newton divided difference interpolation

Numerical Analysis and Statistics

Presented By:Sr. No. Name Enroll. No.

1 Vishal Donga 1408001070672 Lucky Vishwakarma 1408001070683 Jaimin Vitthalpara 1408001070694 Nitu Yadav 140800107070

Guided By:

Prof. Poonam Kumari

Newton’s Divided Difference Polynomial

Method of Interpolation

What is Interpolation ?Given (x0,y0), (x1,y1), …… (xn,yn), 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

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

)()( 0101 xxbbxf

)( 00 xfb

01

011

)()(xxxfxf

b

Divided differences and the coefficients f

ix if x The divided difference of a function,

with respect to is denoted as

It is called as zeroth divided difference and is simply the value of the function, fat ix

ii xfxf

1i if x , x

fThe divided difference of a function,

called as the first divided difference, is denoted

ixwith respect to and 1ix

11

1

i ii i

i i

f x f xf x , x

x x

fThe divided difference of a function,

called as the second divided difference, is denoted as

ixwith respect to and1ix , 2ix

1 2i i if x , x , x

1 2 11 2

2

i i i ii i i

i i

f x , x f x , xf x , x , x

x x

1 2 3

1 2 3 1 2

3

i i i i

i i i i i i

i i

f x , x , x , x

f x , x , x f x , x , xx x

The third divided difference with respect to ix 1ix 2ix 3ix , and ,

The coefficients of Newton’s interpolating polynomial are:

00 xfa

101 x,xfa

2102 x,x,xfa

32103 x,x,x,xfa

432104 x,x,x,x,xfa and so on.

x xf First divided differences

Second divided differences

Third divided differences

0x 0xf

01

0110 xx

xfxfx,xf

1x 1xf 02

1021210 xx

x,xfx,xfx,x,xf

12

1221 xx

xfxfx,xf

03

2103213210 xx

x,x,xfx,x,xfx,x,x,xf

2x 2xf 13

2132321 xx

x,xfx,xfx,x,xf

23

2332 xx

xfxfx,xf

14

3214324321 xx

x,x,xfx,x,xfx,x,x,xf

3x 3xf 24

3243432 xx

x,xfx,xfx,x,xf

34

3443 xx

xfxfx,xf

25

4325435432 xx

x,x,xfx,x,xfx,x,x,xf

4x 4xf 35

4354543 xx

x,xfx,xfx,x,xf

45

4554 xx

xfxfx,xf

5x 5xf

ExampleFind Newton’s interpolating polynomial to approximate a function whose 5 data

points are given below.

f x2.0 0.85467

2.3 0.75682

2.6 0.43126

2.9 0.22364

3.2 0.08567

x

i ix ixf ii x,xf 1 iii x,x,xf 12 ii x,,xf 3 ii x,,xf 4

0 2.0 0.85467

-0.32617

1 2.3 0.75682 -1.26505

-1.08520 2.13363

2 2.6 0.43126 0.65522 -2.02642

-0.69207 -0.29808

3 2.9 0.22364 0.38695

-0.45990

4 3.2 0.08567

The 5 coefficients of the Newton’s interpolating polynomial are:

0 0 0 85467a f x .

1 0 1 0 32617a f x , x .

2 0 1 2 1 26505a f x , x , x .

3 0 1 2 3 2 13363a f x , x , x , x .

4 0 1 2 3 4 2 02642a f x , x , x , x , x .

0 1 0

2 0 1

3 0 1 2

4 0 1 2 3

P x a a x x

a x x x x

a x x x x x x

a x x x x x x x x

0 85467 0 32617 2 0

-1.26505 2 0 2 3

2 13363 2 0 2 3 2 6

2 02642 2 0 2 3 2 6 2 9

P x . . x .

x . x .

. x . x . x .

. x . x . x . x .

P(x) can now be used to estimate the value of the function f(x) say at x = 2.8.

2 8 0 85467 0 32617 2 8 2 0

-1.26505 2 8 2 0 2 8 2 3

2 13363 2 8 2 0 2 8 2 3 2 8 2 6

2 02642 2 8 2 0 2 8 2 3 2 8 2 6 2 8 2 9

P . . . . .

. . . .

. . . . . . .

. . . . . . . . .

2 8 2 8 0 275f . P . .

Thank You