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#This file was created by <bruno> 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.
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