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Curve Fitting and Interpolation
Lecture 5:Polynomial Interpolation
MTH2212 – Computational Methods and Statistics
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 22
Objectives
Introduction Newton’s Divided Difference Method
i. Linear interpolation
ii. Quadratic interpolation
iii. General Form of Newton’s Interpolation
Lagrange Interpolation
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 33
Introduction
Techniques to fit curves to discrete values of data to obtain intermediate estimates.
- Regression (imprecise data)- Interpolation (precise data)
Curve fitting is widely used in engineering
- Trend analysis: extrapolation and interpolation- Hypothesis testing: compare a mathematical
model with measured data.
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 44
Using similar triangles,
And rearranging, we get
01
01
0
01 )()()()(
xx
xfxf
xx
xfxf
)()()(
)()( 001
0101 xx
xx
xfxfxfxf
(1)
Linear Interpolation
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 55
Linear Interpolation
f1(x) is a first order interpolating polynomial.
f1(x) represents the slope of the line connecting the points.
The smaller the interval between data points, the
better the approximation.
01
01 )()(
xx
xfxf
)()()(
)()( 001
0101 xx
xx
xfxfxfxf
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 66
Example 1
Estimate the natural logarithm of 2 using linear
interpolation:
1. Interpolate between ln1 and ln6
2. Use interval ln1 to ln4
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 77
Example1 - Solution
Linear interpolation
1. Using ln1 and ln6
εt=48.3%
2. Using ln1 and ln4
εt=33.3%
0.3583519 )12(16
0791759.10)2(1
f
0.4620981 )12(14
0386294.10)2(1
f
)()()(
)()( 001
0101 xx
xx
xfxfxfxf
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 88
Example 1 - Solution
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 99
Quadratic Interpolation
This is a mean of improving an estimate by introducing a curvature into the line connecting the points.
Using three data points, a second-order polynomial or quadratic polynomial or parabola is used to carry out the estimate:
f2(x) = b0 + b1(x-x0) + b2(x-x0)(x-x1) = b0 + b1x - b1x0 + b2x2 - b2x0x - b2xx1 + b2x0x1
= a0 + a1x + a2x2
Where a0 = b0 - b1x0 + b2x0x1
a1 = b1 - b2x0 - b2x1
a2 = b2
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 1010
Quadratic Interpolation
f (x) = b0 + b1(x-x0) + b2(x-x0)(x-x1)
The values of coefficients b0 , b1 and b2 are computed as follow: Evaluate f at x = x0
Evaluate f at x = x1
Evaluate f at x = x2
)( 00 xfb
01
011
)()(
xx
xfxfb
02
01
01
12
12
2
)()()()(
xx
xxxfxf
xxxfxf
b
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 1111
Quadratic Interpolation
The first two terms in equation (2) are equivalent to linear interpolation from x0 to x1.
b1 represents the slope of the line connecting points x0 and x1.
b2(x-x0)(x-x1) introduces the second-order curvature into the formula.
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 1212
Example 2
Fit a second-order polynomial to the three points used to evaluate the natural logarithm of 2 i.e.
x0 = 1 f (x0) = 0
x1 = 4 f (x1) = 1.386294
x2 = 6 f (x2) = 1.791759
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 1313
Example 2 - Solution
First, let’s compute the coefficients b0 , b1 and b2:
4620981.03
0386294.1
14
1ln4ln)()(
01
011
xx
xfxfb
0518731.0
16
4620981.046
386294.1791759.1)()()()(
02
01
01
12
12
2
xx
xxxfxf
xxxfxf
b
0ln1)( 00 xfb
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 1414
Example 2 - Solution
The quadratic polynomial is then: f2(x) = 0 + 0.4620981(x-1) – 0.0518731(x-1)(x-4)
Let’s now evaluate f2(x) at x=2
f2(x) = 0.5658444
and the relative error εt = 18.4%
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 1515
General Form of Newton’s Interpolation
The analysis used in linear and quadratic interpolation can be generalized to fit an (n-1)th order polynomial to n data points.
