Everybody knows that $\sum_{k=0}^\infty{\frac{1}{2^{2^k}}}$ is transcendental. Is number $\sum_{k=0}^\infty{\frac{1}{2^{k^2}}}$ algebraic or not?


marked as duplicate by Ian Morris, Wolfgang, Stefan Waldmann, Stefan Kohl, Jeremy Rickard Sep 22 '16 at 9:57

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    $\begingroup$ I asked my mother if she knows that $\sum_{k=0}^\infty{\frac{1}{2^{2^k}}}$ is transcendental. She didn't know. You lied. $\endgroup$ – Billy Rubina Sep 22 '16 at 3:46
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    $\begingroup$ may be she lied $\endgroup$ – userded Sep 23 '16 at 11:43

Yes, it is, and even the three numbers $\sum_{k \geq 0}{2^{-k^2}}$, $\sum_{k \geq 1}{k^22^{-k^2}}$ and $\sum_{k \geq 1}{k^42^{-k^2}}$ are algebraically independent. This results from algebraic independence results for theta functions, see Waldschmidt's excellent survey https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/SurveyTrdceEllipt2006.pdf, Corollary 52, and the reference given there: Nesterenko & Philippon, Introduction to algebraic independence theory, Lecture Notes in Mathematics, vol. 1752, Springer-Verlag, Berlin, 2001, Chapter 3.

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    $\begingroup$ I don't know if it is courageous to answer a question like "Is *** algebraic or not" with "yes it is" - mathematically the answer is correct, plus it's a nice Beatles song. $\endgroup$ – Franz Lemmermeyer Sep 21 '16 at 22:14

I can't classify this question as a duplicate, but Is this number already known to be transcendental? Is there a survey about up-to-date trascendence results?

asks the same thing.

Introduction to algebraic Independence Theory, this number is transcendental.


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