Why do local maxima of sinc squared function appear *close* to odd multiples of $\pi/2$?

Question

My question is pretty much the title. I tried two approaches to get local maxima for the $\operatorname{sinc}$ function (I feel it's safe to assume that to know where local minima and maxima of $\operatorname{sinc}$ are the same as to know where the local maxima of $\operatorname{sinc}²$ are).

The first one is by calculating the derivative with respect to the argument, which gives $$ x^{-1}\cos{x}-x^{-2}\sin{x}=0\implies\tan{x}=x, $$ and, as I understand, you can't solve that thing analytically. I figured that, since the minima and maxima of the $\sin$ function are of the form $x_n=\pi/2+ n\pi$, then for $\operatorname{sinc}²$ the maxima would be $x_n$ for $|n|>1$.

However, when solving the first equation numerically (by trial and error) I find that the maxima are close to $x_n$ but are not exactly $x_n$. This page I found tells the same story.

So, why do the maxima of $\operatorname{sinc}²$ not fall exactly on $x_n$?

Answer 1

Looking at the graphs of $\tan(x)$ and $x$, it looks like there are roots near and less than $(n+\frac12)\pi$.

Suppose $x_n$ is close to $(n+\frac12)\pi$, so $x_n = (n+\frac12)\pi-d_n$.

Then $(n+\frac12)\pi-d_n =\tan((n+\frac12)\pi-d_n) =\tan(\pi/2-d_n) =\frac1{\tan(d_n)} $.

If $d_n$ is small, then $\tan(d_n) \approx d_n $ so $(n+\frac12)\pi-d_n =\frac1{\tan(d_n)} \approx \frac1{d_n} $ so, as a first approximation, $d_n \approx \frac1{(n+\frac12)\pi} $.

By iterating, you can get more accurate approximations.

I remember about 50 years ago finding a paper by Hardy in an 1930's English math journal (maybe JLMS) analyzing the roots of $\tan)x) = x$.

This is the kind of analysis there, though, of course, done much better.

Answer 2

Considering the function $$f(x)=\frac{\sin ^2(x)}{x^2}$$ the first and second derivatives are given by $$f'(x)=\frac{2 \sin (x) (x \cos (x)-\sin (x))}{x^3}$$ $$f''(x)=\frac{\left(2 x^2-3\right) \cos (2 x)-4 x \sin (2 x)+3}{x^4}$$ So the extrema correspond to the solutions of $x \cos (x)-\sin (x)=0$ and to the solutions of $\sin(x)=0$. For the last one, it is simply $x=n\pi$.

But $f''(n\pi)=\frac{2}{\pi ^2 n^2} >0$ meaning that the points where $x=n\pi$ are minima and then the maximum values of $f(x)$ correspond to the roots of $x \cos (x)-\sin (x)=0$ that is to say to the solutions of $\tan(x)=x$.

Now, if you look here, you will find the approximation of the solutions $$x=q - \frac 1q-\frac 2{3q^3}-\frac {13}{15q^5}-\frac {146}{105q^7}+\cdots\qquad \text{where} \qquad q=(2n+1)\frac \pi 2$$