- #1

ElPimiento

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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.

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}

$$

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}

$$

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?

1. Homework Statement1. Homework Statement

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}

$$

## Homework 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}

$$

## 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: