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Limits of zeros of polynomial sequences.(English)Zbl 1171.11009

The \(k\)-th order Fibonacci sequence is defined by the recurrence \[G(k,l)=G(k,l-1)+G(k,l-2)+\dots+G(k,l-k)\] with the initial value \(G(k,1)=1\). One may see that, for example, \(G(2,l)=G(2,l-1)+G(2,l-2)=F_l,\) where \(F_l\) is the \(l\)-th Fibonacci number. The characteristic polynomials for these sequences are defined by \(F_k(x)=x^k-x^{k-1}-x^{k-2}-\dots-1.\)
There are some conjectures of G. Grossman and S. Narayan concerning to the zeros of the polynomials \(F_k(x)\) and their derivatives. The authors anwer these in the affirmative. To obtain the results, they take the polynomials \(D_j(x)=F_{j+l}^{(l)}(x)\) (\(l\) is fixed).
In what follows, \(u_j\) denotes the positive root of \(D_j(x)\) and – if \(j\) is even – \(v_j\) denotes the negative root of \(D_j(x)\). (That really one or two real zeros exist, comes from the theorem below.)
Theorem 2.1: (1) Let \(j=k-l\). Then \(\lim_{j\rightarrow\infty}u_j=2.\) All of the other complex roots of \(D_j(x)\) are inside of \(|z|<u_j\). For even \(j\), we have \(\lim_{j\rightarrow\infty}v_j=-1.\)
(2) If \(j\) is odd, then \(D_j(x)\) has one positive root and no negative roots. If \(j\) is even, then \(D_j(x)\) has one positive and one negative roots.
(3) For \(j\geq 2\) we have \(u_{j+1}>u_j\).
(4) There exists a number \(N_0\), such that for even \(n>N_0\), we have \(v_{n+2}<v_n\).
The authors prove the corresponding result for the integrals of the characteristic polynomials, too.

MSC:

11B39 Fibonacci and Lucas numbers and polynomials and generalizations

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