Newton Interpolation polynomial:

Suppose that we are given a data set $(x_{i},f_{i}),i=0,1...n-1$. Let us assume that these are interpolating points of Newton form of interpolating polynomial $p_{n}(x)$ of degree i.e

$$p_{n}(x_{i})=f_{i}, \qquad i=0,1,..n-1$$ (1)

The Newton form of the interpolating polynomial $p_{n}(x)$ is given by

$$p_{n}(x)=a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})(x-x_{1})+ a_{3}(x-x_{0})(x-x_{1})(x-x_{2})+.......+ a_{n}(x-x_{0})(x-x_{1})....(x-x_{n-1}).$$ (2)

For i=0, from (1)-(2) we get

$$f_{0}=P_{n}(x_{0})=a_{0}$$ (3.1)

For $i=1$ from (1)-(2) we get

$$f_{1}=p_{n}(x_{1})=a_{0}+a_{1}(x_{1}-x_{0})$$

$$a_{1}=\frac{f_{1}-f_{0}}{x_{1}-x_{0}}$$ (3.2)

For i=2 from(1)-(2) We get

$$f_{2}=p_{n}(x_{2})=a_{0}+a_{1}(x_{2}-x_{0})+a_{2}(x_{2}-x_{0})(x_{2}-x_{1})$$

using (3.1)-(3.2), we get

$$a_{2}=\frac{[(f_{2}-f_{1})/(x_{2}-x_{1})]-[(f_{1}-f_{0})/(x_{1}-x_{0})]}{(x_{2}-x_{0})}$$ (3.3)

Similarly we can find $a_{3}.....a_{n-1}$. To express $a_{i},i=0......n-1$ in a compact manner let us first define the following notation called divided differences:

$$f[x_{k}]=f_{k}$$ (4.1)

$$\hspace{0.5in}f[x_{k},x_{k+1}]=\frac{f[x_{k+1}]-f[x_{k}]}{x_{k+1}-x_{k}}$$ (4.2)

$$\hspace{0.9in}f[x_{k},x_{k+1},x_{k+2}]=\frac{f[x_{k+1},x_{k+2}]-f[x_{k},x_{k+1}]}{x_{k+2}-x_{k}}$$ (4.3)

$$\hspace{0.5in}f[x_{k},x_{k-1}....x_{i},x_{i+1}]=\frac{f[x_{k+1}...x_{i+1}]-f[x_{k}...x_{i}]}{x_{i+1}-x_{k}}$$ (4.4)

Now the co-efficients $a_{i}, i=_{0}...n-1$ can be expressed in terms of divided differences as fellows:

$$a_{0}=f_{0}=f[x_{0}]$$ (5.1)

$$a_{1}=\frac{f_{1}-f_{0}}{x_{1}-x_{0}}=f[x_{1},x_{0}]$$ (5.2)

$$a_{2}=\displaystyle {\frac{\frac{f_{2}-f_{1}}{x_{2}-x_{1}}-\frac{... ...}{x_{1}-x_{0}}}{x_{2}-x_{0}}}=\frac{f[x_{1},x_{2}]-f[x_{0},x_{1}]}{x_{2}-x_{0}}$$

$$=f[x_{0},x_{1},x_{2}]$$ (5.3)

$$a_{n}=f[x_{0},x_{1}...x_{n}]$$ (5.4)

Note that $a_{1}$ is called as the first divided difference, $a_{2}$ as the second divided difference and so on. Now the polynomial (2) can be rewritten as:

$$p_{n}(x)=f[x_{0}]+f[x_{0},x_{1}](x-x_{0})+f[x_{0},x_{1},x_{2}](x-x_{0})(x-x_{1})+......+f[x_{0}x_{1}......x_{n}](x-x_{0})(x-x_{1})...(x-x_{n-1})$$

$$\hspace{1in}p_{n}(x)=\sum\limits_{k=0}^{n}f[x_{0}......x_{k}]\prod\limits_{i=0}^{k-1}(x-x_{i})$$ (6)

This is called as Newton's Divided Difference interpolation polynomial.

