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Recall that $(a;\,q)_\infty$ is the $q$-Pochhammer symbol: $$(a;\,q)_\infty=\prod_{n=0}^\infty(1-a \, q^n).\tag1$$ Its important special case $(q;\,q)_\infty=\prod_{n=1}^\infty(1-q^n)$ is sometimes called the Euler function. It appears in Euler's pentagonal number theorem, and its reciprocal $(q;\,q)_\infty^{-1}$ is the generating function for the partition numbers. It is also related to Jacobi theta functions and Ramanujan theta functions.

Let $$f(x) = \frac{(-q;\,-q)_\infty}{(q;\,q)_\infty},\quad\text{where}\,\,q=e^{-\pi\sqrt x}.\tag2$$ In the OEIS entry A080054 there is an empirical observation by Simon Plouffe that apparently $f(1)=\sqrt[8]2$.

Empirical (Simon Plouffe, Feb. 20, 2011): $$\sum_{n=0}^{\infty}e^{-\pi n}a(n) =\sqrt[8]2.$$

I did some numerical experiments related to this observation, and the outcomes suggest a fascinating stronger conjecture:

Conjecture: For every $p\in\mathbb Q,\,p>0$, the value $f(p)$ is an algebraic number.

For example, it appears that $$f(3/5) = \sqrt[8]{2} \cdot \sqrt[4]{9 \sqrt{5}+5 \sqrt{15}-11 \sqrt{3}-19},$$ and $f(13/7)$ is an algebraic number of degree $96$ whose minimal polynomial is $$x^{96}-647442063456 \, x^{88}+16702438371168 \, x^{80}-529345497357824 \, x^{72}+4159684203040512 \, x^{64}-12099397290541056 \, x^{56}+16408771708010496 \, x^{48}-10607690933600256 \, x^{40}+2651923007078400 \, x^{32}-367001600 \, x^{24}+257949696 \, x^{16}-100663296 \, x^8+16777216$$ and an isolating rational interval is $(37/36,\,6/5)$.

Is this conjecture new? Is it known to be true? If not, can you suggest any ideas how to (dis-)prove it?

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  • $\begingroup$ A few special cases of this conjecture appear to be known theorems mentioned on the linked pages. $\endgroup$
    – Vladimir Reshetnikov
    Commented Nov 23, 2016 at 4:48
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    $\begingroup$ It appears that your evaluations are upside down. For example, $\frac1{f(1)}=2^{1/8}$. I could be wrong, but check. $\endgroup$
    – T. Amdeberhan
    Commented Nov 23, 2016 at 6:03
  • $\begingroup$ @T.Amdeberhan Thanks. Yes, my $f(x)$ was upside down. Fixed. $\endgroup$
    – Vladimir Reshetnikov
    Commented Nov 23, 2016 at 17:24
  • $\begingroup$ Your $f$ can be written as $\prod_{j=1}^{\infty}(1+q^{2j-1})/(1-q^{2j-1})$ so $f(x) =G_x/g_x$ where $g, G$ denote Ramanujan class invariant. They are algebraic if $x$ is positive rational and Ramanujan gave a list of their values for many positive integers $x$. $\endgroup$
    – Paramanand Singh
    Commented Feb 12, 2021 at 11:19

2 Answers 2

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Yes, it is always algebraic, because it is a modular function evaluated at a CM (complex multiplication) point.

"$(q;q)_\infty$" is $q^{-1/24} \eta(\tau)$ where $q = e^{2\pi i \tau}$, so "$(q;q)_\infty / (-q;-q)_\infty$" is a root of unity times $\eta(\tau) \, / \, \eta(\tau+1/2)$, which is modular for some congruence subgroup of ${\rm SL}_2({\bf Z})$, i.e. a rational function on some modular curve $X$. If $q = e^{-\pi \sqrt x}$ then $\tau = (i/2)\sqrt x$ is an imaginary quadratic irrationality, and thus a CM point on $X$. It is known that every CM point is algebraic, whence the value of $\eta(\tau) \, / \, \eta(\tau+1/2)$ at the point is also algebraic.

The CM theory also provides further information about the degree of such algebraic numbers, their Galois group (always solvable), and conjugates (values of the same function at other CM points).

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  • $\begingroup$ @T.Amdeberhan complex multiplication $\endgroup$
    – Fedor Petrov
    Commented Nov 23, 2016 at 9:24
  • $\begingroup$ Yes, sorry -- I only realized later that I should have expanded the acronym "CM" when I first introduced it. I'll do so next. $\endgroup$
    – Noam D. Elkies
    Commented Nov 23, 2016 at 16:48
  • $\begingroup$ Thanks! Could you recommend a book that would explain how to determine which exactly algebraic number corresponds to a given point? $\endgroup$
    – Vladimir Reshetnikov
    Commented Nov 23, 2016 at 17:15
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    $\begingroup$ You're welcome. But asking to "determine which exactly algebraic number corresponds to a given point" is probably askign too much. Knowing the conjugates will make it easier to calculate the minimal polynomial, but -- as with the "Kronecker Jugendtraum" -- I don't think you can expect a recipe for a general answer that you could specialize to formulas like $f(3/5) = \sqrt[8]{2} \cdot \sqrt[4]{9 \sqrt{5}+5 \sqrt{15}-11 \sqrt{3}-19}$; even the degree of the algebraic number will involve the class number of an imaginary quadratic field. $\endgroup$
    – Noam D. Elkies
    Commented Nov 23, 2016 at 19:54
  • $\begingroup$ I found an interesting paper on this topic: Ramanujan’s Class Invariants With Applications to the Values of q–Continued Fractions and Theta Functions $\endgroup$
    – Vladimir Reshetnikov
    Commented Dec 3, 2016 at 4:58
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This is not meant an answer, instead I wish to list some computational values for $f(x)$. Not sure if these are known.

$$f(1)=\sqrt[8]2\,,$$ $$f(2)=\sqrt[16]2\,\sqrt[4]{\cos\frac{\pi}8}\,, \qquad f(4)=\sqrt[16]2\,\sqrt{\cos\frac{\pi}8}\,,$$ $$f(3)=\sqrt[4]{\sec\frac{\pi}{12}}\,, \qquad f(1/3)=\sqrt[4]{\csc\frac{\pi}{12}}\,,$$ $$f(1/2)=\sqrt[4]{1+\sqrt2}\,, \qquad f(1/4)=\sqrt{1+\sqrt2}\,\,.$$

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