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(adsbygoogle = window.adsbygoogle || []).push({});

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

[tex] \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1 - \cos(n \pi)}{n} \sin(nx) [/tex]

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 [tex] \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\sin(2k-1) x}{2k-1} [/tex]

How would i integrate the series?

f i integrate the answer for the text i get [tex] \frac{-4}{\pi} \sum_{n=1}^{\infty} \frac{\cos(2k-1) x}{(2k-1)^2} [/tex]

but hte answer in the book i

[tex] \frac{-4}{\pi} \sum_{n=1}^{\infty} \frac{1-\cos(2k-1) x}{(2k-1)^2} [/tex]

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

Thank you

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