Search

By using the principle of mathematical induction, it was shown by Singh and Chanam (J. Math. Inequal 15:1663–1675, 2021) that if p(z) is a polynomial of degree n having all its zeros in | z | ≤ 1 , then for all z on | z | = 1 for which p (z) ≠ 0 , ℜ z p ′ (z) p (z) ≥ n + 1 2 - 1 2 | a 0 | | a n | . In this paper, by using simple techniques we generalize the above inequality, thereby give a simple proof of the above inequality. As an application of our result, we obtain improvements of the well-known result due to Malik (J. Lond. Math. Soc 1(2):57–60, 1969). Further, we obtain some sharp refinements of a result due to Aziz and Rather (J. Math. Inequal. Appl 1:231–238, 1998). These results take into account the placement of the coefficients of the underlying polynomial. Moreover, a concrete numerical example is presented in order to graphically illustrate and compare the obtained inequalities with a classical result, showing that in some situations, the bounds obtained by our results can be considerably sharper than the ones previously known. [ABSTRACT FROM AUTHOR]

In this paper, we establish some new polynomial inequalities involving various operators and their relationship with the existing results. All these inequalities have been proved under the assumption that the underlying polynomial is being constrained with respect to its zeros. [ABSTRACT FROM AUTHOR]

If $ P(z)=a_n\prod _{v=1}^{n}(z-z_v) $ P (z) = a n ∏ v = 1 n (z − z v) is a polynomial of degree n having all its zeros in $ |z|\le k, k\ge 1 $ | z | ≤ k , k ≥ 1 then Aziz [Inequalities for the derivative of a polynomial. Proc Am Math Soc. 1983;89(2):259–266] proved that \[ \max_{|z|=1}|P'(z)|\ge \frac{2}{1+k^n}\sum_{v=1}^{n}\frac{k}{k+|z_v|}\max_{|z|=1}|P(z)|. \] max | z | = 1 | P ′ (z) | ≥ 2 1 + k n ∑ v = 1 n k k + | z v | max | z | = 1 | P (z) |. Recently, Kumar [On the inequalities concerning polynomials. Complex Anal Oper Theory. 2020;14(6):1–11 (Article ID 65)] established a generalization of this inequality and proved under the same hypothesis for a polynomial $ P(z)=a_0+a_1z+a_2z^2+\cdots +a_nz^n=a_n\prod _{v=1}^{n}(z-z_v) $ P (z) = a 0 + a 1 z + a 2 z 2 + ⋯ + a n z n = a n ∏ v = 1 n (z − z v) , that $$\begin{align*} & \max_{|z|=1}|P'(z)| \\ & \ge \left(\frac{2}{1+k^n}+\frac{(|a_n|k^n-|a_0|)(k-1)}{(1+k^n)(|a_n|k^n+k|a_0|)}\right)\sum_{v=1}^{n}\frac{k}{k+|z_v|}\max_{|z|=1}|P(z)|. \end{align*} $$ max | z | = 1 | P ′ (z) | ≥ (2 1 + k n + (| a n | k n − | a 0 |) (k − 1) (1 + k n) (| a n | k n + k | a 0 |)) ∑ v = 1 n k k + | z v | max | z | = 1 | P (z) |. In this paper, we sharpen the above inequalities and further extend the obtained results to the polar derivative of a polynomial. As a consequence, our results also sharpens considerably some results of Dewan and Upadhye [Inequalities for the polar derivative of a polynomial. J Ineq Pure Appl Math. 2008;9:1–9 (Article ID 119)]. [ABSTRACT FROM AUTHOR]

The goal of this paper is to obtain some comparison inequalities for a linear operator between polynomials in the plane. The polynomials under study have constraints on their zeros and the estimates obtained turn out to be generalizations of many classical inequalities for polynomials, including the well known Erdos-Lax inequality and inequality of the Ankeny-Rivlin theorem. [ABSTRACT FROM AUTHOR]

Let an be defined recursively by an = an–1 + an–2 + pn + q if n ≥ 2, with a0 = a1 = 1, where p and q are constants. The p = 0 case of an corresponds to what have been termed the generalized Leonardo numbers, the q = 1 case of which are the classical Leonardo numbers. Let L n (x) = ∑ i = 0 n a i x n − i , which we refer to as a generalized Leonardo coefficient polynomial. Here, it is shown, under some mild assumptions on p and q, that Ln(x) has exactly one real zero if n is odd and no real zeros if n is even. This yields a previous result for the Fibonacci coefficient polynomials when p = q = 0 and a new analogous result for the Leonardo coefficient polynomials when p = 0, q = 1. Further, when p = 0 and –1 < q ≤ 1 + ρ, we show that the sequence consisting of the real zeros of the Ln(x) for n odd is strictly decreasing with limit –ρ, where ρ = 1 + 5 2. Finally, under the same assumptions on p and q, we are able to show the convergence in modulus of all the zeros of Ln(x). [ABSTRACT FROM AUTHOR]