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

$I_{1}={\Big [}\int _{a}^{b}l_{\text{o}}(x)dx{\Big ]}f(x_{\text{o}})$ + ${\Big [}\int _{a}^{b}l_{1}(x)dx{\Big ]}f(x_{1})$

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 $I_{1}={\Big [}\int _{a}^{b}{\frac {x-x_{1}}{x_{\text{o}}-x_{1}}}dx{\Big ]}f(x_{\text{o}})$ + ${\Big [}\int _{a}^{b}{\frac {x-x_{\text{o}}}{x_{1}-x_{\text{o}}}}dx{\Big ]}f(x_{1})$

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

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

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

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

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$\neq$j $l_{i}(x_{j})=0$

then only surviving terms are given by

$\displaystyle P_{2}(x_{j})=l_{\text{o}}(x_{\text{o}})f(x_{\text{o}})+l_{1}(x_{1})f(x_{1})+l_{2}(x_{2})f(x_{2})=f(x_{\text{o}})+f(x_{1})+f(x_{2})=f(x_{j})$

(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 $||f(x)||\infty ,||g(x)||\infty and||f(x)-g(x)||\infty$

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');

$||f(x)||\infty =1$

$||g(x)||\infty =1$

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

Abhishekksingh 16:25, 27 January 2010 (UTC)