In this note, we improve upon Cauchy’s classical bound, and upon some recent bounds for the moduli of the zeros of a polynomial. Be a polynomial of degree n, with complex coefficients. Then, according to Cauchy’s classical result, we have the following theorem. | Turk J Math 30 (2006) , 95 – 100. ¨ ITAK ˙ c TUB On Cauchy’s Bound for Zeros of a Polynomial V. K. Jain Abstract In this note, we improve upon Cauchy’s classical bound, and upon some recent bounds for the moduli of the zeros of a polynomial. Key Words: Zeros, polynomials, upper bound, moduli, refinement. 1. Introduction and Statement of Results Let f(z) = z n + an−1 z n−1 + an−2 z n−2 + . . . + a1 z + a0 , ai 6= 0, for at least one i ∈ I, I = {0, 1, 2, . . ., n − 1}, be a polynomial of degree n, with complex coefficients. Then, according to Cauchy’s classical result [1], we have the following theorem. Theorem A Z[f(z)] ⊂ B(η) ⊂ B(1 + a), where η is the unique positive root of the equation Q(x) = 0, Q(x) = xn − |an−1 |xn−1 − |an−2|xn−2 − . . . − |a1 |x − |a0 |, (1) AMS Mathematics Subject Classification: Primary 30C15, 30C10 95 JAIN Z[f(z)] = the set of all zeros of the polynomial f(z), ¯ B(r) = {z : |z| (1 + δ0 )n − |an−1 |(1 + δ0 )n−1 − |an−2 |(1 + δ0 )n−2 (1 + δ0 )n−3 − 1 −|an−3 |(1 + δ0 )n−3 − a , δ0 (1 + δ0 )n−3 (1 + δ0 )3 − |an−1|(1 + δ0 )2 − a , |an−2 |(1 + δ0 ) − |an−3| − δ0 = (1 + δ0 )n−3 Q0 (δ0 ), δ0 = 0, which implies η < 1 + δ0 . Again, Q0 (δ1 ) = Q0 (δ1 ) − δ1 Q1 (δ1 ) − Q1 (δ1 ) = δ1 (a − |an−3 |), (by (3) and (6)), ≥ 0, thereby implying that δ0 ≤ δ1 . And now Theorem 1 follows, by using the fact that η is unique positive root of the equation Q(x) = .
