2.3.1 Gregory-Newton Forward Difference Approach:

Very often it so happens in practice that the given data set $(x_{i},y_{i}),i=0,1,...n$ correspond to a sequence $\{x_{i}\}$ of equally spaced points. Here we can assume that

$$x_{i}=x_{0}+ih,\qquad i=0,1,2..n$$ (1)

where $x_{0}$ is the starting point (sometimes, for convenience, the middle data point is taken as $x_{0}$ and in such a case the integer $i$ is allowed to take both negative and positive values.) and is the step size. Further it is enough to calculate simple differences rather than the divided differences as in the non-uniformly placed data set case. These simple differences can be forward difference $(\Delta f_{i})$ or back differences $(\nabla f_{i})$. We will first look at forward differences and the interpolation polynomial based on forward differences.

The first order forward difference $\Delta f_{i}$ is defined as

$$\Delta f_{i}=f_{i+1}-f_{i}$$ (7.1)

The second order forward difference $\Delta^{2}f_{i}$ is defined as

$$\Delta^{2}f_{i}=\Delta f_{i+1}-\Delta f_{i}$$ (7.2)

The $k^{th}$ order forward difference $\Delta^{k}f_{i}$ is defined as

$$\Delta^{k}f_{i}=\Delta^{k-1}f_{i+1}-\Delta^{k-1}f_{i}$$ (7.3)

Since we already know Newton interpolation polynomial in terms of divided differences, to derive or generate Newton interpolation polynomial in terms of forward differences it is enough to express forward differences in terms of divided differences.

Recall the definition of first divided difference $f[x_{0},x_{1}]$,

$$f[x_{0},x_{1}]=\frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}=\frac{f_{1}-f_{0}}{h}=\frac{\Delta f_{0}}{h}$$

$$\therefore \Delta f_{0}=hf[x_{0},x_{1}]$$ (8.1)

Similarly we can get

$$\Delta f_{1}= hf[x_{1},x_{2}]$$ (8.2)

By the definition second order forward difference $\Delta^{2}f_{0}$, we get

$$\Delta^{2}f_{0} = \Delta f_{1}-\Delta f_{0}$$

$$= hf[x_{1},x_{2}]-hf[x_{0},x_{1}]\quad \quad \quad \quad \quad \quad (using (8.1)(8.2))$$

$$= h\{f[x_{1},x_{2}]-f[x_{0},x_{2}]\}$$

$$= h.2h\{(f[x_{1},x_{2}]-f[x_{0},x_{2}])/2h\}$$

$$= 2h^{2}f[x_{0},x_{1},x_{2}]\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (8.3)$$

In a similar way, in general, we can show that

$$\Delta^{k}f_{i}=k!h^{k}f[x_{i},x_{i+1},x_{i+2}...x_{i+k}]$$ (8.4)

$$\therefore f[x_{i},x_{i+1},...x_{i+k}]=\frac{\Delta^{k}f_{i}}{k!h^{k}}$$ (8.5)

For $i=0$,

$$f[x_{0},x_{1}...x_{k}]=\frac{\Delta^{k}f_{0}}{k!h^{k}}$$ (8.6)

Now using (6) ,(8.6) the Newton forward difference interpolation polynomial may be written as follows.

$$p_{n}(x)=\sum\limits_{k=0}^{n}\quad\frac{\Delta^{k}f_{0}}{k!h^{k}}\prod\limits_{i=0}^{k-1}(x-x_{i})$$ (9)

To rewrite (9) in a simpler way let us set

$$x=x_{0}+sh,\quad p_{n}(s)=p_{n}(x)$$

$$\because x_{k}=x_{0}+kh$$

$$x-x_{k}=(s-k)h$$

$$\therefore \qquad p_{n}(s) = \sum\limits_{k=0}^{n}\quad \frac{\Delta^{k}f_{0}}{k!h^{k}}\prod\limits_{i=0}^{k-1}(s-i)h$$

$$= \sum\limits_{k=0}^{n}\quad\frac{\Delta^{k}f_{0}}{k!h^{k}}\quad[s(s-1).......(s-k+1)]h^{k}$$

i.e.

$$p_{n}(s)=\sum\limits_{k=0}^{n}\left(\begin{array}{c} s \\ k \\ \end{array}\right)\Delta^{k}f_{0}$$ (10)

where

This is known as Gregory-Newton forward difference interpolation polynomial. For Convenience while constructing (10) one can first generate a forward difference table and use the $\Delta^{k}f_{i}$ from the table. Suppose we have data set $(x_{i}f_{i})$, $i=0,1,2,3,4$ then forward difference table looks as follows:

