Solving for Coth Manually: A Challenge for Experts

[ognik]

I must be in the 'optional for experts' section. I can get a series for coth using mathmatica, but would like to do it 'manually'. The power series for cosh & sinh leaves a division I don't want to attempt. I tried $ \frac{cosh}{sinh} = \frac{{e}^{2x+1}+1}{{e}^{2x-1}+1} $ and using Maclauren, but that has a singularity for f(0) ... A clue perhaps?

 

[Prove It]

I must be in the 'optional for experts' section. I can get a series for coth using mathmatica, but would like to do it 'manually'. The power series for cosh & sinh leaves a division I don't want to attempt. I tried $ \frac{cosh}{sinh} = \frac{{e}^{2x+1}+1}{{e}^{2x-1}+1} $ and using Maclauren, but that has a singularity for f(0) ... A clue perhaps?

Maybe you can use a derivative to help you. Warning, this won't be pretty...

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \,\left[ \coth{(x)} \right] &= \frac{\mathrm{d}}{\mathrm{d}x} \, \left[ \frac{\cosh{(x)}}{\sinh{(x)}} \right] \\ &= \frac{\sinh{(x)}\sinh{(x)} - \cosh{(x)}\cosh{(x)}}{\left[ \sinh{(x)} \right] ^2 } \\ &= \frac{- \left[ \cosh^2{(x)} - \sinh^2{(x)} \right] }{\sinh^2{(x)}} \\ &= -\frac{1}{\sinh^2{(x)}} \\ &= -\frac{1}{ \left( \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} \right) ^2 } \\ &= -\frac{4}{ \left( \mathrm{e}^x - \mathrm{e}^{-x} \right) ^2 } \\ &= -\frac{4}{ \left( \mathrm{e}^{-x} \right) ^2 \left( \mathrm{e}^{2x} - 1 \right) ^2 } \\ &= -4\,\mathrm{e}^{2x}\,\left( 1 - \mathrm{e}^{2x} \right) ^{-2} \end{align*}$

Now I would advise you to expand $\displaystyle \begin{align*} \left[ 1 + \left( - \mathrm{e}^{2x} \right) \right] ^{-2} \end{align*}$ using the Binomial series, and then substitute the entire exponential series. Multiply everything out and you will have a series for $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \, \left[ \textrm{coth}\,(x) \right] \end{align*}$, which you can then integrate to get a series for $\displaystyle \begin{align*} \textrm{coth}\,(x) \end{align*}$. Told you it wasn't pretty :P

 

[ognik]

Trying again, the full problem is to expand as a power series P(X), where $ P(x) = c(\frac{cosh(x)}{sinh(x)} - \frac{1}{x}) $ (I'm told the classical Langevin theory of paramagnetism leads to this expression for the magnetic polarization)

So Mathematica gives me $ \frac{cosh(x)}{sinh(x)} = \frac{1}{x} + \frac{x}{3} - \frac{{x}^{3}}{45} + \frac{2x^5}{945} + ... $
and I immediately saw how neatly the $ \frac{1}{x} $ cancels, when the 'series' $ \frac{1}{x} $ is subtracted from the coth series, all good.

But, other than this, I have really struggled for a 'clever' way to expand $ \frac{cosh(x)}{sinh(x)} $. I also tried letting $e^{x} $ =u, giving $ \frac{u+\frac{1}{u}}{u-\frac{1}{u}} $ etc., but that just gave me a series in $ e^{2x} $ or brought me back to the exponential version of $ \frac{cosh(x)}{sinh(x)} = \frac{e^{2x}+1}{e^{2x}-1} $

I thought of using $ ln(1+x) = \sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{n}}{n} $, but again I only managed a series in $ e^{2x} $... need a hint please.

 

[Opalg]

The power series expansion of $\coth x$ involves Bernoulli numbers. The formula $$\coth x = \frac1x + \sum_{n=1}^\infty \frac{2^{2n}B_{2n}x^{2n-1}}{(2n)!}$$ (valid for $0<|x|<\pi$) is given here, but without any details of how to derive it.

 

[ognik]

Thanks Opalg, that works.

I have also found an example of the type of expansion I was looking for - using the power series of Cosh and Sinh, but WITHOUT the long division - see calculus - Power (Laurent) Series of $\coth(x)$ - Mathematics Stack Exchange.

That uses a 'trick' that I didn't know, but would like to understand better - 'reciprocating a polynomial' - the solution in the above article implies there is a formula to find the reciprocal of a polynomial (I expect to a limited power?). For example the author of the solution reciprocates $ \frac{1}{1+ax^2+bx^4+O(x^4)} = 1-ax^2+(a^2-b)x^4 + O(x^4) $
Does anyone know a general formula for reciprocating a polynomial?

 

[I like Serena]

Does anyone know a general formula for reciprocating a polynomial?

That particular post explains immediately after - by writing it as a geometric series.

 

[ognik]

I could follow that, just wondered if there was a generalised formula that could also give the higher power coefficients for the reciprocal? For example if the polynomial was a + bx^2 + cx^3 + dx^4 + ex^5 - it could get messy using that geo-series trick...

 

[I like Serena]

I could follow that, just wondered if there was a generalised formula that could also give the higher power coefficients for the reciprocal? For example if the polynomial was a + bx^2 + cx^3 + dx^4 + ex^5 - it could get messy using that geo-series trick...

Well, if we write your polynomial $a + bx^2 + cx^3 + dx^4 + ex^5 + ...$ as $a\Big(1-a_1x-a_2x^2-a_3x^3-a_4x^4-O(x^5)\Big)$, using geometric expansion or long division, we can find that:
$$\frac{1}{a\Big(1-a_1x-a_2x^2-a_3x^3-a_4x^4-O(x^5)\Big)} = \\
\frac{1}{a}\Big(1+a_1x+(a_1^2+a_2)x^2 + (a_1^3+2a_2a_1+a_3)x^3 + (a_1^4 + 3a_2a_1^2+2a_3a_1+a_2^2+a_4)x^4 + O(x^5)\Big)
$$

That's not too messy, is it? (Wondering)

 

[ognik]

Not too messy - I was looking for a formula because I find I sometimes make mistakes with long, drawn out workings; I need to be more patient I know.
It appears there is no commonly known formula for this, fortunately we always have mathematica. Thanks ILS.

 

[I like Serena]

For the record, in practice it usually suffices to approximate:
$$\frac{1}{1-ax-O(x^2)} = 1 + ax + O(x^2)$$