Let $q_n(x)$ be the minimal polynomial for the area of the regular $n$-gon. What is $q_{11}(x)$?

The following are the simpler cases, and $x$ is given by $16 (area^2) $:

For the corresponding triangle the polynomial is $p_3(x) = x-3$, for the square it is $p_4(x) = x - 16$, for the pentagon it is $p_5(x) = x^2-50x+125$, for the hexagon it is $p_6(x) =x-108$.

What is e.g. $p_{11} (x)$? What is $p_n(x)$ in general? The question follows from a discussion on the Robbins' formula: http://youtu.be/OeZ6LsZHKcA

Robbins' formulas are meant to generalize Heron's and Brahmagrupta's formula to general cyclic n-gons (convex and concave alike). So far the formulas for n up to 9 have been discovered. A viewer of that video suggested we can do regular n-gons first. So for the pentagon case, if you substitute all five $distance^2$ with 1, then you will have the following factorized form: $(x^2-50x+125)(x-3)^5 = 0$. This is the same as $p_5(x)(p_3(x))^5 = 0$. This 5 is a result from combinatoric arguments on the degenerate cases: http://youtu.be/alFkaEZ4cZQ

So how do you generalize?

  • $\begingroup$ It would be good to link to whatever video you're referencing and state the formula you're referring to. It would also be good to state an explicit question rather than a general topic. $\endgroup$ – Milo Brandt Feb 16 '15 at 0:35
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    $\begingroup$ A remark concerning a calculation I did: the minimal polynomial of $\sin(\frac{2\pi}{5})$ is e.g $\frac{T_5(x)}{x}$. This is perhaps true für all odd natural numbers $n$. At least a connection to the theory of Prof Wildbergers spread polynomials user wilsonw prefers. $\endgroup$ – Wolfgang Tintemann Feb 16 '15 at 16:38
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    $\begingroup$ Another formula for the minimal polynomial of $\sin(\frac{2\pi}{n})$ for odd n is conjectured to be $$(x-i\sqrt{1-x^2})^{\frac{\phi(n)}{2}}\Phi_n(x+i\sqrt{1-x^2})$$ with the Euler $\phi$ function and the $n$-th cyclotomic polynomial $\Phi_n(x)$ $\endgroup$ – Wolfgang Tintemann Feb 16 '15 at 21:21
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    $\begingroup$ @Gerry, your formula is not the one for the n-gon in the unit circle I think. Your formula is the one for the inner radius equal to 1 according to the Wikipedia article on regular polygons. Please check. $\endgroup$ – Wolfgang Tintemann Feb 17 '15 at 12:31
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    $\begingroup$ @wolfgang, the formula is for the polygon with sides of length 1, which is what OP wants. $\endgroup$ – Gerry Myerson Feb 17 '15 at 22:09

For the area $A$ of regular $n$-gon with side-length $1$, we have $$A = \frac{n}{4} \cot\frac{\pi}{n} \qquad\to\qquad \cos^2\frac{\pi}{n}= \frac{x}{x+n^2}$$ where $x := 16 A^2$. The Chebyshev polynomial $T_n(\cdot)$ is such that $$T_n(\cos(\pi/n)) = \cos(\pi) = -1$$ Therefore, finding a (not-necessarily-minimal) polynomial satisfied by $x$ is a matter of eliminating $c := \cos(\pi/n)$ from the polynomial system $$\begin{align} (x+n^2)\;c^2 - x &= 0 \tag{1}\\[4pt] T_n(c)+1 &= 0 \tag{2} \end{align}$$ A computer algebra system can do this readily; Mathematica, for instance, has a Resultant[] command for this purpose. The general theory of resultants is a bit of computational overkill here, though. We can use $(1)$ to reduce $(2)$ to a linear equation in $c$, say $P c + Q = 0$; then we can write $c^2 = \frac{Q^2}{P^2}$, so that $P^2 x - Q^2 ( x + n^2 )$ gives the target polynomial. (Of course, unless/until a closed form is determined, the brute-force symbol manipulation remains something to be done with a CAS.)

These are the results of the brute-force operation ...

