University of Florida/Egm6341/s10.team3.aks

(10) Prove Simple trapezoidal rule

Ref. Lecture notes [[[media:Egm6341.s10.mtg8.pdf|p.8-2]]]

Problem Statement

Use (2) in Slide [[[media:Egm6341.s10.mtg8.pdf|8-2]]] to obtain Simple trapezoidal rule.

Solution

Given

$I_1=[\underset{a}{\overset{b}{\int }}{l}_{\text{o}}(x)dx]f(x_{\text{o}})$ + $[\underset{a}{\overset{b}{\int }}{l}_1(x)dx]f(x_1)$

where

$l_{\text{o}}(x)=\frac{x-x_1}{(x_{\text{o}}-x_1)}$,

$l_1(x)=\frac{x-x_{\text{o}}}{(x_1-x_{\text{o}}}$,

$x_{\text{o}}=a$

$x_1=b$

Now $I_1=\left[\underset{a}{\overset{b}{\int }}\frac{x-x_1}{x_{\text{o}}-x_1}dx\right]f(x_{\text{o}})$ + $\left[\underset{a}{\overset{b}{\int }}\frac{x-x_{\text{o}}}{x_1-x_{\text{o}}}dx\right]f(x_1)$

= $[\frac{x^2/2-bx}{a-b}{]}_a^{b}f(a)+[\frac{x^2/2-ax}{b-a}{]}_a^{b}f(b)$

= $[(\frac{(\frac{b^2}{2}-b^2)-(\frac{a^2}{2}-ab))}{(a-b)}]f(a)+[\frac{(\frac{b^2}{2}-ab)-(a^2/2-a^2)}{(b-a)}]f(b)$

= $[\frac{(b-a{)}^2}{2(b-a)}]f(a)+[(\frac{(b-a{)}^2}{2(b-a)}]f(b)$

= $[\frac{(b-a)}{2}](f(a)+f(b))$

which is same as equation (1) Slide p.7-1

This completes the Proof of trapezoidal rule

(11) Expansion of Lagrange functions

Ref: Lecture notes p.8-3 [[[media:Egm6341.s10.mtg8.pdf]]|

Problem Statement

Expand(4) from Slide (8-3 [[[media:Egm6341.s10.mtg8.pdf]]]]) to obtain $P_2(x_j)=\sum _{i=0}^2{l}_i(x_j)f(x_i)=f(x_j)$

Solution

$P_2(x_j)=\sum _{i=0}^2{l}_i(x_j)f(x_i)=l_{\text{o}}(x_j)f(x_{\text{o}})+l_1(x_j)f(x_1)+l_2(x_j)f(x_2)$

where $j=0,1,2$

but when $i=j$ $l_i(x_j)=1$

and when i$\ne$j $l_i(x_j)=0$

then only surviving terms are given by

$P_2(x_j)=l_{\text{o}}(x_{\text{o}})f(x_{\text{o}})+l_1(x_1)f(x_1)+l_2(x_2)f(x_2)=f(x_{\text{o}})+f(x_1)+f(x_2)=f(x_j)$

(3) Plot Functions Sin (x),-cos (x)and Sin(x)+cos(x)

Ref. Lecture notes p.3-3 [[[media:Egm6341.s10.mtg3.pdf]]|

Problem Statement

Plot f(x)= sin(x) and g(x) = - cos(x) in the interval of [0,pi]]

and also find $||f(x)||\mathrm{\infty },||g(x)||\mathrm{\infty }and||f(x)-g(x)||\mathrm{\infty }$

Solution

Plot f(x)= sin(x) in interval [0,pi]

Matlab code :

 
x = 0:pi/100:pi;
y = sin(x);
plot(x,y)
xlabel('x = 0:pi');
ylabel('Sine of x');
title('Plot of the Sine Function');

Plot :

f(x)=y= Sin(x)

Plot g(x)= - cos(x) in interval [0,pi]

Matlab code :

x = 0:pi/100:pi;
y = -cos(x);
plot(x,y)
xlabel('x = 0:pi');
ylabel('cosine of x');
title('Plot of the cosine Function');

Plot :

g(x)= y = - cos(x)

Plot f(x)-g(x)= Sin(x)+cos(x)

Matlab code :

x = 0:pi/100:pi;
y = sin(x)+cos(x);
plot(x,y)
title('Plot of the Sine+Cosine Function');
ylabel('Sine+Cosine of x'); 
xlabel('x = 0:pi');

File:Sin+cos.jpg

$||f(x)||\mathrm{\infty }=1$

$||g(x)||\mathrm{\infty }=1$

$||f(x)-g(x)||\mathrm{\infty }=\sqrt{2}$

Abhishekksingh 16:25, 27 January 2010 (UTC)