Timeline for Modular equations for quasimodular forms
Current License: CC BY-SA 2.5
6 events
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| May 7, 2010 at 9:21 | comment | added | Wadim Zudilin | @Martin: Thanks for the $E_{24}$ news. It really seems that aricthmetically these objects are better than $j(\tau)$ as the coefficients involve small primes only. I would be happier if the things are written with "*" for multiplication and "^" for exponentiation. As for the homogeneous degree, I would expect (for $E_2(q^n)$) to be $2\psi(n)=2n\prod_{p\mid n}(1+1/p)$. So, again six for $n=5$. Have you tried $E_4(q^n)$ and $E_6(q^n)$ as well? | |
| May 7, 2010 at 7:29 | history | edited | Martin Rubey | CC BY-SA 2.5 |
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| May 6, 2010 at 12:10 | comment | added | Martin Rubey | Yes, I did this using my (slightly modified) guessing package. (Unfortunately, I did not yet have the time to modify it properly, and very likely I'll have to wait for Waldek to get full speed) However, with what I have I was able to check E24, and I did not find an equation of degree at most 5 and constant coefficients. I'll see what I can do... | |
| May 6, 2010 at 4:12 | comment | added | Wadim Zudilin | @Martin: Thanks! I just checked your relations (up to $O(q^{101})$, the first one can be compared with the one I know): perfect agreement. But I still wonder how to do the things non-experimentally and in general. I can't believe there were no work in this direction... | |
| May 6, 2010 at 3:50 | comment | added | Wadim Zudilin | I TeXify: in the notation $E_{22} = E_2(q^2)$, $E_{23} = E_2(q^3)$ one has $-E6 + (- 6E_{22} + 3E_2)E_4 + 32E_{22}^3 - 48E_2 E_{22}^2 + 24E_2^2 E_{22} - 4E_2^3 = 0$ and $(- 192E_{23} + 64E_2)E_6 - 16E_4^2 + (- 648E_{23}^2 + 432E_2 E_{23} - 72E_2^2 )E_4 + 2187E_{23}^4 - 2916E_2 E_{23}^3 + 1458E_2^2 E_{23}^2 - 324E_2^3 E_{23} + 27E_2^4 = 0$. Yes, that what I mean to get for $E_{kn}$ ($k=2,4,6$ and $n\ge2$). The coefficients look reasonably small (compared to the ones for level 2 modular polynomial). I guess this is an experimental discovery. How to do for the general $n$? | |
| May 5, 2010 at 20:04 | history | answered | Martin Rubey | CC BY-SA 2.5 |