# Euler's Method Error Term (Big O Notation)

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?

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.
• 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 $h=T$. – LutzL Mar 1 '17 at 19:52