Closed form of $\sin{\frac{\pi}{257}}$ and $\sin{\frac{\pi}{65537}}$

Question

Bit of a soft question. I am quite eager to find a closed form (expression involving nested roots and integers) for the value $\sin{\left(\frac{\pi}{65537}\right)}$ since $65537$ is a Fermat prime and thus there must be a closed form. I tried using Wolframalpha but it didn't provide an adequate answer. I don't know of any other calculators that could give me my desired result. So, does anyone have images or links to these sought-after closed forms?

Edit: I forgot that $257$ is also a Fermat prime number, so I'd like a closed form for $\sin{\frac{\pi}{257}}$ as well, thank you.

Answer 1

As far as I see, the comments should provide the required references.

Case of $\sin\frac{\pi}{257}$.

HackR provided a link to the question by Tito Piezas III on a representation of $\cos\frac{2\pi}{257}$ as a tower of nested real radicals. In his answer is provided a formula for $w_4=4\cos\frac{256\pi}{257}$. Then $$\sin\frac{\pi}{257}=\sqrt{1-\left(\frac{w_4}4\right)^2}.$$

Case of $\sin\frac{\pi}{65537}$.

QC_QAOA provided a link to an old Conway's post about the construction of the $65537$-gon. The construction is very cumbersome, but there is provided the sketch of it. But since I saw deleted link-only answers, and, moreover, QC_QAOA failed to get the link to work, I copied the relevant text below and formatted the formulae.

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From: Antreas P. Hatzipolakis

Posted: Feb 8, 1999 5:25 PM

... I recall that a Conway's student tried once to construct the r. $65537$-gon using a computer.

I located Conway's posting, so let him tell us the story:

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Subject: Re: 17 sides

From: John Conway conway@math.Princeton.EDU

Date: Tue, 4 Nov 1997 12:26:51 -0500 (EST)

To: geometry-research@forum.swarthmore.edu

On Tue, 28 Oct 1997, Bradley Brock wrote:

How could Gauss find the length of the $17$-side polygon side???

On a related note: I remember that one of my friends in grad school showed John Conway the output from a little Mathematica program that gave the sides of the $257$-gon.

Brad

Forgive me for not replying to this before now. It obviously refers to John Steinke, who was a graduate student here some time ago, and is a bit misleading. What happened was that I proposed to him the problem of finding a publishable construction for the $65537$-gon, and suggested various methods, and he did the $257$-gon as a baby-example.

You are probably aware of the fact that Hermes worked for many years on finding a construction for the $65537$-gon. I don't really believe the legend, that this was because his teacher told him to "go away and don't bother me again until you've found how to construct a regular $65537$-gon!", but many years ago I did see in Gottingen a cardboard box that was said to contain Hermes's work on the problem.

However, I can hardly believe that he really got anywhere, because the size of the problem is much bigger than one might suppose. I did work out, though, a way of phrasing the solution that could probably fit into $20$ closely-written pages, and could definitely be found nowadays by computer. However, I think Steinke lost interest in the problem and I never tried to persuade anyone else to do it.

It's obvious from the start that there's a "solution" in the form of a set of 16 quadratic equations, the coefficients of each being functions of the roots of the previous ones. The trouble is that those functions get pretty complicated, and it would take a very large book to write them all down in that form. Let for instance $a,A$ ; $b,B$; $c,C$ be the roots of the first three. Then the general element of the field they generate is a linear combination with rational coefficients of the $8$ numbers

$$abc, abC, aBc, aBC, Abc, AbC, ABc, ABC$$

and so to specify the (two) coefficients of the next equation is to give a list of $16$ rational numbers (which we can actually make be integers). In a similar way, to specify the coefficients of the $14$th equation we'll need $2^{14} = 16384$ integers.

My proposed form of the solution involves going about half-way, to about the $8$th equation, in this manner, thereby setting up names for all elements of the subfield of degree $256$. Then one works from the other end, working out the quadratic satisfied by a particular $65537$th root of unity over the next field down, and so on. Let's call it Z - then this equation is $t^2 - (Z + Z^{-1})t + 1 = 0$, and so we invent a name, $Y$, for $Z + Z^{-1}$. In the same way, we invent names for the coefficients of the quadratic defining Y, and so on. By the time we get half-way, we'll've define $Z, Z^{-1}, Y, \dots$ in terms of about $512$ numbers in the degree $256$ subfield. So the solution reduces to writing out the length $256$ names of each of these $512$ numbers.

I would very much like to see this calculation done, since there might well be some patterns that would enable one to write down the answer more simply than the way sketched above.

John Conway

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Additionally, let me quote the first sentence of the last paragraph of Duane W. DeTemple's paper [1, p. 107]:

4. Remarks on the Construction of the Regular $65537$-gon. We have already observed that $g = 3$ is a primitive root of $p = 65537$. The sum, product, and relative order of period pairs must then be computed (one feels sorry for the computerless Hermes!).

1. Duane W. DeTemple: Carlyle Circles and the Lemoine Simplicity. The Americn Mathematical Monthly 98(1991) 97 - 108.

Note: DDeT refers to several regular heptakaidecagon constructions, but not to that one (Lebesgue's) I posted earlier.

