Timeline for Can we use the Rogers-Ramanujan cfrac to parameterize the Fermat quintic $x^5+y^5=1$?
Current License: CC BY-SA 4.0
24 events
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| Feb 20, 2019 at 17:18 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Re-phrased
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| Feb 20, 2019 at 14:58 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Jacobi theta function and copyedit.
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| Feb 2, 2019 at 4:27 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Revised with more info
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Jan 15, 2017 at 14:32 | vote | accept | Tito Piezas III | ||
| Jan 12, 2017 at 7:08 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Connection of cfrac to quintics
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| Jan 11, 2017 at 15:03 | comment | added | Tito Piezas III | @Nemo: Thanks to that paper you cited, I found some nice relations. See answer below. | |
| Jan 11, 2017 at 15:02 | answer | added | Tito Piezas III | timeline score: 8 | |
| Jan 11, 2017 at 9:41 | comment | added | Tito Piezas III | @Nemo: Thanks, I just saw the link to the quintic case. Theta or eta, that's ok. Just like for $k=3$ you could equivalently use the Borwein theta, or eta quotients, or the cubic continued fraction. I'll peruse that paper now. | |
| Jan 11, 2017 at 9:35 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
b(q) and c(q)
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| Jan 11, 2017 at 9:22 | comment | added | Nemo | @მამუკაჯიბლაძე the parametrization in that paper is not in terms of eta quotients. Moreover it also seems somewhat artificial because this parametrization contains factors such as $(q;q)^{3/5}_\infty$ with fractional powers. If it was at least for integral powers it would have been better. | |
| Jan 11, 2017 at 9:20 | comment | added | მამუკა ჯიბლაძე | Re: $\color{blue}{\textit{Update}}$ - is it obvious what are $b$ and $c$ there? | |
| Jan 11, 2017 at 9:16 | comment | added | მამუკა ჯიბლაძე | @Nemo Now that you said it I believe you are obliged to extract the relevant part in an answer :D | |
| Jan 10, 2017 at 17:30 | comment | added | Nemo | I think an answer to your question about quintic analogs of (1) and (3) is given in this paper arxiv.org/pdf/1304.0684.pdf | |
| Jan 9, 2017 at 14:50 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Borweins
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| Jan 8, 2017 at 13:23 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Identity
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| Jan 8, 2017 at 9:43 | comment | added | Tito Piezas III | @მამუკაჯიბლაძე: Yes, I've checked out the page as well. By the way, you made a small typo. It's $$R(q)^{-5}-R(q)^5 = \left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6+11$$ | |
| Jan 8, 2017 at 8:44 | comment | added | მამუკა ჯიბლაძე | I've just checked on Wikipedia, they have several stunning identities involving fifth powers there for$$R(q)= \cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}},$$e. g. $$R(q)^{-5}-R(q)^5=\left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6$$or$$(v^4-3v^3+4v^2-2v+1)v=(v^4+2v^3+4v^2+3v+1)u^5$$with $u=R(q)$, $v=R(q^5)$, or$$(u^5+\phi^5)(v^5+\phi^5)=5\sqrt5\phi^5$$for $u=R(q^a)$, $v=R(q^b)$ with $5ab=4\pi^2$, or$$R(q)^5=w\left(\frac{1-w}{1+w}\right)^2,\ \ R(q^2)^5=w^2\frac{1+w}{1-w}$$with $w=R(q)R(q^2)^2$... | |
| Jan 8, 2017 at 5:37 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
More details
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| Jan 8, 2017 at 2:06 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Details
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| Jan 7, 2017 at 19:08 | comment | added | Noam D. Elkies | The 4 and 8 are the same ones. For 2, there must be many choices because the curve is rational. | |
| Jan 7, 2017 at 17:18 | comment | added | Noam D. Elkies | (I see that your $p=2$ parametrization is in fact obtained from the one for exponent $8$.) | |
| Jan 7, 2017 at 17:05 | comment | added | Noam D. Elkies | There's even a parametrization for $x^4+y^4=1$ and $x^8+y^8=1$ (or at least $ax^8 + by^8 = 1$ for some rational $a,b$). Also sixth powers, if memory serves. But fifth powers, no. Not with eta products/quotients, anyhow. | |
| Jan 7, 2017 at 13:49 | history | asked | Tito Piezas III | CC BY-SA 3.0 |