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V. Reshetnikov gave the remarkable integral, $$\int_0^1\frac{dx}{\sqrt[3]x\,\sqrt[6]{1-x}\,\sqrt{1-x\left(\sqrt{6}\sqrt{12+7\sqrt3}-3\sqrt3-6\right)^2}}=\frac\pi9(3+\sqrt2\sqrt[4]{27})\tag1$$ More generally, given some integer/rational $N$, we are to find an algebraic number $\alpha$ that solves,

$$\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{3}\,|\alpha|}\tag2$$

and absolute value $|\alpha|$. (Compare to the similar integral in this post.) Equivalently, to find $\alpha$ such that,

$$\begin{aligned} \frac{1}{N} &=I\left(\alpha^2;\ \tfrac12,\tfrac13\right)\\[1.8mm] &= \frac{B\left(\alpha^2;\ \tfrac12,\tfrac13\right)}{B\left(\tfrac12,\tfrac13\right)}\\ &=B\left(\alpha^2;\ \tfrac12,\tfrac13\right)\frac{\Gamma\left(\frac56\right)}{\sqrt{\pi}\,\Gamma\left(\frac13\right)}\end{aligned} \tag3$$

with beta function $\beta(a,b)$, incomplete beta $\beta(z;a,b)$ and regularized beta $I(z;a,b)$. Solutions $\alpha$ for $N=2,3,4,5,7$ are known. Let, $$\alpha=\frac{-3^{1/2}+v^{1/2}}{3^{-1/2}+v^{1/2}}\tag4$$ Then, $$ - 3 + 6 v + v^2 = 0, \quad N = 2\\ - 3 + 27 v - 33v^2 + v^3 = 0, \quad N = 3\\ 3^2 - 150 v^2 + 120 v^3 + 5 v^4 = 0, \quad N = 5\\ - 3^3 - 54 v + 1719 v^2 - 3492v^3 - 957 v^4 + 186 v^5 + v^6 = 0, \quad N = 7$$

and (added later),

$$3^4 - 648 v + 1836 v^2 + 1512 v^3 - 13770 v^4 + 12168 v^5 - 7476 v^6 + 408 v^7 + v^8 = 0,\quad N=4$$

using the largest positive root, respectively. The example was just $N=2$, while $N=4$ leads to,

$$I\left(\tfrac{1-\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\right)=\tfrac{3}{8},\quad\quad I\left(\tfrac{1+\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\right)=\tfrac{5}{8}$$

I found these using Mathematica's FindRoot command, and some hints from Reshetnikov's and other's works, but as much as I tried, I couldn't find prime $N=11$.

Q: Is it true one can find algebraic number $\alpha$ for all $N$? What is it for $N=11$?

