An exact equation to calculate the extrema of $\dfrac{\sin(x)}{x}$

Question

A couple of days ago I calculated that the $m^{th}$ extrema of $\dfrac{\sin(x)}{x}$ denoted by $y_m$ is given by this equation below:

$$ \sqrt{1-y_m^2} +y_m \sin^{-1}(y_m)-y_m(-1)^m \left(2m+1\right)\dfrac{\pi}{2}=0$$

which can be simplified to the following :

$$ \sin^{-1}(y_m)+\cot(\sin^{-1}(y_m))-(-1)^m \left(2m+1\right)\dfrac{\pi}{2}=0$$

I cannot find any reference to this expression I calculated anywhere online in any published work or in general. Could anyone tell me whether it is an already known equation?

References:

1. Locations and amplitudes of the extrema of the sinx/x function

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Edit: Since this is confusing some people. I'll clarify that I do not want to solve for $x_k$ (the location of the extrema) but the extrema itself, there are a lot of papers which already have done that in different manners. I instead was looking for a function that directly gives you the extrema. The function which I mentioned above directly gives you the extrema $y(x_m)$ (and not its location $x_m$) when numerically solved for its root. I wanted to know whether this equation which I found was already known/used before. I cannot find any published work where they have given a equation that directly gives you the extremum values for $\dfrac{\sin(x)}{x}$.

Answer 1

Let's find the first positive one. Where $(x_{1},y_{1}) \approx (4.5,-0.2)$

$$f(x) = \frac{sin(x)}{x}$$

$$\frac{d}{dx} f(x) = \frac{x \cdot cos(x) - sin(x)}{x^2}$$

$$\frac{x_{1} \cdot cos(x_{1}) - sin(x_{1})}{x_{1}^{2}} = 0$$

$$x_{1} \cdot cos(x_{1}) - sin(x_{1}) = 0$$

$$x_{1} - tan(x_{1}) = 0$$

$$x_{1} = tan(x_{1})$$

It cannot to be solved exactly.

Answer 2

Here is the derivation of the OP equation.

The extrema $x_m$ of the function $$ f(x)=\frac{\sin x}{x}\tag1 $$ are given by the fixed point of $\tan x$: $$ x_m=\tan x_m\tag2 $$ and lie in the interval $(m\pi,(m+\frac12)\pi)$.

Substituting $x$ in the denominator of (1) with $\tan x_m$ one obtains: $$ y_m=\frac{\sin x_m}{\tan x_m}=\cos x_m $$ or $$ x_m=s_m\arccos y_m+k_m\pi.\tag3 $$ More detailed analysis reveals: $$ s_m=(-1)^m;\quad k_m=\begin{cases}m,&m\text{ even}\\ m+1,&m\text{ odd}\\ \end{cases}.$$

Substitution of (3) in (1) gives rise to the equation $$ y_m=\frac{\sin((-1)^m\arccos y_m+k_m\pi)}{(-1)^m\arccos y_m+k_m\pi}= \frac{(-1)^m\sqrt{1-y_m^2}}{(-1)^m\arccos y_m+k_m\pi} $$ or $$ \frac{\sqrt{1-y_m^2}}{y_m}-\arccos y_m=(-1)^mk_m\pi,\tag4 $$ which should be equivalent to the OP equation.

The equation (4) can be cast in a simpler form: $$ \frac{\sqrt{1-y_m^2}}{|y_m|}-\arccos |y_m|=m\pi.\tag5 $$ However it delivers only the absolute values of $y_m$ which should be additionally multiplied by $(-1)^m$.