Exercise: Richardson on ln(x)

Find $f'(1.8)$ for $f(x)=\ln x$ with $\mathcal{O}(h)$ forward differences and $h=0.1$, and improve to $\mathcal{O}(h^2)$ with Richardson. The exact value is $f'(1.8)=\frac{1}{1.8}=0.5556$.

1. Forward approximation with $h$ and $h/2$:

   $$N_1(0.1)=\frac{f(1.9)-f(1.8)}{0.1}=0.5407,\qquad N_1(0.05)=\frac{f(1.85)-f(1.8)}{0.05}=0.5480$$

2. Richardson extrapolation (all-terms case):

   $$N_2(0.1)=N_1(0.05)+\left[N_1(0.05)-N_1(0.1)\right]=0.5480+0.0073$$