m1 = a + (b-a)/2
m2 = (a+b)/2
why is it that m2's result is more accurate than m1's result ?
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m1 = a + (b-a)/2
m2 = (a+b)/2
why is it that m2's result is more accurate than m1's result ?
Is that the complete question, is there not some contextual information you have omitted?Quote:
Originally Posted by TGASJOOL
(especially as in any compiled language the optimiser will change the rhs of the first into am form identical to the rhs of the second)
CB
normal floating point arithmetic of a pc applies in this case. I have seen many bisection algorithms do it like m2. Why does m1 produce a lower accurate result?
The more operations the larger the rounding errors will become, similarly one should avoid subtracting near equal quantities. All of which makes m1 probably better than m2.Quote:
Originally Posted by TGASJOOL
CB