Example:
Given the following data table, evaluate $f(2.4)$ using $3^{rd}$ order Newton's Divided Difference interpolation polynomial.

i	0	1	2	3	4
$x_{i}$	0	1	2	3	4
$y_{i}=f(x_{i})$	1	2.25	3.75	4.25	5.81

Solution:
Here $n=5$. For constructing $3^{rd}$ order Newton Divided Difference polynomial we need only four points. Let us use the first four points. The $3^{rd}$ Newton Divided Difference polynomial is given by:
$p_{3}(x)=\sum\limits_{k=0}^{3}a_{k}\prod\limits_{j=0}^{k-1}(x-x_{j})\\ =\sum\limits_{k=0}^{3}f[x_{0}...x_{k}]\prod\limits_{j=0}^{k-1}(x-x_{j})$
$=f[x_{0}]+f[x_{0},x_{1}](x-x_{0})+ f[x_{0},x_{1},x_{2}](x-x_{0})(x-x_{1})+ f[x_{0},x_{1},x_{2},x_{3}](x-x_{0})(x-x_{1})(x-x_{3})$

$\therefore a_{0}=f[x_{0}]=1$

$a_{1}=f[x_{0},x_{1}]=\displaystyle {\frac{f(x_{1})-f(x_{0})}{(x_{1}-x_{0})}}=\frac{2.25-1}{1-0}=1.25$

$${f[x_{1}x_{2}]=\frac{f[x_{2}]-f[x_{1}]}{x_{2}-x_{1}}=\frac{3.75-2.25}{2-1}=1.5}$$


$${\therefore a_{2}=f[x_{0},x_{1},x_{2}]=\frac{f[x_{1},x_{2}]-f[x_{0},x_{1}]}{x_{2}-x_{0}}=\frac{1.5-1.25}{2-0}=0.125}$$

$${f[x_{2},x_{3}]=\frac{f[x_{3}]-f[x_{2}]}{x_{3}-x_{2}}=\frac{4.25-3.75}{3-2}=\frac{0.5}{1}=0.5}$$

$${f[x_{1},x_{2},x_{3}]=\frac{f[x_{2},x_{3}]-f[x_{1},x_{2}]}{x_{3}-x_{1}}=\frac{0.5-1.5}{3-1}=-0.5}$$


$${\therefore a_{3}=f[x_{0},x_{1},x_{2},x_{3}]=\frac{f[x_{1},x_{2},x_{3}]-f[x_{0},x_{1},x_{2}]}{x_{3}-x_{0}}}$$

$$=\frac{-0.5-0.125}{3-0}=\frac{-0.625}{3}=-0.20833$$

$\therefore p_{3}(x)=1+1.25(x-0)+0.125(x-0)(x-1)+(-0.20833)(x-0)(x-1)(x-2)$

$\therefore f(2.4)=p_{3}(2.4)$
$=1+1.25(2.4-0)+0.125(2.4-0)(2.4-1)+(-0.20833)(2.4-0)(2.4-1)(2.4-2)$
$=1+(1.25)(2.4)+0.125(2.4)(1.4)-0.20833(2.4)(1.4)(0.4)$
$=4.2200032$

In this example it may be noted that for calculating the $3rd$ order polynomial, we first start with $P_{0}=f[x_{0}]=1$. To it we add $a_{1}(x-x_{0})$ to get $P_{1}$ and to $P_{1}$ we add $a_{2}(x-x_{0})(x-x_{1})$ to get $P_{2}$. Finally on adding $a_{3}(x-x_{0})(x-x_{1})(x-x_{2})$ to $P_{2}$ we get $P_{3}$.
Some remarks on the Error in the interpolation approach:
Suppose that the given data points $(x_{i},y_{i})$, $i=0,1,..n$ correspond to a real valued function $f(x)$ defined on the internal $I=[a,b]$. Let $p_{n}(x)$ be the interpolating polynomial of degree $\leq n.$ $i.e.$ $P_{n}(x_{i})=f(x_{i})=y_{i}$     $i=0,1...n$. Then the interpolation error $e_{n}(x)$ due to interpolation by $p_{n}(x)$ is given by

$$e_{n}(x)=f(x)-p_{n}(x).$$ (6.2)

An estimate of the error is provided in the following theorem.

Theorem: Let $f(x)$ be a real-valued function define on $[a,b]$ and $n+1$ times differentiable on $(a,b).$ If $p_{n}(x)$ is the polynomial of degree $\leq n$ which interpolates $f(x)$ at the (n+1) distinct points $x_{0}..x_{n}\epsilon[a,b]$, then for all $\overline{x}\epsilon[a,b]$, there exits $\xi=\xi(\overline{x})\epsilon (a,b) s.t.$

$$e_{n}(\overline{x})=f(\overline{x})-P_{n}(\overline{x})$$

$$=\frac{f^{n+1}\vert\xi\vert}{(n+1)!}\prod \limits_{j=0}^{n}(\overline{x}-x_{j})$$ (6.3)

Note:
i) $\xi=\xi(\overline{x})$ i.e. $\xi$ depends on the point $\overline{x}$ at which the error estimate is required.
ii) since $f^{n+1}(x)$ i.e. $(n+1)^{th}$ derivative is seldom known the error formula $(6.3)$ in the above theorem is of limited practical value. But when a bound on $\vert f^{n+1}(x)\vert$ is known over the entire internal $[a,b]$, then the formula $(6.3)$ may be used to get a bound on the interpolation error.