Re: [R] (-8)^(1/3) == NaN?
On Sun, 19 Jul 2009, 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 Yes. 3 is representable exactly as a whole number, so (-2)^3 exists, but (1/3) is represented as a fraction whose denominator is 2^54, an even number, so (-8)^(1/3) does not exist (as a real number). More generally, since all floating point numbers are represented as fractions whose denominator is a power of 2, the only way a floating point number can be a legitimate exponent for a negative base is if it represents a whole number. -thomas -8^(1/3) is parsed as -(8^(1/3)) according to operator precedence. On Sun, Jul 19, 2009 at 4:49 PM, Liviu Androniclandronim...@gmail.com wrote: On Sun, Jul 19, 2009 at 12:28 AM, jim holtmanjholt...@gmail.com wrote: First of all, read FAQ 7.31 to understand that 1/3 is not representable in floating point. Also a^b is actually exp(log(a) * b) and log(-8) is not valid (NaN). If this is so, why would the following evaluate as expected? (-8)^(3) [1] -512 Liviu -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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. Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ 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.
Re: [R] (-8)^(1/3) == NaN?
It' true, but if you type -8^(1/3) returns -2, and if you type -8^1/3 it returns -2.6, maybe there are some rules about parenthesis... regards Víctor De: r-help-boun...@r-project.org en nombre de Dave DeBarr Enviado el: sáb 18/07/2009 05:04 Para: r-help@r-project.org Asunto: [R] (-8)^(1/3) == NaN? Why does the expression (-8)^(1/3) return NaN, instead of -2? This is not answered by http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-powers-of-negative-numbers-wrong_003f Thanks, Dave [[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. __ 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.
Re: [R] (-8)^(1/3) == NaN?
It is the order of operator precedence. Look at what you typed: -8^1/3 is parsed as -(8^1)/3 = -2.6 On Sun, Jul 19, 2009 at 1:12 PM, Victor Manuel Garcia Guerrerovmgar...@colmex.mx wrote: It' true, but if you type -8^(1/3) returns -2, and if you type -8^1/3 it returns -2.6, maybe there are some rules about parenthesis... regards Víctor De: r-help-boun...@r-project.org en nombre de Dave DeBarr Enviado el: sáb 18/07/2009 05:04 Para: r-help@r-project.org Asunto: [R] (-8)^(1/3) == NaN? Why does the expression (-8)^(1/3) return NaN, instead of -2? This is not answered by http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-powers-of-negative-numbers-wrong_003f Thanks, Dave [[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. __ 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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.
Re: [R] (-8)^(1/3) == NaN?
On Sun, Jul 19, 2009 at 7:33 PM, jim holtmanjholt...@gmail.com wrote: -8^1/3 is parsed as -(8^1)/3 = -2.6 However the following is evaluated as one would expect: 8^(1/3) [1] 2 -8^(1/3) [1] -2 Perhaps it is parsed in this way: -(8^(1/3)) [1] -2 Liviu __ 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.
Re: [R] (-8)^(1/3) == NaN?
On Sun, Jul 19, 2009 at 12:28 AM, jim holtmanjholt...@gmail.com wrote: First of all, read FAQ 7.31 to understand that 1/3 is not representable in floating point. Also a^b is actually exp(log(a) * b) and log(-8) is not valid (NaN). If this is so, why would the following evaluate as expected? (-8)^(3) [1] -512 Liviu __ 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.
Re: [R] (-8)^(1/3) == NaN?
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 -8^(1/3) is parsed as -(8^(1/3)) according to operator precedence. On Sun, Jul 19, 2009 at 4:49 PM, Liviu Androniclandronim...@gmail.com wrote: On Sun, Jul 19, 2009 at 12:28 AM, jim holtmanjholt...@gmail.com wrote: First of all, read FAQ 7.31 to understand that 1/3 is not representable in floating point. Also a^b is actually exp(log(a) * b) and log(-8) is not valid (NaN). If this is so, why would the following evaluate as expected? (-8)^(3) [1] -512 Liviu -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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.
