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I'm currently stuck on an explanation that works with a power series expansion of: $ \frac{a}{\eta}\: \frac{1}{e^{\frac{\eta}{a}}-1}$

As you can see, the function diverges around $\eta=0$, which is exactly where I need to investigate its behaviour.

This step is taken in the calculation of the Casimir force (quantum physics).

Sadly, I seem to be unable to find the correct way to expand this. The power series expansion is differentiated twice and multiplied by some prefactor, yielding something $ \propto \left( \frac{3a}{\eta^4} - \frac{1}{2\eta^2} - \frac{1}{720a^3} + \mathscr{O}\left[\eta\right] \right)$.

I hope this provides enough information. Any sort of answer including just an instruction would be greatly appreciated.

Bernard
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Daimonie
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  • Oh, the expansion to second order is sufficient as the limit $\eta \to 0$ is taken after differentating twice. – Daimonie Apr 29 '15 at 12:48
  • I'd write it as $f(\eta/a)$ where $f(w)=\frac{1}{w(e^w-1)}$. Just to avoid all the extra letters, expand this first. There is going to be a $w^{-2}$ and $w^{-1}$ terms. Not sure how you got an $\eta^{-4}$ term. – Thomas Andrews Apr 29 '15 at 12:54
  • The $\eta^{-4}$ term will be because $f(w)$ is differentiated twice. Thanks! – Daimonie Apr 29 '15 at 13:28

1 Answers1

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Letting $f(w)=\frac{1}{w(e^w-1)}$, you are seeking $f(\eta/a)$. So if you find a Maclaurin Series for $f$ you can get on on for $f(\eta/a)$.

But $f(w)=\frac{1}{w^2}\frac{w}{e^w-1}$ and the Taylor series for $\frac{w}{e^w-1}$ is easier to compute. We know $$\frac{e^w-1}{w} = \sum_{n=0}^{\infty} \frac {w^{n}}{(n+1)!}=1+\frac{w}{2!}+\frac{w^2}{3!} + \frac{w^3}{4!} + \frac{w^4}{5!}+O(w^5)$$

So if $$\frac{w}{e^w-1} = a_0+a_1w + \frac{a_2w^2}{2!} + \frac{a_3w^3}{3!}+ \frac{a_4w^4}{4!} + O(w^5)$$

Then you immediately get $a_0=1$. Then $a_1+a_0\frac{1}{2!} = 0$ so $a_1=\frac{-1}{2}$, etc.

The coefficients for the Maclaurin series of $\frac{w}{e^w-1}$ are well know, called the Bernoulli numbers.

Of note: $f(w)+\frac{1}{2}w = \frac{w}{2}\frac{e^w+1}{e^{w}-1}$ is an even function. That makes computing the odd coefficients easier - $0=a_3=a_5=a_7=\cdots$.

Thomas Andrews
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  • Extremely neat! I hadn't encountered the Bernoulli numbers before, which sort of explains why the teacher assumed it to be a known expansion, as undergraduate programs vary. Thanks again :) – Daimonie Apr 29 '15 at 13:28