 # One method consists in computing the numerical solution for an arbitrary h and then 2h. The Richardson extrapolation gives an estimate of e=maxt|y(t,2h)−y(t,h)| of the error. When the error is smaller than the tolerance, we keep the result and start from 2y(t,h)−y(t,h). If the error is larger we restart with h/2 until we reach the tolerance. sublimnes9 2022-08-07 Answered
One method consists in computing the numerical solution for an arbitrary $h$ and then $2h$. The Richardson extrapolation gives an estimate of $e=\underset{t}{max}|y\left(t,2h\right)-y\left(t,h\right)|$ of the error. When the error is smaller than the tolerance, we keep the result and start from $2y\left(t,h\right)-y\left(t,h\right)$. If the error is larger we restart with $h/2$ until we reach the tolerance.
( $y\left(t,2h\right)$ means approximation with $2h$)
I don't understand why the Richardson extrapolation is mentioned. For what do I have to use it? Can I not just calculate $y\left(t,2h\right)$ and $y\left(t,h\right)$ and see the error?
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You can just check the error, and not bother with Richardson extrapolation, that's fine. But once you've paid the computational price of calculating the error, you might as well use Richardson extrapolation to blackuce the error further; it costs you essentially nothing.