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#1 2006-10-11 09:49:53

asv
Member
Registered: 2006-01-18
Posts: 5

Simple Fourier Series Question

Hi all,

Here's the question:

---

f(x)= 0 if -pi≤x≤0
     = sin x if 0<x<pi

Show that the Fourier series of f is:

1/pi + (sinx)/2 - (2/pi)∑((cos(2nx)/(4n²-1)).

---

Now, here's my problem...
I can get the 1/pi, i can get where the -(2/pi)∑((cos(2nx)/(4n²-1)) comes from. But where on earth is the (sinx)/2 from?

Please help!

asv

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#2 2006-10-11 12:25:11

fgarb
Member
Registered: 2006-03-03
Posts: 89

Re: Simple Fourier Series Question

There's no real trick to it, it's just that 1/2 comes out of the formula for the sin(x) coefficient when you evaluate it.

Remember, the formula for the sin coefficients looks like this:

Of course, since f is 0 over [-pi,0], the integral is really only from 0 to pi. Plug everything in and you'll see that you get 1/2. The cos terms require more complicated integration, of course.

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#3 2006-10-11 21:12:44

asv
Member
Registered: 2006-01-18
Posts: 5

Re: Simple Fourier Series Question

Thanks for your reply.

I can't get 1/2 for the value of bn. Here's my approach:

pi*bn=∫f(x)sin(nx)dx= ∫sin(x)sin(nx) dx [evaluated between 0 and pi]


=1/2 ∫ (-cos((1+n)x) + cos((1-n)x)) dx  {using that 2sinAsinB=-cos(A+B)+cos(A-B)}

=1/2 [  (-sin((n+1)x))/(n+1) + (sin((n-1)x))/(n-1) ] evaluated between 0 and pi

and upon substituting the limits (with n a natural number), the sine of 0 and the sine of an integer multiple of pi are both zero, so the whole thing is 0.



Plus, if bn was 1/2, wouldn't the question say:


"Show that the Fourier series of f is:

1/pi + ∑(sin(nx))/2 - (2/pi)∑((cos(2nx)/(4n²-1))."  ?

(with the sums running from n=1 to infinity)

asv

EDIT: Please ignore me. Problem solved. Thanks for your help!

Last edited by asv (2006-10-12 03:41:15)

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#4 2006-10-12 15:15:00

fgarb
Member
Registered: 2006-03-03
Posts: 89

Re: Simple Fourier Series Question

Glad to see you figured it out. If you haven't done so already, I highly recommend going back to make sure you understand how the fourier series formulas are derived - not only will they be more satisfying, you'll also learn tricks that would have helped you avoid the mistake you made here.

One of the most important aspects of the trig functions sin and cos is that they are "orthogonal". That means that



and

Whenever m is not equal to n and the integral is over an integer multiple of the period. If these properties didn't hold, the fourier series would not be correct, and it wouldn't be as easy to use.

Another remarkable aspect of this approach is that it can be used with other sets of functions as well, so long as they are also complete and orthogonal. The fourier expansion is most common because it is probably the easiest to use, but it gets ugly when you work with problems that don't have "cartesian symmetry". There are nasty functions called spherical harmonics that are more useful for problems with spherical symmetry, and there are more nastier functions called Bessel Functions that tend to be used for problems with cylindrical symmetry. But if you understand the fourier series, the techniques are all the same!

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