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				| %% This LaTeX-file was created by <bruno> Sun Feb 16 14:24:52 1997 | |
| %% LyX 0.10 (C) 1995 1996 by Matthias Ettrich and the LyX Team | |
|  | |
| %% Don't edit this file unless you are sure what you are doing. | |
| \documentclass[12pt,a4paper,oneside,onecolumn]{article} | |
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| %% Begin LyX user specified preamble: | |
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| \catcode`@=12 % @ ist ab jetzt wieder ein Sonderzeichen | |
|  | |
| 
 | |
| %% End LyX user specified preamble. | |
| \begin{document} | |
| 
 | |
| The Legendre polynomials  \( P_{n}(x) \) are defined through  | |
| \[ | |
| P_{n}(x)=\frac{1}{2^{n}n!}\cdot \left( \frac{d}{dx}\right) ^{n}(x^{2}-1)^{n}\] | |
| (For a motivation | |
| of the  \( 2^{n} \) in the denominator, look at  \( P_{n}(x) \) modulo an odd prime  \( p \), and | |
| observe that  \( P_{n}(x)\equiv P_{p-1-n}(x)\mod p \) for  \( 0\leq n\leq p-1 \). This wouldn't hold if the  \( 2^{n} \) factor in the denominator | |
| weren't present.) | |
| 
 | |
| \begin{description} | |
| 
 | |
| \item [Theorem:]~ | |
| 
 | |
| \end{description} | |
| 
 | |
|  \( P_{n}(x) \) satisfies the recurrence relation | |
| 
 | |
| 
 | |
| \[ | |
| P_{0}(x)=1\] | |
| 
 | |
| 
 | |
| 
 | |
| \[ | |
| (n+1)\cdot P_{n+1}(x)=(2n+1)x\cdot P_{n}(x)-n\cdot P_{n-1}(x)\] | |
| for  \( n\geq 0 \) and the differential equation  \( (1-x^{2})\cdot P_{n}^{''}(x)-2x\cdot P_{n}^{'}(x)+(n^{2}+n)\cdot P_{n}(x)=0 \) for all  \( n\geq 0 \). | |
| 
 | |
| \begin{description} | |
| 
 | |
| \item [Proof:]~ | |
| 
 | |
| \end{description} | |
| 
 | |
| Let  \( F:=\sum ^{\infty }_{n=0}P_{n}(x)\cdot z^{n} \) be the generating function of the sequence of polynomials. It | |
| is the diagonal series of the power series | |
| \[ | |
| G:=\sum _{m,n=0}^{\infty }\frac{1}{2^{n}m!}\cdot \left( \frac{d}{dx}\right) ^{m}(x^{2}-1)^{n}\cdot y^{m}\cdot z^{n}\] | |
| Because the Taylor series | |
| development theorem holds in formal power series rings (see [1], section | |
| 2.16), we can simplify | |
| \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*} | |
| We take over the terminology from the ``diag\_rational'' | |
| paper; here  \( R=Q[x] \) and  \( M=Q[[x]] \) (or, if you like it better,  \( M=H(C) \), the algebra of | |
| functions holomorphic in the entire complex plane).  \( G\in M[[y,z]] \) is rational; | |
| hence  \( F \) is algebraic over  \( R[z] \). Let's proceed exactly as in the ``diag\_series'' | |
| paper.  \( F(z^{2}) \) is the coefficient of  \( t^{0} \) in | |
| \[ | |
| G(zt,\frac{z}{t})=\frac{2t}{2t-\left( (x+zt)^{2}-1\right) z}=\frac{2t}{-z^{3}\cdot t^{2}+2(1-xz^{2})\cdot t+(z-x^{2}z)}\] | |
