Transcript Newton's Divided Difference Polynomial Power Point

Newton’s Divided Difference
Polynomial Method of
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
Newton’s Divided
Difference 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.
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Interpolants
Polynomials are the most common
choice of interpolants because they
are easy to:
Evaluate
Differentiate, and
Integrate.
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Newton’s Divided Difference
Method
Linear interpolation: Given ( x0 , y0 ), ( x1 , y1 ), pass a
linear interpolant through the data
f1 ( x)  b0  b1 ( x  x0 )
where
b0  f ( x0 )
b1 
5
f ( x1 )  f ( x0 )
x1  x0
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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 Newton Divided Difference
method for linear interpolation.
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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
40.12
T ( R)  b0  b1 ( R  R0 )
42
40
38
R0  911.3, T ( R0 )  30.131
ys
R1  636.0, T ( R1 )  40.120
f x desired
f ( range)


36
34
b0  T ( R0 )
32
30.131
 30 .131
T (R 1 ) T (R 0 ) 40.120  30.131

b1 
636 .0  911 .3
R1  R 0
30
946
x s  10
0
896
846
796
746
696
x s  range x desired
 0.036284
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646
x s  10
1
Linear Interpolation (contd)
40.12
42
40
38
ys
f ( range)


36
f x desired
34
32
30.131
30
946
896
x s  10
0
846
796
746
696
x s  range x desired
646
x s  10
1
T ( R)  b0  b1 ( R  R0 )
 30.131 0.036284( R  911.3),
636 .0  R  911 .3
At R  754 .8
T (754.8)  30.131 0.036284(754.8  911.3)
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 35.809 C
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Quadratic Interpolation
Given ( x0 , y0 ), ( x1 , y1 ), and ( x2 , y 2 ), fit a quadratic interpolant through the data.
f 2 ( x)  b0  b1 ( x  x0 )  b2 ( x  x0 )(x  x1 )
b0  f ( x0 )
f ( x1 )  f ( x0 )
b1 
x1  x0
f ( x2 )  f ( x1 ) f ( x1 )  f ( x0 )

x2  x1
x1  x0
b2 
x 2  x0
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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 Newton Divided Difference
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)
50.128
55
50
ys
45
f ( range)


f x desired
40
35
30.131
30
400
451.1
500
600
700
x s  range x desired
800
900
1000
911.3
T ( R)  b0  b1 ( R  R0 )  b2 ( R  R0 )(R  R1 )
R0  911.3, T ( R0 )  30.131
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R1  636.0, T ( R1 )  40.120
R2  451.1, T ( R2 )  50.128
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Quadratic Interpolation (contd)
b0  T ( R0 )
 30 .131
T ( R1 )  T ( R0 )
40.120  30.131

b1 
636 .0  911 .3
R1  R0
 0.036284
T ( R2 )  T ( R1 ) T ( R1 )  T ( R0 ) 50.128 40.120 40.120 30.131


R2  R1
R1  R0
636.0  911.3
b2 
 451.1  636.0
R2  R0
451.1  911.3
 0.054127  0.036284

 460 .2
 3.8771105
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Quadratic Interpolation (contd)
T ( R)  b0  b1 ( R  R0 )  b2 ( R  R0 )(R  R1 )
 30.131 0.036284( R  911.3)  3.8771105 ( R  911.3)(R  636.0), 451 .1  R  911 .3
At R  754.8,
T (754.8)  30.131 0.036284(754.8  911.3)  3.8771105 (754.8  911.3)(754.8  636.0)
 35.089 C
The absolute relative approximate error a obtained between the results from the first and
second order polynomial is
a 
35.089  35.809
 100
35.089
 2.0543 %
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General Form
f 2 ( x)  b0  b1 ( x  x0 )  b2 ( x  x0 )(x  x1 )
where
b0  f [ x0 ]  f ( x0 )
f ( x1 )  f ( x 0 )
b1  f [ x1 , x0 ] 
x1  x0
f ( x 2 )  f ( x1 ) f ( x1 )  f ( x0 )

f [ x 2 , x1 ]  f [ x1 , x0 ]
x 2  x1
x1  x0
b2  f [ x 2 , x1 , x0 ] 

x 2  x0
x 2  x0
Rewriting
f 2 ( x)  f [ x0 ]  f [ x1 , x0 ](x  x0 )  f [ x2 , x1 , x0 ](x  x0 )(x  x1 )
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General Form
Given (n  1) data points, x0 , y0 , x1 , y1 ,......,xn1 , yn1 , xn , yn  as
f n ( x)  b0  b1 ( x  x0 )  .... bn ( x  x0 )(x  x1 )...(x  xn1 )
where
b0  f [ x0 ]
b1  f [ x1 , x0 ]
b2  f [ x2 , x1 , x0 ]

