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AI InsightBy considering the polynomial p(z) with no zeros in |z| < k, (k \geq 1) and simultaneously thinking of s\rm th derivative (0 \leq s < n) of the polynomial, we have obtained the generalization \begin{displaymath} M(p(s),R) \leq \left\{\begin{array}{l} (1/2){\frac{ds}{dRs}(Rn + kn)}(2/(1+k))nM(p,1), R \geq k,\ (1/(Rs+ks))[{\frac{ds}{dxs}(1+xn)}x=1]((R+k)/(1+k))nM(p,1), 1 \leq R \leq k,\end{array}\right. of Ankeny and Rivlin's result.
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Özet
For an arbitrary entire function f(z), let M(f,r) = max|z|=r|f(z)|. For a polynomial p(z) of degree n, it is known that M(p,R) \leq Rn M(p,1), R > 1. By considering the polynomial p(z) with no zeros in |z| < 1, Ankeny and Rivlin obtained the refinement M(p,R) \leq {(Rn+1)/2}M(p,1), R > 1. By considering the polynomial p(z) with no zeros in |z| < k, (k \geq 1) and simultaneously thinking of s\rm th derivative (0 \leq s < n) of the polynomial, we have obtained the generalization \begin{displaymath} M(p(s),R) \leq \left\{\begin{array}{l} (1/2){\frac{ds}{dRs}(Rn + kn)}(2/(1+k))nM(p,1), R \geq k,\ (1/(Rs+ks))[{\frac{ds}{dxs}(1+xn)}x=1]((R+k)/(1+k))nM(p,1), 1 \leq R \leq k,\end{array}\right. of Ankeny and Rivlin's result.
Abstract
For an arbitrary entire function f(z), let M(f,r) = max|z|=r|f(z)|. For a polynomial p(z) of degree n, it is known that M(p,R) \leq Rn M(p,1), R > 1. By considering the polynomial p(z) with no zeros in |z| < 1, Ankeny and Rivlin obtained the refinement M(p,R) \leq {(Rn+1)/2}M(p,1), R > 1. By considering the polynomial p(z) with no zeros in |z| < k, (k \geq 1) and simultaneously thinking of s\rm th derivative (0 \leq s < n) of the polynomial, we have obtained the generalization \begin{displaymath} M(p(s),R) \leq \left\{\begin{array}{l} (1/2){\frac{ds}{dRs}(Rn + kn)}(2/(1+k))nM(p,1), R \geq k,\ (1/(Rs+ks))[{\frac{ds}{dxs}(1+xn)}x=1]((R+k)/(1+k))nM(p,1), 1 \leq R \leq k,\end{array}\right. of Ankeny and Rivlin's result.