Transcript Newton's Divided Difference Polynomial Power Point

Newton’s Divided Difference
Polynomial Method of
Interpolation
Major: All Engineering Majors
Authors: Autar Kaw, Jai Paul
http://numericalmethods.eng.usf.edu
Transforming 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:
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 
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f ( x1 )  f ( x0 )
x1  x0
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Example
The upward velocity of a rocket is given as a function of
time in Table 1. Find the velocity at t=16 seconds using
the Newton Divided Difference method for linear
interpolation.
Table. Velocity as a
function of time
t (s) v(t ) (m/s)
0
10
15
20
22.5
30
6
0
227.04
362.78
517.35
602.97
901.67
Figure. Velocity vs. time data
for the rocket example http://numericalmethods.eng.usf.edu
Linear Interpolation
517.35
v(t )  b0  b1 (t  t 0 )
t 0  15, v(t 0 )  362.78
t1  20, v(t1 )  517.35
b0  v(t 0 )  362 .78
v(t1 )  v(t 0 )
 30 .914
b1 
t1  t 0
550
500
ys
f ( range)


450
f x desired
400
362.78
350
10
12
x s  10
0
7
14
16
18
x s  range x desired
20
22
24
x s  10
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1
Linear Interpolation (contd)
517.35
550
500
ys
f ( range)


450
f x desired
400
362.78
350
10
12
14
x s  10
v(t )  b0  b1 (t  t 0 )
0
16
18
20
x s  range x desired
22
24
x s  10
1
 362.78  30.914(t  15), 15  t  20
At t  16
v(16)  362.78  30.914(16  15)
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 393 .69 m/s
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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
The upward velocity of a rocket is given as a function of
time in Table 1. Find the velocity at t=16 seconds using
the Newton Divided Difference method for quadratic
interpolation.
Table. Velocity as a
function of time
t (s) v(t ) (m/s)
0
10
15
20
22.5
30
10
0
227.04
362.78
517.35
602.97
901.67
Figure. Velocity vs. time data
for the rocket example http://numericalmethods.eng.usf.edu
Quadratic Interpolation (contd)
517.35
550
500
450
ys
400
f ( range)


f x desired
350
300
250
227.04
200
10
10
12
14
16
x s  range x desired
18
20
20
t 0  10, v(t 0 )  227.04
t1  15, v(t1 )  362.78
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t 2  20, v(t 2 )  517.35
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Quadratic Interpolation (contd)
b0  v(t 0 )
 227.04
v(t )  v(t 0 ) 362.78  227.04
b1  1

t1  t 0
15  10
 27.148
v(t 2 )  v(t1 ) v(t1 )  v(t 0 )

t 2  t1
t1  t 0
b2 
t 2  t0
30.914  27.148

10
517.35  362.78 362.78  227.04

20

15
15  10

20  10
 0.37660
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Quadratic Interpolation (contd)
v(t )  b0  b1 (t  t 0 )  b2 (t  t 0 )(t  t1 )
 227.04  27.148(t  10)  0.37660(t  10)(t  15),
10  t  20
At t  16,
v(16)  227.04  27.148(16  10)  0.37660(16  10)(16  15)  392 .19 m/s
The absolute relative approximate errora
order and second order polynomial is
a

obtained between the results from the first
392 .19  393 .69
x100
392 .19
= 0.38502 %
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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
16
f ( x3 )
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Example
The upward velocity of a rocket is given as a function of
time in Table 1. Find the velocity at t=16 seconds using
the Newton Divided Difference method for cubic
interpolation.
Table. Velocity as a
function of time
t (s) v(t ) (m/s)
0
10
15
20
22.5
30
17
0
227.04
362.78
517.35
602.97
901.67
Figure. Velocity vs. time data
for the rocket example http://numericalmethods.eng.usf.edu
Example
The velocity profile is chosen as
v(t )  b0  b1 (t  t 0 )  b2 (t  t 0 )(t  t1 )  b3 (t  t 0 )(t  t1 )(t  t 2 )
we need to choose four data points that are closest to t  16
t0  10,
v(t 0 )  227.04
t1  15,
v(t1 )  362.78
t 2  20,
v(t 2 )  517.35
t3  22.5, v(t 3 )  602.97
The values of the constants are found as:
b0 = 227.04; b1 = 27.148; b2 = 0.37660; b3 = 5.4347×10−3
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Example
b0
t0  10
227 .04
b1
27 .148
t1  15,
362 .78
b2
0.37660
5.4347103
30 .914
t2  20,
517 .35
b3
0.44453
34 .248
t3  22.5,
602 .97
b0 = 227.04; b1 = 27.148; b2 = 0.37660; b3 = 5.4347×10−3
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Example
Hence
v (t )  b0  b1 (t  t 0 )  b2 (t  t 0 )( t  t1 )  b3 (t  t 0 )( t  t1 )(t  t 2 )
 227.04  27.148( t  10)  0.37660(t  10)(t  15)
 5.4347 * 10 3 (t  10)( t  15)( t  20)
At t  16,
v (16)  227.04  27.148(16  10)  0.37660(16  10)(16  15)
 5.4347 * 10 3 (16  10)(16  15)(16  20)
 392.06 m/s
The absolute relative approximate errora
a

obtained is
392 .06  392 .19
x100
392 .06
= 0.033427 %
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Comparison Table
Order of
Polynomial
v(t=16)
m/s
Absolute Relative
Approximate Error
21
1
2
3
393.69
392.19
392.06
----------
0.38502 %
0.033427 %
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Distance from Velocity Profile
Find the distance covered by the rocket from t=11s to
t=16s ?
v (t)  227.04  27.148(t  10)  0.37660( t  10)( t  15)
 5.4347 * 10 (t  10)( t  15)( t  20)
3
 4.2541  21.265t  0.13204t 2  0.0054347t 3
10  t  22.5
10  t  22.5
So
16
s16  s11   v t dt
11
16
  (  4.2541  21.265t  0.13204t 2  0.0054347t 3 ) dt
11
16

t2
t3
t4 
  4.2541t  21.265  0.13204  0.0054347 
2
3
4  11

22
 1605 m
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Acceleration from Velocity Profile
Find the acceleration of the rocket at t=16s given that
v(t )  4.2541  21.265t  0.13204t 2  0.0054347t 3
a (t ) 
d
d
v(t )   4.2541  21.265t  0.13204t 2  0.0054347t 3 
dt
dt
 21.265  0.26408t  0.016304t 2
a(16)  21.265  0.26408(16)  0.016304(16) 2
 29.664 m / s 2
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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
THE END
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