Difference between revisions of "Mott polynomial"
(Created page with "The Mott polynomials $s_n$ are defined by $$s_n(x) = (-1)^n \dfrac{1}{2^nx^n} (n-1)! \displaystyle\sum_{k=0}^{h(\frac{n}{2})} \dfrac{x^{-2k}}{k! (n-k)!(n-2k-k)!},$$ where $h(\...") |
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Revision as of 05:54, 27 April 2015
The Mott polynomials $s_n$ are defined by
$$s_n(x) = (-1)^n \dfrac{1}{2^nx^n} (n-1)! \displaystyle\sum_{k=0}^{h(\frac{n}{2})} \dfrac{x^{-2k}}{k! (n-k)!(n-2k-k)!},$$
where $h(\frac{n}{2})= \left\{ \begin{array}{ll}
\frac{n}{2} &; n \mathrm{\hspace{2pt} even} \\
\frac{n}{2} - \frac{1}{2} &; n \mathrm{\hspace{2pt} odd}.
\end{array} \right.$
The first few are
$$s_0(x)=1,$$
$$s_1(x)=-\dfrac{x}{2},$$
$$s_2(x)=\dfrac{x^2}{4},$$
$$s_3(x)=\dfrac{-3x}{4} -\dfrac{x^3}{8},$$
and
$$s_4(x)=\dfrac{3x^2}{2} + \dfrac{x^4}{16}.$$
Properties
Theorem: The following formula holds: $$\exp \left(\dfrac{x(\sqrt{1-t^2}-1)}{t} \right)=\displaystyle\sum_{k=0}^{\infty} s_k(x) \dfrac{t^n}{n!}.$$
Proof: █
Theorem: The following formula holds: $$s_n(x) = \dfrac{(-1)^n}{n! x^n} {}_3F_0 \left( -n, \dfrac{1}{2}-\dfrac{n}{2}, 1- \dfrac{n}{2};-\dfrac{4}{x^2} \right).$$
Proof: █
