[R] Exponentiate very large numbers

[Spencer Graves]

Functions "log1p" and "expm1" may be of interest here.  To understand 
them, look at a 2- or 3-term Taylor expansion for log(x) [natural log, 
of course] and exp(x). Spencer


On 2/5/2013 10:48 AM, Albyn Jones wrote:
> I stayed out of this one thinking it was probably a homework exercise.
> After others have responded, I'll go ahead with my gloss on
> Bill's function...
>
> The specific problem is really of the form
>
>      exp(a) - exp(a+eps) = exp(a)*(1-exp(eps))
>
> So even though we can't compute exp(1347), we can
> compute 1-exp(eps) for eps of the desired size.
> The given eps was
>
>> 1351+log(.1) -1347
> [1] 1.697415
>
> So the answer is exp(1347)*(1-exp(1.697415)),
> and as Bill noted, you can compute the log
> of the absolute value of that quantity...
>
>> 1347 + log(abs(1-exp(1.697415)))
> [1] 1348.495
>
> albyn
>
>
>
>
>
>
> On 2013-02-05 10:01, William Dunlap wrote:
>>> 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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>>