Solve linear homogeneous recurrent relation with constant coef using generating functions

Question

I have the following linear homogeneous recurrent relation which I have to solve using generating functions. $a_{n+2}-2a_{n+1}-3a_n = 0$
The generating function for this is: $z^{-2}[A(z)-a_0-a_1z]-2z^{-1}[A(z)-a_0]-3A(z)=0$
Next I have to solve for A(z). Can someone describe the detailed steps in order to do this?
The end result should be: $A(z)=\frac{z(2a_0-a_1)-a0}{3z^2-2z-1}$

Answer 1

Just look at this as a linear equation in $A(z)$: $$ z^{-2}(A(z)-a_0-a_1z)-2z^{-1}[A(z)-a_0]-3A(z)=0\\ A(z)(z^{-2}-2z^{-1}-3) = z^{-2}(a_0+a_1z)-2z^{-1}a_0 $$ Dividing by $z^{-2}-2z^{-1}-3$ and simplifying should yield the correct answer.

Answer 2

As an alternative solution to your original equation:
Given the definition of a generating function:
$A(z)=a_{0}+a_{1}\cdot z+a_{2}\cdot z^{2}\ldots={\displaystyle \sum_{n=0}^{\infty}a_{n}\cdot z^{n}}$
$a_{-n}=0|n>0$
You started with the constraint/condition:
$\left(3\cdot z^{2}-2\cdot z+1\right)A(z)=a_{0}+a_{1}\cdot z$
Which means that all three term sums are zero starting from the first three.
Just multiply one by one and line up the seperate coefficients to matching $z^{n} $. This leaves two “initial conditions”/starters undefined as: $a_{0},a_{1}$
And the solution is:
$A(z)=\frac{a_{0}+a_{1}\cdot z}{\left(3\cdot z^{2}-2\cdot z+1\right)}$