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 [[media:Egm6341.s10.mtg9.pdf|p.9-3]]

Problem Statement
Show Simple Trapazoidal rule[[media:Egm6341.s10.mtg7.pdf|p.7-1]] $$ \Rightarrow $$ Composite Trapazoidal rule [[media:Egm6341.s10.mtg7.pdf|p.7-1]]

and

Show Simple Simpson's rule [[media:Egm6341.s10.mtg7.pdf|p.7-2]] $$ \Rightarrow $$ Composite Simpson's rule[[media:Egm6341.s10.mtg7.pdf|p.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 ,

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

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

Similarly

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

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

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

After Summation of above terms we obtain

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

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

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 [[media:Egm6341.s10.mtg15.pdf|p.15-2]]

Problem Statement
Prove that

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

where

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

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

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

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

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

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

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

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

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

Hence Proved ..

(12) Show Derivation

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

Ref: Lecture Notes [[media:Egm6341.s10.mtg15.pdf|p.15-1]]

Problem Statement
Show derivation that
 * {| style="width:100%" border="0" align="left"

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

Solution
From (4) in [[media:Egm6341.s10.mtg14.pdf|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 ,