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A couple of years ago, I came up with the following question, to which I have no answer to this day. I have asked a few people about this, most of my teachers and some friends, but no one had ever heard of the question before, and no one knew the answer.

I hope this is an original question, but seeing how natural it is, I doubt this is the first time someone has asked it.

First, some motivation. Take $P$ any nonzero complex polynomial. It is an easy and classical exercise to show that the roots of its derivative $P'$ lie in the convex hull of its own roots (I know this as the Gauss-Lucas property). To show this, you simply write $P=a\cdot\prod_{i=1}^{r}(X-\alpha_i)^{m_i}$ where the  $\alpha_i~(i=1,\dots,r)$ are the different roots of $P$, and $m_i$ the corresponding multiplicities, and evaluate $\frac{P'}{P}=\sum_i \frac{m_i}{X-\alpha_i}$ on a root $\beta$ of $P'$ which is not also a root of $P$. You'll end up with an expression of $\beta$ as a convex combination of $\alpha_1,\dots,\alpha_r$. It is worth mentioning that all the convex coefficients are $>0$, so the new root cannot lie on the edge of the convex hull of $P$'s roots.

Now fix $P$ a certain nonzero complex polynomial, and consider $\Pi$, its primitive  (antiderivative) that vanishes at $0:~\Pi(0)=0$ and $\Pi'=P$. For each complex  $\omega$, write $\Pi_{\omega}=\Pi-\omega$, so that you get all the primitives of $P$. Also, define for any polynomial $Q$, $\mathrm{Conv}(Q)$, the convex hull of $Q$'s roots.

$\mathrm{MAIN~QUESTION}$: describe $\mathrm{Hull}(P)=\bigcap_{\omega\in\mathbb{C}}\mathrm{Conv}(\Pi_{\omega})$.

By the property cited above, $\mathrm{Hull}(P)$ is a convex compact subset of the complex plane that contains  $\mathrm{Conv}(P)$, but I strongly suspect that it is in general larger.

Here are some easy observations:

$1)$ replacing $P$ (resp. $\Pi$) by $\lambda P$ (resp. $\lambda \Pi$) will not change the result, and considering $P(aX+b)$ will change $\mathrm{Hull}(P)$ accordingly. Hence we can suppose both $P$ and $\Pi$ to be monic. The fact that $\Pi$ is no longer a primitive of $P$ is of no consequence.

$2)$ the intersection defining $\mathrm{Hull}(P)$ can be taken for $\omega$ ranging in a compact subset of $\mathbb{C}$: as $|\omega|\rightarrow\infty$, the roots of $\Pi_{\omega}$ will tend to become close to the $(\deg (P)+1)$-th roots of $\omega$, so for large enough $\omega$, their convex hull will always contain , say, $\mathrm{Conv}(\Pi)$.

$3)$ $\mathrm{Hull}(P)$ can be explicitly calculated in the following cases: $P=X^n$, $P$ of degree $1$ or $2$. There are only 2 kinds of degree $2$ polynomials: two simple roots or a double root. Using $z\rightarrow az+b$, one only has to consider $P=X^2$ and $P=X(X-1)$. The first one yields {$0$}, which equals $\mathrm{Conv}(X^2)$, the second one gives $[0,1]=\mathrm{Conv}(X(X-1))$.

Also, if $\Pi$ is a real polynomial of odd degree $n+1$ that has all its roots real and simple, say $\lambda_1<\mu_1<\lambda_2<\dots<\mu_n<\lambda_{n+1}$, where I have also placed  $P$'s roots $\mu_1,\dots,\mu_n$, and if you further assume that $\Pi(\mu_{2j})\leq\Pi(\mu_n)\leq\Pi(\mu_1)\leq\Pi(\mu_{2j+1})$ for all suitable $j$  (a condition that is best understood with a picture), then $\mathrm{Hull}(P)=\mathrm{Conv}(P)=[\mu_1,\mu_n]$: just vary $\omega$ between $[\Pi(\mu_n),\Pi(\mu_1)]$; the resulting polynomial $\Pi_{\omega}$ is always split over the real numbers and you get

$[\mu_1,\mu_n]=\mathrm{Conv}(P)\subset\mathrm{Hull}(P)\subset \mathrm{Conv}(\Pi_{\Pi(\mu_1)})\cap \mathrm{Conv}(\Pi_{\Pi(\mu_n)})=$ $[\mu_1,\dots]\cap [\dots,\mu_n]=[\mu_1,\mu_n]$

$4)$ The equation $\Pi_{\omega}(z)=\Pi(z)-\omega=0$ defines a Riemann surface, but I don't see how that could be of any use.

