If a product of polynomials converges, does some product of their zeros also converge?












9












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Suppose ${p_k}$ is a sequence of polynomials with $p_k(0)=1$. Let $a_1,a_2,ldots$ be an enumeration of all of the zeros of the $p_k$. Suppose that
$$prod_{k=1}^infty p_k(z)$$
converges uniformly on compact subsets of $mathbb{C}$. Is there some permutation $sigma:mathbb{N}tomathbb{N}$ such that
$$f_sigma (z)=prod_{k=1}^infty left(1-frac{z}{a_{sigma(k)}}right)$$
converges uniformly on compact subsets of $mathbb{C}$?



This is a follow-up to Factoring a convergent infinite product of polynomials., in which an example of such ${p_k}$ is given along with a permutation $sigma$ for which the product $f_sigma (z)$ does $underline{text{not converge}}$.










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$endgroup$












  • $begingroup$
    Since you have $a_{sigma(k)}$ in a denominator, you probably don't want to include $0$ in your enumeration of the zeros, even though it is a zero of all the $p_k$.
    $endgroup$
    – Robert Israel
    Dec 6 '18 at 15:56










  • $begingroup$
    Thank you. I meant $p_k(0)=1$.
    $endgroup$
    – user122916
    Dec 6 '18 at 15:57






  • 1




    $begingroup$
    My intuition is telling me that the answer is definitely "no". But I need to think more.
    $endgroup$
    – mathworker21
    Dec 9 '18 at 3:00






  • 1




    $begingroup$
    I think this is true, but possibly not that easy to prove. It might be provable with a product version of the "polygonal confinement theorem" by Steinitz: sites.math.washington.edu/~morrow/335_17/levy.pdf
    $endgroup$
    – Lukas Geyer
    Dec 11 '18 at 23:45










  • $begingroup$
    @LukasGeyer Do you know if there is an analog of something like Lemma 3.1 in the paper you linked for holomorphic functions? Something like: if $f_1,...,f_m$ are analytic on $|z|leq R$, $ left| sum_{n=1}^m f_n right|_{|z|leq R}< epsilon$ and $ |f_n|_{|z|leq R} < epsilon$, then there exists a permutation $sigma$ of $1,2,ldots,m$ such that whenever $1leq jleq m$, $$left| sum_{n=1}^j f_{sigma(n)} right|_R< 4epsilon.$$
    $endgroup$
    – user122916
    Jan 2 at 22:16


















9












$begingroup$


Suppose ${p_k}$ is a sequence of polynomials with $p_k(0)=1$. Let $a_1,a_2,ldots$ be an enumeration of all of the zeros of the $p_k$. Suppose that
$$prod_{k=1}^infty p_k(z)$$
converges uniformly on compact subsets of $mathbb{C}$. Is there some permutation $sigma:mathbb{N}tomathbb{N}$ such that
$$f_sigma (z)=prod_{k=1}^infty left(1-frac{z}{a_{sigma(k)}}right)$$
converges uniformly on compact subsets of $mathbb{C}$?



This is a follow-up to Factoring a convergent infinite product of polynomials., in which an example of such ${p_k}$ is given along with a permutation $sigma$ for which the product $f_sigma (z)$ does $underline{text{not converge}}$.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Since you have $a_{sigma(k)}$ in a denominator, you probably don't want to include $0$ in your enumeration of the zeros, even though it is a zero of all the $p_k$.
    $endgroup$
    – Robert Israel
    Dec 6 '18 at 15:56










  • $begingroup$
    Thank you. I meant $p_k(0)=1$.
    $endgroup$
    – user122916
    Dec 6 '18 at 15:57






  • 1




    $begingroup$
    My intuition is telling me that the answer is definitely "no". But I need to think more.
    $endgroup$
    – mathworker21
    Dec 9 '18 at 3:00






  • 1




    $begingroup$
    I think this is true, but possibly not that easy to prove. It might be provable with a product version of the "polygonal confinement theorem" by Steinitz: sites.math.washington.edu/~morrow/335_17/levy.pdf
    $endgroup$
    – Lukas Geyer
    Dec 11 '18 at 23:45










