1

Newton’s Divided Difference Polynomial

Method of Interpolation

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Newton’s Divided Difference Method

Linear interpolation: Given pass a linear interpolant through the data

where

),,( 00 yx ),,( 11 yx

)()( 0101 xxbbxf

)( 00 xfb

01

011

)()(

xx

xfxfb

3

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.

t v(t)

s m/s

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67

Table 1: Velocity as a function of time

Figure 2: Velocity vs. time data for the rocket example

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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

)()(

tt

tvtvb

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

7

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.

t v(t)

s m/s

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67

Table 1: Velocity as a function of time

Figure 2: Velocity vs. time data for the rocket example

8

Quadratic Interpolation (contd)

0104.203.0 623 xxxxf10 12 14 16 18 20

200

250

300

350

400

450

500

550517.35

227.04

y s

f range( )

f x desired

2010 x s range x desired

,100 t 04.227)( 0 tv

,151 t 78.362)( 1 tv

,202 t 35.517)( 2 tv

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Quadratic Interpolation (contd)

)( 00 tvb

04.227

01

011

)()(

tt

tvtvb

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

10

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

01011

)()(],[

xx

xfxfxxfb

02

01

01

12

12

02

01120122

)()()()(

],[],[],,[

xx

xx

xfxf

xx

xfxf

xx

xxfxxfxxxfb

12

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

xxxxxxxxxxf

xxxxxxxfxxxxfxfxf

0b

0x )( 0xf 1b

],[ 01 xxf 2b

1x )( 1xf ],,[ 012 xxxf 3b

],[ 12 xxf ],,,[ 0123 xxxxf

2x )( 2xf ],,[ 123 xxxf

],[ 23 xxf

3x )( 3xf

14

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.

t v(t)

s m/s

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67

Table 1: Velocity as a function of time

Figure 2: Velocity vs. time data for the rocket example

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Example

The 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 = 5.4347*10-3

16

Example

b0 = 227.04; b1 = 27.148; b2 = 0.37660; b3 = 5.4347*10-3

0b

100 t 04.227 1b

148.27 2b

,151 t 78.362 37660.0 3b

914.30 3104347.5 x

,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(148.2704.2273

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 %

Latihan

Misalkan Nilai-nilai fungsi seperti pada tabel berikut, tentukanlah nilai aproksimasi dari f(0,1), f(0,35), dan f(1,16) berturut-turut menggunakan interpolasi beda terbagi Newton linear, kuadratik, kubik.

x f(x)

0,0 0,000

0,2 0,406

0,4 0,846

0,6 1,368

0,8 2,060

1,0 3,114

1,2 5,114

20

Distance from Velocity Profile

Find 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

ttttv 5.2210 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

m 1605

21

Acceleration from Velocity Profile

Find 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