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$F(x)$ can be rewritten as $\displaystyle\;\frac{1}{\frac{2}{P(x)} - x}$ where $\displaystyle\;\def\CF{\mathop{\LARGE\mathrm K}} P(x) = \cfrac{1\cdot 2}{1\cdot 2 x + \cfrac{ (1 \cdot 2)(2\cdot 3)}{2\cdot 3 x + \cfrac{(2\cdot 3)(3\cdot 4)}{3\cdot 4 x + \ddots} }}$. The CF $P(x)$ has the form $\displaystyle\; \CF_{\ell=1}^{\infty} ... 3 Just an example:$\sqrt{7}$. Since$4<7<9$,$\left\lfloor \sqrt{7}\right\rfloor = 2$, so: $$\sqrt{7} = 2+(\sqrt{7}-2) = \color{blue}{2}+\frac{1}{\frac{\sqrt{7}+2}{3}}\tag{1}.$$ Since$2+\sqrt{7}\in (4,5)$,$\left\lfloor\frac{\sqrt{7}+2}{3}\right\rfloor =1$, so: $$\frac{\sqrt{7}+2}{3} = 1+\frac{\sqrt{7}-1}{3} = 1+\frac{1}{\frac{\sqrt{7}+1}{2}}$$ and ... 3 Let$x$be the value of the continued fraction$[0;7]$. We have that $$x = \frac{1}{7 + \frac{1}{7 + \frac{1}{7 +...}}}$$ Now we notice that$x$appears in this continued fraction, so we can write it as: $$x = \frac{1}{7 + x}$$ This simplifies to$x^2+7x-1=0$, and solving for$x$gives$x=\frac{-7}{2}\pm \frac{\sqrt{53}}{2}$2 For simplicity, we will start with the case$z = 1$. We will assume$k > 0$and let$\cot\theta = k$and$\mu_{\pm} = k \pm \sqrt{k^2+1}$. The CF at hand has the form $$\def\CF{\mathop{\LARGE\mathrm K}} \CF_{\ell=1}^{\infty} \frac{\alpha_\ell\gamma_{\ell-1}}{\beta_\ell} \quad\text{ where }\quad \gamma_0 = 1\quad\text { and }\quad \begin{cases} ... 2 The case with z \neq 1 can be obtained from the case z=1, because you can just factor them out of the infinite fraction while changing k accordingly. Consider k,m fixed, and let (f_n) be the sequence of homographies f_n(x) = a_n + b_n / x, with a_n = (2n-1)km and b_n = ((n-1)m-1)((n+1)m+1) so that \Theta = \lim_{n \to \infty} (m+1)/ f_0 ... 2 (This is a comment that got too long for the comment box.) For fun, I decided to implement this function for complex arguments in Mathematica, and plot its real and imaginary parts. Here's the picture I got: That pole fence jives with the OP's observation that the function is only sensible for \Re s > -1. Can it be turned into a series? ... 2 Your continued fraction is a very special case of the general continued fraction for the quotient of gamma functions conjectured in this post,see corollary (iii) Edited: It is also a special case of the hyperbolic ... 1 Never mind, thanks to the comment by J. M. I found the source of this expression. The series are connected to the exponential integral:$$\text{Ei}(t)=-\int_{-t}^{\infty} \frac{e^{-p}}{p} dp=\gamma+\log |t|+\sum_{n=1}^{\infty} \frac{t^n}{n!n}$$The continued fraction turns out to be a particular case of incomplete Gamma function:$$\Gamma ... 1 I believe that for$-\frac{39}{25}$you should first rewrite it as$-2+\frac{11}{25}$. Then$11=0\times25+1125=2\times11+311=3\times3+23=1\times2+12=2\times1+0$This gives$-\frac{39}{25}=[-2+0;2,3,1,2]=[-2;2,3,1,2]$Note that recently there has developed a way of representing negative continued fractions in the form ... 1 Continued fraction (2) can be simplified as $$\tan\left(\alpha\tan^{-1}z\right)=\cfrac{\alpha z}{1+\cfrac{\frac{(1^2-\alpha^2)z^2}{1\cdot 3}} {1+\cfrac{\frac{(2^2-\alpha^2)z^2}{3\cdot 5}}{1+\cfrac{\frac{(3^2-\alpha^2)z^2}{5\cdot 7}}{1+\ddots}}}}\tag{2a}$$ This is a special case of the following continued fraction due to N$\ddot{\text{o}}$rlund (B.Berndt, ... 1 Lehmer's procedure involves solving the Pell equation $$x^2-2qy^2=1$$ for$173$-smooth squarefree$q\neq2$. (This makes$\frac{x}{y}$an approximation for$\sqrt{2q}$.) Then, for a finite number of smallest solutions$(x,y)$of each Pell equation, the integers$n=\frac{x-1}{2}$and$n+1$are tested for smoothness. Therefore, the$x$involved is$x=2n+1$, ... 1 Too long for a comment. If you let$a=-1$and$b=2m+1$of the general continued fraction in this post, it reduces to the first continued fraction in this post (with$k=1$) and is expressible as a quotient of gamma functions, ... 1 (Too long for a comment.) I. Level$2$From your other post, we have, $$G_1(x,n)=\cfrac{1}{2x+\cfrac{(-1)(-1+n)} {6x+\cfrac{(1)(1+n)}{10x+\cfrac{(3)(3+n)}{14x+\cfrac{(5)(5+n)}{18x+\ddots}}}}}\tag1$$ with$2v+1 =-1,1,3,5,\dots$The special case$n=6$, $$G_1(x,\color{brown}6)=\frac{1}{3b}\left(a+\sqrt{c^3}\right)\tag2$$ where, ... 1$\sqrt 2 = (1+1)^{1/2}$by the binomial theorem:$(1+a)^{1/2}$$= 1 + (1/2) 1^{-1/2}a-(1/8) 1^{-3/2}a^2+ (3/48) a^3\cdots coefficient of the n^{th} term: c_0 = 1\\c_n = c_{n-1}\frac{(1/2-n)}{n} when n\ge 3, c_n = (-1)^{n+1}\frac{1*3*5*7...(n-2)}{2*4*6*8\cdots n}  More generally, can you find a series that converges to the alegraic ... 1 Suppose you have a decimal expansion for your irrational number. We'll take \sqrt2=1.41421\dots for example. Then we can write$$\sqrt{2}=1+\frac{4}{10}+\frac{1}{100}+\frac{4}{1000}+\frac{2}{10000}+\frac{1}{100000}+\dots Each term in our sequence is rational (and moreover it is clear how to apply this idea to get an infinite sum of rationals to ...

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Let $\displaystyle x=\sqrt{2}+\frac{b}{\sqrt{2}+\ldots}$, then \begin{align*} \sqrt{2}+\frac{b}{x} &= x \\ x\sqrt{2}+b &= x^{2} \\ x^{2}-\sqrt{2} \, x-b &= 0 \\ x &= \frac{\sqrt{2}+\sqrt{2+4b}}{2} \\ &= \sqrt{ \left( \frac{\sqrt{2}+\sqrt{2+4b}}{2} \right)^{2} } \\ &= \sqrt{b+1+\sqrt{2b+1}} \\ \end{align*} Take ...

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I'll have a stab at this. First, here's a link on how to find the simple continued fraction of a number $x$: http://mathworld.wolfram.com/ContinuedFraction.html And here's a link (see pages 13 - 15) on how to factorize a number $n$, using the simple continued fraction of $x = \sqrt n$: http://wstein.org/edu/2010/414/projects/johnson.pdf From the first ...

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