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residue at essential singularity

pantboio

Member
Nov 20, 2012
45
Consider the function
$$f(z)=\frac{e^{\frac{1}{z-1}}}{e^z -1}$$
$z_0=1$ is an essential singularity, hence
$$f(z)=\displaystyle\sum_{-\infty}^{+\infty}a_n(z-1)^n$$
near to $z_0=1$ and i want to find $a_{-1}$. I can write
$$f(z)=\frac{\sum\frac{1}{n!(z-1)^n}}{e\cdot e^{z-1}-1}=\frac{1+\frac{1}{z-1}+\frac{1}{2!(z-1)^2}+\ldots}{e-1+e(z-1)+e\frac{(z-1)^2}{2!}+\ldots}$$
and now, how can i find the coefficient of $\frac{1}{z-1}$? Can someone help me?
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
How did you expand [tex]e^{z}-1[/tex] around 1 ?
 
Last edited:

pantboio

Member
Nov 20, 2012
45

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
$e^{z}-1=e^{z-1+1}-1=e\cdot e^{z-1}-1=e\cdot\sum\frac{(z-1)^n}{n!} -1$
so you are missing some e's in the denominator in your post

[tex]e(e^{z-1})-1=e( 1+(z-1)+\frac{(z-1)^2}{2!}+\,...\, ) -1= e-1+e(z-1)+\frac{e(z-1)^2}{2!}+\,... [/tex]

Then you can do long division ...
 

pantboio

Member
Nov 20, 2012
45
so you are missing some e's in the denominator in your post

[tex]e(e^{z-1})-1=e( 1+(z-1)+\frac{(z-1)^2}{2!}+\,...\, ) -1= e-1+e(z-1)+\frac{e(z-1)^2}{2!}+\,... [/tex]

Then you can do long division ...
unfortunately, i don't know how to divide one series by another.....
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
The trick when dividing by a series is to use:

${1\over 1-x}=1+x+x^2+...$

So for instance

${x \over \sin x}={x \over x - x^3/3! +...} = {1 \over1 - x^2/3! + ...} = 1 + ({x^2 \over 3!} - {x^4 \over 5!} + ...) + ({x^2 \over 3!} - ...)^2 + ... = 1 + {x^2 \over 3!} + (-{1 \over 5!} + {1 \over 3!^2}) x^4 + ...$
 

pantboio

Member
Nov 20, 2012
45
The trick when dividing by a series is to use:

${1\over 1-x}=1+x+x^2+...$

So for instance

${x \over \sin x}={x \over x - x^3/3! +...} = {1 \over1 - x^2/3! + ...} = 1 + ({x^2 \over 3!} - {x^4 \over 5!} + ...) + ({x^2 \over 3!} - ...)^2 + ... = 1 + {x^2 \over 3!} + (-{1 \over 5!} + {1 \over 3!^2}) x^4 + ...$
using your trick i got what follows
$$f(z)=\frac{ e^{ \frac{1}{z-1} }}{e^z-1}=\frac{ \sum\frac{1}{n!(z-1)^n} }{e-1+e(z-1)+\frac{e}{2}(z-1)^2+\ldots}=\frac{ \sum\frac{1}{n!(z-1)^n}}{(e-1)(1-h)}$$
where $h=\frac{e}{1-e}(z-1)+\frac{e}{2(1-e)}(z-1)^2+\ldots$
So what we have is
$$\ldots=\frac{1}{e-1}\sum_{0}^{\infty}\frac{1}{n!(z-1)^n}\sum_{0}^{\infty}h^n$$
but now i'm at a loss, don't know how to proceed.....