Thanks for your explanation. Now I am quite clear about that.
On Wed, Feb 11, 2009 at 12:05 PM, Thomas Lumley <tlum...@u.washington.edu>wrote:
> On Wed, 11 Feb 2009, Feng Li wrote:
>
> 1, Why both R and Matlab give 0*Inf==NaN? To my knowledge, it should be
>> zero
>> mathematically. Am I right?
>>
>
> No. 0*Inf is NaN according to the floating point arithmetic standards that
> R depends on.
>
> It's true that 0*x==0 for all finite x, but it's also true that x*Inf==Inf
> for all non-zero x, and you can't preserve both of these with 0*Inf.
>
> 2, I need to calculate e.g. exp(a)/(exp(b)+c), where both a and b are very
>> large numbers (>>1000, e.g a=1000, b=1007, and c=5). R gives me NaN when I
>> use the following command:
>>
>> exp(1000)/(exp(1007)+5)
>>>
>> [1] NaN
>>
>> I am pretty sure this should be close to zero.
>>
>
> No. It should be close to 1. Try 1000/(1000+5) for a simpler example.
>
> My question is whether there
>> is a general way to solve this kind of question or should I do some
>> settings
>> before computing?
>>
>>
> exp(1000)/(exp(1007)+5) is 1/(exp(7)+5*exp(-1000)), which is the same as
> 1/exp(7) to more than 400 digits accuracy.
>
> -thomas
>
> Thomas Lumley Assoc. Professor, Biostatistics
> tlum...@u.washington.edu University of Washington, Seattle
>
>
>
--
Feng Li
Department of Statistics
Stockholm University
106 91 Stockholm, Sweden
[[alternative HTML version deleted]]
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
