Differential Equations - Fourier Series: Examples - iii

Example 3 Find the Fourier series for $f\left(x\right)=x$ on $-L\le x\le L$.

Let’s start with the integrals for $A_n$.
$$A_0=\frac{1}{2L}{\int }_{-L}^{L}f\left(x\right)dx=\frac{1}{2L}{\int }_{-L}^{L}xdx=0$$$$A_n=\frac{1}{L}{\int }_{-L}^{L}f\left(x\right)\cos \left(\frac{n\pi x}{L}\right)dx=\frac{1}{L}{\int }_{-L}^{L}x\cos \left(\frac{n\pi x}{L}\right)dx=0$$In both cases note that we are integrating an odd function ($x$ is odd and cosine is even so the product is odd) over the interval $\left[-L,L\right]$ and so we know that both of these integrals will be zero.
Next here is the integral for $B_n$
$$B_n=\frac{1}{L}{\int }_{-L}^{L}f\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dx=\frac{1}{L}{\int }_{-L}^{L}x\sin \left(\frac{n\pi x}{L}\right)dx=\frac{2}{L}{\int }_0^{L}x\sin \left(\frac{n\pi x}{L}\right)dx$$In this case we’re integrating an even function ($x$ and sine are both odd so the product is even) on the interval $\left[-L,L\right]$and so we can “simplify” the integral as shown above. The reason for doing this here is not actually to simplify the integral however. It is instead done so that we can note that we did this integral back in the Fourier sine series section and so don’t need to redo it in this section. Using the previous result we get,
$$B_n=\frac{{\left(-1\right)}^{n+1}2L}{n\pi }n=1,2,3,\dots$$In this case the Fourier series is,
$$f\left(x\right)=\sum _{n=0}^{\infty }{A}_n\cos \left(\frac{n\pi x}{L}\right)+\sum _{n=1}^{\infty }{B}_n\sin \left(\frac{n\pi x}{L}\right)=\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}2L}{n\pi }\sin \left(\frac{n\pi x}{L}\right)$$

If you go back and take a look at Example 1 in the Fourier sine series section, the same example we used to get the integral out of, you will see that in that example we were finding the Fourier sine series for $f\left(x\right)=x$ on $-L\le x\le L$. The important thing to note here is that the answer that we got in that example is identical to the answer we got here.

If you think about it however, this should not be too surprising. In both cases we were using an odd function on $-L\le x\le L$ and because we know that we had an odd function the coefficients of the cosines in the Fourier series, $A_n$, will involve integrating and odd function over a symmetric interval, $-L\le x\le L$, and so will be zero. So, in these cases the Fourier sine series of an odd function on $-L\le x\le L$ is really just a special case of a Fourier series.

Note however that when we moved over to doing the Fourier sine series of any function on $0\le x\le L$ we should no longer expect to get the same results. You can see this by comparing Example 1 above with Example 3 in the Fourier sine series section. In both examples we are finding the series for $f\left(x\right)=x-L$ and yet got very different answers.

So, why did we get different answers in this case? Recall that when we find the Fourier sine series of a function on $0\le x\le L$ we are really finding the Fourier sine series of the odd extension of the function on $-L\le x\le L$ and then just restricting the result down to $0\le x\le L$. For a Fourier series we are actually using the whole function on $-L\le x\le L$instead of its odd extension. We should therefore not expect to get the same results since we are really using different functions (at least on part of the interval) in each case.

So, if the Fourier sine series of an odd function is just a special case of a Fourier series it makes some sense that the Fourier cosine series of an even function should also be a special case of a Fourier series. Let’s do a quick example to verify this.

Example 4 Find the Fourier series for $f\left(x\right)=x^2$ on $-L\le x\le L$.

Here are the integrals for the $A_n$ and in this case because both the function and cosine are even we’ll be integrating an even function and so can “simplify” the integral.
$$A_0=\frac{1}{2L}{\int }_{-L}^{L}f\left(x\right)dx=\frac{1}{2L}{\int }_{-L}^L{x}^{2}dx=\frac{1}{L}{\int }_0^L{x}^{2}dx$$$$A_n=\frac{1}{L}{\int }_{-L}^{L}f\left(x\right)\cos \left(\frac{n\pi x}{L}\right)dx=\frac{1}{L}{\int }_{-L}^L{x}^2\cos \left(\frac{n\pi x}{L}\right)dx=\frac{2}{L}{\int }_0^L{x}^2\cos \left(\frac{n\pi x}{L}\right)dx$$As with the previous example both of these integrals were done in Example 1 in the Fourier cosine series section and so we’ll not bother redoing them here. The coefficients are,
$$A_0=\frac{L^2}{3}{A}_n=\frac{4L^2{\left(-1\right)}^n}{n^2{\pi }^2},n=1,2,3,\dots$$Next here is the integral for the $B_n$
$$B_n=\frac{1}{L}{\int }_{-L}^{L}f\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dx=\frac{1}{L}{\int }_{-L}^L{x}^2\sin \left(\frac{n\pi x}{L}\right)dx=0$$In this case the function is even and sine is odd so the product is odd and we’re integrating over $-L\le x\le L$ and so the integral is zero.
The Fourier series is then,
$$f\left(x\right)=\sum _{n=0}^{\infty }{A}_n\cos \left(\frac{n\pi x}{L}\right)+\sum _{n=1}^{\infty }{B}_n\sin \left(\frac{n\pi x}{L}\right)=\frac{L^2}{3}+\sum _{n=1}^{\infty }\frac{4L^2{\left(-1\right)}^n}{n^2{\pi }^2}\cos \left(\frac{n\pi x}{L}\right)$$

As suggested before we started this example the result here is identical to the result from Example 1 in the Fourier cosine series section and so we can see that the Fourier cosine series of an even function is just a special case a Fourier series.