Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The idea is a more convenient form for $N = 0.01001000100001000001...$ in base $r$, hopefully to show whether it is transcendental.

Sorry for brevity.

share|cite|improve this question
We can factor out an $r^c$ term from the sum. For special values of $a$ and $b$, the sum is expressible in terms of Jacobi theta functions, but I know of no closed form for the general form you have. – J. M. Jul 31 '11 at 1:29
let's say $r=2^{-1/2}$, $a=1$, $b=5$, $c=4$, which I think makes the series equal to $N$ in binary. Also, thanks for fixing my TeX. – jon Jul 31 '11 at 1:47
The $b=5$ portion would be troublesome in figuring a closed form for what you say you have... – J. M. Jul 31 '11 at 2:47
@J.M. I tried to complete the squares hoping to kill the linear term, but it didn't work because $5$ "happened" to be an odd number. Did you anticipate this in your comment? Also, if I gave you just $an^2+c$, then do closed-form solutions exist? – Srivatsan Jul 31 '11 at 3:01
Yeah, something like that @Srivatsan. ;) For your much simpler case, you have a representation in terms of the third Jacobi theta function: $\frac{r^c}{2}(1+\vartheta_3(0,r^a))$ – J. M. Jul 31 '11 at 3:04

Of course $N$ is transcendental. [I said it is, I didn't say I can prove it.] It is conjectured that all irrational algebraic numbers are normal in all bases. If this were not transcendental, it would be a spectacular counterexample to that conjecture.

share|cite|improve this answer

This may come under the heading of "Siegel E-functions" or "Siegel G-functions", for which transcendence results are known.

share|cite|improve this answer

Here's a technique for this sort of sum (though this doesn't address the transcendentality question) I found a while ago which involves incomplete theta functions:

\begin{equation} \sum_{n=0}^{\infty} q^{an^{2}+2akn+ak^{2}+p} = q^{p}\left[\frac{\theta_{3}(0,q^{a})+1}{2} - \sum_{m=0}^{k-1} q^{am^{2}}\right] \end{equation}

for example:

\begin{equation} \sum_{n=0}^{\infty} \frac{1}{e^{7n^{2}+70n+173}} = e^{2}\left[\frac{\theta_{3}(0,1/e^{7})+1}{2} - \sum_{m=0}^{4} (1/e^{7})^{m^2}\right] \end{equation}

I'm not sure it works for arbitrary $a$, $b$, and $c$, though.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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