Timeline for A closed formula for $A_n(X)=\sum\limits_{i=0}^n X^{i^2}$
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 26, 2017 at 9:22 | comment | added | Dattier | A bout the first conjecture, it the same case for $G_n(X)=\sum \limits_{k=0}^{n} X^k$ when n+1 is prime, but there are a closed formula. | |
| Jun 26, 2017 at 5:54 | comment | added | Noam D. Elkies | Looking closer at your plot also suggests the zeros of norm $.96064+$ in the right half-plane near $0.5309172 \pm 0.8005987i$. | |
| Jun 26, 2017 at 4:13 | comment | added | Igor Rivin | @NoamD.Elkies Yes, I was mystified by your triple product commentary because of the two-sided vs one-sided issue, but am glad that I was not confused as much as I might have been. As for the pair of zeros, true. | |
| Jun 26, 2017 at 3:33 | comment | added | Noam D. Elkies | The "other side" is elementary as Will Sawin noted. Unfortunately the product formula I had in mind is for the two-sided sum $\sum_{i=-\infty}^\infty X^{i^2} = 2 A_\infty(X) - 1$. The sum $A_\infty(X)$ itself does have at least the conjugate pair of zeros in $|X|<1$ suggested by your plot, near $-0.6701129 \pm .4274072i$ (you can get many more digits using Newton's method), and might have infinitely many more. | |
| Jun 26, 2017 at 2:09 | comment | added | Igor Rivin | @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. | |
| Jun 26, 2017 at 1:33 | comment | added | Igor Rivin | @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). | |
| S Jun 26, 2017 at 1:33 | history | edited | Igor Rivin | CC BY-SA 3.0 |
fixed typo
|
| Jun 26, 2017 at 1:30 | review | Suggested edits | |||
| S Jun 26, 2017 at 1:33 | |||||
| Jun 26, 2017 at 1:20 | comment | added | Noam D. Elkies | 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: "$<=$" $\neq$ "$\leq$" . . .] | |
| Jun 26, 2017 at 0:53 | comment | added | Will Sawin | $\left( \sum_{i=0}^n X^{i^2} \right) / X^{n^2}$ 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. | |
| Jun 26, 2017 at 0:51 | comment | added | Noam D. Elkies | 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. | |
| Jun 26, 2017 at 0:18 | history | answered | Igor Rivin | CC BY-SA 3.0 |