Derive Richardson's method to central divided difference

Question

I was given this Numerical Analysis assignment but I can't understand it. It says derive Richard's Method to $$f'(X_0) = \frac {f(X_0+h)-f(X_0-h)}{2h}$$

Answer 1

Note that by Taylor's expansion around $x=x_0$

$$f(x_0+h)=f(x_0)+f'(x_0)h+o(h)$$

$$f(x_0-h)=f(x_0)-f'(x_0)h+o(h)$$

subtracting both sides

$$f(x_0+h)-f(x_0-h)=2f'(x_0)h+o(h)$$

thus

$$f'(x_0)=\frac{f(x_0+h)-f(x_0-h)}{2h}+\frac{o(h)}{h}$$

and recall that $\frac{o(h)}{h}\to 0$.

Answer 2

In general Richardson extrapolation proceeds as follows. You want to compute a quantity $I$. You have some approximate scheme for doing so depending on a parameter $h$, denoted by $Q(h)$, satisfying the asymptotic relation

$$Q(h)=I + c h^p + \dots$$

where $\dots$ represents terms that we think of as being smaller than these two. You then have

$$Q(2h)=I + 2^p ch^p + \dots.$$

Therefore:

$$2^pQ(h)- Q(2h)=(2^p-1)I + \dots.$$

Thus we have removed the leading order error term. Dividing both sides by $2^p-1$ gives the improved method.

Thus Richardson extrapolation is almost entirely agnostic about the behavior of $Q$ except that you must know the order of its error, since this order $p$ appears explicitly in the formula for the extrapolated method. But you should be able to find that in your case.

Note that there is no need to consider $h$ and $2h$, we could consider $h$ and any other multiple of $h$ we wanted. The formula only changes a little bit. However, $2h$ (or $h/2$, they are equivalent) is commonly used in differentiation and integration problems because it tends to reduce the total number of function evaluations required for a method which will ultimately be of some given order (since there is a tendency for some function evaluations to be "re-used" between $Q(2h)$ and $Q(h)$).