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.

In reply to Re^2: root function by psini
in thread root function by Anonymous Monk

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