[R] Exponentiate very large numbers

[William Dunlap]

> Are there any tricks I can use to get a real result
> for exp( ln(a) ) - exp( ln(0.1) + ln(b) ), either in logarithm or
> exponential form?

  log.a <- 1347
  log.b <- 1351
  f0 <- function(log.a, log.b) exp(log.a) - exp(log(0.1) + log.b)

will not work because f0(1347,1351) is too big to represent as
a double precision number (abs(result)>10^308)).
If you are satisfied with computing log(result) you can do it
with some helper function:
   addOnLogScale <- function (e1, e2) 
   {
       # log(exp(e1) + exp(e2))
       small <- pmin(e1, e2)
       big <- pmax(e1, e2)
       log1p(exp(small - big)) + big
   }
   subtractOnLogScale <- function (e1, e2) 
   {
      # log(abs(exp(e1) - exp(e2)))
       small <- pmin(e1, e2)
       big <- pmax(e1, e2)
       structure(log1p(-exp(small - big)) + big, isPositive = e1 > e2)
   } 
 
as

   f1 <- function(log.a, log.b)  {
        # log(abs(exp(log.a) - exp( log(0.1) + log.b))), avoiding overflow
        subtractOnLogScale( log.a, log(0.1) + log.b )
   }

E.g.,
  > f0(7,3)
  [1] 1094.625
  > exp(f1(7,3))
  [1] 1094.625
  attr(,"isPositive")
  [1] TRUE

With your log.a and log.b we get
  > f1(1347, 1351)
  [1] 1348.495
  attr(,"isPositive")
  [1] FALSE

You can use the Rmpfr package to compute the results to any desired precision. 
 
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf
> Of francesca casalino
> Sent: Monday, February 04, 2013 8:00 AM
> To: r-help at r-project.org
> Subject: Re: [R] Exponentiate very large numbers
> 
> I am sorry I have confused you, the logs are all base e:
> 
> ln(a) = 1347
> ln(b) = 1351
> 
> And I am trying to solve this expression:
> 
> exp( ln(a) ) - exp( ln(0.1) + ln(b) )
> 
> 
> Thank you.
> 
> 2013/2/4 francesca casalino <francy.casalino at gmail.com>:
> > Dear R experts,
> >
> > I have the logarithms of 2 values:
> >
> > log(a) = 1347
> > log(b) = 1351
> >
> > And I am trying to solve this expression:
> >
> > exp( ln(a) ) - exp( ln(0.1) + ln(b) )
> >
> > But of course every time I try to exponentiate the log(a) or log(b)
> > values I get Inf. Are there any tricks I can use to get a real result
> > for exp( ln(a) ) - exp( ln(0.1) + ln(b) ), either in logarithm or
> > exponential form?
> >
> >
> > Thank you very much for the help
> 
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