Title: Default for 23.6 The Complex form
Course: HELM Workbooks 2023-2024

Introduction

In this Section we show how a Fourier series can be expressed more concisely if we introduce the complex number \(i\) where \(i^2 = -1\). By utilising the Euler relation: \[ e^{i\theta} \equiv \cos \theta + i \sin \theta \] we can replace the trigonometric functions by complex exponential functions. By also combining the Fourier coefficients \(a_n\) and \(b_n\) into a complex coefficient \(c_n\) through \[ c_n = \dfrac{1}{2} \left( a_n - i b_n \right) \] we find that, for a given periodic signal, both sets of constants can be found in one operation.

We also obtain Parseval's theorem which has important applications in electrical engineering.

The complex formulation of a Fourier series is an important precursor of the Fourier transform which attempts to Fourier analyse non-periodic functions.

Complex exponential form of a Fourier series

So far we have discussed the trigonometric form of a Fourier series i.e. we have represented functions of period \(T\) in the terms of sinusoids, and possibly a constant term, using \[ f(t) = \dfrac{a_0}{2} + \sum _{n=1}^\infty \left\{ a_n \cos \left( \dfrac{2n\pi t}{T} \right) + b_n \sin \left( \dfrac{2n\pi t}{T} \right) \right\} \] If we use the angular frequency \( \omega = \dfrac{2\pi}{T}\) we obtain the more concise form \[ f(t) = \dfrac{a_0}{2} + \sum _{n=1}^\infty \left( a_n \cos n\omega_0 t + b_n \sin n\omega_0 t \right) \] We have seen that the Fourier coefficients are calculated using the following integrals: \[ a_n = \dfrac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(t) \cos n\omega_0 t \hspace{0.1cm} dt \hspace{1cm} n = 0, 1, 2, \dots \] \[ b_n = \dfrac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(t) \sin n\omega_0 t \hspace{0.1cm} dt \hspace{1cm} n = 1, 2, \dots \] An alternative, more concise form, of a Fourier series is available using complex quantities. This form is quite widely used by engineers, for example in Circuit Theory and Control Theory, and leads naturally into the Fourier Transform.

Revision of the exponential form of a complex number

Recall that a complex number in Cartesian form which is written as \[z = a + ib \] where \(a\) and \(b\) are real numbers and \(i^2 = -1\), can be written in polar form as \[ z = r \left( \cos \theta + i \sin \theta \right) \] where \(r = \lvert z \rvert =\sqrt{a^2 + b^2} \) and \(\theta\), the argument or phase of \(z\) is such that \[ a = r \cos \theta \hspace{1cm} b = r \sin \theta \] A more concise version of the polar form of \(z\) can be obtained by defining a complex exponential quantity \(e^{i\theta}\) by Euler's relation \[ e^{i\theta} \equiv \cos \theta + i \sin \theta \] The polar angle \(\theta\) is normally expressed in radians. Replacing i by \( - i\) we obtain the alternative form \[ e^{-i\theta} \equiv \cos \theta - i \sin \theta \]

Exercise

Write the following in \( \cos \theta + i\sin \theta \) form and Cartesian form.
1. \( e^{i\pi/6} = \cos\) \( + i\sin \)
2. \( e^{i\pi/6} =\) \( + \) \(i\)
Now write the following in \( \cos \theta + i\sin \theta \) form and Cartesian form.
3. \( e^{-i\pi/6} = \cos\) \( + i\sin \)
4. \( e^{-i\pi/6} =\) \( + \) \(i\)
Feedback: We have by definition, \[ \begin{align} e^{i\pi/6} &= \cos \left( \dfrac{\pi}{6} \right) + i \sin \left( \dfrac{\pi}{6} \right) \\ &= \displaystyle[value] + \displaystyle[value] i \end{align} \] and also \[ \begin{align} e^{-i\pi/6} &= \cos \left( \dfrac{-\pi}{6} \right) + i \sin \left( \dfrac{-\pi}{6} \right) \\ &= \cos \left( \dfrac{\pi}{6} \right) - i \sin \left( \dfrac{\pi}{6} \right) \\ &= \displaystyle[value] - \dfrac{1}{2} i \end{align} \]

Adding the results from the previous task gives \[e^{i\pi/6} + e^{-i\pi/6} = 2\cos \left( \dfrac{\pi}{6} \right) \] or \[ \cos \left( \dfrac{\pi}{6} \right) = \dfrac{1}{2} \left( e^{i\pi/6} + e^{-i\pi/6} \right) .\] Subtracting the results from the previous task gives \[e^{i\pi/6} - e^{-i\pi/6} = 2 i \sin \left( \dfrac{\pi}{6} \right) \] or \[ \sin \left( \dfrac{\pi}{6} \right) = \dfrac{1}{2i} \left( e^{i\pi/6} - e^{-i\pi/6} \right) .\] Don't forget the factor \(i\) in this latter case.

