In an exercise I am asked to find the power series of the function $f(z)=\frac{1}{z^2+1}$ centered at the point $1$.
My approach:
$$ \begin{align} \frac{1}{z^2+1} & = \frac{1}{z^2+ 2 - 1} \\ \\& = \frac{1}{2}\cdot\frac{1}{1-\frac{1}{2}(1-z^2)} \\ \\& = \frac{1}{2} \sum_{n \geq 0} \frac{1}{2^n}(-1)^n(z^2-1)^n \\ \\& = \sum_{n \geq 0} \frac{(-1)^n}{2^{n+1}}(z^2-1)^n \end{align} $$
Using the Binomial theorem we get that:
$$\begin{align} \sum_{n \geq 0} \frac{(-1)^n}{2^{n+1}}(z^2-1)^n & = \sum_{n \geq 0} \frac{(-1)^n}{2^{n+1}}\sum_{k=0}^n {n \choose k} (-1)^{n-k} z^{2k} \\ \\ & = \sum_{n \geq 0}\sum_{k=0}^n \frac{(-1)^{2n-k}}{2^{n+1}} {n \choose k} z^{2k} \\ \\ & = \sum_{k \geq 0}\underbrace{\sum_{n\geq k} \frac{(-1)^{2n-k}}{2^{n+1}} {n \choose k}}_{:=b_k} z^{2k} \\ \\ & = \sum_{k \geq 0} b_k z^{2k} \end{align}$$
I got this result but I think that this is wrong for the following reason:
- A power series of a function centered at $a$ is written as: $\sum a_n (z-a)^n$, and I did not end up with something of that form
So how can I solve this problem and find a power series for the function $f$ centered at $a=1$?
hint
$$f(z)=\frac{1}{1+z^2}$$
$$g(z)=f(z+1)=\frac{1}{1+(z+1)^2}$$
$$=\frac{1}{2+2z+z^2}$$ $$=\frac 12 \frac{1}{1+z+\frac{z^2}{2}}$$ $$=\frac 12\frac{1}{1-X}$$ with $$X=-z-\frac{z^2}{2}$$
expand $ g(z)$ around $ z=0 $ as $$\frac 12(1+X+X^2+...)=\sum_{n=0}^\infty a_nz^n$$ to get the expansion of $ f $ around $ z=1 $ by using $$f(z)=g(z-1)=\sum_{n=0}^\infty a_n(z-1)^n$$