So we know what the Richardson Extrapolation for a first derivative looks like using a recursive formula like this:
$$ D_{m\Delta x}^{1}=\frac{f(x+m\Delta x)-f(x-m\Delta x)}{2m\Delta x}$$
$$ D_{m\Delta x}^{n}=\frac{4^{n-1}D_{m\Delta x}^{n-1}-D_{2m\Delta x}^{n-1}}{4^{n-1}-1}+\mathcal{O}(\Delta x^{2n})$$
for integers $ m,n >0$, where bigger $n$ means greater precision of the derivative.
After learning this formula for a first derivative I tried to reach a similar formula for a second derivative so I started by defining :
$$ \Lambda_{m\Delta x}^{1}=\frac{f(x+m\Delta x)+f(x-m\Delta x)-2f(x)}{m^2\Delta x^2}$$
Assuming this, the next step was to get $\Lambda_{m\Delta x}^{2}$ and see if I could extract some kind of pattern so I could get the most generic formula. After some calculations I got ( what I did here was eliminating the term with $\Delta x^4$ so I could raise the precision to $\Delta x^6$):
$$\Lambda_{m\Delta x}^{2}=\frac{4\Lambda_{m\Delta x}^{1}-\Lambda_{2m\Delta x}^{1}}{4-1}+\frac{-4f(x)}{3m^2\Delta x^2}+ \mathcal{O}(\Delta x^{4}) $$
We can clearly see that the first fraction is given by the exact same formula as the first derivative, but I also got another term which I believe should not be there ( at this point I'm pretty sure I didn't mess up at any intermediate step) and
Furthermore, I made a little script in Matlab where I didn't put the extra term and I got what I was supposed to:a pretty precise value for the second derivative of any function. Also this extra term "destroys" the rest of the function for when $\Delta x$ is small ( which should be if we want more precision) it goes to infinity.
Basically I reached the conclusion : $ \Lambda_{m\Delta x}^{n}=\frac{4^{n-1}\Lambda_{m\Delta x}^{n-1}-\Lambda_{2m\Delta x}^{n-1}}{4^{n-1}-1}$ for a second derivative.
So my question is if this conclusion is correct ( and why does this term appear and if it shouldn't, should we be able to get another formula like this one for a third order derivative or higher)?
Best Answer
There shouldn't be any extra term; the Richardson extrapolation of a method $T_n$ with order $p_n$ achieved by considering a method with one step size and another with double that step size is $T_{n+1}(\Delta x)=\frac{2^{p_n}T_n(\Delta x)−T_n(2 \Delta x)}{2^{p_n}-1}$.
The question that Richardson extrapolation needs answered and cannot answer by itself is "what is $p_n$?" The answer comes from the asymptotic expansion of $T_1$. It turns out that if $T_1$ is the usual centered difference then its asymptotic expansion contains all even powers and no odd powers. Thus $p_n=2n$, so $T_{n+1}(\Delta x)=\frac{4^n T_n(\Delta x)−T_n(2\Delta x)}{4^n−1}$. This is equivalent to the first term of what you wrote (in your notation the base rule is $T_0$ rather than $T_1$).
As it happens, when the base rule is the basic centered difference for the first derivative or the centered difference for the second derivative, you have $p_n=2n$, so the same formulae wind up being used.