| The splitting field of the denominator | |
| is  \( L=Q(x)(z)(\alpha ) \) where  | |
| \[ | |
| \alpha _{1/2}=\frac{1-xz^{2}\pm \sqrt{1-2xz^{2}+z^{4}}}{z^{3}}\] | |
| 
 | |
| \[ | |
| \alpha =\alpha _{1}=\frac{2}{z^{3}}-\frac{2x}{z}+\frac{1-x^{2}}{2}z+\cdots \in Q(x)[[z]][\frac{1}{z}]\] | |
| 
 | |
| \[ | |
| \alpha _{2}=\frac{x^{2}-1}{2}z+\cdots \in Q(x)[[z]][\frac{1}{z}]\] | |
| The partial fraction decomposition of  \( G(zt,\frac{z}{t}) \) reads | |
| \[ | |
| G(zt,\frac{z}{t})=-\frac{2}{z^{3}}\cdot \frac{1}{\alpha _{1}-\alpha _{2}}\cdot \left( \frac{\alpha _{1}}{t-\alpha _{1}}-\frac{\alpha _{2}}{t-\alpha _{2}}\right) \] | |
| It follows | |
| that | |
| \[ | |
| F(z^{2})=-\frac{2}{z^{3}}\cdot \frac{1}{\alpha _{1}-\alpha _{2}}\cdot \left( \frac{\alpha _{1}}{0-\alpha _{1}}-0\right) =\frac{1}{\sqrt{1-2xz^{2}+z^{4}}}\] | |
| hence | |
| \[ | |
| F(z)=\frac{1}{\sqrt{1-2xz+z^{2}}}\] | |
| 
 | |
| 
 | |
| It follows that  \( (1-2xz+z^{2})\cdot \frac{d}{dz}F+(z-x)\cdot F=0 \). This is equivalent to the claimed recurrence. | |
| 
 | |
| Starting from the closed form for  \( F \), we compute a linear relation | |
| for the partial derivatives of  \( F \). Write  \( \partial _{x}=\frac{d}{dx} \) and  \( \Delta _{z}=z\frac{d}{dz} \). One computes | |
| \[ | |
| F=1\cdot F\] | |
| 
 | |
| \[ | |
| \left( 1-2xz+z^{2}\right) \cdot \partial _{x}F=z\cdot F\] | |
| 
 | |
| \[ | |
| \left( 1-2xz+z^{2}\right) ^{2}\cdot \partial _{x}^{2}F=3z^{2}\cdot F\] | |
| 
 | |
| \[ | |
| \left( 1-2xz+z^{2}\right) \cdot \Delta _{z}F=(xz-z^{2})\cdot F\] | |
| 
 | |
| \[ | |
| \left( 1-2xz+z^{2}\right) ^{2}\cdot \partial _{x}\Delta _{z}F=(z+xz^{2}-2z^{3})\cdot F\] | |
| 
 | |
| \[ | |
| \left( 1-2xz+z^{2}\right) ^{2}\cdot \Delta _{z}^{2}F=\left( xz+(x^{2}-2)z^{2}-xz^{3}+z^{4}\right) \cdot F\] | |
| Solve | |
| a homogeneous  \( 5\times 6 \) system of linear equations over  \( Q(x) \) to get  | |
| \[ | |
| \left( 1-2xz+z^{2}\right) ^{2}\cdot \left( (-2x)\cdot \partial _{x}F+(1-x^{2})\cdot \partial _{x}^{2}F+\Delta _{z}F+\Delta _{z}^{2}F\right) =0\] | |
| Divide by | |
| the first factor to get | |
| \[ | |
| (-2x)\cdot \partial _{x}F+(1-x^{2})\cdot \partial _{x}^{2}F+\Delta _{z}F+\Delta _{z}^{2}F=0\] | |
| This is equivalent to the claimed equation | |
|  \( (1-x^{2})\cdot P_{n}^{''}(x)-2x\cdot P_{n}^{'}(x)+(n^{2}+n)\cdot P_{n}(x)=0 \). | |
| 
 | |
| \begin{lyxsectionbibliography} | |
| 
 | |
| \item [1] Bruno Haible: D-finite power series in several variables. \em Diploma | |
| thesis, University of Karlsruhe, June 1989\em . Sections 2.14, 2.15 | |
| and 2.22. | |
| 
 | |
| \end{lyxsectionbibliography} | |
| 
 | |
| \end{document}
 |