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I read that the local truncation error of a method of order $p$ is $O(h^{p+1})$ (thus, one order higher). Yet, the global truncation error is in general $O(h^p)$ for a stable method (thus, one order lower than the local error). But why?

If the global truncation error is the accumulation of local truncation error at all steps $h$, shouldn't it be still order $p+1$?

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up vote 2 down vote accepted

For a fixed interval of integration, the number of steps is $O(h^{-1})$. The global error is the local error times the number of steps. So if $e_i$ is the error at step $i$, and $E$ is the global error, and we're integrating over an interval of length $L$, informally we have:

$$ e_i = O(h^{p+1}) \Leftrightarrow |e_i| \le Ch^{p+1} $$

$$ |E| = \left| \sum_{i=1}^{\lfloor L/h \rfloor}e_i \right| \le \sum_{i=1}^{\lfloor L/h \rfloor}Ch^{p+1} \le CLh^p \Leftrightarrow E = O(h^p) $$

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I edited the number of steps (which was $O(h^{−1})$ before) and accepted the answer. If my edit is correct, I believe it is appropriate to accept this answer –  Sosi Jun 4 '13 at 11:33
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