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316 lines
6.0 KiB
316 lines
6.0 KiB
#This file was created by <bruno> Sun Feb 16 14:05:04 1997
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#LyX 0.10 (C) 1995 1996 Matthias Ettrich and the LyX Team
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\lyxformat 2.10
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\textclass article
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\begin_preamble
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\catcode`@=11 % @ ist ab jetzt ein gewoehnlicher Buchstabe
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\def\ll{\langle\!\langle}
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\def\gg{\rangle\!\rangle}
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\catcode`@=12 % @ ist ab jetzt wieder ein Sonderzeichen
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\end_preamble
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\language default
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\inputencoding latin1
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\fontscheme default
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\epsfig dvips
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\papersize a4paper
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\paperfontsize 12
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\baselinestretch 1.00
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\secnumdepth 3
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\tocdepth 3
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\paragraph_separation indent
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\quotes_language english
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\quotes_times 2
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\paperorientation portrait
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\papercolumns 0
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\papersides 1
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\paperpagestyle plain
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\layout Standard
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The Laguerre polynomials
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\begin_inset Formula \( L_{n}(x) \)
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\end_inset
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are defined through
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\begin_inset Formula
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\[
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L_{n}(x)=e^{x}\cdot \left( \frac{d}{dx}\right) ^{n}(x^{n}e^{-x})\]
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\end_inset
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\layout Description
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Theorem:
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\layout Standard
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\begin_inset Formula \( L_{n}(x) \)
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\end_inset
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satisfies the recurrence relation
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\layout Standard
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\begin_inset Formula
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\[
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L_{0}(x)=1\]
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\end_inset
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\layout Standard
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\begin_inset Formula
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\[
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L_{n+1}(x)=(2n+1-x)\cdot L_{n}(x)-n^{2}\cdot L_{n-1}(x)\]
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\end_inset
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for
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\begin_inset Formula \( n\geq 0 \)
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\end_inset
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and the differential equation
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\begin_inset Formula \( x\cdot L_{n}^{''}(x)+(1-x)\cdot L_{n}^{'}(x)+n\cdot L_{n}(x)=0 \)
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\end_inset
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for all
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\begin_inset Formula \( n\geq 0 \)
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\end_inset
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.
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\layout Description
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Proof:
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\layout Standard
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Let
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\begin_inset Formula \( F:=\sum ^{\infty }_{n=0}\frac{L_{n}(x)}{n!}\cdot z^{n} \)
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\end_inset
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be the exponential generating function of the sequence of polynomials.
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It is the diagonal series of the power series
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\begin_inset Formula
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\[
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G:=\sum _{m,n=0}^{\infty }\frac{1}{m!}\cdot e^{x}\cdot \left( \frac{d}{dx}\right) ^{m}(x^{n}e^{-x})\cdot y^{m}\cdot z^{n}\]
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\end_inset
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Because the Taylor series development theorem holds in formal power series
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rings (see [1], section 2.
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16), we can simplify
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\begin_inset Formula
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\begin{eqnarray*}
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G & = & e^{x}\cdot \sum _{n=0}^{\infty }\left( \sum _{m=0}^{\infty }\frac{1}{m!}\cdot \left( \frac{d}{dx}\right) ^{m}(x^{n}e^{-x})\cdot y^{m}\right) \cdot z^{n}\\
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& = & e^{x}\cdot \sum _{n=0}^{\infty }(x+y)^{n}e^{-(x+y)}\cdot z^{n}\\
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& = & \frac{e^{-y}}{1-(x+y)z}
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\end{eqnarray*}
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\end_inset
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We take over the terminology from the
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\begin_inset Quotes eld
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\end_inset
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diag_rational
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\begin_inset Quotes erd
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\end_inset
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paper; here
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\begin_inset Formula \( R=Q[x] \)
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\end_inset
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and
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\begin_inset Formula \( M=Q[[x]] \)
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\end_inset
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(or, if you like it better,
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\begin_inset Formula \( M=H(C) \)
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\end_inset
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, the algebra of functions holomorphic in the entire complex plane).
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\begin_inset Formula \( G\in M[[y,z]] \)
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\end_inset
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is not rational; nevertheless we can proceed similarly to the
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\begin_inset Quotes eld
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\end_inset
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diag_series
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\begin_inset Quotes erd
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\end_inset
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paper.
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\begin_inset Formula \( F(z^{2}) \)
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\end_inset
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is the coefficient of
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\begin_inset Formula \( t^{0} \)
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\end_inset
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in
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\begin_inset Formula
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\[
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G(zt,\frac{z}{t})=\frac{e^{-zt}}{1-z^{2}-\frac{xz}{t}}\in M[[zt,\frac{z}{t},z]]=M\ll z,t\gg \]
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\end_inset
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The denominator's only zero is
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\begin_inset Formula \( t=\frac{xz}{1-z^{2}} \)
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\end_inset
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.
