#This file was created by Sun Feb 16 00:32:21 1997 #LyX 0.10 (C) 1995 1996 Matthias Ettrich and the LyX Team \lyxformat 2.10 \textclass article \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 1 \papersides 1 \paperpagestyle plain \layout Standard The Tschebychev polynomials (of the 1st kind) \begin_inset Formula \( T_{n}(x) \) \end_inset are defined through the recurrence relation \layout Standard \begin_inset Formula \[ T_{0}(x)=1\] \end_inset \layout Standard \begin_inset Formula \[ T_{1}(x)=x\] \end_inset \layout Standard \begin_inset Formula \[ T_{n+2}(x)=2x\cdot T_{n+1}(x)-T_{n}(x)\] \end_inset for \begin_inset Formula \( n\geq 0 \) \end_inset . \layout Description Theorem: \layout Standard \begin_inset Formula \( T_{n}(x) \) \end_inset satisfies the differential equation \begin_inset Formula \( (x^{2}-1)\cdot T_{n}^{''}(x)+x\cdot T_{n}^{'}(x)-n^{2}\cdot T_{n}(x)=0 \) \end_inset for all \begin_inset Formula \( n\geq 0 \) \end_inset . \layout Description Proof: \layout Standard Let \begin_inset Formula \( F:=\sum ^{\infty }_{n=0}T_{n}(x)z^{n} \) \end_inset be the generating function of the sequence of polynomials. The recurrence is equivalent to the equation \begin_inset Formula \[ (1-2x\cdot z+z^{2})\cdot F=1-x\cdot z\] \end_inset \layout Description Proof \protected_separator 1: \layout Standard \begin_inset Formula \( F \) \end_inset is a rational function in \begin_inset Formula \( z \) \end_inset , \begin_inset Formula \( F=\frac{1-xz}{1-2xz+z^{2}} \) \end_inset . From the theory of recursions with constant coefficients, we know that we have to perform a partial fraction decomposition. So let \begin_inset Formula \( p(z)=z^{2}-2x\cdot z+1 \) \end_inset be the denominator and \begin_inset Formula \( \alpha =x+\sqrt{x^{2}-1} \) \end_inset and \begin_inset Formula \( \alpha ^{-1} \) \end_inset its zeroes. The partial fraction decomposition reads \begin_inset Formula \[ F=\frac{1-xz}{1-2xz+z^{2}}=\frac{1}{2}\left( \frac{1}{1-\alpha z}+\frac{1}{1-\alpha ^{-1}z}\right) \] \end_inset hence \begin_inset Formula \( T_{n}(x)=\frac{1}{2}(\alpha ^{n}+\alpha ^{-n}) \) \end_inset . Note that the field \begin_inset Formula \( Q(x)(\alpha ) \) \end_inset , being a finite dimensional extension field of \begin_inset Formula \( Q(x) \) \end_inset in characteristic 0, has a unique derivation extending \begin_inset Formula \( \frac{d}{dx} \) \end_inset on \begin_inset Formula \( Q(x) \) \end_inset . We can therefore try to construct an annihilating differential operator for \begin_inset Formula \( T_{n}(x) \) \end_inset by combination of annihilating differential operators for \begin_inset Formula \( \alpha ^{n} \) \end_inset and \begin_inset Formula \( \alpha ^{-n} \) \end_inset . In fact, \begin_inset Formula \( L_{1}:=(\alpha -x)\frac{d}{dx}-n \) \end_inset satisfies \begin_inset Formula \( L_{1}[\alpha ^{n}]=0 \) \end_inset , and \begin_inset Formula \( L_{2}:=(\alpha -x)\frac{d}{dx}+n \) \end_inset satisfies \begin_inset Formula \( L_{2}[\alpha ^{-n}]=0 \) \end_inset . A common multiple of \begin_inset Formula \( L_{1} \) \end_inset and \begin_inset Formula \( L_{2} \) \end_inset is easily found by solving an appropriate system of linear equations: \layout Standard \begin_inset Formula \( L=(x^{2}-1)\left( \frac{d}{dx}\right) ^{2}+x\frac{d}{dx}-n^{2}=\left( (\alpha -x)\frac{d}{dx}+n\right) \cdot L_{1}=\left( (\alpha -x)\frac{d}{dx}-n\right) \cdot L_{2} \) \end_inset \layout Standard It follows that both \begin_inset Formula \( L[\alpha ^{n}]=0 \) \end_inset and \begin_inset Formula \( L[\alpha ^{-n}]=0 \) \end_inset , hence \begin_inset Formula \( L[T_{n}(x)]=0 \) \end_inset . \layout Description Proof \protected_separator 2: \layout Standard Starting from the above equation, 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 \layout Standard \begin_inset Formula \[ \left( 1-2xz+z^{2}\right) \cdot F=1-xz\] \end_inset \begin_inset Formula \[ \left( 1-2xz+z^{2}\right) ^{2}\cdot \partial _{x}F=z-z^{3}\] \end_inset \begin_inset Formula \[ \left( 1-2xz+z^{2}\right) ^{3}\cdot \partial _{x}^{2}F=4z^{2}-4z^{4}\] \end_inset \begin_inset Formula \[ \left( 1-2xz+z^{2}\right) ^{2}\cdot \Delta _{z}F=xz-2z^{2}+xz^{3}\] \end_inset \begin_inset Formula \[ \left( 1-2xz+z^{2}\right) ^{3}\cdot \partial _{x}\Delta _{z}F=z+2xz^{2}-6z^{3}+2xz^{4}+z^{5}\] \end_inset \begin_inset Formula \[ \left( 1-2xz+z^{2}\right) ^{3}\cdot \Delta _{z}^{2}F=xz+(2x^{2}-4)z^{2}-(2x^{2}-4)z^{4}-xz^{5}\] \end_inset \layout Standard Solve a \begin_inset Formula \( 6\times 6 \) \end_inset system of linear equations over \begin_inset Formula \( Q(x) \) \end_inset to get \begin_inset Formula \[ x\cdot \partial _{x}F+(x^{2}-1)\cdot \partial _{x}^{2}F-\Delta _{z}^{2}F=0\] \end_inset \layout Standard This is equivalent to the claimed equation \begin_inset Formula \( (x^{2}-1)\cdot T_{n}^{''}(x)+x\cdot T_{n}^{'}(x)-n^{2}\cdot T_{n}(x)=0 \) \end_inset . \layout Bibliography \cursor 137 [1] Bruno Haible: D-finite power series in several variables. \shape italic Diploma thesis, University of Karlsruhe, June 1989. \shape default Sections 2. 12 and 2. 15.