We know $F(h)=a_0 +a_1h + a_2 h^3$
$F(1)=4$; $F(1/2)=21/8$; $F(1/4)=145/64$
Find a approximation of $F(0)=a_0$ with Richardson extrapolation method with an absolute error less than $10^{-2}.$
This is what I have done: In Richardson method, we usually used this fact:
$$F(h)=a_0 +a_1h + a_2 h^3$$ then
$$F(h/2)=a_0 +a_1 \frac{h}{2} + a_2 \frac{h^3}{2^3}$$
So $$F(h)-F(h/2)=\frac{1}{2}a_1 h + \frac{7}{8}a_2h^3$$
Then we have this:
$$\frac{F(h)-F(h/2)}{F(h)-a_0}=\frac{\frac{1}{2}a_1 h +\frac{7}{8}a_2 h^3}{a_1 h+ a_2 h^3}$$
So we are asking for getting $|F(h)-a_0|\leq 10^{-2}$ what $h$ we must take?
Thank you!
Original problem in Spanish:
For the given form of $F(h)=a_0+a_1 h+a_2 h^3$, as you say, you can solve the system
$$F(1)=4,F(\tfrac{1}{2})=\tfrac{21}{8},F(\tfrac{1}{4})=\tfrac{145}{64}$$
for the coefficients, and obtain $(a_0,a_1,a_2)=(2,1,1)$, hence
$$F(h)=2+h+h^3.$$
For $h\geq 0$ the condition $\epsilon=|F(h)-F(0)|$ becomes a third-order equation in $h$,
$$h^3+h-\epsilon=0,$$
which can be solved using standard techniques. The only real solution is
$$h_{crit}=\left(\frac{\sqrt{3^3 \epsilon^2+4}}{2\sqrt{3^3}}+\frac{\epsilon}{2}\right)^{1/3}-\frac{1}{3}\left(\frac{\sqrt{3^3 \epsilon^2+4}}{2\sqrt{3^3}}+\frac{\epsilon}{2}\right)^{-1/3}$$
For you tolerance of $\epsilon=10^{-2}$ this is
$$h<h_{crit}\approx 0.0099990003.$$