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

$$\frac{1}{(-k)!} = \frac{1}{(-(-n))!} = \frac{1}{n!}$$


$$\frac{1}{z^k} = \frac{1}{z^{-n}} = z^n$$

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).

  • $\begingroup$ 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? $\endgroup$ – Pernk Dernets Mar 19 '19 at 8:51
  • $\begingroup$ 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 $n=0$. $\endgroup$ – Eevee Trainer Mar 19 '19 at 8:59
  • $\begingroup$ 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 $\endgroup$ – Pernk Dernets Mar 19 '19 at 9:06
  • $\begingroup$ 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). $\endgroup$ – Eevee Trainer Mar 19 '19 at 9:13
  • $\begingroup$ Oh ! off course. Thank you very much for all your in depth answers $\endgroup$ – Pernk Dernets Mar 19 '19 at 9:14

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