# Two complementary continued fractions that are algebraic numbers

Define the two similar continued fractions,

$$x=\cfrac{1}{km\color{blue}+\cfrac{(m-1)(m+1)} {3km\color{blue}+\cfrac{(2m-1)(2m+1)}{5km\color{blue}+\cfrac{(3m-1)(3m+1)}{7km\color{blue}+\ddots}}}}\tag1$$

$$y=\cfrac{1}{km\color{blue}-\cfrac{(m-1)(m+1)} {3km\color{blue}-\cfrac{(2m-1)(2m+1)}{5km\color{blue}-\cfrac{(3m-1)(3m+1)}{7km\color{blue}-\ddots}}}}\tag2$$

For integer $m>1$, it seems $(1)$ and $(2)$ are algebraic numbers. For $(1)$ and the special case $k=1$, then,

$$x=\tan\Big(\frac{\pi}{4m}\Big)=\frac{1}{4m}\frac{\Gamma\Big(\frac{2m-1}{4m}\Big)\Gamma\Big(\frac{2m+1}{4m}\Big)}{\Gamma\Big(\frac{4m-1}{4m}\Big)\Gamma\Big(\frac{4m+1}{4m}\Big)}\tag3$$

and is adapted from Nicco's cfrac in this post. For $(2)$ and general $k\neq1$,

$$y =\tanh\Big(\frac{1}{2m}\,\log (R)\Big)= \frac{R^{1/m}-1}{R^{1/m}+1};\;\;R=\frac{k+1}{k-1}\tag4$$

and solves,

$$\frac{ (1+y)^{m}-(1-y)^{m}}{(1+y)^{m}+(1-y)^{m}}=\frac{1}{k}\tag5$$

hence $x,y$ are radicals.

Questions:

1. What is the formula for $x$ for general $k>1$?
2. Does $y$ also have a formula in terms of gamma functions?
• Inspired by achilles' proof in this post,one is led to conjecture $$F(k,z,m)=\cfrac{1}{km+\cfrac{z(m-1)(m+1)} {3km+\cfrac{z(2m-1)(2m+1)}{5km+\cfrac{z(3m-1)(3m+1)}{7km+\ddots}}}}$$ Then $$F(k,z,m) = \frac{1}{\sqrt{z}}\tan(\frac{1}{m}\tan^{-1}\left(\frac{\sqrt{z}}{k}\right))$$ May 16, 2016 at 18:04
• @Nicco: That's a very nice generalization. May 17, 2016 at 6:58