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If we have a numerical approximation for a differential equation with an error term that is $O(h^2)$, then in that case it would seem that if the step size is less than $1$ that we would prefer to have an error proportional to, say, $h^4$ than $h^2$? Is that correct?

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The actual size of the step doesn't matter, there is most likely nothing universal about whatever unit you're using. An $h^4$ method will in most cases be better than an $h^2$ method, although it might vary slightly from method to method and application to application. –  Arthur Nov 10 '12 at 16:48
If the error is $O(h^2)$ then as the step size doubles the error will increase by $4$. Is that correct? And then if the error was $O(h^4)$ and we double the step size wouldn't that make the error $16$ times as large? –  sonicboom Nov 10 '12 at 17:43
Well, it is more sensitive to large step sizes. But unless you're working with planck time, there's nothing magical about the time step with size 1. But if you start with a huge step size the $h^2$ method will yield a better estimate (allthough with time steps long enough for the $h^2$ method to be the closest, both are probably way off). If you then halve the time step, then the $h^4$ method will close in a lot more error than the $h^2$ one, as you pointed out by a factor of 16 rather than 4. At some point it will catch up. Hopefully before they get too small for the computer. –  Arthur Nov 10 '12 at 19:21
Cheers I get it now. –  sonicboom Nov 10 '12 at 21:27
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