Recently I asked a question about a possible transcendence of the number $\Gamma\left(\frac{1}{5}\right)\Gamma\left(\frac{4}{15}\right)/\left(\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{15}\right)\right)$, which, to my big surprise, turned out to be an algebraic number, but not some decent algebraic number like $\left(\sqrt{5}-1\right)/2$, but an enormous one with the minimal polynomial of degree 120 and a coefficient exceeding $10^{15}$.

So, my question: are there other interesting examples of numbers occurred in some math problems that were expected likely to be transcendental, but later unexpectedly were proven to be algebraic with a huge minimal polynomial.

Piotr Shatalin
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5 Answers5


I don't know if Conway's constant is quite what you are looking for, as I'm not sure one would expect it initially to be transcendental or not. So, perhaps it's my bad intuition, but I was certainly surprised to learn that it is an algebraic number with minimal polynomial of degree 71.

Ittay Weiss
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Sometimes you can get unexpected algebraic values while working with hypergeometric functions. For example, the following absolute value of a complex-valued $_4F_3$ function: $$\left|\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\sqrt{\phi }\right)\right|,$$ where $\phi$ is the golden ratio, is actually an algebraic number with the minimal polynomial of degree 80 and a coefficient exceeding $10^{55}$: $$340282366920938463463374607431768211456 x^{80}+152118072027387528179604384645120000000000 x^{72}+202824096036516704239472512860160000000000 x^{70}-45334718235548594051481600000000000000000000 x^{62}+15111572745182864683827200000000000000000000 x^{60}-5629499534213120000000000000000000000000000000 x^{56}-24769797950537728000000000000000000000000000000 x^{54}-33776997205278720000000000000000000000000000000 x^{52}-9007199254740992000000000000000000000000000000 x^{50}+1006632960000000000000000000000000000000000000000 x^{46}+3523215360000000000000000000000000000000000000000 x^{44}+74161139200000000000000000000000000000000000000000 x^{40}+300000000000000000000000000000000000000000000000000 x^{38}+675000000000000000000000000000000000000000000000000 x^{36}+1050000000000000000000000000000000000000000000000000 x^{34}-2975290298461914062500000000000000000000000000000000 x^{32}-14701161193847656250000000000000000000000000000000000 x^{30}-37252902984619140625000000000000000000000000000000000 x^{28}-74505805969238281250000000000000000000000000000000000 x^{26}-59604644775390625000000000000000000000000000000000000 x^{24}-22351741790771484375000000000000000000000000000000000 x^{22}+7450580596923828125000000000000000000000000000000000 x^{20}-555111512312578270211815834045410156250000000000000000 x^{16}-1110223024625156540423631668090820312500000000000000000 x^{14}-1665334536937734810635447502136230468750000000000000000 x^{12}-2220446049250313080847263336181640625000000000000000000 x^{10}+82718061255302767487140869206996285356581211090087890625$$

I'm not sure if it is expressible in radicals.

Vladimir Reshetnikov
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    Just an everyday polynomial, then. Looks simple enough.. – Thomas May 10 '13 at 03:55
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    Yes, it would be much more interesting if a number with a pretty simple definition turned out to be unexpectedly algebraic, but with a minimal polynomial so large that we could not explicitly write it (e.g. with degree and coefficients exceeding [TREE(3)](http://mathoverflow.net/questions/93828/how-large-is-tree3)) – Vladimir Reshetnikov May 10 '13 at 20:55
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    @VladimirReshetnikov: It is not a radical. Kindly see [this post](https://math.stackexchange.com/questions/2376309/is-left-4f-3-left-frac15-frac25-frac35-frac45-frac1). – Tito Piezas III Jul 30 '17 at 14:47

Some numbers were found recently algebraic using the LLL and PSLQ algorithms : these algorithms return directly the integer coefficients of a polynomial starting with the numerical value provided with enough precision. They are implemented in many Computer algebra software (for example lindep and algdep of pari/gp).

Broadhurst found that the third and fourth bifurcation points (B3 and B4) of the logistic map were algebraic of degree $12$ and $240$ (page 3 from this paper and 5 from this one and this ps file for B4).

See too this MO thread 'What Are Some Naturally-Occurring High-Degree Polynomials?'.

Raymond Manzoni
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  • Unless I'm missing something, the bifurcation points of a rational function being algebraic (and of high degree) doesn't seem unexpected. My (admittedly naive) take on Broadhurst's paper is that the challenge was in identifying the specific polynomial. – Erick Wong May 10 '13 at 00:36
  • @ErickWong: Yes you are right (even if I found these polynomials pleasant) and it would be more surprising if, for example, one of the Feigenbaum constants was proved algebraic... Anyway perhaps that the [integer relation algorithms](http://en.wikipedia.org/wiki/Integer_relation_algorithm) and the MO thread will be of interest. – Raymond Manzoni May 10 '13 at 08:17

I do not know whether it was expected that this number be transcendental, but it was shown by Firsching in 2014 (so, since this question was asked) that the side length of the largest regular tetrahedon that can be embedded in an icosahedron of unit side length is equal to the root $$1.347442850\!\ldots$$ of the irreducible degree $32$ (even) polynomial

\begin{multline} 5041 t^{32} - 1318386 t^{30} + 60348584 t^{28} - 924552262 t^{26} + 5246771058 t^{24}-15736320636 t^{22} \\ + 29448527368 t^{20} - 37805732980 t^{18} + 35173457839 t^{16}-24298372458 t^{14} \\ + 12495147544 t^{12} - 4717349124 t^{10} + 1256858478 t^8 \\- 217962112 t^6 + 21904868 t^4 - 1536272 t^2 + 160801 \end{multline}

Firsching, M. Computing maximal copies of polytopes contained in a polytope, arXiv:1407.0683.

The analogous number for embedding a dodecahedron in a tetrahedron is an algebraic number of degree $16$. (For the remaining $23$ ordered pairs of the Platonic solids the analogous number is algebraic of degree $\leq 4$.)

Travis Willse
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The areas of cyclic polygons is a great place to look for this sort of number. Given the side lengths of a convex cyclic pentagon, you can find the area independent of how the sides are ordered. But as Robbins found, you have to solve a seventh degree equation which in general cannot be expected to be reducible. Succeed, and you still have to take the square root of your solution. The seventh degree equation can be extended to cover cyclic hexagons, but beyond that you need degree 38 ... or more.

Oscar Lanzi
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