0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.