I have to expand $f(z)$ into a formal power series $f(z) = \sum\limits_{k=0}^\infty a_kz^k$ (for $z$ close to 0)
$f(z)= \frac{z^3}{1-4z+3z^2}$
I know that:
$\frac{1}{1-z} = \sum\limits_{k=0}^\infty z^k$
or:
$\frac{1}{(1-z)^2} = \sum\limits_{k=0}^\infty (k+1)z^k$
or the general formula:
$\frac{1}{(1-z)^n} = \sum\limits_{k=0}^\infty \begin{pmatrix} n \\ k \end{pmatrix} 2^k$
Anybody know how can I do that for my function?
Thanks
On
$\bf hint:$ you can use partial fractions. $$\frac 1{1-4z+3z^2} = \frac A{1-z} + \frac B{ 1 - 3z} = 1 + z + z^2 + \cdots + 1 + 3z + 9z^2 + \cdots $$
yoiu find that $A = -\frac 12, B = \frac 32$ so that $$\begin{align} \frac 2{1-4z+3z^2} &= \frac 3{ 1 - 3z} - \frac 1{1-z}\\ & = 3\left( 1 + 3z + 9z^2 + \cdots \right) - \left(1+z + z^2 + \cdots \right)\\ &=2 +8z + 26z^2+ \cdots + (3^{n+1} - 1)z^n+\cdots \end{align}$$
therefore $$\frac {2z^3}{1-4z+3z^2} = 2z^3 +8z^4 + 26z^5+ \cdots + (3^{n+4} - 1)z^n+\cdots $$
You have that $f(z)= \frac{z^3}{1-4z+3z^2} = \frac{z^3}{(1-z)(1-3z)} = - \frac 1 2 z^3 \left(\frac{1}{1-z} - \frac{3} {1-3z} \right)$.
Since $\frac{1}{(1-z)} = \sum\limits_{k=0}^\infty z^k$ and $\frac{1}{(1-3z)} = \sum\limits_{k=0}^\infty 3^k z^k$, you can plug it in and compute the answer.
You get that the coefficient for $z^{k+3}$ is $\frac{3^k-1}{2}$.