Fourier Series for Periods Other Than $2\pi$

• What if the period of $f(x)$ is not $2\pi$? we just make a change of variable. If $f(x)$ is periodic of period $P$, the function can be considered to have one period between $-P/2$ and $P/2$. The functions $sin(2\pi x/P)$ and $cos(2\pi x/P)$ are periodic between $-P/2$ and $P/2$. The formulae become, for $f(x)$ periodic of period $P$;
  

  $$A_n=\frac{1}{P/2}\int_{-P/2}^{P/2} f(x)cos\left( \frac{\pi }{P/2} nx\right) dx,~n=0,1,2,\ldots$$ (7)

  
  
  
  

  $$B_n=\frac{1}{P/2}\int_{-P/2}^{P/2} f(x)sin\left( \frac{\pi }{P/2} nx\right) dx,~n=1,2,3,\ldots$$ (8)

  
  
  Because a function that is periodic with period $P$ between $-P/2$ and $P/2$ is also periodic with period $P$ between $A$ and $A + P$, the limits of integration in Eqs. 7 and 8 can be from $0$ to $P$.

Figure 2: Left: Plot of $f(x)=x$, periodic of period $2\pi$,Right: Plot of the Fourier series expansion for $N=2,4,8$.

Examples:

1. Let $f(x)=x$ be periodic between $-\pi$ and $\pi$. (See Figure 2left). Find the $A$s and $B$s of its Fourier expansion. For $A_0$;
   
   $$A_0=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)dx=\frac{1}{\pi}\int_{-\pi}^{\pi} xdx=\left[ \frac{x^2}{2\pi}\right] ^{\pi}_{-\pi}=0$$
   
   
   For the other $A$s;
   
   $$A_n=\frac{1}{\pi}\int_{-\pi}^{\pi} xcos(nx)dx=0$$
   
   
   For the other $B$s;
   
   $$B_n=\frac{1}{\pi}\int_{-\pi}^{\pi} xsin(nx)dx=\frac{2(-1)^{n+1}}{n},~n=1,2,3,\ldots$$
   
   
   We then have
   
   $$x\approx 2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} sin(nx),~-\pi<x<\pi$$
   
   
   Figure 2right shows how the series approximates to the function when only two, four, or eight terms are used.
2. Find the Fourier coefficients for $f(x)= \vert x\vert$ on $-\pi$ to $\pi$;
   
   $$A_0=\frac{1}{\pi}\int_{-\pi}^{0} -xdx+\frac{1}{\pi}\int_{0}^{\pi} xdx=\pi$$
   
   
   
   
   $$A_n=\frac{1}{\pi}\int_{-\pi}^{0} (-x)cos(nx)dx+\frac{1}{\pi}... ...c{-4}{(n^2\pi)}, & n=1,3,5,\ldots\\ \end{array}\right\rbrace$$
   
   

   
   

   $$B_n=\frac{1}{\pi}\int_{-\pi}^{0} (-x)sin(nx)dx+\frac{1}{\pi}\int_{0}^{\pi} (x)sin(nx)dx=0$$
   
   
   Because the definite integrals are nonzero only for odd values of $n$, it simplifies to change the index of the summation. The Fourier series is then
   
   $$\vert x\vert\approx\frac{\pi}{2}-\frac{4}{\pi}\sum_{n=1}^{\infty} \frac{cos(2n-1)x}{(2n-1)^2}$$
   
   
   Figure 3 shows how the series approximates the function when two, four, or eight terms are used.

   Figure 3: Plot of Fourier series for $\vert x\vert$ for $N=2,4,8$.

   
3. Find the Fourier coefficients for $f(x) = x(2 - x)=2x - x^2$ over the interval [-2, 2] if it is periodic of period 4. Equations 7 and 8 apply.
   
   $$A_0=\frac{2}{4}\int_{-2}^{2}(2x-x^2)dx=\frac{-8}{3}$$
   
   
   
   
   $$A_n=\frac{2}{4}\int_{-2}^{2} (2x-x^2)cos\left( \frac{n\pi x}{2}\right) dx=\frac{16(-1)^{n+1}}{n^2\pi^2},~n=1,2,3,\ldots$$
   
   
   
   
   $$B_n=\frac{2}{4}\int_{-2}^{2} (2x-x^2)sin\left( \frac{n\pi x}{2}\right) dx=\frac{8(-1)^{n+1}}{n\pi},~n=1,2,3,\ldots$$
   
   
   
   
   $$x(2-x)\approx \frac{-4}{3}+\frac{16}{\pi^2}\sum_{n=1}^{\inft... ...{\infty}\frac{(-1)^{n+1}}{n}sin\left( \frac{n\pi x}{2}\right)$$
   
   
   Figure 4 shows how the series approximates to the function when 40 terms are used.

   Figure 4: Plot of Fourier series for $x(2-x)$ for $N=40$.

   

With MATLAB,