One of the best by-products of freedom and free markets is innovation.

If I can get a little geeky just for a second here. I was modeling Derb's August math problem and I asked my PC to run the following:

int count = (int) Integer.MAX_VALUE;

double mygamma = 1.0;

for ( int i = 1; i < count; i++ ) {

mygamma *= Math.cos(Math.PI / (i + 2));

}

That's 2,147,483,647 loops, each dividing two doubles, taking a cosine, and multiplying by another double.

No optimization, nothing. My decent but unexceptional office computer did this in less than five minutes. I cannot imagine the most expensive computers of ten years ago doing this in a day.

Oh, by the way, it's 0.11494204655022983 and it's of no help solving the problem, although it does validate that a limit exists. The first five digits are the same for count = 1000000.

Back to geeky politics, thanks for the disruption.

UPDATE: My point about innovation and free markets got lost, but c'est la vie.

Here's where I'm at. Each succusive circle has the same radius as the bisected side of the current polygon, creating a right trianngle as shown:

So we need the limit of the product (Gamma) of cos(pi/(n+2). I have tried without success to turn that into a series of 1 - (integral) dr but can't pull it off.

Comments

Of course it has a limit. An N-gon, as N approaches infinity, becomes a circle itself. Since the circle the N-gon becomes is real (while infinity is only a conceptual construct) it will have a finite size.

Posted by: johngalt at September 2, 2004 01:49 PMAgreed. It is algebraically and geometrically obvious that it exists.

I just can't figure out what the hell it is....

Posted by: jk at September 2, 2004 02:20 PMDamn, now you've got me working on it. Is there something to the fact that a polygon with an infinite number of sides equals a polygon with 1 side? I.e. an infinite N-gon equals a circle? Thus the limit of N to infinity is equivalent to N=1?

Posted by: Silence Dogood at September 3, 2004 10:15 AMNot quite. Your equation of an infinite number of sides with one side requires a redefinition of 'side' from linear to curved. This does seem like a fun mathematical problem because it clearly has a real solution. Maybe some day I'll have time to play with it (not that I'm certain my math skills are up to the task anyway.) Regardless, it would be fun to try.

Posted by: johngalt at September 3, 2004 11:28 AMSorry, I am geeking out - puzzles do that to me. I am there too, plus down several other paths all that end up in a circle so far. A couple of people I sent it to claim that it is a sum problem not a limit problem but they each came up with a different sum. Gauss proved that the only regular polygons that are constructable are those with N sides where N is a Fermat prime, the largest known equal to 65537, but this tidbit of info does not seem to be helping me.

Posted by: Silence Dogood at September 3, 2004 01:00 PMI'm think I'm convinced that the answer is 0. r=Rcos(pi/n) where r is the inscribed radius and R is the circumscribed radius and n is the number of sides. But r also = s/2*csc(pi/n) where s is the length of the side of the polygon. Both cos and csc limit to 1 as n -> infinity but s would limit to 0 as the included angle between the radii (= to pi/n) limits to 0. This would be a geometric point with 0 radius and 0 side length.

Posted by: Silence Dogood at September 3, 2004 03:01 PMSorry, man. I don't buy it. Yes, s goes to 0 as our theresa-tax-hike-a-gon goes to a circle. That doesn't mean the radius goes to 0.

Posted by: jk at September 3, 2004 03:13 PM