A New Inequality for a Polynomial
نویسندگان
چکیده
Let p(z) = a0 + ∑n j=t ajz be a polynomial of degree n, having no zeros in |z| < k, k ≥ 1, then it has been shown that for R > 1 and |z| = 1, |p(Rz)−p(z)| ≤ (Rn− 1)(1+AtBtkt+1)/(1+kt+1+AtBt(kt+1+k2t))max|z|=1 |p(z)|−{1− (1+AtBtkt+1)/(1+ kt+1 +AtBt(kt+1 + k2t))}((Rn − 1)m/kn), where m = min|z|=k |p(z)|, 1 ≤ t < n, At = (Rt−1)/(Rn−1), and Bt = |at/a0|. Our result generalizes and improves some well-known results. 2000 Mathematics Subject Classification. 30A10, 30E15.
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