Fourier series

[nizama]

Hi there!!

Can anyone please help me with this one..?

Find the Fourier Sine series of f(x)=sin(x/2) for interval (0,pi)

thanx a lot :)

[vadik]

well, sin(x/2) is already a periodical function with period pi, so even if you construct a furier series of this one you would get the same: sin(x/2)

[Xishan]

get the standard Fourier series for sin(X) and replace X=x/2, and thats it!

[Muzza]

well, sin(x/2) is already a periodical function with period pi

Wouldn't it have a period of 4pi?

[nizama]

Well hmm...not really :rolleyes:

I did get almoust correct answer BUT...
you see b_n is zero but a_n is not...and this is where i stuck ...i cant get correct a_n
by the way...can you tell me is this true?

cos(npi) + cos(0pi) = (-1)^(n+1)

thanx again :smile:

[Muzza]

(Assuming n is an integer). It's not true. cos(npi) is either 1 or -1, and cos(0pi) = cos(0) = 1, so the LHS is either 1 + 1 = 2 or -1 + 1 = 0, while the RHS is either 1 or -1... So you can never get equality.

[nizama]

Yes i just got it..
it is when you use integr. in series
so then it would be ((-1)^n +1)

thanx

:smile:

[Ebolamonk3y]

couldn't you just use the regular Taylor series?

[nizama]

Not really :frown:
the task...as so on my exam..they strictly want it with fourier ...
bcs we do also taylor..and in task they mention which one they want..

[matt grime]

THe answers would be different for a start as well.