zeros of an infinite series
$begingroup$
Let $sum_{j=1}^{infty}a_{j}$ be a convergent series of positive numbers and ${z_{j}}_{j=1}^infty$ a closed discrete subset of the open unit disc $mathbb{D}$. Then $h(z):=sum_{j=1}^{infty}frac{a_{j}}{z-z_{j}}$ is a meromorphic function on $mathbb{D}$.
The question is: If we only consider the case of infinite sum, does $h(z)=sum_{j=1}^{infty}frac{a_{j}}{z-z_{j}}$ always have infinitely many zeros on $mathbb{D}$? Note that $h$ never vanishes outside $mathbb{D}$.
This question comes from the following paper (Example 1.1 and Question 3.3). Any comment is welcome.
https://arxiv.org/pdf/1709.03112.pdf
cv.complex-variables
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$begingroup$
Let $sum_{j=1}^{infty}a_{j}$ be a convergent series of positive numbers and ${z_{j}}_{j=1}^infty$ a closed discrete subset of the open unit disc $mathbb{D}$. Then $h(z):=sum_{j=1}^{infty}frac{a_{j}}{z-z_{j}}$ is a meromorphic function on $mathbb{D}$.
The question is: If we only consider the case of infinite sum, does $h(z)=sum_{j=1}^{infty}frac{a_{j}}{z-z_{j}}$ always have infinitely many zeros on $mathbb{D}$? Note that $h$ never vanishes outside $mathbb{D}$.
This question comes from the following paper (Example 1.1 and Question 3.3). Any comment is welcome.
https://arxiv.org/pdf/1709.03112.pdf
cv.complex-variables
New contributor
$endgroup$
add a comment |
$begingroup$
Let $sum_{j=1}^{infty}a_{j}$ be a convergent series of positive numbers and ${z_{j}}_{j=1}^infty$ a closed discrete subset of the open unit disc $mathbb{D}$. Then $h(z):=sum_{j=1}^{infty}frac{a_{j}}{z-z_{j}}$ is a meromorphic function on $mathbb{D}$.
The question is: If we only consider the case of infinite sum, does $h(z)=sum_{j=1}^{infty}frac{a_{j}}{z-z_{j}}$ always have infinitely many zeros on $mathbb{D}$? Note that $h$ never vanishes outside $mathbb{D}$.
This question comes from the following paper (Example 1.1 and Question 3.3). Any comment is welcome.
https://arxiv.org/pdf/1709.03112.pdf
cv.complex-variables
New contributor
$endgroup$
Let $sum_{j=1}^{infty}a_{j}$ be a convergent series of positive numbers and ${z_{j}}_{j=1}^infty$ a closed discrete subset of the open unit disc $mathbb{D}$. Then $h(z):=sum_{j=1}^{infty}frac{a_{j}}{z-z_{j}}$ is a meromorphic function on $mathbb{D}$.
The question is: If we only consider the case of infinite sum, does $h(z)=sum_{j=1}^{infty}frac{a_{j}}{z-z_{j}}$ always have infinitely many zeros on $mathbb{D}$? Note that $h$ never vanishes outside $mathbb{D}$.
This question comes from the following paper (Example 1.1 and Question 3.3). Any comment is welcome.
https://arxiv.org/pdf/1709.03112.pdf
cv.complex-variables
cv.complex-variables
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asked 3 hours ago
Yu FengYu Feng
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1 Answer
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$begingroup$
This was conjectured by J. Borcea, and a counterexample was constructed by J. Langley:
MR2317957
Langley, J. K.
Equilibrium points of logarithmic potentials on convex domains,
Proc. Amer. Math. Soc. 135 (2007), no. 9, 2821–2826.
His counterexample has an additional property that $z_k$ tend to a limit on the unit circle.
However, the is true under some additional conditions imposed on $z_k$.
Notice that a similar question in the plane (under the assumptions $z_ktoinfty$, and
$$sum_k a_k/|z_k|<infty,$$
$f$ is meromorphic in the plane) is unsolved, despite a lot of research on this question.
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1 Answer
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1 Answer
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$begingroup$
This was conjectured by J. Borcea, and a counterexample was constructed by J. Langley:
MR2317957
Langley, J. K.
Equilibrium points of logarithmic potentials on convex domains,
Proc. Amer. Math. Soc. 135 (2007), no. 9, 2821–2826.
His counterexample has an additional property that $z_k$ tend to a limit on the unit circle.
However, the is true under some additional conditions imposed on $z_k$.
Notice that a similar question in the plane (under the assumptions $z_ktoinfty$, and
$$sum_k a_k/|z_k|<infty,$$
$f$ is meromorphic in the plane) is unsolved, despite a lot of research on this question.
$endgroup$
add a comment |
$begingroup$
This was conjectured by J. Borcea, and a counterexample was constructed by J. Langley:
MR2317957
Langley, J. K.
Equilibrium points of logarithmic potentials on convex domains,
Proc. Amer. Math. Soc. 135 (2007), no. 9, 2821–2826.
His counterexample has an additional property that $z_k$ tend to a limit on the unit circle.
However, the is true under some additional conditions imposed on $z_k$.
Notice that a similar question in the plane (under the assumptions $z_ktoinfty$, and
$$sum_k a_k/|z_k|<infty,$$
$f$ is meromorphic in the plane) is unsolved, despite a lot of research on this question.
$endgroup$
add a comment |
$begingroup$
This was conjectured by J. Borcea, and a counterexample was constructed by J. Langley:
MR2317957
Langley, J. K.
Equilibrium points of logarithmic potentials on convex domains,
Proc. Amer. Math. Soc. 135 (2007), no. 9, 2821–2826.
His counterexample has an additional property that $z_k$ tend to a limit on the unit circle.
However, the is true under some additional conditions imposed on $z_k$.
Notice that a similar question in the plane (under the assumptions $z_ktoinfty$, and
$$sum_k a_k/|z_k|<infty,$$
$f$ is meromorphic in the plane) is unsolved, despite a lot of research on this question.
$endgroup$
This was conjectured by J. Borcea, and a counterexample was constructed by J. Langley:
MR2317957
Langley, J. K.
Equilibrium points of logarithmic potentials on convex domains,
Proc. Amer. Math. Soc. 135 (2007), no. 9, 2821–2826.
His counterexample has an additional property that $z_k$ tend to a limit on the unit circle.
However, the is true under some additional conditions imposed on $z_k$.
Notice that a similar question in the plane (under the assumptions $z_ktoinfty$, and
$$sum_k a_k/|z_k|<infty,$$
$f$ is meromorphic in the plane) is unsolved, despite a lot of research on this question.
answered 3 hours ago
Alexandre EremenkoAlexandre Eremenko
49.7k6137255
49.7k6137255
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Yu Feng is a new contributor. Be nice, and check out our Code of Conduct.
Yu Feng is a new contributor. Be nice, and check out our Code of Conduct.
Yu Feng is a new contributor. Be nice, and check out our Code of Conduct.
Yu Feng is a new contributor. Be nice, and check out our Code of Conduct.
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