Transcript PPT

Newton’s Divided Difference
Polynomial Method of
Interpolation
Industrial Engineering Majors
Authors: Autar Kaw, Jai Paul
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
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1
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
A curve needs to be fit through the given points to fabricate
the cam. If the cam follows a straight line profile between
x = 1.28 to x = 0.66, what is the value of y at x=1.1 ? Find
using the Newton’s divided difference method for linear
interpolation.
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x (in.) y (in.)
2.20
0.00
1.28
0.88
0.66
1.14
0.00
1.20
–0.60 1.04
–1.04 0.60
–1.20 0.00
Cam Profile
4
3
5
2
1.4
1.2
1
6
0.8
y
Point
1
2
3
4
5
6
7
Y
7
0.6
1
0.4
X
0.2
0
-2
-1
0
1
2
x
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Linear Interpolation
1.14
y( x)  b0  b1 ( x  x0 )
x0  1.28, y( x0 )  0.88
x1  0.66, y( x1 )  1.14
1.1
1.05
ys
f ( range)

f x desired

b0  y( x0 )
1
0.95
0.9
 0.88
y (x 1 )  y (x 0 ) 1.14  0.88

b1 
0.66  1.28
x1 x 0
0.88
0.85
5
x s  10
0
0
5
x s  range  x desired
 0.41935
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x s  10
1
Linear Interpolation (contd)
1.14
1.1
1.05
ys
f ( range)

f x desired

1
0.95
0.9
0.88
0.85
5
x s  10
0
0
x s  range  x desired
5
10
x s  10
1
y( x)  b0  b1 ( x  x0 )
 0.88  0.41935( x  1.28),
0.66  x  1.28
At x  1.10
y (1.10)  0.88  0.41935(1.10  1.28)
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 0.95548 in.
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Quadratic Interpolation
Given ( x0 , y0 ), ( x1 , y1 ), and ( x2 , y2 ), 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 ( x 2 )  f ( x1 ) f ( x1 )  f ( x0 )

x 2  x1
x1  x0
b2 
x 2  x0
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Example
A curve needs to be fit through the given points to fabricate
the cam. If the cam follows a straight line profile between
x = 1.28 to x = 0.66, what is the value of y at x=1.1 ? Find
using the Newton’s divided difference method for quadratic
interpolation.
10
x (in.) y (in.)
2.20
0.00
1.28
0.88
0.66
1.14
0.00
1.20
–0.60 1.04
–1.04 0.60
–1.20 0.00
Cam Profile
4
3
5
2
1.4
1.2
1
6
0.8
y
Point
1
2
3
4
5
6
7
Y
7
0.6
1
0.4
X
0.2
0
-2
-1
0
1
2
x
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Quadratic Interpolation (contd)
1.14
y( x)  b0  b1 ( x  x0 )  b2 ( x  x0 )( x  x1 )
x0  2.20, y( x0 )  0.00
x1  1.28, y( x1 )  0.88
1.2
1
0.8
ys
f ( range)

f x desired

x2  0.66, y( x2 )  1.14
0.6
0.4
0.2
0
0
0.6
0.66
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0.8
1
1.2
1.4
1.6
x s  range  x desired
1.8
2
2.2
2.2
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Quadratic Interpolation (contd)
b0  y( x0 )
 0.00
y( x1 )  y ( x0 ) 0.88  0.00

b1 
1.28  2.20
x1  x0
 0.95652
y ( x 2 )  y ( x1 ) y ( x1 )  y ( x0 ) 1.14  0.88 0.88  0.00


