CISE301_Topic5 1

SE301: Numerical Methods

Topic 5: InterpolationLectures 20-22:

KFUPM

Read Chapter 18, Sections 1-5

CISE301_Topic5 2

Lecture 20Introduction to Interpolation

IntroductionInterpolation ProblemExistence and UniquenessLinear and Quadratic Interpolation Newton’s Divided Difference MethodProperties of Divided Differences

CISE301_Topic5 3

Introduction Interpolation was used for

long time to provide an estimate of a tabulated function at values that are not available in the table.

What is sin (0.15)?

x sin(x)

0 0.0000

0.1 0.0998

0.2 0.1987

0.3 0.2955

0.4 0.3894

Using Linear Interpolation sin (0.15) ≈ 0.1493 True value (4 decimal digits) sin (0.15) = 0.1494

CISE301_Topic5 4

The Interpolation Problem Given a set of n+1 points,

Find an nth order polynomial that passes through all points, such that:

)(,....,,)(,,)(, 1100 nn xfxxfxxfx

)(xfn

niforxfxf iin ,...,2,1,0)()(

CISE301_Topic5 5

Example An experiment is used to determine

the viscosity of water as a function of temperature. The following table is generated:

Problem: Estimate the viscosity when the temperature is 8 degrees.

Temperature(degree)

Viscosity

0 1.792

5 1.519

10 1.308

15 1.140

CISE301_Topic5 6

Interpolation ProblemFind a polynomial that fits the data

points exactly.

)V(TV

TaV(T)

ii

n

k

kk

0

tscoefficien

Polynomial:

eTemperatur:

Viscosity:

ka

T

V

Linear Interpolation: V(T)= 1.73 − 0.0422 T

V(8)= 1.3924

CISE301_Topic5 7

Existence and Uniqueness Given a set of n+1 points:

Assumption: are distinct

Theorem:There is a unique polynomial fn(x) of order ≤ n

such that:

,...,n,iforxfxf iin 10)()(

nxxx ,...,, 10

)(,....,,)(,,)(, 1100 nn xfxxfxxfx

CISE301_Topic5 8

Examples of Polynomial InterpolationLinear Interpolation

Given any two points, there is one polynomial of order ≤ 1 that passes through the two points.

Quadratic Interpolation

Given any three points there is one polynomial of order ≤ 2 that passes through the three points.

CISE301_Topic5 9

Linear InterpolationGiven any two points,

The line that interpolates the two points is:

Example :Find a polynomial that interpolates (1,2) and (2,4).

)(,,)(, 1100 xfxxfx

001

0101

)()()()( xx

xx

xfxfxfxf

xxxf 2112

242)(1

CISE301_Topic5 10

Quadratic Interpolation Given any three points: The polynomial that interpolates the three points is:

)(, ,)(,,)(, 221100 xfxandxfxxfx

02

01

01

12

12

2102

01

01101

00

1020102

)()()()(

],,[

)()(],[

)(

:

)(

xx

xx

xfxf

xxxfxf

xxxfb

xx

xfxfxxfb

xfb

where

xxxxbxxbbxf

CISE301_Topic5 11

General nth Order InterpolationGiven any n+1 points:The polynomial that interpolates all points is:

)(,..., ,)(,,)(, 1100 nn xfxxfxxfx

],...,,[

....

],[

)(

......)(

10

101

00

10102010

nn

nnn

xxxfb

xxfb

xfb

xxxxbxxxxbxxbbxf

CISE301_Topic5 12

Divided Differences

0

1102110

02

1021210

01

0110

],...,,[],...,,[],...,,[

............

DDorder Second],[],[

],,[

DDorder First ][][

],[

DDorder Zeroth )(][

xx

xxxfxxxfxxxf

xx

xxfxxfxxxf

xx

xfxfxxf

xfxf

k

kkk

kk

CISE301_Topic5 13

Divided Difference Table

x F[ ] F[ , ] F[ , , ] F[ , , ,]

x0 F[x0] F[x0,x1] F[x0,x1,x2] F[x0,x1,x2,x3]

x1 F[x1] F[x1,x2] F[x1,x2,x3]

x2 F[x2] F[x2,x3]

x3 F[x3]

n

i

i

jjin xxxxxFxf

0

1

010 ],...,,[)(

CISE301_Topic5 14

Divided Difference Table

f(xi)

0 -5

1 -3

-1 -15

ixx F[ ] F[ , ] F[ , , ]

0 -5 2 -4

1 -3 6

-1 -15

Entries of the divided difference table are obtained from the data table using simple operations.

