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Chapter 05.
Newton’s Divided Difference Interpolation
After reading this chapter, you should be able to:
- derive Newton’s divided difference method of interpolation,
- apply Newton’s divided difference method of interpolation, and
- apply Newton’s divided difference method interpolants to find derivatives and
integrals.
What is interpolation?
Many times, data is given only at discrete points such as x 0 , y 0 , x 1 , y 1 , ......, xn 1 , yn 1 ,
xn , yn . So, how then does one find the value of y at any other value of x? Well, a
continuous function f x may be used to represent the n 1 data values with f x passing
through the n 1 points (Figure 1). Then one can find the value of y at any other value of
x. This is called interpolation.
Of course, if x falls outside the range of x for which the data is given, it is no longer
interpolation but instead is called extrapolation.
So what kind of function f x should one choose? A polynomial is a common
choice for an interpolating function because polynomials are easy to
(A) evaluate,
(B) differentiate, and
(C) integrate,
relative to other choices such as a trigonometric and exponential series.
Polynomial interpolation involves finding a polynomial of order n that passes
through the n 1 points. One of the methods of interpolation is called Newton’s divided
difference polynomial method. Other methods include the direct method and the Lagrangian
interpolation method. We will discuss Newton’s divided difference polynomial method in
this chapter.
Newton’s Divided Difference Polynomial Method
To illustrate this method, linear and quadratic interpolation is presented first. Then, the
general form of Newton’s divided difference polynomial method is presented. To illustrate
the general form, cubic interpolation is shown in Figure 1.
05.03.2 Chapter 05.
Figure 1 Interpolation of discrete data.
Linear Interpolation
Given ( x 0 (^) , y 0 ) and ( x 1 (^) , y 1 ),fit a linear interpolant through the data. Noting y f ( x ) and
y 1 (^) f ( x 1 ), assume the linear interpolant f 1 (^) ( x )is given by (Figure 2)
f (^) 1 ( x ) b 0 b 1 ( x x 0 )
Since at x x 0 ,
f (^) 1 ( x 0 ) f ( x 0 ) b 0 b 1 ( x 0 x 0 ) b 0
and at x x 1 ,
f (^) 1 ( x 1 ) f ( x 1 ) b 0 b 1 ( x 1 x 0 )
f ( x 0 ) b 1 ( x 1 x 0 )
giving
1 0
1 0 1
x x
f x f x b
So
b 0 (^) f ( x 0 )
1 0
1 0 1
x x
f x f x b
giving the linear interpolant as
f (^) 1 ( x ) b 0 b 1 ( x x 0 )
1 0
1 0 1 0 x x x x
f x f x f x f x
x 0 , y 0
x 1 , y 1
x 2 , y 2
x 3 , y 3
f x
x
y
05.03.4 Chapter 05.
1 0
1 0 1
t t
vt vt b
Figure 3 Graph of velocity vs. time data for the rocket example.
Hence
v ( t ) b 0 b 1 ( t t 0 )
362. 78 30. 914 ( t 15 ), 15 t 20
At t 16 ,
v ( 16 ) 362. 78 30. 914 ( 16 15 )
393. 69 m/s
If we expand
v ( t ) 362. 78 30. 914 ( t 15 ), 15 t 20
we get
v ( t ) 100. 93 30. 914 t , 15 t 20
and this is the same expression as obtained in the direct method.
Quadratic Interpolation
Given ( x 0 (^) , y 0 ), ( x 1 (^) , y 1 ), and ( x (^) 2 , y 2 ), fit a quadratic interpolant through the data. Noting
y f ( x ), y (^) 0 f ( x 0 ), y 1 (^) f ( x 1 ),and y 2 (^) f ( x 2 ),assume the quadratic interpolant f 2 ( x )
is given by
f (^) 2 ( x ) b 0 b 1 ( x x 0 ) b 2 ( x x 0 )( x x 1 )
Newton’s Divided Difference Interpolation 05.03.
At x x 0 ,
f (^) 2 ( x 0 ) f ( x 0 ) b 0 b 1 ( x 0 x 0 ) b 2 ( x 0 x 0 )( x 0 x 1 )
b 0
b 0 (^) f ( x 0 )
At x x 1
f (^) 2 ( x 1 ) f ( x 1 ) b 0 b 1 ( x 1 x 0 ) b 2 ( x 1 x 0 )( x 1 x 1 )
f ( x 1 ) f ( x 0 ) b 1 ( x 1 x 0 )
giving
1 0
1 0 1
x x
f x f x b
At x x 2
f (^) 2 ( x 2 ) f ( x 2 ) b 0 b 1 ( x 2 x 0 ) b 2 ( x 2 x 0 )( x 2 x 1 )
1 0
1 0 2 0 x x b x x x x x x
f x f x f x f x
Giving
2 0
1 0
1 0
2 1
2 1
2
x x
x x
f x f x
x x
f x f x
b
Hence the quadratic interpolant is given by
f (^) 2 ( x ) b 0 b 1 ( x x 0 ) b 2 ( x x 0 )( x x 1 )
2 0
1 0
1 0
2 1
2 1
0 1 0
1 0 0 x x x x x x
x x
f x f x
x x
f x f x
x x x x
f x f x f x
Figure 4 Quadratic interpolation.
x 0 , y 0
x 1 , y 1
x 2 , y 2
f^ ^ x
2
y
x
Newton’s Divided Difference Interpolation 05.03.
Hence
v ( t ) b 0 b 1 ( t t 0 ) b 2 ( t t 0 )( t t 1 )
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
If we expand
v ( t ) 227. 04 27. 148 ( t 10 ) 0. 37660 ( t 10 )( t 15 ), 10 t 20
we get
2 v ( t ) 12. 05 17. 733 t 0. 37660 t , 10 ^ t ^20
This is the same expression obtained by the direct method.
