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May 5, 2010 at 5:12 comment added Wadim Zudilin In fact, I posted a new question about the functional equations. I wonder whether this problem was ever studied.
May 5, 2010 at 5:03 comment added Martin Rubey The equation for the truncated j (with q^-1 removed) is much larger... (coefficient degree 5, degree in x and degree in y equal to 3). I was too supid to simply try all coefficients! What would you like to investigate with respect to the functional equations?
May 4, 2010 at 21:51 comment added Wadim Zudilin Yes, modular polynomials are large. For example, $\Phi_2(j(q)j(q^2))=0$ for $\Phi_2(x,y)=-(xy)^2+x^3+y^3+2^4\cdot3\cdot31xy(x+y)+3^4\cdot5^3\cdot4027xy-2^4\cdot3^4\cdot5^3(x^2+y^2)+2^8\cdot3^7\cdot5^6(x+y)-2^{12}\cdot3^9\cdot5^9$.
May 4, 2010 at 17:10 comment added Martin Rubey Oh, I think I found it, but it's quite large. (I took the q-expansion of j and omitted the coefficient of q^-1)
May 4, 2010 at 16:59 comment added Martin Rubey Wadim, I couldn't find the functional equation for j. Could you please expand once more...
May 3, 2010 at 23:11 comment added Wadim Zudilin And yes to your last question: mixing differentiation and substitution $q\mapsto q^k$ is exactly what I mean.
May 3, 2010 at 23:09 comment added Wadim Zudilin Hi Martin, the $F_k$'s are presumably not tied by any (nontrivial) algebraic relations, that's why I feel them boring enough. All the monomials in $F_k$'s I suggest are linearly independent. (But weren't you interested in having no algebraic dependencies?!) The Eisenstein series example is phantastic because of possessing both the differential and Mahler-type structure. And the latter is, to my knowledge, not investigated... The functional equations are done for the modular invariant $j$ only (mathoverflow.net/questions/22001).
May 3, 2010 at 16:08 comment added Martin Rubey Wadim, I just checked that E2, E4 and E6 indeed satisfy what I called a Mahler-type functional equation (no idea whether this is good terminology). Thanks for pointing that out - I am surprised! (I did know that they satisfy ADEs, however...) However, I couldn't find any relation for the F_k. Are you saying that they satisfy a "mixed" equation? If so, do you happen to know it? (my program would require some tweaking...) (I guess you really meant I should try to mix differentiation and substitution q +-> q^k)
Apr 30, 2010 at 12:10 comment added Wadim Zudilin In the case of $F_{2m+1}(q)$ I am probably to speculative in adding $D_k$'s: already $D$ gives a rich structure. If we assign weight $2k+2m+1$ to each $D^kF_{2m+1}(q)$, then the set of monomials $D^{k_1}F_{2m_1+1}\dots D^{k_s}F_{2m_s+1}$ of given weight is finite, so one may introduce their ordering. In your example you have a single function which does not satisfy an algebraic differential equation, while here (sorry, presumably) no relations between all those monomials could be possible. Maybe, this is more concrete...
Apr 30, 2010 at 11:54 comment added Wadim Zudilin Martin, I was finally surprised by the prize... The set of differrence-differential monomials in a fixed weight will be finite (if the set of operators is finite); in your starting example $|\lambda|$ could be such a weight. The most natural (to me!) example, the ring of quasimodular forms $\mathbb Q[E_2,E_4,E_6]$, has a different grading. I am trying to write down explicitly those $E_j(q^2)$ by means of $E_j(q)$ (for seeing how $D_2$ acts) but it seems to be nontrivial enough. After submitting the answer I've realized that multiple zeta values might be a better example.
Apr 30, 2010 at 11:04 comment added Martin Rubey Wadim, many thanks for your answer! I admit that I accepted without reading all details, to make sure the bounty is properly awarded. Thus, could you expand a bit? In particular, it is not yet clear to me which monomials I should/could consider, and why this set of monomials would be finite if I only consider (say) D and D_2. Already the set of monomials in the usual differentiation operator is infinite, so surely I misunderstood...
Apr 30, 2010 at 6:49 vote accept Martin Rubey
Apr 30, 2010 at 6:49 history bounty ended Martin Rubey
Apr 30, 2010 at 4:03 comment added Wadim Zudilin Yes Will, I follow that question and have realized that your answer is not there any more. I would be happy to see a clear algorithm but for the moment the best answer is still too theoretical to me to be implemented (even I am not the author of the problem).
Apr 30, 2010 at 1:20 comment added Will Jagy Hi Wadim, did you see the answers to mathoverflow.net/questions/22777 on the lattice point close to an intersection?
Apr 30, 2010 at 0:52 history answered Wadim Zudilin CC BY-SA 2.5