CISE301_Topic5 15

Divided Difference Table

f(xi)

0 -5

1 -3

-1 -15

ixx F[ ] F[ , ] F[ , , ]

0 -5 2 -4

1 -3 6

-1 -15

The first two column of the

table are the data columns.

Third column: First order differences.

Fourth column: Second order differences.

CISE301_Topic5 16

Divided Difference Table

0 -5

1 -3

-1 -15

iyixx F[ ] F[ , ] F[ , , ]

0 -5 2 -4

1 -3 6

-1 -15

201

)5(3

01

0110

][][],[

xx

xfxfxxf

CISE301_Topic5 17

Divided Difference Table

0 -5

1 -3

-1 -15

iyixx F[ ] F[ , ] F[ , , ]

0 -5 2 -4

1 -3 6

-1 -15

611

)3(15

12

1221

][][],[

xx

xfxfxxf

CISE301_Topic5 18

Divided Difference Table

0 -5

1 -3

-1 -15

iyixx F[ ] F[ , ] F[ , , ]

0 -5 2 -4

1 -3 6

-1 -15

4)0(1

)2(6

02

1021210

],[],[],,[

xx

xxfxxfxxxf

CISE301_Topic5 19

Divided Difference Table

0 -5

1 -3

-1 -15

iyixx F[ ] F[ , ] F[ , , ]

0 -5 2 -4

1 -3 6

-1 -15

)1)(0(4)0(25)(2 xxxxf

f2(x)= F[x0]+F[x0,x1] (x-x0)+F[x0,x1,x2] (x-x0)(x-x1)

CISE301_Topic5 20

Two Examples

x y

1 0

2 3

3 8

Obtain the interpolating polynomials for the two examples:

x y

2 3

1 0

3 8

What do you observe?

CISE301_Topic5 21

Two Examples

1

)2)(1(1)1(30)(2

2

x

xxxxP

x Y

1 0 3 1

2 3 5

3 8

x Y

2 3 3 1

1 0 4

3 8

1

)1)(2(1)2(33)(2

2

x

xxxxP

Ordering the points should not affect the interpolating polynomial.

CISE301_Topic5 22

Properties of Divided Difference

],,[],,[],,[ 012021210 xxxfxxxfxxxf

Ordering the points should not affect the divided difference:

CISE301_Topic5 23

Example Find a polynomial to

interpolate the data.x f(x)

2 3

4 5

5 1

6 6

7 9

CISE301_Topic5 24

Example

x f(x) f[ , ] f[ , , ] f[ , , , ] f[ , , , , ]

2 3 1 -1.6667 1.5417 -0.6750

4 5 -4 4.5 -1.8333

5 1 5 -1

6 6 3

7 9

)6)(5)(4)(2(6750.0

)5)(4)(2(5417.1)4)(2(6667.1)2(134

xxxx

xxxxxxf

CISE301_Topic5 25

Summary

.polynomial inginterpolat affect thenot should points theOrdering

methodsOther -

2] 18.[Section ion Interpolat Lagrange -

] 18.1[Section Difference DividedNewton -

itobtain toused becan methodsDifferent *

unique. is Polynomial inginterpolat The *

..., ,2 ,1 ,0)()(:Condition ingInterpolat niforxfxf ini

CISE301_Topic5 26

Lecture 21Lagrange Interpolation

CISE301_Topic5 27

The Interpolation Problem Given a set of n+1 points:

Find an nth order polynomial: that passes through all points, such that:

)(,....,,)(,,)(, 1100 nn xfxxfxxfx

)(xfn

niforxfxf iin ,...,2,1,0)()(

CISE301_Topic5 28

Lagrange InterpolationProblem: Given

Find the polynomial of least order such that:

Lagrange Interpolation Formula:

niforxfxf iin ,...,1,0)()( )(xfn

….

….

1x nx

0y 1y ny

ix

iy

n

ijj ji

ji

n

iiin

xx

xxx

xxfxf

,0

0

)(

)()(

0x

CISE301_Topic5 29

Lagrange Interpolation

ji

jix

x

ji

th

i

1

0)(

:spolynomialorder n are cardinals The

cardinals. thecalled are)(

CISE301_Topic5 30

Lagrange Interpolation Example

x 1/3 1/4 1

y 2 -1 7

)4/1)(3/1(27

)1)(3/1(161)1)(4/1(182)(

4/11

4/1

3/11

3/1)(

14/1

1

3/14/1

3/1)(

13/1

1

4/13/1

4/1)(

)()()()()()()(

2

12

1

02

02

21

2

01

01

20

2

10

10

2211002

xx

xxxxxP

xx

xx

xx

xx

xxx

xx

xx

xx

xx

xxx

xx

xx

xx

xx

xxx

xxfxxfxxfxP

CISE301_Topic5 31

ExampleFind a polynomial to interpolate:

