Answer to Question #253095 in Calculus for Anuj

Question

Expand each of the following functions in a Fourier sine series then a Fourier cosine series on the prescribed interval.

(i) 𝑓(𝑥) = 𝑒^−𝑥 ; 0 < 𝑥 < 1,

(ii) 𝑓(𝑥) = { 𝑥 0 < 𝑥 < 𝑙/2 𝑙 − 𝑥 𝑙/2 < 𝑥 < 𝑙 ; 0 < 𝑥 < 𝑙,

Answer 1

The Fourier sine series for a function f(x) with period 2L is given as

$f(x)=\sum_{n=1}^\infty b_n \sin(\frac{n\pi}{L}x)$ where $b_n=\frac{2}{L}\int_0^Lf(x)\sin(\frac{n\pi}{L}x)~dx$

(i) $f(x)=e^{-x}, 0<x<1$

The period of this function is 1, then L=$\frac{1}{2}$

$b_n=\frac{2}{\frac{1}{2}}\int_0^{\frac{1}{2}} e^{-x} \sin(2n\pi x)~dx\\ b_n=4\left[-\frac{e^{-x}(\sin(2 \pi n x)+2 \pi n \cos(2 \pi n x))}{4\pi^2 n^2+1}\right]\Biggr|_0^\frac{1}{2}\\ b_n=-\frac{8\pi n}{4\pi^2 n^2 +1}(e^{-\frac{1}{2}} (-1)^n-1)$

Therefore, the Fourier sine series for f(x) is ;

f$(x)=\sum_{n=1}^\infty -\frac{8\pi n}{4\pi^2 n^2 +1}(e^{-\frac{1}{2}} (-1)^n-1)\sin(2n\pi x)$

(ii) $f(x)=\begin{cases} x~~~~~~~~~~0<x<\frac{l}{2}\\ -x~~~~~~~~~~\frac{l}{2}<x<l\end{cases}$

f(x) is a periodic function of period $l \implies L=\frac{l}{2}$

$b_n=\frac{4}{l}\int_0^\frac{l}{2}x\sin(\frac{2n\pi}{l}x)~dx\\ b_n=\frac{4}{l}\left[ \frac{l^2 \sin(\frac{2\pi n}{l}x)-2\pi l n x \cos(\frac{2 \pi n}{l}x)}{4\pi^2n^2}\right]_0^\frac{l}{2}\\ b_n=-\frac{l}{n\pi}(-1)^n$

Therefore, the Fourier sine series for f(x) is;

$f(x)=\sum_{n=1}^\infty -\frac{l}{n\pi}(-1)^n\sin(\frac{2n\pi}{l}x)$