# Power (Laurent) Series of $\coth(x)$

I need some help to prove that the power series of $\coth x$ is:

$$\frac{1}{x} + \frac{x}{3} - \frac{x^3}{45} + O(x^5) \ \ \ \ \$$

I don't know how to derive this, should I divide the expansion of $\cosh(x)$ by the expansion of $\sinh(x)$? (I've tried but without good results :( )

Or I have to use residue calculus?

Anyone can suggest me a link where I can find a detailed explanation of this expansion?

Thanks.

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You can find the Taylor series of $x*\mathrm{coth}(x)$ by taking derivatives, and divide by $x$ – Cocopuffs Jul 5 '12 at 17:32
You should be able to adapt the approach here, by an appropriate change of variables... – J. M. Jul 5 '12 at 17:39
Divide series for cosh by series for sinh, yes that works. You need to remember "long division" to apply this, however... – GEdgar Jul 5 '12 at 17:45
@GEdgar But... Do I have to make the long division between the expansion of cosh and sinh without truncate them? – MaJac89 Jul 5 '12 at 17:53
Yes, just as you divide infinite decimals "without truncate them"... At each step you use only finitely many terms. – GEdgar Jul 5 '12 at 17:58

Long division.

The problem:
Now $z$ into $1$ is $z^{-1}$
Multiply
Subtract
$z$ into $(1/3)z^2$ is $(1/3) z$
Multiply
Subtract
$z$ into $(-1/45)z^4$
If we want more terms in the quotent, we will have to fill in more terms in all of them where the dots are now.

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Ok you are too kind.. thanks, I've made some stupid mistake in the division:( – MaJac89 Jul 5 '12 at 18:43

$$\begin{eqnarray} \coth(x) &=& \frac{\cosh(x)}{\sinh(x)} = \frac{1 + \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{o}\left(x^5\right)}{x + \frac{x^3}{6} + \frac{x^5}{120} + \mathcal{o}\left(x^5\right)} = \frac{1}{x} \frac{1 + \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{o}\left(x^5\right)}{1 + \frac{x^2}{6} + \frac{x^4}{120} + \mathcal{o}\left(x^4\right)} \\ &=& \frac{1}{x} \left( 1 + \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{o}\left(x^5\right) \right) \left( 1 - \frac{x^2}{6} + \frac{7 x^4}{360} + \mathcal{o}\left(x^4\right) \right) \\ &=& \frac{1}{x} \left( 1 + \frac{x^2}{3} - \frac{x^4}{45} + \mathcal{o}\left(x^4\right) \right) = \frac{1}{x} + \frac{x}{3} - \frac{x^3}{45} + \mathcal{o}\left(x^3\right) \end{eqnarray}$$ where the reciprocation and multiplication of series used: $$\frac{1}{1 + a x^2 + b x^4 + \mathcal{o}\left(x^4\right)} = 1 -a x^2 + \left( a^2-b \right) x^4 + \mathcal{o}\left(x^4\right)$$ $$\left( 1 + a x^2 + b x^4 + \mathcal{o}\left(x^4\right) \right) \left( 1 + c x^2 + d x^4 \mathcal{o}\left(x^4\right) \right) = 1 + \left(a+c\right) x^2 + \left(b + d + a c\right) x^4 + \mathcal{o}\left(x^4\right)$$

The result for the reciprocation is obtained using the geometric series: $$\frac{1}{1-w} = 1 + w + w^2 + \mathcal{o}(w^2)$$ Now substitute in the above $w = a x^2 + b x^4 + \mathcal{o}(x^4)$, and use $$w^2 = \left( a x^2 + b x^4 + \mathcal{o}(x^4) \right)^2 = a^2 x^4 + \mathcal{o}(x^4)$$

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OMG, thanks Sasha... how did you know that reciprocation? Do you have simply found it looking for a polynomial that multiplied by 1+ax^2+bx^4+o(x^4) gives you back 1 ? – MaJac89 Jul 5 '12 at 18:18
@MaJac89 I have filled in steps of computing the reciprocal. – Sasha Jul 5 '12 at 18:23
thanks for the further explanation ;) – MaJac89 Jul 5 '12 at 18:46

Long division of series with $\cosh(x) = 1 + \dfrac{x^2}{2} + \dfrac{x^4}{24} + \ldots$ and $\sinh(x) = x + \dfrac{x^3}{6} + \dfrac{x^5}{120} + \ldots$. Unfortunately I don't know how to typeset this nicely in LaTeX.

First term is $1/x$, $$1 + \dfrac{x^2}{2} + \dfrac{x^4}{24} + \ldots - \dfrac{1}{x} \left(x + \dfrac{x^3}{6} + \dfrac{x^5}{120} + \ldots\right) = \frac{x^2}{3} + \frac{x^4}{30} + \ldots$$ Next term is $x/3$, ...

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