# Newton Interpolation Formula for Unequal Intervals

When the values of the independent variable occur with unequal spacing, the formula discussed earlier is no longer applicable. In this situation another formula which is based on divided difference is used. Before presenting the formula let us first discuss divided differences.

Divided Differences
The values of the independent variable $\left( X \right)$ are given as ${a_o},{a_{1,}}{a_2},{a_3},…$  etc. and the corresponding values of the function (dependent variable) are $f\left( {{a_o}} \right),f\left( {{a_1}} \right),f\left( {{a_2}} \right),f\left( {{a_3}} \right),…$ etc. Thus, the data look as follows:

 $X$ ${a_o}$ ${a_1}$ ${a_2}$ ${a_3}$ ${a_4}$ $f\left( X \right)$ $f\left( {{a_o}} \right)$ $f\left( {{a_1}} \right)$ $f\left( {{a_2}} \right)$ $f\left( {{a_3}} \right)$ $f\left( {{a_4}} \right)$

Here, $\left( {{a_1} – {a_o}} \right),\left( {{a_2} – {a_1}} \right)$  etc. are not equal. The problem of interpolation requires that $X = {X_o}$ , and the value of $f\left( {{X_o}} \right)$ must be found. In order to determine the value of $f\left( {{X_o}} \right)$ we need to compute the divided difference.

The quantities

$\frac{{f\left( {{a_1}} \right) – f\left( {{a_o}} \right)}}{{{a_1} – {a_o}}}$  denoted by $f\left( {{a_1},{a_o}} \right)$
$\frac{{f\left( {{a_2}} \right) – f\left( {{a_1}} \right)}}{{{a_2} – {a_1}}}$  denoted by $f\left( {{a_2},{a_1}} \right)$
$\frac{{f\left( {{a_3}} \right) – f\left( {{a_2}} \right)}}{{{a_3} – {a_2}}}$  denoted by $f\left( {{a_3},{a_2}} \right)$
etc.

are called divided difference of the first order. Moreover, the quantities

$\frac{{f\left( {{a_2},{a_1}} \right) – f\left( {{a_1},{a_o}} \right)}}{{{a_2} – {a_o}}}$  denoted by $f\left( {{a_2},{a_1},{a_o}} \right)$
$\frac{{f\left( {{a_3},{a_2}} \right) – f\left( {{a_2},{a_1}} \right)}}{{{a_3} – {a_1}}}$  denoted by $f\left( {{a_3},{a_2},{a_1}} \right)$
etc.

are called the divided differences of the second order. Similarly, the divided difference of order three and higher may be computed.

The divided differences may be put in a tabular form as follows:

 DIFFERENCES OF ORDER $X$ $f\left( X \right)$ $I$ $II$ $III$ $IV$ ${a_o}$ $f\left( {{a_o}} \right)$ $f\left( {{a_1},{a_o}} \right)$ $f\left( {{a_2},{a_1},{a_o}} \right)$ $f\left( {{a_3},{a_2},{a_1},{a_o}} \right)$ $f\left( {{a_5},{a_4},{a_3},{a_2},{a_1}} \right)$ ${a_1}$ $f\left( {{a_1}} \right)$ $f\left( {{a_2},{a_1}} \right)$ $f\left( {{a_3},{a_2},{a_1}} \right)$ $f\left( {{a_4},{a_3},{a_2},{a_1}} \right)$ ${a_2}$ $f\left( {{a_2}} \right)$ $f\left( {{a_3},{a_2}} \right)$ $f\left( {{a_4},{a_3},{a_2}} \right)$ ${a_3}$ $f\left( {{a_3}} \right)$ $f\left( {{a_4},{a_3}} \right)$ ${a_4}$ $f\left( {{a_4}} \right)$

It can be seen that the divided differences of higher order either vanish or become negligible. We continue computing these differences until such order where they become more or less constant or significantly different.

THE FORMULA

Newton’s formula for unique intervals may be stated as follows:

$\begin{gathered} f\left( {{X_o}} \right) = f\left( {{a_o}} \right) + \left( {{X_o} – {a_o}} \right)\,f\left( {{a_1},{a_o}} \right) + \left( {{X_o} – {a_o}} \right)\left( {{X_o} – {a_1}} \right)f\left( {{a_2},{a_1},{a_o}} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {{X_o} – {a_o}} \right)\left( {{X_o} – {a_1}} \right)\left( {{X_o} – {a_2}} \right)f\left( {{a_3},{a_2},{a_1},{a_o}} \right) + \cdots + \left( {{X_o} – {a_o}} \right)… \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,……….\left( {{X_o} – {a_{n – 1}}} \right)f\left( {{a_n},{a_{n – 1}},…,{a_o}} \right) \\ \end{gathered}$

Here, ${a_o},{a_1},{a_2}$ etc. are the values of the independent variable, ${X_o}$ the given value corresponding to $f\left( {{X_o}} \right)$ is required, and $f\left( {{a_1},{a_o}} \right),f\left( {{a_2},{a_1},{a_o}} \right)$ etc. are the successive divided differences of the first, second and third orders.