I'm having some trouble understanding a step. It's from this question on math stackexhange. It's about finding the laurent series of $\exp{(z + 1/z)} $ i understand we can do this:
$\exp{(z + 1/z)} = \exp{(z)}\exp{(1/z)} $
Then use:
$\exp {x} = \sum_{n=0}^{\infty} \frac{x^n}{n!}$
So we get:
$ \exp{(z)}\exp{(1/z)} = \sum_{n=0}^{\infty} \frac{z^n}{n!} \sum_{m=0}^{\infty} \frac{1}{m!z^m}$
Now comes where i am confused, somehow according to the math stackexhange question we can define:
$$a_k=\begin{cases} 1/k!&\text{for $k\ge0$}\\ 0&\text{for $k<0$}\,,\end{cases}$$ and $$b_k=\begin{cases} 0&\text{for $k>0$}\\ 1/(-k)!&\text{for $k\le0$}\,.\end{cases}$$
Such that: $\exp(z)=\sum_{n=-\infty}^\infty a_k z^k$ and $\exp(1/z)=\sum_{m=-\infty}^\infty b_k z^k$.
I am confused how we can put negative in the factorial in $b_k$ and then move $z^k$ up from the denominator. Also why we sum from negative infinity..
I would love your input on that
It's because those $k$ values are negative, essentially, for the $1/z$ series. Explicitly, for $n>0,$ let $k = -n$. Then
$$\frac{1}{(-k)!} = \frac{1}{(-(-n))!} = \frac{1}{n!}$$
and
$$\frac{1}{z^k} = \frac{1}{z^{-n}} = z^n$$
We can also simply just replace $k$ with $-k$ to get everything for $k$ positive instead of $k$ negative, which is why the $k$'s are reused. But $k$ is still negative, which is why we still sum from $-\infty$, why we have $(-k)!$ ($k!$ wouldn't make sense), and why $z^k$ is moved "up" (it didn't actually move up, it's just that all $k$ values that are relevant for the series - the nonzero summands - are negative).