http://www.shodor.org/cserd/Resources/Algorithms/NumericalDifferentiation/
How did they get the three points $$y'(x)=\frac{y(x+h)-y(x-h)}{2h}$$ to $$y'(x)=\frac{-3y(x)+4y(x+h)-y(x+2h)}{2h}$$
Where does the $-3y + 4y$ come from? It does not make sense to me.
It arises as follows (based on the method discussed in the answer here:
$$ y(x+h) \approx y(x) + hy'(x) + \frac{h^2}{2}y''(x)$$ $$y(x+2h) \approx y(x) + 2hy'(x) + 2h^2y''(x)$$
To cancel the $y''$ terms, we take
$$4y(x+h) - y(x+2h) \approx 3y(x) + 2hy'(x)$$ The advantage of canceling these terms is that it gives us a better estimate -- the error here is now on the order of $h^3$, since we've canceled the $h^2$ terms.
Thus we get $$-3y(x) + 4y(x+h)-y(x+2h) \approx 2hy'(x)$$ So $$y'(x) \approx \frac{-3y(x) + 4y(x+h)-y(x+2h)}{2h}$$ with error on the order of $\frac{h^3}{h}=h^2$.
