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Why is $\sum_{n=0}^\infty (1/w)^{2n+1} = \sum_{m=-\infty}^0 w^{2m+1}$, using the identity ($m=-n$)? Why shouldn't the exponent be equal to $2m-1$? Also, why is the following correct - $\sum_{n=0}^\infty (1/n!)(1/z)^n = \sum_{m=-\infty}^0 z^{-m}/(-m)!$? Why shouldn't the exponent of $z$ be equal to $m$? And last, why is the $a_{-1}$ term of this Laurent series (i.e. the residue) equal to $1$ at $z = 0$?

Grtv
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    You can see directly from the definition of notation $\Sigma$. The residue at $z=0$ is the coefficient of $z^{-1}$ – Jlamprong Jan 19 '14 at 07:24
  • I am sorry, but I still don't quite understand how the above equalities for the sums hold. Could someone please clarify? – Grtv Jan 19 '14 at 07:26

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Indeed, the exponent is $2m-1$ and not $2m+1$. Respect to the other question it is evident that the exponent of $z$ is $m$, otherwise the equality only holds for $z=0$. In this case the Laurent series at $z=0$ is $$\sum_{n=0}^{\infty}\frac{1}{n!}\frac{1}{z^n}=a_{0}+\frac{a_1}{z}+\frac{a_2}{z^2}+\cdots$$ where the term $a_1$ is the residue by definition, which in this case is 1.

Chilote
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  • Okay, great. Now, one final concern regarding this topic - suppose I have the following series: $\sum_{n=0}^{\infty}\frac{(-1)^{(l+n+1)/2}z^l}{((l+n+1)/2)!}$. Why is $a_{-1}$ for an odd $n$ equal to zero? – Grtv Jan 19 '14 at 07:39
  • Also, doesn't $a_{-1}$ imply $l=-1$? – Grtv Jan 19 '14 at 07:40
  • Are you sure that the exponent of $z$ is $l$ instead of $n$? – Chilote Jan 19 '14 at 07:48
  • Pardon me, it should have been $\sum_{l=-n-1}^{\infty}\frac{(-1)^{(l+n+1)/2}z^l}{((l+n+1)/2)!}$. – Grtv Jan 19 '14 at 07:55
  • Now, wouldn't $a_{-1}$ imply $l = -1$? And, why would the former be equal to zero for odd $n$? – Grtv Jan 19 '14 at 07:57
  • Nope, that's how it is written in my text. – Grtv Jan 19 '14 at 08:10
  • But why would $a_{-1}$ be zero? I realise that if $n$ were odd then $(n/2)!$ has no sense, but is that the only reason why $a_{-1}$ is zero? – Grtv Jan 19 '14 at 08:22
  • well, there is something wrong with that series. For instance, as you can see the denominator is a factorial, so $\frac{l+n+1}{2}$ must be integer (because the factorial only is defined for integers). Then if $l$ is the index, it cannot take all the positive integer values. – Chilote Jan 19 '14 at 08:28
  • The original question was to find the Laurent series for: $$\frac{e^{-z^2}}{z^{n+1}}$$ Does it make more sense now? In other words, do you reckon you could please clarify why $a_{-1}$ would be zero for odd $n$? – Grtv Jan 19 '14 at 08:31
  • The expansion of $e^{-z^2}$ is $\sum_{l=0}^{\infty}\dfrac{(-1)^{l}z^{2l}}{l!}$. Then the expansion of $\dfrac{e^{-z^2}}{z^{n+1}}$ is $$\sum_{l=0}^{\infty}\dfrac{(-1)^{l}z^{2l-n-1}}{l!}$$ where $2l-n-1=-1$ iff $n=2l$. Thus, for $n$ odd $z^{-1}$ is not present in the expansion. Therefore $a_1=0$. – Chilote Jan 19 '14 at 09:01
  • First, doesn't the exponent of a Laurent series have to be a single letter, such as m,n,l etc.? I.e. isn't a change of index requisite here in order to correctly formulate the answer as a Laurent series? Furthermore, were I interested in finding $a_{-1}$ for another series but at z=3 (and not at z=0) let's say. How would that affect the methodology we just delineated, viz. merely finding the coefficient of $z^{-1}$? – Grtv Jan 19 '14 at 09:10
  • You are right, a Laurent series has an only one index, which in this case is $l$. You have to realize that $n$ is just a constant. A change of index is not mandatory here. Why should it be? The residue of a Laurent series at $z=a$ is by definition the coefficient of the term $\dfrac{1}{z-a}$. – Chilote Jan 19 '14 at 19:47
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\begin{align} \sum_{n=0}^\infty (1/w)^{2n+1} &= \sum_{n=0}^\infty w^{-2n-1}\\ &=\sum_{n=\infty}^0 w^{-2n-1}\\ &=\sum_{-m=\infty}^0 w^{-2n-1}\\ &=\sum_{m=-\infty}^0 w^{2m+1}\\ \end{align}

Jlamprong
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