3
$\begingroup$

I want to find the Laurent expansion for $\frac{\exp\left(\frac{1}{z^{2}}\right)}{z-1}$ about $z=0$,

I've tried to apply this formula $\frac{1}{1-\omega}=\sum_{n=0}^{\infty }\omega^{n}$ and the usual Taylor series of the exponential function, but I don't know how to continue:
$$\begin{align}f(z)&=\frac{1}{z-1}\exp\left(\frac{1}{z^{2}}\right)\\ &=-\frac{1}{1-z}\exp\left(\frac{1}{z^{2}}\right)\\&=-\left (\sum_{n=0}^{\infty }z^{n} \right )\left ( \sum_{n=0}^{\infty}\frac{1}{n!z^{2n}} \right )\end{align}$$ Thanks in advance.
Ps: I tried applying a Cauchy product, but I think this is not appropriate.
Edit 1: If it is useful at the end of the text, the authors say that the Laurent expansion is:
$\sum_{k=-\infty }^{\infty }a_{k}z^{k}$ with $a_{k}=-e$ if $k\geq 0$ and $a_{k}=-e+1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{(j-1)!}$if $k=-2$ or $k=-2j+1$ where $j=1,2,...$

$\endgroup$
0

2 Answers 2

2
$\begingroup$

First, we can write two series for $\frac1{z-1}$ in the two regions $|z|<1$ and $|z|>1$ as

$$\frac1{z-1}=\begin{cases} -\sum_{n=0}^\infty z^n&,|z|<1\\\\ \sum_{n=1}^\infty z^{-n}&,|z|>1\tag1 \end{cases}$$


Second, the Laurent series for $e^{1/z^2}$ for $0<|z|$ is given by

$$e^{1/z^2}=\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}\tag2$$

where $a_n$ the sequence such hat

$$a_n=\begin{cases} 1&,n\,\text{even}\\\\ 0&,n\,\text{odd} \end{cases}$$


Putting $(1)$ and $(2)$ together reveals

$$\frac{e^{1/z^2}}{z-1}= \begin{cases} -\sum_{m=0}^\infty z^m \sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}&,0<|z|<1\tag3\\\\ \sum_{m=1}^\infty z^{-m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}&,1<|z| \end{cases} $$



For $|z|>1$, the Laurent series of $\frac{e^{1/z^2}}{z-1}$ can be written

$$\begin{align} \frac{e^{1/z^2}}{z-1}&=\sum_{m=1}^\infty z^{-m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}\\\\ &=\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,\sum_{m=1}^\infty z^{-(n+m)}\\\\ &\overbrace{=}^{p=n+m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{p=n+1}^\infty\,z^{-p}\\\\ &=\sum_{p=1}^\infty\left(\sum_{n=0}^{p-1} \frac{a_n}{(n/2)!}\right)\,z^{-p} \end{align}$$



For $0<|z|<1$, the Laurent series of $\frac{e^{1/z^2}}{z-1}$ can be written

$$\begin{align} \frac{e^{1/z^2}}{z-1}&=-\sum_{m=0}^\infty z^{m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}\\\\ &=-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{m=0}^\infty z^{m-n}\\\\ &\overbrace{=}^{p=m-n}-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{p=-n}^\infty z^{p}\\\\ &=-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\left(\sum_{p=-n}^{0} z^{p}+\sum_{p=1}^\infty z^{p}\right)\\\\ &=-e \sum_{p=1}^\infty z^{p}-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{p=0}^{n} z^{-p}\\\\ &=-e \sum_{p=1}^\infty z^{p}-\sum_{p=0}^{\infty}\left(\sum_{n=p}^\infty \frac{a_n}{(n/2)!} \right)z^{-p}\\\\ &=-e \sum_{p=0}^\infty z^{p}-\sum_{p=1}^{\infty}\left(\sum_{n=p}^\infty \frac{a_n}{(n/2)!} \right)z^{-p} \end{align}$$


