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Richardson extrapolation

In numerical analysis, Richardson extrapolation is a method to improve an approximation that depends on a step size.

Let A(h) be an approximation of A that depends on a positive step size h with an error formula of the form

A - A(h) = a_0h^{k_0} + a_1h^{k_1} + a_2h^{k_2} + \cdots

where the ai are unknown constants and the ki are known constants such that hki > hki+1.

The exact value sought can be given by

A = A(h) + a_0h^{k_0} + a_1h^{k_1} + a_2h^{k_2} + \cdots

which can be simplified with Big O notation to be

A = A(h)+ a_0h^{k_0} + O(h^{k_1}).  \,\!

Using the step sizes h and h / t for some t, the two formulas for A are:

A = A(h)+ a_0h^{k_0} + O(h^{k_1})  \,\!
A = A\left(\frac{h}{t}\right) + a_0\left(\frac{h}{t}\right)^{k_0} + O(h^{k_1}) .

Multiplying the second equation by tk0 and subtracting the first equation gives

(t^{k_0}-1)A = t^{k_0}A\left(\frac{h}{t}\right) - A(h) + O(h^{k_1})

which can be solved for A to give

A = \frac{t^{k_0}A\left(\frac{h}{t}\right) - A(h)}{t^{k_0}-1} + O(h^{k_1}) .

By this process, we have achieved a better approximation of A by subtracting the largest term in the error which was O(hk0). This process can be repeated to remove more error terms to get even better approximations.

A general recurrence relation can be defined for the approximations by

A_{i+1}(h) = \frac{t^{k_i}A_i\left(\frac{h}{t}\right) - A_i(h)}{t^{k_i}-1}

such that

A = A_{i+1}(h) + O(h^{k_{i+1}}) .

A well-known practical use of Richardson extrapolation is Romberg integration, which applies Richardson extrapolation to the trapezium rule.

Example

Using Taylor's theorem,

f(x+h) = f(x) + f'(x)h + \frac{f''(x)}{2}h^2 + \cdots

so the derivative of f(x) is given by

f'(x) = \frac{f(x+h) - f(x)}{h} - \frac{f''(x)}{2}h + \cdots.

If the initial approximations of the derivative are chosen to be

A_0(h) = \frac{f(x+h) - f(x)}{h}

then ki = i+1.

For t = 2, the first formula extrapolated for A would be

A = 2A_0\left(\frac{h}{2}\right) - A_0(h) + O(h^2) .

For the new approximation

A_1(h) = 2A_0\left(\frac{h}{2}\right) - A_0(h)

we can extrapolate again to obtain

A = \frac{4A_1\left(\frac{h}{2}\right) - A_1(h)}{3} + O(h^3) .
01-04-2007 01:18:14
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