Identity for $1/z$, $\cot(z)$, and $\coth(z)$.

complex-analysishyperbolic-functionstrigonometry

I'm trying to verify a seemingly simple identity that I encountered in a paper. After discarding some irrelevant scale factors it boils down to the following. It comes in three variants,

$$
\alpha(z) =
\begin{cases}
1/z & \text{(I),} \\
\cot(z) & \text{(II),} \\
\coth(z) & \text{(III),}
\end{cases}
$$

and says that

$$\alpha(a-b)\alpha(b-c) + \alpha(b-c)\alpha(c-a) + \alpha(c-a)\alpha(a-b) = C,$$

for $a,b,c \in \mathbb C$, where

$$
C =
\begin{cases}
0 & \text{(I),} \\
1 & \text{(II),} \\
-1 & \text{(III).}
\end{cases}
$$

Case (I) is straightforward to verify, but I am having trouble with cases (II) and (III). I have tried to use various trigonometric identities to make progress, but have not managed to get it all the way. I'm sure it's simple, and I'm just missing some key step. Any advice?

Best Answer

All choices of $\alpha$ are odd. In terms of $x:=a-b,\,y:=b-c$, (II) states$$\cot x\cot y-(\cot x+\cot y)\cot(x+y)=1,$$which is trivial. Similarly with (III).

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[Math] fourier expansion of $\coth$ and justifying an identity

Adding a bit of diversity to this discussion we suppose that we seek to prove that $$\coth x = \sum_{n=-\infty}^\infty \frac{x}{x^2+\pi^2 n^2}.$$

Rather than use the standard technique (absolutely nothing wrong with it) of applying a circular contour to $$ f(z) = \pi \cot(\pi z) \frac{x}{x^2+\pi^2 z^2}$$ we use Mellin transforms, which are admittedly a bit more complicated in this case (than summing the residues at the two poles of $f(z)$ on the imaginary axis at $\pm ix/\pi$), but perhaps have some complex variable didactic value.

Rewrite this as $$\coth x = \frac{1}{x} + 2 \sum_{n=1}^\infty \frac{x}{x^2+\pi^2n^2} = \frac{1}{x} + 2x \sum_{n=1}^\infty \frac{1}{x^2+\pi^2n^2} = \frac{1}{x} + 2x \sum_{n=1}^\infty \frac{1}{n^2} \frac{1}{(x/n)^2+\pi^2}.$$

Now the inner sum, call it $S(x),$ is harmonic and may be evaluated by inverting its Mellin transform. Recall the Mellin transform identity for harmonic sums with base function $g(x)$, which is $$\mathfrak{M}\left(\sum_{k\ge 1}\lambda_k g(\mu_k x); s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s}\right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = \frac{1}{k^2}, \quad \mu_k = \frac{1}{k} \quad \text{and} \quad g(x) = \frac{1}{x^2+\pi^2}.$$ The Mellin transform of $g(x)$ is $$\int_0^\infty \frac{1}{x^2+\pi^2} x^{s-1} dx,$$ which we evaluate using a semicircular contour in the upper half plane (no convergence issues as is easily verified), getting $$g^*(s) (1 + e^{\pi i(s-1)})= 2\pi i \operatorname{Res}\left(\frac{1}{z^2+\pi^2} z^{s-1}; z = +i\pi\right).$$

Using the branch of the logarithm with arguments from zero to $2\pi$, this becomes $$g^*(s) = \frac{2\pi i}{1- e^{\pi i s}}\frac{1}{2\pi i} (i\pi)^{s-1} = \pi^{s-1} \frac{e^{i(\pi/2)(s-1)}}{1- e^{i\pi s}} = - \pi^{s-1} \frac{i e^{i(\pi/2)s}}{1- e^{i\pi s}}\\ = - \pi^{s-1} \frac{i}{e^{-i(\pi/2)s}- e^{i(\pi/2)s}} = \frac{1}{2} \pi^{s-1} \frac{2i}{e^{i(\pi/2)s}- e^{-i(\pi/2)s}} = \frac{1}{2} \frac{\pi^{s-1}}{\sin((\pi/2) s)}. $$ Now in the present case we have $$\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} \frac{1}{k^2} \frac{1}{k^{-s}} = \zeta(2-s).$$ It follows that the Mellin transform $Q(s)$ of $S(x)$ is given by $$ Q(s) = \frac{1}{2} \frac{\pi^{s-1}}{\sin((\pi/2) s)} \zeta(2-s).$$ The Mellin inversion integral for an expansion about zero is $$\int_{1/2-i\infty}^{1/2+i\infty} Q(s)/x^s ds,$$ shifted to the left. The poles from the sine term are at the even integers, with those in the right half plane being canceled by the trivial zeros of the zeta function.

