Toronto Math Forum
APM3462012 => APM346 Math => Home Assignment 5 => Topic started by: Levon Avanesyan on October 29, 2012, 12:14:01 PM

In C part, shouldn't it be sinh(Î·x) instead of sin(Î·x)?

In C part, shouldn't it be sinh(Î·x) instead of sin(Î·x)?
Yes, thanks

In part a and b, does the word "exceptional" just mean where the expansion is not defined?

In part a and b, does the word "exceptional" just mean where the expansion is not defined?
You need to figure out what does it meanbut solving the problem it will pop up obviously

Do you want the solution in the complex form or real form of the Fourier series?

Do you want the solution in the complex form or real form of the Fourier series?
Real is preferable where we consider even and odd functions on $[l,l]$. Otherwise does not matter.
We are talking here about full F.s. only

in part a, it seems that if we want the expansion to be real domain, the only way we can do is first manipulate in complex domain by the exponential expansion, then use Euler formula to convert into real domain? Because there is no integral formula for $\int sin(x)exp(x)$

Jinchao: you can integrate (e^x)sin(x) by parts. Set u = e^x, dv = sinx dx, and go through. You'll have to integrate by parts a second time, but you'll end up with (e^x)sinx integrals on both sides. Hope that helps! :)

Jinchao: you can integrate (e^x)sin(x) by parts. Set u = e^x, dv = sinx dx, and go through. You'll have to integrate by parts a second time, but you'll end up with (e^x)sinx integrals on both sides. Hope that helps! :)
One can use a representation of sin and cos via complex exponents which makes integration easier. It does not contradict to real full F.s.

Part (a) solution is attached!

$
f(x) = e^{zx} $ for some $z \in \mathbb{C}
$
Part (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}\big(e^{zl}  e^{zl} \big)\\
\end{equation*}
\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)} + \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*}
b_n = {\color{magenta} }\frac{(1)^n {\color{magenta}\pi }{\color{magenta}n } \big(e^{zl}  e^{zl} \big)}{(zl)^2 + (n \pi)^2 }
\end{equation*}
Note the difference in magenta. Exceptional values are $lz = in\pi $ or $lz = in\pi$.
\begin{equation}
e^{zx} = (e^{zl}  e^{zl})\left[\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}} \right].
\end{equation}
Part (b):
\begin{equation*}
\cos{\omega x} = \frac{(e^{i\omega}  e^{i\omega})}{2}
\end{equation*}
Let $z = i\omega$ or $z=i\omega$. Using the result obtained in (1),
\begin{equation*}
\frac{1}{2}\left[ (e^{i\omega l}  e^{i\omega l})\left(\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}} \right)  \\(e^{i\omega l}  e^{i\omega l})\left(\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}} \right) \right]\\
= \sin{\omega l}\left(\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} \right).
\end{equation*}
Exceptional values here appear to be $(\omega l)^2 = (n \pi)^2$. At $\omega = 0$, we have an indeterminate form which is defined in the limit as $\omega$ approaches $0$.
Similarly, for $\sin{\omega x}$, we have
\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*}
Part (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*}

I consider Problem 1 to be closed. Zarak correctly decomposed for nonexceptional values and indicated that for exceptional values coefficients could be found as limits. I would prefer a bit more explicit answer in the exceptional case
 (a) $z=\frac{n\pi i}{l}$, $n\in \mathbb{Z}$, then $e^{zx}=\underbracket{e^{\frac{n\pi x i}{l}}}$ is one of the basis functions in the complex decomposition (so the r.h.e. is the decomposition), while in the real case we have $\cos (\frac{m\pi x}{l}) \pm \sin( \frac{m\pi x}{l})$ as $n=\pm m$, $m=0,1,2,\ldots$
 (b) Ditto, for $\omega =\frac{\pi n}{l}$ $\cos (\frac{\pi n x}{l})$ and $\sin (\frac{\pi n x}{l})$ are required decompositions(
 (c) Ditto, for $\eta=0$ $1$ is a required decomposition