User:Egm6341.s10.team3.heejun

(4) Prove IMVT

Problem Statement

(i) $\int _{a}^{b}w(x)\cdot f(x)dx=f(\xi )\int _{a}^{b}w(x)dx$ for $w(x)\geq 0$ <br\> (ii) Another version of IMVT $w(x)<0$ for all $x\in [a,b]$

<br\> Ref: Lecture Notes p.5-1 <br\>

Solution

<br\> Ref: Lecture Notes p.2-3 <br\> (i)

$m\cdot w(x)\leq f(x)\cdot w(x)\leq M\cdot w(x)$

<br\> where,

m:=min f(x)

M:=max f(x) <br\> a) Integrating, $m\cdot \int _{a}^{b}w(x)dx\leq \int _{a}^{b}w(x)\cdot f(x)dx\leq M\cdot \int _{a}^{b}w(x)dx$ <br\> b) Dividing by ${\frac {1}{\int _{a}^{b}w(x)dx}}$ $m\leq \underbrace {{\frac {1}{\int _{a}^{b}w(x)dx}}\int _{a}^{b}w(x)\cdot f(x)dx} _{Z}\leq M$ $Z={\frac {1}{\int _{a}^{b}w(x)dx}}\int _{a}^{b}w(x)\cdot f(x)dx$ <br\> c) By Interm. Value Them, There exists $\xi \in [a,b]atf(\xi )=Z$ <br\> $\therefore f(\xi )={\frac {1}{\int _{a}^{b}w(x)dx}}\int _{a}^{b}w(x)\cdot f(x)dx$ <br\> (ii) $m\cdot \left|w(x)\right|\leq f(x)\cdot \left|w(x)\right|\leq M\cdot \left|w(x)\right|$ <br\> where,

m:=min f(x)

M:=max f(x) <br\> a) Integrating, $m\cdot \int _{a}^{b}\left|w(x)\right|dx\leq \int _{a}^{b}\left|w(x)\right|\cdot f(x)dx\leq M\cdot \int _{a}^{b}\left|w(x)\right|dx$ <br\> b) Dividing by ${\frac {1}{\int _{a}^{b}\left|w(x)\right|dx}}$ $m\leq \underbrace {{\frac {1}{\int _{a}^{b}\left|w(x)\right|dx}}\int _{a}^{b}\left|w(x)\right|\cdot f(x)dx} _{Z}\leq M$ $Z={\frac {1}{\int _{a}^{b}\left|w(x)\right|dx}}\int _{a}^{b}\left|w(x)\right|\cdot f(x)dx$ <br\> c) By Interm. Value Them, There exists $\xi \in [a,b]atf(\xi )=Z$ <br\> d) $w(x)$ is strictly negative, so $f(\xi )={\frac {1}{{\cancelto {}{(-1)}}\cdot \int _{a}^{b}w(x)dx}}\cdot {\cancelto {}{(-1)}}\cdot \int _{a}^{b}w(x)\cdot f(x)dx$ <br\> $\therefore f(\xi )={\frac {1}{\int _{a}^{b}w(x)dx}}\cdot \int _{a}^{b}w(x)\cdot f(x)dx$ <br\>

(8) Taylor, Trap and Simpson Rule

Problem Statement

$I=\int _{0}^{1}{\frac {e^{x}-1}{x}}dx$

(i) Taylor series exp. f_n<br\> (ii) Comp. Trap. rule <br\> (iii) Comp. Simpson rule <br\> <br\> Ref: Lecture Notes p.6-5 <br\>

Solution

(i)<br\> a) n=2,

$I_{2}=\left({\frac {1}{(1!\cdot 1)}}+{\frac {1}{(2!\cdot 2)}}\right)=1.25$

a-1) Error for n=2,

${\frac {1}{(2+1)!\cdot (2+1)}}\leq I-I_{2}\leq {\frac {e}{(2+1)!\cdot (2+1))}}=0.055556\leq I-I_{2}\leq 0.151016$

<br\> <br\> b) n=4,

$I_{4}=\left({\frac {1}{(1!\cdot 1)}}+{\frac {1}{(2!\cdot 2)}}+{\frac {1}{(3!\cdot 3)}}+{\frac {1}{(4!\cdot 4)}}\right)=1.3159722222223\approx 1.31597$

b-1) Error for n=4,

${\frac {1}{(4+1)!\cdot (4+1)}}\leq I-I_{2}\leq {\frac {e}{(4+1)!\cdot (4+1))}}=0.001667\leq I-I_{2}\leq 0.00453$

