University of Florida/Egm6321/f09.team1.gzc/Mtg8

Mtg 8: Wed, 19 Jan 11

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Legendre poly. P_n(x):

${\displaystyle {P}_{n}(x)=\sum _{i=0}^{\color {red}[{\color {black}n/2}]}{(-1)}^{i}{\frac {(2n-2i)!{x}^{n-2i}}{{2}^{n}i!(n-i)!(n-2i)!}}}$

(1)

${\displaystyle {\color {red}[{\color {black}n/2}]}=integer\ part\ of\ n/2}$

(2)

e.g., m = 5 , n/2 = 2.5 , [2.5] = 2 ${\displaystyle {\color {red}(3)}\ {P}_{0}=1\color {blue}\in {P}_{0}}$ ${\displaystyle {\color {red}(4)}\ {P}_{1}=x\color {blue}\in {P}_{1}}$ ${\displaystyle {\color {red}(5)}\ {P}_{2}={\frac {1}{2}}(3{x}_{2}-1)\color {blue}\in {P}_{1}}$ ${\displaystyle {\color {red}(6)}\ {P}_{3}={\frac {1}{2}}(5{x}_{3}-3x)\color {blue}\in {P}_{1}}$ ${\displaystyle {\color {red}(7)}\ {P}_{4}={\frac {35}{8}}{x}^{4}-{\frac {15}{4}}{x}^{2}+{\frac {3}{8}}\color {blue}\in {P}_{1}}$ ${\displaystyle {\color {blue}{P}_{n}}\ =\ set\ of\ poly.\ of\ degree\ \leqslant \ n}$ HW 2.6: Verify(3)-(7) using(1)-(2) ${\displaystyle {\color {red}(5)}\Rightarrow \ {P}^{2}=0\Rightarrow {x}_{1,2}={\color {red}-}{\frac {1}{\sqrt {3}}}\ or{\color {red}+}{\frac {1}{\sqrt {3}}}}$ page8-2 Weights wi , i = 1,...,n ((1)  p.7-5) Thm: ${\displaystyle I(f)={I}_{n}(f)+{E}_{n}(f)\ {\color {red}(1)}}$ ${\displaystyle {w}_{i}={\frac {-2}{(n+1){P}_{n}^{'}({x}_{i}){P}_{n+1}({x}_{i})}}\ {\color {red}(2)}}$ ${\displaystyle {E}_{n}(f)={\frac {{2}^{2n+1}{(n)!}^{4}}{(2n+1){[(2n)!]}^{2}}}{\frac {f^{\color {red}(2n)}(\xi )}{(2n)!}}\ {\color {red}(3)}}$ ${\displaystyle \xi \in [-1,1]}$

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

${\displaystyle {\color {blue}{Newton-cotes\ method:}}\ I(f)}$

${\displaystyle History\ Newton\ cotes\ \color {blue}\rightarrow \ Lecture\ plan\ suli+Meyers(2003)}$

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

${\displaystyle Int.\ exactly\ {f}_{n}^{\color {red}L}(x)\ \Rightarrow \ \color {blue}\underbrace {\color {black}I({f}_{n}^{\color {red}L})} }$

${\displaystyle I_{n}(f)\$ :=I({f}_{n}^{\color {red}L})=\int {f}_{n}^{\color {red}L}(x)dx\ \color {red}(1)}

${\displaystyle {f}_{n}^{\color {red}L}(x)=P_{n}(x)=\sum _{i=0}^{\infty }{\color {blue}{\underset {Lagrange\ interp.\ func}{\color {black}\underbrace {l_{i,n}(x)} }}}\ f(x_{i})\ \color {red}(2)}$

NOTE:  Demonstrated Wolfram Alpha(WA) e.g., (debt usa)/(gdp usa) integrate x from 0 to 1 link WA comp. results in HW reportsAvoid plagiarism