Timeline for Analytic continuation of the double sum $\sum_{n,m\ge0}x^ny^mt^{nm}$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 17, 2016 at 20:30 | history | bounty ended | Weather Report | ||
| Feb 13, 2016 at 12:42 | comment | added | Lev Borisov | In my paper, I only consider $|t|<1$. The products would diverge, badly, if $|t|>1$. So the product formula is only valid in this regime. | |
| Feb 13, 2016 at 9:06 | comment | added | Weather Report | I am saying the following: take function $\phi(x,y,t)=\sum_{n\in\mathbb{Z}}\frac{x^n}{1-yt^{n}}$. Now it seems that $\phi(x,y^{-1},t)=\sum_{n\in\mathbb{Z}}\frac{x^n}{1-y^{-1}t^{n}}=-y \sum_{n\in\mathbb{Z}} \frac{(xt^{-1})^n}{1-yt^{-n}}=-y\phi(xt^{-1},y,t^{-1})$. This is in contradiction with the explicit formula from your paper. The poles do agree, but the zeros do not: $(1-xyt^{k-1})(1-x^{-1}y^{-1}t^{k})$ vs $(1-xy^{-1}t^{-k})(1-x^{-1}yt^{-k+1})$. | |
| Feb 12, 2016 at 19:25 | comment | added | Lev Borisov | Not sure what the problem is: you have poles at $y^{-1}=t^{-n}$ on one side and at $y=(t^{-1})^{-n}$ on the other side, which is the same thing (as viewed as a function of $y$, in appropriate ranges for $|x|$ and $|t|$). | |
| Feb 12, 2016 at 16:46 | comment | added | Weather Report | Your arguments look reasonable. However, I remain in some confusion. Concerning my self-contradictory conclusion: I think something is wrong with the property 2 ($f(x,y^{-1},t)=-yf(xt^{-1},y,t^{-1})$). It should also work if the summation is over $n\in\mathbb{Z}$. But equation (11) from your paper does not behave in this way. However, what exactly is wrong I do not understand. It seems enough for convergence of $\sum_{n\in\mathbb{Z}}\frac{x^n}{1-yt^n}$ that $|t|<|x|<1$, irrelevant of $y$. Not sure how can one run into trouble by replacing $y\to y^{-1}$ and making identical transformations. | |
| Feb 11, 2016 at 1:26 | history | answered | Lev Borisov | CC BY-SA 3.0 |