Talk:Midpoint method

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Accuracy

This method looks like it will always be less acurate, and require more computation than the Euler method done with step size h/2. —The preceding unsigned comment was added by Neiljackson (talk • contribs).

That is not correct. The midpoint method gives an accuracy of O(h^3), compared to O(h^2) for Euler's method. So overall you gain by using this instead of Euler's method, even in spite of more computation involved. Oleg Alexandrov (talk) 00:48, 12 February 2007 (UTC)[reply]

Midpoint is indeed of a higher order than simple Euler's method. However, "higher order is not always higher accuracy". KTC 10:35, 18 October 2007 (UTC)[reply]

Comparison Graphs

The graph comparison should be drawn against Euler method using h/2 step size because in terms of evaluation of the system of equations, that is the equivalent computation. 203.206.220.108 (talk) 01:24, 1 May 2010 (UTC)[reply]

+1 I was going to say this myself Tuntable (talk) 00:55, 16 December 2014 (UTC)[reply]