interpolation in numerical methods

8/13/2019 Interpolation in numerical methods

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NUMERICAL METHODS

INTERPOLATION


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INTRODUCTION TO INTERPOLATION

IntroductionPolynomial Forms

Linear and Quadratic interpolationInterpolation problemExistence and uniquenessLagrange interpolation polynomialNewtons Divided Difference method Properties of Divided Differences


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INTERPOL TION

DEFINITION

3

LET BE A CONTINEOUS FUNCTION DEFINED INTHE INTERVAL CONTAINING THE POINTS

THE INTERPOLATION IS USED TO FINDTHE VALUE OF THE FUNCTION AT ANY PONIT INTHE INTERVAL OR

INTERPOLATION MEANS TO FIND (APPROXIMATE)VALUES OF A FUNCTION FOR AN x BETWEENDIFFERENT x VALUES AT WHICH THE

VALUES OFf(x)

ARE GIVEN


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INTERPOL TION

EX MPLE

4

x 0 1 2

f(x) 1 0 -1

FIND:

THE PROCESS OF CONSTRUCTION OFTO FIT A TABLE OF DATA POINTS IS

CALLED CURVE FITTING


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POLYNOMIAL FORMS

POWER FORM

SHIFT POWER FORM

NEWTON FORM

5

n

i

n

i j j

jinn x xb x f x P 0 0

)()()(


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Examples of Polynomial InterpolationLinear Interpolation

Given any two points,there is one polynomial oforder 1 that passesthrough the two points.

Quadratic Interpolation

Given any three points thereis one polynomial of order 2 that passes through thethree points.


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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 x f x x f x

x x x f 211224

2)(1

475


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ExampleAn experiment is used to determinethe viscosity of water as afunction of temperature. Thefollowing table is generated

Problem: Estimate the viscositywhen the temperature is 8 degrees.

Temperature(degree)

Viscosity

0 1.792

5 1.519

10 1.308

15 1.140


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ExampleFind a polynomial that fit the data points

exactly.

tscoefficien

polynomial:

eTemperatur :

viscosity:

k a

T

V

Linear Interpolation V(T)= 1.73 0.0422 TV(8)= 1.3924


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Example

10

THE TABLE BELOW GIVES SQUARE ROOT FOR NTEGERS

x 1 2 3 4 5f(x) 1 1.4142 1.7321 2 2.2361

DETERMINE, THE SQUARE ROOT OF 2.5


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The interpolation ProblemGiven a set of n+1 points,

find an n th order polynomialthat passes through all points

)(,....,,)(,,)(, 1100 nn x f x x f x x f x

)( x f n

ni for x f x f iin ,...,2,1,0)()(


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Existence and UniquenessGiven a set of n+1 points

Assumption are distinct real nos

Theorem:

There is a unique polynomial f n (x) oforder n

such that

ni for x f x f iin ,...,1,0)()(

n x x x ,...,, 10

)(,....,,)(,,)(, 1100 nn x f x x f x x f x


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Quadratic InterpolationGiven any three points ,The polynomial that interpolates the three points is

)(,)(,,)(, 221100 x f xand x f x x f x

202

01

01

12

12

2

101

011

00

1020102

)()()()(

)()(

)(

)(

x xat x x

x x x f x f

x x x f x f

b

x xat x x

x f x f b

x xat x f b

where

x x x xb x xbb x f

477


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LAGRANGE INTERPOLATION

Lagrange Interpolation Formula

n

i j j ji

ji

n

iiin

x x

x x x

x x f x f

,0

0

)(

)()(

486

Let, denote distinct real nos and let bearbitrary real nos. The pts can be imagined to bedata values connected by a curve. Any fn satisfying the conditions:

is called an interpolation fn. An interpolation fn is, therefore, a curvethat passes through the data pts.


