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For a given ODE $y'(t)=f(t,y)$, Euler's method is $$ y(t+h)=y(t)+hf(t,y(t)) + O(h^2) $$

It is said that by using Richardson extrapolation, we can improve it to $$ y(t+h)=y(t)+hf(t+\frac{h}{2},y(t)+\frac{h}{2}f(t,y)) + O(h^3) $$

But I don't know how to deduce it.

To use Richardson extrapolation, we need a quantity of interest $A$ and its approximation $A(h)$, what should them be?

BTW, I know we can use the formula $y'(x+h/2)=\frac{y(x+h)-y(x)}{h}+O(h^3)$ to get improved Euler's method. But here I wonder how to get it from Richardson's method?

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Please, see pag.4 in – Papiro Mar 27 '13 at 12:15

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