Fourier series / Kpler's equation

[suckstobeyou]

By expanding e \sin\psi in a Fourier series in \omega t, show that Kepler's equation has the formal solution
\psi = \omega t + \sum_{n=1}^{\infty}{\frac{2}{n}J_{n}(ne)\sin{\omeg a t}}
where J_{n} 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 \psi in powers of e.
:confused:

[suckstobeyou]

is this question too confusing or something, if you don't understand a part of it please let me know.:uhh:

[suckstobeyou]

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

\omega t = \psi - e \sin \psi

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

\sum_{i=1}^{\infty}e\sin \psi

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...

[Tom Mattson]

is this question too confusing or something, (snip)

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:


\psi = \omega t + \sum_{n=1}^{\infty}{\frac{2}{n}J_{n}(ne)\sin(\omeg a t)


should have been this:


\psi = \omega t + \sum_{n=1}^{\infty}{\frac{2}{n}J_{n}(ne)\sin(n\ome ga t)


* 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? (http://www.physicsforums.com/showthread.php?t=94380) It's there to be read, as are the Guidelines you agreed to.

Now on to your question.


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

\omega t = \psi - e \sin \psi


OK, that helps things along.


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

\sum_{i=1}^{\infty}e\sin \psi


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 e\sin(\omega t) in the Bessel function basis. But you can't expect any help here without showing us something.

[suckstobeyou]

ok fair enough. I'm new to Fourier-Bessel series but I'll try my best...
f(x) = e\sin(\omega t)
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
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
f(x) = \frac{1}{2}a_0 + \sum_{n=1}^{\infty}[a_n \cos(nx) + b_n \sin(nx)]
so that's the Fourier series. Should I solve for a_n and b_n and then substitude them in the series? Or am I completely doing this the wrong way?
I also don't understand how you changed e\sin(\psi) to e\sin(\omega t)

You could try looking into Fourier-Bessel series, as it looks like you are supposed to expand e\sin(\omega t) in the Bessel function basis.

[suckstobeyou]

here's the exapnasion

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)

please help I havent heard back from anybody since my last post! I'm doing as much as I can....

[suckstobeyou]

your incapacity to help out or solve this straight forward problem either classifies you as a bunch of psychopaths or outright retards:yuck:

[suckstobeyou]

is anybody even looking at this post? what am I doing wrong?

[lalbatros]

That was not a homework, but a standard textbook material.
See: http://murison.alpheratz.net/Maple/KeplerSolve/KeplerSolve.pdf

[belliott4488]

This thread is two years old?