The Islamic University of Gaza
Faculty of Engineering
Civil Engineering Department
Numerical Analysis
ECIV 3306
Chapter 18
Interpolation
Associate Prof. Mazen Abualtayef Civil Engineering Department, The Islamic University of Gaza
Introduction
Estimation of intermediate values between precise data points. The most common method is polynomial interpolation:
Polynomial interpolation is used when the point determined are very precise. The curve representing the behavior has to pass through every point.
There is one and only one nth-order polynomial that fits n+1 points
n
nxaxaxaaxf 2
210)(
Introduction
First order (linear) 3rd order (cubic) 2nd order (quadratic)
n = 2 n = 3 n = 4
Introduction
There are a variety of mathematical formats in
which this polynomial can be expressed:
The Newton polynomial (Section 18.1)
The Lagrange polynomial (Section 18.2)
Lagrange Interpolating Polynomials
• The general form for n+1 data points is:
n
ijj ji
j
i
n
i
iin
xx
xxxL
xfxLxf
0
0
)(
)()()(
designates the “product of”
Lagrange Interpolating Polynomials
)()()( 1
01
00
10
11 xf
xx
xxxf
xx
xxxf
• Linear version (n = 1):
Used for 2 points of data: (xo,f(xo)) and (x1,f(x1)),
)(xLo )(1 xL
n
ijj ji
j
i
n
i
iin
xx
xxxL
xfxLxf
0
0
)(
)()()(
Lagrange Interpolating Polynomials
1,)(1 jxL
)(
)(
)()(
2
1202
10
1
2101
20
0
2010
212
xfxxxx
xxxx
xfxxxx
xxxx
xfxxxx
xxxxxf
2,)(2 jxL
• Second order version (n = 2):
0,)( jxLo
n
ijj ji
j
i
n
i
iin
xx
xxxL
xfxLxf
0
0
)(
)()()(
Lagrange Interpolating Polynomials Example 18.6
Use a Lagrange interpolating polynomial of the first and second order to evaluate ln(2) on the basis of the data:
10 x 0)1ln()( 0 xf
41 x
62 x
386294.1)4ln()( 1 xf
791760.1)6ln()( 2 xf
Lagrange Interpolating Polynomials – Example 18.6 (cont’d)
• First order polynomial:
)()()( 1
01
0
0
10
1
1 xfxx
xxxf
xx
xxxf
4620981.0386294.114
120
41
42)2(1
f
Lagrange Interpolating Polynomials – Example (cont’d)
• Second order polynomial:
60
6x
40
4x
xx
xx
xx
xxxL
2o
2
1o
1o
)(
64
6x
04
0x
xx
xx
xx
xxxL
21
2
o1
o1
)(
46
4x
06
0x
xx
xx
xx
xxxL
12
1
o2
o2
)(
Lagrange Interpolating Polynomials – Example (cont’d)
5658444.0 791760.1)46)(16(
)42)(12(
386294.1)64)(14(
)62)(12(
0)61)(41(
)62)(42()2(2
f
n
0i
iin xfxLxf )()()( )()( ijxx
xxxL
n
0j ji
j
i
Lagrange Interpolating Polynomials – Example (cont’d)
Pseudocode – Lagrange interpolation
Coefficients of an Interpolating Polynomial
• Although “Lagrange” polynomials are well suited for determining intermediate values between points, they do not provide a polynomial in conventional form:
• Since n+1 data points are required to determine n+1 coefficients, simultaneous linear systems of equations can be used to calculate “a”s.
n
xxaxaxaaxf 2
210)(
Coefficients of an Interpolating Polynomial (cont’d)
n
nnnnn
n
n
n
n
xaxaxaaxf
xaxaxaaxf
xaxaxaaxf
2
210
1
2
121101
0
2
020100
)(
)(
)(
Where “x”s are the knowns and “a”s are the
unknowns.
16
Interpolantion
Polynomials are the most common choice of interpolation because they are easy to:
Evaluate
Differentiate, and
Integrate.
Possible divergence of an extrapolated production
18
Why Spline Interpolation?
Apply lower-order polynomials to subsets of data points. Spline
provides a superior approximation of the behavior of functions that
have local, abrupt changes.
19
Why Splines ?
2251
1)(
xxf
Table : Six equidistantly spaced points in [-1, 1]
Figure : 5th order polynomial vs. exact function
x 2251
1
xy
-1.0 0.038461
-0.6 0.1
-0.2 0.5
0.2 0.5
0.6 0.1
1.0 0.038461
20
Why Splines ?
Figure : Higher order polynomial interpolation is a bad idea
Original
Function
17th
Order
Polynomial
9th
Order
Polynomial
5th
Order
Polynomial
Spline Interpolation
• Polynomials are the most common choice of interpolants.
• There are cases where polynomials can lead to erroneous results because of round off error and overshoot.
• Alternative approach is to apply lower-order polynomials to subsets of data points. Such connecting polynomials are called spline functions.
