unfortunately in that statement, the (e2/e1) power is applied to the sum of inner terms, so I can't condense it any further than this:
(fabs(f1)**(2/e2) + fabs(f2)**(2/e2))**(e2/e1) + fabs(f3)**(2/e1) On Wed, Sep 30, 2009 at 1:52 AM, Robert Bradshaw <[email protected]> wrote: > 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 > _______________________________________________ Cython-dev mailing list [email protected] http://codespeak.net/mailman/listinfo/cython-dev
