Newton divided difference interpolation

1. Numerical Analysis and
Statistics

2. Presented By:
Sr. No. Name Enroll. No.
1 Vishal Donga 140800107067
2 Lucky Vishwakarma 140800107068
3 Jaimin Vitthalpara 140800107069
4 Nitu Yadav 140800107070
Guided By:
Prof. Poonam Kumari

3. Newton’s Divided Difference
Polynomial Method of
Interpolation

4. What is Interpolation ?
Given (x0,y0), (x1,y1), …… (xn,yn), find the
value of ‘y’ at a value of ‘x’ that is not given.

5. 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
01
1
)()(
xx
xfxf
b
−
−
=

6. 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, f
at ix
[ ] ( )ii xfxf =

7. [ ]1i if x , x +
fThe divided difference of a function,
called as the first divided difference, is
denoted
ixwith respect to and 1ix +
[ ]
[ ] [ ]1
1
1
i i
i i
i i
f x f x
f x , x
x x
+
+
+
−
=
−

8. 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 1
1 2
2
i i i i
i i i
i i
f x , x f x , x
f x , x , x
x x
+ + +
+ +
+
−
=
−

9. [ ]
[ ] [ ]
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 , x
x x
+ + +
+ + + + +
+
−
=
−
The third divided difference with respect to
ix 1ix + 2ix + 3ix +
, and,

10. 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.

11. First
divided differences
Second
divided differences
Third
divided differences

12. Example
Find Newton’s interpolating polynomial to
approximate a function whose 5 data points
are given below.
( )f x
2.0 0.85467
2.3 0.75682
2.6 0.43126
2.9 0.22364
3.2 0.08567
x

13. 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

14. 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 .= = −

15. ( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
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
= + −
+ − −
+ − − −
+ − − − −

16. ( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
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.

17. ( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
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 . .≈ =

18. Thank You