On Sep 29, 2009, at 12:03 PM, Chris Colbert wrote: > Actually, yes. The exponents are done specifically in that way to > force the sign of the end value. > > I suppose i could just call abs() on it though....
I'm sure that abs would be faster than pow, though it's not the squaring that's expensive, but the arbitrary powers. (x^(1/e2))^(e2/ e1) -> x^(1/e1) will always work for you since x is positive. - Robert _______________________________________________ Cython-dev mailing list [email protected] http://codespeak.net/mailman/listinfo/cython-dev
