User:Egm6341.s10.team3.heejun/Meeting 26 transcript

 '''MTG 26: Tue. 2 Mar 10'''


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

[[media: Egm6341.s10.mtg26.djvu | Page 26-1]] 
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Proof of Trapezoidal error: [[media: Egm6341.s10.mtg25.djvu | Page 25-3]] & [[media: Egm6341.s10.mtg21.djvu | Page 21-3]]


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 * $$ \displaystyle A := [ \underbrace{P_{3}(t)}_{\color{blue}_{- \frac{t^{3}}{6}+ \alpha t + \beta}} g^{(2)}(t) ]^{+1}_{-1} = 0 $$
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 * $$ \displaystyle \int P_{2} = \color{blue}{- \frac{t^{3}}{6}+ \alpha t + \beta = C_{1} \frac{t^{3}}{3!} + C_{3}t + C_{4}} $$
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Recall, $$ \displaystyle \color{blue}{C_{2} = 0}\,\,\,\, $$ [[media: Egm6341.s10.mtg21.djvu | Page 21-3]]

Want $$ \displaystyle (\alpha, \beta) $$ such that $$ \displaystyle P_{3} (t) = 0 $$ at $$ \displaystyle t \pm 1 $$

Also set $$ \displaystyle P_{3} = 0 \Rightarrow P_{3} $$ is odd, so we can reduce $$ C_{4},\,\, \color{blue}{C_{4}=0=\beta} $$




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 * $$ \displaystyle P_{3}(-1)= 0 = - \frac{(-1)^{3}}{6} + \alpha(-1) \Rightarrow \color{blue}{C_{3}=\alpha=\frac{1}{6}}$$
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 * $$ \displaystyle P_{3}(+1) \underbrace{=}_{\color{blue}{P_{3}\,\, odd}} - P_{3}(-1) =0 $$
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Summary: Steps 2ab:


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 * $$ \displaystyle P_{2} = C_{1} \frac{t^{2}}{2!} + C_{3} $$
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 * $$ \displaystyle P_{3} = C_{1} \frac{t^{3}}{3!} + C_{3}t $$
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 * $$ \displaystyle C_{1}=-1,\,\, C_{3}= \frac{1}{6} $$
 *   (3)  [[media: Egm6341.s10.mtg21.djvu | Page 21-3]]
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[[media: Egm6341.s10.mtg26.djvu | Page 26-2]] 

Steps 3a:


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 * $$ \displaystyle E = [P_{2}(t)g^{(1)}(t)]^{+1}_{-1} + \underbrace {\int_{-1}^{+1} \underbrace {P_{3}(t)}_{\color{red}{u'}} \underbrace {g^{(3)}(t)}_{\color{red}{v}}dt}_{\color{blue}{:=B}} $$
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&= [P_{4}(t)g^{(3)}(t)]^{+1}_{-1} -  \underbrace {\int^{+1}_{-1}P_{4}(t)g^{(4)}(t)dt}_{\color{blue}{\mathbb{C}}} \\\end{align} $$
 * $$ \displaystyle \begin{align} \color{blue}{B} &= [uv]^{+1}_{-1} - \int uv'  \\
 * $$ \displaystyle \begin{align} \color{blue}{B} &= [uv]^{+1}_{-1} - \int uv'  \\
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&= \color{blue}{C_{1} \frac{t^{4}}{4!} + C_{3} \frac{t^{2}}{2!} + \underbrace{C_{5}}_{\color{green}{\alpha}}} \\\end{align} $$
 * $$ \displaystyle \begin{align} P_{4}(t) &= \int P_{3}(t)dt = - \frac{t^{4}}{4!} + \frac{t^{2}}{12} + \alpha \\
 * $$ \displaystyle \begin{align} P_{4}(t) &= \int P_{3}(t)dt = - \frac{t^{4}}{4!} + \frac{t^{2}}{12} + \alpha \\
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Steps 3b:


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 * $$ \displaystyle \mathbb{C} = [P_{5}g^{(4)}]^{+1}_{-1} - \int^{+1}_{-1} P_{5}g^{(5)}dt $$
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 * $$ \displaystyle P_{5}(t) = \int P_{4}(t)dt = - \frac{t^{5}}{120} + \frac{t^{3}}{36} + \underbrace{\alpha}_{\color{blue}{C_{5}}}t + \underbrace{\beta}_{\color{blue}{C_{6}}}  $$
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Select $$ \displaystyle P_{5} $$ such that $$ \displaystyle \begin{align} & P_{5}(\pm1)=0 \\ & P_{5}(0) = 0 \\\end{align} $$

Recall $$ \displaystyle P_{3} $$ [[media: Egm6341.s10.mtg26.djvu | Page 26-1]]

[[media: Egm6341.s10.mtg26.djvu | Page 26-3]] 


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 * $$ \displaystyle C_{6} = 0,\,\, C_{5} = - \frac{7}{360} $$
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HW: $$\displaystyle P_5(t)=\int P_4(t)dt = -\frac{t^5}{120} + \frac{t^3}{36} + C_5t + C_6 $$ Selecting $$\displaystyle P_5(\pm 1)= 0 $$ and $$\displaystyle P_5(0)= 0 $$, find $$\displaystyle C_5 $$ and $$\displaystyle C_6 $$.
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Summary: Steps 3ab:


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 * $$ \displaystyle P_{4}(t) = - \frac{t^{4}}{24} + \frac{t^{2}}{12} - \frac{7}{360} $$
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 * $$ \displaystyle P_{5}(t) = - \frac{t^{5}}{120} + \frac{t^{3}}{36} - \frac{7}{360}t $$
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 * $$ \displaystyle E = [P_{2}g^{(1)} + P_{4}g^{(3)}]^{+1}_{-1} - \int^{+1}_{-1} P_{5}(t)g^{(5)}dt $$
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