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Newtons Divided DifferencePolynomial Method of
Interpolation
Major: All Engineering Majors
Authors: Autar Kaw, Jai Paul
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Undergraduates
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Newtons DividedDifference Method of
Interpolation
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What is Interpolation ?Given (x0,y0), (x1,y1), (xn,yn), find thevalue of y at a value of x that is not given.
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InterpolantsPolynomials are the most commonchoice of interpolants because they
are easy to:
Evaluate
Differentiate, and
Integrate.
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Newtons Divided Difference
MethodLinear 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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ExampleThe upward velocity of a rocket is given as a function oftime in Table 1. Find the velocity at t=16 seconds usingthe Newton Divided Difference method for linearinterpolation.
Table. Velocity as afunction of time
Figure. Velocity vs. time datafor the rocket example
0 0
10 227.0415 362.78
20 517.35
22.5 602.97
30 901.67
)s(t )m/s()(tv
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Linear Interpolation
10 12 14 16 18 20 22 24350
400
450
500
550517.35
362.78
y s
f range( )
f x desired
x s1
10x s0
10 x s range x desired
,150 t 78.362)( 0 tv
,201 t 35.517)( 1 tv
)( 00 tvb 78.362
01
011
)()(tttvtvb
914.30
)()( 010 ttbbtv
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Linear Interpolation (contd)
10 12 14 16 18 20 22 24350
400
450
500
550517.35
362.78
y s
f range( )
f x desired
x s1
10x s0
10 x s range x desired
)()( 010 ttbbtv
),15(914.3078.362 t 2015 t
At 16t
)1516(914.3078.362)16( v
69.393 m/s
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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
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ExampleThe upward velocity of a rocket is given as a function oftime in Table 1. Find the velocity at t=16 seconds usingthe Newton Divided Difference method for quadraticinterpolation.
Table. Velocity as afunction of time
Figure. Velocity vs. time datafor the rocket example
0 0
10 227.0415 362.78
20 517.35
22.5 602.97
30 901.67
)s(t )m/s()(tv
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Quadratic Interpolation (contd))( 00 tvb
04.227
01
01
1
)()(
tt
tvtv
b
1015
04.22778.362
148.27
02
01
01
12
12
2
)()()()(
tt
tt
tvtv
tt
tvtv
b
1020
1015
04.22778.362
1520
78.36235.517
10
148.27914.30
37660.0
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Quadratic Interpolation (contd)))(()()( 102010 ttttbttbbtv
),15)(10(37660.0)10(148.2704.227 ttt 2010 t
At ,16t )1516)(1016(37660.0)1016(148.2704.227)16( v 19.392 m/s
The absolute relative approximate error a obtained between the results from the first
order and second order polynomial is
a 100x19.392
69.39319.392
= 0.38502 %
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General Form))(()()( 1020102 xxxxbxxbbxf
where
Rewriting
))(](,,[)](,[][)( 1001200102 xxxxxxxfxxxxfxfxf
)(][ 000 xfxfb
01
01
011
)()(],[
xx
xfxfxxfb
02
01
01
12
12
02
01120122
)()()()(
],[],[],,[xx
xx
xfxf
xx
xfxf
xxxxfxxfxxxfb
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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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ExampleThe upward velocity of a rocket is given as a function oftime in Table 1. Find the velocity at t=16 seconds usingthe Newton Divided Difference method for cubicinterpolation.
Table. Velocity as afunction of time
Figure. Velocity vs. time datafor the rocket example
0 0
10 227.0415 362.78
20 517.35
22.5 602.97
30 901.67
)s(t )m/s()(tv
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ExampleThe velocity profile is chosen as
))()(())(()()( 2103102010 ttttttbttttbttbbtv
we need to choose four data points that are closest to 16t ,100 t 04.227)( 0 tv
,151 t 78.362)( 1 tv
,202 t 35.517)( 2 tv
,5.223 t 97.602)( 3 tv
The values of the constants are found as:
b0 = 227.04; b1 = 27.148; b2 = 0.37660; b3 = 5.4347103
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Example
b0 = 227.04; b1 = 27.148; b2 = 0.37660; b3 = 5.4347103
0b
100 t 04.227 1b
148.27 2b
,151 t 78.362 37660.0 3b
914.30 3104347.5
,202 t 35.517 44453.0
248.34
,5.223 t 97.602
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ExampleHence
))()(())(()()( 2103102010 ttttttbttttbttbbtv
)20)(15)(10(10*4347.5
)15)(10(37660.0)10(148.2704.227
3
ttt
ttt
At ,16t
)2016)(1516)(1016(10*4347.5
)1516)(1016(37660.0)1016(148.2704.227)16(
3
v
06.392 m/s
The absolute relative approximate error a obtained is
a 100x06.392
19.39206.392
= 0.033427 %
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Comparison Table
Order of
Polynomial
1 2 3
v(t=16)
m/s
393.69 392.19 392.06
Absolute Relative
Approximate Error
---------- 0.38502 % 0.033427 %
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Distance from Velocity ProfileFind the distance covered by the rocket from t=11s tot=16s ?
)20)(15)(10(10*4347.5
)15)(10(37660.0)10(148.2704.227)(
3
ttt
ttttv5.2210 t
32 0054347.013204.0265.212541.4 ttt 5.2210 t So
16
11
1116 dttvss
dtttt )0054347.013204.0265.212541.4( 3216
11
16
11
432
40054347.0
313204.0
2265.212541.4
tttt
m1605
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Acceleration from Velocity ProfileFind the acceleration of the rocket at t=16s given that
32 0054347.013204.0265.212541.4)( ttttv
32 0054347.013204.0265.212541.4)()( tttdt
dtv
dt
dta
2016304.026408.0265.21 tt
2)16(016304.0)16(26408.0265.21)16( a
2/664.29 sm
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Additional ResourcesFor all resources on this topic such as digital audiovisuallectures, primers, textbook chapters, multiple-choicetests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, pleasevisit
http://numericalmethods.eng.usf.edu/topics/newton_div
ided_difference_method.html
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