MATH-415 Numerical Analysis
INTERPOLATION.
INTERPOLATING POLYNOMIALS
1
What is Interpolation ?
Given (x0,y0), (x1,y1), …… (xn,yn), find the value of y at a
value of x that is not given.
Linear Interpolation
3
Interpolants
Polynomials are the most common choice
of interpolants because they are easy to:
Evaluate
Differentiate
Integrate
4
Interpolation
• Thus interpolation is the estimation of intermediate
values between precise data points. The most
common method is:
f ( x)  a0  a1 x  a2 x 2    an x n
• Although there is one and only one nth-order
polynomial that fits n+1 points, there are different
mathematical formats in which this polynomial can
be expressed. The most popular are:
 The Newton polynomial
 The Lagrange polynomial
5
Direct Polynomial Interpolation
Given n+1 data points (x0,y0), (x1,y1),………….. (xn,yn),
pass a polynomial of order n through the data as given
below:
y  a0  a1 x  ....................  an x .
n
where a0, a1,………………. an are real constants.
• Set up n+1 equations to find n+1 constants.
• To find the value y at a given value of x, simply substitute the
value of x in the above polynomial.
http://numericalmethods.eng.usf.
edu
6
Coefficients of an Interpolating
Polynomial
• Since n+1 data points are required to determine n+1
coefficients, simultaneous linear systems of
equations can be used to calculate “a”s:
f ( x0 )  a0  a1 x0  a2 x02   an x0n
f ( x1 )  a0  a1 x1  a2 x12   an x1n

f ( xn )  a0  a1 xn  a2 xn2   an xnn
• where “x”s are the knowns and “a”s are the
unknowns
by Lale Yurttas, Texas A&M
University
7
Example 1
The upward velocity of a rocket is given as a function
of time in Table 1.
Find the velocity at t=16 seconds using the direct
method for linear interpolation.
Table 1 Velocity as a function
of time.
t , s 
vt , m/s 
0
0
10
227.04
15
362.78
20
517.35
22.5
602.97
30
901.67
Figure 2 Velocity vs. time data for the
rocket example
http://numericalmethods.eng.usf.
edu
8
Example 1: Solution
vt   a0  a1t
y
v15  a0  a1 15  362.78
v20  a0  a1 20  517.35
x1 , y1 
x0 , y0 
f1 x 
Solving the above two equations gives,
a0  100.93
a1  30.914
x
Figure 3 Linear interpolation.
Hence
vt   100.93  30.914t, 15  t  20.
v16  100.93  30.91416  393.7 m/s
http://numericalmethods.eng.usf.
edu
9
Newton’s Divided-Difference
Interpolating Polynomials
BC DE

AB AD
BC  f1 ( x)  f ( x0 )
ABC
E
ADE 
AB  x  x0
DE  f ( x1 )  f ( x0 )
C
AD  x1  x0
A
B
D
f1 ( x)  f ( x0 ) f ( x1 )  f ( x0 )

x  x0
x1  x0
10
Newton’s Divided-Difference
Interpolating Polynomials
Linear Interpolation
• Is the simplest form of interpolation, connecting two data
points with a straight line.
f1 ( x)  f ( x0 ) f ( x1 )  f ( x0 )

x  x0
x1  x0
Slope and a
finite divided
difference
approximation
to 1st derivative
f ( x1 )  f ( x0 )
f1 ( x)  f ( x0 ) 
( x  x0 ) Linear-interpolation
x1  x0
formula
• f1(x) designates that this is a first-order interpolating
polynomial
11
Newton’s Divided-Difference
Interpolating Polynomials
Quadratic Interpolation
• If three data points are available, the estimate is improved by
introducing some curvature into the line connecting the
points: f 2 ( x)  b0  b1 ( x  x0 )  b2 ( x  x0 )( x  x1 )
• A simple procedure can be used to determine the
values of the coefficients:
x  x0 b0  f  x0 
f  x1   f  x0 
xx
b 
1
x  x2
1
x1  x0
f  x2   f  x1  f  x1   f  x0 