Factor[Resultant[ ChebyshevT[n,c] + 1, (x+n^2)c^2 - x, c ]]

... performed by Mathematica for various $n$ (with extraneous factors and powers left in for completeness): $$\begin{align} n = 3 &: \quad 81\;(x - 3)^2 \\[4pt] n = 4 &: \quad 4\;(x - 16)^4 \\[4pt] n = 5 &: \quad 625\;(x^2 - 50 x + 125)^2 \\[4pt] n = 6 &: \quad 4x^2\;(x-108)^4 \\[4pt] n = 7 &: \quad 2401\;(x^3 - 245 x^2 + 7203 x - 16807)^2 \\[4pt] n = 8 &: \quad 4\;(x^2 - 384 x + 4096)^4 \\[4pt] n = 9 &: \quad 6561(x-27)^2\;(x^3 - 729 x^2 + 72171 x -177147)^2 \\[4pt] n = 10 &: \quad 4 x^2\;(x^2 - 1000 x + 50000 )^4 \\[4pt] n = 11 &: \quad 14641\;\left(\; \begin{array}{c} x^5 - 1815 x^4 + 614922 x^3 \\ - 53146830 x^2 + 1071794405 x - 2357947691 \\ \end{array}\;\right)^2 \\[4pt] n = 12 &: \quad 4 (x - 144)^4\;(x^2 - 2016 x + 20736)^4 \\[4pt] n = 13 &: \quad 28561\;\left(\; \begin{array}{c} x^6 - 3718 x^5 + 2827539 x^4 - 637138788 x^3 \\ + 44865189655 x^2 - 827150951094 x + 1792160394037 \end{array} \;\right)^2 \end{align}$$

Edit. Since $n$ itself is part of the elimination process, it figures into the result(ant)s in ways that the above doesn't make clear. Here are the interesting factors, using explicit references to powers of $n$ among the coefficients.

$$\begin{align} n = 3 &: \quad x - n \\[4pt] n = 4 &: \quad x - n^2 \\[4pt] n = 5 &: \quad x^2 - 2 n^2 x + n^3 \\[4pt] n = 6 &: \quad x - 3 n^2 \\[4pt] n = 7 &: \quad x^3 - 5 n^2 x^2 - 3 n^4 x - n^5 \\[4pt] n = 8 &: \quad x^2 - 6 n^2 x + n^4 \\[4pt] n = 9 &: \quad x^3 - n^3 x^2 + 11 n^4 x - 3 n^5 \\[4pt] n = 10 &: \quad x^2 - n^3 x + 5 n^4 \\[4pt] n = 11 &: \quad x^5 - 15 n^2 x^4 + 42 n^4 x^3 - 30 n^6 x^2 + 5 n^8 x - n^9 \\[4pt] n = 12 &: \quad x^2 - 14 n^2 x + n^4 \\[4pt] n = 13 &: \quad x^6 - 22 n^2 x^5 + 99 n^4 x^4 - 132 n^6 x^3 + 55 n^8 x^2 - 6 n^{10} x + n^{11} \end{align}$$

It may be worth noting that $11$ is a common factor of most of the coefficients in the $n=13$ polynomial, and $3$ is a common factor in half of the coefficients in the $n=11$ polynomial.

It may also be worth noting that the degree of $n$ in the polynomials gives the sequence $$1, 2, 3, 2, 5, 4, 5, 4, 9, 4, 11, ...$$ which is an offset of OEIS's A216475: "The number of numbers coprime to and less than $n$, excluding 2." (The OEIS sequence begins at index $1$, so its reference to $n+2$ corresponds to our use of $n$ here.) The pattern continues through at least $n=25$ with minor fiddling in the cases of $n=15$, $21$, and $25$: in these cases, the power of $n$ is too big by $1$; we can "fix" that by writing, $n^{p+1}$ as $n\cdot n^p$ and treating the pulled-off $n$ as "just a number". (In the case of $n=25$, writing $n^{20}$ as $25n^{19}$ allows us to divide the polynomial by its leading $x$-coefficient, $5$, making it monic in $x$.)

By the way, the degree sequence for $x$ ... $$1, 1, 2, 1, 3, 2, 3, 2, 5, 2, 6, ...$$ ... appears as an offset of OEIS's A023022: the "half-totient" function, $\phi(n)/2$.

Edit 2. We can use this explicit formula for $T_n$ ... $$T_n(c) = c^n\;\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k}(1-c^{-2})^k$$ ... to give formulas for the polynomials above (the ones with the extra factors and powers; determining the minimal polynomial from these is an exercise for the reader).

