Lagrangian Interpolation
Computer Engineering Majors
Authors: Autar Kaw, Jai Paul
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
http://numericalmethods.eng.usf.edu
1
Lagrange Method of
Interpolation
http://numericalmethods.eng.usf.edu
What is Interpolation ?
Given (x0,y0), (x1,y1), …… (xn,yn), find the
value of ‘y’ at a value of ‘x’ that is not given.
3
http://numericalmethods.eng.usf.edu
Interpolants
Polynomials are the most common
choice of interpolants because they
are easy to:
Evaluate
Differentiate, and
Integrate.
4
http://numericalmethods.eng.usf.edu
Lagrangian Interpolation
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.
5
http://numericalmethods.eng.usf.edu
Example
A robot arm with a rapid laser scanner is doing a quick quality check on
holes drilled in a rectangular plate. The hole centers in the plate that describe
the path the arm needs to take are given below.
If the laser is traversing from x = 2 to x = 4.25 in a linear path, find the
value of y at x = 4 using the Lagrange method for linear interpolation.
6
x (m)
y (m)
2
4.25
5.25
7.81
9.2
10.6
7.2
7.1
6.0
5.0
3.5
5.0
Figure 2 Location of holes on the
rectangular plate.
http://numericalmethods.eng.usf.edu
Linear Interpolation
7.2
1
y ( x)   Li ( x) y ( xi )
7.2
7.18
i 0
7.16
 L0 ( x) y( x0 )  L1 ( x) y( x1 )
ys
f ( range)


7.14
f x desired
7.12
x0  2.00, yx0   7.2
x1  4.25, yx1   7.1
7
7.1
7.1 7.08
5
x s  10
0
0
5
10
x s  range x desired
http://numericalmethods.eng.usf.edu
x s  10
1
Linear Interpolation (contd)
x  xj
x  x1

L0 ( x)  
x0  x1
x

x
j 0 0
j
1
j 0
1
L1 ( x)  
j 0
j 1
y ( x) 

x  xj
x1  x j

x  x0
x1  x0
x  x0
x  x1
y x0  
y  x1 
x0  x1
x1  x0
x  4.25
7.2  x  2.00 7.1, 2.00  x  4.25
2.00  4.25
4.25  2.00
4.00  4.25
4.00  2.00
(7.2) 
(7.1)
2.00  4.25
4.25  2.00
 0.11111(7.2)  0.88889(7.1)
 7.1111in.
y (4.00) 
8
http://numericalmethods.eng.usf.edu
Quadratic Interpolation
For the second order polynomial interpolation (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 )
9
http://numericalmethods.eng.usf.edu
Example
A robot arm with a rapid laser scanner is doing a quick quality check on
holes drilled in a rectangular plate. The hole centers in the plate that describe
the path the arm needs to take are given below.
If the laser is traversing from x = 2 to x = 4.25 in a linear path, find the
value of y at x = 4 using the Lagrange method for quadratic interpolation.
10
x (m)
y (m)
2
4.25
5.25
7.81
9.2
10.6
7.2
7.1
6.0
5.0
3.5
5.0
Figure 2 Location of holes on the
rectangular plate.
http://numericalmethods.eng.usf.edu
Quadratic Interpolation
xo  2.00, yxo   7.2
x1  4.25, yx1   7.1
7.56258
x2  5.25, yx2   6.0
8
7.5
ys
2
L0 ( x)  
j 0
j 0
x  xj
 x  x1  x  x 2

 
x0  x j  x0  x1  x0  x 2
x  xj
 x  x0
L1 ( x)  
 
j  0 x1  x j
 x1  x0
2
j 1
2
L2 ( x )  
j 0
j 2
11
x  xj



 x  x 2 


x

x
2 
 1
 x  x0  x  x1 


 
x 2  x j  x 2  x 0  x 2  x1 
f ( range)


