This paper studies the problem of functional inequalities for analytic functions in classical geometric function theory.Using the di erential subordination principle and(p,q)-derivative operator,it introduces(p,q)-ana...This paper studies the problem of functional inequalities for analytic functions in classical geometric function theory.Using the di erential subordination principle and(p,q)-derivative operator,it introduces(p,q)-analog of a class of multivalently Bazilevic functions as-sociated with a limacon function,and obtains the corresponding coefficient estimates and the Fekete-Szego inequality,which extend and improve the related results for starlike functions,even q-starlike functions.展开更多
在组合数恒等式中,有一类可以通过对等式x<sup>α</sup>（1+x<sup>β</sup>）<sup>n</sup>=sum form r=0 to n(C<sub>n</sub><sup>r</sup>x<sup>a+rB</sup>),...在组合数恒等式中,有一类可以通过对等式x<sup>α</sup>（1+x<sup>β</sup>）<sup>n</sup>=sum form r=0 to n(C<sub>n</sub><sup>r</sup>x<sup>a+rB</sup>),（1+x）<sup>n</sup>=sum form r=0 to n（C<sub>n</sub><sup>r</sup>x<sup>r</sup>）求导或积分而得,方法简便,且能揭示其数量之间的一般关系。兹举例如下: 1、[（1+x）<sup>n</sup>]<sup>′</sup>=（C<sub>n</sub><sup>o</sup>+C<sub>n</sub><sup>1</sup>X+C<sub>n</sub><sup>2</sup>X<sup>2</sup>+C<sub>n</sub><sup>3</sup>X<sup>4</sup>+…+C<sub>n</sub><sup>r</sup>X<sup>r</sup>+…+C<sub>n</sub><sup>n</sup>X<sup>n</sup>）′,展开更多
基金Supported by Natural Science Foundation of Ningxia(2023AAC 03001)Natural Science Foundation of China(12261068)
文摘This paper studies the problem of functional inequalities for analytic functions in classical geometric function theory.Using the di erential subordination principle and(p,q)-derivative operator,it introduces(p,q)-analog of a class of multivalently Bazilevic functions as-sociated with a limacon function,and obtains the corresponding coefficient estimates and the Fekete-Szego inequality,which extend and improve the related results for starlike functions,even q-starlike functions.
文摘在组合数恒等式中,有一类可以通过对等式x<sup>α</sup>（1+x<sup>β</sup>）<sup>n</sup>=sum form r=0 to n(C<sub>n</sub><sup>r</sup>x<sup>a+rB</sup>),（1+x）<sup>n</sup>=sum form r=0 to n（C<sub>n</sub><sup>r</sup>x<sup>r</sup>）求导或积分而得,方法简便,且能揭示其数量之间的一般关系。兹举例如下: 1、[（1+x）<sup>n</sup>]<sup>′</sup>=（C<sub>n</sub><sup>o</sup>+C<sub>n</sub><sup>1</sup>X+C<sub>n</sub><sup>2</sup>X<sup>2</sup>+C<sub>n</sub><sup>3</sup>X<sup>4</sup>+…+C<sub>n</sub><sup>r</sup>X<sup>r</sup>+…+C<sub>n</sub><sup>n</sup>X<sup>n</sup>）′,
