Chemical engineering Department Assist Lecturer Ghadeer Jassim

Lagrangian Interpolation

After reading this chapter, you should be able to:

1. derive Lagrangian method of interpolation,
2. solve problems using Lagrangian method of interpolation, and
3. use Lagrangian interpolants to find derivatives and integrals of discrete
functions.

Lagrangian Method
The 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 , y0 , x1 , y1 ,......, xn1 , yn1 , xn , yn  ,
and
n x  xj
Li ( x)  
j 0 xi  x j
j i

Li (x) is a weighting function that includes a product of n  1 terms with terms of

j  i omitted. The application of Lagrangian interpolation will be clarified using
an example.

x3 , y3 

x1, y1 

x2 , y2  f x 

x0 , y0 
x

Figure 1 Interpolation of discrete data.

Page | 145
Chemical engineering Department Assist Lecturer Ghadeer Jassim

Example 1
The upward velocity of a rocket is given as a function of time in Table 1.

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

Determine the value of the velocity at t  16 seconds using a first order Lagrange
polynomial.

Solution
For first order polynomial interpolation (also called linear interpolation), the
velocity is given by
1
v(t )   Li (t )v(t i )
i 0

 L0 (t )v(t 0 )  L1 (t )v(t1 )

x1 , y1 

f1 x 

x0 , y0 
x
Figure 2 Linear interpolation.

Since we want to find the velocity at t  16 , and we are using a first order
polynomial, we need to choose the two data points that are closest to t  16 that
also bracket t  16 to evaluate it. The two points are t 0  15 and t1  20 .

Page | 146
Chemical engineering Department Assist Lecturer Ghadeer Jassim

Then
t0  15, vt0   362.78
t1  20, vt1   517.35
gives
1 t tj t  t1
L0 (t )   
j 0 t0  t j t 0  t1
j 0

1 t tj t  t0
L1 (t )   
j 0 1 t tj t1  t 0
j 1

Hence
t  t1 t  t0
v(t )  v(t 0 )  v(t1 )
t 0  t1 t1  t 0
t  20 t  15
 (362.78)  (517.35), 15  t  20
15  20 20  15
16  20 16  15
v(16)  (362.78)  (517.35)  0.8(362.78)  0.2(517.35)  393.69 m/s
15  20 20  15

You can see that L0 (t )  0.8 and L1 (t )  0.2 are like weightages given to the
velocities at t  15 and t  20 to calculate the velocity at t  16 .

Quadratic Interpolation
y

x1 , y1 
x2 , y2 

f 2 x 

x0 , y0 
x
Figure 3 Quadratic interpolation.

Example 2
The upward velocity of a rocket is given as a function of time in Table 2.

Page | 147
Chemical engineering Department Assist Lecturer Ghadeer Jassim

Table 2 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

a) Determine the value of the velocity at t  16 seconds with second order

polynomial interpolation using Lagrangian polynomial interpolation.
b) Find the absolute relative approximate error for the second order
polynomial approximation.

Solution
a) For second order polynomial interpolation (also called quadratic
interpolation), the velocity is 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 )

Since we want to find the velocity at t  16 , and we are using a second order
polynomial, we need to choose the three data points that are closest to t  16 that
also bracket t  16 to evaluate it. The three points are t0  10, t1  15, and t 2  20 .
Then
t0  10, vt0   227.04
t1  15, vt1   362.78
t 2  20, vt 2   517.35
gives
2
t tj  t  t1  t  t 2 
L0 (t )      
t
j 0 0  t j 
 0 1  0 2 
t t t  t
j 0

2 t tj  t  t 0  t  t 2 
L1 (t )      
j 0 t1  t j  t1  t 0  t1  t 2 
j 1

Page | 148
Chemical engineering Department Assist Lecturer Ghadeer Jassim

2 t tj  t  t 0  t  t1 
L2 (t )      
j 0 t 2  t j  t 2  t 0  t 2  t1 
j 2

Hence
 t  t1  t  t 2   t  t 0  t  t 2   t  t 0  t  t1 
v(t )    v(t 0 )    v(t1 )    v(t 2 ), t 0  t  t 2

 0 1  0 2 
t t t  t t 
 1 0  1 2 
t t  t t 
 2 0  2 1 
t t  t

(16  15)(16  20) (16  10)(16  20)

v(16)  (227.04)  (362.78)
(10  15)(10  20) (15  10)(15  20)
(16  10)(16  15)
 (517.35)
(20  10)(20  15)

 (0.08)(227.04)  (0.96)(362.78)  (0.12)(517.35)  392.19 m/s

b) The absolute relative approximate error a for the second order polynomial is
calculated by considering the result of the first order polynomial (Example 1) as
the previous approximation.

392.19  393.69
a  100  0.38410%
392.19

Example 3
The upward velocity of a rocket is given as a function of time in Table 3.

Table 3 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

a) Determine the value of the velocity at t  16 seconds using third order

Lagrangian polynomial interpolation.

