11
$\begingroup$

I want to know if there exists a closed formula for sum $A_n(X)=\sum \limits_{i=0}^n X^{i^2}$.

I have found if $n$ is odd then $(X^n+1)\text{ | } A_n(X)$, but I don't have found a closed formula.

$\endgroup$
6
  • 6
    $\begingroup$ The infinite series is a Jacobi theta function ... $$\frac{\theta_3(0,q)-1}{2} = \sum_{n=1}^\infty q^{n^2}$$ $\endgroup$ Jun 25, 2017 at 12:48
  • $\begingroup$ Just an idea for your second question. Let $n$ odd and let $\zeta^n=-1$. It is enough to show that $A_n(\zeta)=0$. Maybe, using the orthogonality relations for roots of unity in the expansion of $(1+\zeta+\zeta^2+\cdots+\zeta^n)^n$, gives $A_n(\zeta)=0$. $\endgroup$
    – efs
    Jun 25, 2017 at 15:03
  • $\begingroup$ I thought you had "found" and not proved your assertion. My bad. I factored your polynomial for small odd $n$ and it gives: small polynomials times a very large irreducible one (over the integers). $\endgroup$
    – efs
    Jun 25, 2017 at 15:58
  • $\begingroup$ The generating function is $$g(X,z) = \sum_{n=0}^\infty A_n(X) z^n = \frac{1}{1-z} \sum_{n=0}^\infty X^{n^2} z^n$$ This is related to Jacobi theta functions: $$(1-e^{it}) g(X, e^{it}) + (1-e^{-it}) g(X, e^{-it}) = 1 + \theta_3(t/2, X)$$ $\endgroup$ Jun 25, 2017 at 20:46
  • 1
    $\begingroup$ $A_n(\exp(x))=\sum\limits_{k=0}^n \sum\limits_{j=0}^\infty \frac{x^jk^{2j}}{j!}=\sum\limits_{j=0}^\infty \frac{x^j}{j!} \sum\limits_{k=0}^n {k^{2j}}=...$ I don't know how continued $\endgroup$
    – Dattier
    Jun 28, 2017 at 8:10

1 Answer 1

6
$\begingroup$

Not at all an answer, just observations. Firstly, a conjecture, based on experiment.

CONJECTURE $A_{2n}$ is irreducible for all $n.$

(empirically true for all $n\leq20.$)

CONJECTURE 2 $A_{2n+1}/(x^{2n+1}+1)$ has at most one non-cyclotomic factor. (empirically true for all $n\leq 20$).

Finally, the roots of these things cluster around the unit circle (see the graphic for $A_{14}$).

Is it true that the zeros of the infinite series are on the unit circle? (oops, this is not supposed to be a question). Roots of $A_{14}$

$\endgroup$
9
  • $\begingroup$ Neat graphic. There's a product formula (probably a special case of the "Jacobi triple product") that shows that $\sum_{i=0}^\infty X^{i^2}$ has no zero with $|X|<1$. But $|X|=1$ is a natural boundary so it doesn't make sense to ask about zeros on or outside the unit circle. $\endgroup$ Jun 26, 2017 at 0:51
  • $\begingroup$ $\left( \sum_{i=0}^n X^{i^2} \right) / X^{n^2}$ converges to $1$ for $|X|>1$, and this renormalized limit is certainly nonvanishing. Probably one can convert this to an explicit bound on the zeroes of the finite sums. $\endgroup$
    – Will Sawin
    Jun 26, 2017 at 0:53
  • 1
    $\begingroup$ Re the second conjecture: $A_n/(x^n+1)$ is irreducible for $n=3,7,15,31$; for other odd $n \leq 41$ it has at least two factors but sometimes has three ($n=17,35$) or even four ($n=29,41$). Still all but one factor is cyclotomic; maybe that's what you meant to conjecture. [Also: "$<=$" $\neq$ "$\leq$" . . .] $\endgroup$ Jun 26, 2017 at 1:20
  • $\begingroup$ @NoamD.Elkies You are absolutely right on all counts, I fixed the conjecture and the typo (the latter coming through alternating between programming and latex). $\endgroup$
    – Igor Rivin
    Jun 26, 2017 at 1:33
  • $\begingroup$ @NoamD.Elkies Presumably, if the product formula whereof you speak actually gives a lower bound on the infinite series in the disk $|z| < r < 1,$ it will imply at least a one sided clustering, and then something similar after $z\to 1/z$ will give the other side. $\endgroup$
    – Igor Rivin
    Jun 26, 2017 at 2:09

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.