Chapter 05.04
Lagrange Method of Interpolation – More Examples
Electrical Engineering
Example 1
Thermistors are used to measure the temperature of bodies. Thermistors are based on
materials’ change in resistance with temperature. To measure temperature, manufacturers
provide you with a temperature vs. resistance calibration curve. If you measure resistance,
you can find the temperature. A manufacturer of thermistors makes several observations with
a thermistor, which are given in Table 1.
Table 1 Temperature as a function of resistance.
R ohm 
T C
1101.0
911.3
636.0
451.1
25.113
30.131
40.120
50.128
Figure 1 Resistance vs. temperature.
05.04.1
05.04.2
Chapter 05.04
Determine the temperature corresponding to 754.8 ohms using a first order Lagrange
polynomial.
Solution
For first order Lagrange polynomial interpolation (also called linear interpolation), the
temperature is given by
1
T ( R )   Li ( R )T ( Ri )
i 0
 L0 ( R)T ( R0 )  L1 ( R)T ( R1 )
y
(x1, y1)
f1(x)
(x0, y0)
x
Figure 2 Linear interpolation.
Since we want to find the temperature at R  754.8 , we need to choose the two data points
that are closest to R  754.8 that also bracket R  754.8 to evaluate it. The two points are
R0  911.3 and R1  636.0 .
Then
R0  911.3, T R0   30.131
R1  636.0, T R1   40.120
gives
1
L0 ( R)  
j 0
j 0

R  Rj
R0  R j
R  R1
R0  R1
Lagrange Method of Interpolation-More Examples: Electrical Engineering
1
L1 ( R)  
j 0
j 1

05.04.3
R  Rj
R1  R j
R  R0
R1  R0
Hence
R  R0
R  R1
T ( R0 ) 
T ( R1 )
R0  R1
R1  R0
R  636.0
R  911.3

(30.131) 
(40.120), 636.0  R  911.3
911.3  636.0
636.0  911.3
754.8  636.0
754.8  911.3
T (754.8) 
(30.131) 
(40.120)
911.3  636.0
636.0  911.3
 0.43153(30.131)  0.56847(40.120)
 35.809 C
You can see that L0 ( R)  0.43153 and L1 ( R)  0.56847 are like weightages given to the
temperatures at R0  911.3 and R1  636.0 to calculate the temperature at R  754.8 .
T ( R) 
Example 2
Thermistors are used to measure the temperature of bodies. Thermistors are based on
materials’ change in resistance with temperature. To measure temperature, manufacturers
provide you with a temperature vs. resistance calibration curve. If you measure resistance,
you can find the temperature. A manufacturer of thermistors makes several observations with
a thermistor, which are given in Table 2.
Table 2 Temperature as a function of resistance.
R ohm 
T C
1101.0
911.3
636.0
451.1
25.113
30.131
40.120
50.128
Determine the temperature corresponding to 754.8 ohms using a second order Lagrange
polynomial. Find the absolute relative approximate error for the second order polynomial
approximation.
Solution
For second order Lagrange polynomial interpolation (also called quadratic interpolation), the
temperature is given by
05.04.4
Chapter 05.04
2
T ( R )   Li ( R )T ( Ri )
i 0
 L0 ( R)T ( R0 )  L1 ( R)T ( R1 )  L2 ( R)T ( R2 )
y
(x1, y1)
(x2, y2)
f2(x)
(x0, y0)
x
Figure 3 Quadratic interpolation.
Since we want to find the temperature at R  754.8, we need to choose the three data points
that are closest to R  754.8 that also bracket R  754.8 to evaluate it. The three points are
R0  911.3, R1  636.0 and R2  451.1 .
Then
R0  911.3, T ( R0 )  30.131
R1  636.0, T ( R1 )  40.120
R2  451.1, T ( R2 )  50.128
gives
2
L0 ( R)  
j 0
j 0
R  Rj
R0  R j
 R  R1  R  R2 


 
 R0  R1  R0  R2 
2 RR
j
L1 ( R)  
j  0 R1  R j
j 1
 R  R0  R  R2 


 
 R1  R0  R1  R2 
Lagrange Method of Interpolation-More Examples: Electrical Engineering
2
L2 ( R)  
j 0
j 2
05.04.5
R  Rj
R2  R j
 R  R0  R  R1 


 
 R2  R0  R2  R1 
Hence
 R  R1  R  R2

T ( R )  
 R0  R1  R0  R2

 R  R0
T ( R0 )  

 R1  R0
 R  R2

 R1  R2

T ( R1 )

 R  R0  R  R1 

T ( R2 ), R0  R  R2
 
R

R
R

R
0  2
1 
 2
(754.8  636.0)(754.8  451.1)
T (754.8) 
(30.131)
(911.3  636.0)(911.3  451.1)
(754.8  911.3)(754.8  451.1)

