newton’s divided difference polynomial method of interpolation
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Newton’s Divided Difference Polynomial
Method of InterpolationChemical Engineering Majors
Authors: Autar Kaw, Jai Paul
http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM
Undergraduates
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:
EvaluateDifferentiate, and Integrate.
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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
To find how much heat is required to bring a kettle of water to its boiling point, you are asked to calculate the specific heat of water at 61°C. The specific heat of water is given as a function of time in Table 1. Use Newton’s divided difference method with a first order and then a second order polynomial to determine the value of the specific heat at T = 61°C.
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Example
Table 1 Specific heat of water as a function of temperature.
Figure 2 Specific heat of water vs. temperature.
Temperature, Specific heat,
22425282100
41814179418641994217
CT
Ckg
JpC
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Linear Interpolation
50 60 70 80 904185
4190
4195
42004.199 103
4.186 103
y s
f range( )
f x desired
x s110x s0
10 x s range x desired
)()( 010 TTbbTC p
,520 T 4186)( 0 TC p
,821 T 4199)( 1 TC p
)( 00 TCb p 4186
01
011
)()(TT
TCTCb pp
528241864199
43333.0
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Linear Interpolation (contd)
)()( 010 TTbbTC p
),52(43333.04186 T 8252 T
At 61T )5261(43333.04186)61( pC
Ckg
J
9.4189
50 60 70 80 904185
4190
4195
42004.199 103
4.186 103
y s
f range( )
f x desired
x s110x s0
10 x s range x desired
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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
)()()()(
xxxx
xfxfxx
xfxf
b
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Quadratic Interpolation (contd)
40 45 50 55 60 65 70 75 80 854175
4180
4185
4190
4195
42004.199 103
4.179 103
y s
f range( )
f x desired
8242 x s range x desired
))(()()( 102010 TTTTbTTbbTC p
,420 T 4179)( 0 TC p
,521 T 4186)( 1 TC p
,822 T 4199)( 2 TC p
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Quadratic Interpolation (contd)
)( 00 TCb p 4179
01
011
)()(TT
TCTCb pp
425241794186
7.0
02
01
01
12
12
2
)()()()(
TTTT
TCTCTT
TCTC
b
pppp
4282
425241794186
528241864199
40
7.043333.0
3106667.6
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Quadratic Interpolation (contd)
))(()()( 102010 TTTTbTTbbTC p
),52)(42(106667.6)42(7.04179 3 TTT 8242 T
At ,61T )5261)(4261(106667.6)4261(7.04179)61( 3
pC
Ckg
J
2.4191
The absolute relative approximate error a obtained between the results from the first and
second order polynomial is
1002.4191
9.41892.4191
a
%030063.0
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General Form))(()()( 1020102 xxxxbxxbbxf
where
Rewriting))(](,,[)](,[][)( 1001200102 xxxxxxxfxxxxfxfxf
)(][ 000 xfxfb
01
01011
)()(],[
xxxfxf
xxfb
02
01
01
12
12
02
01120122
)()()()(],[],[
],,[xx
xxxfxf
xxxfxf
xxxxfxxf
xxxfb
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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
xxxxxxxxxxfxxxxxxxfxxxxfxfxf
0b
0x )( 0xf 1b
],[ 01 xxf 2b
1x )( 1xf ],,[ 012 xxxf 3b
],[ 12 xxf ],,,[ 0123 xxxxf
2x )( 2xf ],,[ 123 xxxf
],[ 23 xxf
3x )( 3xf
To find how much heat is required to bring a kettle of water to its boiling point, you are asked to calculate the specific heat of water at 61°C. The specific heat of water is given as a function of time in Table 1. Use Newton’s divided difference method with a third order polynomial to determine the value of the specific heat at T = 61°C.
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Example
Table 1 Specific heat of water as a function of temperature.
Figure 2 Specific heat of water vs. temperature.
Temperature, Specific heat,
22425282100
41814179418641994217
CT
Ckg
JpC
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Example The specific heat profile is chosen as
))()(())(()()( 2103102010 TTTTTTbTTTTbTTbbTC p
We need to choose four data points that are closest to CT 61 . ,420 T 4179)( 0 TC p
,521 T 4186)( 1 TC p
,822 T 4199)( 2 TC p
,1003 T 4217)( 3 TC p
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Example 0b
,420 T 4179 1b
7.0 2b ,521 T 4186 3106667.6 3b
43333.0 4101849.3 ,822 T 4199 011806.0
1 ,1003 T 4217
The values of the constants are found to be
0b 4179 1b 7.0 2b 3106667.6 3b 41018493 .
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Example))()(())(()()( 2103102010 TTTTTTbTTTTbTTbbTC p
10042825242101849.3
)52)(42(106667.6)42(7.041794
3
TTTT
TTT
At ,61T
CkgJ
C p
0.4190
826152614261101849.3
)5261)(4261(106667.6)4261(7.04179)61(4
3
The absolute relative approximate error a obtained between the results from the second and third order polynomial is
1000.4190
2.41910.4190
a
%027295.0
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Comparison Table
Order of Polynomial 1 2 3
TCp CkgJ
9.4189 2.4191 0.4190
Absolute Relative Approximate Error ---------- %030063.0 %027295.0
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
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