Spline Interpolation
The concept of spline is using a thin, flexible strip (called a spline) to draw smooth curves through a set of points….natural spline (cubic)
Linear Spline
The first order splines for a group of ordered data points can be defined as a set of linear functions:
ii
iii
xx
xfxfm
1
1 )()(
)()()( 000 xxmxfxf 10 xxx
)()()( 111 xxmxfxf 21 xxx
)()()( 111 nnn xxmxfxf nn xxx 1
Linear spline - Example
Fit the following data with first order splines. Evaluate the function at x = 5.
x f(x)
3.0 2.5
4.5 1.0
7.0 2.5
9.0 0.5
6.05.47
15.2
m
3.1
5.06.00.1
)5.45()5.4()5(
mff
Linear Spline
• The main disadvantage of linear spline is that they are not smooth. The data points where 2 splines meets called (a knot), the changes abruptly.
• The first derivative of the function is discontinuous at these points.
• Using higher order polynomial splines ensure smoothness at the knots by equating derivatives at these points.
Quadric Splines
iiii cxbxaxf 2)(
• Objective: to derive a second order polynomial for each
interval between data points.
• Terms: Interior knots and end points
For n+1 data points:
• i = (0, 1, 2, …n), • n intervals,
• 3n unknown
constants (a’s, b’s and c’s)
Quadric Splines
• The function values of adjacent polynomial must be equal at the interior knots 2(n-1).
• The first and last functions must pass through the end points (2).
nixfcxbxa
nixfcxbxa
iiiiiii
iiiiiii
,...,4,3,2)(
,...,4,3,2)(
11
2
1
1111
2
11
)(
)(
2
0101
2
01
nnnnnn xfcxbxa
xfcxbxa
Quadric Splines
• The first derivatives at the interior knots must be equal (n-1).
• Assume that the second derivate is zero at
the first point (1)
(The first two points will be connected by a straight line)
iiiiii
iii
bxabxa
bxaxf
1111
'
22
2)(
01 a
Quadric Splines - Example
Fit the following data with quadratic splines. Estimate the value at x = 5.
Solutions:
There are 3 intervals (n=3), 9 unknowns.
x 3.0 4.5 7.0 9.0
f(x) 2.5 1.0 2.5 0.5
Quadric Splines - Example
1. Equal interior points:
For first interior point (4.5, 1.0)
The 1st equation:
The 2nd equation:
)( 1221221 xfcbxax
0.15.425.20 111 cba
)( 1111121 xfcbxax
)5.4(5.4)5.4( 1112 fcba
0.15.425.20 222 cba)5.4(5.4)5.4( 2222 fcba
Quadric Splines - Example
For second interior point (7.0, 2.5)
The 3rd equation:
The 4th equation:
5.2749 222 cba
5.2749 333 cba
)( 22222
2
2 xfcbxax
)7(7)7( 222
2 fcba
)( 23323
2
2 xfcbxax
)7(7)7( 333
2 fcba
Quadric Splines - Example
First and last functions pass the end points
For the start point (3.0, 2.5)
For the end point (9, 0.5)
5.239 111 cba
5.0981 333 cba
)( 01101
2
0 xfcbxax
)( 33333
2
3 xfcbxax
Quadric Splines - Example
Equal derivatives at the interior knots.
For first interior point (4.5, 1.0)
For second interior point (7.0, 2.5)
Second derivative at the first point is 0
0)( 10'' axf
2211 99 baba 221111 22 baxbax
3322 1414 baba 333222 22 baxbax
Quadratic Splines - Example
5.51
1
5.2
1
13
15.4 1
1
1
c
b
c
b
46.18
76.6
64.0
1
5.2
1
019
1749
15.425.20
2
2
2
2
2
2
c
b
a
c
b
a
Eq.(1) & Eq.(5)
Eq.(2) , Eq.(3) & Eq. (7)
391c
6024b
601a
22
50
52
c
b
a
0114
1981
1749
3
3
3
3
3
3
.
.
.
.
.
.Eq.(4) , Eq.(6) & Eq. (8)
Quadratic Splines - Example
0
0
5.0
5.2
5.2
5.2
1
1
0114011400
00001901
198100000
00000013
174900000
000174900
00015.425.2000
00000015.4
3
3
3
2
2
2
1
1
c
b
a
c
b
a
c
b
Quadric Splines - Example
Solving these 8 equations with 8 unknowns
3.91,6.24,6.1
46.18,76.6,64.0
5.5,1,0
333
222
111
cba
cba
cba
,5.5)(1 xxf 5.40.3 x
,46.1876.664.0)( 2
2 xxxf 0.75.4 x
,3.916.246.1)( 23 xxxf 0.90.7 x
x 3.0 4.5 7.0 9.0
f(x) 2.5 1.0 2.5 0.5
4 6 8
Cubic Splines
iiiii dxcxbxaxf 23)(
Objective: to derive a third order polynomial for
each interval between data points.
Terms: Interior knots and end points
For n+1 data points:
• i = (0, 1, 2, …n),
• n intervals,
• 4n unknown constants (a’s, b’s ,c’s and d’s)
Cubic Splines
• The function values must be equal at the interior knots (2n-2).
