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A courious thing about orthogonal polynomials

  1. Dec 4, 2009 #1
    given a set of orthogonal polynomials [tex] p_{n} (x) [/tex] with respect to a certain positive measure [tex] \mu (x) > 0 [/tex] on a certain interval (a,b)

    then i have notices for several cases that f(z) defined by the integral transform

    [tex] \int_{a}^{b}dx\mu(x)cos(xz)=f(z)[/tex]

    has ALWAYS only real roots ¡¡

    * Laguerre : measure is exp(-x) on (0,oo) then [tex] \int_{0}^{\infty}dxe^{-x}cos(xz)=(1+z^{2})^{-1}[/tex] , there is a root only as x tends to oo

    * Chebyshev: measure is [tex] (1-x^{2})^{-1/2}[/tex] defined on (-1,1) , then [tex] \int_{-1}^{1}dxcos(xz)(1-x^{2})^{-1/2})=J_{0} (2z)[/tex] ALL the roots are real

    * Hermite : measure is [tex] e^{-x^{2}} [/tex] then the Fourier transform is [tex]\int_{-\infty}^{\infty}dxe^{-x^{2}}cos(xz)= Cexp(-z^{2}/4) [/tex] with ONLY a real root as x tends to oo

    * Legendre: measure is 1 defined on (-1,1) , Fourier transform [tex] \int_{-1}^{1}dxcos(xz)=sin(z)/z [/tex] having ALL the roots to be real

    why does this happen ? , also it seems that the Fourier cosine transform f(z) is always an ENTIRE function having always real roots
  2. jcsd
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