Algebraicity of a generating function and binomial numbers

It is an obvious fact that the sum $$\sum_{n\geq 0} \binom{2n}{n} x^n$$ defines an algebraic function. I am interested in the variation of this sum, namely

$$A(x)=\sum_{n\geq 0} \binom{2n}{n}^2 x^n$$

which is not an algebraic function due to the growth of the coefficients (see Enumerative Combinatorics, Vol. 2 from Stanley, for instance).

My question is the following: can we say something about the algebraicity/transcendence of $$A(x_0)$$, for any real $$x_0$$ in the region of convergence)? If $$x_0$$ is algebraic this is quite obvious I guess, but I do not know if there exists an argument saying something when $$x_0$$ is transcendent.

I assume that this is a very general question, and an answer for any transcendent number should be very difficult. But I would be happy to know if something it can be said for some particular transcendental number ($$\pi$$ or $$e$$, for instance).

Maybe the integral representation

$$A(x)=\sum_{n\geq 0} \binom{2n}{n}^2 x^n=\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{du}{(1-16x\cos^2 u)^{1/2}}$$

could be helpful, and something concerning elliptic integrals can be used here, but I do not know enough on this topic.

1 Answer

It is a straightforward computation to express $$A(x) = \sum_{n=0}^{\infty} \binom{2n}{n}^2x^n$$ in terms of hypergeometric functions, namely $$A(x) = {}_2F_1 (\tfrac12,\tfrac12;1;16x).$$ Then in turn one can express this function in terms of elliptic integrals, $${}_2F_1 (\tfrac12,\tfrac12;1;16x) = \frac{1}{\pi} \int_1^{\infty}\frac{du}{\sqrt{u(u-1)(u-16x)}}. \tag{1}\label{1}$$ The integral on the right is a period of the Legendre elliptic curve $$E_x:y^2 = u(u-1)(u-16x).$$ When $$x$$ is algebraic and $$x \neq 0$$, $$1$$, it is a result of Th. Schneider (1934) that such a period is transcendental over $$\overline{\mathbb{Q}}$$, and it was further proved by Schneider (1937) that the ratio in \eqref{1} is itself transcendental. So if $$x$$ is algebraic and $$0<|x|< \frac{1}{16}$$ (so that $$A(x)$$ converges), the value of $$A(x)$$ is transcendental over $$\overline{\mathbb{Q}}$$.

For an account of the connection between elliptic integrals and hypergeometric functions see Chapter 9 of Husemöller's text "Elliptic Curves." See Waldschmidt's survey article "Elliptic Functions and Transcendence," Dev. Math. 17 (2008) for additional information about connections with transcendence, as well as more recent results in this area.

• that's a very precise answer (with very interesting references)... thanks! Jan 4, 2022 at 18:09