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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 at least),occurs when we take the limit to zero

$$\begin{aligned}\lim_{z\to0} \cfrac{z(z+1)}{2z+1+\cfrac{(0z-1)(4z+3)} {3(2z+1)+\cfrac{(2z+0)(6z+4)}{5(2z+1)+\cfrac{(4z+1)(8z+5)}{7(2z+1)+\cfrac{(6z+2)(10z+6)}{9(2z+1)+\ddots}}}}}=\frac{1}{\pi}\end{aligned}$$ yielding a new limit for $\pi$ from which one obtains the first few convergents $\begin{aligned}0,\frac{3}{8},\underline{\frac{5}{16},\frac{15}{47},\frac{7}{22}},\frac{1365}{4288},\frac{3015}{9472},\frac{1575}{4948},\ddots\end{aligned}$.

Where the underlined convergents appear in the stern-brocot tree for $\pi $ associated to its simple continued fraction.

Q: Is the conjectured continued fraction true (for all complex numbers $z$ with $x\gt0$)?

Update:I initially defined the continued fraction $\displaystyle\cot\left(\frac{z\pi}{4z+2}\right)$ for only natural numbers,but as a matter of fact it holds for all complex numbers $z$ with real part greater than zero.This continued fraction is one special case of the general continued fraction found here.

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  • $\begingroup$ en.wikipedia.org/wiki/… does't apply right? $\endgroup$
    – tired
    Commented Oct 23, 2015 at 10:20
  • $\begingroup$ @tired:Yes it does converge by Pringsheim-theorem ,but I was looking for a proof of the continued fraction.My apologies,I should have made the question clearer. $\endgroup$
    – Nicco
    Commented Oct 24, 2015 at 4:09

1 Answer 1

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O. Perron, Die Lehre von den Kettenbrüchen, Chapter XI, Section 82.

Satz 5 on page 488 (2nd edition, 1927):

Continued fraction $$ d+\underset{k=1}{\overset{\infty }{\mathbf K}}\; \frac{a+bk+ck^2}{d+ek} $$ with $c \ne 0, e \ne 0, e^2+4c \ne 0$ has value $$ \frac{\sqrt{e^2+4c}\;{}_2F_1(\alpha,\beta;\gamma;x)\;\gamma}{{}_2F_1(\alpha+1,\beta+1;\gamma+1;x)}, $$ where $\alpha,\beta$ are the roots of the quadratic equation $cZ^2-bZ+a=0$, $$ \gamma = \frac{b+c}{2c}\left(1-\frac{e}{\sqrt{e^2+4c}\;}\right) +\frac{d}{\sqrt{e^2+4c}\;} \\ x=\frac{1}{2}\left(1-\frac{e}{\sqrt{e^2+4c}\;}\right) $$ Choose the square-root so that $$ \mathrm{Re}\frac{e}{\sqrt{e^2+4c}\;}>0 . $$

So substituting in the values in this problem, the question becomes: Is $$ \cot\left(\frac{z\pi}{4z+2n}\right) = \frac{\displaystyle(z+n)\sqrt{2}\; {}_2F_1\left(\frac{-n}{2z+n},\frac{4z+3n}{2z+n}; \frac{3}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)} {\displaystyle(2z+n)\; {}_2F_1\left(\frac{-(2z+2n)}{2z+n},\frac{2z+2n}{2z+n}; \frac{1}{2}; \frac{1}{2}-\frac{1}{2\sqrt{2}}\right)} ? $$ Writing $z/(4z+2n) = t$ and taking reciprocal, the question becomes: Is $$ \tan(t\pi) = \frac{{}_2F_1\left(4t-2,-4t+2; \frac{1}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)} {\sqrt{2}(1-2t)\;{}_2F_1\left(4t-1,-4t+3; \frac{3}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)} ? $$

proof for this

Use quadratic transformation

$$ {}_2F_1\left(2a,2b;a+b+\frac{1}{2};u\right)= {}_2F_1\left(a,b;a+b+\frac{1}{2};4u(1-u)\right) $$

to get $$ {}_2F_1\left(4t-2,-4t+2; \frac{1}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right) ={}_2F_1\left(2t-1,-2t+1;\frac{1}{2};\frac{1}{2}\right) = \sin(\pi t) $$

$$ {}_2F_1\left(4t-1,-4t+3; \frac{3}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right) ={}_2F_1\left(2t-\frac{1}{2},-2t+\frac{3}{2};\frac{3}{2};\frac{1}{2}\right) =\frac{\cos(\pi t)}{\sqrt{2}(1-2t)} $$ These are known to most CASs.

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