Download - 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