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Feb 15, 2018 at 7:29 history edited Pietro Majer CC BY-SA 3.0
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Feb 12, 2018 at 20:23 comment added Pietro Majer Yes, indeed it can also be proven by induction from the formula I wrote, integrating by parts several times
Feb 12, 2018 at 19:36 comment added Tom Copeland @PietroMajer, your feelings about the integrality of the coefficients of $\psi$ are indeed true. See my answer/comments.
Feb 10, 2018 at 0:10 comment added Jair Taylor This all makes sense to me. I think it should lead to a combinatorial interpretation as well. I'll need to think about that.
Feb 10, 2018 at 0:09 history bounty ended Jair Taylor
Feb 10, 2018 at 0:09 vote accept Jair Taylor
Feb 9, 2018 at 23:08 comment added Pietro Majer Computing the first terms of the sequence I got, $\psi_0=x$, $\psi_1=\phi$, $\psi_2=\phi(\phi'-t_1)$, $\psi_3=\phi({\phi'}^2+\phi''\phi-t_2-3t_1\phi''+3t_1t_2)$. The feeling is that $\psi_n$ could be always a polynomial with integer coefficients in $\phi,\phi',\dots,\phi^{(n-1) }$.
Feb 9, 2018 at 11:28 history edited Pietro Majer CC BY-SA 3.0
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Feb 9, 2018 at 5:39 comment added Jair Taylor Ah, I see! That makes sense, thanks! It would seem this is proven. Allow me to think through the implications of this a bit and then I'll accept your answer.
Feb 9, 2018 at 5:37 comment added Pietro Majer Yes, of course, but by the complete induction hypothesis $\psi_j=\phi \chi_j$ for $1\le j<n$, with $\chi_j$ a formal power series with positive coefficients, and there are at least two factors $\psi_j$ in each product, since it's on $s\in\alpha$ with $|\alpha|>1$...
Feb 9, 2018 at 5:13 comment added Jair Taylor These computations look good to me. But why does the last equation have positive coefficients? The $\phi^{-2}(t)$ may have negative coefficients, no?
Feb 9, 2018 at 4:58 history edited Pietro Majer CC BY-SA 3.0
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Feb 9, 2018 at 4:26 history edited Pietro Majer CC BY-SA 3.0
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Feb 9, 2018 at 4:21 history answered Pietro Majer CC BY-SA 3.0