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We know $\cos(2\pi/p)$ for p a Fermat prime can be expressed in real radicals. The case $p=17$ is a root of an 8th deg eqn, but can be also given as a sequence of nested radicals,

$$\begin{aligned} 4\cos(2\pi/17)&=\frac{1}{x}+\sqrt{x}\,(17+4\sqrt{17})^{1/4}=3.72988\dots\\ x &=\frac{1}{2}\Big(y+\sqrt{y^2+4}\Big)\\ y &=\frac{1}{2}(1-\sqrt{17}) \end{aligned}$$

Question: For curiosity's sake, is there a way to explicitly write down $\cos(2\pi/257)$ as a tower of nested real radicals such that it can fit in an MSE post?

P.S. John Conway described a procedure for $p = 65537$ in this old Mathforum post, so maybe it can be adapted to the "easier" $p = 257$.

Edit (a day later): I managed to answer my own question. See below.

Glorfindel
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Tito Piezas III
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  • It can actually be expressed in terms of square roots, not just radicals. Not sure about the specific expression. though :) – Thomas Andrews Oct 06 '13 at 02:30
  • Maybe Christian Gottlieb, The Simple and Straightforward Construction of the Regular 257-gon, Mathematical Intelligencer 21 (1999) 31–37 would be helpful. – Gerry Myerson Oct 06 '13 at 04:29
  • Springer generously offers the article at only $39.95. :) – Tito Piezas III Oct 06 '13 at 04:36
  • I spoke too soon. The one at http://www.susqu.edu/brakke/constructions/big-gon.htm is actually easy to understand. Some copy-and-paste with Mathematica using Ken Brakke's expressions gives the correct value of $x=2\cos(64\pi/257) = 1.418528\dots$. Finally saw $x$ expressed as a real radical. :) – Tito Piezas III Oct 06 '13 at 05:48
  • The {257} is a class-256 polygon of the simple kind, so somewhere there is a nested set of eight surds. – wendy.krieger Oct 06 '13 at 09:21
  • @ Tito Piezas:$\tan(\frac{64\pi}{257})$ is expressible by my conjectured continued fraction for $\tan(\frac{k\pi}{4k+2})$ for the special case $k=128$,found in this post. – Nicco Sep 22 '15 at 08:40
  • @Nicco: I looked at that post. Very nice. – Tito Piezas III Sep 22 '15 at 12:56

4 Answers4

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Finally! Persistence pays off. This is based on Ken Brakke’s solution. However, I managed to simplify it a bit like using only 6 numbers of deg-32 (the $v_i$ below), while Brakke’s used 7 ($a_{39}, a_{40}, a_{46}, a_{47}, a_{48}, a_{54}, a_{55}$ in his site), and also simplifying the $u_i$ and $x_i$. Thus,

$$\begin{aligned} w_1&=4\cos\Big(\frac{4\,\pi}{257}\Big) = u_1+\sqrt{u_1^2-4u_2} = 3.995219\dots\\ w_2&=4\cos\Big(\frac{16\,\pi}{257}\Big) = u_3+\sqrt{u_3^2-4u_4} = 3.923736\dots\\ w_2&=4\cos\Big(\frac{64\,\pi}{257}\Big) = u_1-\sqrt{u_1^2-4u_2} = 2.837057\dots\\ w_4&=4\cos\Big(\frac{256\,\pi}{257}\Big) = u_3-\sqrt{u_3^2-4u_4} = -3.999701\dots\\ \text{where,}\\ 2u_1,\,2u_3&=v_1\pm\sqrt{v_1^2-4(v_2+v_8)}\\ 2u_2,\,2u_4&=v_9\pm\sqrt{v_9^2-4(v_{10}+v_0)}\\ \text{and,}\\ 2v_0&=x_0-\sqrt{x_0^2-4 (x_0 + x_1 + x_2 + x_5)}\\ 2v_1&=x_1-\sqrt{x_1^2-4 (x_1 + x_2 + x_3 + x_6)}\\ 2v_2&=x_2-\sqrt{x_2^2-4 (x_2 + x_3 + x_4 + x_7)}\\ 2v_8&=x_8-\sqrt{x_8^2-4 (x_8 + x_9 + x_{10} + x_{13})}\\ 2v_9&=x_9-\sqrt{x_9^2-4 (x_9 + x_{10} + x_{11} + x_{14})}\\ 2v_{10}&=x_{10}-\sqrt{x_{10}^2-4 (x_{10} + x_{11} + x_{12}+x_{15})}\\ \text{and,}\\ 2x_1,\,2x_{9}&=y_1\pm\sqrt{y_1^2-4(t_1 + y_1 + y_3 + 2 y_6)}\\ 2x_2,\,2x_{10}&=y_2\pm\sqrt{y_2^2-4(t_2 + y_2 + y_4 + 2 y_7)}\\ 2x_3,\,2x_{11}&=y_3\pm\sqrt{y_3^2-4(t_1 + y_3 + y_5 + 2 y_8)}\\ 2x_4,\,2x_{12}&=y_4\pm\sqrt{y_4^2-4(t_2 + y_4 + y_6 + 2 y_1)}\\ 2x_5,\,2x_{13}&=y_5\pm\sqrt{y_5^2-4(t_1 + y_5 + y_7 + 2 y_2)}\\ 2x_6,\,2x_{14}&=y_6\pm\sqrt{y_6^2-4(t_2 + y_6 + y_8 + 2 y_3)}\\ 2x_7,\,2x_{15}&=y_7\color{blue}{\mp}\sqrt{y_7^2-4(t_1 + y_7 + y_1 + 2 y_4)}\\ 2x_8,\,2x_{0}&=y_8\pm\sqrt{y_8^2-4(t_2 + y_8 + y_2 + 2 y_5)}\\ \text{and,}\\ 2y_1,\,2y_5&=z_1\pm\sqrt{z_1^2+4(5 +t_1 +2 z_1)}\\ 2y_2,\,2y_6&=z_2\color{blue}{\mp}\sqrt{z_2^2+4(5 +t_2 +2 z_2)}\\ 2y_3,\,2y_7&=z_3\pm\sqrt{z_3^2+4(5 +t_1 +2 z_3)}\\ 2y_4,\,2y_8&=z_4\color{blue}{\mp}\sqrt{z_4^2+4(5 +t_2 +2 z_4)}\\ \text{and,}\\ 2z_1,\,2z_3&=t_1\pm\sqrt{t_1^2+64}\\ 2z_2,\,2z_4&=t_2\pm\sqrt{t_2^2+64}\\ \text{and,}\\ t_1,\,t_2&=\frac{-1\pm\sqrt{257}}{2}\\ \end{aligned}$$

