Expert: Paul Klarreich - 11/18/2010

Question
f(x)= 1/4(x) for -pi<=x<pi and f(x+2pi)=f(x)
expand 5 non zero terms..i cannot find bn for this series..

Answer
Questioner:   david
Country:  Singapore
Category:  Advanced Math
Private:  No
 
Subject:  Fourier series
Question:  f(x)= 1/4(x) for -pi<=x<pi and f(x+2pi)=f(x)
expand 5 non zero terms..i cannot find bn for this series..
..................................

What does bn mean?  (Translation: don't send me a question without
defining all your terms and symbols.  I don't have your textbook.)
Your f(x) is an odd function, so its cosine terms will all be zero.
Its sine terms have coefficients:

1/pi times:

{pi
| f(x) sin nx dx =
}-pi


1/pi times:

{pi
| x/4 sin nx dx =
}-pi

1/4pi times:

{pi
| x sin nx dx =
}-pi

IBP gives you (see THE INTEGRATOR site at Integrals.Wolfram.com)

(-nx Cos[nx] + Sin[nx] )/n^2

from -pi to pi.

Now sin(n pi) = 0 for all n, so evaluate:

(-x Cos[nx])/n  from -pi to pi

(-pi cos(n pi)) - (-(-pi) cos(- n pi))

If n is ODD, cos(+- n pi) is -1:

(-pi (-1) - (pi (-1))

(pi + pi) = 2pi

If n is EVEN, cos(+- n pi) = 1:

(-pi cos(2 pi)) - (-(-pi) cos(- 2 pi))

(-pi cos(2 pi)) - (pi cos(- 2 pi))

(-pi (1)) - (pi (1))

(-pi - pi) = - 2pi

So your bn should be  (-1)^n (makes your terms alternate)

times  2pi/n (which we got)  

divided by 4pi  (from before)


I hope I didn't blow too many (-) signs.