$f$ period is $\frac{\pi}{m}$ when $m\in\mathbb{N}$, then its Fourier coefficients $a_n=b_n=0$ for $n\neq 2m\cdot k$.

Question

$f$ is piecewise continuous in $[-\pi,\pi]$ , f's Fourier series is (formally) $\frac{a_0}{2} +\sum_1^\infty a_n\cos(nx)+b_n\sin(nx)$ and $f$ period is $\frac{\pi}{m}$ when $m\in\mathbb{N}$, I wish to demonstrate that $a_n=b_n=0$ for $n\neq 2m\cdot k$.

Well I worked technically, and got the result, yet I don't succeed to justify why and if my method is valid.

$f(x) = \frac{a_0}{2} + \sum_1^\infty a_n\cos(nx) + b_n\sin(nx)$.

$f(x+\pi/m) = \frac{a_0}{2} + \sum_1^\infty a_n\cos(n(x+\frac{\pi}{m})) + b_n\sin(n(x+\frac{\pi}{m})) = $

$\frac{a_0}{2} + \sum_1^\infty a_n(\cos(nx)\cos(\frac{n}{m}\pi) -\sin(nx)\sin(\frac{n}{m}\pi)) + b_n(\cos(nx)\sin(\frac{n}{m}\pi) +\sin(nx)\cos(\frac{n}{m}\pi))$

$=\frac{a_0}{2} + \sum_1^\infty [a_n\cos(\frac{n}{m}\pi) + b_n\sin(\frac{n}{m}\pi)]\cos(nx) +[-a_n(sin(\frac{n}{m}\pi))+b_n\cos(\frac{n}{m}\pi)]sin(nx)$

So we got a second representation for $f(x)$ since $f(x)=f(x+\frac{\pi}{m})$, thus writing the equations:

$a_n =a_n\cos(\frac{n}{m}\pi) + b_n\sin(\frac{n}{m}\pi)$

$b_n = -a_n(\sin(\frac{n}{m}\pi))+b_n\cos(\frac{n}{m}\pi)$

leads to $a_n=b_n=0$ for $n\neq 2m\cdot k$.

My problem is that I don't see how to justify that $f(x)=f(y)$ means that the fourier serieses are the same for $x$ and $y$ especially when I don't know the kind of the series convergence.

Answer 1

Not a complete answer, but a hint and some suggestions

We need three things:

1. "A function has only one trigonometric series"; put differently, there's only one trigonometric series of the form

$$\frac{a_0}{2} +\sum_1^\infty a_n\cos(nx)+b_n\sin(nx)$$

whose sum is everywhere $0$, and that is the one where all coefficients are zero. [This requires proof.]

2. The Fourier series of a (continuous) function (defined by the rules for computing the coefficients $a_i$ and $b_i$) is a trigonometric series for that function (i.e., the series for $f$ actually converges to $f$). [This, too, requires proof.]

3. The Fourier series of a sum of functions (or difference) is the sum of the Fourier series for the individual functions (or the difference of the series). [This follows from the definition of the Fourier coefficients, which shows that the Fourier series is a linear function of the input function.]

If you look at $$ g(x) = f(x) - f(x + \frac{\pi}{m}) $$ you know that this function is everywhere zero. So its Fourier series has all zero coefficients. But you've computed those coefficients in your work, and your conclusion leads to the fact that for the series for $f$, you have $a_n = b_n = 0$ for $n \ne 2m\cdot k$.

So the only missing point is my first claim: that if a trigonometric series (of the kind the Fourier series is) sums to zero, then all the coefficients are zero. Do you perhaps have a theorem that would help you to see this?