Antreas

Answer 2

Partial Answer

Recall DeMoivre's Formula:

$$(\cos(\theta) + i\sin(\theta))^n = \cos(n\theta) + i\sin(n\theta)$$

Let $\theta = \frac{\pi}{n}$. Then:

$$(\cos(\frac{\pi}{n}) + i\sin(\frac{\pi}{n}))^n = -1$$

Let $x = \sin(\frac{\pi}{n})$. If $n \ge 2$, then $\cos$ and $\sin$ are both nonnegative, so:

$$\boxed{(\sqrt{1-x^2} + ix)^n = -1}$$

By the Binomial Theorem:

$$\sum_{k=0}^n {n \choose k} (1-x^2)^{k/2}(ix)^{n-k} = -1$$

Or, breaking it down into separate summations based on the value of $k \bmod 4$,

$$\sum_{k \equiv 0} {n \choose k} (1-x^2)^{k/2}x^{n-k} - i\sqrt{1-x^2}\sum_{k \equiv 1} {n \choose k} (1-x^2)^{(k-1)/2}x^{n-k} - \sum_{k \equiv 2} {n \choose k} (1-x^2)^{k/2}x^{n-k} + i\sqrt{1-x^2}\sum_{k \equiv 3} {n \choose k} (1-x^2)^{(k-1)/2}x^{n-k} = -i^{-n}$$

Move the odd square roots to one side of the equation:

$$i\sqrt{1-x^2} \left( \sum_{k \equiv 3} {n \choose k} (1-x^2)^{(k-1)/2}x^{n-k} - \sum_{k \equiv 1} {n \choose k} (1-x^2)^{(k-1)/2}x^{n-k} \right) = -i^{-n} - \sum_{k \equiv 0} {n \choose k} (1-x^2)^{k/2}x^{n-k} + \sum_{k \equiv 2} {n \choose k} (1-x^2)^{k/2}x^{n-k}$$

And now we can just square it, producing a polynomial equation:

$$(x^2 - 1) \left( \sum_{k \equiv 3} {n \choose k} (1-x^2)^{(k-1)/2}x^{n-k} - \sum_{k \equiv 1} {n \choose k} (1-x^2)^{(k-1)/2}x^{n-k} \right)^2 = \left(-i^{-n} - \sum_{k \equiv 0} {n \choose k} (1-x^2)^{k/2}x^{n-k} + \sum_{k \equiv 2} {n \choose k} (1-x^2)^{k/2}x^{n-k}\right)^2$$

For example, for $n=3$, each summation contains only one item, and we get:

$$(x^2 - 1) \left({3 \choose 3} (1-x^2)^{1}x^{0}- {3 \choose 1} (1-x^2)^{0}x^{2}\right)^2 = \left(-i^{-3} - {3 \choose 0} (1-x^2)^{0}x^{3} + {3 \choose 2} (1-x^2)^{1}x^{1}\right)^2$$ $$(x^2 - 1) ((1-x^2) - 3x^{2})^2 = (-i - x^{3} + 3(1-x^2)x)^2$$ $$(x^2 - 1)(1 - 4x^2)^2 = (-i + 3x - 4x^3)^2$$ $$x^2 - 8x^4 + 16x^6 - 1 + 8x^2 - 16x^4 = -1 - 3ix + 4ix^3 - 3ix + 9x^2 - 12x^4 + 4ix^3 - 12x^4 + 16x^6$$ $$8ix^3 - 6ix = 0$$ $$2ix(4x^2 - 3) = 0$$ $$x\in \{0, \frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}\}$$

But since $\frac{\pi}{3}$ is in Quadrant 1, its sine must be positive, and so $x = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$.

Expanding and solving the polynomials for $n=257$ and $n=65537$ is left as an exercise for the reader

Answer 3

Somewhat speculative answer since actually carrying out this procedure would be a ton of work.

If you represent $t = \frac{1}{2^w + 1}$ in binary you get a repeating decimal, explicitly with the formula:

$$t = \frac{1}{2^w + 1} = t\cdot 2^{2w} - (2^w - 1)$$

so you want

$$s = \sin(\pi\cdot t) = \sin(\pi\cdot t \cdot 2^{2w} - \pi\cdot(2^w - 1))$$

then if you repeatedly apply the transforms

$$\sin(a - b) = \sin a \cdot \cos b - \cos a \cdot \sin b$$ $$\cos(a - b) = \cos a \cdot \cos b + \sin a \cdot \sin b$$ $$\sin(2^k \cdot x) = 2^k \cdot \sin(x) \cdot \prod_{j=1}^{k-1}\cos(2^j \cdot x)$$ $$\cos(2^k \cdot x) = T_{2^k}(\cos(x))$$ $$\sin(\pi) = -1$$ $$\cos(\pi) = 0$$

where $T$ is Chebychev polynomials of the first kind, greatly simplified by $\cos(\pi) = 0$, and labelling $c = \cos(\pi \cdot t)$, you should be able (speculatively, I haven't actually done this) to get some formula of the form:

$$s = g_s(s, c)$$ $$c = g_c(s, c)$$ where $g$s are some elementary function. Then solve for $s$. Or maybe $g$ will end up being some fairly large polynomial with difficult to find roots.