The most compact algorithm for Simpson's rule??

12142015, 09:04 PM
Post: #12




RE: The most compact algorithm for Simpson's rule??
(12122015 11:02 PM)Namir Wrote: This is a slightly different version based on one of Dieter's remarks. I still don't get why you want to compute f(b) twice and generate an additional function call that is not required. Why don't you do it the classic way? Code: h = (b  a) / n If you want to count down to n=0 simply add another n=n1 on top of the do loop. Dieter 

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