Results & Lemmas (4)
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THEOREM 2.1
THEOREM 2.1 Let PE C 1 ([ -1, 1)) and define fimctions <Jn = <Jn[ P] (n = I, 2,... ) on [I, co) by <J.(t):= 2n- 1t- 1 I: P(I - t- 1 + t- 1…
THEOREM 2.1 Let PE C 1 ([ -1, 1)) and define fimctions <Jn = <Jn[ P] (n = I, 2, ... ) on [I, co) by <J.(t) := 2n- 1t- 1 I: P(I - t- 1 + t- 1 cos 0) sin(n8) sin 0 dO. (2.4) Then the fimctions <J" solve (2.1) with initial ualues <J.(1) = 2n- 1 f: P(cos8)sin(n8)sin0d0. (2.5)
PROPOSITION 2.2
PROPOSITION 2.2 Formula (2.8) establishes a 1-l correspondence between admissible solutions an of (2.1) and nonzero polynomials Q such that…
PROPOSITION 2.2 Formula (2.8) establishes a 1-l correspondence between admissible solutions {an} of (2.1) and nonzero polynomials Q such that (an[Q])(t)? 0 for all t? 1, n = 1, 2,. ... If {an} is the solution of (2.1) determined by (2.2) then (2.3) and (2.8) yield [O/Z)(r-ll] sin(r - 2k)e Q(cos B) = I (r - 2k) . e k=o sm [(1 /2)(r-1 l]
Theorem 2 · coeff
Theorem 2] de Branges states that any admissible solution of (6.1) yields an inequality for the coefficients a. in 00 v- 1[(/(z))" -…
Theorem 2] de Branges states that any admissible solution of (6.1) yields an inequality for the coefficients a. in 00 v- 1[(/(z))" - (f'(O)z}"] = L a.zv+n, n= I where f is a univalent analytic function on the unit disk sending 0 to 0. We can formulate a generalization of Theorem 2.1:
THEOREM 6.1
THEOREM 6.1 Let PE C1 ([ -1, 1]) and define functions <1n = a.[ P] (n = 1, 2,... ) on [1, oo) by n!(n - l)!f(v + 2) a.(t):= (2v + l)n(2v +…
THEOREM 6.1 Let PE C1 ([ -1, 1]) and define functions <1n = a.[ P] (n = 1, 2, ... ) on [1, oo) by n!(n - l)!f(v + 2) a.(t) := (2v + l)n(2v + l)n-1n 112f(v + j) x r- 1 f 1 P(l - t- 1(1 - x))C~~~(x)(l - x2 r+(li21 dx -1 (6.2) Then the functions <1n solve (6.1) with initial values n! (n - I)! f(v + 2) a.(l) = (2v + 1).(2v + 1)._ 1n112f(v + j) x f 1 P(x}C~~l(x)(l - x 2r+< 1121 dx.
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