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Coefficients for an exponential Fourier Series

  1. Nov 29, 2017 #1
    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. jcsd
  3. Nov 29, 2017 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  4. Nov 30, 2017 #3
    Oh duh, I should've realized that >~<.
    Thanks though! That cleaned it up nicely:
    $$
    k_n = -\frac{2 \alpha (e^{\pi \alpha} - 1)}{\pi (\alpha^2 + n^2)}
    $$
     
  5. Nov 30, 2017 #4
    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
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