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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
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CISE301_Topic5 KFUPM - T212 - Section 01 2
Lecture 20Introduction to Interpolation
IntroductionInterpolation ProblemExistence and UniquenessLinear and Quadratic Interpolation Newton’s Divided Difference MethodProperties of Divided Differences
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CISE301_Topic5 KFUPM - T212 - Section 01 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
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CISE301_Topic5 KFUPM - T212 - Section 01 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)()(
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CISE301_Topic5 KFUPM - T212 - Section 01 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
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CISE301_Topic5 KFUPM - T212 - Section 01 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
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CISE301_Topic5 KFUPM - T212 - Section 01 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
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CISE301_Topic5 KFUPM - T212 - Section 01 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.
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CISE301_Topic5 KFUPM - T212 - Section 01 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
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CISE301_Topic5 KFUPM - T212 - Section 01 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
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CISE301_Topic5 KFUPM - T212 - Section 01 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
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CISE301_Topic5 KFUPM - T212 - Section 01 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
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CISE301_Topic5 KFUPM - T212 - Section 01 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 ],...,,[)(
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CISE301_Topic5 KFUPM - T212 - Section 01 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.
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CISE301_Topic5 KFUPM - T212 - Section 01 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.
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CISE301_Topic5 KFUPM - T212 - Section 01 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
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CISE301_Topic5 KFUPM - T212 - Section 01 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
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CISE301_Topic5 KFUPM - T212 - Section 01 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
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CISE301_Topic5 KFUPM - T212 - Section 01 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)
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CISE301_Topic5 KFUPM - T212 - Section 01 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?
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CISE301_Topic5 KFUPM - T212 - Section 01 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.
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CISE301_Topic5 KFUPM - T212 - Section 01 22
Properties of Divided Difference
],,[],,[],,[ 012021210 xxxfxxxfxxxf
Ordering the points should not affect the divided difference:
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CISE301_Topic5 KFUPM - T212 - Section 01 23
Example Find a polynomial to
interpolate the data.x f(x)
2 3
4 5
5 1
6 6
7 9
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CISE301_Topic5 KFUPM - T212 - Section 01 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
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CISE301_Topic5 KFUPM - T212 - Section 01 25
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.
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CISE301_Topic5 KFUPM - T212 - Section 01 26
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
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CISE301_Topic5 KFUPM - T212 - Section 01 27
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
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CISE301_Topic5 KFUPM - T212 - Section 01 28
Lecture 21Lagrange Interpolation
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CISE301_Topic5 KFUPM - T212 - Section 01 29
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)()(
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CISE301_Topic5 KFUPM - T212 - Section 01 30
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
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CISE301_Topic5 KFUPM - T212 - Section 01 31
Lagrange Interpolation
ji
jix
x
ji
th
i
1
0)(
:spolynomialorder n are cardinals The
cardinals. thecalled are)(
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CISE301_Topic5 KFUPM - T212 - Section 01 32
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
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CISE301_Topic5 KFUPM - T212 - Section 01 33
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
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CISE301_Topic5 KFUPM - T212 - Section 01 34
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
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CISE301_Topic5 KFUPM - T212 - Section 01 35
Interpolating Polynomial
)3)(2)(1(8
5
)2)(1(6
7)1(
2
3)(21)(4
xxxx
xxxxxxxf
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CISE301_Topic5 KFUPM - T212 - Section 01 36
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
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CISE301_Topic5 KFUPM - T212 - Section 01 37
Lecture 22Inverse Interpolation Error in Polynomial
Interpolation
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CISE301_Topic5 KFUPM - T212 - Section 01 38
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
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CISE301_Topic5 KFUPM - T212 - Section 01 39
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
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CISE301_Topic5 KFUPM - T212 - Section 01 40
Inverse Interpolation
x
y
y
x
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CISE301_Topic5 KFUPM - T212 - Section 01 41
Inverse Interpolation
Question:
What is the limitation of inverse interpolation?
• The original function has an inverse.
• y1, y2, …, yn must be distinct.
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CISE301_Topic5 KFUPM - T212 - Section 01 42
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
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CISE301_Topic5 KFUPM - T212 - Section 01 43
-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
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CISE301_Topic5 KFUPM - T212 - Section 01 44
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.
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CISE301_Topic5 KFUPM - T212 - Section 01 45
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
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CISE301_Topic5 KFUPM - T212 - Section 01 46
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
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CISE301_Topic5 KFUPM - T212 - Section 01 47
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?
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CISE301_Topic5 KFUPM - T212 - Section 01 48
Approximation of the Interpolation Error
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CISE301_Topic5 KFUPM - T212 - Section 01 49
Divided Difference Theorem
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CISE301_Topic5 KFUPM - T212 - Section 01 50
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.