View source for University of Florida/Egm6341/s11.team1.Gong/Mtg8

<b>{{font|size=large|Mtg 8: Wed, 19 Jan 11}}</b>


<b>[http://upload.wikimedia.org/wikiversity/en/9/9b/Nm1.s11.mtg8.djvu page8-1]</b>

{{font|color=blue|<U>Legendre poly. P</U>}}<SUB>{{font|color=blue|n}}</SUB>{{font|color=blue|<U>(x):</U>}}


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<math> {P}_{n}(x)= \sum_{i=0}^{{\color{red}[{\color{black}n/2}]} } {(-1)}^{i} \frac{(2n-2i)! {x}^{n-2i}}{ {2}^{n}i!(n-i)!(n-2i)!}</math>

|| <p style="text-align:right"><span style="color:#FF0000">
      (1)</span></p>
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<math> {\color{red}[ {\color{black} n/2}]} = integer \ part \ of \ n/2</math>

|| <p style="text-align:right"><span style="color:#FF0000">
      (2)</span></p>
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e.g., m = 5 , n/2 = 2.5 , [2.5] = 2

<math> {\color{red}(3)} \ {P}_{0}=1  \color{blue} \in {P}_{0} </math>

<math> {\color{red}(4)} \ {P}_{1}=x  \color{blue} \in {P}_{1} </math>

<math> {\color{red}(5)} \ {P}_{2}=  \frac{1}{2}(3{x}_{2}-1)  \color{blue} \in {P}_{1} </math>

<math> {\color{red}(6)} \ {P}_{3}=  \frac{1}{2}(5{x}_{3}-3x)  \color{blue} \in {P}_{1} </math>  

<math> {\color{red}(7)} \ {P}_{4}=  \frac{35}{8}{x}^{4}-\frac{15}{4}{x}^{2}+\frac{3}{8}  \color{blue} \in {P}_{1} </math>

<math>{\color{blue}{P}_{n}} \ = \ set \ of \ poly. \ of \ degree \ \leqslant \ n</math>


{{font|color=blue|<U>HW 2.6:</U>&nbsp; }}{{font|color=green|{{font|color=black|Verify }}{{font|color=red|(3)-(7)}} {{font|color=black|using}} }}{{font|color=red|(1)-(2)}}

<math>{\color{red}(5)}  \Rightarrow \ {P}^{2}=0 \Rightarrow {x}_{1,2} = {\color{red}-} \frac{1}{\sqrt{3}} \ or {\color{red}+} \frac{1}{\sqrt{3}}</math>


<b>[http://upload.wikimedia.org/wikiversity/en/1/10/Nm1.s11.mtg6.djvu page8-2]</b>

Weights w<SUB>i</SUB> , i = 1,...,n ({{font|color=red|(1)}}{{font|color=blue|&nbsp; p.7-5}})

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| <U>{{font|color=blue|Thm:}}</U>
<math>I(f)= {I}_{n}(f)+{E}_{n}(f) \ {\color{red}(1)}</math>

<math>{w}_{i}= \frac{-2}{(n+1) {P}^{'}_{n}({x}_{i}){P}_{n+1}({x}_{i})} \ {\color{red}(2)}</math>

<math>{E}_{n}(f)= \frac{{2}^{2n+1}{(n)!}^{4}}{(2n+1){[(2n)!]}^{2}} \frac{f^{{\color{red}(2n)}}(\xi)}{(2n)!} \ {\color{red}(3)}</math> 

<math>\xi\in[-1,1]</math>

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<math> \underline { \color{blue} {NOTE:}} \ {E}_{n}(f)=0 \  \forall f \in {P}_{2n-1} \ since \ f^{\color{red}(2n)}(x)=0 \ . \  Only \ need \ to \ use \ n \ int. \ pts\  \left\{x_{i}, \ i=1,...n\right\} \ to \ int. \ exactly \ any \ poly \ in \ P_{2n-1} \ i.e., \ of \ degree \  \leqq \ 2n-1</math>

<math>{\color{blue} {Newton-cotes \ method:}} \ I(f)</math>

<math>History \ Newton \ cotes \  \color{blue}  \rightarrow \ Lecture \ plan \ suli+Meyers(2003)</math>



<b>[http://upload.wikimedia.org/wikiversity/en/1/10/Nm1.s11.mtg6.djvu page8-3]</b>

<math>{\color{blue}1)} \ Approx. \ f(.) \ using \  {\color{blue} { \underline { {\color{black} Lagrange}}}} \ interp. \  
funcs \  \Rightarrow \  {f}^{ {\color{red} L}}_{n}(x) </math>


<math> Int. \ exactly \ {f}^{ {\color{red} L}}_{n}(x) \ \Rightarrow \  \color{blue}  \underbrace{ \color{black} I( {f}^{{\color{red}L}}_{n})}  </math>

<math>I_{n}(f) \ := I( {f}^{{\color{red}L}}_{n})=\int {f}^{{\color{red}L}}_{n}(x)dx \ \color{red} (1) </math>

<math>{f}^{{\color{red}L}}_{n}(x)=P_{n}(x)= \sum_{i=0}^{\infty} {\color{blue}  \underset{Lagrange \ interp. \ func}{  {\color{black} \underbrace{l_{i,n}(x)}}}} \ f(x_{i}) \ \color{red} (2)</math>


<P>{{font|color=blue|<U>NOTE:</U>&nbsp; {{font|color=black|Demonstrated {{font|color=blue|Wolfram Alpha(WA)}} e.g., (debt usa)/(gdp usa) integrate x from 0 to 1 link {{font|color=blue|WA}} comp. results in HW reports }}}}{{font|color=red|Avoid plagiarism}}</P>
[[File:nm1.s11.Mtg8.pg3.fig1.svg|400px]]
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