User:EGM6341.s11.team1.Chiu/Mtg18

Mtg 18: Fri, 11 Feb 11

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More accurate error estimate of Simpson's rule (Simple)


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$$ E_{2}= \int^{ \color{blue}b}_{ \color{blue}a} \left( f-f^{ \color{red}L}_{2} \right)dx = \underbrace{ \alpha_{1}}_{ \color{red}>0} + \underbrace{ \alpha_{2}}_{ \color{blue}<0} \ $$
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=> Cancellation of areas

=> making $$ \color{blue} E_{2} \ $$ even smaller than [[media: Nm1.s11.mtg17.djvu| (1) page17-3]]

=>Simpson integration exactly $$ \color{blue} \forall p \in \mathbb{P}_{3} \ $$ (of degree $$ \leq \ $$ 3)

confer A. p.256 End More accurate error estimate of Simpson's rule

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HW*4.1:


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$$ [a,b]=[0,1]{ \color{blue}s10}=[-2,1]{ \color{blue}s11} \ $$
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$$ f \left( x \right)=P_{3} \left( x \right)=c_{0}+c_{1}x^{1}+c_{2}x^{2}+c_{3}x^{3}= \sum^{3}_{i=0}c_{i}x^{i} \qquad {\color{red}cubic} \ $$
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$$ c_{0}=1, c_{1}=3, c_{2}=-9, c_{3}=12 \ $$
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Find $$ \underset{\color{blue} \underset{exact}{\uparrow}}I, \underset{\color{blue} \underset{Simple \ Simpson}{\uparrow}}{I_{2}} \ $$
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Theorem: Simple Simpson's Rule (SSET)
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$$ E_{2}= \frac{- \left(b-a \right)^{5}}{2880}f^{(4)} \left( \xi \right), \xi \in [a,b] \ $$
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$$ =-\frac{h^{5}}{90}f^{(4)} \left( \xi \right), h:= \frac{b-a}{2} \ $$
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Reference:

1) A. p.257, proof based on divided differential.

2) S&M (Suli & Meyers) p.205, proof similar to that of LIET.

End Reference

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Remember:

If $$ f \in \mathbb{P}_{3} \ $$ (polynomial of degree $$ \leq \ $$ 3)


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$$  \displaystyle \Rightarrow f^{(4)} \equiv 0 \Rightarrow E_{2}=0 $$     (1)
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Simple Simpson integration exactly polynomial in $$ \mathbb{P}_{3} \ $$

Proof: (Simple technique in LIET proof) End Remember

Shift origin of $$ x \ $$ to $$ x_{1} \ $$: Transfer of variant


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$$  \displaystyle x(t)=x_{1}+ht,\ t \in [a,b] $$     (2)
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$$  \displaystyle h:= \frac{(b-a)}{2} $$
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$$  \displaystyle \left.\begin{matrix} t=0, \ x(0)=x_{1} \\ t=+1, \ x(1)=x_{1}+h=x_{2} \\ t=-1, \ x(-1)=x_{0} \end{matrix}\right\} $$     (3)
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$$  \displaystyle E_{2}= I-I_{2}= \underbrace{\frac{b-a}{2}}_{ \color{blue} h}e(1) \underset{ \color{red}\underset{(4)}{\uparrow}}= {\color{blue}h}e(1) $$
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$$  \displaystyle e(t): \underset{ \color{red}\underset{(5)}{\uparrow}}= \int^{+t}_{-t} \underbrace{f(x(t))}_{ \color{blue} F(t)}dt- \underbrace{\frac{t}{3}[F(-t)+4F(0)+F(t)]}_{ {\color{red} (2)} \ { \color{blue} p.7-4}} $$
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[[media: Nm1.s11.mtg18.djvu| page18-4]]


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$$  \displaystyle { \color{red} (2)} \ { \color{blue} p.18-3:} \ dx=h \ dt $$ (1)
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$$  \displaystyle \Rightarrow E_{2}=h \ e(1) $$     (2)
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