Fourier series of $f(x)=\pi -x$ for $x \in [0,2\pi]$

Question

Consider the function

$$f(x)=\begin{cases} \pi -x, \quad0<x<2\pi\\ 0, \quad x=0,x=2\pi\end{cases}$$

I need to find the Fourier Series of $$f(x)=\pi -x, \quad x \in [0,2\pi].$$ First, we can expand the function in a $2\pi$-periodic defined on $\mathbb{R}$

The theory in my textbook is:

For a Reimann-integrable function $f:[-\pi,\pi] \rightarrow \mathbb{R}$. The Fourier series of f is: $$ S[f](x)=\frac{a_0}{2}+ \sum_{k=1}^{\infty}(a_kcos(kx)+b_ksin(kx))$$ where $$ a_k=a_k(f)=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)cos(kx)dx, \quad k=0,1,2, \dots$$ and $$ b_k=b_k(f)=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)sin(kx)dx, \quad k=1,2, \dots$$

Now for $f(x)=\pi -x, \quad x \in [0,2\pi].$

My understanding is that i have to "transform" the function to be defined on $[-\pi,\pi]$ for my theory to work. But i can't see how i can do that.

Hint of the exercise: Make the function odd (symmetric around $0$).
Solution:$$ S[f](x)=2\sum_{k=1}^{\infty}\frac{sin(kx)}{k}$$

Sf=\frac{a_0}{2}+ \sum_{k=1}^{\infty}(a_kcos(kx)+b_ksin(kx))

Edit after answers in comments

We transform the original function:

$$f(x)=\begin{cases} \pi -x, \quad0<x<\pi\\ -x-\pi, \quad -\pi<x<0\end{cases}$$

The new function is odd so: $$a_k=0$$ and $$ b_k=b_k(f)=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)sin(kx)dx = \frac{2}{\pi}\int_{-\pi}^{\pi}(\pi-x)sin(kx)dx= \dots = \frac{2}{k}$$

So we have the result we wanted.

But the original function does not have the same graph as transformed one.
Is this a problem?

Answer 1

Your function $f(x)$ for $-\pi < x <0$ should be $\pi - (x+2\pi) = -\pi-x$ is already correct. Because you want $y$ to be the same as $\pi -x$ on $(\pi, 2\pi)$, but shifted $2\pi$ to the left to be on $(-\pi, 0)$.

Using your textbook's version of Fourier series, then the Fourier series is $2\pi$ periodic. So plotting your series using your computed coefficients $a_{k}$ and $b_{k}$, but for $x \in (0, 2\pi)$ will give the same as the original one. But if you plot the Fourier series on $(-\pi, \pi)$, of course it will be different. To give better understanding, try plot it from $(-10\pi, 10\pi)$ you will see repeated patterns.

Example of Periodic Extension capability of Fourier series (*img from Wikipedia):

Answer 2

FourierSeries means using【sum of complex exponential functions, like $e^{j n\omega x},\quad n\in \mathbb{Z},\omega\in \mathbb{R},x\in \mathbb{R} $】 (one form is $\sum\limits_{n}e^{j n\omega x}$) to approximate a periodic function(here the periodic function can even be a function which transforms a RealNumber to a ComplexNumber).

I don't know what the periodic function here is, so I guessed one. In my figure, their periods are both $2\pi$

f = UnitStep[x]*(Pi - x) + UnitStep[-x]*(-x - Pi);
g = FourierSeries[f, x, 10]
ExpToTrig[g]
Plot[{g, f}, {x, -2 Pi, 2 Pi}, PlotLegends -> {"g", "f"}]
(* They are both periodic function actually, but I only plot them in \
[-2Pi,2Pi]*)