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SOME EXTREMAL PROBLEMS FOR TRIGONOMETRICAL AND COMPLEX POLYNOMIALS

CARL HYLTÉN-CAVALLIUS
Mathematica Scandinavica
Vol. 3, No. 1 (August 31, 1955), pp. 5-20
Stable URL: http://www.jstor.org/stable/24490332
Page Count: 16
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Abstract

Let Πn = Πn(it, cosα), where n ≧ 2 is an integer, t ≠ 0 is real and 0 ≦ α ≦ π, be the class of trigonometrical polynomials Φn with real coefficients and of order ≦ n, such that |Φn(x)| ≦ 1 for all real x and Φn(it) = cosα. For every real x the functions $\mathrm{m}\left(\mathrm{x}\right)={\mathrm{inf}}_{{\mathrm{\Phi }}_{\mathrm{n}\in \mathrm{\Pi }\mathrm{n}}}\mathrm{\Phi }\mathrm{n}\left(\mathrm{x}\right)$ and $\mathrm{M}\left(\mathrm{x}\right)={\mathrm{sup}}_{{\mathrm{\Phi }}_{\mathrm{n}}\in {\mathrm{\Pi }}_{\mathrm{n}}}{\mathrm{\Phi }}_{\mathrm{n}}\left(\mathrm{x}\right)$ are determined. If Tv denotes the vth Tchebycheff polynomial and a = cos(α/2n)/cosh½t, one has, for instance, m(x) = T2n(a cos½x) for every x such that a |cos½x| ≧ cos(π/2n). For all other x one has m(x) = -1. This theorem is generalized to the case when the condition Φn(it) = cosα is replaced by Φn(it) = cos(α+iβ), α and β real. It is also used to solve the following problem: Let Pn(z) be a polynomial of a complex variable z with complex coefficients and of degree ≦ n, n an integer ≧ 2. Suppose that on the circle |z| = 1 the absolute value of Pn(z) takes its maximum at the point z = 1. How near to this point can there be a zero z0 = ϱeiφ of Pn(z)? The answer is that z0 must lie outside or on the curve cos½φ = ½ (ϱ½ + ϱ-½) cos(π/2n), - π/n ≦ φ ≦ π/n. If z0 is a point of the curve there is essentially only one polynomial satisfying the conditions, and this polynomial is explicitly given. If z0 lies outside the curve there are infinitely many such polynomials. The curve is a "limaçon of Pascal".

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