A courious thing about orthogonal polynomials

[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

 

[jvicens]

First of all, it is important to note that the concept of orthogonal polynomials and their associated measures is a fundamental tool in mathematical analysis and has many applications in various fields such as physics, engineering, and statistics. These polynomials are defined as a sequence of polynomials that are orthogonal with respect to a given measure on a certain interval. This means that the inner product of any two different polynomials in the sequence is equal to zero.

Now, let's consider the integral transform given in the forum post:

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

This transform is known as the Fourier cosine transform and it is a fundamental tool in the study of orthogonal polynomials. It is also closely related to the Fourier transform, which is widely used in many areas of mathematics and physics.

The reason why the roots of f(z) are always real can be explained by the properties of the Fourier cosine transform. Firstly, it is important to note that the Fourier cosine transform is an even function, meaning that it is symmetric about the y-axis. This implies that the real and imaginary parts of f(z) are also even functions.

Moreover, the Fourier cosine transform is an entire function, which means that it is defined and analytic for all complex values of z. This property is closely related to the fact that the roots of f(z) are always real. It can be shown that if an entire function has only real coefficients, then its roots are also real. In the case of the Fourier cosine transform, the coefficients of the function are all real, as they are given by the integral of a real-valued function (cosine) over a real interval. Therefore, the roots of f(z) are always real.

Furthermore, the properties of the measures used in the examples given in the forum post also play a crucial role in ensuring that the roots of f(z) are real. In particular, the measures used are symmetric about the origin, which further supports the symmetry of the Fourier cosine transform and its real roots.

In conclusion, the reason why the Fourier cosine transform always has real roots can be explained by the properties of the transform itself, as well as the properties of the measures used in the definition of orthogonal polynomials. This is a fundamental and interesting result that has many applications in various fields of mathematics and science.