How do you expand $\frac{1}{e^{2z}+1}$ about $\frac{\pi i}{2}$? I know this must have some negative powers, but by using binomial theorem, this doesn't work as the function is not small about this point.
Expand $\frac{1}{e^{2z}+1}$ about $\frac{\pi i}{2}$?
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0Please proofread your question; it's rather hard to understand in this ungrammatical form. – 2011-11-11
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1But you know how to expand $\dfrac1{1-e^{2z}}$ about $z=0$, no? – 2011-11-11
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0@J.M.: No, if you could explain that it would be helpful – 2011-11-11
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0Could you expand, say, $\frac{1}{1-z}$ about $z=1$? – 2011-11-11
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0There's a bit of a trick for this one. Setting $w = e^z$, you can use the result of http://en.wikipedia.org/wiki/Laurent_series#Example to expand about $w = i$, then plug in $e^z = w$. – 2011-11-11
2 Answers
Step 1. Let $z = \frac{\pi i}{2} + \epsilon$, then $\frac{1}{ \mathrm{e}^{2z}+1} = \frac{1}{ \mathrm{e}^{i \pi} \mathrm{e}^{2 \epsilon }+1}$.
Step 2. Use the series expansion for exponential $\exp(x) \sim 1 + x + \frac{x^2}{2} + \ldots + \frac{1}{k!} x^k + \ldots$.
Step 3. Find expansion of the denominator $\mathrm{e}^{i \pi} \mathrm{e}^{2 \epsilon }+1$.
Step 4. Find the reciprocal, noting that $$ \left( a_1 x + a_2 x^2 + \ldots \right)^{-1} = \frac{1}{a_1 x} \left( 1 + \frac{a_2}{a_1} x + \ldots \right)^{-1} $$
Expansion of $(1+c_1 x + c_2 x^2 + \ldots)^{-1}$ can be found using geometric series: $$ \frac{1}{1+c_1 x + c_2 x^2 + \ldots} = 1 - \left(c_1 x + c_2 x^2 + \ldots\right) + \left(c_1 x + c_2 x^2 + \ldots\right)^2 - \left(c_1 x + c_2 x^2 + \ldots\right)^3 + \ldots $$
Write $x:={\pi i\over 2}+t$. Then
$$ {1\over e^{2x}+1} = {1\over 1-e^{2t}}=-{1\over 2t}\ {2t\over e^{2t}-1}\ .$$
The last fraction looks like the generating function of the Bernoulli numbers. A look at the page
http://en.wikipedia.org/wiki/Bernoulli_numbers
shows that in fact we have
$${1\over e^{2x}+1}=-{1\over 2t}\ \sum_{m=0}^\infty B_m{(2t)^m\over m!}\qquad\bigl(t:=x-{\pi i\over 2}\bigr)\ .$$