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Fourier series / Kpler's equation

  1. Oct 19, 2005 #1
    GIVE ME A HINT!!!! Fourier series / Kepler's equation

    By expanding [tex]e \sin\psi[/tex] in a Fourier series in [tex]\omega t[/tex], show that Kepler's equation has the formal solution
    [tex]\psi = \omega t + \sum_{n=1}^{\infty}{\frac{2}{n}J_{n}(ne)\sin{\omega t}}[/tex]
    where [tex]J_{n}[/tex] is the Bessel function of order n. For small argument, the Bessel function can be approximated in a power series of the argument. Accordingly, from this result derive the first few terms in the expansion of [tex]\psi[/tex] in powers of [tex]e[/tex].
    :confused:
     
    Last edited: Oct 20, 2005
  2. jcsd
  3. Oct 20, 2005 #2
    whats the matter?

    is this question too confusing or something, if you don't understand a part of it please let me know.:uhh:
     
  4. Oct 20, 2005 #3
    Kepler's equation

    Here's the Kepler's equation for those of you who don't know it:

    [tex]\omega t = \psi - e \sin \psi[/tex]

    and Fourier series I think should be in the following form:

    [tex]\sum_{i=1}^{\infty}e\sin \psi[/tex]

    ANY HELP WOULD BE GREATLY APPRECIATED, YOU DON'T HAVE TO GIVE ME THE ANSWER JUST ANYTHING THAT'S ON YOUR MIND WOULD BE HELPFUL. EVEN THE SLIGHTEST CLUE COULD HELP ME GREATLY.

    THANKS...
     
    Last edited: Oct 20, 2005
  5. Oct 20, 2005 #4

    Tom Mattson

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    Gold Member

    Where to begin?

    * You didn't provide any background info in the opening post. Most of us who browse the Homework Help section are goofing off at work, and we don't have our old textbooks on orbital mechanics laying around.

    * After doing some digging, I've discovered that the information that you did provide is wrong.

    This:

    should have been this:

    * You didn't show any work, which is required by the Physics Forums Guidelines, which you agreed to follow. This process works by show-and-tell. You show you work, we tell you where you went wrong.

    See the notice at the top of this Forum: FAQ: Why hasn't anybody answered my question? It's there to be read, as are the Guidelines you agreed to.

    Now on to your question.

    OK, that helps things along.

    You are just randomly guessing. No Fourier series looks like that.

    You could try looking into Fourier-Bessel series, as it looks like you are supposed to expand [itex]e\sin(\omega t)[/itex] in the Bessel function basis. But you can't expect any help here without showing us something.
     
  6. Oct 21, 2005 #5
    ok fair enough. I'm new to Fourier-Bessel series but I'll try my best...
    [tex]f(x) = e\sin(\omega t)[/tex]
    [tex]a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)\,dx = \frac{1}{\pi}\int_{-\pi}^{\pi}e\sin(\omega t) \cos(nx)\,dx[/tex]
    [tex]b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)\,dx = \frac{1}{\pi}\int_{-\pi}^{\pi}e\sin(\omega t) \sin(nx)\,dx[/tex]
    [tex]f(x) = \frac{1}{2}a_0 + \sum_{n=1}^{\infty}[a_n \cos(nx) + b_n \sin(nx)][/tex]
    so that's the Fourier series. Should I solve for [itex]a_n[/itex] and [itex]b_n[/itex] and then substitude them in the series? Or am I completely doing this the wrong way?
    I also don't understand how you changed [itex]e\sin(\psi)[/itex] to [itex]e\sin(\omega t)[/itex]
     
    Last edited: Oct 21, 2005
  7. Oct 23, 2005 #6
    here's the exapnasion

    [tex]f(x) = \frac{1}{\pi}\int_{-\pi}^{\pi}e\sin(\omega t)dt + \sum_{n=1}^{\infty}\frac{1}{\pi}\int_{-\pi}^{\pi}[e\sin(\omega t)\cos(nt)dt]\cos(nt) + \sum_{n=1}^{\infty}\frac{1}{\pi}\int_{-\pi}^{\pi}[e\sin(\omega t)\sin(nt)dt]\sin(nt)[/tex]

    please help I havent heard back from anybody since my last post! I'm doing as much as I can....
     
  8. Oct 25, 2005 #7
    your incapacity to help out or solve this straight forward problem either classifies you as a bunch of psychopaths or outright retards:yuck:
     
  9. Oct 31, 2005 #8
    is anybody even looking at this post? what am I doing wrong?
     
  10. Dec 31, 2007 #9
  11. Dec 31, 2007 #10
    This thread is two years old?
     
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