Re: HP 42s square/nth root Message #11 Posted by Karl Schneider on 1 Feb 2004, 3:41 p.m., in response to message #10 by Andrés C. Rodríguez (Argentina)
Andres, Robin 
Good, informative discussion. I think we're all on the same page, and have it right. I learned a couple things myself, regarding the enhancements of early units and why one of the complexvalued roots of (8)^(1/3) is considered the "primary" root. (The other root, of course, is the complex conjugate of the primary.)
Robin is right that taking the negatvieinteger root of the absolute value of the argument, then negating the answer, is the workaround. In fact, this is what I suspect is the basis of the algorithm on improved later HP's for integer exponents of negative numbers:
For x > 0.00 (floatingpoint) and n an integer,
(x)^n = exp[n*ln(x)]*[(1)^n] (change sign of result if n is odd)
Of course, this applies only if n is an integer.
Now, if y = 0 in y^x, x must be positive  Negative x yields divide by 0, and x = 0 yields undefined 0^0  READERS: Please don't launch that thread again!.
Makes one appreciate what a complete implementation of mathematical functionality entails, doesn't it?
Edited: 1 Feb 2004, 3:49 p.m.
