lecture 9-1

lecture 9-1 - CE 30125 - Lecture 9 LECTURE 9 HERMITE...

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CE 30125 - Lecture 9 p. 9.1 LECTURE 9 HERMITE INTERPOLATING POLYNOMIALS • So far we have considered Lagrange Interpolation schemes which fit an degree polynomial to data or interpolation points • All these Lagrange Interpolation methods discussed had the general form: • Fitting the data points meant requiring the interpolating polynomial to be equal to the functional values at the data points: , N th N 1 + f 1 f 0 f N x N x 0 f 2 x 1 x 2 x gx  a i x i i 0 = N = a o a 1 xa 2 x 2 a 3 x 3 a N x N ++ + + + = i f i = i 0 N =
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CE 30125 - Lecture 9 p. 9.2 • Hermite Interpolation: Develop an interpolating polynomial which equals the func- tion and its derivatives up to order at data points. • Therefore we require that constraints constraints : constraints • We have a total of constraints • We need to set up a general polynomial which is of degree (number of constraints must equal the number of unknowns in the interpolating polynomial). p th N 1 + f 0 , f 0 ,... , f 0 x N x 0 x 1 x (1) (P) f 1 f 1 f 1 (1) (P) f f f (1) (P) gx i  f i = i 0 N = N 1 + g 1 x i f i 1 = i 0 N = N 1 + g p x i f i p = i 0 N = N 1 + p 1 + N 1 + p 1 + N 1 + 1
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CE 30125 - Lecture 9 p. 9.3 • Setting up a polynomial with a total of unknowns: • Procedure to develop Hermite interpolation: Set up the interpolating polynomial Implement constraints Solve for unknown coefficients, , , • Note that Lagrange interpolation is a special case of Hermite interpolation ( , i.e. no derivatives are matched). • It is also possible to set up specialized Hermite interpolation functions which do not include all functional and/or derivative values at all nodes • There may be some missing functional or derivative values at certain nodes • This lowers the degree of the interpolating function. p 1 +  N 1 + gx a i x i i 0 = p 1 + N 1 + 1 = a i i 0 = p 1 + N 1 + 1 p 0 =
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CE 30125 - Lecture 9 p. 9.4 Cubic Hermite Interpolation • Develop a two data point Hermite interpolation function which passes through the func- tion and its first derivative for the interval [0, 1] . • Therefore and . • We must impose constraint equations (match function and its derivative at two data points). • Therefore we require a 3rd degree polynomial. 0
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This note was uploaded on 03/02/2012 for the course CE 30125 taught by Professor Westerink,j during the Fall '08 term at Notre Dame.

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lecture 9-1 - CE 30125 - Lecture 9 LECTURE 9 HERMITE...

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