### [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
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);
}

__
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 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
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 --

__
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 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

```