ATANH

ATANH

[Damian McGuckin]

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

Re: ATANH

[Szabolcs Nagy]

* Damian McGuckin <damianm@esi.com.au> [2019-02-27 13:59:41 +1100]:
> 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

that's because musl assumes i386 uses FLT_EVAL_METHOD==2
so essentially evaluates everything to long double
precision.

fdlibm in the various BSDs assumes FLT_EVAL_METHOD==0
on all targets i think.

(in c99 the return statement is not required to drop
excess precision, so the different result is observable)

but i think either way is fine, it's not going to cause
huge errors.

note that the largest possible threshold may not be the
best choice: you want that branch to be correctly
predicted, so e.g. if 0x1p-40 input happens a few times,
but inputs below 0x1p-50 are very rare then a x < 0x1p-50
check may be better in practice: a 0x1p-40 input won't
break the branch predictor trained for the common case.

> 
> 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