Use Richardson extrapolation to improve the formula $$ f''(x) \sim \frac{f(x+h)-2f(x)+f(x-h)}{h^2} $$ so that the error is reduced to order $h^4$
I am not sure how to go about doing this problem, if anyone could help it would be much appreciated, thank you for your help in advance.
Richardson extrapolation is based on the fact that if the time-step is decreased to half then the error goes to $\frac{1}{2^p}$. The formula is second order accurate in $h$.
So we can write
$$ f''(x)=\frac{1}{h^2}\left(f(x+h) -2f(x) + f(x-h) \right) + Kh^2 + O(h^4) $$
Now we use half the step
$$ f''(x)=\frac{4}{h^2} \left(f(x+\frac{h}{2}) -2f(x) + f(x-\frac{h}{2})\right) + K\frac{h^2}{4} + O(h^4) $$
Multiply the second equation with $4$ and subtract the first to obtain
$$ f''(x)= \frac{1}{h^2} \left[ 16 \left\{f(x+\frac{h}{2})-2f(x) + f(x-\frac{h}{2})\right\} -\left\{f(x+h) -2f(x) + f(x-h)\right\} \right] + O(h^4) $$ EDIT: I had started incorrectly by assuming first order. The odd powers cancel out. Hope I got the rest of the math right.