A sequence of generating functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:40:48Z http://mathoverflow.net/feeds/question/78700 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78700/a-sequence-of-generating-functions A sequence of generating functions Pierre Lescanne 2011-10-20T20:08:28Z 2011-10-20T21:48:41Z <p>I came on this sequence of generating functions on trying to count lambda terms (in lambda calculus) </p> <p>$T^{\langle m+1\rangle} = \frac{T^{\langle m \rangle}(z)}{z} - (T^{\langle m\rangle}(z))^2$</p> <p>with</p> <p>$T^{\langle m\rangle}(0) = 0.$</p> <p>What can be said about these functions? </p> <p>I would be happy to have an expression for $T^{\langle 0\rangle}(z)$ and say something about the asymptotic behavior of the coefficients $T_{n,m}$ for</p> <p>$T^{\langle m\rangle}(z) = \sum_{n=0}^{\infty}\ T_{n,m} z^n.$</p> http://mathoverflow.net/questions/78700/a-sequence-of-generating-functions/78709#78709 Answer by Pietro Majer for A sequence of generating functions Pietro Majer 2011-10-20T21:48:41Z 2011-10-20T21:48:41Z <p>We want all the (formal) power series $T^m(z)$ to be elements of the ideal $z\mathbb{R}[[z]]$ (which rephrases $T_{0,m}=0$). Also, the relation $T^m(z)=zT^{m+1}(z)+zT^m(z)^2$ implies that if for some $n$ we have $T^m(z)\in z^n\mathbb{R}[[z]]$ for all $m$, then also $T^m(z)\in z^{n+1}\mathbb{R}[[z]]$ for all $m$. By induction, we should conclude that the $T^m(z)$ are identically zero for all $m$..</p>