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Lagrange Interpolation

ji

ji x

x

ji

i

1

0)(

s polynomialordern tharecardinalsThe

cardinalsthecalledare)(


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EXAMPLE

x 1/3 1/4 1y 2 -1 7

)4/1)(3/1(27

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

4/114/1

3/113/1

)(

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3/14/13/1

)(

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4/13/14/1

)(

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

2

12

1

02

02

21

2

01

01

20

2

10

10

2211002

x x

x x x x x P

x x x x x x

x x x x

x

x x x x x x

x x x x

x

x x x x x x

x x x x

x

x x f x x f x x f x P


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EXAMPLE

17

, Where,

EXAMPLE

THE TABLE BELOW GIVES SQUARE ROOT FOR NTEGERS

x 1 2 3 4 5f(x) 1 1.4142 1.7321 2 2.2361

DETERMINE, THE SQUARE ROOT OF 2.5


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NEWTONs INTERPOLATION POLYNOMIAL

Given any n+1 points ,The polynomial that interpolates all points is

)(,,...,)(,,)(, 1100 nn x f x x f x x f x

.

)()()()(

)()()(

......)(

202

01

01

12

12

2

101

011

00

10102010

on soand x xat x x x x

x f x f x x

x f x f

a

x xat x x

x f x f a

x xat x f a

x x x xa x x x xa x xaa x f nnn


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Divided Differences

0

1102110

02

1021210

01

0110

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

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

DDorderSecond],[],[],,[

DDorderfirst][][

],[

DDorderzeroth)(][

x x x x x f x x x f

x x x f

x x x x f x x f x x x f

x x x f x f

x x f

x f x f

k

k k k

k k


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Now

],...,,[

....

],,[

)()()()(

],[)()(

][)(

.. ....)(

10

2102

01

01

12

12

2

1001

01

1

00

10102010

nn

nnn

x x x f b

x x x f x x

x x x f x f

x x x f x f

b

x x f x x x f x f

b

x f x f b

x x x xb x x x xb x xbb x f


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polynomial al erpolation

differencedivided s Newtonasknown

x x x x x f x f

n

i

i

j jin

int

'

],...,,[)( 0

1

010


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EX MPLE

22

CONSTRUCT A POLYNOMIAL OF DEGREE TWO WHICHINTERPOLATES THE FOLLOWING SET OF POINTS.

x 1 2 3

f(x) 0 2 6

ALSO, FIND THE VALUE OF F(2.5)

f(2.5)=3.75


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Divided Difference Tablex F[ ] F[ , ] F[ , , ] F[ , , ,]

x 0 F[x0] F[x 0 ,x 1] F[x 0 ,x 1, x 2 ] F[x 0 ,x 1 ,x 2 ,x 3 ]

x 1 F[x 1 ] F[x 1 ,x 2] F[x 1 ,x 2 ,x 3 ]

x 2 F[x 2 ] F[x 2 ,x 3]

x 3 F[x 3 ]


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Divided Difference Tablef(x i )

0 -5

1 -3-1 -15

i xx F[ ] F[ , ] F[ , , ]

0 -5 2 -4

1 -3 6

-1 -15

Entries of the divided differencetable are obtained from the datatable using simple operations


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Divided Difference Tablef(x i )

0 -51 -3

-1 -15

i xx 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


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Divided Difference Table

0 -5

1 -3-1 -15

i yi xx F[ ] F[ , ] F[ , , ]0 -5 2 -4

1 -3 6

-1 -15

201

)5(3

01

0110

][][],[

x x x f x f

x x f


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0 -5

1 -3-1 -15

i yi xx F[ ] F[ , ] F[ , , ]0 -5 2 -4

1 -3 6

-1 -15

611

)3(15

12

1221

][][],[

x x x f x f

x x f


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0 -5

1 -3-1 -15

i yi xx F[ ] F[ , ] F[ , , ]0 -5 2 -4

1 -3 6

-1 -15

4)0(1

)2(6

02

1021210

],[],[],,[

x x x x f x x f

x x x f


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0 -5

1 -3-1 -15

i yi xx F[ ] F[ , ] F[ , , ]0 -5 2 -4

1 -3 6

-1 -15

)1)(0(4)0(25)(2 x x x x f

f 2 (x)= f[x 0 ]+ f[x 0 ,x 1 ] (x-x 0 )+ f[x 0 ,x 1 ,x 2 ] (x-x 0 )(x-x 1 )


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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?


