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

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

Solution

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

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

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

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

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

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

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

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

Hence Proved ..

(12) Show Derivation

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

Ref: Lecture Notes p.15-1

Problem Statement

Show derivation that

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

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 $g^{1}(t)=F(t)$

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

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

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

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

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

Hence Proved ,