# Infinite sum - rewriting $\exp{1/z}$

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).

• Thank you very much. I am still confsued why chancing $k = -n$ why we have to chance summing from $0$ to $- \infty$. Do you mind elaborating more? – Pernk Dernets Mar 19 '19 at 8:51
• The original sum has $1/z^m$ from $0$ to $+\infty$. If we want to move the $z$ term to the numerator instead (i.e. letting $m=-n$), then that's sort of like saying we then instead sum from $-n=0$ to $-n=+\infty$. But then, solving for $n$ in our indices, that's instead summing from $n = -\infty$ to $n=0$. – Eevee Trainer Mar 19 '19 at 8:59
• Oh i see! But the link i sent sums from $-\infty$ to $\infty$ do you know anything about that as well? Sorry for being annyoing – Pernk Dernets Mar 19 '19 at 9:06
• Honestly I'm not 100% sure why they choose to sum to both infinities. Mathematically it's justified - notice that some of the $a_k,b_k$ constants are defined to be zero, so the terms for $k=1,2,3,...$ (or their negatives for the $a_k$ terms) are "irrelevant." Or phrased in another way, we summed from $-\infty$ to $0$, but we can sum even further by having each extra term be $0$ (which is achieved by having $b_k = 0$ in the $1/z$ series, for example). – Eevee Trainer Mar 19 '19 at 9:13
• Oh ! off course. Thank you very much for all your in depth answers – Pernk Dernets Mar 19 '19 at 9:14