Hot answers tagged

4

$F(x)$ 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 $11=0\times25+11$ $25=2\times11+3$ $11=3\times3+2$ $3=1\times2+1$ $2=2\times1+0$ 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 \begin{equation} ...


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 ...


1

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 ...


1

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 ...



Only top voted, non community-wiki answers of a minimum length are eligible