Numerical and Statistical Methods for Civil Engineering

1. Government Engineering College, Bhavnagar.
Civil Engineering Department

2. Topic:-
Interpolation
NUMERICAL AND STATISTICAL METHODS
FOR CIVIL ENGINEERING (2140606).

3. Contents
 Introduction
 Newton’s Divided-Difference Interpolating Polynomials
 Error Estimation in Newton’s Interpolating
Polynomials
 Lagrange Interpolating Polynomials
 Image Interpolation - Theory

4. Introduction
Interpolation : Estimation of a function value at an intermediate point
that lie between precise data points.
 There is one and only one nth-order polynomial that perfectly fits n+1 data points:
 There are several methods to find the fitting polynomial:
the Newton polynomial and the Lagrange polynomial (unequal
interval)

5. Newton’s Divided-Difference Interpolating
Polynomials
Linear Interpolation
Connecting two data points with a straight line
f1(x) designates a first-order interpolating polynomial.
)()()()(
01
01
0
01
xx
xfxf
xx
xfxf





Linear-
interpolation
formula
Slope
)(
)()(
)()( 0
01
01
01 xx
xx
xfxf
xfxf 




6. Quadratic Interpolation
• If three (3) data points are available, the estimate is improved by introducing some curvature into the line
connecting the points.
A second-order polynomial (parabola) can be used for this purpose
• A simple procedure can be used to determine the values of the coefficients
))(()()( 1020102 xxxxbxxbbxf 
)( 000 xfbxx  Could you
figure out how
to derive this
using the above
equation?
Represents a
second order
polynomial
)()(
01
01
11
xx
xfxf
bxx



02
01
01
12
12
22
xx
xx
xfxf
xx
xfxf
bxx







)()()()(

7. ))(()()( 102010 xxxxbxxbbxf 
))((
)(
)()(
)()(
))((
)()()(
)()(
)(
1202
02
01
01
02
2
1202
02102
22
01
01
100
xxxx
xx
xx
xfxf
xfxf
b
xxxx
xxbxfxf
bxx
xx
xfxf
bxfb












))((
)(
)))(()((
)(
)))(()((
1202
01
1201
12
1212
2
xxxx
xx
xxxfxf
xx
xxxfxf
b







)(
)(
)()(
)(
)()(
02
01
01
12
12
2
xx
xx
xfxf
xx
xfxf
b







))((
)(
)))(()(()))(()((
))()((
)(
)))(()((
1202
01
01011201
01
12
1212
2
xxxx
xx
xxxfxfxxxfxf
xfxf
xx
xxxfxf
b







))((
)(
)))(()((
)(
)))(()()()((
1202
01
011201
12
120112
22
xxxx
xx
xxxxxfxf
xx
xxxfxfxfxf
bxx








8. General Form of Newton’s Interpolating
Polynomials
)()(
],[
ji
ji
ji
xx
xfxf
xxf



Bracketed function
evaluations are finite
divided differences
],,,,[],,[],[)(
)())(())(()()(
011012201100
110102010
xxxxfbxxxfbxxfbxfb
xxxxxxbxxxxbxxbbxf
nnn
nnn






],[],[
],,[
ki
kjji
kji
xx
xxfxxf
xxxf



0
02111
011
xx
xxxfxxxf
xxxxf
n
nnnn
nn


 

],,,[],,,[
],,,,[




9. xi f(xi)
x0 f(x0)
x1 f(x1)
x2 f(x2)
x3 f(x3)
x4 f(x4)
xi f(xi)
x0=0 2
x1=2 14
x2=3 74
x3=4 242
x4=5 602
f1(x) = 2 + 6*(x-0) (based on x0 and x1)
f2(x) = 2 + 6*(x-0)+18(x-0)(x-2) (based on x0, x1 and x2)
f3(x) = 2 + 6*(x-0)+18(x-0)(x-2)+9(x-0)(x-2)(x-3) (based on x0, x1, x2, and x3)
f4(x) = 2 + 6x +18x(x-2) +9x(x-2)(x-3) +1x(x-2)(x-3)(x-4) (based on x0, x1, x2, x3, and x4)
= x4 – x2 + 2
EXAMPLEDIVIDED DIFFERENCE TABLE
f[xi,xj]
6
60
168
360
f[xi,xj,xk]
18
54
96
f[x,x,x,x]
9
14
f[x...x]
1
f[xi,xj]
f[x1,x0]
f[x2,x1]
f[x3,x2]
f[x4,x3]
f[xi,xj,xk]
f[x2,x1,x0]
f[x3,x2,x1]
f[x4,x3,x2]
f[x,x,x,x]
f[x3,x2,x1,x0]
f[x4,x3,x2,x1]

10. Given:
x0=1 f(x0)=ln(1) = 0
x1=e f(x1)=ln(2.72) = 1
x2=e2 f(x2)=ln(7.39) = 2
Estimate ln(2) = ?
using interpolation
Find f(x) first
xi f(xi)
x0=1 0
x1=2.72 1
x2=7.39 2
f[xi,xj]
.58
.214
f[xi,xj xk]
-.057
f(x) = 0.58(x-1)
-0.057(x-1)(x-
2.72)
Then calculate
f(2)=0.58(2-1)-0.057(2-1)(2-
2.72)
= 0.621
[ TRUE ln(2) = 0.6931 ]
Example

11. Lagrange Interpolating Polynomials
• The Lagrange interpolating polynomial is simply a reformulation of the Newton’s polynomial that avoids the
computation of divided differences:
• Above formula can be easily verified by plugging in x0, x1…in the equation one at a time and checking if the
equality is satisfied.









n
ij
j ji
j
i
n
i
iin
xx
xx
xL
xfxLxf
0
0
)(
)()()(
)()()( 1
01
0
0
10
1
1 xf
xx
xx
xf
xx
xx
xf






  
  
  
  
  
  
)(
)(
)()(
2
1202
10
1
2101
20
0
2010
21
2
xf
xxxx
xxxx
xf
xxxx
xxxx
xf
xxxx
xxxx
xf










12. A visual depiction of the rationale behind
the Lagrange polynomial . The figure
shows a second order case:
Each of the three terms passes through
one of the data points and zero at the
other two. The summation of the three
terms must, therefore, be unique second
order polynomial f2(x) that passes exactly
through three points.
  
  
  
  
  
  
)(
)(
)()(
2
1202
10
1
2101
20
0
2010
21
2
xf
xxxx
xxxx
xf
xxxx
xxxx
xf
xxxx
xxxx
xf










13. Image Interpolation - Theory
 [IDEA]
 In order to provide a richer environment we are thinking of using
interpolation methods that will generate “artificial images” thus revealing
hidden information.
 [RADON RECONSTRUCTION]
 Radon reconstruction is the technique in which the object is reconstructed
from its projections. This reconstruction method is based on approximating
the inverse Radon Transform.
 [RADON Transform]
 The 2-D Radon transform is the mathematical relationship which maps the
spatial domain (x,y) to the Radon domain (p,phi). The Radon transform
consists of taking a line integral along a line (ray) which passes through the
object space. The radon transform is expressed mathematically as:



 dxdypyxyxpR )sincos(),(),}({ 

14. Image Interpolation - Graphical
Representation (I)





l
z
dy
y
z
dxzyxyR
dyzyxxR
0
0
0
),,()0,(
),,()90,(
0
0
0
0



15. Image Interpolation - Graphical
Representation (II)

16. BHAVIK SHAH – 130210106049
DIGVIJAY SOLANKI – 130210106055
KARTIK HINGOL – 130210106030
NITIN CHAREL – 130210106011
Thank You For Bearing