NEWTON'S FORWARD DIFFERENCE FORMULA
Making use of forward difference operator and forward difference table ( will be defined a little later) this scheme simplifies the calculations involved in the polynomial approximation of fuctons which are known at equally spaced data points.
Consider the equation of the linear interpolation optained in the earlier section :
 f1 - f0 f0x1 - f1x0 f(x) @ P1(x) = ax-1b = x + x1 - x0 x1 - x0

 = 1 [(x1 - x)f0 + (x-x0)f1] (x1 - x0)

 x1 - x x - x0 x - x0 f0 + (f1- f0)   + f0 x1 - x0 x1 - x0 x1 - x0
 x - x0 = f0 + (f1- f0) x1 - x0
= f0 + r Df0    [ r = (x - x0) / (x1 - x0Df0 = f1 - f0 ]
since x1 - x0 is the step lenght h, r can be written as (x - x0)/h and will be between (0, 1).
Error in the linear interpolation :
If  e(x)  is the error in the linear interpolation then
e(x)  =  P1(x) - f(x)   =   f0 + r(f1 - f0) - f(x)
By Taylor's theorem
f(x)   =   f(x+  r h)   =   f0  +  r h f0'   +  1/2 r2 h2 f0'' (f),       x0 < f < x1
f1       =   f(x1)       =    f(x0 + h)    =    f0 + h f0'  + 1/2 h2 f0''(u),                 x0 < u < x1

Þe(x)  =  f +  r ( f0 + h f0' + 1/2 h2 f0'' (u) - f0 ) - ( f0 + r h f0' + 1/2 s2 h2 f0''(t) )

=  1/2 h2 ( r f0''(u) - r2 f0''(t) )
Let us assume that the second derivative of the function is bounded such that | f ''(x) | < M2 and since r < 1 we have | e(x) |  <    1/2 h2 ( M2+M2)   =   h2M2
The general formula is very convenient to find the function value at various points if forward difference at various points are avilable. Similarly the polynomial approximations of functions of higher degree also can be expressed in terms of   r  and forward differences of higher order. Instead of using the method of solving the system as we did earlier it is convenient to use binomial formulae involving the difference operators to generate the higher order interpolation formuale.
Newton's forward difference formulae :
Let the function  f  is known at  n+1  equally spaced data points  a = x0 < x1 < ... <  =  xn = b  in the interval [a,b] as f0, f1, . . . fn. Then the n  the degree polynomial approximation of  f(x)  can be given as
 n f(x)   @   Pn(x)   = S ( r i ) Di f0 i=0
where r = (x-x0 ) / n   Þ   x  = x+  r h   Þ 0 <r <n
 and ( r ) are the binomial coefficients defined as ( r ) = 1, ( r ) = r(r - 1) . . . (r - i + 1) for any integer i > 0 i 0 i i!
Proof :
To prove that the given result is the n the degree polynomial approximation of f(x) it is sufficeint to prove that at the node i i.e., at x = xi  the polynomial approximation Pn(x) gives fi for any tabulated value xi since the curve f(x) passes through the node points xi,  i = 0, 1, . . ., n
Take r  =  k
Þ     x  =  x0 + r h  =  x0 + k h  = xk

 k Þ  fk = S ( k i ) = f0 + k Df0 + . . . + ( k k ) Dk f0 i=0

 the terms after k need not be considered since ( n ) = 0, for  n < i i
we can prove this result by mathematical induction.
consider k = 0
 0 Þ f0 = S Di f0 = f0 hence the result is true for k = 0 and assume that the result is also true upto k = 1, 2, . . . p i=0

 fp =  f0 +  rDf0  + . . . + ( p ) Dp f0 p
consider
fp+1 =  fp + Dfp
 = ( p )Df0 0
+
 ( p )Df0 1
+ . . . +
 ( p )Dpf0 p
+
 ( p )Df0 0
+ . . . +
 ( p )Dpf0 p-1
+
 ( p )Dp+1f0 p
 k+1 S ( k+1 i ) Di f0 i=0
since
 ( p ) i
+
 ( p ) i-1
=
 ( p+1 ) i
i.e., the nth degree polynomial approximation for f(x) can be written as
 f(x)   @   Pn(x)   = f0 + rDf0 + r(r-1) D2f0 + . . . + r(r-1) . . . (r - n +1) Dnf0 2! n!
The formula is called Newton's (Newton-Gregory) forward interpolation formula.
So if we know the forward difference values of  f at x0 until order n then the above formula is very easy to use to find the function values of  f at any non-tabulated value of x in the internal [a,b].The higher order forward differences can be obtained by making use of forward difference table.
Forward difference table : Consider the function value (xi, fi)  i = 0,1,2,--5 then the forward difference table is

 xi fi Dfi D2fi D3fi D4fi D5fi x0 f0 Df0 = f1- f0 x1 f1 D2f0 = Df1- Df0 Df1 = f2 - f1 D3f0 = D2f1- D2f0 x2 f2 D2f1 = Df2 - Df1 D4f0 = D3f1- D3f0 Df2 = f3 - f2 D3f1 = D2f2 - D2f1 D5f0 = D4f1- D4f0 x3 f3 D2f2 = Df3 - Df2 D4f1 = D3f2 - D3f1 Df3 = f4 - f3 D3f2 = D2f3 - D2f2 x4 f4 D2f3 = Df4 - Df3 Df4 = f5 - f4 x5 f5
Example : If  f(x) is known at the following data points
 xi 0 1 2 3 4 fi 1 7 23 55 109
then find f(0.5) and f(1.5) using Newton's forward difference formula.
Solution :
Forward difference table
 xi fi Dfi D2fi D3fi D4fi 0 1 6 1 7 10 16 6 2 23 16 0 32 6 3 55 22 54 4 109
(Note : The given data satifies f(x) = x3 + 2x2 + 3x +1, i.e the function is a third degree polynomial and hence third forward differences are constant by the result).
By Newton's forward difference formula
 f(x)  = f0 + rDf0 + r(r-1) D2f0 + r(r-1)(r-2) D3f0 2! 3!
at x = 0.5,  r = (x - x0) / h = (0.5 - 0) / 1 = 0.5
 f(0.5) = 1 + 0.5 x 6 + 0.5(0.5 - 1) x 10 5 + 0.5(0.5 - 1)(0.5 - 2) x 6 2 6

= 1 + 3 + 2.5 x (-0.5) + (-0.25)(-1.5)

= 3.125

Exact value is
f(0.5) = (0.5)3 + 2(0.5)2 + 3(0.5) + 1
= 0.125 + 0.5 + 1.5 + 1
= 3.125

Error in the Interpolation :

ÞEn(x) = (x - x0)(x - x1) . . .(x - xn)  f(n+1)(x) / (n+1)!                 x0 < x < xn

So for the Newton's method where the nodel points  xi,  i = 0, 1, . . . n  are equally spaced, the error is   En(x) = (x - x0)(x - x0 - h) . . .(x - x0 - nh)  f(n+1)(x) / (n+1)!

 = r(r-1). . .(r-n) h(n+1)f(n+1)(x) (n+1)!

 = ( r ) h(n+1)f(n+1)(x) n+1

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