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 + f1 x1 - x0 x1 - x0

 x - x1 ( x - x0 ) = f1 - f0 + - 1 f1 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 Ñ2f 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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