%% This LaTeX-file was created by Sun Feb 16 14:24:52 1997 %% LyX 0.10 (C) 1995 1996 by Matthias Ettrich and the LyX Team %% Don't edit this file unless you are sure what you are doing. \documentclass[12pt,a4paper,oneside,onecolumn]{article} \usepackage[]{fontenc} \usepackage[latin1]{inputenc} \usepackage[dvips]{epsfig} %% %% BEGIN The lyx specific LaTeX commands. %% \makeatletter \def\LyX{L\kern-.1667em\lower.25em\hbox{Y}\kern-.125emX\spacefactor1000}% \newcommand{\lyxtitle}[1] {\thispagestyle{empty} \global\@topnum\z@ \section*{\LARGE \centering \sffamily \bfseries \protect#1 } } \newcommand{\lyxline}[1]{ {#1 \vspace{1ex} \hrule width \columnwidth \vspace{1ex}} } \newenvironment{lyxsectionbibliography} { \section*{\refname} \@mkboth{\uppercase{\refname}}{\uppercase{\refname}} \begin{list}{}{ \itemindent-\leftmargin \labelsep 0pt \renewcommand{\makelabel}{} } } {\end{list}} \newenvironment{lyxchapterbibliography} { \chapter*{\bibname} \@mkboth{\uppercase{\bibname}}{\uppercase{\bibname}} \begin{list}{}{ \itemindent-\leftmargin \labelsep 0pt \renewcommand{\makelabel}{} } } {\end{list}} \def\lxq{"} \newenvironment{lyxcode} {\list{}{ \rightmargin\leftmargin \raggedright \itemsep 0pt \parsep 0pt \ttfamily }% \item[] } {\endlist} \newcommand{\lyxlabel}[1]{#1 \hfill} \newenvironment{lyxlist}[1] {\begin{list}{} {\settowidth{\labelwidth}{#1} \setlength{\leftmargin}{\labelwidth} \addtolength{\leftmargin}{\labelsep} \renewcommand{\makelabel}{\lyxlabel}}} {\end{list}} \newcommand{\lyxletterstyle}{ \setlength\parskip{0.7em} \setlength\parindent{0pt} } \newcommand{\lyxaddress}[1]{ \par {\raggedright #1 \vspace{1.4em} \noindent\par} } \newcommand{\lyxrightaddress}[1]{ \par {\raggedleft \begin{tabular}{l}\ignorespaces #1 \end{tabular} \vspace{1.4em} \par} } \newcommand{\lyxformula}[1]{ \begin{eqnarray*} #1 \end{eqnarray*} } \newcommand{\lyxnumberedformula}[1]{ \begin{eqnarray} #1 \end{eqnarray} } \makeatother %% %% END The lyx specific LaTeX commands. %% \pagestyle{plain} \setcounter{secnumdepth}{3} \setcounter{tocdepth}{3} %% Begin LyX user specified preamble: \catcode`@=11 % @ ist ab jetzt ein gewoehnlicher Buchstabe \def\mod#1{\allowbreak \mkern8mu \mathop{\operator@font mod}\,\,{#1}} \def\pmod#1{\allowbreak \mkern8mu \left({\mathop{\operator@font mod}\,\,{#1}}\right)} \catcode`@=12 % @ ist ab jetzt wieder ein Sonderzeichen %% End LyX user specified preamble. \begin{document} The Legendre polynomials \( P_{n}(x) \) are defined through \[ P_{n}(x)=\frac{1}{2^{n}n!}\cdot \left( \frac{d}{dx}\right) ^{n}(x^{2}-1)^{n}\] (For a motivation of the \( 2^{n} \) in the denominator, look at \( P_{n}(x) \) modulo an odd prime \( p \), and observe that \( P_{n}(x)\equiv P_{p-1-n}(x)\mod p \) for \( 0\leq n\leq p-1 \). This wouldn't hold if the \( 2^{n} \) factor in the denominator weren't present.) \begin{description} \item [Theorem:]~ \end{description} \( P_{n}(x) \) satisfies the recurrence relation \[ P_{0}(x)=1\] \[ (n+1)\cdot P_{n+1}(x)=(2n+1)x\cdot P_{n}(x)-n\cdot P_{n-1}(x)\] for \( n\geq 0 \) and the differential equation \( (1-x^{2})\cdot P_{n}^{''}(x)-2x\cdot P_{n}^{'}(x)+(n^{2}+n)\cdot P_{n}(x)=0 \) for all \( n\geq 0 \). \begin{description} \item [Proof:]~ \end{description} Let \( F:=\sum ^{\infty }_{n=0}P_{n}(x)\cdot z^{n} \) be the generating function of the sequence of polynomials. It is the diagonal series of the power series \[ G:=\sum _{m,n=0}^{\infty }\frac{1}{2^{n}m!