• The first and last functions must pass through the end points (2).
• The first derivatives at the interior knots must be equal (n-1).
• The second derivatives at the interior knots must be equal (n-1).
• The second derivatives at the end knots are zero (2), (the 2nd derivative function becomes a straight line at the end points)
Alternative technique to get Cubic Splines
• The second derivative within each interval [xi-1, xi ] is a
straight line. (the 2nd derivatives can be represented by first
order Lagrange interpolating polynomials.
1
1''
1
1
'''')()()(
ii
iii
ii
iiii
xx
xxxf
xx
xxxfxf
A straight line
connecting the first
knot f’’(xi-1) and the
second knot f’’(xi)
The second derivative at any point x within the interval
Cubic Splines
• The last equation can be integrated twice
2 unknown constants of integration can be evaluated by applying the boundary conditions:
1. f(x) = f (xi-1) at xi-1
2. f(x) = f (xi) at xi
1
1
''
1
11
''
1
1
3
1
1
''3
1
1
''
6
)()(
6
)()(
6
)(
6
)()(
iiiii
ii
ii
iiiii
ii
ii
i
ii
iii
ii
iii
xxxxxf
xx
xf
xxxxxf
xx
xf
xxxx
xfxx
xx
xfxf
)('' ixf
Unknowns:
)('' 1ixf
i = 0, 1,…, n
Cubic Splines
)()(6
)()(6
)()(
)()(2)()(
1
1
1
1
1
''
1
''
111
''
1
ii
ii
ii
ii
iii
iiiiii
xfxfxx
xfxfxx
xfxx
xfxxxfxx
• For each interior point xi (n-1):
This equation result with n-1 unknown second derivatives where, for boundary points:
f˝(xo) = f˝(xn) = 0
)()(''
1 iii xfxfi
Cubic Splines – Example 18.10
Fit the following data with cubic splines
Use the results to estimate the value at x=5.
Solution:
Natural Spline:
x 3.0 4.5 7.0 9.0
f(x) 2.5 1.0 2.5 0.5
0)9()(,0)3()( ''
3
''''
0
'' fxffxf
Cubic Splines - Example
For 1st interior point (x1 = 4.5)
-
-
-
Apply the following equation:
)()(6
)()(6
)()()()(2)()(
1
1
1
1
1
''
1
''
111
''
1
ii
ii
ii
ii
iiiiiiiii
xfxfxx
xfxfxx
xfxxxfxxxfxx
x 3.0 4.5 7.0 9.0
f(x) 2.5 1.0 2.5 0.5
5.10.35.4011 xxxx ii
5.25.47121 xxxx ii
40.370211 xxxx ii
Cubic Splines - Example
)15.2(5.1
6)15.2(
5.2
6)7(5.2)5.4(42)3(5.1 '''''' fff
0)3('' f
)1.(..............6.9)7(5.2)5.4(8 '''' eqff
x 3.0 4.5 7.0 9.0
f(x) 2.5 1.0 2.5 0.5
Since
For 2nd interior point (x2 = 7 )
5.25.47121 xxxx ii
5.45.491311 xxxx ii
279231 xxxx ii
Cubic Splines - Example
Apply the following equation:
)()(6
)()(6
)()()()(2)()(
1
1
1
1
1
''
1
''
111
''
1
ii
ii
ii
ii
iiiiiiiii
xfxfxx
xfxfxx
xfxxxfxxxfxx
)5.21(5.2
6)5.25.0(
2
6)9(2)7(5.42)5.4(5.2 '''''' fff
Since 0)9('' f
)2(.............6.9)7(9)5.4(5.2 '''' equff
Cubic Splines - Example
Solve the two equations:
The first interval (i=1), apply for the equation:
53308.1)7(,67909.1)5.4(6.9)7(9)5.4(5.2
6.9)7(5.2)5.4(8''''
''''
''''
ffyeild
ff
ff
i
ii
i
1
1
''
1
11
''
1
1
3
1
1
''3
1
1
''
6
)()(
6
)()(
6
)(
6
)()(
iiiii
ii
iii
iiii
ii
ii
i
ii
iii
ii
iii
xxxxxf
xx
xfxx
xxxf
xx
xf
xxxx
xfxx
xx
xfxf
)3(24689.0)5.4(6667.1)3(186566.0)( 31 xxxxf
)3(6
)5.1(67909.1
5.1
15.4
6
)5.1(0
5.1
5.2)3(
)5.1(6
67909.1)3(0)( 33
1
xxxxxf i
Cubic Splines - Example
)5.4(6
)5.2(53308.1
5.2
5.2
76
)5.2(67909.1
5.2
1)5.4(
)5.2(6
53308.1)7(
)5.2(6
67909.1)( 33
2
x
xxxxf
)5.4(638783.1)7(29962.0)5.4(102205.0)7(111939.0)( 33
2 xxxxxf
)7(25.0)9(761027.1)9(127757.0)( 3
3 xxxxf
102886.1)5()( 22 fxf
The 2nd interval (i =2), apply for the equation:
The 3rd interval (i =3),
For x = 5:
4 6 8
