1
$\begingroup$

I want to find the Fourier series for

$f(x)=\left\{\begin{array}{rcl} 0 & \mbox{ if } & -\pi \leq x \leq 0 \\ \sin x & \mbox{ if } & 0 < x < \pi \end{array}\right.$

The Fourier series is given by $F=a_0+\sum (a_n\cos nx + b_n \sin nx)$

My problem is that both $a_n$ and $b_n$ vanish when I try to calculate them.

$$a_0=\frac{1}{2\pi}\int_{0}^{\pi}\sin x = -\frac{1}{2\pi}(\cos \pi - cos 0)= \frac{1}{\pi}$$

$$a_n=\frac{1}{\pi}\int_{0}^{\pi}\sin x\cos nx dx = \frac{1}{\pi}\left(-\cos x \cos nx - \frac{1}{n}\int \sin nx \cos x dx\right)$$ $$=\frac{1}{\pi}\left(-\cos x\cos nx- \frac{1}{n} \left( - \frac{1}{n}\cos x\cos nx -\frac{1}{n}\int \sin x \cos nx dx \right)\right)$$ $$=\frac{1}{\pi}\left(-\cos x\cos nx+\frac{1}{n^2}\cos x \cos nx + \frac{1}{n^2}\int_{0}^{\pi}\sin x\cos nx dx\right)$$ $$\Rightarrow \frac{1}{\pi}\int_{0}^{\pi}\sin x\cos nx dx = \frac{1}{\pi}\frac{(1/n^2-1)\cos x \cos nx}{1-1/n^2}|^\pi_0$$ $$=-\frac{1}{\pi}\cos x \cos nx |^\pi_0 =0$$

$$b_n=\frac{1}{\pi}\int_0^\pi \sin x \sin nx=\frac{1}{\pi}\frac{(1/n^2)\cos x\sin nx - (1/n)\sin x \cos nx}{1-1/n^2}|_0^\pi=0$$

$\endgroup$
  • $\begingroup$ Note $\cos \pi \cos (n\pi) = (-1)^{n+1}$, $\cos 0 \cos (n0) = 1$. $\endgroup$ – Daniel Fischer Nov 29 '16 at 13:59
  • $\begingroup$ You have the wrong basis elements: you should be working with $\sin \left ( \frac{n \pi(x+\pi)}{2\pi} \right )$, and its $\cos$ counterpart (that can be simplified, but still). Also, you definitely should not find that $\int_0^\pi (\sin(x))^2 dx = 0$. In particular, your very last equation involves a division by $0$ when $n=1$. $\endgroup$ – Ian Nov 29 '16 at 14:06
  • $\begingroup$ In line 9 when you integrate by parts, there is $+\frac{1}{n}\int \sin(nx)\cos(x)$ $\endgroup$ – hamam_Abdallah Nov 29 '16 at 14:35
2
$\begingroup$

$a_0=\frac{1}{\pi}\quad$ : OK.

$a_1=\frac{1}{\pi}\int_{0}^{\pi}\sin(x)\cos(x) dx = 0$

$a_n=-\frac{1}{\pi}\int_{0}^{\pi}\sin(x)\cos(nx) dx = -\frac{1}{\pi}\frac{1+\cos(n\pi)}{n^2-1}= -\frac{1}{\pi}\:\frac{1+(-1)^n}{n^2-1}\qquad \qquad n\geq 2$

$b_1=\frac{1}{\pi}\int_{0}^{\pi}\sin^2(x) dx =\frac{1}{2}$

$b_n=\frac{1}{\pi}\int_{0}^{\pi}\sin(x)\sin(nx) dx =0\qquad \qquad n\neq 1$

$$F(x)=\frac{1}{\pi}+\frac{1}{2}\sin(x)-\frac{1}{\pi}\sum_{n=2}^\infty \frac{1+(-1)^n}{n^2-1}\cos(nx)$$

$$F(x)=\frac{1}{\pi}+\frac{1}{2}\sin(x)-\frac{2}{\pi}\sum_{k=1}^\infty \frac{1}{4\:k^2-1}\cos(2kx)$$

Graphs of incomplete Fourier series :

enter image description here

$\endgroup$
  • $\begingroup$ why is $a_0=1/2$? how did you get rid of the $\pi$ value? $\endgroup$ – Cure Nov 29 '16 at 22:54
  • $\begingroup$ This was a typo. $a_0=\frac{1}{\pi}$ in agreement with your answer. $\endgroup$ – JJacquelin Nov 30 '16 at 6:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.