lecture 19

lecture 19 - CE 30125 - Lecture 19 LECTURE 19 GAUSS...

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CE 30125 - Lecture 19 p. 19.1 LECTURE 19 GAUSS QUADRATURE • In general for Newton-Cotes (equispaced interpolation points/ data points/ integration points/ nodes). 84 • Note that for Newton-Cotes formulae only the weighting coefficients were unknown and the were fixed fx  x d x S x E hw ' o f o w ' 1 f 1 w ' N f N ++ +  E + = f 0 f 1 f 2 f N h = x 0 x s x 1 x 2 N x E closed formula w i x i
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CE 30125 - Lecture 19 p. 19.2 • However the number of and placement of the integration points influences the accuracy of the Newton-Cotes formulae: even degree interpolation function exactly integrates an degree poly- nomial This is due to the placement of one of the data points. odd degree interpolation function exactly integrates an degree polyno- mial. • Concept: Let’s allow the placement of the integration points to vary such that we further increase the degree of the polynomial we can integrate exactly for a given number of integration points. • In fact we can integrate an degree polynomial exactly with only integra- tion points N N th N 1 + N N N 2 N 1 + N 1 +
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CE 30125 - Lecture 19 p. 19.3 • Assume that for Gauss Quadrature the form of the integration rule is 85 •In deriving (not applying) these integration formulae • Location of the integration points, are unknown • Integration formulae weights, are unknown unknowns we will be able to exactly integrate any degree polyno- mial! fx  x d x S x E w o f o w 1 f 1 w N f N ++ +  E + = f 0 f 1 f 2 f N x 0 x s x 1 x 2 x N x E f 3 x 3 x i iO N = w i N = 2 N 1 + 2 N 1 +
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CE 30125 - Lecture 19 p. 19.4 Derivation of Gauss Quadrature by Integrating Exact Polynomials and Matching Derive 1 point Gauss-Quadrature • 2 unknowns , which will exactly integrate any linear function •L e t th e general polynomial be where the coefficients can equal any value • Also consider the integration interval to be such that and (no loss in generality since we can always transform coordinates). • Substituting in the form of w o x o fx  Ax B + = AB 1+ 1  x S 1 = x E + 1 = x d 1 + 1 w o o = B + x d 1 + 1 w o o B + =
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CE 30125 - Lecture 19 p. 19.5 • In order for this to be true for any 1st degree polynomial (i.e. any and ). • Therefore , for 1 point Gauss Quadrature. 86 • We can integrate exactly with only 1 point for a linear function while for Newton-Cotes we needed two points! A x 2 2 ---- Bx + + 1 1 w o Ax o B +  = A 0 B 2 + o w o Bw o + = A B 0 x o w o = 2 w o = x o 0 = w o 2 = N 1 = f 0 f(x) x 0 -1 +1
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CE 30125 - Lecture 19 p. 19.6 Derive a 2 point Gauss Quadrature Formula 87 • The general form of the integration formula is , , , are all unknowns • 4 unknowns we can fit a 3rd degree polynomial exactly • Substituting in for into the general form of the integration rule x 0 x 1 -1 +1 Iw o f o w 1 f 1 + = w o x o w 1 x 1 fx  Ax 3 Bx 2 Cx D ++ + = x d 1 + 1 w o o w 1 1 + =
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CE 30125 - Lecture 19 p. 19.7 • In order for this to be true for any third degree polynomial (i.e. all arbitrary coefficients, , , , ), we must have: Ax 3 Bx 2 Cx D ++ +  x d 1 + 1 w o o 3 2 o o D + w 1 3 1 2 1
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lecture 19 - CE 30125 - Lecture 19 LECTURE 19 GAUSS...

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