1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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)]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Fourier series of e^x
  1. Fourier series (Replies: 11)

  2. Fourier series (Replies: 1)

  3. Fourier Series (Replies: 4)