POLYNOMIAL INTERPOLATION course-6 / numerical method POLYNOMIAL INTERPOLATION Fitting polynomial to given data points Most of numerical method schemes are based on polynomial interpolation, e.g. numerical integration and differentiation.

course-6 / numerical method LINEAR INTERPOLATION The linear interpolation shown in figure previous is given by The maximum error of the linear interpolation is expressed in the form

Can three or more data points be fitted by a curve ?

LAGRANGE INTERPOLATION Suppose N+1 data points are given. The Lagrange interpolation formula of order N-th is written as follows

The maximum error of Lagrange interpolation is expressed in the form There is no guarantee that the interpolation polynomial converges to the exact function when the number of data point is increased. In general, interpolation with a large-order polynomial should be avoided or used with extreme cautions

NEWTON INTERPOLATION The drawback of the Lagrange interpolation: The amount of computation needed for one interpolation is large No part of the previous application can be used to interpolate another value of x When the number of data points has to be increased or decreased, the results of the previous computations cannot be used Evaluation of error is not easy

DIVIDED DIFFERENCE To evaluate a Newton interpolation formula, a forward difference table is necessary

Therefore, the forward difference table is given by (for third order) 1 2 3

Hence, the Newton interpolation formula is written as follows where are obtained from forward difference table The maximum error of Newton interpolation is in the form

Application Consider the data points given in the following table i 1 1 2 3 4 5 0.1 0.2 0.3 0.5 0.7 0.9 0.9975 0.9776 0.9384 0.8812 0.8075 0.7196

Derive the Lagrange and Newton forward interpolation fitted to the data points at a. i = 0, 1, 2 (evaluate for x = 0.21) b. i = 1, 2, 3 (evaluate for x = 0.21) c. i = 1, 2, 3, 4 (evaluate for x = 0.21) Estimate the maximum error for every evaluate of x