When applying Richardson Extrapolation to Euler Method, when isn't the local truncation error of h^2 used to get (4*Euler(h/2) - Euler(h)) / 3 instead of using the global error of h to get 2*Euler(h/2) - Euler(h)?
Answer & Explanation
In general when you are applying Richardson extrapolation to a method , it should be an approximation of a fixed quantity which doesn't depend on . Thus for instance for the Euler method, is the result of running the Euler method with step size h for a fixed ODE up to a fixed finite time. So the error that matters is the global error.
You want to compare two approximations at . The error of two Euler steps at half step size is in the leading order
if the truncation error is locally for a step of size .
Thus you get, if denotes the method application to get to a fixed time with step size ,
and with two steps of half the step size
Eliminating the term
This is incidentally the explicit midpoint method
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