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