Trying to use Taylor series to find a formula for a Richardson extrapolation of order 6

Question

I am trying to use Taylor series to find the formula for the Richardson extrapolation based on $f'(x)$ of order $O(h^6)$. I know the formula should have the form of: $$\Delta_1(h) - 20\Delta_1(h/2) + 64\Delta(h/4) = 45f'(x) + O(h^6) $$ where, $$ \Delta_1(h) = \frac{f(x+h) - f(x-h)}{2h} $$ But the problem I am having is that I am is that I am unsure of how exactly to get the form required. My two thought processes were either

1) Just plug the definition of Taylor series in for each delta and then confirm that it gives $45f'(x)$.

The reason why I figured that wouldn't be the best is because it would take a long time (writing by hand) to do this.

The other thought process is to start with a lower order (say order 2) by using Taylor series and then extrapolate multiple times to get the desired formula, while using an argument on how this is acceptable.

But with this I am unsure if a) that would give the desired formula. and b) how exactly would you do that to get the desired formula ?

Any help would be greatly appreciated.

Answer 1

\begin{eqnarray*} \Delta_1(h)= f^{(1)}(x)+\frac{h^2}{6} f^{(3)}(x)+\frac{h^4}{120}f^{(5)} +O(h^6) \\ -20 \Delta_1(h/2)= -20 f^{(1)}(x)-\frac{5h^2}{6} f^{(3)}(x)-\frac{h^4}{96}f^{(5)} +O(h^6) \\ 64 \Delta_1(h/2)= 64 f^{(1)}(x)+\frac{4h^2}{6} f^{(3)}(x)+\frac{h^4}{480}f^{(5)} +O(h^6) \\ \end{eqnarray*} Add these up & the result follows.

Answer 2

You know from symmetry that $$Δ_1(h)=f'(x)+c_2h^2+c_4h^4+c_6h^6+...$$ The first Richardson extrapolation gives thus $$ E_2(h)=Δ_1(h)-4Δ_1(h/2)=-3f'(x)+\tilde c_4h^4+\tilde c_6h^6+... $$ Eliminate the fourth order term with the next extrapolation $$ E_2(h)-16E_2(h/2)=(-15)(-3)f'(x)+O(h^6) $$ or expanded $$ Δ_1(h)-4Δ_1(h/2)-16Δ_1(h/2)+64Δ_1(h/4)=45f'(x)+O(h^6) $$