# Approximate $\coth(x)$ around $x = 0$

I'm trying to approximate $\coth(x)$ around $x = 0$, up to say, third order in $x$. Now obviously a simple taylor expansion doesn't work, as it diverges around $x = 0$. I'm not quite sure how to proceed from here. I had a look at Series expansion of $\coth x$ using the Fourier transform

but this is a bit technical for my purposes, and I also don't need the entire series. Is there an easy way of getting some terms, or do I really have to solve for the entire series? And is there a simpler way to do so than with the Fourier transform?

• Your function has a pole of order one at $x=0$. Jan 18 '15 at 13:16
• Remove the singularity by expanding $\coth x-1/x$ instead.
– user65203
Jan 18 '15 at 17:01

There is a simple way of approximating $\coth$ by noticing that it is a logarithmic derivative. Since: $$\frac{\sinh z}{z}=\prod_{n=1}^{+\infty}\left(1+\frac{z^2}{\pi^2 n^2}\right)\tag{1}$$ by the Weierstrass product for the (hyperbolic) sine function, we have: $$\log\sinh z-\log z = \sum_{n=1}^{+\infty}\log\left(1+\frac{z^2}{\pi^2 n^2}\right),\tag{2}$$ so, by differentiating both sides: $$\coth z-\frac{1}{z} = \sum_{n=1}^{+\infty}\frac{2z}{\pi^2 n^2+z^2}.\tag{3}$$ By expanding the terms in the RHS of $(3)$ as geometric series we can also derive the full Laurent expansion of $\coth$ in a puntured neighbourhood of the origin, in terms of $\zeta(2m)$. This is also the standard technique for showing that $\zeta(2m)$ just depends on the Bernoulli number $B_{2m}$ (formula $(41)$ here).

Note that $\coth\left(x\right)=\,{\rm i}\cot\left(\,{\rm i}x\right)$. With Euler Reflection Formula we have: \begin{align} \coth\left(x\right)&=\frac{\,{\rm i}}{\pi}\left[\pi\cot\left(\pi\,\frac{\,{\rm i}x}{\pi}\right)\right] =\frac{\,{\rm i}}{\pi} \Psi\left(1 - \frac{\,{\rm i}x}{\pi}\right) -\frac{\,{\rm i}}{\pi}\,\Psi\left(\frac{\,{\rm i}x}{\pi}\right) \end{align} where $\Psi$ is the Digamma Function.

With the Digamma Recurrence Formula $\Psi\left(z\right) = \Psi\left(1 + z\right) - \frac{1}{z}$: \begin{align} \coth\left(x\right)&=\frac{\,{\rm i}}{\pi} \Psi\left(1 - \frac{\,{\rm i}x}{\pi}\right) -\frac{\,{\rm i}}{\pi}\left[\Psi\left(1 + \frac{\,{\rm i}x}{\pi}\right) -\frac{1}{\,{\rm i}x/\pi}\right] =\frac{2}{\pi}\,\Im\Psi\left(1 + \frac{\,{\rm i}x}{\pi}\right) + \frac{1}{x} \end{align} The Digamma Function can be expressed as: $$\Psi\left(1 + z\right)=\Psi\left(1\right) + \sum_{n=2}^{\infty}\left(-1\right)^{n}\,\zeta\left(n\right)z^{n - 1}\,,\qquad \left\vert z\right\vert < 1$$ Then, \begin{align} \color{#66f}{\large\coth\left(x\right)}& =\frac{1}{x} + \frac{2}{\pi}\,\Im\sum_{n=2}^{\infty}\left(-1\right)^{n}\, \zeta\left(n\right)\left(\frac{\,{\rm i}x}{\pi}\right)^{n - 1} =\frac{1}{x} + \frac{2}{\pi}\,\Im\sum_{n=1}^{\infty} \zeta\left(2n\right)\,\frac{\,{\rm i}^{2n}\,x^{2n - 1}/\,{\rm i}}{\pi^{2n - 1}} \\[5mm]&=\color{#66f}{\large\frac{1}{x} - 2\sum_{n=1}^{\infty} \left(-1\right)^{n}\,\frac{\zeta\left(2n\right)}{\pi^{2n}}x^{2n - 1}}\,,\qquad \left\vert x\right\vert<\pi \end{align}

