University of Florida/Egm6341/s11.team1.Gong/Mtg8

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page8-1

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$P_n(x)=\sum _{i=0}^{[n/2]}{(-1)}^i\frac{(2n-2i)!x^{n-2i}}{2^{n}i!(n-i)!(n-2i)!}$

(1)

$[n/2]=integerpartofn/2$

(2)

e.g., m = 5 , n/2 = 2.5 , [2.5] = 2 $(3)P_0=1\in P_0$ $(4)P_1=x\in P_1$ $(5)P_2=\frac{1}{2}(3x_2-1)\in P_1$ $(6)P_3=\frac{1}{2}(5x_3-3x)\in P_1$ $(7)P_4=\frac{35}{8}{x}^4-\frac{15}{4}{x}^2+\frac{3}{8}\in P_1$ $P_n=setofpoly.ofdegree⩽n$ Template:FontTemplate:FontTemplate:Font $(5)\Rightarrow P^2=0\Rightarrow x_{1,2}=-\frac{1}{\sqrt{3}}or+\frac{1}{\sqrt{3}}$ page8-2 Weights wi , i = 1,...,n (Template:FontTemplate:Font) Template:Font $I(f)=I_n(f)+E_n(f)(1)$ $w_i=\frac{-2}{(n+1)P_n^{\text{'}}(x_i)P_{n+1}(x_i)}(2)$ $E_n(f)=\frac{2^{2n+1}{(n)!}^4}{(2n+1){[(2n)!]}^2}\frac{f^{(2n)}(\xi )}{(2n)!}(3)$ $\xi \in [-1,1]$ $\underset{\_}{NOTE:}{E}_n(f)=0\mathrm{\forall }f\in P_{2n-1}since{f}^{(2n)}(x)=0.Onlyneedtousenint.pts\{x_i,i=1,...n\}toint.exactlyanypolyin{P}_{2n-1}i.e.,ofdegree\leqq 2n-1$ $Newton-cotesmethod:I(f)$ $HistoryNewtoncotes\to Lectureplansuli+Meyers(2003)$ page8-3 $1)Approx.f(.)using\underset{\_}{Lagrange}interp.funcs\Rightarrow f_n^L(x)$ $Int.exactly{f}_n^L(x)\Rightarrow \underbrace{I(f_n^L)}$ $I_n(f):=I(f_n^L)=\int f_n^L(x)dx(1)$ $f_n^L(x)=P_n(x)=\sum _{i=0}^{\mathrm{\infty }}\underset{Lagrangeinterp.func}{\underbrace{l_{i,n}(x)}}f(x_i)(2)$ File:Nm1.s11.Mtg8.pg3.fig1.svg