cise301 topic5 t212 sec01

CISE301_Topic5 KFUPM - T212 - Section 01 1

SE301: Numerical Methods

Topic 5: InterpolationLectures 20-22:

KFUPM(Term 212)Section 01

Read Chapter 18, Sections 1-5


Lecture 20Introduction to Interpolation

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


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


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)()(


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


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


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


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.


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


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


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


Divided Differences

0

1102110

02

1021210

01

0110

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

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

DDorder Second],[],[

],,[

DDorder First ][][

],[

DDorder Zeroth )(][

xx

xxxfxxxfxxxf

xx

xxfxxfxxxf

xx

xfxfxxf

xfxf

k

kkk

kk


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 ],...,,[)(



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.



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.



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



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



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



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)


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?


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.


Properties of Divided Difference

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

Ordering the points should not affect the divided difference:


Example Find a polynomial to

interpolate the data.x f(x)

2 3

4 5

5 1

6 6

7 9


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


x 1.6 2 2.5 3.2 4 4.5f(x) 2 8 14 15 8 2

Example

Calculate f (2.8) using Newton’s interpolating polynomials of order 1 through 3. Choose the sequence of the points for your estimates to attain the best possible accuracy.


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

2.5 14 1.4286 -8.8095 1.01203.2 15 5.8333 -7.2917 2 8 0 4 8

The first through third-order interpolations can then be implemented as

Example

428571.14)5.28.2(428571.114)8.2(1 f

485714.15)2.38.2)(5.28.2(809524.8)5.28.2(428571.114)8.2(2 f

15.388571.)28.2)(2.38.2)(5.28.2(1.011905 )2.38.2)(5.28.2(809524.8)5.28.2(428571.114)8.2(3

f


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


Lecture 21Lagrange Interpolation


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)()(


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


Lagrange Interpolation

ji

jix

x

ji

th

i

1

0)(

:spolynomialorder n are cardinals The

cardinals. thecalled are)(


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


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


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


Interpolating Polynomial

)3)(2)(1(8

5

)2)(1(6

7)1(

2

3)(21)(4

xxxx

xxxxxxxf


Interpolating Polynomial Using Lagrange Interpolation Method

34

3

24

2

14

1

04

043

4

23

2

13

1

03

042

4

32

3

12

1

02

041

4

31

3

21

2

01

040

4

30

3

20

2

10

1

4523)()(

4

3

2

1

0

43210

4

04

xxxx

xxxx

xxxx

xxxx

xxxx

xfxfi

ii


Lecture 22Inverse Interpolation Error in Polynomial

Interpolation


Inverse Interpolation

.)( :such that Find

. and tableGiven the

:Problem

ggg

g

yxfx

y

….

….

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:

ggn yxf )(

0x


Inverse InterpolationAlternative Approach

….

….

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

)( gn yf

….

….

iy 0y 1y ny

ix 0x 1x nx

0x



x

y

y

x



Question:

What is the limitation of inverse interpolation?

• The original function has an inverse.

• y1, y2, …, yn must be distinct.


Inverse Interpolation Example

5.2)(such that Find table.Given the

:Problem

gg xfx

x 1 2 3

y 3.2 2.0 1.6

3.2 1 -.8333 1.0416

2.0 2 -2.5

1.6 3

2187.1)5.0)(7.0(0416.1)7.0(8333.01)5.2(

)2)(2.3(0416.1)2.3(8333.01)(

2

2

f

yyyyf


-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


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.


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

:such thatfunction a beLet

n

n)(n

n

ab

n

Mf(x)-P(x)

nf

P(x)

Mf(x)f

f(x)

Errors in polynomial InterpolationTheorem


Example 1

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


Interpolation Error

n

iinn

n

n

xxxxxxfxff(x)

xxxx

nf

010

10

],,...,,[)(

:node anot

any for then ,,...,, nodes at the f(x)function the

esinterpolat that degree of polynomial theis (x) If

Not useful. Why?


Approximation of the Interpolation Error

n

iinnn

nn

n

n

xxxxxxfxff(x)

xfx

xxxf(x)

nf

0110

11

10

],,...,,[)(

:by edapproximat iserror ion interpolat The

point. additionalan be))(,(Let

.,...,, nodes at the function the

esinterpolat that degree of polynomial theis (x) If


Divided Difference Theorem

)(10

10)(

!

1],...,,[

:b][a, somefor then b],[a,in pointsdistinct 1any are

,...,, if and b],[a,on continuous is

nn

nn

fn

xxxf

n

xxxfIf

10],...,,[

:then

,order of polynomial a is )(

10 nixxxf

nxfIf

i

1 4 1 0 0

2 5 1 0

3 6 1

4 7

xxf 3)(


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.

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