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, ); }
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.

      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($_), ); }

      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.

        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...