50
$\begingroup$

Does anyone know if this number is algebraic or transcendental, and why?

$$\sum\limits_{n = 1}^\infty {10}^{ - n(n + 1)/2} = 0.1010010001000010000010000001 \ldots $$

$\endgroup$
7
  • 9
    $\begingroup$ Well, it's almost a Liouville number, but the obvious good approximations are not that good ... $\endgroup$ Commented May 2, 2014 at 11:55
  • 1
    $\begingroup$ @AlfredYerger Consider the first three terms of the sum: $0.1 + 0.001 + 0.000001$ The big chains of zeroes for smaller terms are "hidden" by the larger terms to the left :) $\endgroup$ Commented May 5, 2014 at 0:35
  • 1
    $\begingroup$ IIRC this is a challenge problem in Tools of the Trade by Paul Sally. I remember finding some reference a few years ago (after Sally assigned it to me among a list of problems) to this being open. $\endgroup$ Commented May 5, 2014 at 5:49
  • 2
    $\begingroup$ Yes it must be transcendental since it is the root of a geometric sequence which is not a polynomial. Or wait... is that a valid argument? $\endgroup$ Commented Jun 21, 2015 at 22:49
  • 2
    $\begingroup$ Wait, I was being stupid, it's not a geometric sequence even. $\endgroup$ Commented Jun 21, 2015 at 23:09

1 Answer 1

74
+50
$\begingroup$

The number $0.1010010001000010000010000001\ldots$ is transcendental.

Consider following three Jacobi theta series defined by $$\begin{align} \theta_2(q) &= 2q^{1/4}\sum_{n\ge 0} q^{n(n+1)} = 2q^{1/4}\prod_{n=1}^\infty (1-q^{4n})(1 + q^{2n})\\ \theta_3(q) &= \sum_{n\in\mathbb{Z}} q^{n^2} = \prod_{n=1}^\infty (1-q^{2n})(1+ q^{2n-1})^2\\ \theta_4(q) &= \theta_3(-q) = \sum_{n\in\mathbb{Z}} (-1)^n q^{n^2} = \prod_{n=1}^\infty (1-q^{2n})(1- q^{2n-1})^2\\ \end{align} $$ and for any $m \in \mathbb{Z}_{+}$, $k \in \{ 2, 3, 4 \}$, use $\displaystyle D^m\theta_k(q)$ as a shorthand for $\displaystyle \left( q\frac{d}{dq} \right)^m \theta_k(q)$.

Based on Corollary 52 of a survey article Elliptic functions and Transcendence by M. Waldschmidt in 2006,

Let $i, j$ and $k \in \{ 2,3,4 \}$ with $i \ne j$. Let $q \in \mathbb{C}$ satisfy $0 < |q| < 1$. Then each of the two fields $$ \mathbb{Q}( q, \theta_i(q), \theta_j(q), D\theta_k(q)) \quad\text{ and }\quad \mathbb{Q}( q, \theta_k(q), D\theta_k(q), D^2\theta_k(q)) $$ has transcendence degree $\ge 3$ over $\mathbb{Q}$

We know for any non-zero algebraic $q$ with $|q| < 1$, the three $\theta_k(q)$, in particular $\theta_2(q)$ is transcendental. Since

$$\sum_{n=1}^\infty 10^{-n(n+1)/2} = \frac{\sqrt[8]{10}}{2} \theta_2\left(\frac{1}{\sqrt{10}}\right) - 1$$

and using the fact $\frac{1}{\sqrt{10}}$ and $\frac{\sqrt[8]{10}}{2}$ are both algebraic, we find the number at hand is transcendental.

$\endgroup$
5
  • 1
    $\begingroup$ "Take a prime that's not a prime" What's a prime that isn't a prime? 1? -1? $\endgroup$
    – Joao
    Commented Oct 15, 2014 at 6:24
  • 1
    $\begingroup$ @Joao, 1 and -1 are units and not primes. There is a famous composite number which has been known as "************ prime". $\endgroup$ Commented Oct 15, 2014 at 7:17
  • 1
    $\begingroup$ @achillehui Are you giving the solution to "************ prime" yet :)? $\endgroup$
    – Ovi
    Commented Sep 12, 2016 at 3:55
  • 1
    $\begingroup$ @Ovi it is 57. $\endgroup$ Commented Sep 12, 2016 at 4:14
  • 1
    $\begingroup$ Grothendieck prime ($57$) needs $42$ to make it $99$. $\endgroup$
    – Scene
    Commented Dec 26, 2022 at 23:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .