Forward difference formula extrapolation

[stunner5000pt]

the forward difference formula can be expressed as
$$f'(x_{0}) = \frac{1}{h} [f(x_{0} + h) - f(x_{0})] - \frac{h}{2} f''(x_{0}) - \frac{h^2}{6} f'''(x_{0}) + O(h^3)$$

use extrapolation to derivae an O(h^3) formula for f'(x0)

would i be using the taylor expansion to get the answer here? I knwo this is somehow related to Richardson's extrapolation.
please help
am i going to be using something like this?
$$N_{j} (h) = N_{j-1} (h) (\frac{h}{2}) + \frac{N_{j-1}(h/2) - N_{j-1} (h)}{4^{j-1} -1}$$
but N wouldb e replaced by f?

 

[mighty2000]

Yes, you are correct. To derive an O(h^3) formula for f'(x0), you can use Richardson's extrapolation, which involves using a series of approximations with decreasing step sizes. The formula you mentioned, N_j(h) = N_{j-1}(h)(h/2) + (N_{j-1}(h/2) - N_{j-1}(h))/(4^{j-1} - 1), can be used to extrapolate the forward difference formula to a higher order of accuracy. In this case, N would be replaced by f, and h would be the step size. You can also use the Taylor expansion to derive a similar formula.