Clearly, similar calculations could be carried out for any angle \(\theta\). The general results are summarised in the following Key Point.

Key Point

Euler's Relations

\[ e^{i\theta} \equiv \cos\theta + i \sin \theta, \hspace{1cm} e^{-i\theta} \equiv \cos\theta - i \sin \theta\] \[\cos \theta \equiv \dfrac{1}{2} ( e^{i\theta} + e^{-i\theta} ), \hspace{1cm} \sin \theta \equiv \dfrac{1}{2i} ( e^{i\theta} - e^{-i\theta} )\]

Using these results we can redraft an expression of the form \[ a_n \cos n\theta + b_n \sin n\theta \] in terms of complex exponentials.

(This expression, with \(\omega = \omega_0 t\), is of course the \(n^{ \text{th} }\) harmonic of a trigonometric Fourier series.)

Rewriting in complex exponential form, we have \[ a_n \cos n \theta + b_n \sin n\theta = \dfrac{a_n}{2} ( e^{in\theta} + e^{-in\theta} ) + \dfrac{b_n}{2i} ( e^{in\theta} - e^{-in\theta} ) \] Now collecting the terms in \(e^{ in\theta}\) and in \(e^{ -in\theta}\) and use the fact that \( \dfrac{1}{i} = -i \), we have \[ \dfrac{1}{2} \left( a_n - ib_n \right) e^{ in\theta} + \dfrac{1}{2} \left( a_n + ib_n \right) e^{ -in\theta} \] This can be expressed more concisely by defining

\( c_n = \dfrac{1}{2} (a_n - ib_n) \) which has complex conjugate \( \displaystyle{c_n^{*}} = \dfrac{1}{2} (a_n + ib_n) \) giving us \[ a_n \cos n \theta + b_n \sin n\theta = c_n e^{i n \theta} + c_n^{*} e^{-i n \theta} \]

Clearly, we can now rewrite the trigonometric Fourier series \[ \dfrac{a_0}{2} + \sum_{n=1}^\infty \left( a_n \cos n \omega_0 t + b_n \sin n \omega_0 t \right) \] as \[ \dfrac{a_0}{2} + \sum_{n=1}^\infty \left( c_n e^{i n \omega_0 t} + c_n^* e^{-i n \omega_0 t} \right) \] A neater, and particularly concise, form of this expression can be obtained as follows:

Firstly write \( \dfrac{a_0}{2} = c_0\) (which is consistent with the general definition of \(c_n\) since \(b_0 = 0\)). The second term in the summation \[ \sum_{n=1}^\infty c_n^*e^{-i n \omega_0 t} = c_1^* e^{-i \omega_0 t} + c_2^*e^{-2i \omega_0 t} + \dots \] can be written, if we define \( c_{-n} = c_n^* = \dfrac{1}{2} (a_n + ib_n) \), as \[ c_{-1} e^{-i \omega_0 t} + c_{-2} e^{-2i \omega_0 t} + c_{-3} e^{-3i \omega_0 t} +\dots \sum_{n=-1}^{-\infty} c_n e^{i n \omega_0 t} \]

Hence the trigonometric Fourier series can be written \[ c_0 + \sum_{n=1}^{\infty} c_n e^{i n \omega_0 t} + \sum_{n=-1}^{-\infty} c_n e^{i n \omega_0 t} \] or in the very concise form \[ \sum_{n=-\infty}^{\infty} c_n e^{i n \omega_0 t}. \] The complex Fourier coefficients \(c_n\) can be readily obtained as follows using the usual Fourier series definitions for \(a_n\) and \(b_n\).