$\mathrm{QUESTION}$: is this hexagon equal to $\mathrm{Hull}(X^3-1)$?

$\mathrm{QUESTION (Conjecture)}$: is it true that $\mathrm{Hull}(P)=\bigcap_{\omega\in\mathrm{MR}}\mathrm{Conv}(\Pi_{\omega})$, where $\mathrm{MR}$ is the set of all $\omega_0$ such that $\Pi_{\omega_0}$ has a multiple root, i.e., the set of all $\Pi(\alpha_i)$ where the $\alpha_i$ are the roots of $P$?

A couple of years ago, I came up with the following question, to which I have no answer to this day. I have asked a few people about this, most of my teachers and some friends, but no one had ever heard of the question before, and no one knew the answer.

I hope this is an original question, but seeing how natural it is, I doubt this is the first time someone has asked it.

First, some motivation. Take $P$ any nonzero complex polynomial. It is an easy and classical exercise to show that the roots of its derivative $P'$ lie in the convex hull of its own roots (I know this as the Gauss-Lucas property). To show this, you simply write $P = a \cdot \prod_{i=1}^{r}(X-\alpha_i)^{m_i}$ where the  $\alpha_i~(i=1,\dots,r)$ are the different roots of $P$, and $m_i$ the corresponding multiplicities, and evaluate $\frac{P'}{P}=\sum_i \frac{m_i}{X-\alpha_i}$ on a root $\beta$ of $P'$ which is not also a root of $P$. You'll end up with an expression of $\beta$ as a convex combination of $\alpha_1,\dots,\alpha_r$. It is worth mentioning that all the convex coefficients are $>0$, so the new root cannot lie on the edge of the convex hull of $P$'s roots.

Now fix $P$ a certain nonzero complex polynomial, and consider $\Pi$, its primitive  (antiderivative) that vanishes at $0:~\Pi(0)=0$ and $\Pi'=P$. For each complex  $\omega$, write $\Pi_{\omega}=\Pi-\omega$, so that you get all the primitives of $P$. Also, define for any polynomial $Q$, $\mathrm{Conv}(Q)$, the convex hull of $Q$'s roots.

MAIN QUESTION: describe $\mathrm{Hull}(P)=\bigcap_{\omega\in\mathbb{C}}\mathrm{Conv}(\Pi_{\omega})$.

By the property cited above, $\mathrm{Hull}(P)$ is a convex compact subset of the complex plane that contains  $\mathrm{Conv}(P)$, but I strongly suspect that it is in general larger.

Here are some easy observations:

1. replacing $P$ (resp. $\Pi$) by $\lambda P$ (resp. $\lambda \Pi$) will not change the result, and considering $P(aX+b)$ will change $\mathrm{Hull}(P)$ accordingly. Hence we can suppose both $P$ and $\Pi$ to be monic. The fact that $\Pi$ is no longer a primitive of $P$ is of no consequence.

2. the intersection defining $\mathrm{Hull}(P)$ can be taken for $\omega$ ranging in a compact subset of $\mathbb{C}$: as $|\omega| \rightarrow \infty$, the roots of $\Pi_{\omega}$ will tend to become close to the $(\deg (P)+1)$-th roots of $\omega$, so for large enough $\omega$, their convex hull will always contain, say, $\mathrm{Conv}(\Pi)$.

3. $\mathrm{Hull}(P)$ can be explicitly calculated in the following cases: $P=X^n$, $P$ of degree $1$ or $2$. There are only 2 kinds of degree $2$ polynomials: two simple roots or a double root. Using $z\rightarrow az+b$, one only has to consider $P=X^2$ and $P=X(X-1)$. The first one yields {$0$}, which equals $\mathrm{Conv}(X^2)$, the second one gives $[0,1]=\mathrm{Conv}(X(X-1))$.

