Title: Default for 23.4 Convergence
Course: HELM Workbooks 2023-2024

Introduction

In this Section we examine, briefly, the convergence characteristics of a Fourier series. We have seen that a Fourier series can be found for functions which are not necessarily continuous (there may be jumps in the curve) - the only requirement that we have made is that the function be periodic. We have seen that the more terms we take in the Fourier series the better is the approximation to the given signal. But an obvious question to ask is what happens at the points of discontinuity? What does the Fourier series converge to at these points? It must converge to something (finite) since a Fourier series is a sum of very smooth continuous functions. In this Section we give the answer to this question.

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.

Key Point

The Dirichlet conditions for the convergence of a Fourier series of a periodic function \(f(t)\) are:

1. \(f(t)\) must have only a finite number of finite discontinuities, within one period

2. \(f(t)\) must have a finite number of maxima and minima over one period

3. the integral \( \displaystyle{ \int _{-\frac{T}{2}}^{\frac{T}{2}} } \lvert f(t) \rvert dt \) must be finite

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

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

If the Dirichlet conditions are satisfied at a point \(t = t_0\) where \(f(t)\) is continuous then, as we would expect, the Fourier series at \(t_0\) given by \[ \dfrac{a_0}{2} + \sum _{n=1}^\infty \left\{ a_n \cos \left( \dfrac{2n\pi t_0}{T} \right) + b_n \sin \left( \dfrac{2n\pi t_0}{T} \right) \right\} \hspace{1cm} \text{converges to the function value } f(t_0) \] At a point, say \(t = t_1\), at which \(f(t)\) has a discontinuity then the series \[ \dfrac{a_0}{2} + \sum _{n=1}^\infty \left\{ a_n \cos \left( \dfrac{2n\pi t_1}{T} \right) + b_n \sin \left( \dfrac{2n\pi t_1}{T} \right) \right\} \hspace{1cm} \text{converges to } \dfrac{1}{2} \{ f(t_{1-}) + f(t_{1+})\}   \] where \(f (t_{1-})\) is the limit of \(f(t)\) as \(t\) approaches \(t_1\) from the left, and \(f (t_{1+})\) is the limit of \(f(t)\) as \(t\) approaches \(t_1\) from the right as shown in the Figure below.

[Image: approaching limit from both sides]

Key Point

If Dirichlet conditions are satisfied then at a point of continuity \(t = t_0\)

\[ f(t_0) = \dfrac{a_0}{2} + \sum _{n=1}^\infty \left\{ a_n \cos \left( \dfrac{2n\pi t_0}{T} \right) + b_n \sin \left( \dfrac{2n\pi t_0}{T} \right) \right\} \] whereas at a point of discontinuity \(t = t_1\) the Fourier series converges to the average of the two limiting values \[ \dfrac{1}{2} \{ f(t_{1-}) + f(t_{1+}) \}= \dfrac{a_0}{2} + \sum _{n=1}^\infty \left\{ a_n \cos \left( \dfrac{2n\pi t_1}{T} \right) + b_n \sin \left( \dfrac{2n\pi t_1}{T} \right) \right\} \hspace{1cm} \]

Example

Suppose we consider the square wave

[Image: square-pulse]

