Fourier coefficients of g(x) based on Fourier coefficients of f(x)

Question

Assume $f(x)$ is a periodic function with periodicity of $T_0$. If $C_n$ is Fourier coefficient of $f(x)$ , find complex Fourier coefficient of $g(x)= f(x-1)+f(1-x)$ based on $C_n$

my attempt:

by the definition :$$C_n=\frac{1}{T_0}\int_{\frac{-T_0}{2}}^{\frac{T_0}{2}}f(x)e^\frac{-2in\pi x}{T_0}dx \kern10mm (1)$$ let Fourier coefficient of $g(x)$ be $C_{gn}$, so then we have: $$C_{gn}=\!\!\underbrace{\frac{1}{T_0}\int_{\frac{-T_0}{2}}^{\frac{T_0}{2}}f(x-1)e^\frac{-2in\pi x}{T_0}dx}_{\Large I_1}+\!\!\underbrace{\frac{1}{T_0}\int_{\frac{-T_0}{2}}^{\frac{T_0}{2}}f(1-x)e^\frac{-2in\pi x}{T_0}dx}_{\Large I_2}$$

$I_1$ using $(1)$ can be written in below form:

$$I_1=\frac{e^\frac{-2in\pi}{T_0}}{T_0}\int_{\frac{-T_0}{2}}^{\frac{T_0}{2}}f(x-1)e^\frac{-2in\pi (1-x)}{T_0}dx=\frac{e^\frac{-2in\pi}{T_0}}{T_0}\int_{\frac{-T_0}{2}-1}^{\frac{T_0}{2}-1}f(t)e^\frac{-2in\pi t}{T_0}dt=e^\frac{-2in\pi}{T_0}C_n$$ but I wasn't successful writing $I_2$ in a same form as $I_1$ based on $C_n$

and the other question is, was it true to assume $g(x)$ has same periodicity of $T_0$?

any help is appreciated , thanks!

Answer 1

Let me know if the below looks completely wrong.

We first define the following linear operators:

1. Translation: $\mathcal{T}_ag(x) = g(x - a)$
2. Fourier transform: $(\mathcal{F}g)(\gamma) = \int_{-\infty}^\infty g(x)e^{-2\pi ix\gamma}dx$
3. Modulation: $\mathcal{E}_bg(x)=e^{2\pi ibx}g(x)$

Since $(\mathcal{F}g(-x))(\gamma) = (\mathcal{F}g(x))(\gamma)$ the Fourier transform of $g(x)$ is:

$\hat{g}(\gamma) = (\mathcal{F}g)(\gamma) = 2(\mathcal{F}f(x-1))(\gamma) = 2(\mathcal{F}\mathcal{T}_af(x))(\gamma) = 2\mathcal{E}_b(\mathcal{F}f(x))(\gamma) = 2e^{-2\pi i\gamma}\hat{f}(\gamma)$

The Fourier coefficients of $\hat{g}(\gamma)$ is:

$c_n = \frac{1}{T_0}\hat{g}(\frac{n}{T_0}) = \frac{1}{T_0}2e^{\frac{-2\pi in}{T_0}}\hat{f}(\frac{n}{T_0}) = 2e^{\frac{-2\pi in}{T_0}}c_{n, f}$

where $c_{n,f}$ is the fourrier coefficients of $f$