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
It can be expressed as
G(t) =(1-t)-1 = 1 + t + t2 + t3 + t4 + ⋯[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) = because it can be expressed as
G(t) =(1-t)-2 = 1 + 2t + 3t2 + 4t3 +⋯ +(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 t2 + Z3 t3 +⋯+Zr tr
G(t) = [Assume |Zt|<1]
So, G(t) = 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.
Generating functions can be used for the following purposes -
Example: Solve the recurrence relation ar + 2-3ar+1 + 2ar = 0
By the method of generating functions with the initial conditions a0 = 2 and a1 = 3.
Let us assume that
Multiply equation (i) by tr and summing from r = 0 to ∞, we haveNow, put a0 = 2 and a1 = 3 in equation (ii) and solving, we get
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) = . Hence, ar = 1 + 2r.
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