<br\> <br\> c) n=8,

$I_{4}=\left({\frac {1}{(1!\cdot 1)}}+{\frac {1}{(2!\cdot 2)}}+{\frac {1}{(3!\cdot 3)}}+{\frac {1}{(4!\cdot 4)}}+{\frac {1}{(5!\cdot 5)}}+{\frac {1}{(6!\cdot 6)}}+{\frac {1}{(7!\cdot 7)}}+{\frac {1}{(8!\cdot 8)}}\right)=1.3179018152401\approx 1.3179$

c-1) Error for n=8,

${\frac {1}{(8+1)!\cdot (8+1)}}\leq I-I_{2}\leq {\frac {e}{(8+1)!\cdot (8+1))}}=3.06192\times 10^{-7}\leq I-I_{2}\leq 8.32317\times 10^{-7}$

<br\>

(ii)<br\> This one can be calculated by the Comp. Trap. rule or the Corrected Trap. rule but it is hard to define the first term, X₀.

Thus, some math-codes are required. <br\>

=Find the true value of I= MATLAB Code clc; clear all; format long F=@(x)(exp(x)-1)./x; I=quad(F,0,1) I = 1.317902151956861 %Assumption: This is the true valuse of I

<br\>

a) the Comp. Trap. rule

$\int _{a}^{b}f(x)dx={\frac {b-a}{2n}}*[f(x_{0})+2f(x_{1})+2f(x_{2})+.........+2f(x_{n-1})+f(x_{n})]$

b) the Corrected Trap. rule

$\int _{a}^{b}f(x)dx={\frac {b-a}{2n}}*[f(x_{0})+2f(x_{1})+2f(x_{2})+.........+2f(x_{n-1})+f(x_{n})]-{\frac {h^{2}}{12}}*[f^{'}(b)-f^{'}(a)]$

Where

$h={\frac {(b-a)}{n}}$ <br\> <br\> c) Results of Comp. Trap. rule

n	In	En
2	1.328291728	-0.010389576
4	1.320504619	-0.002602468
8	1.318553087	-0.000650935
16	1.318064905	-0.000162753
32	1.317942841	-4.06892*10-5
64	1.317912324	-1.0172*10-5
128	1.317904695	-2.54263*10-6
256	1.317902787	-6.3528*10-7

Where

E_n = The true value of I (=1.317902152) - I_n <br\> Finally, the value of n=128 is closer to the true value with about 10^-6 order. <br\> d) Example of MATLAB Code (n=2)

clc;
clear all;
format long
X=eps:0.5:1;
Y=(exp(X)-1)./X;
Z=trapz(X,Y)

<br\>

(iii)<br\> This one can be calculated by the Comp. Simpson rule but it is also hard to define the first term, X0.

Thus, some math-codes are required.

<br\> <br\>

a) the Comp. Simpson rule

$\int _{a}^{b}f(x)dx={\frac {b-a}{3n}}*[f(x_{0})+4f(x_{1})+2f(x_{2})+.........+2f(x_{n-2})+4f(x_{n-1})+f(x_{n})]$

<br\> <br\>

b) Results of Comp. Simpson rule

n	In	En
2	1.318008666	-0.000106514
4	1.317908917	-6.76473*10-6
8	1.317902576	-4.24046*10-7
16	1.317902178	-2.60589*10-8

Where

E_n = The true value of I (=1.317902152) - I_n <br\> Finally, the value of n=4 is closer to the true value with about 10^-6 order. <br\> c) MATLAB Source Code

function y = simpson(f,a,b,n)
%SIMPSON Simpson's rule integration with equally spaced points
%
% y=SIMPSON(f,a,b,n) returns the Simpson's rule approximation to
% the integral of f(x) over the interval [a,b] using n+1 equally
% spaced points. The input variable f is a string containing the
% name of a function of one variable. The function f(x) must accept
% a vector argument and return the vector of values of the function.
%
% NOTE: n must be even.

h=(b-a)/n;
x=linspace(a,b,n+1);
fx=feval(f,x);

y=h/3*(fx(1)+4*sum(fx(2:2:n))+2*sum(fx(3:2:n-1))+fx(n+1));

<br\>

<br\> d) Example of MATLAB Code (Run, n=2)

clear all
format long
z=simpson(@(x) (exp(x)-1)./x,eps,1,2)

<br\>

--Heejun Chung 17:43, 27 January 2010 (UTC)