Both Newton’s interpolation method and Lagrange interpolation method must give the same answer.

x y

0 1

1 3

2 2

3 5

4 4

CISE301_Topic5 32

Newton’s Interpolation Method0 1 2 -3/2 7/6 -5/8

1 3 -1 2 -4/3

2 2 3 -2

3 5 -1

4 4

CISE301_Topic5 33

Interpolating Polynomial

4324

4

8

5

12

59

8

95

12

1151)(

)3)(2)(1(8

5

)2)(1(6

7)1(

2

3)(21)(

xxxxxf

xxxx

xxxxxxxf

CISE301_Topic5 34

Interpolating Polynomial Using Lagrange Interpolation Method

24

)3)(2)(1(

)34(

)3(

)24(

)2(

)14(

)1(

)04(

)0(

6

)4)(2)(1(

)43(

)4(

)23(

)2(

)13(

)1(

)03(

)0(

4

)4)(3)(1(

)42(

)4(

)32(

)3(

)12(

)1(

)02(

)0(

6

)4)(3)(2(

)41(

)4(

)31(

)3(

)21(

)2(

)01(

)0(

24

)4)(3)(2)(1(

)40(

)4(

)30(

)3(

)20(

)2(

)10(

)1(

4523)()(

4

3

2

1

0

43210

4

0

4

xxxxxxxx

xxxxxxxx

xxxxxxxx

xxxxxxxx

xxxxxxxx

xfxf

i

ii

CISE301_Topic5 35

Lecture 22Inverse Interpolation Error in Polynomial

Interpolation

CISE301_Topic5 36

Inverse Interpolation

given is where,)( : thatsuch Find

valuesof a table Given :Problem

kk yyxfx

….

….

1xnx

0y 1y nyix

iy

One approach:

Use polynomial interpolation to obtain fn(x) to interpolate the data then use Newton’s method to find a solution to x

kn yxf )(

0x

CISE301_Topic5 37

Inverse Interpolation

….

….

1x nx

0y 1y nyix

iy

Inverse interpolation:

1. Exchange the roles

of x and y.

2. Perform polynomial

Interpolation on the

new table.

3. Evaluate

)( kn yfx

….

….

iy 0y 1y ny

ix 0x 1x nx

0x

CISE301_Topic5 38

Inverse Interpolation

x

y

y

x

CISE301_Topic5 39

Inverse Interpolation

Question:

What is the limitation of inverse interpolation?

• The original function has an inverse.

• y1, y2, …, yn must be distinct.

CISE301_Topic5 40

Inverse Interpolation Example

5.2)( that such Find table. theGiven

:Problem

xfx

x 1 2 3

y 3.2 2.0 1.6

3.2 1 -.8333 1.0417

2.0 2 -2.5

1.6 3

2187.1)5.0)(7.0(0417.1)7.0(8333.01)5.2(

)2)(2.3(0417.1)2.3(8333.01)(

2

2

fx

yyyyfx

CISE301_Topic5 41

Errors in polynomial Interpolation

Polynomial interpolation may lead to large errors (especially for high order polynomials).

BE CAREFUL

When an nth order interpolating polynomial is used, the error is related to the (n+1)th order derivative.

CISE301_Topic5 42

-5 -4 -3 -2 -1 0 1 2 3 4 5-0.5

0

0.5

1

1.5

2

true function

10 th order interpolating polynomial

10th Order Polynomial Interpolation

CISE301_Topic5 43

1

)1()1(

)1(4)()(

:Then points). end the(including b][a, in

points spacedequally 1at esinterpolatthat

n degree of polynomialany beLet

.)( and b],[a, on continuous is )(

: thatsuch functiona be)(Let

n

nn

n

ab

n

Mx-Pxf

nf

P(x)

Mxfxf

xf

Errors in polynomial InterpolationTheorem

CISE301_Topic5 44

Example

910

1

)1(

th

1034.19

6875.1

)10(4

1

)1(4

9 ,1

01

.[0,1.6875] interval in the points) spacedequally 01 (using

f(x) einterpolat topolynomialorder 9 use want toWe

sin

f(x)-P(x)

n

ab

n

Mf(x)-P(x)

nM

nforf

(x) f(x)

n

n

CISE301_Topic5 45

Summary The interpolating polynomial is unique. Different methods can be used to obtain it.

Newton’s divided difference Lagrange interpolation Others

Polynomial interpolation can be sensitive to data.

BE CAREFUL when high order polynomials are used.