# Generalized Bessel Functions of the First Kind by Árpád Baricz (auth.)

By Árpád Baricz (auth.)

In this quantity we learn the generalized Bessel services of the 1st sort through the use of a few classical and new findings in complicated and classical research. Our target is to provide fascinating geometric houses and useful inequalities for those generalized Bessel services. additionally, we expand many recognized inequalities regarding round and hyperbolic services to Bessel and changed Bessel functions.

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N≥0 ∑ 1 t q+n+ 0 b−1 2 (1−t) p−q−1 dt . 32) implies 1 0 uq (tz)t q+ b−1 2 (1 − t) p−q−1 dt = Γ (κ p − κq)Γ (κq ) (−c/4)n zn ∑ Γ (κ p ) n≥0 (κ p )n n! = Γ (κ p − κq)Γ (κq ) u p (z). e. 31) holds. e. μ p,q,b is a probability measure on [0, 1]. Thus the proof is complete. 3. 33) where d ν p,q,b (t) = ν p,q,b (t) dt and ν p,q,b (t) = 2t 2q+b (1 − t 2) p−q−1 B q + b+1 2 ,p−q is also √ a probability measure on [0, 1]. 26). In particular, when c = 1 and b = 1, the function λ p reduces to the function J p .

L. Duren [90, p. 11]): if U is a simply connected domain which is a proper subset of the complex plane and ζ is a given point in U, then there is a unique function f which maps U conformally onto the unit disk D and has the properties f (ζ ) = 0 and f (ζ ) > 0. Thus, the investigation of the analytic functions which are univalent in a simply connected domain with more than one boundary point can be confined to the investigation of analytic functions which are univalent in D. Hence without loss of generality we assume that E is the unit disk D.

This value will be strictly negative for all real ρ , because the discriminant Δ of Q1 (ρ ) satisfies Δ = c2 y2 − 16(κ − 1)2 < c2 − 16(κ − 1)2 ≤ 0 whenever y ∈ (−1, 1). 5. 5 we conclude Re h(z) = Re u p (z) > 0 for all z ∈ D. When κ ≥ |c|/4 and c = 0, then the above result implies Re u p+1 (z) > 0 for all z ∈ D. 2 we conclude that Re − 4κ u (z) = Re u p+1 (z) > 0 for all z ∈ D. c p This means that u p is close-to-convex with respect to the convex function ϕ : D → C, defined by ϕ (z) = −(cz)/(4κ ).

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