x2  x1
x1  x0
b2 
x2  x0
12
General Form of Newton’s Interpolating
Polynomials
f n ( x)  f ( x0 )  ( x  x0 ) f [ x1 , x0 ]  ( x  x0 )( x  x1 ) f [ x2 , x1 , x0 ]
   ( x  x0 )( x  x1 )  ( x  xn 1 ) f [ xn , xn 1 , , x0 ]
b0  f ( x0 )
b1  f [ x1 , x0 ]
b2  f [ x2 , x1 , x0 ]

bn  f [ xn , xn 1 ,  , x1 , x0 ]
f [ xi , x j ] 
f ( xi )  f ( x j )
f [ xi , x j , xk ] 
xi  x j
f [ xi , x j ]  f [ x j , xk ]
Bracketed function
evaluations are finite
divided differences
xi  xk

f [ xn , xn 1 ,  , x1 , x0 ] 
f [ xn , xn 1 , , x1 ]  f [ xn 1 , xn  2 , , x0 ]
xn  x0
13
Lagrange Interpolating Polynomials
• The Lagrange interpolating polynomial is simply a
reformulation of the Newton’s polynomial that
avoids the computation of divided differences:
n
f n ( x)   Li ( x) f ( xi )
i 0
n
Li ( x)  
j 0
j i
by Lale Yurttas, Texas A&M
University
x  xj
xi  x j
14
Lagrange Interpolating Polynomials
Lagrangian interpolating polynomial is given by
n
f n ( x)   Li ( x) f ( xi )
i 0
where ‘ n ’ in f n (x) stands for the n th order polynomial that approximates the function y  f (x)
given at (n  1) data points as  x0 , y 0 , x1 , y1 ,......,  x n 1 , y n 1 ,  x n , y n  , and
n
Li ( x)  
j 0
j i
x  xj
xi  x j
Li (x) is a weighting function that includes a product of (n  1) terms with terms of j  i
omitted.
http://numericalmethods.eng.usf.
edu
15
Lagrange Interpolating Polynomials
x  x0
x  x1
f1 ( x) 
f ( x0 ) 
f ( x1 )
x0  x1
x1  x0
L0
x  x1  x  x2 
x  x0  x  x2 


f 2 ( x) 
f ( x0 ) 
f ( x1 )
 x0  x1  x0  x 2 
 x1  x0  x1  x 2 
x  x0  x  x1 


f ( x2 )
 x2  x0  x2  x 1 
L
1
L2
16
Lagrange Interpolating Polynomials
17
Error of Interpolation
• For an nth-order interpolating polynomial, an
analogous relationship for the error is:
f ( n1) (x )
Rn 
( x  x0 )( x  x1 ) ( x  xn )
(n  1)!
• x is located in some interval containing the
unknown and the data:  x, x0 , x1 ,..., xn 
18
Example
The upward velocity of a rocket is given as a function of time
in Table 1. Find the velocity at t=16 seconds using the
Lagrangian method for linear interpolation.
Table Velocity as a
function of time
t (s) v(t ) (m/s)
0
10
15
20
22.5
30
0
227.04
362.78
517.35
602.97
901.67
Figure. Velocity vs. time data
for the rocket example
http://numericalmethods.eng.usf.
edu
19
Linear Interpolation
517.35
550
1
v(t )   Li (t )v(ti )
500
i 0
 L0 (t )v (t 0 )  L1 (t )v (t1 )
ys
f ( range)

f x desired

t 0  15, t 0   362.78
t1  20, t1   517.35
450
400
362.78
350
10
12
x s  10
0
http://numericalmethods.eng.usf.
edu
14
16
18
20
22
24
x s  range  x desired
x s  10
1
20
Linear Interpolation (contd)
1
L0 (t )  
j 0
j 0
1
L1 (t )  
j 0
j 1
v(t ) 
v(16) 
t tj
t0  t j
t tj
t1  t j