Firstly, \[ c_0 = \dfrac{a_0}{2} = \dfrac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(t) \hspace{0.1cm} dt \] For \( n = 1, 2, 3, \dots\) we have \[ \begin{align} c_n &= d\frac{1}{2}(a_n- i b_n)=d\frac{1}{T}\int_{-\frac{T}{2}}^\frac{T}{2} f(t)(\cos n\omega_0 t - i \sin n\omega_0 t) dt \\ &= \dfrac{1}{T}\int_{-\frac{T}{2}}^\frac{T}{2} f(t) e^{- i n\omega_0t} dt \end{align} \] Also for \( n = 1, 2, 3, \dots\) we have \[ c_{-n}={c^*_n}=\dfrac{1}{2}(a_n+ i b_n)=\dfrac{1}{T}\int_{-\frac{T}{2}}^\frac{T}{2} f(t) e^{i n\omega_0 t} dt \] This last expression is equivalent to stating that for \(n=-1,-2,-3,\dots\) \[ c_n=\dfrac{1}{T}\int_{-\frac{T}{2}}^\frac{T}{2} f(t) e^{-i n\omega_0t}dt \] The three equations above can be contained in the one expression \[ c_n=\dfrac{1}{T}\int_{-\frac{T}{2}}^\frac{T}{2} f(t) e^{-i n\omega_0t} dt\qquad \text{ for}\quad n=0,\pm 1,\pm 2,\pm 3,\dots \] The results of this discussion are summarised in the following Key Point.

Key Point

Fourier Series in Complex Form

A function \(f(t)\) of period \(T\) has a complex Fourier series \[ f(t)=\sum_{n=-\infty}^\infty c_n e^{ i n \omega_0 t}\qquad\text{ where}\qquad c_n=\frac{1}{T}\int_{-\frac{T}{2}}^\frac{T}{2} f(t) e^{- i n\omega_0t}dt\] For the special case \(T=2\pi\), so that \( \omega_0=1\), these formulae become particularly simple: \[f(t)=\sum_{n=-\infty}^\infty c_n e^{i nt}\qquad c_n=\frac{1}{2\pi}\int_{-\pi}^\pi f(t) e^{-\text{ i}nt} dt\]

Properties of the complex Fourier coefficients

Using properties of the trigonometric Fourier coefficients \(a_n\), \(b_n\) we can readily deduce the following results for the \(c_n\) coefficients:

1. \(c_0=\dfrac{a_0}{2}\) is always real

2a. Suppose the periodic function \(f(t)\) is even so that all \(b_n\) are zero. Then, since in the complex form the \(b_n\) arise as the imaginary part of \(c_n\), it follows that for \(f(t)\) even the coefficients \(c_n\) (\(n=\pm 1,\pm 2,\ldots\)) are wholly real.

2b. Suppose the periodic function \(f(t)\) is odd so that all \(a_n\) are zero Since, for an odd periodic function the Fourier coefficients \(a_n\) (which constitute the real part of \(c_n\)), then in this case the complex coefficients \(c_n\) are wholly imaginary.

3. Since \[ c_n=\dfrac{1}{T}\int_{-\frac{T}{2}}^\frac{T}{2}f(t) e^{-i n\omega_0t}dt \] then if \(f(t)\) is even, \(c_n\) will be real, and we have two possible methods for evaluating \(c_n\):

a) Evaluate the integral above as it stands i.e. over the full range \(\left(-\dfrac{T}{2},\ \dfrac{T}{2}\right)\). Note carefully that the second term in the integrand is neither an even nor an odd function so the integrand itself is \[ (\text{ even function})\times (\text{ neither even nor odd function})=\ \text{ neither even nor odd function.}\] Thus we cannot write \[ c_n=\frac{2}{T}\int_0^{T/2} f(t) e^{- i n\omega_0 t} \ dt\]

b) Put \(e^{- i n\omega_0t}=\cos n\omega_0t- i\sin n\omega_0t\) so \[ \begin{eqnarray*}f(t) e^{- i n\omega_0t}=f(t)\cos n\omega_0 t- i f(t)\sin n\omega_0 t&=&(\text{ even})(\text{ even})- i (\text{ even})(\text{ odd}) \\ &=&(\text{ even})-i (\text{ odd}). \end{eqnarray*} \] Hence \[c_n=\dfrac{2}{T}\int_{0}^\frac{T}{2}f(t)\cos n\omega_0t\ dt=\dfrac{a_n}{2}.\]

4. If \(f(t+\dfrac{T}{2})=-f(t)\) then of course only odd harmonic coefficients \(c_n \left(n=\pm1,\pm 3,\pm 5,\ldots\right)\) will arise in the complex Fourier series just as with trigonometric series.

