2.3.2 Newton-Gregory Backward Difference Interpolation polynomial:

If the data size is big then the divided difference table will be too long. Suppose the desired intermediate value $(\widetilde{x})$ at which one needs to estimate the function $(i.e.f(\widetilde{x}))$ falls towards the end or say in the second half of the data set then it may be better to start the estimation process from the last data set point. For thus we need to use backward-differences and backward difference table.
Let us first define backward differences and generate backward difference table, say for the data set $(x_{i},f_{i}),i=0,1,2,3,4.$
First order backward difference $\nabla f_{i}$ is defined as:

$$\nabla f_{i}=f_{i}-f_{i-1}$$ (11.1)

Second order backward difference $\nabla^{2}f_{i}$ is defined as:

$$\nabla^{2}f_{i}=\nabla f_{i}-\nabla f_{i-1}$$ (11.2)

In general, the $k^{th}$ order backward difference is defined as

$$\nabla^{k}f_{i}=\nabla^{k-1}f_{i}-\nabla^{k-1}f_{i-1}$$ (11.3)

In this case the reference point is $x_{n}$ and therefore we can derive the Newton-Gregory backward difference interpolation polynomial as:

$$p_{n}(b)=f_{n}+s\nabla f_{n}+\frac{s(s+1)}{2!}\nabla^{2}f_{n}+.... ..+\frac{s(s+1)...(s+n-1)}{n!}\nabla^{n}f_{n}$$ (12)

Where $$s=\frac{x-x_{n}}{h}$$
For constructing $p_{n}(s)$ as given in $Eqn.(12)$ it will be easier if we first generate backward-difference table. The backward difference table for the data $(x_{i},f_{i}),\quad i,=0,1,2,3,4$ is given below:

Newton Backward Difference Table:
Now let us apply Newton Backward difference approach to the second example solved earlier following the Newton forward difference approach i.e.
Example:
Given the following data estimate $f(4.12)$ using Newton-Gregory backward 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:
Here

$$x_{n}=5,\quad x=4.12,\qquad h=1$$

$$\therefore s=\frac{x-x_{n}}{h}=\frac{4.12-5}{1}=-0.88$$

$\therefore$ Newton Backward Difference polynomial $P_{5}(x)$ is given by

$$P_{5}(s)=f_{5}+s\nabla f_{5}+\frac{s(s+1)}{2!}\nabla^{2}f_{5}+\frac{s(s+1)(s+2)}{3!}\nabla^{3}f_{5} +\frac{s(s+1)(s+2)(s+3)}{4!}\nabla^{4}f_{5}+$$

$$\frac{s(s+1)(s+2)(s+3)(s+4)}{5!}\nabla^{5}f_{5}$$

Let us first generate backward difference table:

$$\therefore p_{5}(-0.88)=32+(-0.88)16+\frac{(-0.88)(-0.88+1)}{2}{8} +\frac{(-0.88)(-0.88+1)(-0.88+2)}{6}\quad(4)$$

$$+\frac{(-0.88)(-0.88+1)(-0.88+2)(-0.88+3)}{24}\quad(2) \frac{+(-0.88)(-0.88+1)(-0.88+2)(-0.88+3)(-0.88+3)}{120}$$

$$= 32-14.08-0.4224-0.07885-0.0209-0.0065$$

$$= 17.92-0.4229-0.7885-0.0209-0.0065$$

$$= 17.4976-0.07885-0.0209-0.0065$$

$$= 17.41875-0.0209-0.0065$$

                                        =17.39135                                                                                                        (13.5)

Now for comparison with the earlier solution i.e. the one obtained by forward Newton Divided Difference approach we may look at the above solution in stages similar to that provided earlier i.e.

$$P_{1}(-0.88) = f_{5}+s\nabla f_{5}$$

$$= 32+(-0.88)16=17.92\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(13.1)$$

$$P_{2}(-0.88) = P_{1}(-0.88)+\frac{(-0.88)(-0.88+1)}{2!}8$$

$$= 17.92-0.4224$$

$$= 17.4976\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(13.2)$$

$$P_{3}(-0.88) = P_{2}(-0.88)+\frac{(-0.88)(-0.88+1)(-0.88+2)}{3!}$$

$$= 17.4976-0.07885$$

$$= 17.41875\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(13.3)$$

$$P_{4}(-0.88) = P_{3}(-0.88)+\frac{(-0.88)(-0.88+1)(-0.88+2)(-0.88+3)}{4!}$$

$$= 17.41875-0.0209$$

$$= 17.39785\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(13.4)$$

$$P_{5}(-0.88) = 17.39785-\frac{(-0.88)(-0.88+1)(-0.88+2)(-0.88+3)(-0.88+4)}{5!}$$

$$= 17.39785-0.0065$$

$$= 17.39135\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(13.5)$$

Now one may note from (13.2) and (10a.2) that it is definitely advantageous of use backward difference approach here, as in exactly the same number of steps we are relatively more close to the approximate solution.