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Laurent series |
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| Apr6-06, 02:01 AM | #1 |
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Laurent series
Just wondering where to go with this one..
calculate the laurent series of [tex]\frac{1}{e^z-1}[/tex] don't even know where to start on it I know [tex]e^z={{\sum^{\infty}}_{j=0}}\frac{z^j}{j!}[/tex] but not much else... |
| Apr6-06, 05:47 AM | #2 |
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The first thing you need to do is figure out where you're taking the Laurent expansion about (presumably zero since that is what your expression for e^z is. Why not put that in to the expression and play around with it?
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| Apr6-06, 02:36 PM | #3 |
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Well, I only assumed that I knew that the expansion of e^z was about 0. It only specifies "calculate the laurent expansion of [tex]\frac{1}{e^z-1}[/tex] for [tex]0 < |z| < 2\pi[/tex]"
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| Apr6-06, 07:31 PM | #4 |
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Laurent series
I tried that but couldn't really come up with anything..
[tex]\frac{1}{e^z-1}=\frac{1}{(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+...)-1}=\frac{1}{z+\frac{z^2}{2!}+\frac{z^3}{3!}+...}=\frac{1}{z(1+\frac{z}{ 2!}+\frac{z^2}{3!}+...)}[/tex] no idea where to go with this.. I can't see how I could turn the series into a useful series that converges to a algebraic expresion that I could actually rearrange to continue.... |
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