Laurent Series $e^{\frac1{1-z}}$, $|z|>1$.

Question

I tried to find Laurent expansion for:

$e^{\frac1{1-z}}$, $|z|>1$.

I tried next: $\frac1{1-z}=-\sum_{n=1}^{\infty}\frac1{z^n}$,

then using $e^{\frac1{1-z}}=1+\frac1{1-z}+\frac{1}{{2!(1-z)}^2}+\frac{1}{{3!(1-z)}^3}+...$

Then I have $e^{\frac1{1-z}}$ = $1$ + $(-\frac1{z}-\frac1{z^2}-\frac1{z^3}-...) + (\frac1{2!})(\frac1{z^2}+\frac1{z^4}+…)+...$ = $1 - \frac1z-\frac1{2z^2}-...$.

The result is wrong, since in the book the answer is: $1 - \frac1z+ \frac1{2z^2}-\frac1{6z^3}+\frac1{24z^4}-\frac{19}{120z^5}...$

What I did wrong?

@DavidC.Ullrich ... Yes, using invalid steps can make any problems easier. — GEdgar, Jul 30 '18 at 16:56

score 3 · Answer 1 · answered Jul 30 '18 at 16:42

3

Note that$$\left(-\frac1z-\frac1{z^2}-\frac1{z^3}-\frac1{z^4}-\cdots\right)^2=\frac1{z^2}+\frac2{z^3}+\frac3{z^4}+\cdots,$$that$$\left(-\frac1z-\frac1{z^2}-\frac1{z^3}-\frac1{z^4}-\cdots\right)^3=-\frac1{z^3}-\frac3{z^4}-\frac6{z^5}-\cdots,$$and so on. This explains why you got the wrong result.

answered Jul 30 '18 at 16:42

José Carlos Santos

427,504

sorry, but I can't get the answer from the book. I have again $1 - \frac1z-\frac1{2z^2}-...$, but not $1 - \frac1z+\frac1{2z^2}-...$ – Metso Jul 30 '18 at 17:48
What's the smallest $n$ for which your coefficient of $\frac1{z^n}$ is different from the one from the book? – José Carlos Santos Jul 30 '18 at 17:51
the answer: $1 - \frac1z+ \frac1{2z^2}-\frac1{6z^3}+\frac1{24z^4}-\frac{19}{120z^5}…$ As you can see it starts from $\frac1{2z^2}$, $n=2$ – Metso Jul 30 '18 at 17:52
@Metso You forgot the $\frac1{2!}$ that you wrote yourself in the original question. – José Carlos Santos Jul 30 '18 at 18:02
result is $1-\frac1z -\frac1{2!z^2}$ – Metso Jul 30 '18 at 18:07
@Metso No, you don't. The only $\frac1z$ in the Laurent expansion is the one coming from $\left(-\frac1z-\frac1{z^2}-\frac1{z^3}-\cdots\right)^1$. So, the coefficient of $\frac1z $ in the Laurent expansion is $-1$. – José Carlos Santos Jul 30 '18 at 18:10
can you get $ 1-\frac1z+ \frac1{2z^2}-\frac1{6z^3}+...$? – Metso Jul 30 '18 at 18:12
No. I see now that there's a problem. The coefficient of $\frac1{z^2}$ is $-\frac12$, not $\frac12$. – José Carlos Santos Jul 30 '18 at 18:18
I have several editions of this book and the answer is the same. It's very old book. Strange. – Metso Jul 30 '18 at 18:25
@Metso Which book is that? – José Carlos Santos Jul 30 '18 at 18:26
I. I. Privalov, “Introduction to the Theory of Functions of Complex Variables,” Nauka, Moscow, 1984. – Metso Jul 30 '18 at 18:29
@Metso There's nothing I can do. I stand by my answer. – José Carlos Santos Jul 30 '18 at 18:34

Nosrati · Accepted Answer · 2018-07-30T19:27:04.903

2

With $z=\dfrac{1}{w}$ we have $|w|<1$ then by $e^z$ expansion \begin{align} e^{\frac{1}{1-z}} &= e^{\frac{-w}{1-w}} \\ &= \sum_{n\geq0}\dfrac{1}{n!}\left(\frac{-w}{1-w}\right)^n \\ &= \sum_{n\geq0}\dfrac{(-w)^n}{n!}\left(1-w\right)^{-n} \\ &= \sum_{n\geq0}\dfrac{(-w)^n}{n!}\left(1+nw+\dfrac{n(n+1)}{2}w^2+\cdots\right) \\ &= \sum_{n\geq0}\dfrac{(-1)^n}{n!}\left(w^n+nw^{n+1}+\dfrac{n(n+1)}{2}w^{n+2}+\cdots\right) \end{align} after calculation some terms we let $w=\dfrac1z$.

