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        riddles >> medium >> Circumscribing Circles
        
(Message started by: ThudanBlunder on Nov 26^th, 2008, 8:51am)

Title: Circumscribing Circles
Post by ThudanBlunder on Nov 26^th, 2008, 8:51am

Draw a circle of radius 1 and circumscribe it with an equilateral triangle. Now draw the circumscribing circle of the triangle and then circumscribe this circle with a square. Continue in this fashion, drawing a circumscribing n-gon. then its circumscribing circle, and then the (n+1)-gon which circumscribes the circle.

Do the circumscribing circles increase without limit?

Title: Re: Circumscribing Circles
Post by towr on Nov 26^th, 2008, 9:20am

[hide]The size of the circles, prod_k=3..inf 1/cos(pi/(2k)), seems to converge, but I don't know how to make sure.[/hide]

Title: Re: Circumscribing Circles
Post by ThudanBlunder on Nov 28^th, 2008, 10:56am

on 11/26/08 at 09:20:54, towr wrote:

[hide]The size of the circles, prod_k=3..inf 1/cos(pi/(2k)), seems to converge, but I don't know how to make sure.[/hide]

Yes, it converges, even with 2k replaced by k. :P

Can you give an answer to a few decimal places?

Title: Re: Circumscribing Circles
Post by towr on Nov 28^th, 2008, 11:50am

on 11/28/08 at 10:56:51, ThudanBlunder wrote:

Yes, it converges, even with 2k replaced by k. :P

Ah, yes, I meant 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/(2k); I was supposed to be dividing the whole circle in 2k angles, not just half.

Quote:

Can you give an answer to a few decimal places?

Seems to be [hide]8.700[/hide] Of course the precision leaves to be desired, since any small rounding errors at the machine-word level explode in multiplications.

Title: Re: Circumscribing Circles
Post by towr on Nov 28^th, 2008, 11:56am

http://mathworld.wolfram.com/PolygonCircumscribing.html

Title: Re: Circumscribing Circles
Post by Eigenray on Nov 28^th, 2008, 11:56am

[hideb]log[1/cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/n)] = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif²/(2n²) + O(1/n⁴),
so the sum coverges.

Moreover, for any finite N, we can find both
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/prod.gif_{n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif N} 1/cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/n) and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif_n>N http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif²/(2n²)
in closed form. Taking N=100 gives 8.70001. But if we take the first 4 terms in the Taylor series, then N=100 gives 8.7000366252081943, accurate to 15 digits.
r = 8.7000366252081945032224098591130...[/hideb]
Edit: Ah, this is more or less equivalent to formula 7 at Mathworld I guess, once one knows the [link=http://www.research.att.com/~njas/sequences/A000182]Taylor series of log sec[/link].