Mtg 8: Wed, 19 Jan 11
page8-1
Legendre poly. Pn(x):
P n ( x ) = ∑ i = 0 [ n / 2 ] ( − 1 ) i ( 2 n − 2 i ) ! x n − 2 i 2 n i ! ( n − i ) ! ( n − 2 i ) ! {\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 ] = i n t e g e r p a r t o f 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 ) P 0 = 1 ∈ P 0 {\displaystyle {\color {red}(3)}\ {P}_{0}=1\color {blue}\in {P}_{0}}
( 4 ) P 1 = x ∈ P 1 {\displaystyle {\color {red}(4)}\ {P}_{1}=x\color {blue}\in {P}_{1}}
( 5 ) P 2 = 1 2 ( 3 x 2 − 1 ) ∈ P 1 {\displaystyle {\color {red}(5)}\ {P}_{2}={\frac {1}{2}}(3{x}_{2}-1)\color {blue}\in {P}_{1}}
( 6 ) P 3 = 1 2 ( 5 x 3 − 3 x ) ∈ P 1 {\displaystyle {\color {red}(6)}\ {P}_{3}={\frac {1}{2}}(5{x}_{3}-3x)\color {blue}\in {P}_{1}}
( 7 ) P 4 = 35 8 x 4 − 15 4 x 2 + 3 8 ∈ 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}}
P n = s e t o f p o l y . o f d e g r e e ⩽ n {\displaystyle {\color {blue}{P}_{n}}\ =\ set\ of\ poly.\ of\ degree\ \leqslant \ n}
HW 2.6: Verify(3)-(7) using(1)-(2)
( 5 ) ⇒ P 2 = 0 ⇒ x 1 , 2 = − 1 3 o r + 1 3 {\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)
I ( f ) = I n ( f ) + E n ( f ) ( 1 ) {\displaystyle I(f)={I}_{n}(f)+{E}_{n}(f)\ {\color {red}(1)}}
w i = − 2 ( n + 1 ) P n ′ ( x i ) P n + 1 ( x i ) ( 2 ) {\displaystyle {w}_{i}={\frac {-2}{(n+1){P}_{n}^{'}({x}_{i}){P}_{n+1}({x}_{i})}}\ {\color {red}(2)}}
E n ( f ) = 2 2 n + 1 ( n ) ! 4 ( 2 n + 1 ) [ ( 2 n ) ! ] 2 f ( 2 n ) ( ξ ) ( 2 n ) ! ( 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]}
N O T E : _ E n ( f ) = 0 ∀ f ∈ P 2 n − 1 s i n c e f ( 2 n ) ( x ) = 0 . O n l y n e e d t o u s e n i n t . p t s { x i , i = 1 , . . . n } t o i n t . e x a c t l y a n y p o l y i n P 2 n − 1 i . e . , o f d e g r e e ≦ 2 n − 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}
N e w t o n − c o t e s m e t h o d : I ( f ) {\displaystyle {\color {blue}{Newton-cotes\ method:}}\ I(f)}
H i s t o r y N e w t o n c o t e s → L e c t u r e p l a n s u l i + M e y e r s ( 2003 ) {\displaystyle History\ Newton\ cotes\ \color {blue}\rightarrow \ Lecture\ plan\ suli+Meyers(2003)}
page8-3
1 ) A p p r o x . f ( . ) u s i n g L a g r a n g e _ i n t e r p . f u n c s ⇒ f n L ( x ) {\displaystyle {\color {blue}1)}\ Approx.\ f(.)\ using\ {\color {blue}{\underline {\color {black}Lagrange}}}\ interp.\ funcs\ \Rightarrow \ {f}_{n}^{\color {red}L}(x)}
I n t . e x a c t l y f n L ( x ) ⇒ I ( f n L ) ⏟ {\displaystyle Int.\ exactly\ {f}_{n}^{\color {red}L}(x)\ \Rightarrow \ \color {blue}\underbrace {\color {black}I({f}_{n}^{\color {red}L})} }
I n ( f ) := I ( f n L ) = ∫ f n L ( x ) d x ( 1 ) {\displaystyle I_{n}(f)\ :=I({f}_{n}^{\color {red}L})=\int {f}_{n}^{\color {red}L}(x)dx\ \color {red}(1)}
f n L ( x ) = P n ( x ) = ∑ i = 0 ∞ l i , n ( x ) ⏟ L a g r a n g e i n t e r p . f u n c f ( x i ) ( 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
![{\displaystyle {E}_{n}(f)={\frac {{2}^{2n+1}{(n)!}^{4}}{(2n+1){[(2n)!]}^{2}}}{\frac {f^{\color {red}(2n)}(\xi )}{(2n)!}}\ {\color {red}(3)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf2e6fc29a6745b20860730c730c945fd402ac7f)
![{\displaystyle \xi \in [-1,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df3cb69ecb0c91502f24f927c941372006d26211)
![{\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)!}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bea1b6b3cf6ecd899da0cb7507c35e08f0ff112)
![{\displaystyle {\color {red}[{\color {black}n/2}]}=integer\ part\ of\ n/2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa62578f6794ad7c7ae9b2218577ed03ce2df12d)
