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Pn(x)=∑i=0[n/2](−1)i(2n−2i)!xn−2i2ni!(n−i)!(n−2i)!
(1)
[n/2]=integer part of n/2
(2)
e.g., m = 5 , n/2 = 2.5 , [2.5] = 2
(3) P0=1∈P0
(4) P1=x∈P1
(5) P2=12(3x2−1)∈P1
(6) P3=12(5x3−3x)∈P1
(7) P4=358x4−154x2+38∈P1
Pn = set of poly. of degree ⩽ n
(5)⇒ P2=0⇒x1,2=−13 or+13
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Weights wi , i = 1,...,n (Template:FontTemplate:Font)
I(f)=In(f)+En(f) (1)
wi=−2(n+1)Pn'(xi)Pn+1(xi) (2)
En(f)=22n+1(n)!4(2n+1)[(2n)!]2f(2n)(ξ)(2n)! (3)
ξ∈[−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
Newton−cotes method: I(f)
History Newton cotes → Lecture plan suli+Meyers(2003)
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1) Approx. f(.) using Lagrange_ interp. funcs ⇒ fnL(x)
Int. exactly fnL(x) ⇒ I(fnL)⏟
In(f) :=I(fnL)=∫fnL(x)dx (1)
fnL(x)=Pn(x)=∑i=0∞li,n(x)⏟Lagrange interp. func f(xi) (2)
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File:Nm1.s11.Mtg8.pg3.fig1.svg
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