Determine the sequence generated by a generating function

Question

$A(z)=2z-1+\frac{1}{2z-2z^2}$

I have no clue how to solve this, I tried looking at other examples but I am just stuck, could anyone be so kind and explain how to solve this step by step?

Answer 1

If you want to see the generating function:

$\begin{array}\\ A(z) &=2z-1+\frac{1}{2z-2z^2}\\ &=2z-1+\frac1{2z}\frac{1}{1-z}\\ &=2z-1+\frac1{2z}\sum_{n=0}^{\infty} z^n\\ &=2z-1+\frac1{2z}+\sum_{n=1}^{\infty} \frac{z^{n-1}}{2}\\ &=2z-1+\frac1{2z}+\sum_{n=0}^{\infty} \frac{z^{n}}{2}\\ &=2z-1+\frac1{2z}+\frac12+\frac{z}{2}+\sum_{n=2}^{\infty} \frac{z^{n}}{2}\\ &=\frac53 z-\frac12+\frac1{2z}+\sum_{n=2}^{\infty} \frac{z^{n}}{2}\\ \end{array} $

If you want to invert the function, that is, find $z$ in terms of $A$:

$\begin{array}\\ A(z) &=2z-1+\frac{1}{2z-2z^2}\\ &=\frac{(2z-1)(2z-2z^2)+1}{2z-2z^2}\\ &=-\frac{4 z^3 - 6 z^2 + 2 z - 1 }{2z-2z^2}\\ \end{array} $

so $-A(2z-2z^2) =4 z^3 - 6 z^2 + 2 z - 1 $ or $4 z^3 - (2A+6) z^2 + (2a+2) z - 1 =0 $.

Wolfy gives some very complicated expressions for $z$ in terms of $A$.