Finding Fourier Series f(x)=1 & Integrating for g(x)=x on 0<=x<=pi

[stunner5000pt]

Find the Fourier series for f(x)=1 on the interval 0 <=x <= pi in temrs of phi = sin nx. By integrating thi series find a convergent series for hte function g(x) =x oin this interval assuming that the set {sin nx} is complete

i can find the Fourier series for f(x) =1. But i would like to know why n is supposed to be odd becuase the answer i get is
$$\frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1 - \cos(n \pi)}{n} \sin(nx)$$
the text ssumes n is odd. Is it becuase if n was even then the summand would be zero?
the answer in the text is $$\frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\sin(2k-1) x}{2k-1}$$

How would i integrate the series?
f i integrate the answer for the text i get $$\frac{-4}{\pi} \sum_{n=1}^{\infty} \frac{\cos(2k-1) x}{(2k-1)^2}$$
but hte answer in the book i
$$\frac{-4}{\pi} \sum_{n=1}^{\infty} \frac{1-\cos(2k-1) x}{(2k-1)^2}$$

why or where is that 1 / (2k-1)^2 term coming from? Can someone please help?

Thank you

 

[quasar987]

Find the Fourier series for f(x)=1 on the interval 0 <=x <= pi in temrs of phi = sin nx. By integrating thi series find a convergent series for hte function g(x) =x oin this interval assuming that the set {sin nx} is complete
i can find the Fourier series for f(x) =1. But i would like to know why n is supposed to be odd becuase the answer i get is
$$\frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1 - \cos(n \pi)}{n} \sin(nx)$$
the text ssumes n is odd. Is it becuase if n was even then the summand would be zero?

Precisely. Try plugging different values of n in cos(n pi). You will see that cos(n pi) = (-1)^n. Hence, for n even, cos(n pi)=1 and the summand vanishes. Now your sum is over all odd positive integers. But you can restore the sum to be over all positive integer by noting that the change of variable n-->2k-1 (with k running over the naturals) spans exactly every odd positive integers. This is what the book did.

How would i integrate the series?

If you've seen how to prove the uniform convergence of Fourier series, do it. If not, assume it does converge uniformly such that the integral of the sum is the serie of the integrals.

f i integrate the answer for the text i get $$\frac{-4}{\pi} \sum_{n=1}^{\infty} \frac{\cos(2k-1) x}{(2k-1)^2}$$
but hte answer in the book i
$$\frac{-4}{\pi} \sum_{n=1}^{\infty} \frac{1-\cos(2k-1) x}{(2k-1)^2}$$
why or where is that 1 / (2k-1)^2 term coming from? Can someone please help?
Thank you

I think it comes from the constants of integration and then evaluating the serie at x=0, but I don't understand the mechanism exactly. It is true that setting one constant 0 and the other 1 / (2k-1)^2 makes the serie work when evaluating at x = 0, but how can we be sure that these are the right constant?