Here \(f(t)\) has finite discontinuities at \(-\pi\), \(0\) and \(\pi\). The Fourier series of \(f(t)\) is \[ \dfrac{1}{2} + \dfrac{2}{\pi} \left( \sin t + \dfrac{1}{3} \sin 3t + \dfrac{1}{5} \sin 5t + \dots \right) .\] At \(t = \dfrac{\pi}{2}\), for example, where \(f(t)\) is continuous the square wave converges to \( f \left( \dfrac{\pi}{2}\right) = 1\). On the other hand at \(t = \pi\) the Fourier series clearly has the value \( \dfrac{1}{2}\) (since all the sine terms are zero here). This value \( \dfrac{1}{2} \) agrees with the average of the two limiting values of \(f(t)\) at \(t = \dfrac{\pi}{2} : \dfrac{1}{2} (1+0) = \dfrac{1}{2} \). If we actually put \(t = \dfrac{\pi}{2}\) in the Fourier series we obtain \[ \dfrac{1}{2} + \dfrac{2}{\pi} \left( \sin \left( \dfrac{\pi}{2} \right) + \dfrac{1}{3} \sin \left( \dfrac{3\pi}{2} \right) + \dfrac{1}{5} \sin \left( \dfrac{5\pi}{2} \right) + \dots \right) \] \[ = \dfrac{1}{2} + \dfrac{2}{\pi} \left( 1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \dots \right) \] Since the series converges, as we have seen, to \(f \left( \dfrac{\pi}{2} \right) = 1\), we obtain the interesting result \[ \dfrac{1}{2} + \dfrac{2}{\pi} \left( 1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \dots \right) = 1 \] or \[ 1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \dots = \dfrac{\pi}{4} \]

Example

Another interesting result can be discovered by considering the function \[ f(t) = \begin{cases} 0 \hspace{1cm} - \pi < t < 0 \\ t^2 \hspace{1cm} 0 < t < \pi \end{cases} \] \[ f(t + 2\pi ) = f(t) \] shown in the Figure below

[Image: piece-wise function]

This has Fourier series \[ \begin{align} \dfrac{\pi^2}{6} &- 2 \left( \cos t - \dfrac{ \cos 2t}{4} + \dfrac{ \cos 3t}{9} - \dots \right) \\ &+ \left\{ \left( \pi -\dfrac{4}{\pi} \right) \sin t - \dfrac{\pi}{2} \sin 2t + \left( \dfrac{\pi}{3} - \dfrac{4}{9\pi} \right) \sin 3t - \dfrac{\pi}{4} \sin 4t + \dots \right\} \end{align} \] If we now set \(t = \pi\), all of the sine terms become zero, and we are left with \[ \begin{align} \dfrac{\pi^2}{6} - &2 \left( \cos \pi - \dfrac{ \cos 2\pi}{4} + \dfrac{ \cos 3\pi}{9} - \dots \right) \\ &= \dfrac{\pi^2}{6} - 2 \left( -1 - \dfrac{1}{4} - \dfrac{1}{9} - \dfrac{1}{16} - \dots \right) = \dfrac{\pi^2}{6} + 2 \left( 1 + \dfrac{1}{4} + \dfrac{1}{9} + \dfrac{1}{16} + \dots \right) \end{align} \] At \(t = \pi\) the Fourier series will converge to \[ \dfrac{1}{2} (\pi^2 + 0) = \dfrac{\pi^2}{2} \] (the average of the left and right hand limits).

So \[ \dfrac{\pi^2}{6} + 2 \left( 1 + \dfrac{1}{4} + \dfrac{1}{9} + \dfrac{1}{16} + \dots \right) = \dfrac{\pi^2}{2} \] \[ \therefore 1 + \dfrac{1}{4} + \dfrac{1}{9} + \dfrac{1}{16}+ \dots = \dfrac{1}{2} \left( \dfrac{\pi^2}{2} - \dfrac{\pi^2}{6} \right) = \dfrac{\pi^2}{6} \]

Note also that if we substitute \(t = 0\) in the Fourier series (which converges to \(f(0) = 0\)) we obtain another infinite series but with alternating signs: \[ \dfrac{\pi^2}{6} - 2 \left( 1 - \dfrac{1}{4} + \dfrac{1}{9} - \dfrac{1}{16} + \dots \right) = 0\] or \[ 1 - \dfrac{1}{4} + \dfrac{1}{9} - \dfrac{1}{16} + \dots = \dfrac{\pi^2}{12} \]

Exercise

Obtain the Fourier series of \[ f(t) = \lvert t \rvert \hspace{1cm} \pi \leq t \leq \pi \hspace{1cm} f(t + 2\pi) = f(t) \] and use it to calculate the summation \( \displaystyle{\sum _{n = 1}^\infty }\dfrac{1}{(2n - 1)^2} \) by putting \( t = 0\).

