User:Egm6341.s10.team3.heejun/Meeting 20 transcript

 '''MTG 20: Tue. 16 Feb 10'''


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Note:  Typeset of Transparencies, not lecture transcript  Heejun Chung 20:59, 11 August 2010 (UTC)

[[media: Egm6341.s10.mtg20.djvu | Page 20-1]] 
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Continue from [[media: Egm6341.s10.mtg19.djvu | Page 19-3]]


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 * $$ \displaystyle CT_{1}(n) := \underbrace{CT_{0}(n)}_{\color{blue}_{T_{0}(n)}} + a_{1}h^{2} $$
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 * $$ \displaystyle I = CT_{1}(n) + O(h^{4}) $$
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 * $$ \displaystyle CT_{2}(n) := CT_{1}(n) + a_{2} h^{\cancelto{\color{blue}{=\,4}}{2 \cdot 2}} $$
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 * $$ \displaystyle I = CT_{2}(n) + O( h^{\cancelto{\color{blue}{=\,6}}{2 \cdot 3}} ) $$
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HW: Develop higher-order corrected Trapezoidal rule: $$ \displaystyle CT_{k}(n) = CT_{k-1}(n) + a_{k}h^{2 \cdot k} $$ Refer to HWs on [[media: Egm6341.s10.mtg6.djvu | Page 6-5]] and [[media: Egm6341.s10.mtg19.djvu | Page 19-2]] Find $$ \displaystyle I_{n} $$ using $$ \displaystyle CT_{1}, CT_{2}, CT_{3} $$ for n=2, 4, 8, 16,... until $$ \displaystyle I-I_{n} = O(10^{-16}) $$
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[[media: Egm6341.s10.mtg20.djvu | Page 20-2]] 

Another application of theorem [[media: Egm6341.s10.mtg19.djvu | Page 19-3]]:

Periodic functions (HW [[media: Egm6341.s10.mtg18.djvu | Page 18-1]])


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 * $$ \displaystyle f\,\, periodic\,\, on\,\, [a,b]\,\, \Leftrightarrow\,\, f_{(k)}(a) = f_{(k)}(b),\,\,\,\, k=0,1,2,... $$
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 * $$ \displaystyle Clearly\,\, \color{red}{(1)}\,\, \color{black}{valid\,\, for\,\, k=2i-1,\,\,\,\, i.e.,\,\, odd\,\, number} $$
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Theorem [[media: Egm6341.s10.mtg19.djvu | Page 19-3]]: $$ \displaystyle a_{i}=0\,\, \forall\,\, i $$

Theorem [[media: Egm6341.s10.mtg18.djvu | Page 18-2]] and [[media: Egm6341.s10.mtg18.djvu | Page 18-3]]: $$ \displaystyle E_{n}^{1}=0 $$


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HW: The error in the higher order trapezoidal rule is given by, $$ \displaystyle E_n^{1} = \sum_{i=1}^{\infty}a_i \cdot h^{2i} $$ Demonstrate that for which n, the $$ \displaystyle E_{n}^{1}=0 $$ would be achieved?
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HW: Discuss Pros (primary positive aspects) and Cons (primary negative aspects): (1) Taylor series (2) Composite Trapezoidal rule (3) Composite Simpson rule (4) Romberg (Richardson) (5) $$ \displaystyle CT_{k}(n) $$
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[[media: Egm6341.s10.mtg20.djvu | Page 20-3]] 

Proof of Theorem [[media: Egm6341.s10.mtg18.djvu | Page 18-2]] and [[media: Egm6341.s10.mtg19.djvu | Page 19-3]]:

$$ \displaystyle E_{n}^{1} = \underbrace{\int_{a}^{b}f(x)dx}_{\color{blue}_{I}} - \underbrace{T_{0}(n)}_{\color{blue}_{I_{n}}} $$