General Form of Newton’s Divided Difference Polynomial
In the two previous cases, we found linear and quadratic interpolants for Newton’s divided
difference method. Let us revisit the quadratic polynomial interpolant formula
f (^) 2 ( x ) b 0 b 1 ( x x 0 ) b 2 ( x x 0 )( x x 1 )
where
b 0 (^) f ( x 0 )
1 0
1 0 1
x x
f x f x b
2 0
1 0
1 0
2 1
2 1
2
x x
x x
f x f x
x x
f x f x
b
Note that b 0 , b 1 , and b 2 are finite divided differences. b 0 , b 1 ,and b 2 are the first, second,
and third finite divided differences, respectively. We denote the first divided difference by
f [ x 0 ] f ( x 0 )
the second divided difference by
1 0
1 0 1 0
[ , ]
x x
f x f x f x x
and the third divided difference by
2 0
2 1 1 0 2 1 0
[ , ] [ , ]
[ , , ]
x x
f x x f x x f x x x
2 0
1 0
1 0
2 1
x x
x x
f x f x
x x
f x f x
where f [ x 0 ], f [ x 1 , x 0 ], and f [ x 2 , x 1 , x 0 ] are called bracketed functions of their variables
enclosed in square brackets.
Rewriting,
f (^) 2 ( x ) f [ x 0 ] f [ x 1 , x 0 ]( x x 0 ) f [ x 2 , x 1 , x 0 ]( x x 0 )( x x 1 )
05.03.8 Chapter 05.
This leads us to writing the general form of the Newton’s divided difference polynomial for
n 1 data points, x 0 , y 0 , x 1 , y 1 ,......, xn 1 , yn 1 , xn , yn , as
fn ( x ) b 0 b 1 ( x x 0 ).... bn ( x x 0 )( x x 1 )...( x xn 1 )
where
b 0 (^) f [ x 0 ]
b 1 (^) f [ x 1 , x 0 ]
b 2 (^) f [ x 2 , x 1 , x 0 ]
b (^) n 1 f [ xn 1 , xn 2 ,...., x 0 ]
b (^) n f [ xn , xn 1 ,...., x 0 ]
where the definition of the
th m divided difference is
b (^) m f [ xm ,........, x 0 ]
0
[ ,........, 1 ] [ 1 ,........, 0 ]
x x
f x x f x x
m
m m
From the above definition, it can be seen that the divided differences are calculated
recursively.
For an example of a third order polynomial, given ( x 0 (^) , y 0 ), ( x 1 (^) , y 1 ), ( x (^) 2 , y 2 ),and ( x 3 , y 3 ),
( ) [ ] , , , ( )
3 2 1 0 0 1 2
3 0 1 0 0 2 1 0 0 1
f x x x x x x x x x x
f x f x f x x x x f x x x x x x x
Figure 5 Table of divided differences for a cubic polynomial.
Example 3
The upward velocity of a rocket is given as a function of time in Table 3.
x 0 f x 0
b 0
x 1 f x 1
x 2 f x 2
x 3 f x 3
b 1
b 2
b 3
f x 1 , x 0
f x 2 , x 1
f x 3 , x 2
f x 2 , x 1 , x 0
f x 3 , x 2 , x 1
f x 3 , x 2 , x 1 , x 0
05.03.10 Chapter 05.
2 1
2 1 2 1
[ , ]
t t
vt vt v t t
v [ t 1 , t 0 ] 27. 148
2 0
2 1 1 0 2
[ , ] [ , ]
t t
vt t vt t b
b 3 (^) v [ t 3 , t 2 , t 1 , t 0 ]
3 0
[ 3 , 2 , 1 ] [ 2 , 1 , 0 ]
t t
vt t t vt t t
3 1
3 2 2 1 3 2 1
[ , ] [ , ]
[ , , ]
t t
vt t vt t v t t t
3 2
3 2 3 2
[ , ]
t t
vt vt v t t
2 1
2 1 2 1
[ , ]
t t
vt vt v t t
3 1
3 2 2 1 3 2 1
[ , ] [ , ]
[ , , ]
t t
vt t vt t v t t t
v [ t 2 , t 1 , t 0 ] 0. 37660
3 0
3 2 1 2 1 0 3
[ , , ] [ , , ]
t t
vt t t vt t t b
3
- 4347 10
Hence
v ( t ) b 0 b 1 ( t t 0 ) b 2 ( t t 0 )( t t 1 ) b 3 ( t t 0 )( t t 1 )( t t 2 )
Newton’s Divided Difference Interpolation 05.03.
3
t t t
t t t
At t 16 ,
3
v
392. 06 m/s
b) The distance covered by the rocket between t 11 s and t 16 s can be calculated from
the interpolating polynomial
3
t t t
vt t t t
2 3 t t t 10 t 22. 5
Note that the polynomial is valid between t 10 and t 22.5 and hence includes the limits
of t^ ^11 and t^ ^16.
So
16
11
s 16 s 11 vtdt
( 4. 2541 21. 265 t 0. 13204 t 0. 0054347 t ) dt
2 3
16
11
16
11
2 3 4
t t t t
1605 m
c) The acceleration at t 16 is given by
( 16 ) v ( t ) t 16 dt
d a
( ) v ( t ) dt
d a t
2 3
- 2541 21. 265 t 0. 13204 t 0. 0054347 t dt
d
2 21. 265 0. 26408 t 0. 016304 t 2 a ( 16 ) 21. 265 0. 26408 ( 16 ) 0. 016304 ( 16 ) 2 29. 664 m/s
INTERPOLATION
Topic Newton’s Divided Difference Interpolation
