Problem Statement

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

and

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

Solution

Composite Trapezoidal rule

From Simple trapezoidal rule we have ,

${\displaystyle I_{1}={\frac {h}{2}}[f_{\text{o}}+f_{1}]}$

${\displaystyle =h[{\frac {1}{2}}f_{\text{o}}+{\frac {1}{2}}f_{1}]}$

similarly we have

${\displaystyle I_{2}=h[{\frac {1}{2}}f_{1}+{\frac {1}{2}}f_{2}]}$

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

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

Summation of all of above expression gives

${\displaystyle I_{\text{n}}=h[{\frac {1}{2}}f_{\text{o}}+({\frac {1}{2}}f_{1}+{\frac {1}{2}}f_{1})+({\frac {1}{2}}f_{2}+{\frac {1}{2}}f_{2})+.....+({\frac {1}{2}}f_{\text{n-1}}+{\frac {1}{2}}f_{\text{n-1}})+{\frac {1}{2}}f_{\text{n}}]}$

${\displaystyle =h[{\frac {1}{2}}f_{\text{o}}+f_{1}+f_{2}+.......+f_{\text{n-1}}+{\frac {1}{2}}f_{\text{n}}]}$

Hence Proved..

Composite Simpson's Rule

From Simple Simpson's rule we obtain ,

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

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

Similarly

${\displaystyle I_{4}={\frac {h}{3}}[f_{2}+4f_{3}+f_{4}]}$

${\displaystyle I_{6}={\frac {h}{3}}[f_{4}+4f_{5}+f_{6}]}$ . . . . ${\displaystyle I_{\text{n-2}}={\frac {h}{3}}[f_{\text{n-4}}+4f_{\text{n-3}}+f_{\text{n-2}}]}$

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

After Summation of above terms we obtain

${\displaystyle I_{\text{n}}={\frac {h}{3}}[f_{\text{o}}+4f_{1}+(f_{2}+f_{2})+4f_{\text{3}}+(f_{4}+f_{4})+4f_{5}+f_{6}.......+(f_{\text{n-2}}+f_{\text{n-2}})+4f_{\text{n-1}}+f_{\text{n}}]}$

${\displaystyle I_{\text{n}}={\frac {h}{3}}[f_{\text{o}}+4f_{1}+2f_{2}+4f_{\text{3}}+2f_{4}+4f_{5}+f_{6}.......+2f_{\text{n-2}}+4f_{\text{n-1}}+f_{\text{n}}]}$

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

Hence Proved

Problem Statement

Prove that

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

where

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

Solution

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

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

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

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

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

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

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

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

Hence Proved ..

(12) Show Derivation

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

Ref: Lecture Notes p.15-1