Transcript Lagrangian Method Power Point Interpolation

Lagrangian Interpolation
Electrical Engineering Majors
Authors: Autar Kaw, Jai Paul
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Transforming Numerical Methods Education for STEM
Undergraduates
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1
Lagrange Method of
Interpolation
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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
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Interpolants
Polynomials are the most common
choice of interpolants because they
are easy to:
Evaluate
Differentiate, and
Integrate.
4
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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
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Example
Thermistors are based on materials’ change in resistance with
temperature. A manufacturer of thermistors makes the following
observations on a thermistor. Determine the temperature
corresponding to 754.8 ohms using the Lagrangian method for
linear interpolation.
6
R (Ω)
T(°C)
1101.0
911.3
636.0
451.1
25.113
30.131
40.120
50.128
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Linear Interpolation
1
T ( R)   Li ( R)T ( Ri )
40.12
i 0
 L0 ( R)T ( R0 )  L1 ( R)T ( R1 )
42
40
38
ys
f ( range)
R0  911.3, T R0   30.131
R1  636.0, T R1   40.120


36
f x desired
34
32
30.131
30
946
x s  10
0
7
896
846
796
746
696
x s  range x desired
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646
x s  10
1
Linear Interpolation (contd)
1
L0 ( R)  
j 0
j 0
1
L1 ( R)  
j 0
j 1
T ( R) 
R  Rj

R0  R j
R  Rj
R1  R j

R  R1
R0  R1
R  R0
R1  R0
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)
8
 35.809 C
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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
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Example
Thermistors are based on materials’ change in resistance with
temperature. A manufacturer of thermistors makes the following
observations on a thermistor. Determine the temperature
corresponding to 754.8 ohms using the Lagrangian method for
quadratic interpolation.
10
R (Ω)
T(°C)
1101.0
911.3
636.0
451.1
25.113
30.131
40.120
50.128
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Quadratic Interpolation (contd)
Ro  911.3, T Ro   30.131
50.128
R1  636.0, T R1   40.120
55
50
R2  451.1, T R2   50.128
ys
45
f ( range)

R  Rj
 R  R1  R  R2

L0 ( R)  
 
R

R
R

R
j 0
0
j
1  R0  R2
 0
j 0
2
2
L1 ( R)  
j 0
j 1
2
L2 ( R)  
j 0
j 2
11
R  Rj
 R  R0
 
R1  R j  R1  R0
R  Rj

f x desired
 R  R2

 R1  R2
 R  R0
 
R2  R j  R2  R0






40
35
30.131
30
400
451.1
500
600
700
800
900
x s  range x desired
 R  R1 


R

R
1 
 2
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1000
911.3
Quadratic Interpolation (contd)
 R  R1  R  R2 
 R  R0  R  R2 
 R  R0  R  R1 

T ( R0 )  


T ( R1 )  
T ( R2 )
T ( R)  
R

R
R

R
R

R
R

R
R

R
R

R
1  0
2 
0  1
2 
0  2
1 
 0
 1
 2
(754.8  636.0)(754.8  451.1)
(754.8  911.3)(754.8  451.1)
(30.131) 
(40.120)
(911.3  636.0)(911.3  451.1)
(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)
T (754.8) 
 (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
a 
12
35.089  35.809
 100
35.089
 2.0543 %
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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
13
200
10
10
12
14
16
18
x s  range x desired
20
22
24
22.5
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Example
Thermistors are based on materials’ change in resistance with
temperature. A manufacturer of thermistors makes the following
observations on a thermistor. Determine the temperature
corresponding to 754.8 ohms using the Lagrangian method for
cubic interpolation.
14
R (Ω)
T(°C)
1101.0
911.3
636.0
451.1
25.113
30.131
40.120
50.128
http://numericalmethods.eng.usf.edu
Cubic Interpolation (contd)
Ro  1101.0, T Ro   25.113
R1  911.3, T R1   30.131
R2  636.0, T R2   40.120
R3  451.1, T R3   50.128
50.128
3
L0 ( R)  
j 0
j 0
3
L1 ( R)  
j 0
j 1
3
L2 ( R)  
j 0
j 2
R  Rj
 R  R1  R  R2  R  R3 



 
R0  R j  R0  R1  R0  R2  R0  R3 
15
50
45
R  Rj
 R  R0
 
R1  R j  R1  R0
 R  R2

 R1  R2
ys
 R  R3 


R

R
3 
 1
f ( range)

