Raise a Laurent series to a power

Question

I want to know if there is a general rule by which a Laurent series can be raised to a power; in other words,I have the Laurent series for $f(z) =\dfrac{1}{z(z-1)(z-2)}$ about a certain pole, and I want to find the series for $f(z)$ and cube it, to get a series for $f(z)^3$, $f(z)$ cubed, about that pole. Does anyone know if there is a formula for this?

Answer 1

As copper.hat remarks, you can find the coefficients by convolution. That is, if $f(z)=\sum_{n\in\mathbb Z}a_n z^n$ and $g(z)=\sum_{n\in\mathbb Z}b_n z^n$ are two Laurent series that converge in some annulus $A=\{r<|z|<R\}$, then $f(z)g(z)=\sum_{n\in\mathbb Z}c_n z^n$ where $c_n=\sum_{k\in\mathbb Z} a_{k}b_{n-k}$. It's worth noting that the series defining $c_n$ actually converges. For example, as $k\to +\infty$, we have $a_k=O((R-\epsilon)^{-k})$ and $b_{n-k}=O((r+\epsilon)^{n-k })$ for any $\epsilon>0$. Multiply to get $a_kb_k=O(((r+\epsilon)/(R-\epsilon))^{k})$, which converges as a geometric series.

So, you can find $f^3 = (f\cdot f)\cdot f$ by convolving the coefficients twice.

But convolution is usually not something you can do by hand and feel happy about the result you got. In your situation it seems more practical to cube $f$ using the formula you have, and then get the Laurent series out of the formula for $f^3$.