bn1  f [ xn1 , xn2 ,....,x0 ]
bn  f [ xn , xn1 ,....,x0 ]
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General form
The third order polynomial, given ( x0 , y0 ), ( x1 , y1 ), ( x2 , y 2 ), and ( x3 , y3 ), is
f 3 ( x)  f [ x0 ]  f [ x1 , x0 ](x  x0 )  f [ x2 , x1 , x0 ](x  x0 )(x  x1 )
 f [ x3 , x2 , x1 , x0 ](x  x0 )(x  x1 )(x  x2 )
b0
x0
b1
f ( x0 )
b2
f [ x1 , x0 ]
x1
f ( x1 )
f [ x2 , x1 , x0 ]
f [ x2 , x1 ]
x2
f ( x2 )
b3
f [ x3 , x2 , x1 , x0 ]
f [ x3 , x2 , x1 ]
f [ x3 , x 2 ]
x3
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f ( x3 )
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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 Newton Divided Difference
method for cubic interpolation.
17
R (Ω)
T(°C)
1101.0
911.3
636.0
451.1
25.113
30.131
40.120
50.128
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Example
For the third order polynomial, we choose the temperature given by
T ( R)  b0  b1 ( R  R0 )  b2 ( R  R0 )(R  R1 )  b3 ( R  R0 )(R  R1 )(R  R2 )
R0  1101.0,
R1  911.3,
T ( R0 )  25.113
T ( R1 )  30.131
R2  636.0, T ( R2 )  40.120
R3  451.1, T ( R3 )  50.128
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Example
b0
R0  1101.0,
25 .113
b1
 0.026452
R1  911.3,
b2
2.1144105
30 .131
 2.7124108
 0.036284
R2  636.0,
b3
3.8771105
40 .120
 0.054127
R3  451.1,
50 .128
The values of the constants are found as:
b0  25.113
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b1  0.026452 b2  2.1144105
b3  2.7124108
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Example
T R   b0  b1 R  R0   b2 R  R0 R  R1   b3 R  R0 R  R1 R  R2 
 25.113 0.026452R  1101.0   2.1144105 R  1101.0 R  911.3
 2.7124108 R  1101.0R  911.3R  636.0 , 451.1  R  1101.0
At R  754.8,
T (754.8)  25.113 0.026452(754.8  1101.0)  2.1144105 (754.8  1101.0)(754.8  911.3)
 2.7124108 754.8  1101.0754.8  911.3754.8  626.0
 35.242C
The absolute percentage relative approximate error, a for the value obtained for T(754.8)
between second and third order polynomial is
a 
35.242  35.089
 100
35.242
 0.43458 %
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Comparison Table
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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
1
 b0  b1 (ln R  ln R0 )  b2 (ln R  ln R0 )(ln R  ln R1 )  b3 (ln R  ln R0 )(ln R  ln R1 )(ln R  ln R2 )
T
1
substituting y  , and x  ln R, the calibration curve is given by
T
y( x)  b0  b1 ( x  x0 )  b2 ( x  x0 )(x  x1 )  b3 ( x  x0 )(x  x1 )(x  x2 )
R (Ω)
1101.0
911.3
636.0
451.1
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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
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 ?
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Actual Calibration
y( x)  b0  b1 ( x  x0 )  b2 ( x  x0 )(x  x1 )  b3 ( x  x0 )(x  x1 )(x  x2 )
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
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Actual Calibration
b0
xo  7.0040, 0.039820
b1
0.035069
x1  6.8149, 0.033188
b2
0.022040
0.022974
x2  6.4552, 0.024925
b3
0.011173
0.012070
0.014487
x3  6.1117, 0.019949
The values of the constants are found as:
b1  0.035069
b0  0.039820
b2  0.022040
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b3  0.011173
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Actual Calibration
y( x)  b0  b1 ( x  x0 )  b2 ( x  x0 )(x  x1 )  b3 ( x  x0 )(x  x1 )(x  x2 )
 0.039820 0.035069x  7.0040  0.022974x  7.0040x  6.8149
 0.011173x  7.0040x  6.8149x  6.4552, 6.1117 x  7.0040
Since x  ln 754 .8  6.6265
At x  6.6265,
y6.6265  0.039820 0.0350716.6265 7.0040  0.0229726.6265 7.00406.6265 6.8149
 0.0111826.6265 7.00406.6265 6.81496.6265 6.4552
 0.028285
T
1
y

1
0.028285
 35.355 C
25
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Actual Calibration
Since the actual measured value at 754.8 ohms is 35.285 C , the absolute relative true error
between the value used for Cubic Interpolation is
t 
35.285  35.242
 100
35.285
 0.12253 %
and for actual calibration is
t 
35.285 35.355
100
35.285
 0.19825 %
Therefore, the calibration curve obtained more accurate results than a cubic polynomial
interpolant given by Newton’s Divided Difference method
26
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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/newton_div
ided_difference_method.html
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