\( \displaystyle{ \sum _{n = 1}^\infty } \dfrac{1}{(2n - 1)^2} = \)
Feedback: As \( f(t) \) is an even function, \( b_n = 0 \). For \( a_0\) we have \[ \begin{align} a_0 &= \dfrac{1}{\pi} \displaystyle{ \int _{-\pi}^\pi } \lvert t \rvert \hspace{0.1cm} dt \\ &= \dfrac{1}{\pi} \displaystyle{ \int _{-\pi}^0 } -t \hspace{0.1cm} dt + \dfrac{1}{\pi} \displaystyle{ \int _{0}^\pi } t \hspace{0.1cm} dt \\ &= \pi \end{align} \] By using integration by parts, we then find \[ \begin{align} a_n &= \dfrac{1}{\pi} \displaystyle{ \int _{-\pi}^\pi } \lvert t \rvert \cos nt \hspace{0.1cm} dt \\ &= \dfrac{1}{\pi} \displaystyle{ \int _{-\pi}^0 } - t \cos nt \hspace{0.1cm} dt + \dfrac{1}{\pi} \displaystyle{ \int _{0}^\pi } t \cos nt \hspace{0.1cm} dt \\ &= \dfrac{1}{\pi} \left( \left[ -\dfrac{t}{n} \sin nt \right]_{-\pi}^0 + \dfrac{1}{n} \displaystyle{ \int _{-\pi}^0 } \sin nt \hspace{0.1cm} dt + \left[ \dfrac{t}{n} \sin nt \right]_{0}^\pi - \dfrac{1}{n} \displaystyle{ \int _{0}^\pi } \sin nt \hspace{0.1cm} dt \right) \\ &= \dfrac{1}{\pi} \left( \left[ - \dfrac{1}{n^2} \cos nt \right]_{-\pi}^0 + \left[ \dfrac{1}{n^2} \cos nt \right]_{0}^\pi \right) \\ &= \dfrac{2}{n^2 \pi} \left( \cos n\pi - 1 \right) \end{align} \] \[ = \begin{cases} 0 \hspace{1cm} n = 2, 4, 6, \dots \\ -\dfrac{4}{n^2 \pi} \hspace{1cm} n = 1, 3, 5, \dots \end{cases} \] Now rewriting \(n\) to take the form of an odd number as \( 2n - 1\) (where \(n = 1, 2, 3, 4, \dots\)), we have the Fourier series \[ f(t) = \dfrac{\pi}{2} + \sum _{n=1}^\infty \dfrac{-4}{(2n-1)^2\pi} \cos \left[ (2n - 1)t \right] \] Now setting \(t = 0\) we have \(f(t) = 0\), and therefore \[ 0 = \dfrac{\pi}{2} + \sum _{n=1}^\infty \dfrac{-4}{(2n-1)^2\pi} \] which can be rearranged to give \[ \sum _{n=1}^\infty \dfrac{1}{(2n-1)^2} = \dfrac{\pi^2}{8} \]

Exercise

Obtain the Fourier series of \[ f(t) = t \hspace{1cm} \pi \leq t \leq \pi \hspace{1cm} f(t + 2\pi) = f(t) \] and use it to find the answer to \( 1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \dots\)

