[Rd] 1/tan(-0) != 1/tan(0)
Testing the code that Morten Welinder suggested for improving
extreme tail behavior of qcauchy(),
I found what you can read in the subject.
namely that the tan() + floating-point implementation on all
four different versions of Redhat linux, I have access to on
i686 and amd64 architectures,
1/tan(c(-0,0))
gives
-Inf Inf
and of course, that can well be considered a feature, since
after all, the tan() function does jump from -Inf to +Inf at 0.
I was still surprised that this even happens on the R level,
and I wonder if this distinction of -0 and 0 shouldn't be
mentioned in some place(s) of the R documentation.
For the real problem, the R source (in C), It's simple
to work around the fact that
qcauchy(0, log=TRUE)
for Morten's code proposal gives -Inf instead of +Inf.
Martin
MM == Martin Maechler [EMAIL PROTECTED]
on Wed, 1 Jun 2005 08:57:18 +0200 (CEST) writes:
Morten == Morten Welinder [EMAIL PROTECTED]
on Fri, 27 May 2005 20:24:36 +0200 (CEST) writes:
.
Morten Now that pcauchy has been fixed, it is becoming
Morten clear that qcauchy suffers from the same problems.
Morten
Morten qcauchy(pcauchy(1e100,0,1,FALSE,TRUE),0,1,FALSE,TRUE)
Morten should yield 1e100 back, but I get 1.633178e+16.
Morten The code below does much better. Notes:
Morten 1. p need not be finite. -Inf is ok in the log_p
Morten case and R_Q_P01_check already checks things.
MM yes
Morten 2. No need to disallow scale=0 and infinite
Morten location.
MM yes
Morten 3. The code below uses isnan and finite directly.
Morten It needs to be adapted to the R way of doing that.
MM I've done this, and started testing the new code; a version will
MM be put into the next version of R.
MM Thank you for the suggestions.
double
qcauchy (double p, double location, double scale, int lower_tail, int
log_p)
{
if (isnan(p) || isnan(location) || isnan(scale))
return p + location + scale;
R_Q_P01_check(p);
if (scale 0 || !finite(scale)) ML_ERR_return_NAN;
if (log_p) {
if (p -1)
lower_tail = !lower_tail, p = -expm1 (p);
else
p = exp (p);
}
if (lower_tail) scale = -scale;
return location + scale / tan(M_PI * p);
}
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RE: [Rd] 1/tan(-0) != 1/tan(0)
On 01-Jun-05 Martin Maechler wrote:
Testing the code that Morten Welinder suggested for improving
extreme tail behavior of qcauchy(),
I found what you can read in the subject.
namely that the tan() + floating-point implementation on all
four different versions of Redhat linux, I have access to on
i686 and amd64 architectures,
1/tan(c(-0,0))
gives
-Inf Inf
and of course, that can well be considered a feature, since
after all, the tan() function does jump from -Inf to +Inf at 0.
I was still surprised that this even happens on the R level,
and I wonder if this distinction of -0 and 0 shouldn't be
mentioned in some place(s) of the R documentation.
Indeed I would myself consider this a very useful feature!
However, a query: Clearly from the above (ahich I can reproduce
too), tan() can distinguish between -0 and +0, and return
different results (otherwise 1/tan() would not return different
results).
But how can the user tell the difference between +0 amnd -0?
I've tried the following:
sign(c(-0,0))
[1] 0 0
sign(tan(c(-0,0)))
[1] 0 0
sign(1/tan(c(-0,0)))
[1] -1 1
so sign() is not going to tell us. Is there a function which can?
Short of wrting one's own:
sign0 -
function(x){
if(abs(x)0) stop(For this test x must be +0 or -0)
return(sign(1/tan(x)))
}
;)
Best wishes,
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 01-Jun-05 Time: 10:50:06
-- XFMail --
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Re: [Rd] 1/tan(-0) != 1/tan(0)
On Jun 1, 2005, at 5:50 AM, (Ted Harding) wrote: However, a query: Clearly from the above (ahich I can reproduce too), tan() can distinguish between -0 and +0, and return different results (otherwise 1/tan() would not return different results). But how can the user tell the difference between +0 amnd -0? That's indeed a good question - by definition (-0)==(+0) is true, -00 is false and signum of both -0 and 0 is 0. I don't see an obvious way of distinguishing them at R level. Besides computational ways (like the 1/tan trick) the only (very ugly) way coming to my mind is something like: a==0 substr(sprintf(%f,a),1,1)==- Note that print doesn't display the sign, only printf does. At C level it's better - you can use the signbit() function/macro there. Any other ideas? Cheers, Simon __ R-devel@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
Re: [Rd] 1/tan(-0) != 1/tan(0)
On 6/1/05, Simon Urbanek [EMAIL PROTECTED] wrote: On Jun 1, 2005, at 5:50 AM, (Ted Harding) wrote: However, a query: Clearly from the above (ahich I can reproduce too), tan() can distinguish between -0 and +0, and return different results (otherwise 1/tan() would not return different results). But how can the user tell the difference between +0 amnd -0? That's indeed a good question - by definition (-0)==(+0) is true, -00 is false and signum of both -0 and 0 is 0. I don't see an obvious way of distinguishing them at R level. Besides computational ways (like the 1/tan trick) the only (very ugly) way coming to my mind is something like: a==0 substr(sprintf(%f,a),1,1)==- Note that print doesn't display the sign, only printf does. On my XP machine running R 2.1.0 patched 2005-05-14 sprintf(%f,-0) [1] 0.00 does not print the sign. however, the tan trick can be done without tan using just division: R sign0 - function(x) if (x != 0) stop(x not zero) else sign(1/x) R sign0(0) [1] 1 R sign0(-0) [1] -1 __ R-devel@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
