1
$\begingroup$

I'm new to Fourier expansions and transforms, and I'm not sure how to proceed with this question.

I know a function f(x) can be expressed as an infinite sum of $c_ne^{in \pi x/L}$, and that $c_n = (\frac1{2L})\int_{-L}^{L}e^{-in \pi x/L}f(x) $

However I have to find $c_n$ for the function $f(x) = e^{|x|}$ for $-\pi < x < \pi$ (period $2\pi$).

In the question I am also given Parseval's identity.

I can't put $f(x)$ into $c_n$ (at least it doesn't seem like I can solve it), and I'm not sure how Parseval's identity helps because it has $\sum|c_n|^2$, not $c_n$.

So,how do you calculate $c_n$?

$\endgroup$

2 Answers 2

1
$\begingroup$

Hint: split each integral in two pieces, from $-\pi$ to $0$ and from $0$ to $\pi$. The absolute value disappears and you can integrate by parts.

$\endgroup$
0
$\begingroup$

$c_n = (\frac{1}{2\pi})\int_{-\pi}^{\pi}e^{-in \pi x/L} e^{|x|} =(\frac{1}{2\pi})\int_{-\pi}^{0}e^{-in \pi x/L} e^{-x}+(\frac{1}{2\pi})\int_{0}^{\pi}e^{-in \pi x/L} e^{x}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .