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I have just been looking at the error in Euler's Method, and I noticed something strange.

I understand that the error is proportional to $O(h)$ via the argument that the local truncation error is proportional to $O(h^2)$, and there are n of these errors, where n is proportional to 1/h.

Given this, I agree that as we gradually increase the number of terms (n) we will obtain a smaller and smaller value of h and hence the error will decrease. It makes intuitive sense that the error for one approximation will be smaller than the error for multiple successive approximations. $O(h)>O(h^2)$ when $h<1$ and $h\rightarrow0$ (or $n\rightarrow\infty$).

However, what if we vary n in the opposite direction. Once h increases above 1, it seems to me that the local truncation error will actually be larger than the global error since $O(h)<O(h^2)$ (as $h\rightarrow\infty$ or $n\rightarrow0$).

How can this be?

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It may be possible, but it isn't of terrible interest to numerical analysts. The issue of convergence is really a question about what happens when you refine/enrich your approximation space. You want to know if you put more effort in to the solution, you get more information out. One could also argue that $n$ should be a positive integer (it will be, say, the number of sub-intervals in your approximation), and so the statement $n\rightarrow 0$ doesn't make sense.

Edit to add: with the big-oh statements, these usually carry with them the statement as $n\rightarrow \infty$, so you'd have to go back and start from scratch if you wanted to see the behavior as $n$ shrinks or as $h$ grows.

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  • $\begingroup$ Oh yes, could we not argue that if n were sufficiently small and the interval sufficiently large, that the local truncation error will be larger than the overall error. If so is this something to do with multiple errors cancelling when h is greater than 1? $\endgroup$ – Resquiens Mar 1 '17 at 17:50
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    $\begingroup$ As formulated in the answer, most $O$ classifications only kick in for $n>n_0$, and the constant that is implicit in the notation also can heavily depend on $n_0$. -- For a local error of $C·h^2$ and Lipschitz constant $L$ of $f$ in $\dot x=f(x)$ you get a global error bound of $\frac{e^{LT}-1}L·Ch$ which is larger than $Ch^2$ even for $h=T$. $\endgroup$ – LutzL Mar 1 '17 at 19:52

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