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If you have log available, you can use Newton method to easyly find the solution to the equation f(z) = log(z) - y * log x = 0 :
#!/usr/bin/perl use strict; use warnings; my ($X, $Y, $n) = @ARGV; $n ||= 20; my $Z = $X ** $Y; # f(z) = log z - y * log x = 0; # f'(z) = f1(z) = 1/z my $ylogx = $Y * log $X; my $z = 1; for my $i (1..$n) { my $fz = log($z) - $ylogx; my $f1z = 1 / $z; $z = $z - $fz / $f1z; my $e = abs($z - $Z); printf "%3d: z=%8f, e=%8f\n", $i, $z, $e; }
This method does not handle correctly values of y < 0 but this is not a problem because z(y) = 1 / z(-y).

Also, note that even if the algorithm converges in very few iterations, it uses log that is usually an expensive operation and so, other methods requiring more iterations may be actually faster.

In reply to Re: [OT] Function to do x**y for |y|<1 w/o ** operator. by salva
in thread [OT] Function to do x**y for |y|<1 w/o ** operator. by swampyankee

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