Finding x in $\frac{\,_2F_1(\frac{1}{5},\frac{4}{5},\,1,\,1-x)}{\,_2F_1(\frac{1}{5},\frac{4}{5},\,1,\,x)} = \sqrt{n}$

Question

I was trying to find a closed-form for $0<x<1$ in,

$$\frac{\,_2F_1(\frac{1}{m},\,1-\frac{1}{m},\,1,\,1-x)}{\,_2F_1(\frac{1}{m},\,1-\frac{1}{m},\,1,\,x)} = \sqrt{n}$$

where $\,_2F_1(a,b,c,z)$ is the hypergeometric function. There are formulas for $m = 2,3,4,6$, so I was wondering if there are for other m as well. However, one thing I observed was that, let,

$$q = \exp\left(\frac{-\,\pi\sqrt{n}}{\sin(\pi/m)}\right)$$

Conjecture:

$$\lim_{n\to \infty}\frac{x}{q} = \text{constant}$$

namely,

$$\begin{array}{cc} m&\lim_{n\to \infty}\frac{x}{q}\\\\ 2&16\\ 3&27\\ 4&64\\ 5&25\sqrt{5}\left(\frac{1+\sqrt{5}}{2}\right)^\sqrt{5}=163.95\dots\\ 6&432\\ 7&1152.795095384373\dots\\ 8&2^8\left(1+\sqrt{2}\right)^{2\sqrt{2}}=3096.65\dots\\ \end{array}$$

and so on. This implies a good approximation to x in,

$$\frac{\,_2F_1(\frac{1}{5},\frac{4}{5},\,1,\,1-x)}{\,_2F_1(\frac{1}{5},\frac{4}{5},\,1,\,x)} = \sqrt{n}$$

is given by,

$$x \approx 25\sqrt{5}\left(\frac{1+\sqrt{5}}{2}\right)^\sqrt{5} \exp\left(\frac{-\,\pi\sqrt{n}}{\sin(\pi/5)}\right)$$

(One can numerically solve for x for a given n using Mathematica's FindRoot command.)

Question: Is the conjecture true? And what is the closed-form for the constant $1152.79509\dots$ when $m=7$?

EDIT:

Courtesy of Sasha’s answer below, then the closed-form for m = 7, as radicals raised to radical powers is,

$$ (14)^2 \prod_{k=1}^{3} \frac{1}{\sin(\pi k/7)^{4\cos(2\pi k/7)}} = 1152.79509\dots$$

In general,

$$\lim_{n\to\infty}\frac{x}{q} = (2m)^2 \prod_{k=1}^{\lfloor (m-1)/2 \rfloor} \frac{1}{\sin(\pi k/m)^{4\cos(2\pi k/m)}} $$

Answer 1

The hypergeometric function ${}_2F_1\left(\frac{1}{m}, 1-\frac{1}{m} ; 1; y\right)$ is an increasing from of $y$, starting at 1 for $y=0$ and increasing boundlessly as $y$ approaches one.

Thus, for large $n$ we should expect $x$ to be small. Taking the first to terms of the series expansion at unity: $$ {}_2F_1\left(\frac{1}{m}, 1-\frac{1}{m} ; 1; 1 - x \right) = -\frac{1}{\pi} \sin\left(\frac{\pi}{m} \right) \left( \log x + 2 \left( \gamma + \psi\left(\frac{1}{m} \right) \right) \right) - \cos\left(\frac{\pi}{m} \right) + \mathcal{O}(x) $$ which coincides with the expansion of the ratio of hypergeometric functions. $$ \frac{ {}_2F_1\left(\frac{1}{m}, 1-\frac{1}{m} ; 1; 1 - x \right) }{ {}_2F_1\left(\frac{1}{m}, 1-\frac{1}{m} ; 1; x \right)} = -\frac{1}{\pi} \sin\left(\frac{\pi}{m} \right) \left( \log x + 2 \left( \gamma + \psi\left(\frac{1}{m} \right) \right) \right) - \cos\left(\frac{\pi}{m} \right) + \mathcal{O}(x) $$ This gives a large $n$ approximation of the root: $$ x_n = \exp\left( -\frac{\pi}{\sin\left(\frac{\pi}{m}\right)} \left( \sqrt{n} + \cos\left(\frac{\pi}{m}\right) \right) - 2 \left( \gamma + \psi\left(\frac{1}{m}\right) \right) \right) $$ where $\gamma$ denotes the Euler-Mascheroni constant. From here, denoting $q_n = \exp\left( -\frac{\pi \sqrt{n} }{\sin\left(\frac{\pi}{m}\right)} \right)$ $$ \lim_{n \to \infty} \frac{x_n}{q_n} = \exp\left( -\frac{\pi}{\tan\left(\frac{\pi}{m}\right)} - 2 \left( \gamma + \psi\left(\frac{1}{m}\right) \right) \right) $$

The expression for $m=5$ agrees numerically with your expression, implying $$ \psi\left(\frac{1}{5}\right) = -\gamma - \frac{1}{2} \left( \pi \sqrt{1 + \frac{2}{\sqrt{5}}} + \ln\left( 25 \sqrt{5} \left(\frac{1+\sqrt{5}}{2} \right)^{\sqrt{5}} \right) \right) \tag{1} $$ I checked this identity using Mathematica, and found it to agree to 50,000 significant decimal points:

In[136]:= N[(-(Log[25*Sqrt[5]*((1 + Sqrt[5])/2)^Sqrt[5]] + 
               Sqrt[(1/5)*(5 + 2*Sqrt[5])]*Pi))*(1/2) - EulerGamma, 
  50000] - N[PolyGamma[1/5], 50000]

Out[136]= 0``49998.97559316131

I wonder if this identity is a known one. This identity is the Gauss's digamma theorem for $k=5$ and $m=1$ (thanks Edgar): $$ \psi\left(\frac{m}{k}\right) = -\gamma -\ln(2k) -\frac{\pi}{2}\cot\left(\frac{m\pi}{k}\right) +2\sum_{n=1}^{\lfloor (k-1)/2\rfloor} \cos\left(\frac{2\pi nm}{k} \right) \ln\left(\sin\left(\frac{n\pi}{k}\right)\right) $$