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333 lines
5.7 KiB
333 lines
5.7 KiB
#This file was created by <bruno> Sun Feb 16 00:32:21 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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\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 1
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\papersides 1
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\paperpagestyle plain
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\layout Standard
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The Tschebychev polynomials (of the 1st kind)
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\begin_inset Formula \( T_{n}(x) \)
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\end_inset
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are defined through the recurrence relation
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\layout Standard
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\begin_inset Formula
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\[
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T_{0}(x)=1\]
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\end_inset
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\layout Standard
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\begin_inset Formula
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\[
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T_{1}(x)=x\]
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\end_inset
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\layout Standard
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\begin_inset Formula
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\[
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T_{n+2}(x)=2x\cdot T_{n+1}(x)-T_{n}(x)\]
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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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.
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\layout Description
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Theorem:
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\layout Standard
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\begin_inset Formula \( T_{n}(x) \)
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\end_inset
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satisfies the differential equation
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\begin_inset Formula \( (x^{2}-1)\cdot T_{n}^{''}(x)+x\cdot T_{n}^{'}(x)-n^{2}\cdot T_{n}(x)=0 \)
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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 \( F:=\sum ^{\infty }_{n=0}T_{n}(x)z^{n} \)
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\end_inset
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be the generating function of the sequence of polynomials.
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The recurrence is equivalent to the equation
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\begin_inset Formula
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\[
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(1-2x\cdot z+z^{2})\cdot F=1-x\cdot z\]
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\end_inset
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\layout Description
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Proof
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\protected_separator
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1:
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\layout Standard
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\begin_inset Formula \( F \)
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\end_inset
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is a rational function in
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\begin_inset Formula \( z \)
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\end_inset
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,
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\begin_inset Formula \( F=\frac{1-xz}{1-2xz+z^{2}} \)
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\end_inset
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.
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From the theory of recursions with constant coefficients, we know that
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we have to perform a partial fraction decomposition.
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So let
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\begin_inset Formula \( p(z)=z^{2}-2x\cdot z+1 \)
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\end_inset
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be the denominator and
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\begin_inset Formula \( \alpha =x+\sqrt{x^{2}-1} \)
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\end_inset
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and
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\begin_inset Formula \( \alpha ^{-1} \)
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\end_inset
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its zeroes.
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The partial fraction decomposition reads
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\begin_inset Formula
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\[
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F=\frac{1-xz}{1-2xz+z^{2}}=\frac{1}{2}\left( \frac{1}{1-\alpha z}+\frac{1}{1-\alpha ^{-1}z}\right) \]
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\end_inset
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hence
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\begin_inset Formula \( T_{n}(x)=\frac{1}{2}(\alpha ^{n}+\alpha ^{-n}) \)
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\end_inset
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.
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Note that the field
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\begin_inset Formula \( Q(x)(\alpha ) \)
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\end_inset
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, being a finite dimensional extension field of
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\begin_inset Formula \( Q(x) \)
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\end_inset
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in characteristic 0, has a unique derivation extending
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\begin_inset Formula \( \frac{d}{dx} \)
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\end_inset
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on
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\begin_inset Formula \( Q(x) \)
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\end_inset
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.
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We can therefore try to construct an annihilating differential operator
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for
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\begin_inset Formula \( T_{n}(x) \)
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\end_inset
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by combination of annihilating differential operators for
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\begin_inset Formula \( \alpha ^{n} \)
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\end_inset
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and
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\begin_inset Formula \( \alpha ^{-n} \)
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\end_inset
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.
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In fact,
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\begin_inset Formula \( L_{1}:=(\alpha -x)\frac{d}{dx}-n \)
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\end_inset
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satisfies
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\begin_inset Formula \( L_{1}[\alpha ^{n}]=0 \)
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\end_inset
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, and
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\begin_inset Formula \( L_{2}:=(\alpha -x)\frac{d}{dx}+n \)
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\end_inset
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satisfies
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\begin_inset Formula \( L_{2}[\alpha ^{-n}]=0 \)
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\end_inset
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.
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A common multiple of
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\begin_inset Formula \( L_{1} \)
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\end_inset
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and
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\begin_inset Formula \( L_{2} \)
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\end_inset
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is easily found by solving an appropriate system of linear equations:
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\layout Standard
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\begin_inset Formula \( L=(x^{2}-1)\left( \frac{d}{dx}\right) ^{2}+x\frac{d}{dx}-n^{2}=\left( (\alpha -x)\frac{d}{dx}+n\right) \cdot L_{1}=\left( (\alpha -x)\frac{d}{dx}-n\right) \cdot L_{2} \)
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\end_inset
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\layout Standard
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It follows that both
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\begin_inset Formula \( L[\alpha ^{n}]=0 \)
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\end_inset
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and
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\begin_inset Formula \( L[\alpha ^{-n}]=0 \)
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\end_inset
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, hence
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\begin_inset Formula \( L[T_{n}(x)]=0 \)
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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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\protected_separator
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2:
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\layout Standard
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Starting from the above equation, we compute a linear relation for the partial
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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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\layout Standard
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\begin_inset Formula
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\[
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\left( 1-2xz+z^{2}\right) \cdot F=1-xz\]
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\end_inset
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\begin_inset Formula
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\[
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\left( 1-2xz+z^{2}\right) ^{2}\cdot \partial _{x}F=z-z^{3}\]
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\end_inset
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\begin_inset Formula
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\[
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\left( 1-2xz+z^{2}\right) ^{3}\cdot \partial _{x}^{2}F=4z^{2}-4z^{4}\]
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\end_inset
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\begin_inset Formula
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\[
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\left( 1-2xz+z^{2}\right) ^{2}\cdot \Delta _{z}F=xz-2z^{2}+xz^{3}\]
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\end_inset
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\begin_inset Formula
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\[
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\left( 1-2xz+z^{2}\right) ^{3}\cdot \partial _{x}\Delta _{z}F=z+2xz^{2}-6z^{3}+2xz^{4}+z^{5}\]
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\end_inset
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\begin_inset Formula
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\[
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\left( 1-2xz+z^{2}\right) ^{3}\cdot \Delta _{z}^{2}F=xz+(2x^{2}-4)z^{2}-(2x^{2}-4)z^{4}-xz^{5}\]
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\end_inset
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\layout Standard
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Solve a
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\begin_inset Formula \( 6\times 6 \)
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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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\[
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x\cdot \partial _{x}F+(x^{2}-1)\cdot \partial _{x}^{2}F-\Delta _{z}^{2}F=0\]
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\end_inset
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\layout Standard
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This is equivalent to the claimed equation
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\begin_inset Formula \( (x^{2}-1)\cdot T_{n}^{''}(x)+x\cdot T_{n}^{'}(x)-n^{2}\cdot T_{n}(x)=0 \)
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\end_inset
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.
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\layout Bibliography
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\cursor 137
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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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Sections 2.
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12 and 2.
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15.
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