Do Orthogonal Polynomials have always real zeros ??

[zetafunction]

the idea is , do orthogonal polynomials p_{n} (x) have always REAl zeros ?

for example n=2 there is a second order polynomial with 2 real zeros

if we consider that there is a self-adjoint operator L so L[p_{n} (x)]= \mu _{n} p_{n} (x) if the orthogonal POLYNOMIALS are eigenfunctions of an operator with a real spectrum are ALL the zeros real ? , and if all the zeros are REAL can they be related to the spectrum of L ??

[HallsofIvy]

I'm a bit confused. What do you mean by "orthogonal polynomials"? I know what "orthogonal sets of polynomials" are but there are many different such sets. I can't think of any polynomials that would qualify as simply "orthogonal polynomials" without further requirements.

[zetafunction]

i meant

p_{n} (x) is a Polynomial.

let be 'n' and 'm' integers then , there is a weight function so

\int_{a}^{b} dx w(x) p_{n}(x).p_{m}(x)= \delta _{m}^{n}

this polynomial p_{n} (x) or set of polinomial dependent on the index 'n' are eigenfunctions of a certain operator L with Real spectrum

for example , Legendre, Chebyshev, Polynomials seem to have real roots only and are used to compute quadrature formulae

http://en.wikipedia.org/wiki/Gaussian_quadrature

[g_edgar]

For example, 5.3 here:

http://en.wikipedia.org/wiki/Orthogonal_polynomials#Existence_of_real_roots

and much more here:

http://mathworld.wolfram.com/OrthogonalPolynomials.html

[mathman]

Many ortho polynomial sets have p0(x)=1.

[HallsofIvy]

i meant

p_{n} (x) is a Polynomial.

let be 'n' and 'm' integers then , there is a weight function so

\int_{a}^{b} dx w(x) p_{n}(x).p_{m}(x)= \delta _{m}^{n}

this polynomial p_{n} (x) or set of polinomial dependent on the index 'n' are eigenfunctions of a certain operator L with Real spectrum

for example , Legendre, Chebyshev, Polynomials seem to have real roots only and are used to compute quadrature formulae

http://en.wikipedia.org/wiki/Gaussian_quadrature
That's what I thought you meant. "Orthogonal polynomials" are "orthogonal sets of polynomials". Your question doesn't make sense because there is no such thing as an "orthogonal polynomial". Given any function f(x) over some interval, there exists a set of functions, including f(x), that is "orthogonal".