Interpolation. Gaussian Quadrature. September 25, 2011

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1 Gussin Qudrture September 25, 2011

2 Approximtion of integrls Approximtion of integrls by qudrture Mny definite integrls cnnot be computed in closed form, nd must be pproximted numericlly. Bsic building block 1 u(ξ)w(ξ) dξ 1 ξ i re the qudrture points ω i re the qudrture weights w(ξ) > 0 is weight function N u(ξ i )ω i (1) Integrtion cn be done with function evlutions t ξ i How do I pick ξ i nd ω i to minimum errors nd/or mximize efficiency

3 Approximtion of integrls Exmple Qudrture: Trpezoidl integrtion w = 1 Qudrture points re eqully spced:ξ i = 2i N 1 Qudrture weights re ω i = { 2 2N 2 N for i = 0, N for 1 i < N (2) Formul tkes the form, with u i = u(ξ i ), 1 1 u(ξ) dξ 2 N [ u0 2 + u u N 1 + u ] N 2 (3) Cn we do better?

4 Approximtion of integrls Possible route Choose ξ i nd vry ω i to mximize order of exct integrtion. I cn impose (N + 1) constrints on ω i, e.g. mximize the degree of polynomils tht qudrture pproximtes exctly: 1 ξi k ω k = ξ k dξ, 0 k N (4) 1 Guss qudrture djust ξ i so tht formul bove pplies for k > (N + 1). Relies on properties of orthogonl polynomils.

5 Approximtion of integrls Guss qudrture Let p m (x) be the set of orthogonl polynomils on the intervl x b w.r.t. weight function w(x) nd of degree m p m (x)p n (x)w(x) dx = δ m,n p m 2 (5) Let the colloction points be the roots of p N+1 (x i ) = 0 nd the weights: xi k ω k = x k w(x) dx, 0 k N (6) then...

6 Approximtion of integrls Guss qudrture The weights re ll positive ω i > 0. Guss qudrture is exct for ll polynomils, q, of degree less or equl to 2N + 1 q(x i )ω k = q(x)w(x) dx (7) It is not possible to find x i, ω i combintion where the integrtion is exct for polynomils of degree 2N + 2.

7 Approximtion of integrls Guss qudrture flvors Guss qudrture comes in different flvors depending on whether the qudrture points include none, one, or both end points. Guss qudrture: ll qudrture points re strictly inside the intervl: < x i < b Guss-Lobtto qudrture: x 0 = nd x N = b. Useful for imposing BC s we will see lter. Guss-Rdu qudrture: x 0 = or x N = b but not both. Useful for integrls or PDEs on semi-infinite intervls

8 Approximtion of integrls Guss-Lobtto qudrture Formul is exct for polynomils of degree 2N 1. For Jcobi type polynomils, qudrture points re the roots of (1 x 2 )p N (x) = 0 Jcobi polynomils include Chebyshev, Legendre nd ssocited Legendre polynomils.

9 Exmples Exmple: Chebyshev Guss Qudrture Roots re known nlyticlly since T N+1 (x) = cos(n + 1)θ, with x = cos θ. The roots re then ξ i = cos θ i = Qudrture weights re (2i + 1)π 2(N + 1), 0 i N (8) ω i = π N + 1 The formul is exct for polynomil of degree 2N + 1: 1 1 (9) q(x) dx = N q(x i )ω i (10) 1 x 2

10 Exmples Exmple: Chebyshev Guss-Lobtto Qudrture Roots re known nlyticlly since T N N sin Nθ (x) = sin θ The roots of (1 x 2 )p N (x) = sin θ sin Nθ Qudrture weights re ω i = ξ i = iπ N, 0 i N (11) π 2N for i = 0, N, nd ω i = π N, 1 i < N (12) The formul is exct for polynomil of degree 2N 1: 1 1 q(x) dx = N q(x i )ω i (13) 1 x 2

11 Exmples Exmple: Legendre Guss Qudrture Roots must be computed numericlly L N+1 (x i ) = 0 Qudrture weights re ω i = 2 (1 x i ) 2 L N+1 (x i) (14) The formul is exct for polynomil of degree 2N + 1: 1 1 q(x) dx = N q(x i )ω i (15)

12 Exmples Exmple: Legendre Guss-Lobtto Qudrture Roots must be computed numericlly (1 x 2 )L N (x) = 0 Qudrture weights re ω i = 2 N(N + 1)[L N (ξ)] 2 (16) The formul is exct for polynomil of degree 2N 1: 1 1 q(x) dx = N q(x i )ω i (17)

13 Some Mth Proofs Some mthemticl proofs for Gussin Qudrture How do we know tht the orthogonl polynomil p N+1 hs N + 1 roots? How cn we prove tht formul is exct for polynomils of degree 2N + 1? Strting point is tht p m is the set of orthogonl polynomils: p m (x) p n (x) w(x) dx = δ m,n p m 2 (18)

14 Some Mth Proofs Multiplicity nd loction of roots Assume p N hs only k roots in [, b]: p N cn be written s p N = (x r 1 )(x r 2 )... (x r k )h(x) (19) where h k is polynomil of degree N k tht must be single-signed in [, b]. If it chnges sign, then there must be nother root r k+1 Consider the following polynomil: z(x) = (x r 1 )(x r 2 )... (x r k )p N (x) (20) z(x) = [(x r 1 )(x r 2 )... (x r k )] 2 h(x) (21) z(x) must lso be single-signed.

15 Some Mth Proofs Multiplicity nd loction of roots p N is orthogonl to ll polynomils of degree k < N then (x r 1 )(x r 2 )... (x r k ) p }{{} N (x)w(x) dx = δ N,k polynomil of degree k [(x r 1 )(x r 2 )... (x r k )] 2 h(x)w(x) dx = δ N,k Since z is single signed this is possible only if k = N, then p N hs N roots in [, b] Since p N is of degree N nd hs N roots, the roots must be isolted. Likewise p N (x) must hve N 1 roots corresponding to the extrem of p N (x).

16 Some Mth Proofs Exctness of qudrture for polynomils Let f (x) be polynomil of degree 2N + 1, it cn lwys be written s f (x) = p N+1 (x)q(x) + r(x) (22) where q(x), nd r(x) re polynomils of degree t most N. f (x)w(x) dx = = = p N+1 (x)q(x)w(x) dx + r(x)w(x) dx }{{}}{{} 0 by orthogonlity Exct qudrture N N p N+1 (x i ) q(x }{{} i )ω i + r(x i )ω i 0 N N [p N+1 (x i )q(x i ) + r(x i )] ω }{{} i = f (x i )ω i f (x i )

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