Author Topic: Problem 1  (Read 9937 times)

Levon Avanesyan

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Problem 1
« on: October 29, 2012, 12:14:01 PM »
In C part, shouldn't it be sinh(ηx) instead of sin(ηx)?

Victor Ivrii

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Re: Problem 1
« Reply #1 on: October 29, 2012, 12:29:42 PM »
In C part, shouldn't it be sinh(ηx) instead of sin(ηx)?

Yes, thanks

Hanqing Liu

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Re: Problem 1
« Reply #2 on: October 30, 2012, 07:22:56 PM »
In part a and b, does the word "exceptional" just mean where the expansion is not defined?

Victor Ivrii

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Re: Problem 1
« Reply #3 on: October 30, 2012, 07:30:00 PM »
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 mean--but solving the problem it will pop up obviously

Djirar

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Re: Problem 1
« Reply #4 on: October 30, 2012, 07:42:37 PM »
Do you want the solution in the complex form or real form of the Fourier series?

Victor Ivrii

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Re: Problem 1
« Reply #5 on: October 30, 2012, 07:48:06 PM »
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

Jinchao Lin

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Re: Problem 1
« Reply #6 on: October 31, 2012, 02:07:42 AM »
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 $\int sin(x)exp(x)$

Ian Kivlichan

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Re: Problem 1
« Reply #7 on: October 31, 2012, 02:27:03 AM »
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! :)

Victor Ivrii

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Re: Problem 1
« Reply #8 on: October 31, 2012, 07:48:02 AM »
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.

Aida Razi

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Re: Problem 1
« Reply #9 on: October 31, 2012, 09:30:03 PM »
Part (a) solution is attached!

Zarak Mahmud

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Re: Problem 1
« Reply #10 on: October 31, 2012, 09:30:10 PM »

$
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*}
« Last Edit: November 12, 2012, 09:07:33 PM by Zarak Mahmud »

Victor Ivrii

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Re: Problem 1
« Reply #11 on: November 01, 2012, 03:04:15 AM »
I consider Problem 1 to be closed. Zarak correctly decomposed for non-exceptional 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