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in reply to Re: Exponential Function Programming
in thread Exponential Function Programming

$ret - $d != $ret is interesting. One would think it's the same as $d (in boolean context), but it isn't. By testing the effect of $d ($ret - $d != $ret) instead of $d itself, the loop can be ended sooner. 
my $ret = 1;

for my $d (1e-15, 1e-20) {
   printf("%g %d %d\n",
      $d,
      ( $d              )?1:0,
      ( $ret-$d != $ret )?1:0,
   );
}

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1e-015 1 1   # Often equivalent.
1e-020 1 0   # But not when there's an underflow.

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Re^3: Exponential Function Programming
by pajout (Curate) on Nov 23, 2007 at 18:21 UTC
    Of course, underflow behaves friendly in this case. But my prime goal was to demonstrate no necessity for evaluating both factorial and power in each loop.
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      But such a demonstrations wasn't necessary for me since I had finished writing the following from scratch before you posted your solution. (I didn't post it since I didn't want to do the OP's homework for him.) 

      sub ikegami_exp {
   my ($x) = @_;
   my $last   = 0;
   my $result = 1;
   my $acc    = 1;
   for (my $n = 1; abs($result-$last)>0; $n++) {
      $last = $result;
      $acc *= $x/$n;
      $result += $acc;
   }    
   return $result;
}

for (1, 0.1, 10, 3.14, 0.1234, 100) {
   printf("%.16e: exp() = %.16e  ikegami_exp() = %.16e\n",
      $_,
      exp($_),
      ikegami_exp($_),
   );
}

      [download]

      The only difference was the terminating condition, so naturally, that's what interested me :) Mine goes into an infinite loop if you try to find ikegami_exp(1000), but that's easily fixed by replacing abs($result-$last)>0 with $result != $last.

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[d/l]
[select]
        Yep :>)
        I have decided to post it because my experience is that such tiny codes could initiate interest. Additionally, if somebody will present it as homework, there is a big risk that question "How it works" will appear...
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