Why this function can't be used in Fourier expansion?

Question

$$y = \arccos(\sin(2x))$$

I can't see why it can't be used in Fourier expansion series. It seems to me that it satisfies all the Dirichlet properties:

Periodic ? Yes, $\pi $.

Continuous ? Yes

Finite number of max/mín in a period? Yes, there is one maximum and zero mínimum.

Module of the integral converges? Yes,$\frac{\pi^2}{2}$

So, what is the problem with the function? Why can't it be used for Fourier expansion?

Where is the information that it cannot be expanded into a Fourier series coming from? — , Nov 28 '21 at 19:45
I am fairly certain that your function can be Fourier expanded, my go to reference on Fourier Analysis [Olver Intro to PDE's] states that a Fourier series of a function f(x) converges uniformly if it is 2$\pi$ periodic, piecewise $C^1$ (differentiable with continuous derivative, except at countably many points), and continuous all of which your function satisfies. — Sam, Nov 28 '21 at 19:51
The information is here: Mathematical Methods for physics and engineering, Riley. — Lac, Nov 28 '21 at 19:52
Oh, and the range indicated to the expansion is from negative infinite to positive infinite — Lac, Nov 28 '21 at 19:53
This is just a shifted sawtooth wave which has a Fourier expansion and is often analysed in control & communication courses. Why did you declare that cannot be used? — copper.hat, Nov 28 '21 at 20:07
“… used in Fourier expansion series” is an odd choice of words. It has a Fourier expansion, but the terms used in a Fourier expansion are all of the form $\sin nx,$ $\cos nx.$ — Thomas Andrews, Nov 28 '21 at 20:21

score 1 · Answer 1 · answered Nov 28 '21 at 21:25

The book is available here.

This is Exercise 12.3 on page 429 (12.9 Exercises). However in the book the function is written slightly differently:

(d) $\cos^{-1}(\sin 2x)$, $-\infty < x < \infty$;

Note that the author wrote $\cos^{-1}$ instead of $\arccos$.

The answer is given on page 431 (12.10 Hints and answers).

According to the answer, the "Dirichlet condition" for which it fails is the following:

(ii) it must be single-valued and continuous, except possibly at a finite number of finite discontinuities;

Thus I guess what the author has in mind is that the "function" $\cos^{-1}$ is not single-valued.

1 Answers1