\( 1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \dots = \)
Feedback: As \( f(t) \) is an odd function, \( a_n = 0 \). For \( b_n\) we have \[ \begin{align} b_n &= \dfrac{1}{\pi} \displaystyle{ \int _{-\pi}^\pi } t \sin nt \hspace{0.1cm} dt \\ &= \dfrac{1}{\pi} \left( \left[ -\dfrac{t}{n} \cos nt \right]_{-\pi}^\pi + \dfrac{1}{n} \displaystyle{ \int _{-\pi}^\pi } \cos nt \hspace{0.1cm} dt \right) \\ &= \dfrac{1}{\pi} \left( -\dfrac{\pi}{n} \cos n \pi -\dfrac{\pi}{n} \cos n \pi + \dfrac{1}{n^2} \left[ \sin n\pi \right] _{-\pi}^\pi \right) \\ &= -\dfrac{2\cos n \pi}{n} \end{align} \] \[ = \begin{cases} -\dfrac{2}{n} \hspace{1cm} n = 2, 4, 6, \dots \\ \dfrac{2}{ni} \hspace{1cm} n = 1, 3, 5, \dots \end{cases} \] which can be expressed as \( \dfrac{-2(-1)^n}{n} \). So the Fourier series is given by \[ f(t) =-2 \sum _{n=1}^\infty \dfrac{(-1)^n}{n} \sin nt \] Now setting \(t = \dfrac{\pi}{2} \) we have \(f(t) = \dfrac{\pi}{2}\). The Fourier series is \[ -2 \sum _{n=1}^\infty \dfrac{(-1)^n}{n} \sin \dfrac{n\pi}{2} \] which is equal to zero for even values of \(n\). Thus we have which can be rearranged to give \[ \begin{align} \dfrac{\pi}{2} &= -2 \sum _{n=1}^\infty \dfrac{(-1)^n}{n}\sin \dfrac{n\pi}{2} \\ &= 2 \left( 1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \dots \right) \end{align} \] Thus \[ 1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \dots = \dfrac{\pi}{4} \]

Exercise

Verify that the Fourier series of \[ f(t) = \dfrac{ t^2}{4} \hspace{1cm} \pi \leq t \leq \pi \hspace{1cm} f(t + 2\pi) = f(t) \] is \[ \dfrac{\pi^2}{12} + \sum_{n=1}^\infty \dfrac{\cos n\pi}{n^2} \cos nt \] and use it to calculate
1. \( 1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + \dots = \)
2. \( 1 - \dfrac{1}{2^2} + \dfrac{1}{3^2} - \dfrac{1}{4^2} + \dots = \)
3. \( 1 + \dfrac{1}{3^2} + \dfrac{1}{5^2} + \dfrac{1}{7^2} + \dots = \)
Feedback: We have the Fourier series \[ \dfrac{\pi^2}{12} + \sum_{n=1}^\infty \dfrac{\cos n\pi}{n^2} \cos nt \] 1. Now setting \(t = \pi\) we have \(f(t) = \dfrac{\pi^2}{4}\), and therefore \[ \begin{align} \dfrac{\pi^2}{4} &= \dfrac{\pi^2}{12} + \sum _{n=1}^\infty \dfrac{\cos n\pi}{n^2} \cos n\pi \\ &= \dfrac{\pi^2}{12} + 1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + \dots \end{align} \] which can be rearranged to give \[ 1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + \dots = \dfrac{\pi^2}{6} \] 2. Now setting \(t = 0\) we have \(f(t) = 0 \), and therefore \[ \begin{align} 0 &= \dfrac{\pi^2}{12} + \sum _{n=1}^\infty \dfrac{\cos n\pi}{n^2} \\ &= \dfrac{\pi^2}{12} - 1 + \dfrac{1}{2^2} - \dfrac{1}{3^2} + \dfrac{1}{4^2} - \dots \end{align} \] which can be rearranged to give \[ 1 - \dfrac{1}{2^2} + \dfrac{1}{3^2} - \dfrac{1}{4^2} + \dots = \dfrac{\pi^2}{12} \] 3. Adding the results from part (a) and (b) together gives \[ 1 + \dfrac{1}{3^2} + \dfrac{1}{5^2} + \dfrac{1}{7^2} + \dots = \dfrac{\pi^2}{8} \]