Edit: One may find \begin{align} &1 \\ &-(w+w^2+w^3+w^4+w^5+\cdots) \\ &+\frac12(w^2+2w^3+3w^4+4w^5+\cdots) \\ &-\frac16(w^3+3w^4+6w^5+\cdots) \\ &+\cdots\\ &=1-w-\frac{w^2}{2}-\frac{w^3}{6}+\frac{w^4}{24}+\frac{19 w^5}{120}+\cdots\\ &=\color{blue}{1-\dfrac1z-\frac{1}{2z^2}-\frac{1}{6z^3}+\frac{1}{24z^4}+\frac{19}{120z^5}+\cdots} \end{align}

edited Jul 30 '18 at 19:27

answered Jul 30 '18 at 18:02

Nosrati

29,995

the answer in the book: $1 - \frac1z+ \frac1{2z^2}-\frac1{6z^3}+\frac1{24z^4}-\frac{19}{120z^5}…$. I can't get it. – Metso Jul 30 '18 at 18:27
See the series and let $n=0$ – Nosrati Jul 30 '18 at 18:32
so the first is $1$ right? – Nosrati Jul 30 '18 at 18:33
yes, the first is 1. I'll try other numbers – Metso Jul 30 '18 at 18:37
now for $n=1$ it is $-(w+w^2)$ – Nosrati Jul 30 '18 at 18:39
exactly, i use only $\frac1{z}$, but also can't get $1 - \frac1z+ \frac1{2z^2}$ – Metso Jul 30 '18 at 18:41
and for $n=2,3$ – Nosrati Jul 30 '18 at 18:42
I found $1-(w+w^2)+\frac12(w^2+2w^3+w^4)-\frac16(w^3+3w^4+3w^5+w^6)+\cdots$ – Nosrati Jul 30 '18 at 18:46
exactly, then we have $1-w-\frac1{2}*w^2+...$, but in the book it should be $+\frac1{2}$ – Metso Jul 30 '18 at 18:50
Sorry. my answer has a typo. I'll correct it. Do the calculations again! – Nosrati Jul 30 '18 at 18:55
$1-w-\frac1{2}*w^2+...$ instead of +$\frac1{2}$. – Metso Jul 30 '18 at 18:58
it fixed. . . . – Nosrati Jul 30 '18 at 18:59
the coefficient of $\frac1{z^2}$ negative but in the book is positive – Metso Jul 30 '18 at 19:03
I found $$1-\frac1z-\frac{1}{2z^2}-\frac{1}{6z^3}+\cdots$$ – Nosrati Jul 30 '18 at 19:04
perhaps it is wrong. I'll check with wolfram now! – Nosrati Jul 30 '18 at 19:05
$$1-w-\frac{w^2}{2}-\frac{w^3}{6}+\frac{w^4}{24}+\frac{19 w^5}{120}+\frac{151 w^6}{720}+\frac{1091 w^7}{5040}+\frac{7841 w^8}{40320}+\frac{56519 w^9}{362880}+$$ – Nosrati Jul 30 '18 at 19:09
that means typo problem in the book? – Metso Jul 30 '18 at 19:10
$|z|>1$, not oppsite – Metso Jul 30 '18 at 19:12
Yes. a typo. no matter $$\frac1w=z$$ – Nosrati Jul 30 '18 at 19:13
type in wolfram alpha 'Laurent Expansion [e^(-w/(1 - w)), {w, 0, 10}] ' – Nosrati Jul 30 '18 at 19:14
could you please provide a link how to use Wolfram, I'll be appreciated. Thank you – Metso Jul 30 '18 at 19:14
ok, thank you again – Metso Jul 30 '18 at 19:15
you are welcome! – Nosrati Jul 30 '18 at 19:16

Laurent Series $e^{\frac1{1-z}}$, $|z|>1$.

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