We have $$\sum_{q\ge 0} \operatorname{Res}(Q(s)/x^s; s=-2q) = \sum_{q\ge 0} \frac{(-1)^q}{\pi^{2q+2}} \zeta(2+2q) x^{2q}\\ = \sum_{q\ge 0} \frac{(-1)^q}{\pi^{2q+2}} \frac{(-1)^q B_{2q+2} (2\pi)^{2q+2}}{2(2q+2)!} x^{2q} = \sum_{q\ge 0} \frac{2^{2q+2} B_{2q+2}}{2(2q+2)!} x^{2q}.$$ We thus have for the initial sum $$\frac{1}{x} + 2x S(x) = \frac{1}{x} + 2x \sum_{q\ge 0} \frac{2^{2q+2} B_{2q+2}}{2(2q+2)!} x^{2q} = \frac{1}{x} + \frac{1}{x} \sum_{q\ge 0} \frac{2^{2q+2} B_{2q+2}}{(2q+2)!} x^{2q+2}.$$

Recall that the generating function of the Bernoulli numbers is precisely $$\frac{x}{e^x-1}.$$ Adapting the expression from the harmonic sum calculation to match this generating function (note that the odd index $B_m$ are zero when $m> 1$) we get $$\frac{1}{x} - \frac{1}{x} + 1 + \frac{1}{x} \sum_{m\ge 0} \frac{B_m}{m!} (2x)^m.$$ Simplify one last time, getting $$ 1 + \frac{1}{x}\frac{2x}{e^{2x}-1} = \frac{e^{2x}+1}{e^{2x}-1} = \frac{e^x+e^{-x}}{e^x-e^{-x}} = \coth x,$$ QED.

Just to recapitulate, the direction of this computation was as follows. We started by conjecturing that a certain harmonic sum represents the hyperbolic cotangent in a neighborhood of zero. Then we computed the Mellin transform of this sum. We inverted that transform for an expansion about zero. The expansion that we obtained was precisely the generating function of the Bernoulli numbers with two exceptions. Substituting that generating function into the sum and correcting for the exceptions we obtained the defining equation of the hyperbolic cotangent in terms of exponentials.

[Math] A conformal mapping onto a region bounded by convex contours (Ahlfors)

Let $\gamma_k$ be the image of $C_k$ under $p+q$.

Since $\gamma_k$ is convex, there are only three types of points on the $w$-plane.

On-curve: $w$ is on $\gamma_k$.
External: We can draw a straight line connecting $w$ and infinity without intersecting $\gamma_k$. Obviously the winding number of $\gamma_k$ with respect to an external point is 0.
Internal: We cannot draw a straight line connecting $w$ and infinity without intersecting $\gamma_k$.

Now let us assume that $w$ is a value of $p+q$, i.e. $p(z) + q(z) = w$ for a certain $z \in \Omega$.

$w$ cannot be internal, because this would be in contradiction with the fact that $\Omega$ is connected and $z_0 \in \Omega$ is mapped to infinity.

If $w$ is external, we are done.

If $w$ is on-curve, either it is on a slit, or it has internal points in its neighborhood.

In the first case, we can make the same argument under (17) on Page 260 in Ahlfors that the principle part of the following integral must vanish.

$$\int_{C_k} \frac{p'(z) + q'(z)}{p(z)+q(z)-w} dz$$

The second case is impossible because of the following corollary (Ahlfors, Page 132).

Corollary 1. A nonconstant analytic function maps open sets onto open sets.

Best Answer

Related Solutions

[Math] fourier expansion of $\coth$ and justifying an identity

[Math] A conformal mapping onto a region bounded by convex contours (Ahlfors)

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