Example

Find the complex Fourier series of the saw-tooth wave shown in the Figure below

[Image: saw tooth function]

Solution

We have \[ f(t) = \dfrac{At}{T} \hspace{1cm} 0 < t < T \hspace{1cm} f(t+T)=f(t) \] The period is \(T\) in this case so \(\omega_0=\dfrac{2\pi}{T}\).

Looking at the graph of \(f(t)\) we can say immediately

(a) the Fourier series will contain a constant term \(c_0\)

(b) if we imagine shifting the horizontal axis up to \(\dfrac{A}{2}\) the signal can be written \(f(t) = \dfrac{A}{2} + g(t)\), where \(g(t)\) is an odd function with complex Fourier coefficients that are purely imaginary.

Hence we expect the required complex Fourier series of \(f(t)\) to contain a constant term \( \dfrac{A}{2}\) and complex exponential terms with purely imaginary coefficients. We have, from the general theory, and using \(0 < t < T\) as the basic period for integrating, \[ c_n = \dfrac{1}{T} \int _0^T \dfrac{At}{T} e^{-i n \omega_0 t} dt = \dfrac{A}{T^2} \int_0^T t e^{-i n \omega_0 t} dt \] We can evaluate the integral using parts: \[ \begin{align} \int_0^T t e^{-i n \omega_0 t} dt &= \left[ \dfrac{ t e^{-i n \omega_0 t} }{(-in\omega_0)} \right]_0^T + \dfrac{1}{i n \omega_0} \int_0^T e^{-i n \omega_0 t} dt \\ &= \dfrac{T e^{i n \omega_0 T}}{(-i n \omega_0)} - \dfrac{1}{(i n \omega_0)^2} \Big[ e^{-i n \omega_0 t} \Big]_0^T \end{align} \] But \( \omega_0 = \dfrac{2\pi}{T}\) so \[ \begin{align} e^{-i n \omega_0 T} = e^{-i n 2 \pi} &= \cos 2n\pi - i\sin 2n\pi \\ &= 1 - 0i = 1 \end{align} \] Hence the integral becomes \[ \dfrac{T}{-in\omega_0} - \dfrac{1}{(in\omega_0)^2} \left( e^{-i n \omega_0 T} - 1 \right) \] Hence \[ c_n = \dfrac{A}{T^2} \left( \dfrac{T}{-in\omega_0} \right) = \dfrac{iA}{2\pi n} \hspace{1cm} n = \pm 1, \pm 2, \dots \] Note that \[ c_{-n} = \dfrac{iA}{2\pi(-n)} = \dfrac{-iA}{2\pi n} = c_n^* \] Also \[ c_0 = \dfrac{1}{T} \int _0^T \dfrac{At}{T}dt = \dfrac{A}{2} \] as expected.

Hence the required complex Fourier series is \[ f(t) = \dfrac{A}{2} + \dfrac{iA}{2\pi} \sum _{{n=-\infty\atop n\neq 0}} ^ \infty \dfrac{e^{i n \omega_0 t}}{n} \] which could be written, showing only the constant and the first two harmonics, as \[ f(t) = \dfrac{A}{2\pi} \left\{ \dots - i \dfrac{e^{-i 2 \omega_0 t}}{2} - i e^{-i\omega_0 t} + \pi + i e^{i\omega_0 t} + i \dfrac{e^{i 2 \omega_0 t}}{2} + \dots \right\} . \]

The corresponding trigonometric Fourier series for the function can be readily obtained from this complex series by combining the terms in \( \pm n\), \(n = 1, 2, 3, \dots \)

For example this first harmonic \[ \begin{eqnarray*} \dfrac{A}{2\pi}\left\{- i e^{-i \omega_0t}+ i e^{ i\omega_0t}\right\} &=&\dfrac{A}{2\pi}\left\{- i (\cos \omega_0t - i\sin \omega_0t)+ i (\cos\omega_0t +i \sin \omega_0t)\right\}\qquad\qquad\\ \\ &=&\dfrac{A}{2\pi}(-2\sin\omega_0t)=-\dfrac{A}{\pi}\sin\omega_0t \end{eqnarray*} \] Performing similar calculations on the other harmonics we obtain the trigonometric form of the Fourier series \[ f(t)=\frac{A}{2}-\frac{A}{\pi}\sum_{n=1}^\infty \frac{\sin n\omega_0t}{n}.\]

Exercise

Find the complex Fourier series of the periodic function: \[ f(t)=e^t \qquad -\pi < t < \pi \hspace{1cm} f(t+2\pi)=f(t)\]