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We can write
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\begin_inset Formula
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\[
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e^{-zt}=e^{-\frac{xz^{2}}{1-z^{2}}}+\left( zt-\frac{xz^{2}}{1-z^{2}}\right) \cdot P(z,t)\]
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\end_inset
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with
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\begin_inset Formula \( P(z,t)\in Q[[zt,\frac{xz^{2}}{1-z^{2}}]]\subset Q[[zt,x,z]]=M[[zt,z]]\subset M\ll z,t\gg \)
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\end_inset
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.
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This yields -- all computations being done in
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\begin_inset Formula \( M\ll z,t\gg \)
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\end_inset
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--
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\begin_inset Formula
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\begin{eqnarray*}
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G(zt,\frac{z}{t}) & = & \frac{e^{-\frac{xz^{2}}{1-z^{2}}}}{1-z^{2}-\frac{xz}{t}}+\frac{zt}{1-z^{2}}\cdot P(z,t)\\
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& = & \frac{1}{1-z^{2}}\cdot e^{-\frac{xz^{2}}{1-z^{2}}}\cdot \sum _{j=0}^{\infty }\left( \frac{x}{1-z^{2}}\frac{z}{t}\right) ^{j}+\frac{zt}{1-z^{2}}\cdot P(z,t)
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\end{eqnarray*}
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\end_inset
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Here, the coefficient of
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\begin_inset Formula \( t^{0} \)
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\end_inset
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is
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\begin_inset Formula
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\[
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F(z^{2})=\frac{1}{1-z^{2}}\cdot e^{-\frac{xz^{2}}{1-z^{2}}}\]
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\end_inset
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hence
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\begin_inset Formula
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\[
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F(z)=\frac{1}{1-z}\cdot e^{-\frac{xz}{1-z}}\]
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\end_inset
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\layout Standard
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It follows that
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\begin_inset Formula \( (1-z)^{2}\cdot \frac{d}{dz}F-(1-x-z)\cdot F=0 \)
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\end_inset
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.
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This is equivalent to the claimed recurrence.
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\layout Standard
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Starting from the closed form for
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\begin_inset Formula \( F \)
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\end_inset
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, we compute a linear relation for the partial derivatives of
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\begin_inset Formula \( F \)
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\end_inset
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.
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Write
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\begin_inset Formula \( \partial _{x}=\frac{d}{dx} \)
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\end_inset
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and
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\begin_inset Formula \( \Delta _{z}=z\frac{d}{dz} \)
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\end_inset
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.
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One computes
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\begin_inset Formula
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\[
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F=1\cdot F\]
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\end_inset
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\begin_inset Formula
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\[
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\left( 1-z\right) \cdot \partial _{x}F=-z\cdot F\]
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\end_inset
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\begin_inset Formula
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\[
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\left( 1-z\right) ^{2}\cdot \partial _{x}^{2}F=z^{2}\cdot F\]
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\end_inset
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\begin_inset Formula
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\[
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\left( 1-z\right) ^{2}\cdot \Delta _{z}F=((1-x)z-z^{2})\cdot F\]
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\end_inset
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\begin_inset Formula
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\[
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\left( 1-z\right) ^{3}\cdot \partial _{x}\Delta _{z}F=(-z+xz^{2}+z^{3})\cdot F\]
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\end_inset
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Solve a homogeneous
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\begin_inset Formula \( 4\times 5 \)
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\end_inset
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system of linear equations over
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\begin_inset Formula \( Q(x) \)
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\end_inset
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to get
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\begin_inset Formula
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\[
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\left( 1-z\right) ^{3}\cdot \left( (1-x)\cdot \partial _{x}F+x\cdot \partial _{x}^{2}F+\Delta _{z}F\right) =0\]
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\end_inset
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Divide by the first factor to get
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\begin_inset Formula
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\[
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(1-x)\cdot \partial _{x}F+x\cdot \partial _{x}^{2}F+\Delta _{z}F=0\]
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\end_inset
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This is equivalent to the claimed equation
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\begin_inset Formula \( x\cdot L_{n}^{''}(x)+(1-x)\cdot L_{n}^{'}(x)+n\cdot L_{n}(x)=0 \)
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\end_inset
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.
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\layout Bibliography
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\cursor 123
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[1] Bruno Haible: D-finite power series in several variables.
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\shape italic
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Diploma thesis, University of Karlsruhe, June 1989
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\shape default
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.
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Sections 2.
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15 and 2.
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22.
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