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Polygon Circumscribing Constant

\begin{figure}\begin{center}\BoxedEPSF{PolygonCircumscribe.epsf}\end{center}\end{figure}

If a Triangle is Circumscribed about a Circle, another Circle around the Triangle, a Square outside the Circle, another Circle outside the Square, and so on. From Polygons, the Circumradius and Inradius for an $n$-gon are

$\displaystyle R$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}s\csc\left({\pi\over n}\right)$ (1)
$\displaystyle r$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}s\cot\left({\pi\over n}\right),$ (2)

where $s$ is the side length. Therefore,
\begin{displaymath}
{R\over r}={1\over \cos\left({\pi\over n}\right)} = \sec\left({\pi\over n}\right),
\end{displaymath} (3)

and an infinitely nested set of circumscribed polygons and circles has
\begin{displaymath}
K\equiv {r_{\rm final\ circle}\over r_{\rm initial\ circle}}...
...c\left({\pi\over 4}\right)\sec\left({\pi\over 5}\right)\cdots.
\end{displaymath} (4)

Kasner and Newman (1989) and Haber (1964) state that $K=12$, but this is incorrect. Write
\begin{displaymath}
K= \prod_{n=3}^\infty {1\over\cos\left({\pi\over n}\right)}
\end{displaymath} (5)


\begin{displaymath}
\ln K =-\sum_{n=3}^\infty \ln(\cos x).
\end{displaymath} (6)

Define
\begin{displaymath}
y_0(x) \equiv -\ln(\cos x)={\textstyle{1\over 2}}x^2+{\texts...
...xtstyle{1\over 45}} x^6+{\textstyle{17\over 2520}} x^8+\ldots.
\end{displaymath} (7)

Now define
\begin{displaymath}
y_1(x)={\textstyle{1\over 2}}a x^2,
\end{displaymath} (8)

with
\begin{displaymath}
y_1({\textstyle{\pi\over 3}})=y_0({\textstyle{\pi\over 3}})
\end{displaymath} (9)


\begin{displaymath}
{\textstyle{1\over 2}}a\left({\pi\over 3}\right)^2 =\ln 2,
\end{displaymath} (10)

so
\begin{displaymath}
a=2\left({3\over \pi}\right)^2\ln 2,
\end{displaymath} (11)

and
\begin{displaymath}
y_2(x)={9\ln 2\over \pi^2} x^2.
\end{displaymath} (12)

But $y_2(x)>y_1(x)$ for $x\in (0,\pi/3)$, so
\begin{displaymath}
\sum_{n=3}^\infty y_2\left({\pi\over n}\right)> -\sum_{n=3}^\infty \ln\left[{\cos\left({\pi\over n}\right)}\right]
\end{displaymath} (13)


$\displaystyle \ln K$ $\textstyle <$ $\displaystyle \sum_{n=3}^\infty y_2\left({\pi\over n}\right){9\ln 2\over \pi^2}...
...n=3}^\infty \left({\pi\over n}\right)^2 = 9\ln 2 \sum_{n=3}^\infty {1\over n^2}$  
  $\textstyle =$ $\displaystyle 9\ln 2\left({\sum_{n=1}^\infty {1\over n^2}-\sum_{n=1}^2 {1\over n^2}}\right)= 9\ln 2[\zeta(2)-{\textstyle{5\over 4}}]$  
  $\textstyle =$ $\displaystyle 9\ln 2\left({{\pi^2\over 6}-{5\over 4}}\right)= 2.4637$ (14)


\begin{displaymath}
K < e^{2.4637} =11.75.
\end{displaymath} (15)

If the next term is included,
\begin{displaymath}
y_2(x)=a({\textstyle{1\over 2}}x^2+{\textstyle{1\over 12}} x^4).
\end{displaymath} (16)

As before,
\begin{displaymath}
y_2({\textstyle{\pi\over 3}})=y_0({\textstyle{\pi\over 3}})
\end{displaymath} (17)


\begin{displaymath}
a={972\ln 2\over \pi^2(54+\pi^2)},
\end{displaymath} (18)

so
\begin{displaymath}
y_2(x) = {972\ln 2\over \pi^2(54+\pi^2)}({\textstyle{1\over 2}}x^2+{\textstyle{1\over 12}} x^4)
\end{displaymath} (19)


