What's the Legendre of zero?

Question

I know that this question may look like meaningless, but through solving a question, I encountered with this form of Legendre. $ P_{l}(0)$ or a sum over it. You have any idea about it?

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Use the generating function, with $\ds{\verts{h}\ <\ 1}$, $\ds{{1 \over \root{1 - 2xh + h^{2}}} =\sum_{\ell\ =\ 0}^{\infty}h^{\ell}\,{\rm P}_{\ell}\pars{x}}$ such that:

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\begin{align} \sum_{\ell\ =\ 0}^{\infty}h^{\ell}\,{\rm P}_{\ell}\pars{0} &={1 \over \root{1 + h^{2}}}=\sum_{\ell\ =\ 0}^{\infty}{-1/2 \choose \ell}h^{2\ell} =\sum_{\ell\ =\ 0}^{\infty}{\ell - 1/2 \choose \ell}\pars{-1}^{\ell}h^{2\ell} \\[5mm]&=\sum_{{\vphantom{\LARGE A}\ell\ =\ 0 }\atop \ell\ \mbox{even}}^{\infty} {\ell/2 - 1/2 \choose \ell/2}\pars{-1}^{\ell/2}h^{\ell} \end{align}

Then,

$$ \color{#66f}{\large\,{\rm P}_{\ell}\pars{0}} =\color{#66f}{\large\left\{\begin{array}{lcl} 0 & \mbox{if} & \ell\ \mbox{is odd} \\[2mm] \pars{-1}^{\ell/2}{\bracks{\ell - 1}/2 \choose \ell/2} & \mbox{if} & \ell\ \mbox{is even} \end{array}\right.} $$

Answer 2

It is possible to prove that

$$P_{{L}} \left( 0 \right) ={\frac { \left( -1 \right) ^{L} {_2F_1(-L,-L;\,1;\,-1)}}{{2}^{L}}} $$

Please look the equation (35) at Legendre Polynomials

It is possible to obtain the simplified form

$$P_{{L}} \left( 0 \right) = \left( -1 \right) ^{L}{-1/2\choose L/2}$$