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#This file was created by <bruno> Sun Feb 16 14:24:48 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\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_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\cursor 47 The Legendre polynomials \begin_inset Formula \( P_{n}(x) \)\end_inset
are defined through \begin_inset Formula \[P_{n}(x)=\frac{1}{2^{n}n!}\cdot \left( \frac{d}{dx}\right) ^{n}(x^{2}-1)^{n}\]
\end_inset
(For a motivation of the \begin_inset Formula \( 2^{n} \)\end_inset
in the denominator, look at \begin_inset Formula \( P_{n}(x) \)\end_inset
modulo an odd prime \begin_inset Formula \( p \)\end_inset
, and observe that \begin_inset Formula \( P_{n}(x)\equiv P_{p-1-n}(x)\mod p \)\end_inset
for \begin_inset Formula \( 0\leq n\leq p-1 \)\end_inset
. This wouldn't hold if the \begin_inset Formula \( 2^{n} \)\end_inset
factor in the denominator weren't present.)\layout Description
Theorem:\layout Standard
\begin_inset Formula \( P_{n}(x) \)\end_inset
satisfies the recurrence relation\layout Standard
\begin_inset Formula \[P_{0}(x)=1\]
\end_inset
\layout Standard
\begin_inset Formula \[(n+1)\cdot P_{n+1}(x)=(2n+1)x\cdot P_{n}(x)-n\cdot P_{n-1}(x)\]
\end_inset
for \begin_inset Formula \( n\geq 0 \)\end_inset
and the differential equation \begin_inset Formula \( (1-x^{2})\cdot P_{n}^{''}(x)-2x\cdot P_{n}^{'}(x)+(n^{2}+n)\cdot P_{n}(x)=0 \)\end_inset
for all \begin_inset Formula \( n\geq 0 \)\end_inset
.
\layout Description
Proof:\layout Standard
Let \begin_inset Formula \( F:=\sum ^{\infty }_{n=0}P_{n}(x)\cdot z^{n} \)\end_inset
be the generating function of the sequence of polynomials. It is the diagonal series of the power series\begin_inset Formula \[G:=\sum _{m,n=0}^{\infty }\frac{1}{2^{n}m!}\cdot \left( \frac{d}{dx}\right) ^{m}(x^{2}-1)^{n}\cdot y^{m}\cdot z^{n}\]
\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 & = & \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*}
\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 rational; hence \begin_inset Formula \( F \)\end_inset
is algebraic over \begin_inset Formula \( R[z] \)\end_inset
. Let's proceed exactly as in 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 \[G(zt,\frac{z}{t})=\frac{2t}{2t-\left( (x+zt)^{2}-1\right) z}=\frac{2t}{-z^{3}\cdot t^{2}+2(1-xz^{2})\cdot t+(z-x^{2}z)}\]
\end_inset
The splitting field of the denominator is \begin_inset Formula \( L=Q(x)(z)(\alpha ) \)\end_inset
where \begin_inset Formula \[\alpha _{1/2}=\frac{1-xz^{2}\pm \sqrt{1-2xz^{2}+z^{4}}}{z^{3}}\]
\end_inset
\begin_inset Formula \[\alpha =\alpha _{1}=\frac{2}{z^{3}}-\frac{2x}{z}+\frac{1-x^{2}}{2}z+\cdots \in Q(x)[[z]][\frac{1}{z}]\]
\end_inset
\begin_inset Formula \[\alpha _{2}=\frac{x^{2}-1}{2}z+\cdots \in Q(x)[[z]][\frac{1}{z}]\]
\end_inset
The partial fraction decomposition of \begin_inset Formula \( G(zt,\frac{z}{t}) \)\end_inset
reads\begin_inset Formula \[G(zt,\frac{z}{t})=-\frac{2}{z^{3}}\cdot \frac{1}{\alpha _{1}-\alpha _{2}}\cdot \left( \frac{\alpha _{1}}{t-\alpha _{1}}-\frac{\alpha _{2}}{t-\alpha _{2}}\right) \]
\end_inset
It follows that\begin_inset Formula \[F(z^{2})=-\frac{2}{z^{3}}\cdot \frac{1}{\alpha _{1}-\alpha _{2}}\cdot \left( \frac{\alpha _{1}}{0-\alpha _{1}}-0\right) =\frac{1}{\sqrt{1-2xz^{2}+z^{4}}}\]
\end_inset
hence\begin_inset Formula \[F(z)=\frac{1}{\sqrt{1-2xz+z^{2}}}\]
\end_inset
\layout Standard
It follows that \begin_inset Formula \( (1-2xz+z^{2})\cdot \frac{d}{dz}F+(z-x)\cdot F=0 \)\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 \[\left( 1-2xz+z^{2}\right) \cdot \partial _{x}F=z\cdot F\]
\end_inset
\begin_inset Formula \[\left( 1-2xz+z^{2}\right) ^{2}\cdot \partial _{x}^{2}F=3z^{2}\cdot F\]
\end_inset
\begin_inset Formula \[\left( 1-2xz+z^{2}\right) \cdot \Delta _{z}F=(xz-z^{2})\cdot F\]
\end_inset
\begin_inset Formula \[\left( 1-2xz+z^{2}\right) ^{2}\cdot \partial _{x}\Delta _{z}F=(z+xz^{2}-2z^{3})\cdot F\]
\end_inset
\begin_inset Formula \[\left( 1-2xz+z^{2}\right) ^{2}\cdot \Delta _{z}^{2}F=\left( xz+(x^{2}-2)z^{2}-xz^{3}+z^{4}\right) \cdot F\]
\end_inset
Solve a homogeneous \begin_inset Formula \( 5\times 6 \)\end_inset
system of linear equations over \begin_inset Formula \( Q(x) \)\end_inset
to get \begin_inset Formula \[\left( 1-2xz+z^{2}\right) ^{2}\cdot \left( (-2x)\cdot \partial _{x}F+(1-x^{2})\cdot \partial _{x}^{2}F+\Delta _{z}F+\Delta _{z}^{2}F\right) =0\]
\end_inset
Divide by the first factor to get\begin_inset Formula \[(-2x)\cdot \partial _{x}F+(1-x^{2})\cdot \partial _{x}^{2}F+\Delta _{z}F+\Delta _{z}^{2}F=0\]
\end_inset
This is equivalent to the claimed equation \begin_inset Formula \( (1-x^{2})\cdot P_{n}^{''}(x)-2x\cdot P_{n}^{'}(x)+(n^{2}+n)\cdot P_{n}(x)=0 \)\end_inset
.
\layout Bibliography
[1] Bruno Haible: D-finite power series in several variables. \shape italic Diploma thesis, University of Karlsruhe, June 1989\shape default . Sections 2.14, 2.15 and 2.22.
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