Also, if $\Pi$ is a real polynomial of odd degree $n+1$ that has all its roots real and simple, say $\lambda_1 < \mu_1 < \lambda_2 < \dots < \mu_n < \lambda_{n+1}$, where I have also placed  $P$'s roots $\mu_1, \dots, \mu_n$, and if you further assume that $\Pi(\mu_{2j}) \leq \Pi(\mu_n) \leq\Pi(\mu_1) \leq\Pi(\mu_{2j+1})$ for all suitable $j$  (a condition that is best understood with a picture), then $\mathrm{Hull}(P)=\mathrm{Conv}(P)=[\mu_1,\mu_n]$: just vary $\omega$ between $[\Pi(\mu_n), \Pi(\mu_1)]$; the resulting polynomial $\Pi_{\omega}$ is always split over the real numbers and you get

$$[\mu_1,\mu_n]=\mathrm{Conv}(P)\subset\mathrm{Hull}(P)\subset \mathrm{Conv}(\Pi_{\Pi(\mu_1)})\cap \mathrm{Conv}(\Pi_{\Pi(\mu_n)}) = \\= [\mu_1,\dots]\cap [\dots,\mu_n]=[\mu_1,\mu_n]$$

4. The equation $\Pi_{\omega}(z)=\Pi(z)-\omega=0$ defines a Riemann surface, but I don't see how that could be of any use.

QUESTION: is this hexagon equal to $\mathrm{Hull}(X^3-1)$?

QUESTION (Conjecture): is it true that $\mathrm{Hull}(P)=\bigcap_{\omega\in\mathrm{MR}}\mathrm{Conv}(\Pi_{\omega})$, where $\mathrm{MR}$ is the set of all $\omega_0$ such that $\Pi_{\omega_0}$ has a multiple root, i.e., the set of all $\Pi(\alpha_i)$ where the $\alpha_i$ are the roots of $P$?

$1)$ replacing $P$ (resp. $\Pi$) by $\lambda P$ (resp. $\lambda \Pi$) will not change the result, and considering $P(aX+b)$ will change $\mathrm{Hull}(P)$ accordingly. Hence we can suppose both $P$ and $\Pi$ be unitary. The fact that $\Pi$ is no longer a primitive of $P$ is of no consequence.

$1)$ replacing $P$ (resp. $\Pi$) by $\lambda P$ (resp. $\lambda \Pi$) will not change the result, and considering $P(aX+b)$ will change $\mathrm{Hull}(P)$ accordingly. Hence we can suppose both $P$ and $\Pi$ to be monic. The fact that $\Pi$ is no longer a primitive of $P$ is of no consequence.

A couple of years ago, I came up with the following question, to which I have no answer to this day. I have asked a few people about this, most of my teachers and some friends, but noone had ever heard of the question before, and noone knew the answer.

First, some motivation. Take $P$ any non zero complex polynomial. It is an easy and classical exercise to show that the roots of its derivative $P'$ lie in the convex hull of its own roots (I know this as the Gauss-Lucas property). To show this, you simply write $P=a\cdot\prod_{i=1}^{r}(X-\alpha_i)^{m_i}$ where the $\alpha_i~(i=1,\dots,r)$ are the different roots of $P$, and $m_i$ the corresponding multiplicities, and evaluate $\frac{P'}{P}=\sum_i \frac{m_i}{X-\alpha_i}$ on a root $\beta$ of $P'$ which is not also a root of $P$. You'll end up with an expression of $\beta$ as a convex combination of $\alpha_1,\dots,\alpha_r$. It is worth mentioning that all the convex coefficients are $>0$, so the new root cannot lie on the edge of the convex hull of $P$'s roots.

Now fix $P$ a certain non zero complex polynomial, and consider $\Pi$ it's primitive (antiderivative) that vanishes at $0:~\Pi(0)=0$ and $\Pi'=P$. For each complex $\omega$, write $\Pi_{\omega}=\Pi-\omega$ so that you get all the primitives of $P$. Also, define for any polynomial $Q$, $\mathrm{Conv}(Q)$, to be the convex hull of $Q$'s roots.

By the above quoted property, $\mathrm{Hull}(P)$ is a convex compact subset of the complex plane that contains $\mathrm{Conv}(P)$, but I strongly suspect that it is in general larger.

$1)$ replacing $P$ (resp. $\Pi$) by $\lambda P$ (resp. $\lambda \Pi$) will not change the result, and considering $P(aX+b)$ will change $\mathrm{Hull}(P)$ accordingly. Hence we can suppose both $P$ and $\Pi$ be unitary. The fact that $\Pi$ is no longer a primitive of $P$ is of no consequence.

$3)$ $\mathrm{Hull}(P)$ can be explicitely calculated in the following cases: $P=X^n$, $P$ of degree $1$ or $2$. There are only 2 kinds of degree $2$ polynomials: 2 simple roots or a double root. Using $z\rightarrow az+b$, one only has to consider $P=X^2$ and $P=X(X-1)$. The first one yields {$0$}, which equals $\mathrm{Conv}(X^2)$, the second one gives $[0,1]=\mathrm{Conv}(X(X-1))$.

