# Global Error Analysis of Euler's Method

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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This question would be more appropriate on Math.SE, as it pertains to undergraduate numerical analysis. –  David Ketcheson Oct 12 '12 at 11:55
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. –  Ryan Budney Oct 12 '12 at 16:06