1
$\begingroup$

I have asked the following question on Math.SE some time ago and offered a bounty, yet received no answers nor comments, so I'm posting it here.


The Prouhet-Thue-Morse constant, defined as

$$ \tau =\sum _{{i=0}}^{{\infty }}{\frac {t_{i}}{2^{{i+1}}}}=0.412454033640\ldots $$

where the $t_i$ are elements of the Thue-Morse sequence, is transcendental. But is

$$ \tau_b =\sum _{{i=0}}^{{\infty }}{\frac {t_{i}}{b^{{i+1}}}} $$

also transcendental, for $b>2$?

$\endgroup$
3
  • 2
    $\begingroup$ Did you try to adapt the proof for the transcendence of $\tau$ to $\tau_b$? $\endgroup$ Mar 29, 2019 at 10:13
  • $\begingroup$ No, as I was hoping there were already results of that nature available $\endgroup$
    – Klangen
    Mar 29, 2019 at 10:19
  • 1
    $\begingroup$ I believe there are papers of Allouche that address this. I will try to check later. $\endgroup$ Mar 29, 2019 at 13:53

1 Answer 1

3
$\begingroup$

Michel Waldschmidt, Words and transcendence, writes in Section 3.1, on page 461, "Mahler also proved in 1929 that the so-called Prouhet–Thue–Morse–Mahler number in base $g\ge2$, given by $$\xi_g=\sum_{n\ge0}{a_n\over g^n}$$ where $(a_n)_{n\ge􏰀0}$ is the Prouhet–Thue–Morse sequence, is transcendental; see [52] and [15, Section 13.4]." [15] is Allouche and Shallit, Automatic sequences: Theory, Applications, Generalizations (Cambridge University Press, 2003). [52] is K. Nishioka, Mahler Functions and Transcendence (Lecture Notes in Mathematics 1631, Springer- Verlag, 1996).

Waldschmidt gives "the idea of proof," and refers to [15, Section 13.4] and [52, Example 1.3.1] for the full proof.

$\endgroup$
1
  • $\begingroup$ Fantastic, thanks a lot! $\endgroup$
    – Klangen
    Mar 31, 2019 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.