Linked Questions

22
votes
3answers
4k views

Proving $2 ( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} )$ is a root of$ \sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$

How can one show that the number $2 \left( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} \right)$ is a root of the equation $\sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$?
17
votes
2answers
1k views

Constructive Proof of Kronecker-Weber?

This question is motivated by my attempt at solving Proving $2 ( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} )$ is a root of$ \sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$ Consider ...
32
votes
2answers
1k views

Something strange about $\sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$ and its friends

This post can be generalized to, $$\begin{align} \sqrt{ 2+ \sqrt{ 2 + \sqrt{ 2-x}}}=x&,\quad\quad x = -2\cos\left(\frac{8\pi}{9}\right)=1.8793\dots\quad\quad\quad \\ \\ \sqrt{ 4+ \sqrt{ 4 + \sqrt{...
13
votes
2answers
463 views

What's the explanation for these (infinitely many?) Ramanujan-type identities?

Define the function, $$F(\beta) := \sqrt[3]{\beta+x_1}+\sqrt[3]{\beta+x_2}+\sqrt[3]{\beta+x_3}\tag1$$ where, $$x_1 =2\cos\big(\tfrac{2\pi }{7}\big),\;x_2 =2\cos\big(\tfrac{4\pi }{7}\big),\; x_3 = ...
23
votes
1answer
1k views

What comes after $\cos\left(\tfrac{2\pi}{7}\right)^{1/3}+\cos\left(\tfrac{4\pi}{7}\right)^{1/3}+\cos\left(\tfrac{6\pi}{7}\right)^{1/3}$?

We have, $$\big(\cos(\tfrac{2\pi}{5})^{1/2}+(-\cos(\tfrac{4\pi}{5}))^{1/2}\big)^2 = \tfrac{1}{2}\left(\tfrac{-1+\sqrt{5}}{2}\right)^3\tag{1}$$ $$\big(\cos(\tfrac{2\pi}{7})^{1/3}+\cos(\tfrac{4\pi}{7})...
3
votes
4answers
623 views

An infinite nested radical problem

From this link, problem 36, I found that $$\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}=2\left(\cos{\dfrac{4\pi}{19}}+\cos{\dfrac{6\pi}{19}}+\cos{\dfrac{10\pi}{19}}\right).$$ The signs : ...
6
votes
1answer
457 views

Gauss-type sums for cube roots

(Quadratic) Gauss sums express square root of any integer as a sum of roots of unity (or of cosines of rational multiples of $2\pi$, if you will) with rational coefficients. But Kronecker-Weber ...
11
votes
1answer
531 views

A little more on $\sqrt[3]{\cos\bigl(\tfrac{2\pi}7\bigr)}+\sqrt[3]{\cos\bigl(\tfrac{4\pi}7\bigr)}+\sqrt[3]{\cos\bigl(\tfrac{8\pi}7\bigr)}$

Using a special case of an identity by Ramanujan, we find that given the roots $x_i$ of $$x^3 + x^2 - (3 n^2 + n)x + n^3=0\tag1$$ which, since its discriminant is negative, always has three real ...
9
votes
1answer
394 views

What's so special about primes $x^2+27y^2 = 31,43, 109, 157,\dots$ for cubics?

While trying to find a closed-form solution for particular cubics as sums of cosines (related to this question), I came across this family with all roots real. Given a prime $p=6m+1$. Define, $$F(x) = ...
5
votes
2answers
163 views

Evaluate this Trigonometric Expression

Evaluate $$ \sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$$ I found the following $\large{\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7} + \cos \frac{...
2
votes
2answers
256 views

A question by Ramanujan about a relational expression of a triangle

I found the following question in a book without any proof: Question : Suppose that each length of three edges of a triangle $ABC$ are $BC=a, CA=b, AB=c$ respectively. If $$\frac1a=\frac1b+\frac1c, \...
8
votes
0answers
262 views

A Ramanujan-type trigonometric identity

At the end of the following article: http://www.ijpam.eu/contents/2013-85-1/15/15.pdf It is asserted that the russian mathematician, Sergey Markelov, in private communication, told them that he ...
5
votes
0answers
181 views

More on primes $p=u^2+27v^2$ and roots of unity

Given, $$p=u^2+27v^2=6m+1\tag1$$ and the cubic, $$x^3+x^2-2mx+N=0\tag2$$ with its constant expressed in terms of $(1)$ as, $$N = \frac{1}{27}(1-3p\pm2pu)\tag3$$ and the sign $\pm u$ chosen ...