Re^2: root function

Yes it does. (-3)**(1/3) results in nan; But so does the log approach, too.

The odd roots of negative numbers must be treated separately as they are limit cases.

I'd try something like:

if ($base>0) {
  return $base**(1/$root);
} elsif ($base==0) {
  return 0; ## or do manage the special case 0**0;
} elsif ($root)==int($root) && $root%2==1) {
  return -((-$base)**($root));
} else {
  return 'nan';
}
[download]

Not tested, so probably broken :)

Careful with that hash Eugene.

Comment on Re^2: root function Download Code

Replies are listed 'Best First'.
Re^3: root function by swampyankee (Parson) on Jun 21, 2008 at 21:40 UTC
The logarithms of negative numbers are defined; it's just that they require complex numbers to represent. It's not my fault that Perl (and C) can't cope with them ;-). All will become clear when one remembers that e^πi = -1 Information about American English usage here and here. Floating point issues? Please read this before posting. — emc	[reply]
Re^4: root function by psini (Deacon) on Jun 21, 2008 at 21:47 UTC
++ But I said that would not work, not that it is wrong. BTW, my math teacher preferred to write e^πi+1=0; she said that that form contains all the five really important numbers! Update: Well, she was a good teacher, now I realize she made that remark 31 years ago... and I still remember it Careful with that hash Eugene.	[reply]
Re^5: root function by swampyankee (Parson) on Jun 21, 2008 at 23:39 UTC
I spoke too soon! There is, of course, a module for complex arithmetic. Information about American English usage here and here. Floating point issues? Please read this before posting. — emc	[reply]
Re^4: root function by ysth (Canon) on Jun 22, 2008 at 06:53 UTC
Not complex numbers, series of complex numbers. e^(2n+1)πi = -1, for any integer n. Which makes them undefined, for at least some purposes. -- Online Fortune Cookie Search	[reply]
Re^5: root function by psini (Deacon) on Jun 22, 2008 at 07:14 UTC
Following your interpretation of "undefined", sqrt(2) should be undefined too (two solution), or arctan(x) for any x in the interval `[-1,1]` (infinite periodic). Careful with that hash Eugene.	[reply] [d/l]