in reply to Stirling Approx to N! for large Number?
There was a more accurate formula than Stirling's in tilly's post. Wolfram's Mathematica attributes it to Gosper:
sub factorial_gosper {
# sqrt(2*n*PI + 1/3)*n**n/exp(n)
my $n = shift;
return exp($n*log($n)  $n + 0.5*log(2*$n*PI + 1.0/3.0));
}
This gives better accuracy for small numbers. For 8!, I get 40034.48, compared to the correct 40320.
There are several problems with your implemenation of Stirling's formula. For one thing, log(e^(n)), which you give as e*log(n). Since the log is just the exponent base e, the correct answer is n.
Tilly's formula is closer to being right; there was only one parenthesis mistake. I corrected that and tried them all in a modified version of your code:
#!/usr/bin/perl w
use strict;
+
my $h =8;
my $pure = factorial_pure($h);
my $stirling = factorial_stirling($h);
my $real_stirling = real_stirling($h);
my $gosper = factorial_gosper($h);
+
print "PURE : $pure\n";
print "STIRLING : $stirling\n";
print "REAL : $real_stirling\n";
print "GOSPER : $gosper\n";
+
#Subroutines
+
sub factorial_stirling{
my $n = shift;
use constant PI => 4*atan2 (1,1);
use constant e => exp(1);
+
my $log_nfact =
$n * log ($n)  e* log($n) + 0.5 * (log(2*PI)+log($n));
+
# Below is Tilly's suggestion, with less accuracy
# $n * log ($n)  ($n) + (0.5 * (log(2+ log(PI))+log($n)));
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return exp($log_nfact);
}
+
+
sub factorial_pure {
+
my ($n,$res) = (shift,1);
return undef unless $n>=0 and $n == int($n);
$res *= $n while $n>1;
return $res;
+
}
+
sub real_stirling {
my $n = shift;
my $log_nfact = $n*log($n)  $n + 0.5*(log(2) + log(PI)+ log($n));
return exp($log_nfact);
}
+
sub factorial_gosper {
# sqrt(2*n*PI + 1/3)*n**n/exp(n)
my $n = shift;
return exp($n*log($n)  $n + 0.5*log(2*$n*PI + 1.0/3.0));
}
+
I get the answers below:
PURE : 40320
STIRLING : 417351.253768542
REAL : 39902.3954526567
GOSPER : 40034.4823215513
