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Consider Legendre polynomials $p_n (x)$ on $[-1,1]$. For each $n \in \mathbb{N}$ we denote the zeros of $p_n (x)$ by $\left( x_j ^{(n)} \right) _{j=1} ^n$.

We know that these roots are distinct, and we know that for all $n\in \mathbb{N}$ then $x_j ^{(n)} \neq x_i ^{(n+1)}$ for all $i$ and $j$. In fact, we even now that they are interlaced i.e. $x_i ^{(n+1)} < x_i ^{(n)} < x_{i+1} ^{(n+1)}$.

On the other hand, we know for example that for all odd $n=2k+1$ we have that $x_{k+1} ^{(2k+1)} = 0$, and so I arrive at my question:

Question 1: Are there any rules as to if, when, and how roots recur, i.e. $x_i ^{(n)} = x_j ^{(m)}$ for some relevant indices?

Question 2: Apart from the well known Chebyshev polynomials, are there any other orthogonal polynomials with high rate of roots recurrence?

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  • $\begingroup$ Does not recurrence relation among Legendre polynomial's (or Chebyshev, Hermite etc) and ODE they satisfy give you an information about a relation between consecutive zeros? $\endgroup$ – Paata Ivanishvili Sep 5 '16 at 2:42
  • $\begingroup$ I assume I don't know ALL the relevant recurrence relations, but from the ones I do I couldn't deduce to recurrence of zeros. I'd be happy if you will prove me wrong. $\endgroup$ – Amir Sagiv Sep 5 '16 at 5:14
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It is a conjecture of Stieltjes, apparently still open, see

T.J. Stieltjes, Letter No. 275 of Oct. 2, 1890, in Correspondance d'Hermite et de Stieltjes, vol 2, Gauthier-Villars, Paris, 1905.

that Legendre polynomials of different degrees have no common roots, except $x=0$ when both degrees are odd.

Laguerre polynomials $L_{n}^{\alpha}$, $\alpha>-1$, orthogonal with respect to the weight function $x^{\alpha}e^{-x}$ on $(0,\infty)$, may have common roots. For instance, as observed in

K. Driver, K. Jordaan, Stieltjes interlacing of zeros of Laguerre polynomials from different sequences. Indag. Math. (N.S.) 21 (2011), 204-211.

$L_{2}^{23}(x)$ and $L_{4}^{23}(x)$ have $x=30$ as a common root. It seems also to be an open question how many common zeros are possible in general for $L_{n}^{\alpha}$ and $L_{n+m}^{\alpha}$, $n\geq 0$ and $m>1$. The corresponding question for Hermite polynomials about a common root other than zero seems to be unanswered as well.

Finally, a classical theorem of Stieltjes states that if $p_{0},p_{1},p_{2},\ldots$ is any sequence of orthogonal polynomials then the zeros of $p_{k}$ and $p_{n}$, $k<n$, are interlacing in the sense that each open interval of the form $$(-\infty, z_{1}), (z_{1}, z_{2}), . . . , (z_{k-1}, z_{k}), (z_{k} , \infty)$$ where $z_{1}<z_{2}<\cdots<z_{k}$ are the zeros of $p_{k}$, contains at least one zero of $p_{n}$. It implies that at least $k+1$ zeros of $p_{n}$ are distinct from the zeros of $p_{k}$, or equivalently $p_{k}$ and $p_{n}$ have at most $\min(k,n-k-1)$ zeros in common. It is proved in

P.C. Gibson, Common zeros of two polynomials in an orthogonal sequence, J. Approx. Theory 105 (2000) 129-132.

that this bound is sharp in the sense that for any $k<n$, there exists a sequence of orthogonal polynomials $q_{0},\ldots,q_{n}$ such that $q_{k}$ and $q_{n}$ have precisely $\min(k,n-k-1)$ zeros in common.

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Numerical evidence suggests that the nonzero roots of the Legendre polynomials do not repeat. The numerical experiment I performed is simple: I gathered all of the nonzero roots of the first 100 Legendre polynomials. I sorted them in ascending order. I then graphed the first order difference of this sorted vector with logarithmic scaling in the vertical direction. Here is the graphical output.

enter image description here

Its a bit hard to see, but there are no holes in this figure, which suggests that all 5000 nonzero roots of the first 100 Legendre polynomials are distinct.

ADD

For completeness, here is an eight line MATLAB code that verifies Stieltjes' 126 year old conjecture up to the first 100 Legendre polynomials (check out user111's answer for a detailed reference to Stieltjes' paper).

all_roots=[];
for i=1:100
    sums x;
    roots = vpasolve(legendreP(i,x) == 0);
    all_roots(end+1:end+length(roots))=roots;
end
all_roots=all_roots(~(all_roots==0));
length(unique(all_roots))
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