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Entire function sharing a small function with its mixed-operators

Entire function sharing a small function with its mixed-operators AbstractIn this article, we investigate the uniqueness problem on a transcendental entire function f⁢(z){f(z)}with its linear mixed-operators Tf, where T is a linear combination of differential-difference operators Dην:=f(ν)⁢(z+η){D^{\nu}_{\eta}:=f^{(\nu)}(z+\eta)}and shift operators Eζ:=f⁢(z+ζ){E_{\zeta}:=f(z+\zeta\/)}, where η,ν,ζ{\eta,\nu,\zeta}are constants.We obtain that if a transcendental entire function f⁢(z){f(z)}satisfies λ⁢(f-α)<σ⁢(f)<+∞{\lambda(f-\alpha)<\sigma(f\/)<+\infty}, where α⁢(z){\alpha(z)}is an entire function with σ⁢(α)<1{\sigma(\alpha)<1}, and if f and Tf share one small entire function a⁢(z){a(z)}with σ⁢(a)<σ⁢(f){\sigma(a)<\sigma(f\/)}, thenT⁢f-a⁢(z)f⁢(z)-a⁢(z)=τ,{\frac{Tf-a(z)}{f(z)-a(z)}=\tau,}where τ is a non-zero constant.Furthermore, we obtain the value τ and the expression of fby imposing additional conditions. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Georgian Mathematical Journal de Gruyter

Entire function sharing a small function with its mixed-operators

Georgian Mathematical Journal , Volume 26 (1): 16 – Mar 1, 2019

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References (29)

Publisher
de Gruyter
Copyright
© 2019 Walter de Gruyter GmbH, Berlin/Boston
ISSN
1572-9176
eISSN
1572-9176
DOI
10.1515/gmj-2017-0024
Publisher site
See Article on Publisher Site

Abstract

AbstractIn this article, we investigate the uniqueness problem on a transcendental entire function f⁢(z){f(z)}with its linear mixed-operators Tf, where T is a linear combination of differential-difference operators Dην:=f(ν)⁢(z+η){D^{\nu}_{\eta}:=f^{(\nu)}(z+\eta)}and shift operators Eζ:=f⁢(z+ζ){E_{\zeta}:=f(z+\zeta\/)}, where η,ν,ζ{\eta,\nu,\zeta}are constants.We obtain that if a transcendental entire function f⁢(z){f(z)}satisfies λ⁢(f-α)<σ⁢(f)<+∞{\lambda(f-\alpha)<\sigma(f\/)<+\infty}, where α⁢(z){\alpha(z)}is an entire function with σ⁢(α)<1{\sigma(\alpha)<1}, and if f and Tf share one small entire function a⁢(z){a(z)}with σ⁢(a)<σ⁢(f){\sigma(a)<\sigma(f\/)}, thenT⁢f-a⁢(z)f⁢(z)-a⁢(z)=τ,{\frac{Tf-a(z)}{f(z)-a(z)}=\tau,}where τ is a non-zero constant.Furthermore, we obtain the value τ and the expression of fby imposing additional conditions.

Journal

Georgian Mathematical Journalde Gruyter

Published: Mar 1, 2019

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