ee3561_unit 5(c)al-dhaifallah14361 ee 3561 : computational methods unit 5 : interpolation dr....

EE3561_Unit 5 (c)AL-DHAIFALLAH1436 1

EE 3561 : Computational MethodsUnit 5:

Interpolation

Dr. Mujahed AlDhaifallah (Term 351)

Read Chapter 18, Sections 1-5


Introduction 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

)(,....,,)(,,)(, 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 fit 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

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

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


Summary

polynomial inginterpolat affect thenot should points theOrdering

methodsOther -

2] 18. [Section ionInterpolat Lagrange-

] 18.1 [Section e Differenc DividedNewton-

it obtain toused be can methodst Differen*

unique. is l Polynomiainginterpolat The *

,...,2,1,0)()( that such

)( polynomialorder n an Findpoints, 1)(n ofset a Given

: ProblemionInterpolatth

niforxfxf

xf

ini

in


Inverse interpolation

Error in polynomial interpolation

Dr. Mujahed Al-Dhaifallah (Term 351)


Inverse Interpolation

ggg

g

yxfx

y

)(such that Find

and tableGiven the

:Problem 0x ….

….

1x nx

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 )(


Inverse InterpolationAlternative Approach

0x ….

….

1x nx

0y 1y ny

ix

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



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


Errors in polynomial Interpolation Polynomial interpolations may lead to large

error (especially when high order polynomials are used)

BE CAREFUL

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


10 th order Polynomial Interpolation

-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

21

1)(

xxf


Error in polynomial interpolation

Dr. Mujahed Al-Dhaifallah (Term 351)


Errors in polynomial InterpolationTheorem

1

)1(1)(n

)1(4P(x)-f(x)

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

points spacedequally 1nat f esinterpolatthat

n degree of polynomialany beLet P(x)

b],[a, on continuous is (x)f

thatsuch functiona be f(x)Let

n

n

n

ab

n

M

Mf


Example 1

910

1

)1(

1034.19

6875.1

)10(4

1P(x)-f(x)

)1(4P(x)-f(x)

9,1

01

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

f(x) einterpolat topolynomialorder th 9 use want toWe

sin(x)f(x)

n

n

n

ab

n

M

nM

nforf


Interpolation Error

0 1

0 10

If (x) is the polynomial of degree n that interpolates

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

not a node

( ) [ , ,..., , ]

n

n

n

n n ii

f

x x x

f(x) f x f x x x x x x

Not useful. Why?


Approximation of the Interpolation Error

0 1

1 1

0

If (x) is the polynomial of degree n that interpolates

the function f(x) at the nodes , ,...,

( , ( )) an additional point

The interpolation error is approximated by

( ) [ ,

n

n

n n

n

f

x x x

Let x f x be

f(x) f x f x x

1 10

,..., , ]n

n n ii

x x x x

Example Given the following table of data



Divided Difference Theorem

)(10

10)(

!

1],...,,[

b][a, somefor thenb][a, in pointsdistinct 1nany are

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

nn

nn

fn

xxxf

xxxfIf

10],...,,[

then

n order of polynomial a s )(

10 nixxxf

ixfIf

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

ee3561_unit 5(c)al-dhaifallah14361 ee 3561 : computational methods unit 5 : interpolation dr....

Documents

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