}\cdot \left( \frac{d}{dx}\right) ^{m}(x^{2}-1)^{n}\cdot y^{m}\cdot z^{n}\] Because the Taylor series development theorem holds in formal power series rings (see [1], section 2.16), we can simplify \begin{eqnarray*} G & = & \sum _{n=0}^{\infty }\frac{1}{2^{n}}\cdot \left( \sum _{m=0}^{\infty }\frac{1}{m!}\cdot \left( \frac{d}{dx}\right) ^{m}(x^{2}-1)^{n}\cdot y^{m}\right) \cdot z^{n}\\ & = & \sum _{n=0}^{\infty }\frac{1}{2^{n}}\cdot \left( (x+y)^{2}-1\right) ^{n}\cdot z^{n}\\ & = & \frac{1}{1-\frac{1}{2}\left( (x+y)^{2}-1\right) z} \end{eqnarray*} We take over the terminology from the ``diag\_rational'' paper; here \( R=Q[x] \) and \( M=Q[[x]] \) (or, if you like it better, \( M=H(C) \), the algebra of functions holomorphic in the entire complex plane). \( G\in M[[y,z]] \) is rational; hence \( F \) is algebraic over \( R[z] \). Let's proceed exactly as in the ``diag\_series'' paper. \( F(z^{2}) \) is the coefficient of \( t^{0} \) in \[ G(zt,\frac{z}{t})=\frac{2t}{2t-\left( (x+zt)^{2}-1\right) z}=\frac{2t}{-z^{3}\cdot t^{2}+2(1-xz^{2})\cdot t+(z-x^{2}z)}\] The splitting field of the denominator is \( L=Q(x)(z)(\alpha ) \) where \[ \alpha _{1/2}=\frac{1-xz^{2}\pm \sqrt{1-2xz^{2}+z^{4}}}{z^{3}}\] \[ \alpha =\alpha _{1}=\frac{2}{z^{3}}-\frac{2x}{z}+\frac{1-x^{2}}{2}z+\cdots \in Q(x)[[z]][\frac{1}{z}]\] \[ \alpha _{2}=\frac{x^{2}-1}{2}z+\cdots \in Q(x)[[z]][\frac{1}{z}]\] The partial fraction decomposition of \( G(zt,\frac{z}{t}) \) reads \[ G(zt,\frac{z}{t})=-\frac{2}{z^{3}}\cdot \frac{1}{\alpha _{1}-\alpha _{2}}\cdot \left( \frac{\alpha _{1}}{t-\alpha _{1}}-\frac{\alpha _{2}}{t-\alpha _{2}}\right) \] It follows that \[ F(z^{2})=-\frac{2}{z^{3}}\cdot \frac{1}{\alpha _{1}-\alpha _{2}}\cdot \left( \frac{\alpha _{1}}{0-\alpha _{1}}-0\right) =\frac{1}{\sqrt{1-2xz^{2}+z^{4}}}\] hence \[ F(z)=\frac{1}{\sqrt{1-2xz+z^{2}}}\] It follows that \( (1-2xz+z^{2})\cdot \frac{d}{dz}F+(z-x)\cdot F=0 \). This is equivalent to the claimed recurrence. Starting from the closed form for \( F \), we compute a linear relation for the partial derivatives of \( F \). Write \( \partial _{x}=\frac{d}{dx} \) and \( \Delta _{z}=z\frac{d}{dz} \). One computes \[ F=1\cdot F\] \[ \left( 1-2xz+z^{2}\right) \cdot \partial _{x}F=z\cdot F\] \[ \left( 1-2xz+z^{2}\right) ^{2}\cdot \partial _{x}^{2}F=3z^{2}\cdot F\] \[ \left( 1-2xz+z^{2}\right) \cdot \Delta _{z}F=(xz-z^{2})\cdot F\] \[ \left( 1-2xz+z^{2}\right) ^{2}\cdot \partial _{x}\Delta _{z}F=(z+xz^{2}-2z^{3})\cdot F\] \[ \left( 1-2xz+z^{2}\right) ^{2}\cdot \Delta _{z}^{2}F=\left( xz+(x^{2}-2)z^{2}-xz^{3}+z^{4}\right) \cdot F\] Solve a homogeneous \( 5\times 6 \) system of linear equations over \( Q(x) \) to get \[ \left( 1-2xz+z^{2}\right) ^{2}\cdot \left( (-2x)\cdot \partial _{x}F+(1-x^{2})\cdot \partial _{x}^{2}F+\Delta _{z}F+\Delta _{z}^{2}F\right) =0\] Divide by the first factor to get \[ (-2x)\cdot \partial _{x}F+(1-x^{2})\cdot \partial _{x}^{2}F+\Delta _{z}F+\Delta _{z}^{2}F=0\] This is equivalent to the claimed equation \( (1-x^{2})\cdot P_{n}^{''}(x)-2x\cdot P_{n}^{'}(x)+(n^{2}+n)\cdot P_{n}(x)=0 \). \begin{lyxsectionbibliography} \item [1] Bruno Haible: D-finite power series in several variables. \em Diploma thesis, University of Karlsruhe, June 1989\em . Sections 2.14, 2.15 and 2.22. \end{lyxsectionbibliography} \end{document}