POLYNOMIAL POLYNOMIAL INTERPOLATIONINTERPOLATION

• Fitting polynomial to given data points

• Most of numerical method schemes are

based on polynomial interpolation, e.g.

numerical integration and differentiation.

LINEAR INTERPOLATION

• The linear interpolation shown in figure previous is given by

• The maximum error of the linear interpolation is expressed in the form

)()()( 101

00

01

1 xfxx

xxxf

xx

xxxf

, )(''2

1)( 10 mxfxxxxxe

210 xx

xm

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

)()())((

)())(()( 0

02010

21 xfxxxxxx

xxxxxxxf

N

N

)()())((

)())((1

12101

20 xfxxxxxx

xxxxxx

N

N

)()())((

)())((

120

110N

NNNN

N xfxxxxxx

xxxxxx

• 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

, )()()!1(

1)( )1(

10 mN

N xfxxxxxxN

xe

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

ii

iiii xx

xfxfxxf

1

11

)()(,

ii

iiiiiii xx

xxfxxfxxxf

2

12121

],[],[,,

0

121021210

,,,,,,,,,,,

xx

xxxxfxxxfxxxxf

N

NNN

Ni ,,2,1,0

• Therefore, the forward difference table is given by (for third order)

i

0

1

2

3

ix

0x

1x

2x

)( ixf

)( 0xf

)( 1xf

)( 2xf

],[ 1ii xxf

],[ 10 xxf

],[ 21 xxf

3x )( 3xf

],[ 32 xxf

],,[ 21 iii xxxf

],,[ 210 xxxf

],,[ 321 xxxf

],,[ 3ii xxf

],,,[ 3210 xxxxf

• 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

],,[))((],[)()( 210101000 xxxfxxxxxxfxxxfxf

],,,[)())(( 10110 NN xxxfxxxxxx

],,,[ , , ],,[ ],,[ 1021010 Nxxxfxxxfxxf

],,,[)()( 11010 NN xxxfxxxxxxxe

ApplicationApplication

Consider the data points given in the following table

i 0 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

ix

)( ixf

• 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