Maxima and minima of $\operatorname{sinc}$ function

Question

The function $\operatorname{sinc}{\pi x}$ has maxima and minima given by the function's intersections with $\cos \pi x$, or alternatively by $\frac {d}{dx}\operatorname{sinc}{\pi x}=0$.

Mathematica tells me that

$$\frac {d}{dx}\operatorname{sinc}{\pi x}=\pi \Bigl(\frac {\cos \pi x}{\pi x}-\frac {\sin \pi x}{\pi^2 x^2}\Bigr)$$

So question 1, how do I prove this?

And question 2, how do I derive an equation for all maxima and minima?

Answer 1

Set derivative equal to 0.

You will after some manipulation like multiplying $(\pi x)^2$ and dividing by $\pi$ get $$\pi x \cos(\pi x) = \sin(\pi x)$$ and equivalently by dividing both sides by $\pi x$ $$\cos(\pi x) = \frac{\sin(\pi x)}{\pi x}$$ Now the right hand side is $\text{sinc}(\pi x)$ and left hand side is the function you want to show it should intersect.

So we are done showing where the extrema are.

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Now to show which are max and which are min.

Sinc as a function is a multiplication between $\frac{1}{\pi x}$ and $\sin(\pi x)$

On $\mathbb R^+$ the first of these is monotonically decreasing and positive. Sin is periodic and alternating +1 -1. Both functions are continuous. We can now use an argument with theorem of intermediate value to show it will be alternatingly max and min with as many maxes as mins.

Answer 2

Here is a tested scheme for getting the maxima and minima of the sinc function. First of all, we all agree that we should set the derivative of the sinc function to zero. (Note that here I am taking $\text{sinc}(x)=\sin(x)/x$. You can always replace $x$ by $\pi x$). Now, the derivative of $\text{sinc}(x)$ is

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

and setting it to zero will lead to the condition

$$x-\tan x=0$$

This problem comes up in many areas of mathematics. The book, An Atlas of Functions, $2^{nd}$ Edition, by Oldham, Myland, and Spanier, Springer, 2009, shows that the the roots of the above equation, call them $r_n$, can be found efficiently by the equivalent relation

$$r_n=n\pi+\arctan(r_n)$$

where $(n-1/2)\pi<r_n<(n+1/2)\pi$. The beauty of this relation is that if always returns a value for $r_n$ that is less than the starting value. Hence, you can start an iteration at the upper limit of $r_n$ and proceed monotonically to the correct result. However, in my own program I simply used a root finder that I am already familiar with. Once you have the roots, you can get the values of the maxima/minima as $y_n=\text{sinc}(r_n)$.

Now, as for which roots are minima and maxima, well, odd-$n$ are minima and even-$n$ are maxima. These are shown as red and blue dots, respectively, in the figure below. Also by symmetry, we also have $y_n=\text{sinc}(-r_n)$ (not shown in the figure).