Existence of a complex polynomial with given zeros and critical points?
Suppose we are given two families of complex numbers (need not to be distinct) $z_1, z_2, cdots, z_n$ and $w_1, w_2, cdots, w_{n-1}$ such that
- the second family lies in the convex hull of first family and
- both families have the same center of mass.
Now, can we find a polynomial $P$ of degree $n$ that has first family as its collection of zeros and second family as its collection of critical points (counting with multiplicities) ?
Hermite interpolation gives a polynomial of degree $2n-1$ with these properties introducing new zeros and critical points, but I hope conditions 1. and 2. are enough to guarantee the existence of such a polynomial of degree $n.$ If not, what additional conditions do we need to impose of those two families?
complex-analysis algebraic-geometry polynomials lagrange-interpolation
edited 2 days ago
Andrés E. Caicedo
64.8k8158246
asked 2 days ago
BumblebeeBumblebee
9,62812551
add a comment |
Suppose we are given two families of complex numbers (need not to be distinct) $z_1, z_2, cdots, z_n$ and $w_1, w_2, cdots, w_{n-1}$ such that
- the second family lies in the convex hull of first family and
- both families have the same center of mass.
Now, can we find a polynomial $P$ of degree $n$ that has first family as its collection of zeros and second family as its collection of critical points (counting with multiplicities) ?
Hermite interpolation gives a polynomial of degree $2n-1$ with these properties introducing new zeros and critical points, but I hope conditions 1. and 2. are enough to guarantee the existence of such a polynomial of degree $n.$ If not, what additional conditions do we need to impose of those two families?
complex-analysis algebraic-geometry polynomials lagrange-interpolation
edited 2 days ago
Andrés E. Caicedo
64.8k8158246
asked 2 days ago
BumblebeeBumblebee
9,62812551
1
You need more conditions. Consider the case where $n>2$ is even, half of the $z_k$ are 0, half are 2, and the $w_k$ are all in $(0,2)$ and evenly spaced around 1.
– Andrés E. Caicedo
2 days ago
4
The point is that once you fix $n$ zeros, you determine the polynomial up to a multiplicative constant, you have absolutely no leeway on the location of the critical points.
– Andrés E. Caicedo
2 days ago
@AndrésE.Caicedo: Thank you very much for your explanation.
– Bumblebee
2 days ago
add a comment |
Suppose we are given two families of complex numbers (need not to be distinct) $z_1, z_2, cdots, z_n$ and $w_1, w_2, cdots, w_{n-1}$ such that
- the second family lies in the convex hull of first family and
- both families have the same center of mass.
Now, can we find a polynomial $P$ of degree $n$ that has first family as its collection of zeros and second family as its collection of critical points (counting with multiplicities) ?
Hermite interpolation gives a polynomial of degree $2n-1$ with these properties introducing new zeros and critical points, but I hope conditions 1. and 2. are enough to guarantee the existence of such a polynomial of degree $n.$ If not, what additional conditions do we need to impose of those two families?
complex-analysis algebraic-geometry polynomials lagrange-interpolation
edited 2 days ago
Andrés E. Caicedo
64.8k8158246
asked 2 days ago
BumblebeeBumblebee
9,62812551
Suppose we are given two families of complex numbers (need not to be distinct) $z_1, z_2, cdots, z_n$ and $w_1, w_2, cdots, w_{n-1}$ such that
- the second family lies in the convex hull of first family and
- both families have the same center of mass.
Now, can we find a polynomial $P$ of degree $n$ that has first family as its collection of zeros and second family as its collection of critical points (counting with multiplicities) ?
Hermite interpolation gives a polynomial of degree $2n-1$ with these properties introducing new zeros and critical points, but I hope conditions 1. and 2. are enough to guarantee the existence of such a polynomial of degree $n.$ If not, what additional conditions do we need to impose of those two families?
complex-analysis algebraic-geometry polynomials lagrange-interpolation
complex-analysis algebraic-geometry polynomials lagrange-interpolation
edited 2 days ago
Andrés E. Caicedo
64.8k8158246
asked 2 days ago
BumblebeeBumblebee
9,62812551
edited 2 days ago
Andrés E. Caicedo
64.8k8158246
asked 2 days ago
BumblebeeBumblebee
9,62812551
edited 2 days ago
Andrés E. Caicedo
64.8k8158246
edited 2 days ago
Andrés E. Caicedo
64.8k8158246
edited 2 days ago
Andrés E. Caicedo
64.8k8158246
64.8k8158246
asked 2 days ago
BumblebeeBumblebee
9,62812551
asked 2 days ago
BumblebeeBumblebee
9,62812551
asked 2 days ago
BumblebeeBumblebee
9,62812551
9,62812551
1
You need more conditions. Consider the case where $n>2$ is even, half of the $z_k$ are 0, half are 2, and the $w_k$ are all in $(0,2)$ and evenly spaced around 1.
– Andrés E. Caicedo
2 days ago
4
The point is that once you fix $n$ zeros, you determine the polynomial up to a multiplicative constant, you have absolutely no leeway on the location of the critical points.
– Andrés E. Caicedo
2 days ago
@AndrésE.Caicedo: Thank you very much for your explanation.
– Bumblebee
2 days ago
add a comment |
1
You need more conditions. Consider the case where $n>2$ is even, half of the $z_k$ are 0, half are 2, and the $w_k$ are all in $(0,2)$ and evenly spaced around 1.
– Andrés E. Caicedo
2 days ago
4
The point is that once you fix $n$ zeros, you determine the polynomial up to a multiplicative constant, you have absolutely no leeway on the location of the critical points.
– Andrés E. Caicedo
2 days ago
@AndrésE.Caicedo: Thank you very much for your explanation.
– Bumblebee
2 days ago
1
1
You need more conditions. Consider the case where $n>2$ is even, half of the $z_k$ are 0, half are 2, and the $w_k$ are all in $(0,2)$ and evenly spaced around 1.
– Andrés E. Caicedo
2 days ago
You need more conditions. Consider the case where $n>2$ is even, half of the $z_k$ are 0, half are 2, and the $w_k$ are all in $(0,2)$ and evenly spaced around 1.
– Andrés E. Caicedo
2 days ago
4
4
The point is that once you fix $n$ zeros, you determine the polynomial up to a multiplicative constant, you have absolutely no leeway on the location of the critical points.
– Andrés E. Caicedo
2 days ago
The point is that once you fix $n$ zeros, you determine the polynomial up to a multiplicative constant, you have absolutely no leeway on the location of the critical points.
– Andrés E. Caicedo
2 days ago
@AndrésE.Caicedo: Thank you very much for your explanation.
– Bumblebee
2 days ago
@AndrésE.Caicedo: Thank you very much for your explanation.
– Bumblebee
2 days ago
add a comment |
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1
You need more conditions. Consider the case where $n>2$ is even, half of the $z_k$ are 0, half are 2, and the $w_k$ are all in $(0,2)$ and evenly spaced around 1.
– Andrés E. Caicedo
2 days ago
4
The point is that once you fix $n$ zeros, you determine the polynomial up to a multiplicative constant, you have absolutely no leeway on the location of the critical points.
– Andrés E. Caicedo
2 days ago
@AndrésE.Caicedo: Thank you very much for your explanation.
– Bumblebee
2 days ago