interpolation in numerical methods
Post on 04-Jun-2018
237 Views
Preview:
TRANSCRIPT
-
8/13/2019 Interpolation in numerical methods
1/51
NUMERICAL METHODS
INTERPOLATION
-
8/13/2019 Interpolation in numerical methods
2/51
INTRODUCTION TO INTERPOLATION
IntroductionPolynomial Forms
Linear and Quadratic interpolationInterpolation problemExistence and uniquenessLagrange interpolation polynomialNewtons Divided Difference method Properties of Divided Differences
-
8/13/2019 Interpolation in numerical methods
3/51
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
-
8/13/2019 Interpolation in numerical methods
4/51
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
-
8/13/2019 Interpolation in numerical methods
5/51
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
)()()(
-
8/13/2019 Interpolation in numerical methods
6/51
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.
-
8/13/2019 Interpolation in numerical methods
7/51
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
-
8/13/2019 Interpolation in numerical methods
8/51
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
-
8/13/2019 Interpolation in numerical methods
9/51
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
-
8/13/2019 Interpolation in numerical methods
10/51
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
-
8/13/2019 Interpolation in numerical methods
11/51
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)()(
-
8/13/2019 Interpolation in numerical methods
12/51
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
-
8/13/2019 Interpolation in numerical methods
13/51
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
-
8/13/2019 Interpolation in numerical methods
14/51
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.
-
8/13/2019 Interpolation in numerical methods
15/51
Lagrange Interpolation
ji
ji x
x
ji
i
1
0)(
s polynomialordern tharecardinalsThe
cardinalsthecalledare)(
-
8/13/2019 Interpolation in numerical methods
16/51
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
)(
14/11
3/14/13/1
)(
13/11
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
-
8/13/2019 Interpolation in numerical methods
17/51
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
-
8/13/2019 Interpolation in numerical methods
18/51
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
-
8/13/2019 Interpolation in numerical methods
19/51
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
-
8/13/2019 Interpolation in numerical methods
20/51
NEWTONs INTERPOLATION POLYNOMIAL
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
-
8/13/2019 Interpolation in numerical methods
21/51
NEWTONs INTERPOLATION POLYNOMIAL
polynomial al erpolation
differencedivided s Newtonasknown
x x x x x f x f
n
i
i
j jin
int
'
],...,,[)( 0
1
010
-
8/13/2019 Interpolation in numerical methods
22/51
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
-
8/13/2019 Interpolation in numerical methods
23/51
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 ]
-
8/13/2019 Interpolation in numerical methods
24/51
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
-
8/13/2019 Interpolation in numerical methods
25/51
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
-
8/13/2019 Interpolation in numerical methods
26/51
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
-
8/13/2019 Interpolation in numerical methods
27/51
Divided Difference Table
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
-
8/13/2019 Interpolation in numerical methods
28/51
Divided Difference Table
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
-
8/13/2019 Interpolation in numerical methods
29/51
Divided Difference Table
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 )
-
8/13/2019 Interpolation in numerical methods
30/51
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?
-
8/13/2019 Interpolation in numerical methods
31/51
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
-
8/13/2019 Interpolation in numerical methods
32/51
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
-
8/13/2019 Interpolation in numerical methods
33/51
ExampleFind a polynomial tointerpolate the data.
x f(x)
2 3
4 5
5 1
6 6
7 9
-
8/13/2019 Interpolation in numerical methods
34/51
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
-
8/13/2019 Interpolation in numerical methods
35/51
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
-
8/13/2019 Interpolation in numerical methods
36/51
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
-
8/13/2019 Interpolation in numerical methods
37/51
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
-
8/13/2019 Interpolation in numerical methods
38/51
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
-
8/13/2019 Interpolation in numerical methods
39/51
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 )(
-
8/13/2019 Interpolation in numerical methods
40/51
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
-
8/13/2019 Interpolation in numerical methods
41/51
-
8/13/2019 Interpolation in numerical methods
42/51
-
8/13/2019 Interpolation in numerical methods
43/51
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
-
8/13/2019 Interpolation in numerical methods
44/51
-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
-
8/13/2019 Interpolation in numerical methods
45/51
-
8/13/2019 Interpolation in numerical methods
46/51
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
-
8/13/2019 Interpolation in numerical methods
47/51
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
-
8/13/2019 Interpolation in numerical methods
48/51
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?
-
8/13/2019 Interpolation in numerical methods
49/51
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
-
8/13/2019 Interpolation in numerical methods
50/51
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)(
-
8/13/2019 Interpolation in numerical methods
51/51
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
top related