Firstly what is an $O(h^3)$ formula? Also I am not quite sure how to answer the question?

Question

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 Richardson's extrapolation to derive an $O(h^3)$ formula for $f'(x_0).$

Answer 1

It seems in this case you are expanding $f'(x)$, so $h$ is some infinitesimal variable. These variables are of different orders, i.e. they tend to $0$ at different rates. In this specific example $O(g(h))$ means that all other terms tend to $0$ as $h^3$ (i.e. $h^3, h^4$ etc). In some non-rigorous way you can see them as 'unimportant' terms.

EDIT: $g(h) = O(f(h))$ means $\exists \ c_1 \ s.t. g(h) \leq c_1 f(h) \ \forall \ h <H$. Hence, if I understand the problem correctly, you need to derive this constant.

Answer 2

Hint: The given forward-difference formula is an example of an $O(h)$ formula. To improve on this formula, define $a = \frac{-f''(x_0)}{2}$ and let: $$ g_0(h) = \frac{f(x_0 + h) - f(x_0)}{h} $$ so that: $$ f'(x_0) = g_0(h) + ah + O(h^2) \tag{1} $$ Then by replacing $h$ with $h/2$, observe that: $$ f'(x_0) = g_0(h/2) + \frac{1}{2}ah + O(h^2) \tag{2} $$ Notice that we can eliminate the $O(h)$ terms by taking $2 \cdot (2) - (1)$, which yields the following $O(h^2)$ formula: $$ f'(x_0) = 2g_0(h/2) - g_0(h) + O(h^2) $$ All you have to do now is recursively repeat this procedure another time (what do you think your new $g_1(h)$ should be?) in order to improve the formula so that it is now $O(h^3)$.