7
f x desired
6.5
6
6
2
2
2.5
3
3.5
4
x s  range x desired
4.5
5
5.5
5.25
http://numericalmethods.eng.usf.edu
Quadratic Interpolation (contd)
 x  x1  x  x2 
 x  x0  x  x2 
 x  x0  x  x1 

 y( x0 )  


 y( x1 )  
 y( x2 )
y( x)  
x

x
x

x
x

x
x

x
x

x
x

x
1  0
2 
0  1
2 
0  2
1 
 0
 1
 2
y4.00 
4.00  4.254.00  5.25 7.2  4.00  2.004.00  5.25 7.1  4.00  2.004.00  4.25 6.0
2.00  4.252.00  5.25
4.25  2.004.25  5.25
5.25  2.005.25  4.25
 0.0427357.2  1.11117.1   0.153856.0
 7.2735 in.
The absolute relative approximate error a obtained between the results from the first and second order
polynomial is
7.2735 7.1111
a 
 100
7.2735
 2.2327 %
12
http://numericalmethods.eng.usf.edu
Comparison Table
13
Order of
Polynomial
1
2
Location (in.)
7.1111
7.2735
Absolute Relative
Approximate Error
----------
2.2327%
http://numericalmethods.eng.usf.edu
Example
A robot arm with a rapid laser scanner is doing a quick quality check on
holes drilled in a rectangular plate. The hole centers in the plate that describe
the path the arm needs to take are given below.
If the laser is traversing from x = 2 to x = 4.25 in a linear path, find the
value of y at x = 4 using a fifth order Lagrange polynomial.
x (m)
y (m)
2
4.25
5.25
7.81
9.2
10.6
7.2
7.1
6.0
5.0
3.5
5.0
Figure 2 Location of holes on the
rectangular plate.
14
http://numericalmethods.eng.usf.edu
Fifth Order Interpolation
5
y ( x)   Li ( x) y ( xi )
i 0
 L0 ( x) y( x0 )  L1 ( x) y( x1 )  L2 ( x) y( x2 )  L3 ( x) y( x3 )  L4 ( x) y( x4 )  L5 ( x) y( x5 )
xo  2.00, yxo   7.2
x1  4.25, yx1   7.1
x2  5.25, yx2   6.0
x3  7.81, yx3   5.0
x4  9.20, yx4   3.5
x5  10.60, yx5   5.0
15
http://numericalmethods.eng.usf.edu
Fifth Order Interpolation (contd)
5
L0 ( x)  
j 0
j 0
5
L1 ( x)  
j 0
j 1
5
L2 ( x)  
j 0
j 2
5
L3 ( x)  
j 0
j 3
5
L4 ( x)  
j 0
j 4
5
L5 ( x)  
j 0
j 5
16
x  xj
 x  x1  x  x2  x  x3  x  x4  x  x5 





 
x0  x j  x0  x1  x0  x2  x0  x3  x0  x4  x0  x5 
x  xj
 x  x0  x  x2  x  x3  x  x4  x  x5 





 
x1  x j  x1  x0  x1  x2  x1  x3  x1  x4  x1  x5 
x  xj
 x  x0  x  x1  x  x3  x  x4  x  x5 





 
x 2  x j  x2  x0  x2  x1  x2  x3  x2  x4  x2  x5 
x  xj
 x  x0  x  x1  x  x2  x  x4  x  x5 





 
x3  x j  x3  x0  x3  x1  x3  x2  x3  x4  x3  x5 
x  xj
 x  x0  x  x1  x  x2  x  x3  x  x5 





 
x 4  x j  x4  x0  x4  x1  x4  x2  x4  x3  x4  x5 
x  xj
 x  x0  x  x1  x  x2  x  x3  x  x4 





 
x5  x j  x5  x0  x5  x1  x5  x2  x5  x3  x5  x4 
http://numericalmethods.eng.usf.edu
Fifth Order Polynomial (contd)
 x  x1  x  x2  x  x3  x  x4  x  x5 




