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Message-ID: <alpine.LRH.2.02.1902241254030.6973@key0.esi.com.au>
Date: Sun, 24 Feb 2019 13:53:31 +1100 (AEDT)
From: Damian McGuckin <damianm@....com.au>
To: musl@...ts.openwall.com
cc: Shane Seelig <stseelig@...l.com>
Subject: Re: x87 asin and acos
On Sat, 23 Feb 2019, Rich Felker wrote:
>> They don't seem to be numerically equal. For example, if x is smaller
>> than sqrt(LDBL_EPSILON/2), 1-x**2 is 1, but (1-x)*(1+x) is not.
>> don't recall the process of writing the function in detail, but I'm
>> pretty sure this matters to the result, especially since sqrt then
>> expands the magnitude of the error.
>
> After some discussion on irc, I think the above may be wrong.
Yes and no.
Interestingly, if you to look at double (or float) for now
1 - x**2 == 1 if |x| < sqrt(DBL_EPSILON/2)
whereas
(1 - x) is not 1 nor is (1 + x)
although interesting, to the precision (a
(1 - x) * (1 + x) == 1 if DBL_EPSILON/2 < |x| < sqrt(DBL_EPSILON/2)
But at |x| == DBL_EPSILON/2 (== the round bit)
(1 + x) == 1 but (1 - x) != 1 and so
(1 + x) * (1 - x) != 1
because DBL_EPSILON/2 affects the round bit. That said
sqrt((1 - x) * (1 + x)) = 1 + 0.5 * x == 1
because (DBL_EPSILON/2) * 0.5 is half the round bit and does not affect
it. So, the formula does not affect the result of the sqrt().
The same happens for floats.
I think the same can be said for long double.
Regards - Damian
Pacific Engineering Systems International, 277-279 Broadway, Glebe NSW 2037
Ph:+61-2-8571-0847 .. Fx:+61-2-9692-9623 | unsolicited email not wanted here
Views & opinions here are mine and not those of any past or present employer
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