Is it possible to express $\displaystyle \frac{1}{e^t+e^{-t}}$ as a power series in $\displaystyle e^{-t}$?

So can we find $\displaystyle a_n$ with $\displaystyle \frac{1}{e^t+e^{-t}} = \sum_{n=0}^\infty a_n (e^{-t})^n$ ?

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- Apr 7th 2010, 02:10 AMEinStoneExpress as power series
Is it possible to express $\displaystyle \frac{1}{e^t+e^{-t}}$ as a power series in $\displaystyle e^{-t}$?

So can we find $\displaystyle a_n$ with $\displaystyle \frac{1}{e^t+e^{-t}} = \sum_{n=0}^\infty a_n (e^{-t})^n$ ? - Apr 7th 2010, 02:48 AMchisigma
Setting $\displaystyle e^{-t} = x$ we obtain...

$\displaystyle \frac{1}{e^{t} + e^{-t}} = \frac{1}{x+\frac{1}{x}} = \frac{x}{1+x^{2}} = $

$\displaystyle = x\cdot (1-x^{2} + x^{4} - \dots) = x - x^{3} + x^{5} - \dots$ (1)

... so that is...

$\displaystyle \frac{1}{e^{t} + e^{-t}} = e^{-t} - e^{-3t} + e^{-5t} - \dots$ (2)

Because the series (1) converges for $\displaystyle |x|<1$, the series (2) converges for $\displaystyle t>0$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Apr 7th 2010, 02:55 AMHallsofIvy
Didn't we just have this exact problem recently?