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I got the number $$\frac{\Gamma\left(\frac{1}{5}\right)\Gamma\left(\frac{4}{15}\right)}{\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{15}\right)}=0.824326275998351470388591998726842...$$ in the process of some quite long calculations that are not pertinent to my question.

Question 1: Is this number a rational, algebraic irrational or transcendental?

I performed some numeric calculations and they did not find a close match among roots of polynomials of degree less than $100$ and integer coefficients of absolute value less than $10^{12}$.

Some people I've asked this question told me that it is very likely unknown and probably will be unknown (in the rigorous sense) for a long time. But many of them were ready to bet that this number is transcendental. Their reasoning was that there are $2^{\aleph_0}$ transcendental numbers and only $\aleph_0$ algebraics. So, if the number with a simple definition is not explicitly constructed to be algebraic, and simple tests do not suggest that it is algebraic, then this just would be highly unlikely coincidence for it to be algebraic. And mathematical intuition and common sense says that there are no such coincidences. Actually, I do not hope very much to get the answer to my first question, and would like to answer another (more philosophical) one:

Question 2: Is this common sense reasoning valid? What is the role of such intuition in mathematics?

Piotr Shatalin
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    WolframAlpha says "unknown": [Is (Gamma(1/5) Gamma(4/15))/(Gamma(1/3) Gamma(2/15)) an algebraic number?](http://www.wolframalpha.com/input/?i=Is+%28Gamma%281%2F5%29+Gamma%284%2F15%29%29%2F%28Gamma%281%2F3%29+Gamma%282%2F15%29%29+an+algebraic+number%3F) – Zakharia Stanley May 09 '13 at 02:56
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    I would love to see that polynomial if it *is* algebraic! – Liu Jin Tsai May 09 '13 at 03:05
  • By Cantor's theorems, there are uncountable trascendental numbers and countable algebraic numbers. To betting types this says that they should bet that any random number grabbed out of thin air is trascendental. Is this valid? Hell, no! – vonbrand May 09 '13 at 03:05
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    @Liu You can see it now! – Piotr Shatalin May 09 '13 at 04:47

1 Answers1

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I'll lead off with the bombshell, and follow with some notes.

Your constant is algebraic.

Indeed, it satisfies the (irreducible) degree $120$ equation:

$$729 + 914166000 x^{30} + 3529576586250 x^{60} - 1259674334325000 x^{90} + 3125 x^{120}.$$

It is also expressible via radicals:

$$\frac{\sqrt{2}\cdot 3^{1/20}}{5^{1/6} {\left(5-\frac{7}{\sqrt{5}}+\sqrt{6-\frac{6}{\sqrt{5}}}\right)^{1/4}}}.$$

All of this follows from a recent paper, Expessions for Values of the Gamma Function, in which expressions for $\Gamma(m/n)$ with $n$ either dividing $24$ or $60$ are given in terms of some algebraic constants, $\pi$, and the following ten $\Gamma$-values:

$$\Gamma\left(\frac{1}{3}\right),\Gamma\left(\frac{1}{4}\right),\Gamma\left(\frac{1}{5}\right),\Gamma\left(\frac{2}{5}\right),\Gamma\left(\frac{1}{8}\right),\Gamma\left(\frac{1}{15}\right),\Gamma\left(\frac{1}{20}\right),\Gamma\left(\frac{1}{24}\right),\Gamma\left(\frac{1}{60}\right),\Gamma\left(\frac{7}{60}\right).$$

It is conjectured (Lang) that these constants are algebraically independent over $\mathbb{Q}(\pi)$.

awwalker
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