$\displaystyle \ln K$ $\textstyle <$ $\displaystyle {972\ln 2\over \pi^2(54+\pi^2)} \sum_{n=3}^\infty \left[{{1\over 2}\left({\pi\over n}\right)^2 +{1\over 12}\left({\pi\over n}\right)^4}\right]$  
  $\textstyle =$ $\displaystyle {972\ln 2\over \pi^2(54+\pi^2)}\left\{{{1\over 2}\left[{\zeta(2)-{5\over 4}}\right]
+{\pi^2\over 12}\left[{\zeta(4)-1-{1\over 2^4}}\right]}\right\}$  
  $\textstyle =$ $\displaystyle {972\ln 2\over \pi^2(54+\pi^2)}\left[{{1\over 2}\left({{\pi^2\ove...
...}}\right)
+{\pi^2\over 12}\left({{\pi^4\over 90}-1-{1\over 2^4}}\right)}\right]$  
  $\textstyle =$ $\displaystyle {9(8\pi^6-45\pi^2-5400)\ln 2\over 80(\pi^2+54)} = 2.255,$ (20)

and
\begin{displaymath}
K<e^{2.255} = 9.535.
\end{displaymath} (21)

The process can be automated using computer algebra, and the first few bounds are 11.7485, 9.53528, 8.98034, 8.8016, 8.73832, 8.71483, 8.70585, 8.70235, 8.70097, and 8.70042. In order to obtain this accuracy by direct multiplication of the terms, more than 10,000 terms are needed. The limit is
\begin{displaymath}
K=8.700036625\ldots.
\end{displaymath} (22)


Bouwkamp (1965) produced the following Infinite Product formulas

$\displaystyle K$ $\textstyle =$ $\displaystyle {2\over\pi}\prod_{m=1}^\infty\prod_{n=1}^\infty \left[{1-{1\over m^2(n+{\textstyle{1\over 2}})^2}}\right]$ (23)
  $\textstyle =$ $\displaystyle 6\mathop{\rm exp}\nolimits \left\{{\sum_{k=1}^\infty {[\lambda(2k)-1]2^{2k}[\zeta(2k)-1-2^{-2k}]\over k}}\right\},$  
      (24)

where $\zeta(x)$ is the Riemann Zeta Function and $\lambda(x)$ is the Dirichlet Lambda Function. Bouwkamp (1965) also produced the formula with accelerated convergence
$K = {\textstyle{1\over 12}}\sqrt{6}\pi^4 (1-{\textstyle{1\over 2}}\pi^2+{\texts...
...{1\over 24}}\pi^4)(1-{\textstyle{1\over 8}}\pi^2+{\textstyle{1\over 384}}\pi^4)$
$ \times \csc\left({\pi^2\over\sqrt{6+2\sqrt{3}}}\right)\csc\left({\pi^2\over\sqrt{6-2\sqrt{3}}}\right)B,\quad$ (25)
where
\begin{displaymath}
B\equiv \prod_{n=3}^\infty \left({1-{\pi^2\over 2n^2}+{\pi^4\over 24n^4}}\right)\sec\left({\pi\over n}\right)
\end{displaymath} (26)

(cited in Pickover 1995).

See also Polygon Inscribing Constant


References

Bouwkamp, C. ``An Infinite Product.'' Indag. Math. 27, 40-46, 1965.

Finch, S. ``Favorite Mathematical Constants.'' http://www.mathsoft.com/asolve/constant/infprd/infprd.html

Haber, H. ``Das Mathematische Kabinett.'' Bild der Wissenschaft 2, 73, Apr. 1964.

Kasner, E. and Newman, J. R. Mathematics and the Imagination. Redmond, WA: Microsoft Press, pp. 311-312, 1989.

Pappas, T. ``Infinity & Limits.'' The Joy of Mathematics. San Carlos, CA: Wide World Publ./Tetra, p. 180, 1989.

Pickover, C. A. ``Infinitely Exploding Circles.'' Ch. 18 in Keys to Infinity. New York: W. H. Freeman, pp. 147-151, 1995.

Pinkham, R. S. ``Mathematics and Modern Technology.'' Amer. Math. Monthly 103, 539-545, 1996.

Plouffe, S. ``Product(cos(Pi/n),n=3..infinity).'' http://www.lacim.uqam.ca/piDATA/productcos.txt.



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© 1996-9 Eric W. Weisstein
1999-05-25