NUMERICAL METHODS IN ENGINEERING ENGR 391
Transcript of NUMERICAL METHODS IN ENGINEERING ENGR 391
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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
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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)(
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Introduction
First order (linear) 3rd order (cubic) 2nd order (quadratic)
n = 2 n = 3 n = 4
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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)
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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”
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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
)(
)()()(
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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
)(
)()()(
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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
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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
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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
)(
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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
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Lagrange Interpolating Polynomials – Example (cont’d)
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Pseudocode – Lagrange interpolation
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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)(
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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.
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16
Interpolantion
Polynomials are the most common choice of interpolation because they are easy to:
Evaluate
Differentiate, and
Integrate.
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Possible divergence of an extrapolated production
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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.
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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
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Why Splines ?
Figure : Higher order polynomial interpolation is a bad idea
Original
Function
17th
Order
Polynomial
9th
Order
Polynomial
5th
Order
Polynomial
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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.
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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)
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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
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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
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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.
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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)
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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)
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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)
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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
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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
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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
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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
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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
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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
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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
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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
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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