[Image: e^t]

\(f(t) = \displaystyle{\sum_{n=-\infty}^\infty}\)
Feedback: We have, since \(T=2\pi\), so \(\omega_0=1\) \[ \begin{align} c_n & =\frac{1}{2\pi}\int_{-\pi}^\pi e^te^{-i nt} dt \\ &= \frac{1}{2\pi}\int_{-\pi}^\pi e^{(1- in)t}dt=\frac{1}{2\pi}\left[\frac{e^{(1- in)t}}{(1- in)}\right]_{-\pi}^\pi \\ &=\frac{1}{2\pi}\frac{1}{(1- in)}\left(e^{(1-in)\pi}-e^{-(1- in)\pi}\right) \end{align} \] Now simplify this as far as possible and write out the Fourier series: \[ e^{(1- in)\pi}=e^\pi\ e^{- in\pi}=e^\pi (\cos n\pi- i\sin n\pi)=e^\pi\cos n\pi \] \[ e^{-(1-in)\pi}=e^{-\pi}e^{ in\pi}=e^{-\pi}\cos n\pi \] Hence \( c_n=\dfrac{1}{2\pi}\dfrac{1}{(1- in)}(e^\pi-e^{-\pi})\cos n\pi=\dfrac{\sinh\pi}{\pi}\dfrac{(1+ in)}{(1+n^2)}\cos n\pi\). Note that the coefficients \(c_n\) \(n=\pm 1,\pm 2,\ldots\) have both real and imaginary parts in this case as the function being expanded is neither even nor odd. Also \( c_{-n}=\dfrac{\sinh \pi}{\pi}\dfrac{(1-in)}{(1+(-n)^2)}\cos (-n\pi)=\dfrac{\sinh \pi}{\pi}\dfrac{(1-in)}{(1+n^2)}\cos n\pi={c^*_n}\) as required. This includes the constant term \(c_0=\dfrac{\sinh \pi}{\pi}.\) Hence the required Fourier series is \[ f(t)=\frac{\sinh \pi}{\pi}\sum_{n=-\infty}^\infty (-1)^n\frac{(1+\text{ i}n)}{(1+n^2)} e^{\text{ i}nt}\qquad \text{ since}\quad \cos n\pi=(-1)^n \]

Parseval’s theorem

This is essentially a mathematical theorem but has, as we shall see, an important engineering interpretation particularly in electrical engineering. Parseval's theorem states that if \(f(t)\) is a periodic function with period \(T\) and if \(c_n\) \( (n=0,\pm 1,\pm 2,\ldots)\) denotes the complex Fourier coefficients of \(f(t)\), then \[ \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} f^2(t)\, dt=\sum_{n=-\infty}^\infty |c_n|^2.\] In words the theorem states that the mean square value of the signal \(f(t)\) over one period equals the sum of the squared magnitudes of all the complex Fourier coefficients.

Proof of Parseval’s theorem

Assume \(f(t)\) has a complex Fourier series of the usual form: \[ f(t)=\sum_{n=-\infty}^{\infty} c_n e^{ in\omega_0t}\qquad \left(\omega_0=\frac{2\pi}{T}\right) \] where \[ c_n=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} f(t)e^{- in\omega_0t}dt \] Then \[ f^2(t)=f(t)f(t)=f(t)\sum c_n e^{ in\omega_0t}=\sum c_n f(t) e^{in\omega_0t} \] Hence \[ \begin{eqnarray*} \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} f^2(t)\,dt&=& \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}\sum c_n f(t) e^{ in\omega_0t}dt\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\\ \\ &=&\frac{1}{T}\sum c_n \int_{-\frac{T}{2}}^{\frac{T}{2}} f(t) e^{ in\omega_0t}dt\\ \\ &=&\sum c_n {c^*_n}\\ \\ &=&\sum_{n=-\infty}^\infty |c_n|^2 \end{eqnarray*} \] which completes the proof.

