[R] (-8)^(1/3) == NaN?

[Rolf Turner]

On 20/07/2009, at 9:13 AM, jim holtman wrote:

> If the power that a number is being raised to is integer, then is does
> evaluate honoring the unary minus.
>
>> (-2) ^ 5  #integer power
> [1] -32
>> (-2) ^ 5.1
> [1] NaN

	<snip>

I was vaguely aware of this ... but it now triggers in my mind the
question of how the ^ function decides when the exponent is an integer.

A bit of experimentation seems to indicate that, e.g., (-2)^x ``works''
if (and only if?) round(x)==x returns TRUE.

Note that (-2)^x may NOT ``work'' in some cases were all.equal(x,round 
(x))
returns TRUE.

Young players should also be aware of the following trap.  It
can happen that n + epsilon ``is an integer'' according to my
rule, but m + epsilon is NOT an integer according to this rule.
Where m and n are both integers.

E.g.:

 > eps <- 0.4e-15
 > x <- 5+eps
 > x==round(x)
[1] TRUE
 > y <- 3+eps
 > y==round(y)
[1] FALSE

This is of course due to the exigencies of how n and m are represented
in floating point arithmetic.  Not too deep once you're aware of the
problem, but it can still be a ``gotcha'' if one is not alert.

(Be alert.  The world needs more lerts!)

	cheers,

		Rolf Turner

P. S.  Perhaps young players should be reminded at this point that  
is.integer() is
no help here.  This function tells you about the ***storage mode***  
of its argument.
Only.

		R. T.


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