Fourier Series for f on the Interval [-π, π) | Homework Statement

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SUMMARY

The Fourier series for the function defined on the interval [-π, π) is accurately represented as ##\frac{4}{π} \sum_{k=0}^\infty \frac{1}{(2k+1)} \sin((2k+1)x)##. The Fourier coefficients are calculated as follows: ##a_0/2 = 1##, ##a_n = 0##, and ##b_n = \frac{2}{πn} - 2\frac{(-1)^n}{n}##. Corrections were noted regarding the values of ##a_0## and ##b_n##, indicating that a ##\pi## was omitted in the second term of ##b_n##.

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Homework Statement


Define ##f : [−π, π) → \mathbb R ## by
##f(x)## = ##−1## if ##− π ≤ x < 0##, ##1## if ##0 ≤ x < π.##
Show that the Fourier series of f is given by
##\frac{4}{π} \sum_{n=0}^\infty \frac{1}{(2k+1)} . sin(2k+1)x##

Homework Equations



The Fourier series for ##f## on the interval ##[−π, π)## is given by:

##\frac{a0}{2} + \sum_{n=0}^\infty ancos(nx) + \sum_{n=0}^\infty bnsin(nx)##
(Not quite sure why the LaTex isn't working here, I'm new at this.)
Where a0/2, an, bn are the Fourier coefficients of ##f##.

The Attempt at a Solution



I have attempted to find the Fourier coefficients, however I don't think that they're correct.

I have found
a0/2 = 1
an = 0.
bn = ##\frac {2}{πn} -2\frac {(-1)^n}{n}##

Could somebody tell me if these are correct? If not, I'll post up how I reached those answers.
 
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I think you lost a ##\pi## in the second term of your expression for ##b_n##.
 
##a_0## isn't correct either.
 

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