User:EGM6341.s11.TEAM1.Yoon.HW4

=1= Given:
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 * $$\displaystyle P_3(x) = 3+ 8x -2x^2 + 6x^3 $$
 * $$\displaystyle (Eq. 1) $$
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EXACT INTEGRAL Integrating Eq (1),


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 * $$\displaystyle I = \int_{a}^{b} \Bigg(3+ 8x -2x^2 + 6x^3\Bigg)\,dx $$
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$$\displaystyle = \int_{0}^{1} \Big(3+ 8x -2x^2 + 6x^3\Big)\,dx $$
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$$\displaystyle =\Big[ 3x + 4x^2 - \frac{2x^3}{3} + \frac{3x^4}{2} \Big]_0^1 $$
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$$\displaystyle = \Big[ 3 +4 - \frac{2}{3} + \frac{3}{2}\Big]$$
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$$\displaystyle = \frac{47}{6} $$
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 * $$\displaystyle \therefore I = 7.8333$$
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SIMPSON'S RULE We know from lecture slide Eq(2) [[media:Egm6341.s10.mtg7.pdf|p.7-2]] that the Simple Simpson's Rule is given by,


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 * $$\displaystyle I_2 = \frac{h}{3}\bigg[ f(x_0)+ 4f(x_1)+ f(x_2)\bigg]$$
 * $$\displaystyle (Eq. 2) $$
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where,


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 * $$\displaystyle x_0\, = a\,=0 $$
 * $$\displaystyle (Eq. 3) $$
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 * $$\displaystyle x_2\, = b\,=1 $$
 * $$\displaystyle (Eq. 4) $$
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 * $$\displaystyle x_1\, = \frac{a+b}{2}\,= \frac{1}{2} $$
 * $$\displaystyle (Eq. 5) $$
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 * $$\displaystyle h\, = \frac{b-a}{2}\,= \frac{1}{2} $$
 * $$\displaystyle (Eq. 6) $$
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 * $$\displaystyle f(x_0)\, = \Big[ 3 + 4(0) - 2\cdot(0)^2 + 6\cdot(0)^3 \Big]\,= 3 $$
 * $$\displaystyle (Eq. 7) $$
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 * $$\displaystyle f(x_1)\, = \Big[ 3 + 4(0.5) - 2\cdot(0.5)^2 + 6\cdot(0.5)^3 \Big]\,= 7.25 $$
 * $$\displaystyle (Eq. 8) $$
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 * $$\displaystyle f(x_2)\, = \Big[ 3 + 4(1) - 2\cdot(1)^2 + 6\cdot(1)^3 \Big]\,= 15 $$
 * $$\displaystyle (Eq. 9) $$
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 * $$\displaystyle \therefore\, Substituting\, Eq (6)\, through\, Eq (9)\, in\, Eq(2)\, we\, have\,$$
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 * $$\displaystyle I_2 = \frac{1}{6}\bigg[ 3+ 29 + 15 \bigg] =\, \frac{47}{6} $$
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$$\displaystyle \therefore I_2 = 7.8333$$
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 * $$\displaystyle Since\, I = I_2\, it\, is\, very\, clear\, that\, the\, Simpson\, rule\, integrates\, a\, 3^{rd}\, order\, polynomial\, exactly $$
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=7= $$ \begin{align} \displaystyle E_{n}^{2} &:= I - I_{n} \\ &=\int_{a}^{b}f(x)dx - \frac{h}{3}[f(x_{0})+4f(x_{1})+2f(x_{2})+...+2f(x_{n-2})+4f(x_{n-1})+f(x_{n})] \\ &=\sum_{i=1}^{n} [ \int_{x_{i-1}}^{x_{i}}f(x)dx - \frac{h}{3}[f(x_{i-1}) + 4f(\frac{(x_{i-1}+x_{i})}{2}) + f(x_{i})] \\\end{align}\ $$

$$ \left| E_{n}^{2} \right| \leq \left| \frac{h^{5}}{90} \underbrace {\sum_{n}^{i=1}max\left(F^{(4)}(\xi )\right)}_{Call \bar{M}_{4}} \right| $$

$$ \displaystyle \xi \in [x_{i-1},x_{i}] $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg14.pdf|p.14-1]]

$$ \Rightarrow M_{4} := max \left| f^{(4)}(\xi)\right| $$

$$ \displaystyle \xi \in [a,b] $$

$$ \bar{M}_{4} \leq n \times M_{4} $$

$$ \Rightarrow \left| E_{n}^{2} \right| \leq \left| \frac{(b-a)^{5}}{2880n^{5}}nM_{4} \right| = \left| \frac{(b-a)(b-a)^{4}}{2880n^{4}} \right| = \frac{(b-a)h^{4}}{2880}M_{4} $$

where $$ \displaystyle h= \frac{(b-a)}{n} $$

Therefore,
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$$\displaystyle \displaystyle \left| {E}_{n}^{2} \right| \leq \frac{(b-a)^{5}}{2880 n^{4}} M_{4} = \frac{(b-a)h^{4}}{2880} M_{4} $$
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