Newton’s Divided Difference Polynomial Method of Interpolation
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Newton’s Divided Difference Polynomial
Method of Interpolation
Mechanical Engineering Majors
Authors: Autar Kaw, Jai Paul
http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM
Undergraduates
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Newton’s Divided Difference Method of
Interpolation
http://numericalmethods.eng.usf.edu
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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:
EvaluateDifferentiate, and Integrate.
![Page 5: Newton’s Divided Difference Polynomial Method of Interpolation](https://reader034.fdocuments.in/reader034/viewer/2022051416/56812ef0550346895d948d44/html5/thumbnails/5.jpg)
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Newton’s Divided Difference Method
Linear interpolation: Given pass a linear interpolant through the data
where
),,( 00 yx ),,( 11 yx
)()( 0101 xxbbxf
)( 00 xfb
01
011
)()(
xx
xfxfb
![Page 6: Newton’s Divided Difference Polynomial Method of Interpolation](https://reader034.fdocuments.in/reader034/viewer/2022051416/56812ef0550346895d948d44/html5/thumbnails/6.jpg)
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Example A trunnion is cooled 80°F to − 108°F. Given below is the table of
the coefficient of thermal expansion vs. temperature. Determine the value of the coefficient of thermal expansion at T=−14°F using the direct method for linear interpolation.
Temperature (oF)
Thermal Expansion Coefficient (in/in/oF)
80 6.47 × 10−6
0 6.00 × 10−6
−60 5.58 × 10−6
−160 4.72 × 10−6
−260 3.58 × 10−6
−340 2.45 × 10−6
![Page 7: Newton’s Divided Difference Polynomial Method of Interpolation](https://reader034.fdocuments.in/reader034/viewer/2022051416/56812ef0550346895d948d44/html5/thumbnails/7.jpg)
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Linear Interpolation
5045403530252015105.5
5.6
5.7
5.8
5.9
6
5.58
y s
f range( )
f x desired
x s1
10x s0
10 x s range x desired
)()( 010 TTbbT
,00 T 60 10006 .Tα
,601 T 61 10585 .Tα
)( 00 Tb
610006 .
01
011
)()(
TT
TTb
060
1000610585 66
..
6100070 .
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Linear Interpolation (contd)
5045403530252015105.5
5.6
5.7
5.8
5.9
6
5.58
y s
f range( )
f x desired
x s1
10x s0
10 x s range x desired
010 TTbbTα
,T.. 010007010006 66
060 T
Fin/in/109025
01410007010006146
66
.
..α
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Quadratic InterpolationGiven ),,( 00 yx ),,( 11 yx and ),,( 22 yx fit a quadratic interpolant through the data.
))(()()( 1020102 xxxxbxxbbxf
)( 00 xfb
01
011
)()(
xx
xfxfb
02
01
01
12
12
2
)()()()(
xx
xx
xfxf
xx
xfxf
b
![Page 10: Newton’s Divided Difference Polynomial Method of Interpolation](https://reader034.fdocuments.in/reader034/viewer/2022051416/56812ef0550346895d948d44/html5/thumbnails/10.jpg)
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Example A trunnion is cooled 80°F to − 108°F. Given below is the table of
the coefficient of thermal expansion vs. temperature. Determine the value of the coefficient of thermal expansion at T=−14°F using the direct method for quadratic interpolation.
Temperature (oF)
Thermal Expansion Coefficient (in/in/oF)
80 6.47 × 10−6
0 6.00 × 10−6
−60 5.58 × 10−6
−160 4.72 × 10−6
−260 3.58 × 10−6
−340 2.45 × 10−6
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Quadratic Interpolation (contd)
60 40 20 0 20 40 60 805.4
5.6
5.8
6
6.2
6.4
6.66.47
5.58
y s
f range( )
f x desired
8060 x s range x desired
102010 TTTTbTTbbTα
,800 T 60 10476 .Tα
,01 T 61 10006 .Tα
,602 T 62 10585 .Tα
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Quadratic Interpolation (contd)
600 10476 .Tαb
9
66
01
011 108755
800
1047610006
...
TT
TαTαb
12
66
6666
02
01
01
12
12
2
1003578
140
100058750100070
8060800
1047610006060
1000610585
.
..
....
TT
TTTαTα
TTTαTα
b
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Quadratic Interpolation (contd)
102010 TTTTbTTbbTα
,TT.T.. 08010035788010875510476 1296 8060 T
At ,14T
Fin/in/1090725
01480141003578801410875510476146
1296
.
