Newton's Divided Difference Method for Interpolation: Linear and Quadratic, Slides of Mathematical Methods for Numerical Analysis and Optimization

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Newton’s Divided Difference

Polynomial Method of

Interpolation

What is Interpolation?

Given (x

,y

), (x

,y

), …… (x

n

,y

n

), find the

value of ‘y’ at a value of ‘x’ that is not given.

Newton’s Divided Difference Method

Linear interpolation: Given pass a linear

interpolant through the data

where

( x 0 , y 0 ), ( x 1 , y 1 ),

f 1 (^) ( x )  b 0  b 1 ( x  x 0 )

b 0 (^)  f ( x 0 )

1 0

1 0 1

( ) ( )

x x

f x f x b 

 

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

Figure. Velocity vs. time data

for the rocket example

t (s) v ( t )(m/s)

Linear Interpolation (contd)

(^35010 12 14 16 18 20 22 )

400

450

500

517.35^550

y (^) s

f range( ) f x (^) de sire d

x (^) s 0  10 x (^) s  rangex (^) de sire d x (^) s 1  10 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

Quadratic Interpolation

Given (^ x 0 , y 0 ), (^ x 1 , y 1 ), and (^ x 2 , y 2 ), fit a quadratic interpolant through the data.

f (^) 2 ( x ) b 0  b 1 ( x  x 0 ) b 2 ( x  x 0 )( x  x 1 )

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 

Quadratic Interpolation (contd)

(^20010 12 14 16 18 )

250

300

350

400

450

500

517.35^550

y (^) s

f range( ) f x (^) de sire d

10 x (^) s  rangex (^) de sire d 20

t 0  10 , v ( t 0 )  227. 04

t 1  15 , v ( t 1 )  362. 78

t 2  20 , v ( t 2 )  517. 35

Quadratic Interpolation (contd)

b 0 (^)  v ( t 0 )

 227. 04

1 0

1 0 1

t t

v t v t b



 27. 148

2 0

1 0

1 0

2 1

2 1

2

t t

t t

v t v t

t t

v t v t

b



 0. 37660

General Form

2 0 1 0 2 0 1

f x  b  b x  x  b x  x x  x

where

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 )

b 0 (^)  f [ x 0 ] f ( x 0 )

1 0

1 0 1 1 0

[ , ]

x x

f x f x b f x x 

2 0

1 0

1 0

2 1

2 1

2 0

2 1 1 0 2 2 1 0

[ , ] [ , ]

[ , , ]

x x

x x

f x f x

x x

f x f x

x x

f x x f x x b f x x x 

General Form

Given ( 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 ]

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

Figure. Velocity vs. time data

for the rocket example

t (s) v ( t )(m/s)

Example

The velocity profile is chosen as

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 )

we need to choose four data points that are closest to t ^16

t 0  10 , v ( t 0 ) 227. 04

t 1  15 , v ( t 1 ) 362. 78

t 2  20 , v ( t 2 ) 517. 35

t 3  22. 5 , v ( t 3 ) 602. 97

The values of the constants are found as:

b 0 = 227.04; b 1 = 27.148; b 2 = 0.37660; b 3 = 5.4347×

− 3

Example

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 )

5. 4347 * 10 ( 10 )( 15 )( 20 )

6. 04 27. 148 ( 10 ) 0. 37660 ( 10 )( 15 )

3    

     

 t t t

t t t

At t  16 ,

5. 4347 * 10 ( 16 10 )( 16 15 )( 16 20 )

( 16 ) 227. 04 27. 148 ( 16 10 ) 0. 37660 ( 16 10 )( 16 15 )

3    

     



v

 392. 06 m/s

The absolute relative approximate error  a obtained is

 a x 100

392. 06

Comparison Table

Order of

Polynomial

1 2 3

v(t=16)

m/s

393.69 392.19 392.

Absolute Relative

Approximate Error

---------- 0.38502 % 0.033427 %