Fourier Series and Convergence for f(x) = 1 and g(x) = x on the Interval [0, pi]

[stunner5000pt]

Find the Fourier Series for f(x) =1 on the interval 0<= x <= pi in terms of \phi_{n} = \sin{nx}. Bu integrating this series find a convergent series for the function g(x) =x on this interval assuming the set {sin nx} is complete
Ok for the FOurier Coefficients
c_{n} = \frac{\int_{0}^{\pi} f \phi \rho dx}{\int_{0}^{\pi} \phi^2} dx
this is how it is in my test
rho is suppsod to be the weight
ok for the numerator
\int_{0}^{\pi} \sin{nx} dx = \frac{1}{n} [- \cos{nx}]_{0}^{\pi} = (-1)^n + 1
for hte denominator
\int_{0}^{\pi} (\sin{nx})^2 dx = \frac{\pi}{2} - \frac{\sin{2n \pi}}{4} =\frac{\pi}{2}
so the Fourier series is
\frac{2}{\pi} \sum_{n=1}^{\infty} \frac{\sin{nx}}{n} (-1^n + 1)
what do they mean by integrate the series? Does it mena i should integrate the argument of this sum? ANd how would one find a convergent series for the function g(x) =x??
Please help!

 

[dextercioby]

Yep, under certain conditions you can integrate a series term by term. E.g.

Taylor series for \frac{1}{1+x} with \mod{x}&lt;1 is

\frac{1}{1+x}=1-x+x^{2}-x^{3}+x^{4}-...

and converges. Integrating it term by term one gets a convergent series for \ln\left(1+x\right).

Daniel.

BTW, where din that \frac{1}{n} from integrating \sin nx go?

 

[stunner5000pt]

but how would the ln(1/1+x) relate to my problem?? DO you mena i am supposed to expand the sin or cos by taylor series?>?

i have edited to include that 1/n of the sine


is my FOurier Series correct? The back of my book states something else:
\frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\sin{2k-1}x}{2k-1}

im not sure how they got this...

 

[stunner5000pt]

how does what is in the text post #3 relate to waht i got?

It is similar in that i do have the sine part except what about the -1^n +1 part ? It appears the book has assumed n to be even ... why??
Also why has the book gotten 2k-1 instead of n? If n is supposed to be even then why is 2k-1, an odd number, used as n in teh same argument?