Page | 149
Chemical engineering Department Assist Lecturer Ghadeer Jassim

b) Find the absolute relative approximate error for the third order polynomial
approximation.
c) Using the third order polynomial interpolant for velocity, find the distance
covered by the rocket from t  11 s to t  16 s .
d) Using the third order polynomial interpolant for velocity, find the acceleration
of the rocket at t  16 s .

Solution
a) For third order polynomial interpolation (also called cubic interpolation), the
velocity is given by
3
v(t )   Li (t )v(t i )
i 0

 L0 (t )v(t0 )  L1 (t )v(t1 )  L2 (t )v(t2 )  L3 (t )v(t3 )

x3 , y3 

x1, y1 

x2 , y2  f 3 x 

x0 , y0 
x

Figure 4 Cubic interpolation.

Since we want to find the velocity at t  16 , and we are using a third order
polynomial, we need to choose the four data points closest to t  16 that also
bracket t  16 to evaluate it. The four points are t0  10, t1  15, t 2  20 and t 3  22.5 .
Then
t0  10, vt0   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

Page | 150
Chemical engineering Department Assist Lecturer Ghadeer Jassim

gives
3 t tj  t  t1  t  t 2  t  t 3 
L0 (t )       
j 0 t 0  t j  t 0  t1  t 0  t 2  t 0  t 3 
j 0

3 t tj  t  t 0  t  t 2  t  t 3 
L1 (t )       
j 0 t1  t j  t1  t 0  t1  t 2  t1  t 3 
j 1

3 t tj  t  t 0  t  t1  t  t 3 
L2 (t )       
j 0 t 2  t j  t 2  t 0  t 2  t1  t 2  t 3 
j 2

3 t tj  t  t 0  t  t1  t  t 2 
L3 (t )       
j 0 t 3  t j  t 3  t 0  t 3  t1  t 3  t 2 
j 3

Hence
 t  t1  t  t 2  t  t3   t  t0  t  t 2  t  t3 
v(t )     v(t0 )     v(t1 )
 t0  t1  t0  t 2  t0  t3   t1  t0  t1  t 2  t1  t3 
 t  t0  t  t1  t  t3   t  t0  t  t1  t  t 2 
    v(t 2 )     v(t3 ), t0  t  t3
 t 2  t 0  t 2  t1  t 2  t 3   t 3  t 0  t 3  t1  t 3  t 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)
 (517.35)
(20  10)(20  15)(20  22.5)
(16  10)(16  15)(16  20)
 (602.97)
(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

b) The absolute percentage relative approximate error, a for the value obtained
for v(16) can be obtained by comparing the result with that obtained using the
second order polynomial (Example 2)

392.06  392.19
a   100  0.033269%
392.06

Page | 151
Chemical engineering Department Assist Lecturer Ghadeer Jassim

c) The distance covered by the rocket between t  11 s to t  16 s can be calculated

from the interpolating polynomial as
(t  15)(t  20)(t  22.5) (t  10)(t  20)(t  22.5)
v(t )  (227.04)  (362.78)
(10  15)(10  20)(10  22.5) (15  10)(15  20)(15  22.5)
(t  10)(t  15)(t  22.5)
 (517.35)
(20  10)(20  15)(20  22.5)
(t  10)(t  15)(t  20)
 (602.97), 10  t  22.5
(22.5  10)(22.5  15)(22.5  20)

(t 2  35t  300)(t  22.5) (t 2  30t  200)(t  22.5)

 (227.04)  (362.78)
(5)(10)(12.5) (5)(5)(7.5)
(t 2  25t  150)(t  22.5) (t 2  25t  150)(t  20)
 (517.35)  (602.97)
(10)(5)(2.5) (12.5)(7.5)(2.5)

 (t 3  57.5t 2  1087.5t  6750)(0.36326)  (t 3  52.5t 2  875t  4500)(1.9348)

 (t 3  47.5t 2  712.5t  3375)(4.1388)  (t 3  45t 2  650t  3000)(2.5727)
 4.245  21.265t  0.13195t 2  0.00544t 3 , 10  t  22.5

Note that the polynomial is valid between t  10 and t  22.5 and hence includes the
limits of t  11 and t  16 .
So
16 16
s(16)  s(11)   v(t )dt   (4.245  21.265t  0.13195t 2  0.00544t 3 )dt
11 11
16
 t2 t3 t4 
  4.245t  21.265  0.13195  0.00544   1605 m
 2 3 4 11

d) The acceleration at t  16 is given by

a16  vt  t 16
d
dt
Given that
v(t )  4.245  21.265t  0.13195t 2  0.00544t 3 , 10  t  22.5

at  
d
dt
vt  
d
dt

 4.245  21.265t  0.13195t 2  0.00544t 3 
 21.265  0.26390t  0.01632t 2 , 10  t  22.5

a(16)  21.265  0.26390(16)  0.01632(16) 2  29.665 m/s 2

Page | 152