Also if $\Pi$ is a real polynomial of odd degree $n+1$, that has all its roots real and simple, say $\lambda_1<\mu_1<\lambda_2<\dots<\mu_n<\lambda_{n+1}$, where I have also placed $P$'s roots $\mu_1,\dots,\mu_n$, and if you further assume that $\Pi(\mu_{2j})\leq\Pi(\mu_n)\leq\Pi(\mu_1)\leq\Pi(\mu_{2j+1})$ for all suitable $j$ (a condition that is best understood with a picture), then $\mathrm{Hull}(P)=\mathrm{Conv}(P)=[\mu_1,\mu_n]$: just vary $\omega$ between $[\Pi(\mu_n),\Pi(\mu_1)]$, the resulting polynomial $\Pi_{\omega}$ is always split over the real numbers and you get

Computing $\mathrm{Hull}(X^3-1)$ requires factorizing degree 4 polynomials, so one naturally tries to look for good values of $\omega$, the $\omega$ that allow for easy factorisation of $\Pi_{\omega}=X^4-4X-\omega$. For instance the $\omega$ that produce a double root. All that remains to be done afterwards is to factor a polynomial of degree $2$. The problem is symmetric, and you can focus on the case where 1 is the double root (i.e. $\omega=-3$). Plugging the result in the intersection, and rotating twice, you obtain the following superset of $\mathrm{Hull}(X^3-1)$: a hexagon that is the intersection of 3 similar isocele triangles with their main vertex located on the three third roots of unity $1,j,j^2$

Consider the question of how the convex hulls of the roots of $\Pi_{\omega}$ vary as $\omega$ varies. When $\omega_0$ is such that all roots of $\Pi_{\omega_0}$ are simple, then the inverse function theorem shows that the roots of $\Pi_{\omega}$ with $\omega$ in a small neighborhood of $\omega_0$ vary holomorphically $\sim$ linearly in $\omega-\omega_0$: $z(\omega)-z(\omega_0)\sim \omega-\omega_0$. If however $\omega_0$ is such that $\Pi_{\omega_0}$ has a multiple root $z_0$ of multiplicity $m>1$, then a small variation of $\omega$ about $\omega_0$ will split the multiple root $z_0$ into $m$ distinct roots of $\Pi_{\omega}$ that will spread out roughly as $z_0+c(\omega-\omega_0)^{\frac{1}{m}}$, where $c$ is some non zero coefficient. This means that for small variations, these roots will move at much higher velocities than the simple roots, and they will do the major contribution to the variation of $\mathrm{Conv}(\Pi_{\omega})$, also, they spread evenly out, and (at least if the multiplicity is greater or equal to $3$) they will tend to increase the convex hull around $z_0$. Thus it seems not too unreasonable to conjecture that the convex hull $\mathrm{Conv}(\Pi_{\omega})$ has what one can only describe as critical points at the $\omega_0$ that produce roots with multiplicities. I'm fairly certain there is a sort of calculus on convex sets that would allow one to make the above statement precise, but I don't know it.

Back to $X^3-1$: explicit calculations suggest that up to second order the double root $1$ of $X^4-4X+3-h$ for $|h|<<1$ splits in half nicely (here $\omega=-3+h$) and the convex hull will continue to contain the aforementioned hexagon.

$\mathrm{QUESTION\Conjecture}$: is it true that $\mathrm{Hull}(P)=\bigcap_{\omega\in\mathrm{MR}}\mathrm{Conv}(\Pi_{\omega})$, where $\mathrm{MR}$ is the set of all $\omega_0$ such that $\Pi_{\omega_0}$ has a multiple root, i.e. the set of all $\Pi(\alpha_i)$ where the $\alpha_i$ are the roots of $P$?

Are you aware of a solution? Is this a classical problem? Is anybody brave enough to make a computer program that would compute some intersections of convex hulls obtained from the roots to see if my conjecture is any good?

A couple of years ago, I came up with the following question, to which I have no answer to this day. I have asked a few people about this, most of my teachers and some friends, but no one had ever heard of the question before, and no one knew the answer.

