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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

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 $\dfrac{\sin(x)}{x}$.

diffusiondiver11
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  • I succeeded to derive your equation (+1). – user Mar 05 '21 at 15:30
  • @user How did you derive it? Is it pretty straight forward? – diffusiondiver11 Mar 05 '21 at 15:48
  • It was not complicated. I can post it if you'd like. – user Mar 05 '21 at 15:52
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    @user Please do. Thank you. I spent a lot of time deriving that equation I'd surely like to see a shorter way. – diffusiondiver11 Mar 05 '21 at 15:55
  • You might include a link to your answer at https://math.stackexchange.com/questions/1272624/maxima-and-minima-of-sinx-x – Barry Cipra Mar 05 '21 at 16:02
  • @YvesDaoust if you are so insistent on knowing a use, then here it is: the ODE $\dot{x}=rx-\sin(x)$ undergoes bifurcation for many different values of $r$. These $r$ values at which bifurcations occur are given by this exact equation I posted above. Obviously there must have been a reason I am asking for the extrema and not their locations. This very dynamical systems problem of calculating the bifurcating points led me to calculate this equation. – diffusiondiver11 Mar 05 '21 at 16:40
  • @YvesDaoust the equation is partly derived using what you wrote. To account for all $r$ values at once you need to anyway find the equation I did (maybe in a different but equivalent form). I was required to find an approximate (or exact) equation for $r$. The way you have written it, it's independent of $r$ (or a number like $m$ which indexes it) which isn't what I was looking for. I know exactly what I am looking for and I understand what you are trying to suggest as well but it's just not what helps me. But I thank you for your suggestions. – diffusiondiver11 Mar 05 '21 at 17:14

2 Answers2

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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 $(-1)^m$.

user
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    Now it's excatly what I have! Thank you. In your original answer only the maxima were correctly extracted but the minima were somehow wrong. My original way of deriving this equation was through perturbation method which was pretty heafty and time consuming. Good to see there is an even shorter way and better way. – diffusiondiver11 Mar 06 '21 at 02:44
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Let's find the first positive one. Where $(x_{1},y_{1}) \approx (4.5,-0.2)$

function

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

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

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

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

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

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

It cannot to be solved exactly.