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