By Edwin F. Beckenbach, Richard Bellman (auth.)

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45. Starting with one hyperbolic polynomial, we can form new hyperbolic polynomials by means of the following resuIt: § 38. Examples 37 Lemma 2. If P(x) is hyperbolie with respeet to a, and m > 1, then Õ m Q(x) =k~lak ÕXk P(x) (2) is also hyperbolie with respeet to a. The proof follows from ROLLE'S theorem as in § 12. Repeated application of this lemma shows that the polynomials {Pk} defined by m P(sa+x)= :EskPk(X) (3) k =1 are hyperbolic. § 37. Garding's Inequality Let M x m ), where each of the Xi is an m-dimensional vector Xi = (x~, x~, ...

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