Mtg 8: Wed, 19 Jan 11
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Legendre poly. Pn(x):
Pn(x)=∑i=0[n/2](−1)i(2n−2i)!xn−2i2ni!(n−i)!(n−2i)!{\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)
[n/2]=integer part of n/2{\displaystyle {\color {red}[{\color {black}n/2}]}=integer\ part\ of\ n/2}
(2)
e.g., m = 5 , n/2 = 2.5 , [2.5] = 2
(3) P0=1∈P0{\displaystyle {\color {red}(3)}\ {P}_{0}=1\color {blue}\in {P}_{0}}
(4) P1=x∈P1{\displaystyle {\color {red}(4)}\ {P}_{1}=x\color {blue}\in {P}_{1}}
(5) P2=12(3x2−1)∈P1{\displaystyle {\color {red}(5)}\ {P}_{2}={\frac {1}{2}}(3{x}_{2}-1)\color {blue}\in {P}_{1}}
(6) P3=12(5x3−3x)∈P1{\displaystyle {\color {red}(6)}\ {P}_{3}={\frac {1}{2}}(5{x}_{3}-3x)\color {blue}\in {P}_{1}}
(7) P4=358x4−154x2+38∈P1{\displaystyle {\color {red}(7)}\ {P}_{4}={\frac {35}{8}}{x}^{4}-{\frac {15}{4}}{x}^{2}+{\frac {3}{8}}\color {blue}\in {P}_{1}}
Pn = set of poly. of degree ⩽ n{\displaystyle {\color {blue}{P}_{n}}\ =\ set\ of\ poly.\ of\ degree\ \leqslant \ n}
HW 2.6: Verify(3)-(7) using(1)-(2)
(5)⇒ P2=0⇒x1,2=−13 or+13{\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}}}}
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Weights wi , i = 1,...,n ((1) p.7-5)
I(f)=In(f)+En(f) (1){\displaystyle I(f)={I}_{n}(f)+{E}_{n}(f)\ {\color {red}(1)}}
wi=−2(n+1)Pn′(xi)Pn+1(xi) (2){\displaystyle {w}_{i}={\frac {-2}{(n+1){P}_{n}^{'}({x}_{i}){P}_{n+1}({x}_{i})}}\ {\color {red}(2)}}
En(f)=22n+1(n)!4(2n+1)[(2n)!]2f(2n)(ξ)(2n)! (3){\displaystyle {E}_{n}(f)={\frac {{2}^{2n+1}{(n)!}^{4}}{(2n+1){[(2n)!]}^{2}}}{\frac {f^{\color {red}(2n)}(\xi )}{(2n)!}}\ {\color {red}(3)}}
ξ∈[−1,1]{\displaystyle \xi \in [-1,1]}
NOTE:_ En(f)=0 ∀f∈P2n−1 since f(2n)(x)=0 . Only need to use n int. pts {xi, i=1,...n} to int. exactly any poly in P2n−1 i.e., of degree ≦ 2n−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}
Newton−cotes method: I(f){\displaystyle {\color {blue}{Newton-cotes\ method:}}\ I(f)}
History Newton cotes → Lecture plan suli+Meyers(2003){\displaystyle History\ Newton\ cotes\ \color {blue}\rightarrow \ Lecture\ plan\ suli+Meyers(2003)}
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1) Approx. f(.) using Lagrange_ interp. funcs ⇒ fnL(x){\displaystyle {\color {blue}1)}\ Approx.\ f(.)\ using\ {\color {blue}{\underline {\color {black}Lagrange}}}\ interp.\ funcs\ \Rightarrow \ {f}_{n}^{\color {red}L}(x)}
Int. exactly fnL(x) ⇒ I(fnL)⏟{\displaystyle Int.\ exactly\ {f}_{n}^{\color {red}L}(x)\ \Rightarrow \ \color {blue}\underbrace {\color {black}I({f}_{n}^{\color {red}L})} }
In(f) :=I(fnL)=∫fnL(x)dx (1){\displaystyle I_{n}(f)\ :=I({f}_{n}^{\color {red}L})=\int {f}_{n}^{\color {red}L}(x)dx\ \color {red}(1)}
fnL(x)=Pn(x)=∑i=0∞li,n(x)⏟Lagrange interp. func f(xi) (2){\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