t  t1
t 0  t1

t  t0
t1  t 0
t  t0
t  t1
t  20
t  15
v(t 0 ) 
v(t1 ) 
(362.78) 
(517.35)
t 0  t1
t1  t 0
15  20
20  15
16  20
16  15
(362.78) 
(517.35)
15  20
20  15
 0.8(362.78)  0.2(517.35)
 393.7
m/s.
http://numericalmethods.eng.usf.
edu
21
Quadratic Interpolation
For the second order polynomial interpolatio n (also called quadratic interpolation), we
choose the velocity given by
2
v (t )   Li ( t ) v(t i )
i 0
 L0 (t )v (t 0 )  L1 (t ) v( t1 )  L2 (t ) v( t 2 )
http://numericalmethods.eng.usf.
edu
22
Example
The upward velocity of a rocket is given as a function of time
in Table 1. Find the velocity at t=16 seconds using the
Lagrangian method for quadratic interpolation.
Table Velocity as a
function of time
t (s) v(t ) (m/s)
0
10
15
20
22.5
30
0
227.04
362.78
517.35
602.97
901.67
Figure. Velocity vs. time data
for the rocket example
http://numericalmethods.eng.usf.
edu
23
Quadratic Interpolation (contd)
t 0  10, v(t 0 )  227.04
517.35
t1  15, v(t1 )  362.78
550
500
t 2  20, v(t 2 )  517.35
450
ys
400
2
L0 (t )  
j 0
j 0
2
L1 (t )  
j 0
j 1
2
L2 (t )  
j 0
j 2
t tj
t0  t j
t t j
t1  t j
t tj
t2  t j
 t  t1  t  t 2 


 
t

t
t

t
 0 1  0 2 
 t  t0
 
 t1  t 0
 t  t 2

 t1  t 2
f ( range)

f x desired

300



 t  t 0  t  t1 


 
t

t
t

t
 2 0  2 1 
350
250
227.04
200
10
12
10
http://numericalmethods.eng.usf.
edu
14
16
18
20
x s  range  x desired
20
24
Quadratic Interpolation (contd)
 t  t1  t  t2 
 t  t0   t  t 2 
 t  t0   t  t1 
v t   
v
t

v
t






 0 


 1 
 v  t2 
t

t
t

t
t

t
t

t
t

t
t

t
 0 1  0 2 
 1 0  1 2 
 2 0  2 1 
 16  15  16  20 
 16  10  16  20 
 16  10   16  15 
v 16   
227.04

362.78












  517.35 
 10  15  10  20 
 15  10  15  20 
 20  10  20  15 
  0.08  227.04    0.96  362.78    0.12  527.35 
 392.19 m/s
The absolute relative approximate error a obtained between the
results from the first and second order polynomial is
392.19  393.70
 100
392.19
 0.38410%
a 
http://numericalmethods.eng.usf.
edu
25
Cubic Interpolation
For the third order polynomial (also called cubic interpolation), we choose the velocity given by
3
v (t )   Li ( t ) v(t i )
i 0
 L0 (t ) v( t 0 )  L1 ( t ) v(t 1 )  L2 ( t ) v(t 2 )  L3 ( t ) v(t 3 )
602.97
700
600
ys
500
f ( range)

f x desired

400
300
227.04
200
10
10
12
14
16
18
20
22
x s  range  x desired
http://numericalmethods.eng.usf.
edu
24
22.5
26
Cubic Interpolation (contd)
t o  10, v t o   227.04
t1  15, v t1   362.78
t 2  20, v t 2   517.35
t 3  22.5, v t 3   602.97
3
L0 (t )  
j 0
j 0
t tj
t0  t j
 t  t 1  t  t 2  t  t 3 


 ;
 
t

t
t

t
t

t
 0 1  0 2  0 3 
602.97
700
600
3
L1 (t )  
j 0
j 1
3
L2 (t )  
j 0
j 2
3
L3 ( t )  
j 0
j 3
t t j
t1  t j
 t  t0
 
 t1  t 0
 t  t 2  t  t 3


t

t
 1 2  t1  t 3



ys
500
f ( range)
t tj
t2  t j
 t  t 0  t  t1  t  t 3


 
t

t
t

t
 2 0  2 1  t 2  t 3

 ;


