The Fourier coeficients of the function $f ( x ) = x$ on $[ - 1,1 ]$ are
A) $a _ { 0 } = - 2 / \pi , b _ { n } = 0 , a _ { n } = 0$
B) $a _ { 0 } = 0 , b _ { n } = ( - 1 ) ^ { n } 2 / ( n \pi ) , a _ { n } = 0$
C) $a _ { 0 } = 0 , b _ { n } = ( - 1 ) ^ { n + 1 } 2 / ( n \pi ) , a _ { n } = 0$
D) $a _ { 0 } = 0 , b _ { n } = 0 , a _ { n } = ( - 1 ) ^ { n } 2 / ( n \pi )$
E) $a _ { 0 } = 0 , b _ { n } = 0 , a _ { n } = ( - 1 ) ^ { n + 1 } 2 / ( n \pi )$

Question

Quizplus · Accepted Answer

The Fourier series of a function $f(x) = x$ on the interval $[-1,1]$ consists of sine terms only because the function is odd. The coefficients $a_n = 0$ for all $n$ because there are no cosine terms (which represent the even part of the function). The $a_0$ term, which represents the average value of the function over the interval, is also 0 because the function is symmetric about the origin and its positive and negative values cancel out. The $b_n$ coefficients are calculated using the integral of $f(x) \sin(n\pi x)$ over the interval $[-1,1]$, leading to $b_n = (-1)^{n+1}2/(n\pi)$, indicating the alternating sign and decreasing magnitude of the sine coefficients.

The Fourier Coeficients of the Function f(x)=xf ( x ) = xf(x)=x

The Fourier Coeficients of the Function $f ( x ) = x$