40

f x desired
35
R  Rj
 R  R0
 
R2  R j  R2  R0
R  Rj
 R  R0
L3 ( R)  
 
j 0 R3  R j
 R3  R0
j 3
3
55
 R  R1  R  R3 



R

R
R

R
1  2
3 
 2
 R  R1  R  R2


 R  R  R  R
1  3
2
 3




30
25.113
25
400
451.1
500
600
700
800
900
x s  range x desired
1000
1100
1200
1.10110
55
50
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3
Cubic Interpolation (contd)
 R  R1  R  R2  R  R3 
 R  R0  R  R2  R  R3 


T ( R0 )  

T ( R1 )

T ( R)  
 R0  R1  R0  R2  R0  R3 
 R1  R0  R1  R2  R1  R3 
 R  R0  R  R1  R  R3 
 R  R0  R  R1  R  R2 

T ( R2 )  


T ( R3 )

 
 R2  R0  R2  R1  R2  R3 
 R3  R0  R3  R1  R3  R2 
R0  R  R0
(754.8  911.3)(754.8  636.0)(754.8  451.1)
(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
T (754.8) 
16
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Cubic Interpolation
The absolute percentage relative approximate error, a between second
and third order polynomial is
35.242 35.089
100
a 
35.242
 0.43458%
17
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Comparison Table
18
Order of
Polynomial
1
2
3
Temperature C
35.809
35.089
35.242
Absolute Relative
Approximate Error
----------
2.0543%
0.43458%
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Actual Calibration
The actual calibration curve used by industry is given by
3
1
1
1
1
1
1
 L0 (ln R)
 L1 (ln R)
 L2 (ln R)
 L3 (ln R)
  Li (ln R)
T (ln R0 )
T (ln R1 )
T (ln R2 )
T (ln R3 )
T i 0
T (ln Ri )
substituting y 
1
, 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 )
Find the calibration curve and find the temperature corresponding to 754.8 ohms. What is the difference
between the results from cubic interpolation? In which method is the difference larger, if the actual
measured value at 754.8 ohms is 35.285 C ?
R (Ω)
1101.0
911.3
636.0
451.1
19
T (°C)
25.113
30.131
40.120
50.128
x (ln R)
7.0040
6.8149
6.4552
6.1117
y (1/T)
0.039820
0.033188
0.024925
0.019949
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Actual Calibration
y( x)  L0 ( x) y( x0 )  L1 ( x) y( x1 )  L2 ( x) y( x2 )  L3 ( x) y( x3 )
xo  7.0040, yxo   0.039820
x1  6.8149, yx1   0.033188
x2  6.4552, yx2   0.024925
x3  6.1117, yx3   0.019949
x  xj
 x  x1  x  x 2  x  x3 



L0 ( x)  
 
x

x
j 0 0
j
 x0  x1  x0  x 2  x0  x3 
j 0
3
x  xj
 x  x0  x  x1  x  x3 



L2 ( x)  
 
x

x
x

x
x

x
x

x
j 0 2
j
0  2
1  2
3 
 2
j 2
3
20
3
L1 ( x)  
j 0
j 1
3
L3 ( x)  
j 0
j 3
x  xj
 x  x0  x  x2  x  x3 



 
x1  x j  x1  x0  x1  x2  x1  x3 
x  xj
 x  x0  x  x1  x  x2 



 
x3  x j  x3  x0  x3  x1  x3  x2 
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Actual Calibration
 x  x1  x  x 2  x  x3 
 x  x0  x  x 2  x  x3 


 y ( x0 )  

 y ( x1 )

y ( x)  
x

x
x

x
x

x
x

x
x

x
x

x
1  0
2  0
3 
0  1
2  1
3 
 0
 1
, xo ≤ x ≤ x3
 x  x0  x  x1  x  x3 
 x  x0  x  x1  x  x 2 

 y ( x 2 )  


 y ( x3 )

 
x

x
x

x
x

x
x

x
x

x
x

x
0  2
1  2
3 
0  3
1  3
2 
 2
 3
x  ln754.8  6.6265
(6.6265 6.8149)(6.6265 6.4552)(6.6265 6.1117)
(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)
y (6.6265) 
 (0.17938)(0.039820)  (0.69585)(0.033188)  (0.54005)(0.024925)
 (0.056519)(0.019949)
21
 0.028285
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Finally, since y 
Actual
Calibration
1
1
1
T
, T
y

 35.355C
0.028285
Since the actual measured value at 754.8 ohms is 35.285°C, the
absolute relative true error for the third order polynomial
approximation is
35.285 35.242
t 
35.285
 0.12253%
100
and for the calibration curve used by industry is
35.285 35.355
t 
100
35.285
 0.19825%
Therefore, a cubic polynomial interpolant given by Lagrangian
method obtained more accurate results than the calibration curve.
22
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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
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