cln_mirror/doc/polynomial/tchebychev.lyx

#This file was created by <bruno> Sun Feb 16 00:32:21 1997
#LyX 0.10 (C) 1995 1996 Matthias Ettrich and the LyX Team
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\quotes_language english 
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\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.