the generating function G2(t), then λ1 a(1)r + λ2 a(2)r has the generating function λ1 G1(t)+ λ2 G2(t). Here λ1 and λ2 are constants.

Application Areas

Generating functions can be used for the following purposes -

  • For solving recurrence relations
  • For proving some of the combinatorial identities
  • For finding asymptotic formulae for terms of sequences

Example: Solve the recurrence relation ar + 2-3ar+1 + 2a= 0
By the method of generating functions with the initial conditions a0 = 2 and a1 = 3.

Let us assume that

Generating Functions | Engineering Mathematics - Civil Engineering (CE)
Multiply equation (i) by tr and summing from r = 0 to ∞, we have

Generating Functions | Engineering Mathematics - Civil Engineering (CE)

Now, put a0 = 2 and a= 3 in equation (ii) and solving, we get

Generating Functions | Engineering Mathematics - Civil Engineering (CE)
Put t = 1 on both sides of equation (iii) to find A. Hence
-1=- A       ∴ A = 1
Put t = 1/2  on both sides of equation (iii) to find B. Hence
1/2 = 1/2 B       ∴ B = 1
Thus G (t) =  Generating Functions | Engineering Mathematics - Civil Engineering (CE). Hence, a= 1 + 2r.

The document Generating Functions | Engineering Mathematics - Civil Engineering (CE) is a part of the Civil Engineering (CE) Course Engineering Mathematics.
All you need of Civil Engineering (CE) at this link: Civil Engineering (CE)
65 videos|120 docs|94 tests

Top Courses for Civil Engineering (CE)

65 videos|120 docs|94 tests
Download as PDF
Explore Courses for Civil Engineering (CE) exam

Top Courses for Civil Engineering (CE)

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Important questions

,

Generating Functions | Engineering Mathematics - Civil Engineering (CE)

,

Extra Questions

,

video lectures

,

ppt

,

Exam

,

MCQs

,

Free

,

Generating Functions | Engineering Mathematics - Civil Engineering (CE)

,

study material

,

Semester Notes

,

past year papers

,

practice quizzes

,

Objective type Questions

,

pdf

,

Sample Paper

,

mock tests for examination

,

Generating Functions | Engineering Mathematics - Civil Engineering (CE)

,

shortcuts and tricks

,

Previous Year Questions with Solutions

,

Viva Questions

,

Summary

;