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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?

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I would love to see that polynomial if it is algebraic! –  Liu Jin Tsai May 9 '13 at 3:05
    
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 9 '13 at 3:05
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@Liu You can see it now! –  Piotr Shatalin May 9 '13 at 4:47

1 Answer 1

up vote 46 down vote accepted

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 $\mathbb{Q}(\pi)$.

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It is really amazing! The polynomial seems to be a tiny bit beyond a reasonably large search area chosen by Piotr. –  Vladimir Reshetnikov May 9 '13 at 4:24
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If only Piotr looked at $30$th powers! –  A Walker May 9 '13 at 4:25
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This result also follows from formulae (76), (77) listed in mathworld.wolfram.com/GammaFunction.html - divide (76) by (77) and take a square root. –  Vladimir Reshetnikov May 9 '13 at 4:37
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Amazing answer! I am astonished... –  Piotr Shatalin May 9 '13 at 4:45
    
Brilliant answer! This is what I love about MSE. –  Clive Newstead May 9 '13 at 23:50

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