Exponential type of a product of entire functions
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Let ${a_n}_{n=1}^infty$ and ${b_m}_{m=1}^infty$ be two sequences of points in $mathbb{C}$ such that
$$
f(z)=prod_{n=1}^inftyleft(1-frac{z}{a_n}right)quadmbox{and}quad g(z)=prod_{m=1}^inftyleft(1-frac{z}{b_m}right)
$$
are entire functions of finite exponential types $0<A_f<infty$ and $0<A_g<infty$ (growth orders are $rho_f=rho_g=1$), respectively. Is it true that the exponential type $A_{fg}$ of the entire function $f(z)g(z)$ satisfies
$$
max{A_f,A_g}<A_{fg}
$$
(strict inequality)?
More generally, given that the growth of $f$ as above is $rho_f=1$, is there a way to establish the exponential type from the sequence ${a_n}$?
Thank you.
Edit: Question answered on mathoverflow here.
complex-analysis
asked Nov 27 at 20:07
Bedovlat
591311
add a comment |
up vote
1
down vote
favorite
Let ${a_n}_{n=1}^infty$ and ${b_m}_{m=1}^infty$ be two sequences of points in $mathbb{C}$ such that
$$
f(z)=prod_{n=1}^inftyleft(1-frac{z}{a_n}right)quadmbox{and}quad g(z)=prod_{m=1}^inftyleft(1-frac{z}{b_m}right)
$$
are entire functions of finite exponential types $0<A_f<infty$ and $0<A_g<infty$ (growth orders are $rho_f=rho_g=1$), respectively. Is it true that the exponential type $A_{fg}$ of the entire function $f(z)g(z)$ satisfies
$$
max{A_f,A_g}<A_{fg}
$$
(strict inequality)?
More generally, given that the growth of $f$ as above is $rho_f=1$, is there a way to establish the exponential type from the sequence ${a_n}$?
Thank you.
Edit: Question answered on mathoverflow here.
complex-analysis
asked Nov 27 at 20:07
Bedovlat
591311
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let ${a_n}_{n=1}^infty$ and ${b_m}_{m=1}^infty$ be two sequences of points in $mathbb{C}$ such that
$$
f(z)=prod_{n=1}^inftyleft(1-frac{z}{a_n}right)quadmbox{and}quad g(z)=prod_{m=1}^inftyleft(1-frac{z}{b_m}right)
$$
are entire functions of finite exponential types $0<A_f<infty$ and $0<A_g<infty$ (growth orders are $rho_f=rho_g=1$), respectively. Is it true that the exponential type $A_{fg}$ of the entire function $f(z)g(z)$ satisfies
$$
max{A_f,A_g}<A_{fg}
$$
(strict inequality)?
More generally, given that the growth of $f$ as above is $rho_f=1$, is there a way to establish the exponential type from the sequence ${a_n}$?
Thank you.
Edit: Question answered on mathoverflow here.
complex-analysis
asked Nov 27 at 20:07
Bedovlat
591311
Let ${a_n}_{n=1}^infty$ and ${b_m}_{m=1}^infty$ be two sequences of points in $mathbb{C}$ such that
$$
f(z)=prod_{n=1}^inftyleft(1-frac{z}{a_n}right)quadmbox{and}quad g(z)=prod_{m=1}^inftyleft(1-frac{z}{b_m}right)
$$
are entire functions of finite exponential types $0<A_f<infty$ and $0<A_g<infty$ (growth orders are $rho_f=rho_g=1$), respectively. Is it true that the exponential type $A_{fg}$ of the entire function $f(z)g(z)$ satisfies
$$
max{A_f,A_g}<A_{fg}
$$
(strict inequality)?
More generally, given that the growth of $f$ as above is $rho_f=1$, is there a way to establish the exponential type from the sequence ${a_n}$?
Thank you.
Edit: Question answered on mathoverflow here.
complex-analysis
complex-analysis
asked Nov 27 at 20:07
Bedovlat
591311
asked Nov 27 at 20:07
Bedovlat
591311
edited Nov 29 at 22:16
asked Nov 27 at 20:07
Bedovlat
591311
asked Nov 27 at 20:07
Bedovlat
591311
asked Nov 27 at 20:07
Bedovlat
591311
591311
add a comment |
add a comment |
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