LECTURE 5 HERMITE INTERPOLATING POLYNOMIALS

CE 30125 - LECTURE 5p. 5 HERMITE INTERPOLATING POLYNOMIALS So far we have considered Lagrange interpolation schemes which fit an degreepolynomial 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 thefunctional values at the data points:, NthN1+ f1 f0 fN xN x0 f2 x1 x2 xgx aixii0=N =gx aoa1xa2x2a3x3 aNxN++ + ++=gxi fi=i0N =CE 30125 - LECTURE 5p. 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 interpolatingpolynomial).

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 x f f 1

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  Interpolation, Hermite, Hermite interpolation

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