Improving accuracy of f''(x) by Richardson extrapolation

Question

I have been given the following Question, I literally have no Idea where to start with it, I understand that the original expression comes from central difference approximation, that makes sense. I know what Richardson extrapolation is, but I am just struggling to understand how to combine the two.

Answer 1

I will derive the necessary error expansion as well as a formula for the 4th order accurate approximation.

Let $I = (a,b) \subseteq \mathbb R$ be an open interval. Let $f : I \rightarrow \mathbb R$ by six times differentiable. Let $x \in (a,b)$ and let $h_0 > 0$ be such that $(x-h_0,x+h_0) \subseteq I$. Let $0 < |h| \leq h_0$. By Taylor's theorem there exists $\xi$ between $x$ and $x+h$ such that $$ f(x+h) = f(x) + f'(x)h + \frac{1}{2} f''(x) h^2 + \frac{1}{6} f^{(3)}(x)h^3 + \frac{1}{24} f^{(4)}(x)h^4 + \frac{1}{120} f^{(5)}(x)h^5 + \frac{1}{720} f^{(6)}(\xi)h^6. $$ Similiarly, there exists $\nu$ between $x$ and $x-h$ such that $$ f(x-h) = f(x) - f'(x)h + \frac{1}{2} f''(x) h^2 - \frac{1}{6} f^{(3)}(x)h^3 + \frac{1}{24} f^{(4)}(x)h^4 - \frac{1}{120} f^{(5)}(x)h^5 + \frac{1}{720} f^{(6)}(\nu)h^6. $$ It follows, that $$ \frac{f(x+h)-2f(x)+f(x-h)}{h^2} = f''(x) + \frac{1}{12} f^{(4)}(x)h^2 + \frac{1}{360} \frac{f^{(6)}(\xi) + f^{(6)}(\nu)}{2} h^4$$ Equivalently, the difference between the target $T = f''(x)$ and the approximation $A_h$ given by $$A_h = \frac{f(x+h)-2f(x)+f(x-h)}{h^2} $$ satisfies $$ \left | T - A_h - \frac{1}{12} f^{(4)}(x)h^2 \right | \leq \frac{1}{360} \sup \left\{ |f^{(6)} (z)| \: : x - h_0 \leq z \leq x + h_0 \right \} h^4. $$ If $f^{(6)}$ is continuous, then we are certain that the right hand side is finite. This rigorously establishes that $$ T - A_h = \alpha h^2 + O(h^4), \quad 0 \leq |h| \leq h_0$$ where $\alpha = \frac{1}{12} f^{(4)}(x)$ is independent of $h$.

Now the path is clear. We have $$ T - A_{2h} = 4 \alpha h^2 + O(h^4), \quad 0 \leq |h| \leq \frac{1}{2} h_0. $$ It follows that $$ A_{h} - A_{2h} = 3 \alpha h^2 + O(h^4), \quad 0 \leq |h| \leq \frac{1}{2} h_0.$$ Equivalently, the dominant error term $\alpha h^2$ satisfies $$ \alpha h^2 = \frac{A_h - A_{2h}}{3} + O(h^4), \quad 0 \leq |h| \leq \frac{1}{2} h_0.$$ It follows $$ T - A_h = \frac{A_h - A_{2h}}{3} + O(h^4), \quad 0 \leq |h| \leq \frac{1}{2} h_0.$$ This shows that, the approximation $$ M_h = A_h + \frac{A_h - A_{2h}}{3}$$ is 4th order accurate.

It may or may not be interesting to express this approximation directly in terms of $f$, but I would always evaluate $M_h$ as stated, i.e., a good approximation $A_h$ plus a small correction, i.e., $\frac{A_h - A_{2h}}{3}$.