Re: [R] over/under flow

From: Martin Maechler <maechler_at_stat.math.ethz.ch>
Date: Mon 04 Jul 2005 - 17:33:31 EST

>>>>> "William" == William H Asquith <wasquith@austin.rr.com> >>>>> on Sun, 3 Jul 2005 16:00:40 -0500 writes:

    William> Great, but to followup, how do I select the bounds (B,E) on the root     William> for an arbitrary machine?

    William> OVER = uniroot(function(x) lgamma(x)-log(.Machine$double.xmax),     William> c(B,E))$root

    William> If uniroot() is fast enough, is it appropriate for me to set B at say 1 
    William> and E at log(.Machine$double.max)?  Suggestions on do this the proper 
    William> "R way"?  Perhaps this . . .

    William> OVER = uniroot(function(x) lgamma(x)-log(.Machine$double.xmax),     William> c(1,log(.Machine$double.xmax)))$root

    William> I am working on a package so different machines will be involved thus     William> simple 171,172 might not be the best idea for the root?

Well,

  1. We nowadays *require* compliance with the usual ``IEEE arithmetic standard'' {put a bit too simplistically} and I'd bet that those compiler / runtime library combinations that can build R properly, would also give cutoffs in this same interval
  2. lgamma(x) is very linear ``out there'' -- try curve(lgamma(x) - log(.Machine$double.xmax), 170, 180)
    • so you can easily enlarge the interval a bit, e.g. to (170,175) or even to (100,200)

Martin

    William> On Jul 3, 2005, at 3:43 PM, Peter Dalgaard wrote:

>> "William H. Asquith" <wasquith@austin.rr.com> writes:
>>

    >>> I am porting some FORTRAN to R in which an Inf triggers an if().  The
    >>> trigger is infinite on exp(lgamma(OVER)).  What is the canonical R
    >>> style of determining OVER when exp(OVER)== Inf?  The code structure
    >>> that I am
    >>> porting is best left intact--so I need to query R somehow to the value
    >>> of OVER that causes exp(lgamma(OVER)) to equal Inf.
    >>> 
    >>> On my system,
    >>> exp(lgamma(171)) is about first to equal Inf.
    >>> 
    >>> I asked similar question a few weeks ago on exp(OVER) and got the
    >>> answer back as log(.Machine$double.xmax).  I now have the lgamma
    >>> involved.  I think that answer is what is OVER such the
    >>> 
    >>> .Machine$double.xmax = lgamma(OVER),

>>
>> Not quite... (see below)
>>
    >>> but I am not sure how to invert or solve for OVER

>>
>>
    >>> uniroot(function(x) lgamma(x)-log(.Machine$double.xmax), c(171,172))

>> $root
>> [1] 171.6244
>>
>> $f.root
>> [1] -1.462051e-07
>>
>> $iter
>> [1] 3
>>
>> $estim.prec
>> [1] 6.103516e-05
>>
>>
>> --
>> O__ ---- Peter Dalgaard ุster Farimagsgade 5, Entr.B
>> c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
>> (*) \(*) -- University of Copenhagen Denmark Ph: (+45)
>> 35327918
>> ~~~~~~~~~~ - (p.dalgaard@biostat.ku.dk) FAX: (+45)
>> 35327907
>>
    William> ______________________________________________
    William> R-help@stat.math.ethz.ch mailing list
    William> https://stat.ethz.ch/mailman/listinfo/r-help     William> PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html

    William> !DSPAM:42c852c3295961260279199!



R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Mon Jul 04 17:36:44 2005

This archive was generated by hypermail 2.1.8 : Fri 03 Mar 2006 - 03:33:11 EST