Author Topic: HA5 Problem 1  (Read 2641 times)

Victor Ivrii

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HA5 Problem 1
« on: February 26, 2015, 10:50:47 PM »
Decompose into full Fourier series on interval $[-l,l]$:

a.  $e^{z x}$ where $z\in \mathbb{C}$; find "exceptional" values of $z$;
b.  $\cos(\omega x)$, $\sin (\omega x)$ where $0<\omega\in \mathbb{R}$; fins "exceptional" values of $\omega$;
c.  $\cosh (\eta x)$, $\sinh (\eta x)$ where $0<\eta\in \mathbb{R}$;

Biao Zhang

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Re: HA5 Problem 1
« Reply #1 on: February 27, 2015, 07:15:34 PM »
[a:]
\begin{equation*}
a_0 = \frac{1}{l} \int_{-l}^{l} e^{zx}dx\\
= \frac{1}{zl} e^{zx}\big|_{-l}^{l}\\
= \frac{1}{zl}(e^{zl} - e^{-zl} )
\end{equation*}
Then we have
\begin{equation*}
a_n = \frac{1}{2l} \int_{-l}^{l} e^{zx}\big( \exp{(\frac{in\pi x}{l})} + \exp{(-\frac{in\pi x}{l})} \big) dx \\
= \frac{1}{2l} \int_{-l}^{l} \exp{((z + \frac{in\pi }{l})x)} dx +\frac{1}{2l} \int_{-l}^{l} \exp{((z -\frac{in\pi }{l})x)} dx \\
= \frac{1}{2l} \frac{\exp{((z +\frac{in\pi}{l})x)}}{z +\frac{in\pi }{l}} \big|_{-l}^{l} + \frac{1}{2l} \frac{\exp{((z -\frac{in\pi}{l})x)}}{z -\frac{in\pi }{l}} \big|_{-l}^{l}\\
= \frac{1}{2l} \frac{e^{zl + in\pi }}{z +\frac{in\pi }{l}} -\frac{1}{2l} \frac{e^{-zl - in\pi }}{z +\frac{in\pi }{l}} + \frac{1}{2l} \frac{e^{zl - in\pi }}{z -\frac{in\pi }{l}} - \frac{1}{2l} \frac{e^{-zl + in\pi }}{z -\frac{in\pi }{l}}\\
=\frac{1}{2l}(-1)^n \big[e^{zl} - e^{-zl} \big]\big(\frac{1}{z +\frac{in\pi }{l}} + \frac{1}{z -\frac{in\pi }{l}}\big)\\
=\frac{(-1)^n}{l} \big(e^{zl} - e^{-zl} \big) \frac{z}{z^2 + (\frac{n \pi}{l})^2}\\
= \frac{(-1)^nzl \big(e^{zl} - e^{-zl} \big)}{(zl)^2 + (n \pi)^2}
\end{equation*}
Similarly, for $b_n$:\\
\begin{equation*}
= {- }\frac{(-1)^n {\pi }{n } \big(e^{zl} - e^{-zl} \big)}{(zl)^2 + (n \pi)^2 }
\end{equation*}
So, exception values are $lz=in\pi$ or $lz=-in\pi$

e^{zx} = (e^{zl} - e^{-zl}) [\frac{1}{2l} + \sum_{n=1}^{\infty} \frac{(-1)^nzl }{(zl)^2 + (n \pi)^2 } \cos{\frac{n\pi x}{l}} - \sum_{n=1}^{\infty} \frac{(-1)^n n\pi }{(zl)^2 + (n \pi)^2 } \sin{\frac{n\pi x}{l}}  ].

[b:]

\cos{\omega x} = \frac{(e^{i\omega} - e^{-i\omega})}{2}

Let $z = i\omega$ or $z=-i\omega$, from Part (a) we have:
\begin{equation*}
\frac{1}{2} [ (e^{i\omega l} - e^{-i\omega l}) (\frac{1}{2i \omega l} + \sum_{n=1}^{\infty} \frac{(-1)^n i\omega l }{-(\omega l)^2 + (n \pi)^2 } \cos{\frac{n\pi x}{l}} - \sum_{n=1}^{\infty} \frac{(-1)^n n\pi }{-(\omega l)^2 + (n \pi)^2 } \sin{\frac{n\pi x}{l}}  )  - \\
(e^{i\omega l} - e^{-i\omega l}) (\frac{1}{-2i \omega l} + -\sum_{n=1}^{\infty} \frac{(-1)^n i\omega l }{-(\omega l)^2 + (n \pi)^2 } \cos{\frac{n\pi x}{l}} - \sum_{n=1}^{\infty} \frac{(-1)^n n\pi }{-(\omega l)^2 + (n \pi)^2 } \sin{\frac{n\pi x}{l}}  )  ]\\
= \sin{\omega l} (\frac{1}{\omega l} -  \sum_{n=1}^{\infty} 2(-1)^n  \frac{\omega l \cos{\frac{n\pi x}{l}}}{(n\pi)^2 -( \omega l)^2}   ).
\end{equation*}
Exception values occur when $(\omega l)^2=(n\pi)^2$. When $\omega =0$, we have an indeterminate form which is defined in the limit as $\omega$ approaches 0.
Similarly, for sin $\omega x$:
\begin{equation*}
\sin{\omega x} = -2\sin{\omega l} \sum_{n=1}^{\infty} 2(-1)^n\frac{n \pi \sin{\frac{n\pi x}{l}}}{(n\pi)^2 -( \omega l)^2}
\end{equation*}
[c:]
Just as in (b), we have $\cosh{\eta x} = \frac{e^{\eta x} + e^{-\eta x}}{2}$ and can make the substitution $z = \eta$ or $z = -\eta$. Then  \begin{equation*}  \cosh{\eta x} = \sin{\eta l}\left(\frac{1}{\eta l} -  \sum_{n=1}^{\infty} 2(-1)^n  \frac{\eta l \cos{\frac{n\pi x}{l}}}{(n\pi)^2 +( \eta l)^2}  \right) \end{equation*} And for $\sinh{\eta x}$, we have \begin{equation*} \sinh(\eta x) = \sinh{\eta l}\left(- 2 \sum_{n=1}^{\infty} (-1)^n  \frac{n\pi  \sin{\frac{n\pi x}{l}}}{(n\pi)^2 +( \eta l)^2}  \right). \end{equation*}