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What is the typical trick for finding the taylor series of a common function that is in the denominator when adding a constant.

eg:

$$f(x)=\frac{1}{e^x-c}$$

I know you can write $f(x)=\frac{e^x}{e^{2x}-ce^x}$ and then invoke

$$Taylor(\frac{f(x)}{g(x)})=\frac{Taylor{f(x)}}{Taylor(g(x))}$$

but I feel that there might be an easier way to evaluate $f(x)$

Any hint would be appreciated

Edit:

the origin of this question is to do integrals like

$$\int_a^b (\frac{8\pi hc}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda kt}}-1})\,d\lambda$$

and limits for the same function

$$\lim_{\lambda \to +\lambda_0}{\frac{8\pi hc}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda kt}}-1}}$$

The Phenotype
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user2654176
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    Long division ... see http://math.stackexchange.com/a/167172/442 or http://math.stackexchange.com/a/342389/442 I'm not writing this one out for you. – GEdgar Sep 12 '13 at 18:13
  • The case $c=1$ is special, we are then close to the generating function of the Bernoulli numbers. For $c\ne 1$, one should be able to get a recurrence. – André Nicolas Sep 12 '13 at 18:22

1 Answers1

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If the problem for you is that the function is in the denominator, why not just take it as the numerator, i.e. re-write

$f(x) = \frac{1}{e^{x} - 2}$

as

$f(x) = (e^{x} - 2)^{-1}$

and proceed as

$f'(x) = -(e^{x} -2)^{-2}\cdot e^{x} \\ f''(x) = 2(e^{x} -2)^{-3}\cdot e^{2x} - (e^{x} -2)^{-2}\cdot e^{x} \\ f'''(x) = -6(e^{x}-2)^{-4}\cdot e^{3x} + 4(e^{x} -2)^{-3}\cdot e^{2x} + 2(e^{x} -2)^{-3}\cdot e^{2x} - (e^{x} -2)^{-2}\cdot e^{x} \\ \ldots$

where you should be able to write the derivatives more compactly

Keeran Brabazon
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  • well, x is a complicated function with the boltzmann factor, so I would have to do an taylor expansion in this case – user2654176 Sep 12 '13 at 17:49