G'day Gregory,
On Tue, 26 Oct 2010 19:05:03 -0400
Gregory Ryslik <rsa...@comcast.net> wrote:
> Hi,
>
> This might be me missing something painfully obvious but why does the
> cube root of the following produce an NaN?
>
> > (-4)^(1/3)
> [1] NaN
1/3 is not exactly representable as a binary number. My guess is that
the number that is closest to 1/3 and representable cannot be used as
the exponent for negative numbers, hence the NaN.
Essentially, don't expect finite precision arithmetic to behave like
infinite precision arithmetic, it just doesn't. The resources
mentioned in FAQ 7.31 can probably shed more light on this issue.
Cheers,
Berwin
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