Whew! The $w_i, u_i, v_i, x_i, y_i, z_i, t_i$ of course are algebraic numbers of deg $2^7, 2^6, 2^5, 2^4, 2^3, 2^2, 2$, respectively. One can see some patterns, like how the 16 $x_i$ are so orderly expressed by the 8 $y_i$. This solution uses 24 quadratic equations $(1+2+6+8+4+2+1 = 24)$, while the one by W. Bishop involves $25$. I do not know if it can be reduced even further.

Glorfindel
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Tito Piezas III
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  • You can express $\cos\left(\frac{2\pi}{257}\right)$ using Galois-theory by solving $7$ quadratic equations. – Anton Odina Jun 29 '23 at 15:49
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    FYI, Chapter 6 (pp. 31-35) of my manuscript A detailed and elementary solution to $x^{17} = 1$ summarizes Wayne Bishop's 1978 list of $25$ quadratic polynomials for a solution to $x^{257} = 1.$ (I also give $11$ references on pp. 46-47 for the $n = 257$ case.) Aside from the typo you mentioned in a comment to @Gerry Myerson's answer (and maybe other typos I silently corrected), the actual published equations given in Bishop's paper are correct. I spent 7-8 hours back in 2011 (continued) – Dave L. Renfro Nov 08 '23 at 20:11
  • setting up and calculating the details I give in my manuscript (Scientific Workplace 2.5 set for 500-digit calculations "defining new variables" at each stage rather than copying and pasting lengthy decimal approximations, using a 1999 computer with Windows 98 and 64 MB RAM; I think I ran the calculations 3 separate times during those 7-8 hours), and they were confirmed again in 2013 but this time it took less than a minute even with 1000-digit calculations (I had my "set-up" template SWP document with all the "defined variable" settings in place (continued) – Dave L. Renfro Nov 08 '23 at 20:11
  • so that part didn't have to be redone, and I was using this computer I had purchased in 2013 with Windows 7 and Scientific Workplace 5.5). Incidentally, I exchanged some emails with Wayne Bishop about this back in 2011, and he mentioned how he needed to use (via a former department graduate student's access) Lookheed Martin's (areospace company) mainframe to do the $n = 257$ case, and (continued) – Dave L. Renfro Nov 08 '23 at 20:11
  • although at the end of the paper he mentions that by the time of publication of the paper he will have done the $n = 65537$ case, that turned out not to be true -- it took too long despite repeated trials. By the way, I happened by accident to come across your question/answer just now, which is why this comment is 10 years later! – Dave L. Renfro Nov 08 '23 at 20:12
  • @DaveL.Renfro Thanks for the informative historical background. Imagine that Wayne Bishop back then had to use a major aerospace company's mainframe to do the calculations, while now I can easily do it on my laptop! A small question though. Bishop had to use 25 quadratic equations, but my solution (based on Ken Brakke's) uses only 24. And a deleted 2013 answer by Wendy Krieger says one may only need 8 surds, while Anton Odina's 2023 comment says only 7 quadratic equations are to be solved. I believe Krieger and Odina meant nested surds, correct? – Tito Piezas III Nov 09 '23 at 05:23
  • I don't know what is meant by nested surds in this context, unless it has something to do with how far back roots can be used for coefficients (i.e. in forming the coefficients for Equation $n,$ are we only allowed to use rationals and roots from Equation $n-1$ and maybe Equation $n-2,$ or can we use rationals and roots from any of the equations before Equation $n).$ I guess at least the latter, since looking back at my manuscript for Bishop's equations, I see that many go back much further than $n-1$ (e.g. Equation 21 uses a root from Equation 8). – Dave L. Renfro Nov 09 '23 at 08:06
  • @DaveL.Renfro I figured as much. For example, the solution above uses $1+2+6+8+4+2+1=24$ quadratics, and there are 7 addends on the $\text{LHS}$. Thus, the explicit solution would have 7 levels of square roots nested within each other, like those Russian matryoshka dolls. – Tito Piezas III Nov 09 '23 at 08:26
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The relevant quadratic polynomials are given explicitly by Wayne Bishop, How to construct a regular polygon, American Math Monthly, March 1978, pp 186-188, available at http://poncelet.math.nthu.edu.tw/disk5/js/geometry/bishop.pdf

