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      <text>23.3.2.1 Fourier series implications</text>
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      <text><![CDATA[<h4>Fourier series implications</h4>
Since a sum of even functions is itself an even function it is not unreasonable to suggest that a Fourier
series containing only cosine terms (and perhaps a constant term which can also be considered as
an even function) can only represent an even periodic function. Similarly a series of sine terms (and
no constant) can only represent an odd function. These results can readily be shown more formally
using the expressions for the Fourier coefficients \(a_n\) and \(b_n\).
<p></p>
Recall that for a \(2\pi\)-periodic function,
\[ b_n = \dfrac{1}{\pi} \displaystyle{ \int_{-\pi}^\pi }f(t) \sin nt \hspace{0.1cm} dt \]
Then, if \(f(t)\) is even,
\[ f(t) \sin nt = \text{ (even) } \times \text{ (odd) } = \text{ (odd)} \]
hence
\[ b_n = \dfrac{1}{\pi} \displaystyle{ \int_{-\pi}^\pi } \text{ (odd function)} \hspace{0.1cm} dt = 0\]
Thus an even function has no sine terms in its Fourier series.
<p></p>
Also, 
\[ f(t) \cos nt = \text{ (even) } \times \text{ (even) } = \text{ (even)} \]
\[ \therefore \hspace{1cm} a_n =  \dfrac{1}{\pi} \displaystyle{ \int_{-\pi}^\pi } \text{ (even function)} \hspace{0.1cm} dt =  \dfrac{2}{\pi} \displaystyle{ \int_{-\pi}^\pi }f(t) \cos nt \hspace{0.1cm} dt \]
It should be obvious that, for an odd function \(f(t)\),
\[ 
\begin{align}
a_n &amp;= \dfrac{1}{\pi} \displaystyle{ \int_{-\pi}^\pi }f(t) \cos nt \hspace{0.1cm} dt = \dfrac{1}{\pi} \displaystyle{ \int_{-\pi}^\pi } \text{ (odd function)} \hspace{0.1cm} dt = 0 \\
b_n &amp;= \dfrac{2}{\pi} \displaystyle{ \int_{-\pi}^\pi }f(t) \sin nt \hspace{0.1cm} dt
\end{align}
\]
Analogous results hold for functions of any period, not necessarily \(2\pi\).]]></text>
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