# 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 ,