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My research has brought me to the following, very general problem.

Given a fixed, but arbitrary, natural number, $\displaystyle v$, consider the following family of polynomials: The $\displaystyle (n-1)^{th}$ derivative of

$\displaystyle (1-x^2)^{v+n} \ \ \forall n \in \mathbb{N} $

I would like to prove (or disprove) that the roots of this entire family of polynomials forms a dense subset of the interval $\displaystyle [0,1]$ for any value of $\displaystyle v$ (I am not interested in roots outside the interval $\displaystyle [0,1]$).

In other words, given any subinterval, $\displaystyle [a,b]$,no mater how small, at least one of these polynomials has at least one root in the interval $\displaystyle [a,b]$ (for any fixed value of $\displaystyle v$).

I realize my question is very general and will happily accept any partial solutions.

  • 0
    There's a proof of the interlacing of Legendre polynomial roots in Chihara; have you by any chance been able to see the book?2010-12-03

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First note that this family of polynomials is orthogonal, on the interval $[-1,1]$, with the weight factor $(1-x^2)^{-(v+1)}$. This is not much of a surprise since the definition is very similar to that of the traditional Legendre Polynomials, which are orthogonal. Next, we use the following deep result involving orthogonal polynomials:

If $\{p_n\}$ is a family of orthogonal polynomials with roots in $[-1,1]$ and $N(a,b,n)$ represents the number of roots of $p_n$ in [$\cos(b),\cos(a)$] then

$\lim_{n\to \infty}\frac1{n} N(a,b,n)=\frac{b-a}{\pi}$

Thus for any small subinterval [$\cos(b),\cos(a)$], there exists $n$ sufficiently large such that $N(a,b,n)>1$ implying that the roots of these polynomials do form a dense subset of $[-1,1]$.

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    Where does this result come from?2017-12-31