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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: More Fourier Series

**Physics Forums | Science Articles, Homework Help, Discussion**