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Tito Piezas III
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  • Do you know about the RootApproximant function? It tries to find the simplest algebraic number matching a numeric approximation. It likes to be given quite a lot of correct decimal digits, though. From my experience, if the degree $>8$, then you need to provide $>200$ digits. – Vladimir Reshetnikov Dec 01 '16 at 00:14
  • I used RootApproximant[N[InverseBetaRegularized[1/7, 1/2, 1/3], 250]] to find my conjectures. Had no luck with 1/11. I got results for dozens of other fractions though. – Vladimir Reshetnikov Dec 01 '16 at 00:41
  • @VladimirReshetnikov: These "dozens" of fractions, did you get unit fractions other than $p=2,3,5,7$? – Tito Piezas III Dec 01 '16 at 02:35
  • With InverseBetaRegularized[1/n, 1/2, 1/3] -- no. But InverseBetaRegularized[1/6,1/3,1/3], InverseBetaRegularized[1/4,1/4,1/2] and others seem to be algebraic. No fraction with a denominator $>7$ produced an apparently algebraic number. – Vladimir Reshetnikov Dec 01 '16 at 03:11
  • @VladimirReshetnikov: There seems to be a denominator $>7$ that yields an algebraic number, namely $InverseBetaRegularized[\color{blue}{1/8},1/4,1/2]$ which is a palindromic 32-deg eqn with discriminant $d=2^{2064}$. – Tito Piezas III Dec 01 '16 at 03:49
  • @VladimirReshetnikov: InverseBetaRegularized[1/8,1/3,1/3] also seems to be an algebraic number, deg-16. – Tito Piezas III Dec 01 '16 at 06:07
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    Hasn't achille hui in his answer already showed that $\alpha$ is algebraic for all rational $N$? In his notations we have $\sqrt[3]{1-\alpha^2}=\frac{4}{\eta^2}\wp\left(i\frac{\sqrt{3}}{2N}\right)$. We also know that $\wp\left(i\frac{\sqrt{3}}{2N}\right)=\wp\left(\frac{2\omega_2-\omega_1}{N}\right)$. Now addition theorem for $\wp$ shows that $\wp\left(\frac{2\omega_2-\omega_1}{N}\right)/\eta^2$ is algebraic for all rational $N$, which means that $\alpha$ is also algebraic. – Martin Nicholson Dec 01 '16 at 06:34
  • @Nemo: I guess I missed that part. However, half of the question is to find the explicit minimal polynomial of the tricky $N=11$ and perhaps provide an explanation of its presumably high degree. – Tito Piezas III Dec 01 '16 at 06:53
  • @TitoPiezasIII Can you give an explicit polynomial for InverseBetaRegularized[1/8, 1/4, 1/2]? – Vladimir Reshetnikov Dec 01 '16 at 19:56
  • I found an explicit form for InverseBetaRegularized[1/8, 1/3, 1/3]: http://mathb.in/109745 – Vladimir Reshetnikov Dec 02 '16 at 00:20
  • It appears that InverseBetaRegularized[1/11, 1/3, 1/3] is $\left(1-\sqrt{1-4\alpha}\right)/2,$ where $\alpha$ is the root of the polynomial $\small 11 \alpha ^{20}-23221 \alpha ^{19}-1153603 \alpha ^{18}+62045313 \alpha ^{17}+66133914 \alpha ^{16}+1596123771 \alpha ^{15}-8579472693 \alpha ^{14}+4760052033 \alpha ^{13}-22319781 \alpha ^{12}-8054721004 \alpha ^{11}+10595519759 \alpha ^{10}-4869514969 \alpha ^9+1106263389 \alpha ^8-189881835 \alpha ^7+59389374 \alpha ^6-17393277 \alpha ^5+2270301 \alpha ^4-102729 \alpha ^3+605 \alpha ^2+242 \alpha -1$ in the interval $0<\alpha<1/6$. – Vladimir Reshetnikov Dec 02 '16 at 01:01
  • It also appears that InverseBetaRegularized[1/11, 1/2, 1/3] is $\alpha^2,$ where $\alpha$ is an algebraic number of degree $40$ whose minimal polynomial and isolating interval are given here. – Vladimir Reshetnikov Dec 02 '16 at 03:13
  • @VladimirReshetnikov: Beautiful! Would you like to put that as an answer? – Tito Piezas III Dec 02 '16 at 03:26
  • Oh, I have no proof yet. – Vladimir Reshetnikov Dec 02 '16 at 03:27
  • @VladimirReshetnikov: I tested your deg 40 and it involves the solvable quintic $\small 16671744 - 36246528 x - 12153024 x^2 - 198432 x^3 + 2808 x^4 + x^5=0$ which is cyclic and whose discriminant is divisible by $11^4\times23^4$. I guess $\alpha$, in addition to being algebraic numbers (by achille), are in fact also radicals hence their minpoly has a solvable Galois group. – Tito Piezas III Dec 02 '16 at 03:41
  • Let us continue this discussion in chat. – Vladimir Reshetnikov Dec 02 '16 at 03:42
  • @VladimirReshetnikov: I was about to post the explicit polynomial for InverseBetaRegularized[1/8, 1/4, 1/2] when I noticed an interesting pattern. Kindly see this post. – Tito Piezas III Dec 02 '16 at 14:06

2 Answers2

2

(Too long for a comment. And courtesy of V. Reshetnikov's result here, though as he points out it is tentative.)

