Intuitive explanation of the maximas of the function $\sin(\pi x)/x^2$

Question

I came across this function while playing around in desmos: https://www.desmos.com/calculator/jp0kirirgk

What I cannot understand is :

1. Why are the maximas of $\sin(\pi x)/x^2$ shifted from the maximas of $\sin(\pi x)$?
2. Why do these maximas approach those of $\sin(\pi x)$ as $x$ tends to infinity?

I can understand it analytically by computing the derivative of the function which comes out to be

$$ f'(x)= \frac{\pi x^2\cos(\pi x)-2x\sin(\pi x)}{x^4}=\frac{2\cos(\pi x)(\pi x/2 - \tan(\pi x))}{x^3} $$

and then using the first derivative test so the maximas only exist where $\tan(\pi x)=\pi x/2$; but I am looking for a moral justification.

Answer 1

The function is best understood as a product of two functions, one being $f_1(x)=\sin(\pi x)$, the other being $f_2(x)=\frac{1}{x^2}$.

The answer to your two questions is:

1. The maximas shift because if $x_0$ is the maximum of $f_1$, then going a bit to the left, to $x_0-\epsilon$, will cause $f_1$ to decrease very slightly, while $f_2$ will increase more (because the absolute value of its derivative is bigger), so their product will increase. The more to the left you move, however, the stronger $f_1$'s decrease becomes, until it eventually overpowers $f_2$'s increase.
2. As $x$ becomes bigger, the absolute value of $f_2$'s derivative decreases, so the effect described above becomes smaller. $f_2$ increases less as you move left of $f_1$'s maximum, and therefore its increase can more easily be overpowered by $f_1$'s decrease (which remains the same).

Answer 2

$$f(x)=\frac{\sin (\pi x)}{x^2}\quad \implies \quad f'(x)=\frac{\pi x \cos (\pi x)-2 \sin (\pi x)}{x^3}$$ Because if the $x$ term in front of the derivative, the extrema take place closer and closer to $m=n+\frac 12$.

Using series expansion around this point and continuing with series reversion $$x_f=m-\frac{2}{\pi ^2 m}-\frac{4}{3 \pi ^4 m^3}+O\left(\frac{1}{m^5}\right)$$ whiel for $g(x)=\sin(\pi x)$ the extrema are at $x_g=m$. So, more or less, the shift is given by $$x_g-x_f \sim \frac{2}{\pi ^2 m}$$ which tends to zero when $m$ increases.