With RK4 the expressions become $$ \tilde{x}=x_2+\epsilon+O((h)^{6}), $$ with $$ \epsilon = \frac{|x_1-x_2|}{15}. $$ The estimate is one order higher than the original RK4. But this method is normally rather inefficient since it requires a lot of computations. We solve typically the equation three times at each time step. However, we can compare the estimate \( \epsilon \) with some by us given accuracy \( \xi \). We can then ask the question: what is, with a given \( x_j \) and \( t_j \), the largest possible step size \( \tilde{h} \) that leads to a truncation error below \( \xi \)? We want $$ C\tilde{h} \le \xi, $$ which leads to $$ \left(\frac{\tilde{h}}{h}\right)^{M+1}\frac{|x_1-x_2|}{(1-2^{-M})}\le \xi, $$ meaning that $$ \tilde{h}=h\left(\frac{\xi}{\epsilon}\right)^{1+1/M}. $$