(40.120)
(636.0  911.3)(636.0  451.1)
(754.8  911.3)(754.8  636.0)

(50.128)
(451.1  911.3)( 451.1  636.0)
 (0.28478)(30.131)  (0.93372)( 40.120)  (0.21850)(50.128)
 35.089 C
The absolute relative approximate error a obtained between the results from the first and
second order polynomial is
35.089  35.809
a 
 100
35.089
 2.0543%
Example 3
Thermistors are used to measure the temperature of bodies. Thermistors are based on
materials’ change in resistance with temperature. To measure temperature, manufacturers
provide you with a temperature vs. resistance calibration curve. If you measure resistance,
you can find the temperature. A manufacturer of thermistors makes several observations with
a thermistor, which are given in Table 3.
Table 3 Temperature as a function of resistance.
R ohm 
T C
1101.0
911.3
636.0
451.1
25.113
30.131
40.120
50.128
05.04.6
Chapter 05.04
a) Determine the temperature corresponding to 754.8 ohms using a third order Lagrange
polynomial. Find the absolute relative approximate error for the third order
polynomial approximation.
b) The actual calibration curve used by industry is given by
3
1
1
  Li (ln R)
T i 0
T (ln Ri )
1
1
1
1
 L0 (ln R)
 L1 (ln R)
 L2 (ln R)
 L3 (ln R)
T (ln R0 )
T (ln R1 )
T (ln R2 )
T (ln R3 )
1
substituting y  and x  ln R, the calibration curve is given by
T
y( x)  L0 ( x) y( x0 )  L1 ( x) y( x1 )  L2 ( x) y( x2 )  L3 ( x) y( x3 )
Table 4 Manipulation for the given data.
1
y 
R ohm  T C x ln R 
T 
1101.0 25.113 7.0040 0.039820
911.3
30.131 6.8149 0.033188
636.0
40.120 6.4552 0.024925
451.1
50.128 6.1117 0.019949
Find the calibration curve and find the temperature corresponding to 754.8 ohms. What is
the difference between the results from part (a)? Is the difference larger using the results
from part (a) or part (b), if the actual measured value at 754.8 ohms is 35.285C ?
Solution
a) For third order Lagrange polynomial interpolation (also called cubic interpolation), we
choose the temperature given by
3
T ( R )   Li ( R )T ( Ri )
i 0
 L0 ( R)T ( R0 )  L1 ( R)T ( R1 )  L2 ( R)T ( R2 )  L3 ( R)T ( R3 )
Since we want to find the temperature at R  754.8, we need to choose the four data points
that are closest to R  754.8 that also bracket R  754.8 to evaluate it. The four data points
are R0  1101.0, R1  911.3, R2  636.0 and R3  451.1 .
Then
R0  1101.0, T ( R0 )  25.113
R1  911.3, T ( R1 )  30.131
R2  636.0, T ( R2 )  40.120
R3  451.1, T ( R3 )  50.128
gives
Lagrange Method of Interpolation-More Examples: Electrical Engineering
3
L0 ( R)  
j 0
j 0
05.04.7
R  Rj
R0  R j
 R  R1  R  R2  R  R3 



 



 R0  R1  R0  R2  R0  R3 
3 RR
j
L1 ( R)  
j  0 R1  R j
j 1
 R  R0  R  R2  R  R3 



 
R

R
R

R
R

R
0  1
2  1
3 
 1
3
R  Rj
L2 ( R)  
j  0 R2  R j
j 2
 R  R0  R  R1  R  R3 



 
 R2  R0  R2  R1  R2  R3 
3
R  Rj
L3 ( R)  
j 0 R3  R j
j 3
 R  R0  R  R1  R  R2 



 
 R  R  R  R 
R

R
0  3
1  3
2 
 3
Hence
 R  R1  R  R2

T ( R)  
R

R
1  R0  R2
 0
 R  R0
 
 R2  R0
 R  R3 
 R  R0

T ( R0 )  
 R0  R3 
 R1  R0
 R  R2

 R1  R2
 R  R3 
T ( R1 )

R

R
3 
 1
 R  R1  R  R3 
 R  R0

T ( R2 )  

 R2  R1  R2  R3 
 R3  R0
 R  R1  R  R2 


T ( R3 )
 R3  R1  R3  R2 
R0  R  R3
(754.8  911.3)(754.8  636.0)(754.8  451.1)
T (754.8) 
(25.113)
(1101.0  911.3)(1101.0  636.0)(1101.0  451.1)
(754.8  1101.0)(754.8  636.0)(754.8  451.1)

