{"record_type":"pith_number_record","schema_url":"https://pith.science/schemas/pith-number/v1.json","pith_number":"pith:2018:PPHE5TPLHSRUMTFCX4PKWZNS4Q","short_pith_number":"pith:PPHE5TPL","schema_version":"1.0","canonical_sha256":"7bce4ecdeb3ca3464ca2bf1eab65b2e402d794751e87d57ee49d1baa1c6b7603","source":{"kind":"arxiv","id":"1805.06927","version":1},"attestation_state":"computed","paper":{"title":"Estimating the Koebe radius for polynomials","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.CV","authors_text":"Alex Stokolos, Andrey Smorodin, Dmitriy Dmitrishin","submitted_at":"2018-05-17T19:00:16Z","abstract_excerpt":"For a pair of conjugate trigonometrical polynomials $C (t) = \\sum_ { j = 1 } ^N { { a_j}\\cos jt }, S(t) = \\sum_ { j = 1 } ^N { { a_j}\\sin jt }$ with real coefficients and normalization ${a_1} = 1 $ we solve the extremal problem \\[\n  \\sup_ {a_2,...,a_N} \\left ({ \\min_t \\left\\{ {\\Re \\left ({ F\\left ({ { e^ {it} } } \\right) } \\right): \\Im \\left ({ F\\left ({ { e^ {it} } } \\right) } \\right) = 0 } \\right\\} } \\right) = -\\frac14 \\sec ^2\\frac\\pi{N + 2}. \\] We show that the solution is unique and is given by \\[ a_j^ {(0)} = \\frac {1} { { { U'_N}\\left ({\\cos \\frac{\\pi } { { N + 2 } } } \\right) } } { U' 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the Koebe radius for polynomials","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.CV","authors_text":"Alex Stokolos, Andrey Smorodin, Dmitriy Dmitrishin","submitted_at":"2018-05-17T19:00:16Z","abstract_excerpt":"For a pair of conjugate trigonometrical polynomials $C (t) = \\sum_ { j = 1 } ^N { { a_j}\\cos jt }, S(t) = \\sum_ { j = 1 } ^N { { a_j}\\sin jt }$ with real coefficients and normalization ${a_1} = 1 $ we solve the extremal problem \\[\n  \\sup_ {a_2,...,a_N} \\left ({ \\min_t \\left\\{ {\\Re \\left ({ F\\left ({ { e^ {it} } } \\right) } \\right): \\Im \\left ({ F\\left ({ { e^ {it} } } \\right) } \\right) = 0 } \\right\\} } \\right) = -\\frac14 \\sec ^2\\frac\\pi{N + 2}. \\] We show that the solution is unique and is given by \\[ a_j^ {(0)} = \\frac {1} { { { U'_N}\\left ({\\cos \\frac{\\pi } { { N + 2 } } } \\right) } } { U' 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