x 2  x1
x1  x0
b2 
 0.66  1.28 1.28  2.20
0.66  2.20
x2  x0
 0.41935  0.95652

 1.54
 0.34881
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Quadratic Interpolation (contd)
y ( x)  b0  b1 ( x  x0 )  b2 ( x  x0 )( x  x1 )
 0  0.95652( x  2.20)  0.34881( x  2.20)( x  1.28), 0.66  x  2.20
At x  1.10,
y (1.10)  0  0.95652(1.10  2.20)  0.34881(1.10  2.20)(1.10  1.28)
 0.98311 in.
The absolute relative approximate error a obtained between
the results from the first and second order polynomial is
0.98311  0.95548
 100
0.98311
 2.8100%
a 
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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 , y n1 , xn , y n  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 , y2 ), and ( x3 , y3 ), is
f 3 ( x)  f [ x0 ]  f [ x1 , x0 ]( x  x0 )  f [ x2 , x1 , x0 ]( x  x0 )( x  x1 )
 f [ x3 , x 2 , x1 , x0 ]( x  x0 )( x  x1 )( x  x 2 )
b0
x0
f ( x0 )
b1
f [ x1 , x0 ]
x1
b2
f [ x2 , x1 , x0 ]
f ( x1 )
f [ x3 , x2 , x1 , x0 ]
f [ x2 , x1 ]
x2
b3
f [ x3 , x2 , x1 ]
f ( x2 )
f [ x3 , x 2 ]
x3
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f ( x3 )
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Example
A curve needs to be fit through the given points to fabricate
the cam. If the cam follows a straight line profile between
x = 1.28 to x = 0.66, what is the value of y at x=1.1 ? Find
using the Newton’s divided difference method for a sixth
order polynomial.
17
x (in.) y (in.)
2.20
0.00
1.28
0.88
0.66
1.14
0.00
1.20
–0.60 1.04
–1.04 0.60
–1.20 0.00
Cam Profile
4
3
5
2
1.4
1.2
1
6
0.8
y
Point
1
2
3
4
5
6
7
Y
7
0.6
1
0.4
X
0.2
0
-2
-1
0
1
2
x
http://numericalmethods.eng.usf.edu
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Example
The value of y profile is chosen as
y ( x)  b0  b1 ( x  x0 )  b2 ( x  x0 )( x  x1 )  b3 ( x  x0 )( x  x1 )( x  x 2 )
 b4 ( x  x0 )( x  x1 )( x  x 2 )( x  x3 )  b5 ( x  x0 )( x  x1 )( x  x 2 )( x  x3 )( x  x 4 )
 b6 ( x  x0 )( x  x1 )( x  x 2 )( x  x3 )( x  x 4 )( x  x5 )
x0  2.20,
y( x0 )  0.00
x1  1.28,
y( x1 )  0.88
x2  0.66,
y( x2 )  1.14
x3  0.00,
y( x3 )  1.20
x4  0.60, y( x4 )  1.04
x5  1.04, y( x5 )  0.60
x6  1.20, y( x6 )  0.00
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Example
The values of the constants are found as:
b0  0.00
b1  0.95652
b2  0.34881
b3  0.041914
b4  0.020135
b5  0.024834
b6  0.17103
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Example
y ( x)  b0  b1 ( x  x0 )  b2 ( x  x0 )( x  x1 )  b3 ( x  x0 )( x  x1 )( x  x 2 )
 b4 ( x  x0 )( x  x1 )( x  x 2 )( x  x3 )  b5 ( x  x0 )( x  x1 )( x  x 2 )( x  x3 )( x  x 4 )
 b6 ( x  x0 )( x  x1 )( x  x 2 )( x  x3 )( x  x 4 )( x  x5 )
 0  0.95652( x  2.2)  0.34881( x  2.2)( x  1.28)
 0.041914( x  2.2)( x  1.28)( x  0.66)
 0.020135( x  2.2)( x  1.28)( x  0.66)( x  0)
 0.024834( x  2.2)( x  1.28)( x  0.66)( x  0)( x  0.6)
 0.17103( x  2.2)( x  1.28)( x  0.66)( x  0)( x  0.6)( x  1.04)
y( x)  1.2  0.25112 x  0.27255x 2  0.56765 x 3
 0.072013x 4  0.45241x 5  0.17103x 6 ,  1.20  x  2.20
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Example
y ( x)  0  0.95652( x  2.2)  0.34881( x  2.2)( x  1.28)  0.041914( x  2.2)( x  1.28)( x  0.66)
 0.020135( x  2.2)( x  1.28)( x  0.66)( x  0)
 0.024834( x  2.2)( x  1.28)( x  0.66)( x  0)( x  0.6)
 0.17103( x  2.2)( x  1.28)( x  0.66)( x  0)( x  0.6)( x  1.04)
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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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