Gerry Myerson
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  • Thanks. Those 25 polynomials (I counted) can make someone go blind, and I think there are some typos. For example, in p.188 the fifth polynomial is given as $x^2 - \alpha_{2, 1}x + (5x_{2, 0} + 2\alpha_{2, 1} + 5\alpha_{2, 2} + 4\alpha_{2, 3})$, but I think that stray $x_{2,0}$ variable should be $\alpha_{2,0}$. If there are less obvious typos, it would take a whole lot of time to find it. I wonder if it can be simplified using Conway's suggestion. – Tito Piezas III Oct 06 '13 at 05:02
  • Ken Brakke's version at http://www.susqu.edu/brakke/constructions/big-gon.htm has no typos. (I spoke too soon about it being difficult. It was actually easy to implement in Mathematica.) – Tito Piezas III Oct 06 '13 at 05:50
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Although a solution has been given and more than six years have passed, I will give a solution that is a little different from the desired. I used analytical methods, but for the above reasons I will only give the final result.

$2 \cos \left (\frac{2 \pi}{2^8 +1} \right ) =$

$\sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2- \sqrt{2+ 2 \cos \left (\frac{2 \pi}{2^8 +1} \right )}}}}}}}}$

Γιώργος Πλούσος
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    This looks quite interesting. How did you arrive about? And how will you prove? Is it infinite nested square roots of 2? – Sivakumar Krishnamoorthi Sep 08 '20 at 17:26
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    I have learned how to handle such mathematical expressions since 2004, but I have not been particularly involved with the proofs. The reliability of my methods is based on experience and intuition. It is difficult to describe these methods here, but I have published related works, the most recent of which can be found in the following links; have been written in an amateurish way and in bad English.

    https://vixra.org/abs/2004.0456

    https://vixra.org/author/george_plousos

     – Γιώργος Πλούσος Sep 10 '20 at 03:26
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Refer here for solving cyclic infinite nested square roots of 2

Steps to get cyclic infinite nested square roots of 2

$x =\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+x}}}}$, where $x = 2\cos(\frac{8}{17}\cdot\pi)$

$\sqrt{2+2\cos(\frac{8}{17}\cdot\pi)} = 2\cos(\frac{4}{17}\cdot\pi)$ Next step is to substitute $x =\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+x}}}} = \sqrt{2-\sqrt{2+\sqrt{2+2\cos(\frac{4}{17}\cdot\pi)}}}$

$=\sqrt{2-\sqrt{2+2\cos(\frac{2}{17}\cdot\pi)}}$ $=\sqrt{2-\cos(\frac{1}{17}\cdot\pi)}$ $=2\sin(\frac{1}{34}\cdot\pi)$ $=2\cos(\frac{8}{17}\cdot\pi)$

Therefore

$2\cos(\frac{8}{17}\cdot\pi)$ can be expanded was infinite cyclic nested square roots of 2 as follows

$$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+...}}}}$$ repetition of (- + + +) i.e 1-3+ signs are repeated infinitely in the nested square roots of 2 in a cyclic manner

With available programming language like python it is easy to calculate the value of $2\cos\frac{8\pi}{17}$ for desired number of digits

Same principle may be applied to evaluate $$2\cos\frac{128π}{257}$$ similar steps will lead to

$$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}}}}$$ The pattern of repetition is (- + + + + + + +) i.e. 1 - & 7 + signs repeated infinitely in cyclical pattern

Therefore $(\cos\frac{2π}{257})$ can be evaluated with Half angle cosine formula which will be expressed (+ + + + + +) (128 = 2^7 to 2^1 will lead to 6 + signs initially and then it will continue as infinite cyclic nested square roots of 2 containing 1 - & 7 + signs

$(\cos\frac{2π}{257})$ will look like this as follows expressed in real radicals

$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}}}}}}}}}}}$$

Sivakumar Krishnamoorthi
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