I am reading something that considers the following power series in $\mathbb{C}$:
$$ f(z) = z + a_2z^2 + a_3z^3 + a^4z^4 + ... \\ |z| < 1 $$
They define:
$$g(z) = z + b_0 + b_1z^{-1} + b_2z^{-2} + b_3z^{-3} + ... \\ |z| > 1 $$
And assert: $g(z) = \frac{1}{f(z^{-1})}$ and $f(z) = \frac{1}{g(z^{-1})}$. I'm having trouble seeing this though. Ideas?
On
Below is a relatively simple way of looking at the sums as geometric. I assumed that $a_{0}$ and $b_{0}$ are equal to 1. This is a more general way of looking at the problem, just to give you an idea!
This is a geometric power series $\sum_{i=0}^{n}a_{i}z^{i}$. The common ratio is $z^{}$.
From a known formula, we can say that $\sum_{i=0}^{n}a_{i}z^{i}$ = $\frac{1}{1-z_{}}$. Although we cannot necessarily compute the inverse in finite terms, we can say that the inverse, $f(z^{-1})$, converges to
$\frac{-1}{z^{}}+1$.
Then, $\frac{1}{f(z^{-1})}$ = $\frac{1}{\frac{-1}{z^{}}+1}$= $\frac{z^{}}{z^{}-1}$.
Now all we have to do is to prove that $g(z)$ converges to $\frac{z^{}}{z^{}-1}$. The common ratio for $g(z)$ is $z^{-1}$. Using the previous formula, $g(z)$ converges to
$\frac{1}{1-z^{-1}}$=$\frac{z^{}}{z^{}-1}$.
The other argument is similar to the one above. I hope this helps!
By direct calculation, we see that \begin{align} f(z^{-1}) =&\ \frac{1}{z}+a_2\frac{1}{z^2}+a_3\frac{1}{z^3}+\ldots=\ \frac{1}{z}\left(1+a_2\frac{1}{z}+a_3\frac{1}{z^2}+\ldots \right) \end{align} which means \begin{align} \frac{1}{f(z^{-1})} =&\ \frac{z}{1+\left(a_2\frac{1}{z}+a_3\frac{1}{z^2}+\ldots \right)} = z\sum^\infty_{n=0}(-1)^n\left( a_2\frac{1}{z}+a_3\frac{1}{z^2}+\ldots \right)^n\\ =&\ z\left( 1+\sum^\infty_{k=1} b_{k-1}\frac{1}{z^k}\right). \end{align}