Up to third order it becomes:

$$\color{#66f}{\large% \coth\left(x\right)}\approx\frac{1}{x} + 2\,\frac{\zeta\left(2\right)}{\pi^{2}}\,z -2\,\frac{\zeta\left(4\right)}{\pi^{4}}\,z^{3} =\color{#66f}{\large\frac{1}{x} + \frac{1}{3}\,x - \frac{1}{45}\,x^{3}}$$ because $\zeta\left(2\right) = {\pi^{2} \over 6}$ and $\zeta\left(4\right) = {\pi^{4} \over 90}$.

$$\coth x=\frac{e^x+e^{-x}}{e^x-e^{-x}}\\\sim\frac{1+x^2/2!+x^4/4!+...}{x+x^3/3!+x^5/5!+...}\\\sim\frac1x(1+x^2/2!+x^4/4!+...)(1-(x^2/3!+x^4/5!+..))\\=\frac1x+\frac x3+...$$

• Care to elaborate on the meaning of $\sim$, especially in the third line? Jan 18 '15 at 13:18
• @RonGordon I remove last line $\sim$ because i just multiplied the terms, surely i'm just 1% of you Jan 18 '15 at 13:54
• Actually, about $8.8 \%$, but given time, that should increase significantly. ;) Jan 18 '15 at 14:00

The series representation must reflect that divergence at $x=0$. Thus,

\begin{align}\coth{x} &= \frac{1+\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)}{\frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}+O(x^7)} \\ &= \frac1{x} \frac{1+\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)}{1+\frac{x^2}{3!}+\frac{x^4}{5!}+O(x^6)}\\&= \frac1{x} \left (1+\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6) \right ) \left [1 - \left (\frac{x^2}{3!}+\frac{x^4}{5!}+O(x^6) \right )+\left (\frac{x^2}{3!}+O(x^4) \right )^2+O(x^6) \right ] \\ &= \frac1{x} \left [1+\left (\frac1{2!}-\frac1{3!} \right ) x^2 +\left (\frac1{(3!)^2} + \frac1{4!}-\frac1{5!} - \frac1{2! 3!}\right ) x^4+O(x^6) \right ]\\ &= \frac1{x}+\frac{x}{3} - \frac{x^3}{45}+O(x^5) \end{align}

Note that the divergence at $x=0$ is reflected in the $1/x$ term.

• shouldn't using a $=$ sign be misleading? Jan 18 '15 at 13:14
• I have to say, I don't completely understand how to go from the second line to the third line. Is that a common technique? Jan 18 '15 at 13:32
• @user3183724: Yes. It is the equivalent of the expansion of $$\frac1{1+z} = 1 - z + z^2 +O(z^3)$$ However, note that I am tracking the errors in the expansion carefully. This is how we are able to use the equal sign meaningfully. Jan 18 '15 at 13:49
• Thank you, I see it now! All answers are good, but I have to choose one of course, and this use of the equality sign makes it a little easier to follow. But, thanks to you all! Jan 18 '15 at 13:54

Here's a somewhat different approach, which can be extended by using the power series of $\cosh$ and $\sinh$ as others have.

$$\coth(x) = \frac{\cosh(x)}{\sinh(x)} = \frac{1}{x} \frac{\cosh(x)}{\frac{\sinh(x)}{x}} \\ = \frac{1}{x} \frac{\cosh(x)}{1+\left ( \frac{\sinh(x)}{x} - 1 \right )} \\ = \frac{1}{x} \cosh(x) \sum_{n=0}^\infty \left ( 1-\frac{\sinh(x)}{x} \right )^n$$

provided $\left | 1 - \frac{\sinh(x)}{x} \right | < 1$ (and of course provided $x \neq 0$, otherwise neither side is defined). The last line applies the geometric series. You can turn this into a power series by substituting the power series of $\cosh$ and $\sinh$ and then pulling some tricks (a Cauchy convolution product, among others).

A similar approach to this one would be to write

$$\coth(x) = \frac{1}{x} \left ( x \coth(x) \right )$$

and then work out the Maclaurin series of $x \coth(x)$ by directly calculating derivatives. Of course this, too, quickly becomes a complicated mess as the number of terms grows.

My copy of the classic "Handbook of Mathematical Functions" by Abramowitz and Stegun says

$$\mathrm{coth}(z) = \frac{1}{z} + \frac{z}{3} - \frac{z^3}{45} + \frac{2}{945} z^5 - \ldots + \frac{2^{2n}B_{2n}}{(2n)!} z^{2n-1}$$

where $B_n$ is the $n^{th}$ Bernoulli number.

Note in particular the difference in the $z^3$ term compared with the answer above.

• That should be an infinite series. Sep 12 '17 at 19:14