A courious thing about orthogonal polynomials

1. Dec 4, 2009

zetafunction

given a set of orthogonal polynomials $$p_{n} (x)$$ with respect to a certain positive measure $$\mu (x) > 0$$ on a certain interval (a,b)

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

$$\int_{a}^{b}dx\mu(x)cos(xz)=f(z)$$

has ALWAYS only real roots ¡¡

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

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

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

* Legendre: measure is 1 defined on (-1,1) , Fourier transform $$\int_{-1}^{1}dxcos(xz)=sin(z)/z$$ 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

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