The Forward difference formula can be expressed as:
$$f'(x_0)={1 \over h}[f(x_0+h)-f(x_0)]-{h\over 2}f''(x_0)-{h^2 \over 6}f'''(x_0)+O(h^3)$$
Use Extrapolation to derive an an $O(h^3)$ formula for $f'(x_0)$
I am unsure how to begin but From what I have seen in the textbook, extrapolating is replacing the h with 2h for example, multiplying the equation by some value, then subtracting it from its original equation. I am not exactly sure what the process is but this is the answer:
$$f'(x_0)={1 \over 12h}[f(x_0+4h)-12f(x_0+2h)+32f(x_0+h)-21f(x_0)]$$
Maybe this will be helpful. Here is the example in the textbook. You gotta go to the previous page which is page 190. Start reading from "which implies that". Its the third line from the top of the page.
Best Answer
Given: $$f'(x_0)={1 \over h}[f(x_0+h)-f(x_0)]-{h\over 2}f''(x_0)-{h^2 \over 6}f'''(x_0) + O(h^3) \tag{1}$$
Replace $h$ with $2h$ and simplify: $$f'(x_0)={1 \over {2h}}[f(x_0+2h)-f(x_0)]-hf''(x_0)-{{2}\over{3}h^2}f'''(x_0) + O(h^3) \tag{2}$$
Subtract $\frac{1}{2} (2)$ from $(1)$: $$f'(x_0) - \frac{1}{2}f'(x_0)={1 \over h}[f(x_0+h)-f(x_0)] - {1 \over {4h}}[f(x_0+2h)-f(x_0)] + \frac{1}{6}h^2f'''(x_0) + O(h^3) \tag{3}$$
Multiply $(3)$ by $2$ and rewrite: $$f'(x_0) = \frac{-f(x_0+2h) + 4f(x_0 + h) - 3f(x_0)}{2h} + \frac{4}{3}h^2 f'''(x_0) + O(h^3) \tag{4}$$
You now have $O(h^2)$
Replace $h$ with $2h$ in $(4)$, multiply and subtract. You now have $O(h^3)$.