When $n=2m$, there's no appreciable work, since $c$ appears in even powers throughout, and $(2)$ simply becomes (after clearing denominators) $$\left(x+n^2\right)^m + \sum_{k=0}^{m} \binom{n}{2k}(-1)^k x^{m-k} n^{2k}$$

When $n=2m+1$, there's an extra power of $c$, so our manipulations involve squaring to get $$\left(x+n^2\right)^{n} - x\;\left(\;\sum_{k=0}^{m} \binom{n}{2k}(-1)^k x^{m-k} n^{2k} \; \right)^2 $$

  • $\begingroup$ Please give a hint how you calculated the general formulas of the "interesting factors" in your "Edit" section above. $\endgroup$ – Wolfgang Tintemann Oct 4 '15 at 15:10
  • $\begingroup$ @WolfgangTintemann: At this point, I'm afraid I don't remember much about what I was thinking. :) $\endgroup$ – Blue Oct 4 '15 at 15:57

Edited and corrected

If I set $$\kappa_n(x)=(x-i\sqrt{1-x^2})^{\frac{\phi(n)}{2}}\Phi_n(x+i\sqrt{1-x^2})$$ and then calculate $$f_n(x)=\kappa_{4n}(\frac{x}{2n})\kappa_{4n}(-\frac{x}{2n})$$ then I get at least for odd primes $n$ $$f_n(x)=\frac{(-1)^{\frac{\phi(n)}{2}}}{n^{\phi(n)}}\prod_{k=1}^{\frac{\phi(n)}{2}}(x^2-(2n\sin(\frac{2\pi}{n}))^2)$$ So the 16*area^2 should be a root of $f(\sqrt{x})$. I have not yet done some CAS program checks though.

The correction is necessary as then $\frac{\phi(4n)}{2}=\phi(n)$ and so $f_n(x)$ is a polynomial then because of the degree correpondence of the cyclotomic polynomials. I will test this for e.g. n=5 asap.

Edit 1 The product formula for $f_n(x)$ above is for $n=p$ odd prime. For odd $n$ I set $E_n=\{k|1\le k<n,\ \text{k being coprime to $n$}\ (k,,)=1\}$ and then ( not yet fully proved ) $$f_n(x)=\frac{(-1)^{\frac{\phi(n)}{2}}}{n^{\phi(n)}}\prod_{k\in E_n}(x^2-(2n\sin(\frac{2\pi}{n}))^2)$$ Edit 2 I hope to have proved correctly that $\kappa_{4n}(x)$ is an even polynomial. So it results in $$f_n(x)=\kappa_{4n}^2(\frac{x}{2n})$$ being a square of the irreducible polynomial.

Edit 3 Please excuse that I work contrary to the problem posted on the case with the n-gon lying on the unit circle. Just now I dont know whether there are connections to the case of side length 1. I hope to be able to give further results on my work if available.

  • $\begingroup$ I checked the case n=5 with Maple. Its OK. I got $$f_5(x)=25 - 2 x^2 + 7/125 x^4 - 2/3125 x^6 + 1/390625 x^8$$ which is factorized to $$1/390625 (3125 - 125 x^2 + x^4 )^2$$ One root y=x^2 is $\frac{25}{2}(5+\sqrt{5})$ which according to formulas in Wikepedia for outer radius 1 is exactly the 16*(area)^2. qed. $\endgroup$ – Wolfgang Tintemann Feb 17 '15 at 13:28
  • $\begingroup$ May I know who made this conjecture and how did he/she come up with this formula? $\endgroup$ – wilsonw Feb 17 '15 at 14:03
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    $\begingroup$ I did and I am working on the general case for even n too. But this will be much harder. The original idea goes back to the work of Mr. Heierman who defined some valuable polynomials in his article on his Website. I just hacked the case n=17 : quite correcly in the set of roots of $$827240261886336764177 - 34349076618117789516 x + 415992277382049354 x^2 - 2261945156303778 x^3 + 6522333207335 x^4 - 10668805498 x^5 + 9938999 x^6 - 4913 x^7 + x^8$$ appears the value for the quadrea $$150.8528458464590309956191688936093948707676158087195160151694...$$ Mathematica doesnt show radicals. $\endgroup$ – Wolfgang Tintemann Feb 17 '15 at 14:11
  • $\begingroup$ Would you mind posting the link to the website so I could have an idea why this formula works? $\endgroup$ – wilsonw Feb 19 '15 at 14:08
  • $\begingroup$ I suppose you want the paper of William E Heierman. But be prepared that it is not immediately clear how I came to the formula. It took me a while to analyze his work. Here 2 links ( one in French ) 1. fr.wikipedia.org/wiki/… 2. corunduminium.com/Trigpolys.html Hope it helps to understand how it fits all together. $\endgroup$ – Wolfgang Tintemann Feb 19 '15 at 17:19

Meanwhile I also did work on the case of the side of the polygonal being equal to 1. My results agree with the ones given by user Blue. I use the polynomial given as


Then one gets the expression/"interesting factors" of user Blue when one calculates

$$n^{\phi(n)}p_n(\frac xn)$$

where $\phi(n)$ is the Euler totient function and $\Phi_n(x)$ the n-th cyclotomic polynomial.


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