In this particular case it is part of the C99 stanadrd (7.12.7.4) for
the 'pow' function R uses.
On Wed, 27 Oct 2010, Berwin A Turlach wrote:
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
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
As we can see:
(-1.587401)^3
[1] -4
Thanks!
Greg
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R-help@r-project.org mailing list
Look at this:
x - as.complex(-4)
x
[1] -4+0i
x^(1/3)
[1] 0.793701+1.37473i
(-4)^(1/3)
[1] NaN
It seems that R gives you the principal root, which is complex, and
not the real root.
Kjetil
On Tue, Oct 26, 2010 at 8:05 PM, Gregory Ryslik rsa...@comcast.net wrote:
Hi,
This might be me
hmm interesting. When I did -4^(1/3) got the correct answer, but then again
that's because it processes the negative later. i.e. -4^(1/2) gave me -2
instead of the 2i I expected. Also when I did (-4+0i)^(1/3) it gave me
0.793701+1.37473i. Possible bug?
Sachin
--- Please consider the environment
)
[1] -2
which comes as rather a surprise!
Bill.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Kjetil Halvorsen
Sent: Wednesday, 27 October 2010 9:17 AM
To: Gregory Ryslik
Cc: r-help Help
Subject: Re: [R] cube root of a negative
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
which comes as rather a surprise!
Bill.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Kjetil Halvorsen
Sent: Wednesday, 27 October 2010 9:17 AM
To: Gregory Ryslik
Cc: r-help Help
Subject: Re: [R] cube root of a negative number
Look
Because it is implemented as
antilog((1/3)*log(-4))
most likely using base 2 for the log/antilog functions.
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
As
G'day Bill,
On Wed, 27 Oct 2010 10:34:27 +1100
bill.venab...@csiro.au wrote:
[...]
It is no surprise that this does not work when working in the real
domain, except by fluke with something like
-4^(1/3)
[1] -1.587401
where the precedence of the operators is not what you might
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