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Touchard polynomials
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== Properties == === Basic properties === The value at 1 of the ''n''th Touchard polynomial is the ''n''th [[Bell numbers|Bell number]], i.e., the number of [[partition of a set|partitions of a set]] of size ''n'': :<math>T_n(1)=B_n.</math> If ''X'' is a [[random variable]] with a [[Poisson distribution]] with expected value λ, then its ''n''th moment is E(''X''<sup>''n''</sup>) = ''T''<sub>''n''</sub>(λ), leading to the definition: :<math>T_{n}(x)=e^{-x}\sum_{k=0}^\infty \frac {x^k k^n} {k!}.</math> Using this fact one can quickly prove that this [[polynomial sequence]] is of [[binomial type]], i.e., it satisfies the sequence of identities: :<math>T_n(\lambda+\mu)=\sum_{k=0}^n {n \choose k} T_k(\lambda) T_{n-k}(\mu).</math> The Touchard polynomials constitute the only polynomial sequence of binomial type with the coefficient of ''x'' equal 1 in every polynomial. The Touchard polynomials satisfy the Rodrigues-like formula: :<math>T_n \left(e^x \right) = e^{-e^x} \frac{d^n}{dx^n}\;e^{e^x}.</math> The Touchard polynomials satisfy the [[recurrence relation]] :<math>T_{n+1}(x)=x \left(1+\frac{d}{dx} \right)T_{n}(x)</math> and :<math>T_{n+1}(x)=x\sum_{k=0}^n{n \choose k}T_k(x).</math> In the case ''x'' = 1, this reduces to the recurrence formula for the [[Bell numbers]]. A generalization of both this formula and the definition, is a generalization of Spivey's formula<ref>{{Cite web |title=Implications of Spivey's Bell Number Formula |url=https://cs.uwaterloo.ca/journals/JIS/VOL11/Gould/gould35.html |access-date=2023-05-28 |website=cs.uwaterloo.ca}}</ref> <math display="block">T_{n+m}(x) = \sum_{k=0}^n \left\{ {n \atop k} \right\} x^k \sum_{j=0}^m \binom{m}{j} k^{m-j} T_j(x)</math> Using the [[Umbral calculus|umbral notation]] ''T''<sup>''n''</sup>(''x'')=''T''<sub>''n''</sub>(''x''), these formulas become: :<math>T_n(\lambda+\mu)=\left(T(\lambda)+T(\mu) \right)^n,</math>{{clarification needed|date=April 2024}} :<math>T_{n+1}(x)=x \left(1+T(x) \right)^n.</math> The [[generating function]] of the Touchard polynomials is :<math>\sum_{n=0}^\infty {T_n(x) \over n!} t^n=e^{x\left(e^t-1\right)},</math> which corresponds to the [[Stirling numbers of the second kind#Generating functions|generating function of Stirling numbers of the second kind]]. Touchard polynomials have [[contour integral]] representation: :<math>T_n(x)=\frac{n!}{2\pi i}\oint\frac{e^{x({e^t}-1)}}{t^{n+1}}\,dt.</math> === Zeroes === All zeroes of the Touchard polynomials are real and negative. This fact was observed by L. H. Harper in 1967.<ref name = 'Harper'>{{Cite journal | last = Harper | first = L. H. | title = Stirling behavior is asymptotically normal | year = 1967 | volume = 38 | issue = 2 | pages = 410–414 | journal = The Annals of Mathematical Statistics | doi=10.1214/aoms/1177698956 | doi-access = free }}</ref> The absolute value of the leftmost zero is bounded from above by<ref name = 'MC'>{{Cite journal | last1 = Mező | first1 = István | last2 = Corcino | first2 = Roberto B. | title = The estimation of the zeros of the Bell and r-Bell polynomials | year = 2015 | volume = 250 | pages = 727–732 | journal = Applied Mathematics and Computation | doi=10.1016/j.amc.2014.10.058 }}</ref> :<math>\frac1n\binom{n}{2}+\frac{n-1}{n}\sqrt{\binom{n}{2}^2-\frac{2n}{n-1}\left(\binom{n}{3}+3\binom{n}{4}\right)},</math> although it is conjectured that the leftmost zero grows linearly with the index ''n''. The [[Mahler measure]] <math>M(T_n)</math> of the Touchard polynomials can be estimated as follows:<ref>{{cite web|last1=István|first1=Mező|title=On the Mahler measure of the Bell polynomials|url=https://sites.google.com/site/istvanmezo81/others|accessdate=7 November 2017}}</ref> :<math> \frac{\lbrace\textstyle{n\atop \Omega_n}\rbrace}{\binom{n}{\Omega_n}}\le M(T_n)\le\sqrt{n+1}\left\{{n\atop K_n}\right\}, </math> where <math>\Omega_n</math> and <math>K_n</math> are the smallest of the maximum two ''k'' indices such that <math>\lbrace\textstyle{n\atop k}\rbrace/\binom{n}{k}</math> and <math>\lbrace\textstyle{n\atop k}\rbrace</math> are maximal, respectively.
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