Date: Fri, 27 Jul 2012 09:45:02 -0500 From: Stephen Montgomery-Smith <stephen@missouri.edu> To: Bruce Evans <brde@optusnet.com.au> Cc: freebsd-bugs@freebsd.org, FreeBSD-gnats-submit@freebsd.org, Stephen Montgomery-Smith <stephen@freebsd.org> Subject: Re: bin/170206: complex arcsinh, log, etc. Message-ID: <5012A96E.9090400@missouri.edu> In-Reply-To: <20120727233939.A7820@besplex.bde.org> References: <201207270247.q6R2lkeR021134@wilberforce.math.missouri.edu> <20120727233939.A7820@besplex.bde.org>
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On 07/27/2012 09:26 AM, Bruce Evans wrote: > % hm1 = -1; > % for (i=0;i<12;i++) hm1 += val[i]; > % return (cpack(0.5 * log1p(hm1), atan2(y, x))); > > It is the trailing terms that I think don't work right here. You sort > them and add from high to low, but normally it is necessary to add > from low to high (consider terms [1, DBL_EPSILON/2, DBL_EPSILON/4]). > Adding from high to low cancels with the -1 term, but then only > particular values work right. Also, I don't see how adding the low > terms without extra precision preserves enough precision. I understand what you are saying. But in this case adding in order of smallest to largest (adding -1 last) won't work. If all the signs in the same direction, it would work. But -1 has the wrong sign. But I can also tell you that I haven't thought my algorithm through every special case. I can tell you it seems to work in all the examples I tried. But I don't have a mathematical proof.
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