 y ( x0 )
y ( x)  
 x0  x1  x0  x2  x0  x3  x0  x4  x0  x5 
 x  x0  x  x2  x  x3  x  x4  x  x5 


 y ( x1 )


 
 x1  x0  x1  x2  x1  x3  x1  x4  x1  x5 
 x  x0  x  x1  x  x3  x  x4  x  x5 


 y ( x2 )


 
 x2  x0  x2  x1  x2  x3  x2  x4  x2  x5 
 x  x0  x  x1  x  x2  x  x4  x  x5 




 y ( x3 )
 
 x3  x0  x3  x1  x3  x2  x3  x4  x3  x5 
 x  x0  x  x1  x  x2  x  x3  x  x5 


 y ( x4 )


 
 x4  x0  x4  x1  x4  x2  x4  x3  x4  x5 
 x  x0  x  x1  x  x2  x  x3  x  x4 




 y ( x5 )
 
 x5  x0  x5  x1  x5  x2  x5  x3  x5  x4 
17
http://numericalmethods.eng.usf.edu
Fifth Order Polynomial (contd)
( x  4.25)(x  5.25)(x  7.81)(x  9.20)(x  10.60)
( 7.2 )
(2.00  4.25)(2.00  5.25)(2.00  7.81)(2.00  9.20)(2.00  10.60)
( x  2.00)(x  5.25)(x  7.81)(x  9.20)(x  10.60)

(7.1)
(4.25  2.00)(4.25  5.25)(4.25  7.81)(4.25  9.20)(4.25  10.60)
( x  2.00)(x  4.25)(x  7.81)(x  9.20)(x  10.60)

( 6.0)
(5.25  2.00)(5.25  4.25)(5.25  7.81)(5.25  9.20)(5.25  10.60)
( x  2.00)(x  4.25)(x  5.25)(x  9.20)(x  10.60)

(5.0)
(7.81 2.00)(7.81 4.25)(7.81 5.25)(7.81 9.20)(7.81 10.60)
( x  2.00)(x  4.25)(x  5.25)(x  7.81)(x  10.60)

(3.5)
(9.20  2.00)(9.20  4.25)(9.20  5.25)(9.20  7.81)(9.20  10.60)
( x  2.00)(x  4.25)(x  5.25)(x  7.81)(x  9.20)

(5.0)
(10.60  2.00)(10.60  4.25)(10.60  5.25)(10.60  7.81)(10.60  9.20)
y ( x) 
18
http://numericalmethods.eng.usf.edu
Fifth Order Polynomial (contd)
x 5  37.11x 4  536.77x 3  3773.2 x 2  12862x  16994
yx  
 365.38
x 5  34.86x 4  462.83x 3  2879.7 x 2  8169.5 x  7997.1

35.461
x 5  33.86x 4  433.22x 3  2572.3 x 2  6903.5 x  6473.9

 29.304
x 5  31.3 x 4  366.53x 3  1984.1x 2  4912.4 x  4351.8

41.069
x 5  29.91x 4  335.81x 3  1757.2 x 2  4241.6 x  3694.3

 78.273
x 5  28.51x 4  308.78x 3  1573.7 x 2  3727.5 x  3206.4

228.24
y( x)  30.898 41.344x  15.855x 2  2.7862x 3  0.23091x 4  0.0072923x 5 ,
2  x  10.6
19
http://numericalmethods.eng.usf.edu
Fifth Order Polynomial (contd)
y( x)  30.900 41.351x  15.858x 2  2.7865x 3  0.23093x 4  0.0072929x 5 ,
2  x  10.6
20
http://numericalmethods.eng.usf.edu
Additional Resources
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice
tests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, please
visit
http://numericalmethods.eng.usf.edu/topics/lagrange_
method.html
THE END
http://numericalmethods.eng.usf.edu