  • $begingroup$
    @LukasGeyer Do you know if there is an analog of something like Lemma 3.1 in the paper you linked for holomorphic functions? Something like: if $f_1,...,f_m$ are analytic on $|z|leq R$, $ left| sum_{n=1}^m f_n right|_{|z|leq R}< epsilon$ and $ |f_n|_{|z|leq R} < epsilon$, then there exists a permutation $sigma$ of $1,2,ldots,m$ such that whenever $1leq jleq m$, $$left| sum_{n=1}^j f_{sigma(n)} right|_R< 4epsilon.$$
    $endgroup$
    – user122916
    Jan 2 at 22:16
















9












9








9


4



$begingroup$


Suppose ${p_k}$ is a sequence of polynomials with $p_k(0)=1$. Let $a_1,a_2,ldots$ be an enumeration of all of the zeros of the $p_k$. Suppose that
$$prod_{k=1}^infty p_k(z)$$
converges uniformly on compact subsets of $mathbb{C}$. Is there some permutation $sigma:mathbb{N}tomathbb{N}$ such that
$$f_sigma (z)=prod_{k=1}^infty left(1-frac{z}{a_{sigma(k)}}right)$$
converges uniformly on compact subsets of $mathbb{C}$?



This is a follow-up to Factoring a convergent infinite product of polynomials., in which an example of such ${p_k}$ is given along with a permutation $sigma$ for which the product $f_sigma (z)$ does $underline{text{not converge}}$.










share|cite|improve this question











$endgroup$




Suppose ${p_k}$ is a sequence of polynomials with $p_k(0)=1$. Let $a_1,a_2,ldots$ be an enumeration of all of the zeros of the $p_k$. Suppose that
$$prod_{k=1}^infty p_k(z)$$
converges uniformly on compact subsets of $mathbb{C}$. Is there some permutation $sigma:mathbb{N}tomathbb{N}$ such that
$$f_sigma (z)=prod_{k=1}^infty left(1-frac{z}{a_{sigma(k)}}right)$$
converges uniformly on compact subsets of $mathbb{C}$?



This is a follow-up to Factoring a convergent infinite product of polynomials., in which an example of such ${p_k}$ is given along with a permutation $sigma$ for which the product $f_sigma (z)$ does $underline{text{not converge}}$.







complex-analysis infinite-product






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 6 '18 at 15:58







user122916

















asked Dec 6 '18 at 15:50









user122916user122916

431315




431315












  • $begingroup$
    Since you have $a_{sigma(k)}$ in a denominator, you probably don't want to include $0$ in your enumeration of the zeros, even though it is a zero of all the $p_k$.
    $endgroup$
    – Robert Israel
    Dec 6 '18 at 15:56










  • $begingroup$
    Thank you. I meant $p_k(0)=1$.
    $endgroup$
    – user122916
    Dec 6 '18 at 15:57






  • 1




    $begingroup$
    My intuition is telling me that the answer is definitely "no". But I need to think more.
    $endgroup$
    – mathworker21
    Dec 9 '18 at 3:00






  • 1




    $begingroup$
    I think this is true, but possibly not that easy to prove. It might be provable with a product version of the "polygonal confinement theorem" by Steinitz: sites.math.washington.edu/~morrow/335_17/levy.pdf
    $endgroup$
    – Lukas Geyer
    Dec 11 '18 at 23:45










  • $begingroup$
    @LukasGeyer Do you know if there is an analog of something like Lemma 3.1 in the paper you linked for holomorphic functions? Something like: if $f_1,...,f_m$ are analytic on $|z|leq R$, $ left| sum_{n=1}^m f_n right|_{|z|leq R}< epsilon$ and $ |f_n|_{|z|leq R} < epsilon$, then there exists a permutation $sigma$ of $1,2,ldots,m$ such that whenever $1leq jleq m$, $$left| sum_{n=1}^j f_{sigma(n)} right|_R< 4epsilon.$$
    $endgroup$
    – user122916
    Jan 2 at 22:16




















  • $begingroup$
    Since you have $a_{sigma(k)}$ in a denominator, you probably don't want to include $0$ in your enumeration of the zeros, even though it is a zero of all the $p_k$.
    $endgroup$
    – Robert Israel
    Dec 6 '18 at 15:56