Parseval's theorem can also be written in terms of the Fourier coefficients \(a_n\), \(b_n\) of the trigonometric Fourier series. Recall that \[ \begin{array}{lll} c_0=\dfrac{a_0}{2}\qquad c_n={\frac{a_n- ib_n}{2}}\quad n=1,2,3,\ldots\qquad\qquad c_n={\frac{a_n+ ib_n}{2}}\quad n=-1,-2,-3,\ldots \end{array} \] so \[ |c_n|^2=\frac{a_n^2+b_n^2}{4}\qquad\quad n=\pm 1,\pm 2,\pm 3,\ldots \] so \[ \sum_{n=-\infty}^\infty |c_n|^2=\frac{a_0^2}{4}+2\sum_{n=1}^\infty \frac{a_n^2+b_n^2}{4} \] and hence Parseval's theorem becomes \[ \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} f^2(t)dt =\frac{a_0^2}{4}+\frac{1}{2}\sum_{n=1}^\infty (a_n^2+b_n^2) \]

The engineering interpretation of this theorem is as follows. Suppose \(f(t)\) denotes an electrical signal (current or voltage), then from elementary circuit theory \(f^2(t)\) is the instantaneous power (in a 1 ohm resistor) so that \[ \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f^2(t)\,dt \] is the energy dissipated in the resistor during one period.

Now a sinusoid wave of the form \[ A\cos \omega t\qquad (\text{ or}\ A\sin\omega t) \] has a mean square value \(\dfrac{A^2}{2}\) so a purely sinusoidal signal would dissipate a power \(\dfrac{A^2}{2}\) in a 1 ohm resistor. Hence Parseval's theorem states that the average power dissipated over 1 period equals the sum of the powers of the constant (or d.c.) components and of all the sinusoidal (or alternating) components.

Example

The triangular signal shown below has trigonometric Fourier series \[ f(t)=\frac{\pi}{2}-\frac{4}{\pi} \sum_{{n=1\atop(\text{ odd}\ n)}}^\infty \frac{\cos nt}{n^2}.\]

[Image: triangle function]

Use Parseval's theorem to show that \[\sum_{{n=1\atop (n\ \text{ odd})}}^\infty\frac{1}{n^4}=\frac{\pi^4}{96}.\]

Solution

First, identify \(a_0\), \(a_n\) and \(b_n\) for this situation and write down the definition of \(f(t)\) for this case:

We have \(\dfrac{a_0}{2}=\dfrac{\pi}{2}\) \[a_n=\left\{\begin{array}{cll}-\dfrac{4}{n^2\pi}&&n=1,3,5,\ldots\\ \\ 0&&n=2,4,6,\ldots\end{array}\right.\] \[ b_n=0\qquad\qquad n=1,2,3,4,\ldots\] Also \[ f(t)=|t|\qquad -\pi<t<\pi\] \[\qquad\displaystyle f(t+2\pi)=f(t)\]

Now evaluate the integral on the left hand side of Parseval’s theorem and hence complete the problem:

We have \(f^2(t)=t^2\) so \[\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} f^2(t)\, dt=\frac{1}{2\pi}\int_{-\pi}^\pi t^2dt=\frac{1}{2\pi}\left[\frac{t^3}{3}\right]_{-\pi}^\pi =\frac{\pi^2}{3}\] The right-hand side of Parseval’s theorem is \[ \frac{a_0^2}{4}+\sum_{n=1}^\infty a_n^2=\frac{\pi^2}{4}+\frac{1}{2}\sum_{{n=1\atop (n\ \text{ odd})}}^\infty \frac{16}{n^4\pi^2}\] Hence \[ \frac{\pi^2}{3}=\frac{\pi^2}{4}+\frac{8}{\pi^2}\sum_{{n=1\atop (n\ \text{ odd})}}^\infty \frac{1}{n^4}\qquad\quad\therefore\qquad \frac{8}{\pi^2}\sum_{{n=1\atop (n\ \text{ odd})}}^\infty \frac{1}{n^4}=\frac{\pi^2}{12}\qquad\therefore\qquad \sum_{{n=1\atop (n\ \text{ odd})}}^\infty \frac{1}{n^4}=\frac{\pi^4}{96}.\]

Exercise

Obtain the complex Fourier series for each of the following functions of period \(2\pi\).

1. \(f(t)=t,\qquad -\pi \le t\le \pi\):
\(\displaystyle f(t) = \sum_{n=-\infty \atop {n\neq 0}}^\infty\)
2. \(f(t)=t,\qquad 0\le t\le 2\pi\):
\(f(t) =\) \( + \displaystyle\sum_{n=-\infty \atop {n\neq 0}}^\infty\)
3. \(f(t)=\text{e}^t,\qquad -\pi\le t\le \pi\):
\( f(t) =\) \( \displaystyle\sum_{n=-\infty}^\infty\)