University of Florida/Egm6341/s10.team3.aks/HW2

(4) Derive the composite trapazoidal rule and composite simpson's rule from simple trapazoidal and simple simpson's rule

Ref Lecture notes p.9-3

Problem Statement

Show Simple Trapazoidal rulep.7-1 $\Rightarrow$ Composite Trapazoidal rule p.7-1

and

Show Simple Simpson's rule p.7-2 $\Rightarrow$ Composite Simpson's rulep.7-2

Solution

Composite Trapezoidal rule

From Simple trapezoidal rule we have ,

$I_1=\frac{h}{2}[f_{\text{o}}+f_1]$

$=h[\frac{1}{2}{f}_{\text{o}}+\frac{1}{2}{f}_1]$

similarly we have

$I_2=h[\frac{1}{2}{f}_1+\frac{1}{2}{f}_2]$

$I_3=h[\frac{1}{2}{f}_2+\frac{1}{2}{f}_3]$ . . . . $I_{\text{n-1}}=h[\frac{1}{2}{f}_{\text{n-2}}+\frac{1}{2}{f}_{\text{n-1}}]$

$I_{\text{n}}=h[\frac{1}{2}{f}_{\text{n-1}}+\frac{1}{2}{f}_{\text{n}}]$

Summation of all of above expression gives

$I_{\text{n}}=h[\frac{1}{2}{f}_{\text{o}}+(\frac{1}{2}{f}_1+\frac{1}{2}{f}_1)+(\frac{1}{2}{f}_2+\frac{1}{2}{f}_2)+.....+(\frac{1}{2}{f}_{\text{n-1}}+\frac{1}{2}{f}_{\text{n-1}})+\frac{1}{2}{f}_{\text{n}}]$

$=h[\frac{1}{2}{f}_{\text{o}}+f_1+f_2+.......+f_{\text{n-1}}+\frac{1}{2}{f}_{\text{n}}]$

Hence Proved..

Composite Simpson's Rule

From Simple Simpson's rule we obtain ,

$I_2=\frac{h}{3}[f_{\text{o}}+4f_1+f_2]$

where $h=\frac{b-a}{2}$

Similarly

$I_4=\frac{h}{3}[f_2+4f_3+f_4]$

$I_6=\frac{h}{3}[f_4+4f_5+f_6]$ . . . . $I_{\text{n-2}}=\frac{h}{3}[f_{\text{n-4}}+4f_{\text{n-3}}+f_{\text{n-2}}]$

$I_{\text{n}}=\frac{h}{3}[f_{\text{n-2}}+4f_{\text{n-1}}+f_{\text{n}}]$

After Summation of above terms we obtain

$I_{\text{n}}=\frac{h}{3}[f_{\text{o}}+4f_1+(f_2+f_2)+4f_{\text{3}}+(f_4+f_4)+4f_5+f_6.......+(f_{\text{n-2}}+f_{\text{n-2}})+4f_{\text{n-1}}+f_{\text{n}}]$

$I_{\text{n}}=\frac{h}{3}[f_{\text{o}}+4f_1+2f_2+4f_{\text{3}}+2f_4+4f_5+f_6.......+2f_{\text{n-2}}+4f_{\text{n-1}}+f_{\text{n}}]$

where n = 2k and k = 1,2,3,4.....

Hence Proved

(14) Prove

$e^{(3)}(t)=-\frac{t}{3}[F^{(3)}(t)-F^{(3)}(-t)]$

Ref Lecture p.15-2

Problem Statement

Prove that

$e^{(3)}(t)=-\frac{t}{3}[F^{(3)}(t)-F^{(3)}(-t)]$

where

$e(t)=\underset{-t}{\overset{t}{\int }}f(x(t))dt-\frac{t}{3}[F(-t)+4F(0)+F(t)]$

Solution

$e(t)=\underset{-t}{\overset{t}{\int }}f(x(t))dt-\frac{t}{3}[F(-t)+4F(0)+F(t)]$

$e(t)=\underset{-t}{\overset{k}{\int }}f(x(t))dt+\underset{k}{\overset{t}{\int }}f(x(t))dt-\frac{t}{3}[F(-t)+4F(0)+F(t)]$

$e^{(1)}(t)=(F(-t)+F(t))-\frac{t}{3}[-F^1(-t)+F^1(t)]-\frac{1}{3}[F(-t)+4F(0)+F(t)]$

$e^{(2)}(t)=(-F^1(-t)+F^1(t))-\frac{t}{3}[F^2(-t)+F^2(t)]-\frac{1}{3}[-F^1(-t)+F^1(t)]-\frac{1}{3}[-F^1(-t)+F^1(t)]$

$e^{(2)}(t)=(-F^1(-t)+F^1(t))-\frac{t}{3}[F^2(-t)+F^2(t)]-\frac{2}{3}[-F^1(-t)+F^1(t)]$

$e^{(3)}(t)=(F^2(-t)+F^2(t))-\frac{t}{3}[-F^3(-t)+F^3(t)]-\frac{1}{3}[F^2(-t)+F^2(t)]-\frac{2}{3}[F^2(-t)+F^2(t)]$ $e^{(3)}(t)=(F^2(-t)+F^2(t))-\frac{t}{3}[-F^3(-t)+F^3(t)]-(F^2(-t)+F^2(t))$

$e^{(3)}(t)=-\frac{t}{3}[-F^3(-t)+F^3(t)]$

$e^{(3)}(t)=-\frac{t}{3}[F^3(t)-F^3(-t)]$

Hence Proved ..

(12) Show Derivation

${\alpha }^{(1)}(t)$

Ref: Lecture Notes p.15-1

Problem Statement

Show derivation that

${\alpha }^{(1)}(t)=F(-t)+F(t)$

Solution

From (4) in p.14-2, we can write down the expression of $\alpha (t)$

Given :

f(x(t)) = F(t)

Let there exists $g^1(t)=F(t)$

$\alpha (t)=\underset{-t}{\overset{t}{\int }}f(x(t))dt=\underset{-t}{\overset{t}{\int }}F(t))dt=\underset{-t}{\overset{t}{\int }}{g}^1(t))dt=g(t)-g(-t)$

${\alpha }^{(1)}(t)=\frac{d}{dt}[g(t)-g(-t)]$

$\Rightarrow {\alpha }^{(1)}(t)=g^1(t)-(-g^1(-t))$

but $g^1(t)=F(t)$

$\Rightarrow {\alpha }^{(1)}(t)=F(t)+F(-t))$

Hence Proved ,