University of Florida/Egm4313/s12.team11.gooding/R6

R6.3

Problem Statement

Page 491, #15,17. Plot the truncated Fourier series for n=2,4,8.

Solution

15.

$f(x)=\{\begin{array}{c}-x-\pi ,-\pi <x<-\pi /2\\ x,-\pi /2<x<\pi /2\\ \pi -x,\pi /2<x<\pi \end{array}$

Since $f(-x)=-f(x)$, the function is odd.

$a_0=(1/\pi )*[-1\underset{-\pi }{\overset{-\pi /2}{\int }}(x+\pi )dx+\underset{-\pi /2}{\overset{\pi /2}{\int }}(x)dx+\underset{\pi /2}{\overset{\pi }{\int }}(\pi -x)dx]$
After simplification, $a_0=0$
. Using Equation (5) on page 490, the exapansion becomes;
$f(x)=8k/{\pi }^2[sin(\pi *x/L)-sin(3\pi *x/L)/9...)$
With $L=\pi ,k=\pi /2$ this becomes;
$f(x)=4/\pi (sinx-sin(3x)/9+sin(5x)/25-sin(6x)/36....)$ for n is "odd".
This means when n is even, $f(x)=0$

File:R63a.jpg

17.
$f(x)=1-|x|,-1\le x\le 1$

$f(x)=\{\begin{array}{c}1+x-\pi ,-1<x<0\\ 1-x,0<x<1\end{array}$

$a_0=1/2[\underset{-1}{\overset{0}{\int }}(1+x)dx+\underset{0}{\overset{1}{\int }}(1-x)dx]$
$a_0=1/2[1-1/2+1-1/2]$

 $a_0=1/2$

$a_n=1/1[\underset{-1}{\overset{0}{\int }}(1+x)cos(n\pi *x/1)dx+\underset{0}{\overset{1}{\int }}(1-x)cos(n\pi *x/1)dx]$
Simplifying results in;
$a_n=2[1/(n^2{\pi }^2{)}_{(}-1{)}^n/(n^2{\pi }^2)]$
So,

$a_n=4/(n^2{\pi }^2)$ 

when n is odd.
$b_n=1/1[\underset{-1}{\overset{0}{\int }}(1+x)sin(n\pi *x/1)dx+\underset{0}{\overset{1}{\int }}(1-x)sin(n\pi *x/1)dx]$
Simplifying results in;

$b_n=0$

$f(x)=1/2+\sum _{n=0}^{\mathrm{\infty }}4/(n^2{\pi }^2)*cos(n\pi *x)$

Since $a_n=0$ for even n, $f(x)$ becomes $f(x)=1/2$ for even n.

File:R63b.jpg