Re: [R] (-8)^(1/3) == NaN?
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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Re: [R] (-8)^(1/3) == NaN?
It also works for raising a number to a negative integer: (-3)^(-3) [1] -0.03703704 On Sun, Jul 19, 2009 at 6:23 PM, Rolf Turnerr.tur...@auckland.ac.nz wrote: 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. ## Attention:This e-mail message is privileged and confidential. If you are not theintended recipient please delete the message and notify the sender.Any views or opinions presented are solely those of the author. This e-mail has been scanned and cleared by MailMarshalwww.marshalsoftware.com ## -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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.
[R] (-8)^(1/3) == NaN?
Why does the expression (-8)^(1/3) return NaN, instead of -2? This is not answered by http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-powers-of-negative-numbers-wrong_003f Thanks, Dave [[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.
Re: [R] (-8)^(1/3) == NaN?
First of all, read FAQ 7.31 to understand that 1/3 is not representable in floating point. Also a^b is actually exp(log(a) * b) and log(-8) is not valid (NaN). You expression is not really taking the cube root; it is taking values to the 1/3 power. If you want to cube root function, then try: cubeRoot - function(x) sign(x) * exp(log(abs(x)) / 3) cubeRoot(8) [1] 2 cubeRoot(-8) [1] -2 On Sat, Jul 18, 2009 at 6:04 PM, Dave DeBarrdave.deb...@microsoft.com wrote: Why does the expression (-8)^(1/3) return NaN, instead of -2? This is not answered by http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-powers-of-negative-numbers-wrong_003f Thanks, Dave [[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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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.
Re: [R] (-8)^(1/3) == NaN?
The correct mathematical answer is really one (or perhaps all three?) of three complex numbers that are the solutions to x^3+8=0. Here is one of the others: as.complex(-8)^(1/3) [1] 1+1.732051i I suspect there is a reason why R is willing to produce this particular solution and not the other two after being told to use the complex range, but I don't know the reason. You can get all three with: polyroot(c(8,0,0,1)) [1] 1+1.732051i -2+0.00i 1-1.732051i -- David. On Jul 18, 2009, at 6:04 PM, Dave DeBarr wrote: Why does the expression (-8)^(1/3) return NaN, instead of -2? This is not answered by http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-powers-of-negative-numbers-wrong_003f Thanks, Dave David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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.
Re: [R] (-8)^(1/3) == NaN?
On 18-Jul-09 22:04:57, Dave DeBarr wrote: Why does the expression (-8)^(1/3) return NaN, instead of -2? This is not answered by http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-powers-of-negative- numbers-wrong_003f Thanks, Dave Because R does not try to evaluate (-8)^(1/3), but (-8)^x, where x is a very close approximation to 1/3 but is not exactly 1/3 (which is impossible in a finite binary representation). But even if it could exactly represent 1/3, R would still need to have a special look-up for certain fractional powers (1/3, 1/5, ... ) to enable it to recognise that these are odd-integer-roots of negatgive numbers, and therefore can be evaulated as -(nth_root(abs(x))). It doesn't help, either, to try to do it in complex numbers, since (-8) will then be seen as 8*exp(i*pi) whose cube root will be found as 2*exp(i*pi/3) = 2*(cos(pi/3) + i*sin(pi/3)) = 2*(1/2 + i*sqrt(3)/2): (complex(1,-8,0)) # [1] -8+0i complex(1,-8,0)^(1/3) # [1] 1+1.732051i (8*exp(complex(1,0,pi)))^(1/3) # [1] 1+1.732051i sqrt(3) # [1] 1.732051 I'm not sure what best to suggest for your situation. Basically, if it is in a context where it can only be (negative number)^(1/(odd integer)) then you are better off modifying the logic of your program so as to ensure the result you want. Hoping this helps, Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 18-Jul-09 Time: 23:54:11 -- XFMail -- __ 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.