$\endgroup$
7
  • $\begingroup$ thanks, but I don't understand why took that series for $exp(\frac{1}{z^{2}})$. $\endgroup$ Commented Nov 18, 2020 at 23:45
  • $\begingroup$ @BrigitteEliana What is it that you don't understand? $\endgroup$
    – Mark Viola
    Commented Nov 19, 2020 at 2:51
  • $\begingroup$ Yes, I don't understand how did you determine that the Laurent series for $exp(\frac{1}{z^{2}})$ is the given in equation (2)? $\endgroup$ Commented Nov 19, 2020 at 3:32
  • $\begingroup$ The Laurent expansion for $|z|>0$ of $e^{1/z^2}$ is given by $$e^{1/z^2}=\sum_{n=0}^\infty \frac1{n!}\,\frac1{z^{2n}}$$Now, note that all of the terms are of even inverse powers of $z$. That is, all of the coefficients of odd inverse powers are equal to $0$. So, we define $a_n$ to be $1$ when $n$ is even and $0$ when $n$ is odd. Then, $\sum_{n=0}^\infty \frac1{n!}\,\frac1{z^{2n}}=\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,\frac1{z^n}$. Is that clear now? $\endgroup$
    – Mark Viola
    Commented Nov 19, 2020 at 3:40
  • 1
    $\begingroup$ @BrigitteEliana And feel free to up vote an answer as you see fit of course. ;-) $\endgroup$
    – Mark Viola
    Commented Nov 22, 2020 at 17:45
0
$\begingroup$

Starting with your $=-\left (\sum\limits_{m=0}^{\infty }z^{m} \right )\left ( \sum\limits_{n=0}^{\infty}\frac{1}{n!z^{2n}} \right )$ changing one of the $n$ to $m$, you can say the coefficient of $z^k$ is

  • $-\sum\limits_{n=0}^{\infty} \frac1{n!} =-e$ when $k\le 0$
  • $-\sum\limits_{n=k/2}^{\infty} \frac1{n!} =\sum\limits_{n=0}^{n=(k-2)/2} \frac1{n!}-e$ when $k\gt 0$ and even
  • $-\sum\limits_{n=(k+1)/2}^{\infty} \frac1{n!} =\sum\limits_{n=0}^{(k-1)/2} \frac1{n!}-e$ when $k\gt 0$ and even

But that looks wrong to me: I do not think $$\cdots -e z^{-5} -e z^{-4} -e z^{-3} -e z^{-2} -e z^{-1} -e z^{0}+ \\(1-e)z^1 +(1-e)z^2 +(2-e)z^3 +(2-e)z^4+\left(\frac52-e\right)z^5+\cdots$$ converges when $|z| \le 1$.

Meanwhile for the same question asked elsewhere, a suggested answer was in effect $$z^{-1}+z^{-2}+2 z^{-3}+2 z^{-4}+\frac{5 }{2}z^{-5}+\frac{5}{2}z^{-6}+\cdots$$ but I do not think that converges either when $|z|\le 1$

$\endgroup$
10
  • $\begingroup$ The series you wrote for $\frac1{1-z}$ fails to converge for $|z|>1$. $\endgroup$
    – Mark Viola
    Commented Nov 18, 2020 at 21:43
  • $\begingroup$ @MarkViola - You may be correct if you say it fails to converge for $|z|\le 1$ $\endgroup$
    – Henry
    Commented Nov 18, 2020 at 21:49
  • 1
    $\begingroup$ Henry, $\frac1{1-z}=\sum_{n=0}^\infty z^n$ for $|z|<1$, and fails to converge for $|z|\ge 1$. $\endgroup$
    – Mark Viola
    Commented Nov 18, 2020 at 21:51
  • $\begingroup$ @MarkViola I am more worried that $\cdots -e z^{-5} -e z^{-4} -e z^{-3} -e z^{-2} -e z^{-1} -e z^{0}$ does not converge than I am about $(1-e)z^1 +(1-e)z^2 +(2-e)z^3 +(2-e)z^4+(\frac52-e)z^5+\cdots$ $\endgroup$
    – Henry
    Commented Nov 18, 2020 at 21:54
  • $\begingroup$ I've posted a complete solution that includes the Laurent series for $|z|>1$ and the Laurent series for $0<|z|<1$. $\endgroup$
    – Mark Viola
    Commented Nov 18, 2020 at 23:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .