Editing Richardson extrapolation (section)

==Example of Richardson extrapolation==
Suppose that we wish to approximate <math>A^*</math>, and we have a method <math>A(h)</math> that depends on a small parameter <math>h</math> in such a way that
<math display="block">A(h) = A^\ast + C h^n + O(h^{n+1}).</math>

Let us define a new function<math display="block"> R(h,t) := \frac{ t^n A(h/t) - A(h)}{t^n-1} </math>where <math>h</math> and <math>\frac{h}{t}</math> are two distinct step sizes.

Then
<math display="block"> R(h, t) = \frac{ t^n ( A^* + C \left(\frac{h}{t}\right)^n + O(h^{n+1}) ) - ( A^* + C h^n + O(h^{n+1}) ) }{ t^n - 1} = A^* + O(h^{n+1}). </math>
<math> R(h,t) </math> is called the Richardson [[extrapolation]] of ''A''(''h''), and has a higher-order error estimate <math> O(h^{n+1}) </math> compared to <math> A(h) </math>.

Very often, it is much easier to obtain a given precision by using ''R''(''h'') rather than ''A''(''h&prime;'') with a much smaller ''h&prime;''. Where ''A''(''h&prime;'') can cause problems due to limited precision ([[rounding error]]s) and/or due to the increasing [[Computational cost|number of calculations]] needed (see examples below).