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I know that the local error at each step of Euler's method is O(t^2), where t is the time step. And since there are (b-a)/t steps, the order of the global error is O(t).

However, I saw a derivation of the global error by saying:

[f(x+t) - f(x)] / t = f'(x) + O(t)

Where O(t) represents the rest of the Taylor series expansion for f. My question is: how does this show that the global error is O(t)? Isn't this just showing that the slope's error is O(t)?

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    $\begingroup$ This question would be more appropriate on Math.SE, as it pertains to undergraduate numerical analysis. $\endgroup$
    – David Ketcheson
    Commented Oct 12, 2012 at 11:55
  • $\begingroup$ Some versions of the Gronwall inequality apply to Euler's method, giving you a global error estimate. You could Google that term, "Gronwall inequality". It's exponential in nature. $\endgroup$
    – Ryan Budney
    Commented Oct 12, 2012 at 16:06

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Any analysis of global error must include information about how local errors are amplified in subsequent steps. So your statement

I know that the local error at each step of Euler's method is O(t^2), where t is the time step. And since there are (b-a)/t steps, the order of the global error is O(t).

isn't accurate without some assumption of stable error propagation.

You are correct that the "derivation of the global error" given does not say anything about global error.

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