Example:
Given the following data, estimate $f(1.85)$ using Newton-Gregory forward difference interpolation polynomial:

i	0	1	2	3	4
$x_{i}$	1.0	3.0	5.0	7.0	9.0
$f_{i}$	0	1.0986	1.6094	1.9459	2.1972

Solution:
Here we have five data points i.e $i=0,1,2,3,4.$ Let us first generate the forward difference table.

$$h=2,\quad x=1.83,\quad x_{0}=1.0,\quad s=\frac{x-x_{0}}{h}=0.415$$

Newton Gregory forward difference interpolation polynomial is given by:

$$p_{n}(s)=\sum\limits_{k=0}^{n}\left(\begin{array}{c} s \\ k \\ \end{array}\right)\Delta^{k}f_{0}$$

$$p_{1}(0.415) = \left(\begin{array}{c} 0.415 \\ 0 \\ \end{array}\right)f_{0}+\left(\begin{array}{c} 0.415 \\ 1 \\ \end{array}\right)\Delta f_{0}$$

$$= 0+(0.415)(1.0986)=0.455919$$

$$p_{2}(0.415) = p_{1}(0.415)+\left(\begin{array}{c} 0.415 \\ 2 \\ \end{array}\right)\Delta^{2}f_{0}$$

$$= 0.455919+\frac{0.415(0.415-1)}{2}(-0.5878)$$

$$= 0.455919+0.071352$$

$$= 0.527271$$

$$p_{3}(0.415) = p_{2}(0.415)+\left(\begin{array}{c} 0.415 \\ 3 \\ \end{array}\right)\Delta^{3} f_{0}$$

$$= 0.527271+\frac{0.415(0.415-1)(0.415.2)}{6}(0.4135)$$

$$= 0.527271+0.026519$$

$$= 0.554157$$

$$P_{4}(0.415) = P_{3}(0.415)+\left(\begin{array}{c} 0.415 \\ 4 \\ \end{array}\right)\Delta^{4} f_{0}$$

$$= 0.554157+\frac{0.415(0.415-1)(0.415-2)(0.415-3)}{24}(-0.3244)$$

$$= 0.554157+0.013445$$

$$= 0.567602$$

$$\therefore f(1.83)=0.567602$$

Example 2:
Given the following data estimate f(4.12) using Newton-Gregory forward difference interpolation polynomial:

i	0	1	2	3	4	5
$x_{i}$	0	1	2	3	4	5
$f_{i}$	1	2	4	8	16	32

Solution:
Let us first generate the Newton-Gregory forward difference Table:

Here $i=0,1,2,3,4,5;$
$x=4.12,\quad x_{0}=0,\quad h=1, \quad \displaystyle{s=\frac{x-x_{0}}{h}=\frac{4.12}{1}=4.12}$

We know that the forward difference interpolation polynomial is given by:

$$\therefore P_{1} (4.12) =$$

$$= f_{0}+4.12\quad \Delta f_{0}$$

$$= 1+(4.12)(1)=5.12\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \qquad (10a.1)$$

$$P_{2}(4.12) =$$

$$= 5.12+\frac{4.12(4.12-1)}{2}1$$

$$= 5.12+6.4272=11.5472\qquad \qquad \qquad \qquad \qquad \qquad \quad \qquad (10a.2)$$

$$P_{3}(4.12) =$$

$$= 11.5472+\frac{4.12(4.12-1)(4.12-2)}{6}1$$

$$= 11.5472+4.5419$$

$$= 16.0891 \qquad \qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \qquad \qquad(10a.3)$$

$$p_{4}(4.12) =$$

$$= 16.0891+\frac{4.12(4.12-1)(4.12-2)(4.12-3)}{24}1$$

$$= 16.0891+1.2717$$

$$= 17.3608 \qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad\qquad \qquad \qquad \qquad(10a.4)$$

$$p_{5}(4.12) = p_{4}(4.12)+\frac{4.12(4.12-1)(4.12-2)(4.12-3)(4.12-4)}{120}$$

$$= 17.3608+0.0305$$

$$= 17.3913\qquad \qquad \quad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad(10a.5)$$

$$\therefore f(4.12)=17.3913$$