Persiflage

I had no intention to discuss the Ramanujan Machine again, but over the past few days there has been a flurry of (attempted) trollish comments on that post, so after taking a brief look at the latest version, I thought I would offer you my updates. (I promise for the last time.)

Probably the nicest thing I have to say about the updated paper is that it is better than the original. My complaints about the tone of the paper remain the same, but I don’t think it is necessary for me to revisit them here.

Concerning the intellectual merit, I think it is worth making the following remarks. First, I am only address the contributions to mathematics, Second, what counts as a new conjecture is not really as obvious as it sounds. Since continued fractions are somewhat recherché, it might be more helpful to give an analogy with infinite series. Suppose I claimed it was a new result that

\( \displaystyle{ 2G = \sum_{n=0}^{\infty} a_n = 1 + \frac{1}{2} + \frac{5}{36} + \frac{5}{72} + \frac{269}{3600} – \frac{1219}{705600} + \ldots } \)

where for \(n \ge 4\) one has

\(2 n^2 a_n = n^2 a_{n-1} – 2 (n-2)^2 a_{n-2} + (n-2)^2 a_{n-3}.\)

How can you evaluate this claim? Quite probably this is the first time this result has been written down, and you will not find it anywhere in the literature. But it turns out that

\( \displaystyle{ \left(\sum_{n=0}^{\infty} \frac{x^n}{2^n} \right) \times \left(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)^2} \right)
= \sum_{n=0}^{\infty} a_n x^n}\)

and letting \(x=1\) recovers the identity above and immediately explains how to prove it. To a mathematician, it is clear that the proof explains not only why the originally identity is true, but also why it is not at all interesting. It arises as more or less a formal manipulation of a definition, with a few minor things thrown in like the sum of a geometric series and facts about which functions satisfy certain types of ordinary differential equations. The point is that the identities produced by the Ramanujan Machine have all been of this type. That is, upon further scrutiny, they have not yet revealed any new mathematical insights, even if any particular example, depending on what you know, may be more or less tricky to compute.

What then about the improved irrationality measures for the Catalan constant? I think that is a polite way of describing a failed attempt to prove that Catalan’s constant was irrational. It’s something that would be only marginally publishable in a mathematics journal even with a proof. Results about the irrationality measure in the range where they are irrational have genuine implications about the arithmetic of the relevant numbers, but these results do not.

What then about the new continued fractions developed over the last year — maybe these are now deeper? Here you have to remember that continued fractions, especially of the kind considered in this paper, are more or less equivalent to questions about certain types of ordinary differential equations and their related periods. (But importantly, not conversely: most of these interesting ODEs have nothing to do with continued fractions since they are associated with recurrences of length greater than two.) For your sake, dear reader, I voluntarily chose to give up an hour or two of my life and took a closer look at one of their “new conjectures.” I deliberately chose one that they specifically highlighted in their paper, namely:

Where \(G\) here is Catalan’s constant \(L(2,\chi_4)\). As you might find unsurprising, once you start to unravel what is going on you find that, just as in the example above, the mystery of these numbers goes away. This example can be generalized in a number of ways without much change to the argument. Let \(p_0=1\) and \(q_0 = 0\), and otherwise let

\(\displaystyle{\frac{p_n}{q_n} = \frac{3}{1}, \frac{33}{13}, \frac{765}{313}, \frac{30105}{12453}, \frac{1790775}{743403}, \ldots} \)

denote the (non-reduced) partial fraction convergents. If

\( \displaystyle{ P(z) = \sum \frac{4^n p_n z^n}{n!^2} = 1 + 12z + 132 z^2 + \ldots
\quad Q(z) = \sum \frac{4^n q_n z^n}{n!^2} = 4z +52 z^2 + \ldots} \)

Then, completely formally, \(DP(z) = 0\) where

\( \displaystyle{ D = z(8z-1)(4z-1) \frac{d^2}{dz^2} + (160 z^2 – 40 z + 1) \frac{d}{dz} + 12(8z – 1)}\)

and \(DQ(z) = 4\). If \(K\) and \(E\) denote the standard elliptic functions, one observes that \(P(z)\) is nothing but the hypergeometric function

But now one is more or less done! The argument is easily finished with a little help from mathematica. Another solution to \(DF(z) = 0\) is of course

\( \displaystyle{ R(z) = \frac{ 2 E((1-8z)^2) -2 K((1-8z)^2) }{(1 – 8z)^2} = \log(z) + 2 + \ldots } \)

and knowing both homogenous solutions allows one to write \(Q(z) = u(z) P(z) + v(z) R(z)\) and then easily compute that

\(\displaystyle{ \lim_{n \rightarrow \infty} \frac{p_n}{q_n}
= \lim_{z \rightarrow 1/8} \frac{P(z)}{Q(z)} = \frac{2}{-1 + 2G}.}\)

as desired. For those playing at home, note that a convenient choice of \(u(z)\) and \(v(z)\) can be given by

\( \displaystyle{ v(z) = \int \frac{ E(16 z(1-4z))}{\pi} = 4 z – 8 z^2 + \ldots }\)