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 : 

= f₀ + r Df₀    [ r = (x - x₀) / (x₁ - x₀)  Df₀ = 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(x₀  +  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)  =  f₀  +  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 < ... <  =  x_n = b  in the interval [a,b] as f₀, f₁, . . . f_n. Then the n  the degree polynomial approximation of  f(x)  can be given as 

where r = (x-x₀ ) / n   Þ   x  = x₀  +  r h   Þ 0 <r <n

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 = x_i  the polynomial approximation P_n(x) gives f_i for any tabulated value x_i since the curve f(x) passes through the node points x_i,  i = 0, 1, . . ., n

Take r  =  k

we can prove this result by mathematical induction.

consider k = 0

consider

f_p+1 =  f_p + Df_p

i.e., the nth degree polynomial approximation for f(x) can be written as

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 (x_i, f_i)  i = 0,1,2,--5 then the forward difference table is

Example : If  f(x) is known at the following data points 

then find f(0.5) and f(1.5) using Newton's forward difference formula.

Solution :

Forward difference table

(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

at x = 0.5,  r = (x - x₀) / h = (0.5 - 0) / 1 = 0.5

         = 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 :

ÞE_n(x) = (x - x₀)(x - x₁) . . .(x - x_n)  f⁽ⁿ⁺¹⁾(x) / (n+1)!                 x₀ < x < x_n

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