Global Error Analysis of Euler's Method - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T14:10:52Z http://mathoverflow.net/feeds/question/109429 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109429/global-error-analysis-of-eulers-method Global Error Analysis of Euler's Method math2316 2012-10-12T04:59:31Z 2012-10-12T11:54:43Z <p>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).</p> <p>However, I saw a derivation of the global error by saying:</p> <pre><code>[f(x+t) - f(x)] / t = f'(x) + O(t) </code></pre> <p>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)?</p> http://mathoverflow.net/questions/109429/global-error-analysis-of-eulers-method/109454#109454 Answer by David Ketcheson for Global Error Analysis of Euler's Method David Ketcheson 2012-10-12T11:54:43Z 2012-10-12T11:54:43Z <p>Any analysis of global error must include information about how local errors are amplified in subsequent steps. So your statement </p> <blockquote> <p>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).</p> </blockquote> <p>isn't accurate without some assumption of stable error propagation.</p> <p>You are correct that the "derivation of the global error" given does not say anything about global error.</p>