Find the Fourier series for the function f(x) = sin(4x)

1. Sep 16, 2004

broegger

I have to find the Fourier series for the function f(x) = sin(4x), but no matter what I find _all_ the Fourier coefficients to be zero; i.e. $$(2\pi)^{-1}\int_{-\pi}^{\pi}sin(4x)e^{-inx}dx = 0$$ for all n.

I can't see the point in finding the Fourier series for sin(4x) anyway, since the function is simple harmonical - but shouldn't some of the coefficients be non-zero?

I'm new to the subject, so please excuse me if the answer is obvious...

Last edited: Sep 16, 2004
2. Sep 16, 2004

matt grime

The integral of sin(4x)sin(4x) is an integral of a positive not identically zero conintuous function over some interval possibly [0,2pi], so it can' t be zero. it is a silly question, but it has shown you that there's something you're doing wrong, so it's served some purpose, surely?

3. Sep 16, 2004

HallsofIvy

Yes, ONE of the coefficients is not 0!

The point of this problem is that you should be able to recognize that it is trivial and simply write down the answer without doing any calculations!

4. Sep 16, 2004

matt grime

HallsofIvy, did you look at the latex source for that tag? seems like they're doing fourier series the pure way not the applied way, in that they're using exp{inx} rather than sins and cosines individually, so the question isn't totally vacuous.

5. Sep 16, 2004

broegger

I don't know why my latex isn't being generated.

We're using the system {exp(inx)} - which coefficient is the non-zero one?

6. Sep 16, 2004

HallsofIvy

Thanks, Matt, I didn't look at that before.

broegger, there seems to a general problem with the latex lately.

What I meant was, assuming you were writing the Fourier series as a sum of sin(nx), cos(nx) (Matt, YOU were the one who mislead me, by talking about integrating sin(4x)sin(4x)!). In that case, sin(4x) IS the Fourier series.

Okay, since you are writing the Fourier series as a sum of einx, there will be two non-zero coefficients. Do you know how to write sine as exponentials?

7. Sep 16, 2004

matt grime

I missed it the first time too, and only checked cos i was wondering exactly over what interval we were working. Sorry for sending you off on the slightly wrong tangent, HallsofIvy.

Last edited: Sep 16, 2004
8. Sep 16, 2004

broegger

Oh, of course.. sin(nx) = 1/(2i)[exp(inx)-exp(-inx)].. It turns out that the Fourier series is just sin(4x) itself.. thanks for your help!

I have a lot of exercises so I'll probably be back with more questions in this thread ;)

9. Sep 16, 2004

matt grime

In this case, expanding using exponentials the fourier series is:

(1/2i)*(exp{4inx} - exp{-4inx})

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