# Homework Help: Coefficients for an exponential Fourier Series

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1. Nov 29, 2017

### ElPimiento

I'm kinda just hoping someone can look over my work and tell me if I'm solving the problem correctly. Since my final answer is very messy, I don't trust it.

1. The problem statement, all variables and given/known data

We're asked to find the Fourier series for the following function:
$$f(\theta)=e^{−\alpha \lvert \theta \rvert}}, \ \ \ \text{for} −\pi<\theta<\pi,$$
where $f(\theta + \pi) = f(\theta)$. Except our Fourier series has the form:
$$f(\theta) = \sum_{-\infty}^{\infty} c_n e^{in\theta}$$

2. Relevant equations
Variation on Euler:
$$\cos{\theta} = \frac{1}{2}(e^{i\theta} + e^{-i\theta})$$
Vector projection:
$$proj_\vec{v}(\vec{u}) = \frac{\langle\vec{v}, \vec{u}\rangle}{\langle\vec{v}, \vec{v}\rangle}$$

3. The attempt at a solution
So, supposing $n$ is an Integer, I First realized that every positive $n$ has a negative partner, hence our special Fourier series formula can be rewritten as
\begin{align*} \sum_{-\infty}^{\infty} c_n e^{in\theta} & = c_0 + \sum_{-\infty}^{-1} (c_{n^{(-)}} e^{in\theta}) + \sum_{1}^{\infty} (c_{n^{(+)}} e^{in\theta}) \\ & = c_0 + \sum_{-\infty}^{\infty }k_n (e^{in\theta} + e^{-in\theta}) \\ & = \sum_{-\infty}^{\infty} k_n \cos(n\theta) \\ \end{align*}
Where I have used the variation on Euler's Equation (above) and set $k_n = (c_{n^{(-)}} + c_{n^{(+)}})/2$.
Then using the inner product:
$$\langle f(x), g(x) \rangle = \int_{-\pi}^{\pi} dx f(x) g(x)$$
I found that the coefficients should be:
\begin{align*} k_n & = \frac{\langle e^{-\alpha \lvert \theta \rvert}, \cos(n \theta) \rangle}{\langle \cos(n \theta), cos(n \theta)\rangle} \\ & = \frac{\int_{-\pi}^{\pi} e^{-\alpha \lvert \theta \rvert} cos(n \theta)} {\int_{-\pi}^{\pi} cos^2(n \theta)} \\ & = \frac{\int_{0}^{\pi} e^{-\alpha \theta} cos(n \theta)} {\int_{0}^{\pi} cos^2(n \theta)} \ \ \ \ \ (\because \text{all the functions are even}) \\ & = - \frac{4 n (\frac{\alpha \cos(\pi n) - n \sin(\pi n)}{(\alpha^2 + n^2)e^{(\pi \alpha)}} - \frac{\alpha}{\alpha^2 + n^2} )} {2 \pi n + \sin(2 \pi n)} \\ \end{align*}
Which is a mess.

I used cosines hoping it would simplify the problem (since they're even and the resulting series starts at 0 instead of $-\infty$), but it doesn't look like it did much good. So is the method at least right?

Last edited: Nov 29, 2017
2. Nov 29, 2017

### Ray Vickson

Why don't you simplify $\cos(\pi n), \sin(\pi n)$ and $\sin (2 \pi n)$? Remember that $n$ is an integer.

When you do that you get a result that agrees with that obtained using Maple.

3. Nov 30, 2017

### ElPimiento

Oh duh, I should've realized that >~<.
Thanks though! That cleaned it up nicely:
$$k_n = -\frac{2 \alpha (e^{\pi \alpha} - 1)}{\pi (\alpha^2 + n^2)}$$

4. Nov 30, 2017

### ElPimiento

Woops! got a little too excited with the cancelation:
\begin{align*} k_n & = -\frac{2 \, {\left(\frac{\left(-1\right)^{n} \alpha}{\alpha^{2} e^{\left(\pi \alpha\right)} + n^{2} e^{\left(\pi \alpha\right)}} - \frac{\alpha}{\alpha^{2} + n^{2}}\right)}}{\pi} \\ & = - \frac{2 \alpha ((-1)^n e^{- \pi \alpha} - 1)}{\pi (\alpha^2 + n^2)} \\ \end{align*}

Last edited: Nov 30, 2017