LECTURE 5 HERMITE INTERPOLATING POLYNOMIALS

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CE 30125 - LECTURE 5p. 5HERMITE 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

  Interpolation, Hermite, Hermite interpolation