The data points are used to evaluate the coefficients :
))...()((...)()( 21010101 nnn xxxxxxbxxbbxf
],,...,,[
.
.
],,[
],[
)(
01211
0122
011
00
xxxxfb
xxxfb
xxfb
xfb
nnn
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 1616
General Form of Newton’s Interpolation
The bracketed function evaluations are finite divided differences
The general form of Newton’s interpolating polynomial:
01
0321210121
],...,,[],...,,[],,...,,[
],[],[],,[
)()(],[
xx
xxxfxxxfxxxxf
xx
xxfxxfxxxf
xx
xfxfxxf
n
nnnnnn
ki
kjjikji
ji
jiji
0121210
0121001001
,,...,,))...()((...
,,))((,)()(
xxxxfxxxxxx
xxxfxxxxxxfxxxfxf
nnn
n
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 1717
General Form of Newton’s Interpolation
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 1818
Example 3
Fit a third-order Newton’s interpolating polynomial to the four points used to evaluate the natural logarithm of 2
i.e.
x0 = 1 f (x0) = 0
x1 = 4 f (x1) = 1.386294
x2 = 6 f (x2) = 1.791759
x3 = 5 f (x3) = 1.609438
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 1919
Example 3 - Solution
The third-order polynomial is))()(())(()()( 210310201014 xxxxxxbxxxxbxxbbxf
i xi f(xi) First Second Third
0 1 0 f[x1,x0]= 0.462098 f[x2,x1,x0]= -0.0518731 f[x3,x2,x1,x0]= 0.00786541 4 1.386294 f[x2,x1]= 0.2027325 f[x3,x2,x1]= -0.02041152 6 1.791759 f[x3,x2]= 0.1823213 5 1.609438
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 2020
Example 3 - Solution
The values of the coefficients of the polynomial are:
Therefore, the third order polynomial is
Then
0078655.0],,,[
051873.0],,[
462098.0],[
0)(
01232
0122
011
00
xxxxfb
xxxfb
xxfb
xfb
))()((0078655.0))((051873.0)(462098.00)( 2101003 xxxxxxxxxxxxxf
%3.9 6287686.0)2(3 tf
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 2121
Lagrange Interpolating Polynomials
Lagrange interpolating polynomial is a reformulation of the Newton polynomial without the computation of divided differences.
where
Π represents “the product of”.
)()()(0
i
n
iin xfxLxf
n
ijj ji
ji xx
xxxL
0
)(
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 2222
Lagrange Interpolating Polynomials
For n = 1 i.e. linear (1st order) version:
For n = 2 i.e. quadratic (2nd order) version:
)()()( 101
00
10
11 xf
xx
xxxf
xx
xxxf
)())((
))((
)())((
))((
)())((
))(()(
21202
10
12101
20
02010
212
xfxxxx
xxxx
xfxxxx
xxxx
xfxxxx
xxxxxf
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 2323
Example 4
Use a Lagrange interpolating polynomial of the first and second order to evaluate ln2 given the following data:
x0 = 1 f (x0) = 0
x1 = 4 f (x1) = 1.386294
x2 = 6 f (x2) = 1.791759
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 2424
Example 4 - Solution
Using first order Lagrange polynomial:
Using second order Lagrange polynomial:
The results are similar to those of Newton’s interpolation.
4620981.0386294.114
120
41
42)(1
xf
5658444.0791759.1)46)(16(
)42)(12(
386294.1)64)(14(
)62)(12(
0)61)(41(
)62)(42()(2
xf
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 2525
Quiz #2 (Section 1)
Given the data
Calculate f (4) using Newton’s interpolating polynomials of order 3.
x 1 2 3 5 6
f (x) 4.75 4 5.25 19.75 36
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 2626
Quiz #5 (Section 2)
Given the data
Calculate f (4) using 1- Newton’s interpolating polynomials of order 1 and 2.2- The Lagrange polynomial of order 1 and 2.
x 1 2 3 5 6
f (x) 4.75 4 5.25 19.75 36
Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 2727
Assignment #3
Computational Methods
Statistics