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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". KTC10:35, 18 October 2007 (UTC)[reply]