(30.131)
(911.3  1101.0)(911.3  636.0)(911.3  451.1)
(754.8  1101.0)(754.8  911.3)(754.8  451.1)

(40.120)
(636.0  1101.0)(636.0  911.3)(636.0  451.1)
(754.8  1101.0)(754.8  911.3)(754.8  636.0)

(50.128)
(451.1  1101.0)( 451.1  911.3)( 451.1  636.0)
 (0.098494)( 25.113)  (0.51972)(30.131)  (0.69517 )(40.120 )
 (0.11639 )(50.128 )
 35.242 C
05.04.8
Chapter 05.04
The absolute relative approximate error a obtained between the results from the second
and third order polynomial is
35.242  35.089
a 
 100
35.242
 0.43458%
b) Finding the cubic interpolant using the Lagrangian method of interpolation for
y( x)  L0 ( x) y( x0 )  L1 ( x) y( x1 )  L2 ( x) y( x2 )  L3 ( x) y( x3 )
requires that we first calculate the new values of x and y .
x ln R 
7.0040
6.8149
6.4552
6.1117
Then
1
y 
T 
0.039820
0.033188
0.024925
0.019949
x0  7.0040, yx0   0.039820
x1  6.8149, yx1   0.033188
x2  6.4552, yx2   0.024925
x3  6.1117, yx3   0.019949
gives
3
L0 ( x)  
j 0
j 0
x  xj
x0  x j
 x  x1  x  x2  x  x3 



 
x

x
x

x
x

x
1  0
2  0
3 
 0
3 x x
j
L1 ( x)  
j 0 x1  x j
j 1
 x  x0  x  x2  x  x3 



 
 x1  x0  x1  x2  x1  x3 
3
x  xj
L2 ( x)  
j 0 x 2  x j
j 2
 x  x0  x  x1  x  x3 



 
x

x
x

x
x

x
0  2
1  2
3 
 2
3
x  xj
L3 ( x)  
j  0 x3  x j
j 3
Lagrange Method of Interpolation-More Examples: Electrical Engineering
05.04.9
 x  x0  x  x1  x  x2 



 
 x3  x0  x3  x1  x3  x2 
Hence
 x  x1  x  x 2

y ( x)  
 x0  x1  x0  x 2
 x  x0
 
 x 2  x0
 x  x3 
 x  x0  x  x 2

 y ( x0 )  

 x0  x3 
 x1  x0  x1  x 2
 x  x1  x  x3 
 x  x0

 y ( x 2 )  

 x 2  x1  x 2  x3 
 x3  x 0
 x  x3 
 y ( x1 )

 x1  x3 
 x  x1  x  x 2 


 y ( x3 )
 x3  x1  x3  x 2 
x 0  x  x3
Since we’re looking for the temperature at R  754.8 , we will be using
x  ln( 754.8)
 6.6265
At x  6.6265,
(6.6265  6.8149)(6.6265  6.4552)(6.6265  6.1117)
y (6.6265) 
(0.039820)
(7.0040  6.8149)(7.0040  6.4552)(7.0040  6.1117)
(6.6265  7.0040)(6.6265  6.4552)(6.6265  6.1117)

(0.033188)
(6.8149  7.0040)(6.8149  6.4552)(6.8149  6.1117)
(6.6265  7.0040)(6.6265  6.8149)(6.6265  6.1117)

(0.024925)
(6.4552  7.0040)(6.4552  6.8149)(6.4552  6.1117)
(6.6265  7.0040)(6.6265  6.8149)(6.6265  6.4552)

(0.019949)
(6.1117  7.0040)(6.1117  6.8149)(6.1117  6.4552)
 (0.17938)(0.039820)  (0.69585)(0.033188)
 (0.54005)(0.024925)  (0.056519)(0.019949)
 0.028285
1
Finally, since y  ,
T
1
T
y
1

0.028285
 35.355 C
Since the actual measured value at 754.8 ohms is 35.285C, the absolute relative true error
between the value used for part (a) is
35.285  35.242
t 
 100
35.285
 0.12253%
and for part (b) is
35.285  35.355
t 
 100
35.285
 0.19825%
05.04.10
Chapter 05.04
Therefore, a cubic polynomial interpolant given by the Lagrangian method of interpolation,
that is,
3
T ( R )   Li ( R )T ( Ri )
i 0
 L0 ( R)T ( R0 )  L1 ( R)T ( R1 )  L2 ( R)T ( R2 )  L3 ( R)T ( R3 )
obtained more accurate results than the calibration curve of
3
1
1
  Li (ln R)
T i 0
T (ln Ri )
1
1
1
1
 L0 (ln R)
 L1 (ln R)
 L2 (ln R)
 L3 (ln R)
T (ln R0 )
T (ln R1 )
T (ln R2 )
T (ln R3 )