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Re^2: Exponential Function Programming

by ikegami (Patriarch)
 on Nov 22, 2007 at 20:21 UTC ( #652444=note: print w/replies, xml ) Need Help??

in reply to Re: 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...

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