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Date: Wed, 27 Feb 2019 13:59:41 +1100 (AEDT)
From: Damian McGuckin <damianm@....com.au>
To: musl@...ts.openwall.com
Subject: ATANH
The comments for this routine say:
* atanh(x) = log((1+x)/(1-x))/2 = log1p(2x/(1-x))/2 ~= x + x^3/3 + o(x^5)
There is a point where atanh(x) can be approximated by just x to machine
precision. This is where the exponent of x is less than some number, or
where x itself is less than some number.
In MUSL, for the 80-bit version, directly from the code, this is
2^(-LDBL_MANT_DIG/2) = 0x1.0p-32;
Interestingly, this same number, is used double atanh and float atanhf.
Note that in fdlibm, 0x1.0p-28 is used for all types
In fact, isn't
x * (1 + x^2/3) == x (after roundoff)
if
x < 2^(p/2),
i.e. 0x1.0p-26 for double, 0x1.0p-12 for float.
Note that for 80-bit wide, it is 0x1.0p-32 agreeing with MUSL!!
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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