  • $begingroup$
    Thank you. I meant $p_k(0)=1$.
    $endgroup$
    – user122916
    Dec 6 '18 at 15:57






  • 1




    $begingroup$
    My intuition is telling me that the answer is definitely "no". But I need to think more.
    $endgroup$
    – mathworker21
    Dec 9 '18 at 3:00






  • 1




    $begingroup$
    I think this is true, but possibly not that easy to prove. It might be provable with a product version of the "polygonal confinement theorem" by Steinitz: sites.math.washington.edu/~morrow/335_17/levy.pdf
    $endgroup$
    – Lukas Geyer
    Dec 11 '18 at 23:45










  • $begingroup$
    @LukasGeyer Do you know if there is an analog of something like Lemma 3.1 in the paper you linked for holomorphic functions? Something like: if $f_1,...,f_m$ are analytic on $|z|leq R$, $ left| sum_{n=1}^m f_n right|_{|z|leq R}< epsilon$ and $ |f_n|_{|z|leq R} < epsilon$, then there exists a permutation $sigma$ of $1,2,ldots,m$ such that whenever $1leq jleq m$, $$left| sum_{n=1}^j f_{sigma(n)} right|_R< 4epsilon.$$
    $endgroup$
    – user122916
    Jan 2 at 22:16


















$begingroup$
Since you have $a_{sigma(k)}$ in a denominator, you probably don't want to include $0$ in your enumeration of the zeros, even though it is a zero of all the $p_k$.
$endgroup$
– Robert Israel
Dec 6 '18 at 15:56




$begingroup$
Since you have $a_{sigma(k)}$ in a denominator, you probably don't want to include $0$ in your enumeration of the zeros, even though it is a zero of all the $p_k$.
$endgroup$
– Robert Israel
Dec 6 '18 at 15:56












$begingroup$
Thank you. I meant $p_k(0)=1$.
$endgroup$
– user122916
Dec 6 '18 at 15:57




$begingroup$
Thank you. I meant $p_k(0)=1$.
$endgroup$
– user122916
Dec 6 '18 at 15:57




1




1




$begingroup$
My intuition is telling me that the answer is definitely "no". But I need to think more.
$endgroup$
– mathworker21
Dec 9 '18 at 3:00




$begingroup$
My intuition is telling me that the answer is definitely "no". But I need to think more.
$endgroup$
– mathworker21
Dec 9 '18 at 3:00




1




1




$begingroup$
I think this is true, but possibly not that easy to prove. It might be provable with a product version of the "polygonal confinement theorem" by Steinitz: sites.math.washington.edu/~morrow/335_17/levy.pdf
$endgroup$
– Lukas Geyer
Dec 11 '18 at 23:45




$begingroup$
I think this is true, but possibly not that easy to prove. It might be provable with a product version of the "polygonal confinement theorem" by Steinitz: sites.math.washington.edu/~morrow/335_17/levy.pdf
$endgroup$
– Lukas Geyer
Dec 11 '18 at 23:45












$begingroup$
@LukasGeyer Do you know if there is an analog of something like Lemma 3.1 in the paper you linked for holomorphic functions? Something like: if $f_1,...,f_m$ are analytic on $|z|leq R$, $ left| sum_{n=1}^m f_n right|_{|z|leq R}< epsilon$ and $ |f_n|_{|z|leq R} < epsilon$, then there exists a permutation $sigma$ of $1,2,ldots,m$ such that whenever $1leq jleq m$, $$left| sum_{n=1}^j f_{sigma(n)} right|_R< 4epsilon.$$
$endgroup$
– user122916
Jan 2 at 22:16






$begingroup$
@LukasGeyer Do you know if there is an analog of something like Lemma 3.1 in the paper you linked for holomorphic functions? Something like: if $f_1,...,f_m$ are analytic on $|z|leq R$, $ left| sum_{n=1}^m f_n right|_{|z|leq R}< epsilon$ and $ |f_n|_{|z|leq R} < epsilon$, then there exists a permutation $sigma$ of $1,2,ldots,m$ such that whenever $1leq jleq m$, $$left| sum_{n=1}^j f_{sigma(n)} right|_R< 4epsilon.$$
$endgroup$
– user122916
Jan 2 at 22:16












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