1 Convergence of a Fourier series

We have now shown how to obtain a Fourier series for periodic functions. We have suggested that we would expect to be able to approximate such functions by using a few terms of the Fourier series.

The detailed question of the convergence or otherwise of Fourier series has not been discussed. The reason for this is that the great majority of functions likely to be encountered in practice have Fourier series that do indeed converge and can therefore be safely used as approximations.

The precise conditions that have to be fulfilled for a Fourier series to converge are known as Dirichlet conditions after the French mathematician who investigated the matter. The three conditions are listed in the following Key Point.

It follows, for example, that if $f\left(t\right)$ is defined over $\left(-\pi ,\pi \right)$ as one of the following functions $t^3$ or $1∕\left(t-4\right)$ or $3t+2$ and $f\left(t+2\pi \right)=f\left(t\right)$ , then $f\left(t\right)$ can indeed be represented as a Fourier series as each function satisfies the Dirichlet conditions.

On the other hand, if, over $\left(-\pi ,\pi \right)$ , $f\left(t\right)$ is $\frac{1}{t}$ or $\frac{1}{t-2}$ or $tant$ then $f\left(t\right)$ cannot be expanded in a Fourier series because each of these functions has an infinite discontinuity within $\left(-\pi ,\pi \right)$ .

If the Dirichlet conditions are satisfied at a point $t=t_0$ where $f\left(t\right)$ is continuous then, as we would expect, the Fourier series at $t_0$ given by

$\frac{a_0}{2}+{\sum }_{n=1}^{\infty }\left\{a_{n}cos\left(\frac{2n\pi t_0}{T}\right)+b_{n}sin\left(\frac{2n\pi t_0}{T}\right)\right\}\text{converges to the function value}f\left(t_0\right)$

At a point, say $t=t_1$ , at which $f\left(t\right)$ has a discontinuity then the series

$\frac{a_0}{2}+{\sum }_{n=1}^{\infty }\left\{a_{n}cos\left(\frac{2n\pi t_1}{T}\right)+b_{n}sin\left(\frac{2n\pi t_1}{T}\right)\right\}\text{converges to}\frac{1}{2}\left\{f\left(t_{1-}\right)+f\left(t_{1+}\right)\right\}$

where $f\left(t_{1-}\right)$ is the limit of $f\left(t\right)$ as $t$ approaches $t_1$ from the left and $f\left(t_{1+}\right)$ is the limit as $t$ approaches $t_1$ from the right (Figure 19).

Figure 19

Example 2

  Suppose we consider the square wave

Figure 20

Here $f\left(t\right)$ has finite discontinuities at $-\pi ,0$ and $\pi$ . The Fourier series of $f\left(t\right)$ is (see Section 23.3, subsection 2)

$\frac{1}{2}+\frac{2}{\pi }\left(sint+\frac{1}{3}sin3t+\frac{1}{5}sin5t+\dots \right).$

At $t=\frac{\pi }{2}$ , for example, where $f\left(t\right)$ is continuous the square wave converges to $f\left(\frac{\pi }{2}\right)=1.$ On the other hand at $t=\pi$ the Fourier series clearly has the value $\frac{1}{2}$ (since all the sine terms are zero here). This value $\frac{1}{2}$ agrees with the average of the two limiting values of $f\left(t\right)$ at $t=\frac{\pi }{2}$ : $\frac{1}{2}\left(1+0\right)=\frac{1}{2}$ . If we actually put $t=\frac{\pi }{2}$ in the Fourier series we obtain

$\frac{1}{2}+\frac{2}{\pi }\left(sin\left(\frac{\pi }{2}\right)+\frac{1}{3}sin\left(\frac{3\pi }{2}\right)+\frac{1}{5}sin\left(\frac{5\pi }{2}\right)+\dots \right)$

$=\frac{1}{2}+\frac{2}{\pi }\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots \right)$

Since the series converges, as we have seen, to $f\left(\frac{\pi }{2}\right)=1$ , we obtain the interesting result

$\frac{1}{2}+\frac{2}{\pi }\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots \right)=1\text{or}1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots =\frac{\pi }{4}$

Task!

The function

$f\left(t\right)=\left\{\begin{array}{ccc}0\hfill & \hfill & \hfill -\pi <t<0\\ t^2\hfill & \hfill & \hfill 0<t<\pi \end{array}\right.$

$f\left(t+2\pi \right)=f\left(t\right)$

has Fourier series (see Exercise 4 at the end of Section 23.2)

By using a suitable value of $t$ show that

$1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\dots =\frac{{\pi }^2}{6}$

First decide on the appropriate value of $t$ to use:

Answer

Now put $t=\pi$ in the series and decide what the Fourier series will converge to at this value. Hence complete the question:

Answer

Note that in the last Task if we substitute $t=0$ in the Fourier series (which converges to $f\left(0\right)=0$ ) we obtain another infinite series but with alternating signs:

$\frac{{\pi }^2}{6}-2\left(1-\frac{1}{4}+\frac{1}{9}-\dots \right)=0$ or $1-\frac{1}{4}+\frac{1}{9}-\frac{1}{16}+\dots =\frac{{\pi }^2}{12}$

Exercises

1. Obtain the Fourier series of

   $f\left(t\right)=\left|t\right|-\pi \le t\le \pi$ $f\left(t+2\pi \right)=f\left(t\right)$

   By putting $t=0$ show that

   ${\sum }_{n=1}^{\infty }\frac{1}{{\left(2n-1\right)}^2}=\frac{{\pi }^2}{8}$

2. 1. Obtain the Fourier series of the $2\pi$ periodic function

      $f\left(t\right)=\frac{t^2}{4}-\pi \le t\le \pi$

      and use it to obtain the following identities:

      $\text{(i) }1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots =\frac{{\pi }^2}{6}$ $\text{(ii)}1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots =\frac{{\pi }^2}{12}$

   2. Show that $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}\cdots =\frac{{\pi }^2}{8}$
3. Obtain the Fourier series of the $2\pi$ periodic function

   $f\left(t\right)=t-\pi \le t\le \pi$

   Use the series to show that

   $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots =\frac{\pi }{4}$

Answer