f x desired

400
300
t tj
t3  t j
 t  t 0  t  t1  t  t 2 



 
t

t
t

t
t

t
 3 0  3 1  3 2 
227.04
200
10
12
10
http://numericalmethods.eng.usf.
edu
14
16
18
x s  range  x desired
20
22
24
22.5
27
Cubic Interpolation (contd)
 t  t1  t  t 2  t  t3 
 t  t0  t  t 2  t  t3 


vt1   

vt 2 

vt   
t

t
t

t
t

t
t

t
t

t
t

t
 0 1  0 2  0 3 
 1 0  1 2  1 3 
 t  t0  t  t1  t  t3 
 t  t1  t  t1  t  t 2 

vt 2   


vt3 

 
t

t
t

t
t

t
t

t
t

t
t

t
 2 0  2 1  2 3 
 3 1  3 1  3 2 
 16  15  16  20  16  22.5 
 16  10  16  20  16  22.5 
v16   


227.04   


362.78
 10  15  10  20  10  22.5 
 15  10  15  20  15  22.5 
 16  10  16  15  16  22.5 
 16  10  16  15  16  20 



517.35  


602.97 
20

10
20

15
20

22
.
5
22
.
5

10
22
.
5

15
22
.
5

20








  0.0416 227.04   0.832 362.78  0.312 517.35   0.1024602.97 
 392.06 m/s
The absolute relative approximate error a obtained between the
results from the second and third order polynomial is
392.06  392.19
a 
100
392.06
 0.033269%
http://numericalmethods.eng.usf.
edu
28
Comparison Table
Order of Polynomial
1
2
3
v(t=16) m/s
393.69
392.19
392.06
Absolute Relative
Approximate Error
--------
0.38410%
0.033269%
http://numericalmethods.eng.usf.
edu
29
Lagrangian Interpolation:
Algorithm
Function Lagrange(z,x,f)
% Returns an interpolated value at z
% x is an (n+1)-dimensional array of the given data points
% f is an (n+1)-dimensional array of the given function values
Interpolated_Value=0
for i=1:n+1
Lagrangian=1
for j=1:n+1
if(j~=i)
then Lagrangian=Lagrangian*(z-x(j))/(x(i)-x(j))
end if
end for j
Interpolated_Value=Interpolated_Value+Lagrangian*f(i)
end for I
Return Interpolated_Value
30
Spline Interpolation
• There are cases where polynomials can lead to
erroneous results because of round-off errors
or significant distinction between an
interpolating polynomial and an interpolated
function
• Alternative approach is to apply lower-order
polynomials to subsets of data points. Such
connecting polynomials are called spline
functions
31
Why Splines ?
1
f ( x) 
1  25 x 2
Table : Six equidistantly spaced points in [-1, 1]
x
y
1
1  25 x 2
-1.0
0.038461
-0.6
0.1
-0.2
0.5
0.2
0.5
0.6
0.1
1.0
0.038461
Figure : 5th order polynomial vs. exact function
http://numericalmethods.eng.usf.
edu
32
Why Splines ?
1.2
0.8
y
0.4
0
-1
-0.5
0
0.5
1
-0.4
-0.8
x
19th Order Polynomial
f (x)
5th Order Polynomial
Figure : Higher order polynomial interpolation is a bad idea
http://numericalmethods.eng.usf.
edu
33
Linear Spline Interpolation
Given  x0 , y0 , x1 , y1 ,......, x n 1 , y n1  x n , y n  , fit linear splines to the data. This simply involves
forming the consecutive data through straight lines. So if the above data is given in an ascending
order, the linear splines are given by  yi  f ( xi ) 
Figure : Linear splines
http://numericalmethods.eng.usf.
edu
34
Quadratic Spline Interpolation
Given  x0 , y0 ,  x1 , y1 ,......, x n 1 , y n 1 ,  x n , y n  , fit quadratic splines through the data. The splines
are given by
f ( x )  a1 x 2  b1 x  c1 ,
 a 2 x 2  b2 x  c2 ,
x 0  x  x1
x1  x  x 2
.
.
.
 a n x 2  bn x  cn ,
x n 1  x  x n
Find a i , bi , ci , i  1, 2, …, n
http://numericalmethods.eng.usf.
edu
35
Spline Interpolation
36
Spline Interpolation
37
38