Problem 1

[Levon Avanesyan]

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

[Victor Ivrii]

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

Yes, thanks

[Hanqing Liu]

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

[Victor Ivrii]

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]

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

[Victor Ivrii]

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]

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]

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]

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]

Part (a) solution is attached!

[Zarak Mahmud]

$
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*}

[Victor Ivrii]

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