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Homework Help: Fourier series of e^x

  1. May 11, 2005 #1
    Hi all,

    I've been having little problems getting Fourier series of [itex]e^x[/itex].

    I have given

    f(x) = e^{x}, x \in [-\pi, \pi)


    a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi} e^{x}\ dx = \frac{2\sinh \pi}{\pi}

    a_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi} e^{x}\cos (nx)\ dx =

    \frac{1}{\pi}\left\{\left[ \frac{1}{n}\ e^{x} \sin (nx)\right]_{-\pi}^{\pi} -

    \frac{1}{n}\int_{-\pi}^{\pi} e^{x}\sin (nx)\ dx\right\} =

    \frac{(-1)^{n+1}\left(e^{-\pi} - e^{\pi}\right)}{\pi(1+n^2)}

    b_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi}e^{x}\sin (nx)\ dx = \frac{1}{\pi}\left\{

    \left[ -\frac{1}{n}\ e^{x}\cos (nx)\right]_{-\pi}^{\pi} +

    \frac{1}{n}\int_{-\pi}^{\pi} e^{x}\cos(nx)\ dx\right\} =

    \frac{n(-1)^{n}\left(e^{-\pi} - e^{\pi}\right)}{\pi(1+n^2)}

    So the Fourier series looks like this:

    S(f,x) = \frac{\sinh \pi}{\pi}\ +\ \frac{2\sinh \pi}{\pi}\sum_{n=1}^{\infty}

    \frac{(-1)^{n+1}}{1+n^2}\left(\cos(nx) - n\sin(nx))

    Anyway, our professor gave us another right (I hope so) result:

    S(f,x) = \frac{\sinh \pi}{\pi}\ +\ \frac{2\sinh \pi}{\pi}\sum_{n=1}^{\infty}

    \frac{(-1)^{n}}{1+n^2}\left(\cos(nx) - n\sin(nx))

    Obviously my series is just of opposite sign than it should be, but I can't find
    the mistake, could you help me please?
    Last edited: May 11, 2005
  2. jcsd
  3. May 11, 2005 #2

    Am I missing something? The two solutions look the same to me!
  4. May 11, 2005 #3
    jdavel, twoflower's solution has (-1)^(n + 1) inside the sum, while the professor's solution has (-1)^n.
    Last edited: May 11, 2005
  5. May 11, 2005 #4

  6. May 11, 2005 #5


    User Avatar
    Homework Helper

    You must have made a mistake in the integrals. It's hard to tell because you've skipped 3 or 4 steps in your post, but it looks like at least your bottom integral is wrong (wrong sign) for the case where n is odd (the only case I felt like doing in my head).
  7. Jan 24, 2010 #6
    as far as i can tell, when you have taken (e^-pi - e^pi) out from the summation you have left it as sinh(pi) whereas it should be -sinh(pi), this should then match with the professors solution.
    I have a similar question, using this series i must show that [tex]\sum1/(1 + n^2)[/tex] = 1/2(picoth(pi) -1), however when i calculate it to be 1/2(picoth(pi) + pi - 1), can anyone help?
  8. Apr 7, 2011 #7
    Why is integration of e^x=2sinh (pie)/pie ?
    i am not able to understand this concept can some body enlighten this for me !!
  9. Aug 11, 2011 #8
    Sinh(x)= 1/2 [e^x - e^(-x)]
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