In general, x^y is evaluated as exp(y*log(x)). In your case, x is negative, so log(x) is NaN. Note also that 1/3 is not represented exactly in your computer anyway, so you would not get an exact cube root this way; e.g.:
R> format((1234567891112^3)^(1/3),digits=16) [1] "1234567891112.001" (Probably a bad example, but you get the idea.) In general, sign(x)*(abs(x)^(1/3)) is the way to go for cube roots. David L. Reiner > -----Original Message----- > From: [EMAIL PROTECTED] [mailto:r-help- > [EMAIL PROTECTED] On Behalf Of > [EMAIL PROTECTED] > Sent: Tuesday, July 12, 2005 8:54 AM > To: Duncan Murdoch > Cc: r-help@stat.math.ethz.ch > Subject: Re: [R] R: to the power > > hi all > > i simply wanted to work with real numbers and thought that (-8)^(1/3) > should > work. > > sorry for not making the question clearer. > > / > allan > > Quoting Duncan Murdoch <[EMAIL PROTECTED]>: > > > On 7/12/2005 9:29 AM, Robin Hankin wrote: > > > Hi > > > > > > I find that one often needs to keep reals real and complexes complex. > > > > > > Try this: > > > > > > "cuberooti" <- > > > function (x) > > > { > > > if (is.complex(x)) { > > > return(sqrt(x + (0+0i))) > > > } > > > sign(x)* abs(x)^(1/3) > > > } > > > > > > > > > best wishes > > > > > > [see that (0+0i) sitting there!] > > > > I don't understand this. > > > > 1. I don't think you meant to use sqrt() there, did you?? > > > > 2. What effect does the 0+0i have? x has already been determined to be > > complex. > > > > Duncan Murdoch > > > > ______________________________________________ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! http://www.R-project.org/posting- > guide.html ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