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Two Examples

1

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

2

x

x x x x P

x Y

1 0 3 1

2 3 53 8

x Y

2 3 3 1

1 0 43 8

1

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

2

x

x x x x P

Ordering the points should not affect the interpolating polynomial


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Properties of divided Difference

],,[],,[],,[ 012021210 x x x f x x x f x x x f

Ordering the points should not affect the divided difference


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ExampleFind a polynomial tointerpolate the data.

x f(x)

2 3

4 5

5 1

6 6

7 9


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Example

x f(x) f[ , ] f[ , , ] f[ , , , ] f[ , , , , ]2 3 1 -1.6667 1.5417 -0.6750

4 5 -4 4.5 -1.83335 1 5 -16 6 37 9

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

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

x x x x x x f


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Examplefind a polynomial to interpolate

both Newtons

interpolation methodand Lagrangeinterpolation method

must give the sameanswer

x y

0 1

1 3

2 2

3 5

4 4


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Example 0 1 2 -3/2 1/6 -9/24

1 3 -1 2 -4/3

2 2 3 -2

3 5 -1

4 4


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Interpolating Polynomial

)3)(2)(1(249

)2)(1(61

)1(23

)(21)(4

x x x x

x x x x x x x f


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Interpolating Polynomial UsingLagrange Interpolation method

343

242

141

040

434

232

131

030

424

323

121

020

414

313

212

010

404

303

202

101

4523)()(

4

3

2

1

0

43210

4

04

x x x x

x x x x

x x x x

x x x x

x x x x

x f x f i

ii

491


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Inverse Interpolation

g g g

g

y x f x

y

)(such thatFind

andtableGiven the

:Problem 0 x . .

1 x n x

0 y 1 y n yi x

i y

One approach :

Use polynomial interpolation to obtain f n(x) to interpolate thedata then use Newtons method to find a solution to

g g n y x f )(


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Inverse Interpolation Alternative Approach

0 x .

.

1 x n x

0 y 1 y n y

i x

i y

Inverse interpolation:

1. Exchange the roles

of x and y

2. Perform polynomial

Interpolation on the

new table.

3. Evaluate

)( g n y f

. .

i y 0 y 1 y n y

i x 0 x 1 x n x


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Inverse InterpolationExample

5.2)(such thatFind table.Given the

:Problem

g g x f x

x 1 2 3y 3.2 2.0 1.6

3.2 1 -.8333 1.04162.0 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

y y y y f


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-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

10 th order Polynomial Interpolation


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Errors in polynomial Interpolation Theorem

1

)1(1)(n

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

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

pointsspacedequally1natf esinterpolatthatndegreeof polynomialany beP(x)Let

b],[a,oncontinuousis(x)f

such thatfunctiona bef(x)Let

n

n

nab

n M

M f


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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]intervalin the points)spacedequally01(using

f(x)einterpolatto polynomialorderth9usewant toWe sin(x)f(x)

n

n

nab

n M

n M

n for f


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Interpolation Error

n

iinn

n

n

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

x x x

f

010

10

],,...,,[)(nodeanot

any xforthen,...,,nodesat thef(x)functionthe

esinterpolatn thatdegreeof polynomialtheis(x)If

Not useful. Why?


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Approximation of the Interpolation Error

n

iinnn

nn

n

n

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

be x f x Let

x x x f

0110

11

10

],,...,,[)(

byedapproximatiserrorioninterpolatThe

pointadditionalan))(,(

,...,,nodesat thef(x)functiontheesinterpolatn thatdegreeof polynomialtheis(x)If


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Divided Difference Theorem

)(10

10)(

!

1],...,,[

b][a,someforthen b][a,in pointsdistinct1nanyare,...,,if and b][a,oncontinuousis

nn

nn

f n

x x x f

x x x f If

10],...,,[then

norderof polynomialas)(

10 ni x x x f

i x f If

i

1 4 1 0 02 5 1 03 6 14 7

x x f 3)(


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SummaryThe interpolating polynomial is uniqueDifferent methods can be used to obtain it

Newtons Divided difference

Lagrange interpolationothers

Polynomial interpolation can be sensitiveto data.BE CAREFUL when high orderpolynomials are used

interpolation in numerical methods

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