...α ---
The absolute relative approximate error a obtained between the results from the
first and second order polynomial is
%087605.0
1001090725
10902510907256
66
-
--
a .
..
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General Form
))(()()( 1020102 xxxxbxxbbxf
where
Rewriting))(](,,[)](,[][)( 1001200102 xxxxxxxfxxxxfxfxf
)(][ 000 xfxfb
01
01011
)()(],[
xx
xfxfxxfb
02
01
01
12
12
02
01120122
)()()()(
],[],[],,[
xx
xx
xfxf
xx
xfxf
xx
xxfxxfxxxfb
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General FormGiven )1( n data points, nnnn yxyxyxyx ,,,,......,,,, 111100 as
))...()((....)()( 110010 nnn xxxxxxbxxbbxf
where
][ 00 xfb
],[ 011 xxfb
],,[ 0122 xxxfb
],....,,[ 0211 xxxfb nnn
],....,,[ 01 xxxfb nnn
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General formThe third order polynomial, given ),,( 00 yx ),,( 11 yx ),,( 22 yx and ),,( 33 yx is
))()(](,,,[
))(](,,[)](,[][)(
2100123
1001200103
xxxxxxxxxxf
xxxxxxxfxxxxfxfxf
0b
0x )( 0xf 1b
],[ 01 xxf 2b
1x )( 1xf ],,[ 012 xxxf 3b
],[ 12 xxf ],,,[ 0123 xxxxf
2x )( 2xf ],,[ 123 xxxf
],[ 23 xxf
3x )( 3xf
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Example A trunnion is cooled 80°F to − 108°F. Given below is the table of
the coefficient of thermal expansion vs. temperature. Determine the value of the coefficient of thermal expansion at T=−14°F using the direct method for cubic interpolation.
Temperature (oF)
Thermal Expansion Coefficient (in/in/oF)
80 6.47 × 10−6
0 6.00 × 10−6
−60 5.58 × 10−6
−160 4.72 × 10−6
−260 3.58 × 10−6
−340 2.45 × 10−6
![Page 18: Newton’s Divided Difference Polynomial Method of Interpolation](https://reader034.fdocuments.in/reader034/viewer/2022051416/56812ef0550346895d948d44/html5/thumbnails/18.jpg)
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Example
The coefficient of thermal expansion profile is chosen as
2103102010 TTTTTTbTTTTbTTbbTα
,800 T 60 10476 .Tα
,01 T 61 10006 .Tα
,602 T 62 10585 .Tα
,1603 T 63 10724 .Tα
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Example 0b
,800 T 610476 . 1b
9108755 . 2b
,01 T 610006 . 121003578 . 3b
6100070 . 151018458 .
,602 T 610585 . 1110
61000860 .
,1603 T 610724 .
The values of the constants are 6
0 10476 .b 91 108755 .b 12
2 1003578 .b 153 1018458 .b
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Example
600801018458
0801003578801087551047615
1296
2103102010
TT-T-.
TT.T..
TTTTTTbTTTTbTTbbTα
At ,14T
Fin/in/1090775
601401480141018458
0148014100357880141087551047614
6
15
1296
.
.
...α
The absolute relative approximate error a obtained between the results from the second and
third order polynomial is
%0083867.0
1001090775
109072510907756
66
.
..a
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Comparison Table
Order of Polynomial 1 2 3
Thermal Expansion Coefficient (in/in/oF)
5.902 × 10−6 5.9072 × 10−6 5.9077 × 10−6
Absolute Relative Approximate Error
---------- 0.087605% 0.0083867%
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Reduction in DiameterThe actual reduction in diameter is given by
where Tr = room temperature (°F)Tf = temperature of cooling medium (°F)
Since Tr = 80 °F and Tr = −108 °F,
Find out the percentage difference in the reduction in the diameter by the above integral formula and the result using the thermal expansion coefficient from the cubic interpolation.
108
80
dTDD
Tf
Tr
dTDD
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Reduction in DiameterWe know from interpolation that
80160
,101845.8101944.8104786.61000.6 31521296
T
TTTT
Therefore,
6
180
80
415
312
296
108
80
31521296
109.1105
4101845.8
3101944.8
2104786.61000.6
101845.8101944.8104786.61000.6
TTTT
dTTTT
dTD
D f
r
T
T
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Reduction in diameterUsing the average value for the coefficient of thermal expansion from cubic interpolation
6
6
106.1110
80108109077.5
rf TT
TD
D
The percentage difference would be
%42775.0
100109.1105
106.1110109.11056
66
a
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Additional ResourcesFor 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_divided_difference_method.html