newton’s divided difference polynomial method of interpolation
DESCRIPTION
Newton’s Divided Difference Polynomial Method of Interpolation. Chemical Engineering Majors Authors: Autar Kaw, Jai Paul http://numericalmethods.eng.usf.edu Transforming Numerical Methods Education for STEM Undergraduates. - PowerPoint PPT PresentationTRANSCRIPT
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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
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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
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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
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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.
http://numericalmethods.eng.usf.edu16
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
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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