All the roots of $\lambda x+\cot(x)=0$ are real. Looking for an alternative proof

Question

I am looking for an alternative (possibly simpler) proof of the following fact, that has some relevance in finding the eigenfunctions for the laplacian operator.

For any $\lambda\in\mathbb{R}^+$, $\lambda\geq 1$, all the solutions of $$ \cot(x) + \lambda x = 0$$ belong to $\mathbb{R}$.

My proof goes as follows: we just have to prove that all the roots of $f(x)=\cot(x)+\lambda x$ are real. If we assume the opposite, then $f'(x)=\lambda-\frac{1}{\sin^2(x)}$ must have some complex root by the Gauss-Lucas theorem. That is the same as stating that for some $z\in\mathbb{C}\setminus\mathbb{R}$ we have $\sin(z)=\pm\frac{1}{\sqrt{\lambda}}$, or, by setting $z=\sigma+it$, $$ e^{i\sigma-t}-e^{-i\sigma+t} = \pm\frac{2i}{\sqrt{\lambda}}. $$ However, if we set $w=e^{iz}$, the equation $w-\frac{1}{w}=\frac{2i}{\sqrt{\lambda}}$ is solved only by $$ w = \frac{i\pm\sqrt{\lambda-1}}{\sqrt{\lambda}} $$ that is a number on the unit circle. That leads to $z\in\mathbb{R}$, i.e. $t=0$.

Extra question. What can we say about the distribution of the roots if $\lambda\in\mathbb{R}$ but $\lambda < 1$?

Probably the methods indicated on pp. 9-10 of the slides from my April 2006 MAA talk The Remarkable Equation $\tan x = x$ can be applied, but I don't have time now to look into this. — Dave L. Renfro, Apr 12 '16 at 16:12

Mark Viola · Accepted Answer · 2016-04-12T18:46:33.213

4

Following the methodology outlined in the paper referenced by David Renfro, let $z=x+iy$ with $x\in \mathbb{R}$ and $y\in \mathbb{R}$. Then, assume that $x\ne 0$ and $y\ne 0$. The equation $\cot (z)+\lambda z=0$ implies that

$$\frac{\sin(2x)}{\cosh(2y)-\cos(2x)}-i\frac{\sinh(2y)}{\cosh(2y)-\cos(2x)}+\lambda (x+iy)=0 \tag 1$$

Equating real and imaginary parts of $(1)$ reveals

$$\begin{align} \frac{\sin(2x)}{\cosh(2y)-\cos(2x)}&=-\lambda x \tag 2\\\\\ \frac{\sinh(2y)}{\cosh(2y)-\cos(2x)}&=\lambda y \tag 3 \end{align}$$

Since neither $x$ nor $y$ is zero, the denominator of the left-hand sides of $(2)$ and $(3)$ are strictly positive. We may divide, therefore, $(2)$ into $(3)$ to obtain

$$\frac{\sin(2x)}{\sinh(2y)}=-\frac{x}{y}$$

for $\lambda \ne 0$, whereupon rearranging yields

$$- 2 < \frac{\sin(2x)}{x}=-\frac{\sinh(2y)}{y} < -2\tag 4$$

We arrive at the desired contradiction. Therefore, there can only be purely real or purely imaginary solutions to $\cot(z)+\lambda z=0$, $\lambda \ne 0$.

If $x=0$, then solutions of the equation of interest are solutions to

$$\coth(y)=-\lambda y \tag 5$$

If we restrict $\lambda\ge 1$, then we see that there are no solutions to $\cot(iy)+\lambda (iy)=0$.

NOTE:

Special thanks to Martin R for identifying several errata in the original edit.

edited Apr 12 '16 at 18:46

answered Apr 12 '16 at 17:01

Mark Viola

179,405

The left-hand side of (4) is not positive for all $x$. – Shouldn't one at least mention that this uses the methods from the paper cited by Dave L. Benfro in above comment? – Martin R Apr 12 '16 at 17:28
Also I think that the equation (5) should be $\coth (y) = \lambda y$. I might have an error in my calculations though. – Martin R Apr 12 '16 at 17:37
@JackD'Aurizio Thank you Jack! Much appreciative, -Mark – Mark Viola Apr 12 '16 at 17:46
The left-hand side of (4) can be negative! – Martin R Apr 12 '16 at 17:47
Well, I had the same solution in preparation :) – I still think that $(5)$ is wrong, you should double-check that. – Martin R Apr 12 '16 at 17:55
No. $\cos(0)$ is $1$, not $0$. – Martin R Apr 12 '16 at 17:58
Your final conclusion does not fit to equation (5). – My calculations give $\coth(y) = - \lambda y$ as equation (5), which has no solution for $\lambda \ge 0$ and two solutions for $\lambda < 0$. – Martin R Apr 12 '16 at 18:35
@MartinR We are restricting $\lambda\ge 1$. So, there are no solutions upon such a restricition. – Mark Viola Apr 12 '16 at 18:46
Jack, in the OP, isn't $\lambda \ge 1$ a constraint? – Mark Viola Apr 12 '16 at 19:14
@JackD'Aurizio: As I commented above, equation $(5)$ has no solution for $\lambda \ge 0$ and two solutions for $\lambda < 0$ (easy to see from the graph of $\coth$). That covers all cases. – Martin R Apr 12 '16 at 19:36
@MartinR: sure, thanks. – Jack D'Aurizio Apr 12 '16 at 19:47

All the roots of $\lambda x+\cot(x)=0$ are real. Looking for an alternative proof

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