The algebraic number $\alpha$ that solves,

$$\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}=\frac{1}{11}\,\frac{2\pi}{\sqrt{3}\;\alpha}$$

seems to have a $40$-deg minpoly. However, it turns out we can also reduce its degree and express it using the common form above. Let,

$$\alpha=\frac{3^{1/2}-v^{1/2}}{3^{-1/2}+v^{1/2}}$$

where $v$ is a the second largest positive root ($r_9$ in Mathematica syntax) of,

$$\small P(v)=-3^{10} + 23816430 v^2 - 323903448 v^3 + 2177615583 v^4 - 9297934272 v^5 + 25869358152 v^6 - 37475802144 v^7 - 16459141842 v^8 + 180065426112 v^9 - 338100745356 v^{10} + 329418595440 v^{11} - 211367836746 v^{12} + 102243404736 v^{13} - 8162926200 v^{14} - 9999738144 v^{15} + 1006439643 v^{16} - 134177472 v^{17} - 2246706 v^{18} + 30888 v^{19} + 11 v^{20} = 0$$

Also,

$$I\big(\alpha^2;\tfrac{1}{2},\tfrac{1}{3}\big)=\tfrac{1}{5}\,I\big(\tfrac{1-\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\big)=\tfrac{1}{6}\,I\big(\tfrac{1+\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\big)=\frac{1}{11}$$

with regularized beta function $I(z;a,b)$. Furthermore, if

$$y =\frac{r_1+r_9+r_{13}+r_{14}}{12}$$

then $y$ is a root of the solvable quintic,

$$67 - 1748 y - 7033 y^2 - 1378 y^3 + 234 y^4 + y^5=0$$

with discriminant divisible by $11^4\times23^4$. Using the other quartic symmetric polynomials show that the $20$-deg is just a quartic in disguise, hence is solvable. All these suggest that $P(v)$ is the correct polynomial for $N=11$.

Tito Piezas III
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2

I. Duplication

Following Nemo's lead in this answer, we find the formula, $$\frac{1}{2}I(p^2;\tfrac{1}{2},\tfrac{1}{3})=I(1+q^3;\tfrac{1}{2},\tfrac{1}{3})$$ where $p,q$ are related by the $12$-deg, $$p^2(-2 + 2 q + q^2)^6 = 36(1 + q^3) (4 + 4 q + 6 q^2 - 2 q^3 + q^4)^2$$ This then enables us to find infinitely many $\displaystyle\frac{1}{2^n N}$.

For example, since $I(p^2;\tfrac{1}{2},\tfrac{1}{3})=\frac{1}{3}$ is known, then solving for $I(\alpha^2;\tfrac{1}{2},\tfrac{1}{3})=\frac{1}{6}$ turns out to involve a $36$-deg equation.

II. Triplication

(Courtesy of Nemo.) Starting with, $$B\left(z;\frac{1}{2},\frac{1}{3}\right)=2 \sqrt{z} \, _2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};z\right). $$ The transformation $$ \, _2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};-\frac{3 z \left(1-\frac{z}{9}\right)^2}{(1-z)^2}\right)=\frac{(1-z) \, }{1-\frac{z}{9}}{}_2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};z\right) $$ applied two times gives $$ \frac{1}{3} B\left({\frac{(9-z)^2 z \left(z^3+225 z^2-405 z+243\right)^2}{729 (1-z)^2 (z+3)^6}};\frac{1}{2},\frac{1}{3}\right)=B\left(z;\frac{1}{2},\frac{1}{3}\right). $$

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