I need to find the Fourier series of $$ f(x)=\begin{cases}a,& 0<x<\frac{\pi}{3}\\ 0,&\frac{\pi}{3}<x<\frac{2\pi}{3}\\-a,& \frac{2\pi}{3}<x<\pi\\ \end{cases} $$ $$ f(x)= a_0+\sum_{n=1}^{\infty}a_n\cos\frac{n\pi x}{L}+\sum_{n=1}^{\infty}b_n\sin\frac{n\pi x}{L}, $$ $$ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx $$ $$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)cos(nx)dx $$ $$ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)sin(nx)dx $$ I found That An and A0 are both equal to 0 which meant i only had Bn to find $$ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)sin(nx)dx $$ $$ b_n = \frac{1}{\pi} \int_{0}^{\frac{\pi}{3}} asin(nx)dx + \frac{1}{\pi} \int_\frac{2\pi}{3}^{{\pi}} -asin(nx)dx $$ $$ b_n = \frac{a}{n\pi}\left[-\cos(nx)\right]_0^{\frac{\pi}3} + \frac{a}{n\pi}\left[\cos(nx)\right]_\frac{2\pi}3^{\pi} $$ When n is odd $$ b_n = 0 $$ When n = 2 , 4 $$ b_n = \frac{a}{n\pi}\left[1-\cos(n\frac{\pi}{3}) + \cos(n\pi) -\cos(\frac{2n\pi}{3})\right] $$ $$ b_n = \frac{3a}{n\pi} $$ When n = 6 bn = 0 So this gives me a Fourier Series of $$ f(x)=\frac{3a}{\pi}\left[\frac{1}{2}\sin(2x)+\frac{1}{4}\sin(4x)+\frac{1}{8}\sin(8x).... \right] $$ But the book has a solution of $$ f(x)=\frac{3a}{\pi}\left[\sin(2x)+\frac{1}{2}\sin(4x)+\frac{1}{4}\sin(8x).... \right] $$ Not sure where im going wrong !


The standard Fourier series is on an interval of length $2\pi$, whereas your problem is posed on $[0,\pi]$. $L$ has to figure into the coefficients, which you can see just by looking at the constant term.

In other words, if you are going to expand $f$ on $[0,\pi]$, then the standard expansion will involve a constant, and $\sin(2nx)$, $\cos(2nx)$. Assuming you have such an expansion $$ f(x) = a_{0} + \sum_{n=1}^{\infty} a_{n}\cos(2nx)+\sum_{n=1}^{\infty}b_{n}\cos(2nx), $$ then you multiply by one of these functions, on both sides, integrate over $[0,\pi]$, use the fact that the integrals of unlike terms are zero, in order to obtain $$ \begin{align} \int_{0}^{\pi}f(x)dx & = a_{0}\int_{0}^{\pi}dx =\pi a_{0}\\ \int_{0}^{\pi}f(x)\cos(2nx)dx & = a_{n}\int_{0}^{\pi}\cos^{2}(2nx)\,dx=a_{n}\frac{\pi}{2},\;\;\; n > 0,\\ \int_{0}^{\pi}f(x)\sin(2nx)dx & = b_{n}\int_{0}^{\pi}\sin^{2}(2nx)\,dx=b_{n}\frac{\pi}{2}. \end{align} $$ The values of integrals of $\sin^{2}$ and $\cos^{2}$ are easily determined: both types of integrals are over a full period and, so must be equal; and $\sin^{2}(2nx)+\cos^{2}(2nx)=1$, which means that the integrals of $\sin^{2}$ and $\cos^{2}$ are half of the period (period=$\pi$, so integrals of $\sin^{2}$ and $\cos^{2}$ are $\pi/2$.) So your series becomes $$ \begin{align} f(x) & \sim \frac{1}{\pi}\int_{0}^{\pi}f(x)\,dx + \\ & + \sum_{n=1}^{\infty}\left(\frac{2}{\pi}\int_{0}^{\pi}f(x)\cos(2nx)\,dx\right)\cos(2nx) \\ & +\sum_{n=1}^{\infty}\left(\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(2nx)\,dx\right)\sin(2nx). \end{align} $$ The normalization constants $\frac{2}{\pi}$ and ${1}{\pi}$ come from dividing by the integrals of the squares of $1$, $\sin(2nx)$ and $\cos(2nx)$.

Your function $f$ is $a$ times a part which is odd about the $x=\pi/2$. The $\sin$ terms are odd about $x=\pi/2$ and the $\cos$ terms are even about $x=\pi/2$. That helps simplify. The cosine terms require integrating only the part of $f$ which is even about $x=\pi/2$, while the sine terms require integrating only the part of $f$ which is odd about $x=\pi/2$. So the $\cos$ terms drop out. In this particular case, you need only integrate the correct part over $[0,\pi/2]$ and double the value. So, $$ a_{n} = 0,\;\; b_{n}=2a\int_{0}^{\pi/3}\sin(2nx)\,dx. $$

  • $\begingroup$ so because Ive done it over half the interval i need to double it? $\endgroup$ – Steve Jan 10 '14 at 19:17
  • $\begingroup$ @Steve: The normalization comes from the coefficient equation $a_{n}\int_{0}^{\pi}e_{n}^{2}(x)\,dx = \int_{0}^{\pi}f(x)e_{n}(x)\,dx$, where $e_{n}$ are the eigenfunctions (sin and cos). Take a look at the constant term, for example. Does that help? $\endgroup$ – DisintegratingByParts Jan 10 '14 at 19:27
  • $\begingroup$ Sorry Im not really following that if Im honest $\endgroup$ – Steve Jan 10 '14 at 19:39
  • $\begingroup$ @Steve: You are trying to expand $f=a_{0}+\sum_{n=1}^{\infty}a_{n}\cos(2\pi x/L)+\sum_{n=1}b_{n}\sin(2\pi x/L)$. The trick which leads to the coefficients is to ignore convergence issues, multiply by one of the functions (constant, cos, sin) and integrate on $[0,L]$. Integrals of non-like terms are 0 (this is the orthogonality condition). So you end up with, for example, $\int f(x)\cos(2\pi m x/L)dx=a_{m}\int \cos(2\pi m x/L)\cos(2\pi m x/L)dx$, which means you have to divide by $\int \cos(2\pi mx/L)^{2}dx$. The same type of equation holds for the constant term, too. That's how you normalize. $\endgroup$ – DisintegratingByParts Jan 10 '14 at 19:46
  • $\begingroup$ Soo from ( really struggling with editing) $\int f(x)\cos(2\pi m x/L)dx=a_{m}\int \cos(2\pi m x/L)\cos(2\pi m x/L)dx$ $\int f(x)\cos(2\pi m x/L)dx=a_{m}\int \cos^2(2\pi m x/L)dx$ $\int f(x)\cos(2\pi m x/L)dx=a_{m}[\frac{x}{2}+..sin] $ and that function of sin would be 0 when evaluated between 0 and pi. So $\int f(x)\cos(2\pi m x/L)dx=a_{m}*\frac{\pi}{2}$ $a_{m}=\frac{2}{\pi}\int f(x)\cos(2\pi m x/L)dx$ If it was evaluated between 0 and 2pi it would be $a_{m}=\frac{1}{\pi}\int f(x)\cos(2\pi m x/L)dx$ Correct or not? $\endgroup$ – Steve Jan 10 '14 at 20:17

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