# Asymmetrical step size to apply Richardson Extrapolation to improve Runge-Kutta order 2 solution

I'm trying to solve a series of problems related to approximations of ODEs with Runge-Kutta that have their approximation to values improved by using the Richardson Extrapolation.

Some of these problems propose the use of step sizes different than $$h/2$$ when comparing the solution obtained with a step-size $$h$$. For example, one of the problems asks me to apply Richardson Extrapolation with two steps: one of $$0.25h$$ and then another of $$0.75h$$, which then are to be compared with one of the $$h$$ step iterations.

I've been given the error formula $$\epsilon= 2^r \frac{y^\frac{h}{2}_{m+1} - y^h_{m+1}}{1 - 2^r}$$ to calculate the error for an iteration of step-size $$h$$ when applying Richadson Extrapolation with a secondary step-size of $$h/2$$, with $$r$$ being the order of the used method and $$m+1$$ representing the number of the given iteration.

I'm completely lost in finding out how to adapt that formula for the $$0.25h$$ and $$0.75h$$ step-sizes or how to derive the formula starting from the Taylor formula.

This means that the error of step size $$h$$ is $$ch^{p+1}+O(h^{p+2})$$, $$p$$ the order of the method, and the errors of the partial steps are thus scaled $$ch_k^{p+1}+O(h_k^{p+2})$$. This gives $$y_1=y_*+ch^{p+1}+O(h^{p+2})\\ y_2=y_*+ch^{p+1}[(\tfrac14)^{p+1}+(\tfrac34)^{p+1}]+O(h^{p+2})$$ Now the extrapolated value is obtained by eliminating $$ch^{p+1}$$, that is $$[(\tfrac14)^{p+1}+(\tfrac34)^{p+1}]y_1-y_2=[(\tfrac14)^{p+1}+(\tfrac34)^{p+1}-1]y_* \\~\\ y_*=\frac{4^{p+1}y_2-[3^{p+1}+1]y_1}{4^{p+1}-[3^{p+1}+1]}$$