NEWTON'S BACKWARD DIFFERENCE FORMULA
This is another way of approximating a function with an nth degree polynomial passing through (n+1) equally spaced points.
As a particular case, lets again consider the linear approximation to f(x)
f1 - f0
f0x1 - f1x0
f(x) @ P1(x) =
x + 
x1 - x0
x1 - x0

 
x1 - x
x - x0

 f0
f
x1 - x0
x1 - x0

 
x - x1
(
x - x0
)
f1
f0 +
 - 1 f
x0 - x1
x - x0

 
x - x1
f1
(f1- f0
x1 - x0
= f1 + sÑf1
where s = (x - x1) / (x1 - x0) and Ñf1 is the backward difference of  at x1.
The same can be obtained from the difference operators as follows.
f(x)  =  f (xn
x - xn
 h)

h
 
=  f(xn + sh)   =   Es f(xn),      s  =  ( x - xn )/ h

 
=  ( 1 - Ñ )s f( xn )

 
         = fn + sÑfn +
 s(s + 1) 
Ñ2fn + . . . +
 s(s + 1) . . . (s + n -1)
Ñnfn + . . .
2!
n!
Neglecting the (n+1) difference as onwards we get
 f(x)  @ Pn(x)  = fn + sÑfn +
 s(s + 1) 
Ñ2fn + . . . +
 s(s + 1) . . . (s + n -1)
Ñnfn
2!
n!
This polynomial is called the Newton - Gregory backward difference formula.
The error in this interpolation
 En(x)  = 
 s(s + 1) . . . (s + n)
hn+1f(n+1)(x) ( x0 < (x) < xn )
(n + 1)!
Example :

Find f(0.15) using Newton backward difference table from the data

x
f(x)
 Ñf
Ñ2
 Ñ3f
 Ñ4f
0.1
0.09983
 
 
 
 
 
 
0.09884
 
 
 
0.2
0.19867 
 
-0.00199
 
 
 
 
0.09685
 
-0.00156
 
0.3
0.29552
 
-0.00355
 
0.00121
 
 
0.0939 
 
-0.00035
 
0.4
0.38942
 
 -0.0039
 
 
 
 
0.09
 
   
0.5
0.47943
 
     
s  =  ( x - xn ) / h  =  (0.15 - 0.5) / 0.1  =  -3.5
f(0.15)  = fn + sÑfn +
 s(s + 1) 
Ñ2fn s(s + 1)(s + 2) Ñ3fn
 s(s + 1)(s + 2)(s + 3)
Ñ4fn
2!
3!
              = .97943 + (-3.5)*.09 +
 (-3.5)(-3.5 + 1) 
(-.0039) (-3.5)(-3.5 + 1)(-3.5 + 2) (-.00035)
2!
3!
 
(-3.5)(-3.5 + 1)(-3.5 + 2)(-3.5 + 3)
(.00121) 
4!
=  0.97943 - 0.315 - 0.01706 + 0.000765625 + 0.00033086
= 0.14847
Error in the approximation
 E5(x)  = 
 -3.5(-3.5 +1)(-3.5 + 2)(-3.5 + 3)(-3.5 + 4)
h5f5(x) ( 0.1 < x < 0.5 )
5!
= 2.734375 x 10-7 f5(x)
 

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