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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?

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    $\begingroup$ What denotes $sinc(\pi x)$? $\endgroup$ Commented Jul 8, 2018 at 18:15
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    $\begingroup$ Sorry, assumed it was common notation. $\operatorname{sinc}{\pi x}=\frac{\sin \pi x}{\pi x}$ - is this not in common use? Mathematica recognises it, Wikipedia has a page, etc... $\endgroup$ Commented Jul 8, 2018 at 18:22
  • $\begingroup$ Yes, @RichardBurke-Ward, the notation is standard, but not universal among mathematicians. I did not pay attention to it myself until about 30 years after my PhD. :) $\endgroup$ Commented Aug 22, 2023 at 22:17
  • $\begingroup$ Use the inverse $\frac{\tan(x)}x$ series or the Bessel J zero function $\endgroup$ Commented Aug 22, 2023 at 22:20
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    $\begingroup$ @Dan, right! Signal processing. My acquaintance was belatedly precipitated by doing Fourier transform examples, and accidentally discovering such stuff. Perhaps the "Borwein integrals" (a fun thing to see) were a significant link in some random chain of example-hunting. :) $\endgroup$ Commented Aug 22, 2023 at 22:36

2 Answers 2

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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).

Sinc with minimax

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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.


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.

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  • $\begingroup$ But that just takes us back to the original statement, that the maxima and minima are defined by $\cos (\pi x)=\operatorname{sinc}{(\pi x)}$... I'm after a formula that gives the maxima and minima in terms of $x$ - as in "$min=f(x)$ and $max=g(x)$... $\endgroup$ Commented Jul 8, 2018 at 20:42
  • $\begingroup$ Ah, ok I thought you just wanted to derive that every point that is either max or min is an intersection with cos. Well, do you know about continouity and theorem of intermediate value? $\endgroup$ Commented Jul 8, 2018 at 20:45
  • $\begingroup$ Oops. No. Can you help me with that? $\endgroup$ Commented Jul 8, 2018 at 20:48
  • $\begingroup$ Apologies if I don't reply for a while; it's getting late in this part of the world. But I would really appreciate some pointers, or a solution - and I promise to respond to comments / answers tomorrow. G'night. $\endgroup$ Commented Jul 8, 2018 at 20:53
  • $\begingroup$ @RichardBurke-Ward I sketched a proof for which are min and which are max. You can probably fill in the blanks if you are taking a first calculus course for engineers or something like that. $\endgroup$ Commented Jul 8, 2018 at 20:57

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