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      <text>23.5.2.3 ex3a cosine series</text>
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      <text><![CDATA[<p class="HELM_exercise">Exercise</p>
Obtain the half-range Fourier Cosine series for the functions
\[f(t) = e^{2t} \hspace{1cm} 0 \leq t \leq 1\]
<p></p>
Series is
<p>[[input:ans1]]&nbsp;[[validation:ans1]] \(+ \displaystyle{ \sum_{n=1}^\infty} \) [[input:ans2]]&nbsp;[[validation:ans2]]</p>[[feedback:prt1]] [[feedback:prt2]]&nbsp;]]></text>
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      <text><![CDATA[We extend as an even function, so 
\[ f(t) = e^{-2t} \hspace{1cm} -1 \leq t \leq 0 \]
Therefore we define 
\[ F(t) = 
\begin{cases} 
 e^{-2t} \hspace{1cm} -1 \leq t \leq 0 \\
 e^{2t} \hspace{1cm} 0 \leq t \leq 1
\end{cases}
\]
As \( F(t)\) is an even function, \(b_n = 0\). Then we have
\[
\begin{align}
a_n &amp;= \int_{-1}^1 F(t) \cos n\pi t \hspace{0.1cm} dt = 2\int_{0}^1 e^{2t} \cos n\pi t \hspace{0.1cm} dt  \\
&amp;= 2\left\{ \left[  \dfrac{e^{2t}}{n\pi} \sin n\pi t \right]_0^1 - \dfrac{2}{n\pi} \int_0^1 e^{2t} \sin n\pi t \hspace{0.1cm} dt \right\} \\
&amp;= 2\left\{ -  \dfrac{2}{n\pi} \left( \left[ - \dfrac{e^{2t}}{n\pi} \cos n\pi t \right]_0^1 + \dfrac{2}{n\pi} \int_0^1 e^{2t} \cos n\pi t \hspace{0.1cm} dt  \right) \right\} \\
&amp;= -\dfrac{4}{n\pi} \left(\dfrac{1}{n\pi} - \dfrac{e^2}{n\pi} \cos n\pi \right) - \dfrac{4}{n^2\pi^2}a_n
\end{align}
\]
Rearranging now for \(a_n\) gives
\[a_n = \dfrac{4}{4 + n^2\pi^2} \left(  e^2 \cos n\pi -1 \right) \]
Also, \(a_0 = e^2 - 1\), so \( \dfrac{a_0}{2} = \dfrac{e^2 - 1}{2}\).
<p></p>
Finally the half range Fourier cosine series is given by
\[ F(t) = \dfrac{e^2 - 1}{2} +  \sum_{n=1}^\infty \dfrac{4}{4 + n^2\pi^2} \left(  e^2 \cos n\pi -1 \right) \cos n\pi t\]]]></text>
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      <text>ta1: (e^2-1)/2;
ta2: (4)/(4+n^2*pi^2)*(e^2*cos(n*pi) - 1)*cos(n*pi*t);</text>
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