Articles of continued fractions

Fundamental unit in the ring of integers $\mathbb Z$

Find a fundamental unit in the ring of integers $\mathbb Z[\frac{1+\sqrt{141}}{2}]$ of $\mathbb Q(\sqrt{141})$ I have different corollaries for different numbers, the most appropriate for $141$ is the one below. I used an algorithm (don’t know if you know this, but $\beta_0=\sqrt{141}+\lfloor\sqrt{141}\rfloor, \quad\beta_{n+1}=\frac{1}{\beta_n-\lfloor\beta_n\rfloor}$ $a_n=\lfloor\beta_n\rfloor$ $p_n=p_{n-1}a_n+p_{n-2}, \quad q_n=q_{n-1}a_n+q_{n-2} $) to determine the continued fraction expansion of […]

Solution to $x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}$

Possible Duplicate: Why does this process, when iterated, tend towards a certain number? (the golden ratio?) Please post your favorit solution to the following Compute $x=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ldots}}}$ Thank you

Can every transcendental number be expressed as an infinite continued fraction?

Every infinite continued fraction is irrational. But can every number, in particular those that are not the root of a polynomial with rational coefficients, be expressed as a continued fraction?

Smallest number $N$, such that $\sqrt{N}-\lfloor\sqrt{N}\rfloor$ has a given continued fraction sequence

How can I find the smallest positive integer $N$, such that the continued fraction of $\sqrt{N}-\lfloor\sqrt{N}\rfloor$ begins with a given finite sequence containing a zero followed by positive integers ? For example, the sequence $[0,1,2,3,4]$ is given. We have to find the smallest number $N$, such that $\sqrt{N}-\lfloor\sqrt{N}\rfloor$ begins with $[0,1,2,3,4]$. The number $\sqrt{388}-\lfloor\sqrt{388}\rfloor$ has […]

Is it right that the fundamental recurrence of an arbitrary continued fraction cannot be proved without induction?

Let $\dfrac{A_{n}}{B_{n}}$ be the $n^{th}$ convergent (approximant) $$ \frac{A_{n}}{B_{n}}=b_{0}+\dfrac{a_{1}}{b_{1}+\dfrac{a_{2}}{b_{2}+\dfrac{a_{3}}{\begin{array}{c} b_{3}+ \\ \\ \end{array} \begin{array}{cc} \ddots & \\ & \end{array} +\dfrac{a_{n-1}}{b_{n-1}+\dfrac{a_{n}}{b_{n}}}}}} $$ of a continued fraction. $A_{n}$, $A_{n-1}$, and $A_{n-2}$ satisfy the recurrence $$ \begin{eqnarray*} \begin{pmatrix} A_{n} \\ B_{n} \end{pmatrix} &=& \begin{pmatrix} A_{n-1} & A_{n-2} \\ B_{n-1} & B_{n-2} \end{pmatrix} \begin{pmatrix} b_{n} \\ a_{n} \end{pmatrix} \quad […]

What is the probability that $250$ random digits contain $7777$ , $8888$ and $9999$?

First, how I arrived at a number having the property that the first $250$ digits after the decimal point contain $7777$ , $8888$ and $9999$ I wanted to construct a number which can be shown to be transcendental using the irrationality measure Consider the sequence $$a_1=0\ ,\ a_2=1\ ,\ a_n=a_{n-1}^2+a_{n-2}\ for \ n>2$$ Then, […]

Extract a Pattern of Iterated continued fractions from convergents

I have been working on an article at where I posted a table of “convergents constants” (defined at for a few numbers. It would be nice to support the article with some quality analysis. Before June 9, 2011, was starting to extract and clearly define a pattern to these constants cf the article. […]

a conjectured continued-fraction for $\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)$ that leads to a new limit for $\pi$

Given a complex number $\begin{aligned}\frac{z}{n}=x+iy\end{aligned}$ and a gamma function $\Gamma(z)$ with $x\gt0$, it is conjectured that the following continued fraction for $\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)$ is true $$\begin{aligned}\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)=\frac{\displaystyle\Gamma\left(\frac{z}{4z+2n}\right)\Gamma\left(\frac{3z+2n}{4z+2n}\right)}{\displaystyle\Gamma\left(\frac{z+n}{4z+2n}\right)\Gamma\left(\frac{3z+n}{4z+2n}\right)}=\cfrac{2z+2n}{2z+n+\cfrac{(0z-n)(4z+3n)} {3(2z+n)+\cfrac{(2z+0n)(6z+4n)}{5(2z+n)+\cfrac{(4z+n)(8z+5n)}{7(2z+n)+\cfrac{(6z+2n)(10z+6n)}{9(2z+n)+\ddots}}}}}\end{aligned}$$ Or in gauss’s notation $$\begin{aligned}\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)=-\frac{1}{2z+2n}\underset{m=0}{\overset{\infty}{\mathbf K}}\frac{((2m-2)z+(m-2)n)((2m+2)z+(m+2)n)}{((2m+1)(2z+n)}\end{aligned}$$ Corollaries: 1):let $z=1$ and $n=2$,then we obtain a beautiful continued fraction for square root 2 $$\begin{aligned}{-1+\cfrac{3}{2+\cfrac{\frac{(-1)(5)}{(1)(3)}} {2+\cfrac{\frac{(1)(7)}{(3)(5)}}{2+\cfrac{\frac{(3)(9)}{(5)(7)}}{2+\cfrac{\frac{(5)(11)}{(7)(9)}}{2+\ddots}}}}}}=\sqrt{2}\end{aligned}$$ 2):However the most interesting case(for me […]

Cantor set as a set of continued fractions?

Does the classical cantor set have a nice description as a set of continued fractions? I made a (superficial) search and didn’t find anything, but I’m very tired right now, so please forgive me that I don’t put more effort in asking this.

A trigonometric identity for special angles

Prove that for a natural number $n$, $$\prod_{k=1}^n \tan\left(\frac{k\pi}{2n+1}\right) = 2^n \prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}.$$