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We have, $$\arg z_1 = \frac{k\,\pi}3, \quad z_1 = \left(\tfrac{1+\sqrt{-3}}{2}\right)^k\tag1$$ $$\arg z_2=\frac{k\,\pi}3, \quad z_2 = \left( B\Big(\color{blue}{\tfrac{-5+8\,\sqrt{-11}}{27}};\,\tfrac12,\tfrac13\Big)\right)^k \tag2$$ for $k=1,2,3$ and incomplete beta function $B(z;a,b)$. The second with $k=1$ is by V. Reshetnikov. But that cannot be an isolated result.

Reshetnikov states (without giving details) that $(2)$ is equivalent to, $$\begin{aligned}B\left(\frac19;\,\frac16,\frac13\right) &=\frac{\Gamma\big(\tfrac16\big)\,\Gamma\big(\tfrac13\big)}{2\sqrt{\pi}}\\ &=\frac12\,B\big(\tfrac16,\tfrac13\big) \end{aligned}\tag3$$ Thus, $(3)$ is equivalently, $$I\left(\color{blue}{\frac19};\,\frac16,\frac13\right)=\frac12 \tag4$$ with regularized beta function $I(z;a,b)$. The equation $$I\left(z;\,\frac16,\frac13\right)=\frac1n \tag5$$ is quite easy to solve. For example, the smallest real root of, $$-1 + 99 z - 243 z^2 + 81 z^3= 0\tag6$$ will yield $n=3$.

Q: But how do we get from $\color{blue}{\frac19}$ of $(4)$ to $\color{blue}{\frac{-5+8\,\sqrt{-11}}{27}}$ of $(2)$? (I assume a hypergeometric transformation is involved.)

If the answer can be found, then we can use the appropriate root $z$ of $(6)$ to find another, $$\arg\,B\left( P(z);\,\frac12,\frac13\right)=\beta\,\pi\tag7$$ where $\beta$ is a rational/algebraic number.

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  • $\begingroup$ The exponent in $(2)$ should be inside $\arg$, no? $\endgroup$ – J. M. is a poor mathematician Dec 11 '16 at 9:14
  • $\begingroup$ @J.M.: I rephrased it for clarity. :) $\endgroup$ – Tito Piezas III Dec 11 '16 at 11:48

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