Timeline for $q$-analog of an integral from quantum field theory?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Nov 24, 2017 at 15:34 | vote | accept | Nemo | ||
| Nov 24, 2017 at 15:34 | answer | added | Nemo | timeline score: 5 | |
| Nov 23, 2017 at 8:40 | history | edited | Nemo | CC BY-SA 3.0 |
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| Nov 23, 2017 at 8:38 | history | undeleted | Nemo | ||
| Nov 21, 2017 at 23:38 | history | deleted | Nemo | via Vote | |
| Nov 21, 2017 at 14:39 | history | edited | Nemo | CC BY-SA 3.0 |
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| Nov 21, 2017 at 13:59 | history | edited | Nemo | CC BY-SA 3.0 |
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| Nov 21, 2017 at 13:55 | comment | added | Nemo | @fedja I see your point now, thanks. | |
| Nov 21, 2017 at 13:50 | comment | added | fedja | Ah, sorry. Missed the symmetry assumption. Then (assuming my expression is $F(x,y)$), use $F(x/3,y/3)F(y/3,z/3)F(x/3,z/3)$. It is still $0$ with two points fixed. (actually you want more symmetries, but it should get clear that one can do that too) | |
| Nov 21, 2017 at 13:47 | comment | added | Nemo | @fedya I hope I'm not missing anything in your comments, but $q^{239}(e^{10x}+ve^{10y})$ is not symmetric in $x,y,z$, so this term do not satisfy the above conditions no matter what the value of $v$ is. | |
| Nov 21, 2017 at 13:41 | comment | added | fedja | The problem is that unless you restrict the dependence on $q$, you can always add $q^{239}(e^{x}+ve^{y})$ with appropriately chosen $v$ without affecting any of your conditions and that is a big one. | |
| Nov 21, 2017 at 13:37 | comment | added | Nemo | @fedja I really don't see how this is a problem? As you see the base of $f$ in $(5)$ is $q_1$, so if $f$ is a combination of theta functions one can apply imaginary transformation to all theta functions and write them in base $q$, so the RHS of $5$ is a q-series with base $q$. Did this answer your question? | |
| Nov 21, 2017 at 13:30 | comment | added | fedja | Erm... This doesn't quite agree with the equation you posted in the "motivation", which also includes powers of $q$, $1-q$, and even $\theta_q(q)$. What am I missing? | |
| Nov 21, 2017 at 13:27 | comment | added | Nemo | @fedja yes, as you see I used the notation $\theta_q(x)=\theta(x;q)$ to simplify the formulas. Here $q$ is the base of theta function, not to be confused with the index of the usual definition of Jacobi theta functions. | |
| Nov 21, 2017 at 13:23 | comment | added | fedja | Are you assuming that the only dependence on $q$ is in the index of $\theta_q$? | |
| Nov 21, 2017 at 13:21 | history | edited | Nemo | CC BY-SA 3.0 |
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| Nov 21, 2017 at 12:06 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Nov 21, 2017 at 12:00 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Nov 21, 2017 at 10:48 | history | asked | Nemo | CC BY-SA 3.0 |