If a product of polynomials converges, does some product of their zeros also converge?
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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 $underline{text{not converge}}$.
complex-analysis infinite-product
asked Dec 6 '18 at 15:50
user122916user122916
431315
$endgroup$
|
show 3 more comments
$begingroup$
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 $underline{text{not converge}}$.
complex-analysis infinite-product
asked Dec 6 '18 at 15:50
user122916user122916
431315
$endgroup$
$begingroup$
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 $p_k$.
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– Robert Israel
Dec 6 '18 at 15:56
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Thank you. I meant $p_k(0)=1$.
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– user122916
Dec 6 '18 at 15:57
1
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My intuition is telling me that the answer is definitely "no". But I need to think more.
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– mathworker21
Dec 9 '18 at 3:00
1
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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
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– Lukas Geyer
Dec 11 '18 at 23:45
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@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 $1leq jleq m$, $$left| sum_{n=1}^j f_{sigma(n)} right|_R< 4epsilon.$$
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– user122916
Jan 2 at 22:16
|
show 3 more comments
$begingroup$
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 $underline{text{not converge}}$.
complex-analysis infinite-product
asked Dec 6 '18 at 15:50
user122916user122916
431315
$endgroup$
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
complex-analysis infinite-product
asked Dec 6 '18 at 15:50
user122916user122916
431315
asked Dec 6 '18 at 15:50
user122916user122916
431315
edited Dec 6 '18 at 15:58
user122916
asked Dec 6 '18 at 15:50
user122916user122916
431315
asked Dec 6 '18 at 15:50
user122916user122916
431315
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
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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
|
show 3 more comments
$begingroup$
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
|
show 3 more comments
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$begingroup$
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