Editing Newton polynomial (section)

===Examples===
The divided differences can be written in the form of a table. For example, for a function ''f'' is to be interpolated on points <math>x_0, \ldots, x_n</math>. Write

:<math>\begin{matrix}
   x_0 & f(x_0) &                                 & \\
       &        & {f(x_1)-f(x_0)\over x_1 - x_0}  & \\
   x_1 & f(x_1) &                                 & {{f(x_2)-f(x_1)\over x_2 - x_1}-{f(x_1)-f(x_0)\over x_1 - x_0} \over x_2 - x_0} \\
       &        & {f(x_2)-f(x_1)\over x_2 - x_1}  & \\
   x_2 & f(x_2) &                                 & \vdots \\
       &        & \vdots                          & \\
\vdots &        &                                 & \vdots \\
       &        & \vdots                          & \\
   x_n & f(x_n) &                                 & \\
\end{matrix}</math>
Then the interpolating polynomial is formed as above using the topmost entries in each column as coefficients.

For example, suppose we are to construct the interpolating polynomial to ''f''(''x'') = tan(''x'') using divided differences, at the points
{| cellpadding=5px class="wikitable" style="text-align: right;"
! <math>n</math> !! <math>x_n</math> !! <math>f(x_n)</math>
|-
| <math>0</math> || <math>-\tfrac{3}{2}</math> || <math>-14.1014</math>
|-
| <math>1</math> || <math>-\tfrac{3}{4}</math> || <math>-0.931596</math>
|-
| <math>2</math> || <math>0</math> || <math>0</math>
|-
| <math>3</math> || <math>\tfrac{3}{4}</math> || <math>0.931596</math>
|-
| <math>4</math> || <math>\tfrac{3}{2}</math> || <math>14.1014</math>
|}

Using six digits of accuracy, we construct the table
: <math>\begin{matrix}
-\tfrac{3}{2} & -14.1014  &         &          &          &\\
      &           & 17.5597 &          &          &\\
-\tfrac{3}{4} & -0.931596 &         & -10.8784 &          &\\
      &           & 1.24213 &          & 4.83484  &  \\
0     & 0       &               & 0        &          & 0\\
      &           & 1.24213 &          & 4.83484  &\\
\tfrac{3}{4}  & 0.931596  &         & 10.8784  &          &\\
      &          & 17.5597 &          &          &\\
\tfrac{3}{2} & 14.1014   &         &          &          &\\
\end{matrix}</math>
Thus, the interpolating polynomial is
:<math>\begin{align}
&-14.1014+17.5597(x+\tfrac{3}{2})-10.8784(x+\tfrac{3}{2})(x+\tfrac{3}{4}) +4.83484(x+\tfrac{3}{2})(x+\tfrac{3}{4})(x)+0(x+\tfrac{3}{2})(x+\tfrac{3}{4})(x)(x-\tfrac{3}{4}) \\
={}&-0.00005-1.4775x-0.00001x^2+4.83484x^3
\end{align}</math>
Given more digits of accuracy in the table, the first and third coefficients will be found to be zero.

Another example:

The sequence <math>f_0</math> such that <math>f_0(1) = 6, f_0(2) = 9, f_0(3) = 2</math> and <math>f_0(4) = 5</math>, i.e., they are <math>6, 9, 2, 5</math> from <math>x_0 = 1</math> to <math>x_3 = 4</math>.

You obtain the slope of order <math>1</math> in the following way:
* <math>f_1(x_0, x_1) = \frac{f_0(x_1) - f_0(x_0)}{x_1 - x_0} = \frac{9 - 6}{2 - 1} = 3</math>
* <math>f_1(x_1, x_2) = \frac{f_0(x_2) - f_0(x_1)}{x_2 - x_1} = \frac{2 - 9}{3 - 2} = -7</math>
* <math>f_1(x_2, x_3) = \frac{f_0(x_3) - f_0(x_2)}{x_3 - x_2} = \frac{5 - 2}{4 - 3} = 3</math>

As we have the slopes of order <math>1</math>, it is possible to obtain the next order:
* <math>f_2(x_0, x_1, x_2) = \frac{f_1(x_1, x_2) - f_1(x_0, x_1)}{x_2 - x_0} = \frac{-7 - 3}{3 - 1} = -5</math>
* <math>f_2(x_1, x_2, x_3) = \frac{f_1(x_2, x_3) - f_1(x_1, x_2)}{x_3 - x_1} = \frac{3 - (-7)}{4 - 2} = 5</math>

Finally, we define the slope of order <math>3</math>:
* <math>f_3(x_0, x_1, x_2, x_3) = \frac{f_2(x_1, x_2, x_3) - f_2(x_0, x_1, x_2)}{x_3 - x_0} = \frac{5 - (-5)}{4 - 1} = \frac{10}{3}</math>

Once we have the slope, we can define the consequent polynomials:
* <math>p_0(x) = 6</math>.
* <math>p_1(x) = 6 + 3(x - 1)</math>
* <math>p_2(x) = 6 + 3(x - 1) - 5(x - 1)(x - 2)</math>.
* <math>p_3(x) = 6 + 3(x - 1) - 5(x - 1)(x - 2) + \frac{10}{3} (x - 1)(x - 2)(x - 3)</math>