First, some motivation. Take $P$ any nonzero complex polynomial. It is an easy and classical exercise to show that the roots of its derivative $P'$ lie in the convex hull of its own roots (I know this as the Gauss-Lucas property). To show this, you simply write $P=a\cdot\prod_{i=1}^{r}(X-\alpha_i)^{m_i}$ where the $\alpha_i~(i=1,\dots,r)$ are the different roots of $P$, and $m_i$ the corresponding multiplicities, and evaluate $\frac{P'}{P}=\sum_i \frac{m_i}{X-\alpha_i}$ on a root $\beta$ of $P'$ which is not also a root of $P$. You'll end up with an expression of $\beta$ as a convex combination of $\alpha_1,\dots,\alpha_r$. It is worth mentioning that all the convex coefficients are $>0$, so the new root cannot lie on the edge of the convex hull of $P$'s roots.

Now fix $P$ a certain nonzero complex polynomial, and consider $\Pi$, its primitive (antiderivative) that vanishes at $0:~\Pi(0)=0$ and $\Pi'=P$. For each complex $\omega$, write $\Pi_{\omega}=\Pi-\omega$, so that you get all the primitives of $P$. Also, define for any polynomial $Q$, $\mathrm{Conv}(Q)$, the convex hull of $Q$'s roots.

By the property cited above, $\mathrm{Hull}(P)$ is a convex compact subset of the complex plane that contains $\mathrm{Conv}(P)$, but I strongly suspect that it is in general larger.

$1)$ replacing $P$ (resp. $\Pi$) by $\lambda P$ (resp. $\lambda \Pi$) will not change the result, and considering $P(aX+b)$ will change $\mathrm{Hull}(P)$ accordingly. Hence we can suppose both $P$ and $\Pi$ be unitary. The fact that $\Pi$ is no longer a primitive of $P$ is of no consequence.

$3)$ $\mathrm{Hull}(P)$ can be explicitly calculated in the following cases: $P=X^n$, $P$ of degree $1$ or $2$. There are only 2 kinds of degree $2$ polynomials: two simple roots or a double root. Using $z\rightarrow az+b$, one only has to consider $P=X^2$ and $P=X(X-1)$. The first one yields {$0$}, which equals $\mathrm{Conv}(X^2)$, the second one gives $[0,1]=\mathrm{Conv}(X(X-1))$.

Also, if $\Pi$ is a real polynomial of odd degree $n+1$ that has all its roots real and simple, say $\lambda_1<\mu_1<\lambda_2<\dots<\mu_n<\lambda_{n+1}$, where I have also placed $P$'s roots $\mu_1,\dots,\mu_n$, and if you further assume that $\Pi(\mu_{2j})\leq\Pi(\mu_n)\leq\Pi(\mu_1)\leq\Pi(\mu_{2j+1})$ for all suitable $j$ (a condition that is best understood with a picture), then $\mathrm{Hull}(P)=\mathrm{Conv}(P)=[\mu_1,\mu_n]$: just vary $\omega$ between $[\Pi(\mu_n),\Pi(\mu_1)]$; the resulting polynomial $\Pi_{\omega}$ is always split over the real numbers and you get

Computing $\mathrm{Hull}(X^3-1)$ requires factorizing degree 4 polynomials, so one naturally tries to look for good values of $\omega$, the $\omega$ that allow for easy factorization of $\Pi_{\omega}=X^4-4X-\omega$---for instance, the $\omega$ that produces a double root. All that remains to be done afterwards is to factor a quadratic polynomial. The problem is symmetric, and you can focus on the case where 1 is the double root (i.e., $\omega=-3$). Plugging in the result in the intersection, and rotating twice, you obtain the following superset of $\mathrm{Hull}(X^3-1)$: a hexagon that is the intersection of three similar isoceles triangles with their main vertex located on the three third roots of unity $1,j,j^2$

Consider the question of how the convex hulls of the roots of $\Pi_{\omega}$ vary as $\omega$ varies. When $\omega_0$ is such that all roots of $\Pi_{\omega_0}$ are simple, then the inverse function theorem shows that the roots of $\Pi_{\omega}$ with $\omega$ in a small neighborhood of $\omega_0$ vary holomorphically $\sim$ linearly in $\omega-\omega_0$: $z(\omega)-z(\omega_0)\sim \omega-\omega_0$. If however $\omega_0$ is such that $\Pi_{\omega_0}$ has a multiple root $z_0$ of multiplicity $m>1$, then a small variation of $\omega$ about $\omega_0$ will split the multiple root $z_0$ into $m$ distinct roots of $\Pi_{\omega}$ that will spread out roughly as $z_0+c(\omega-\omega_0)^{\frac{1}{m}}$, where $c$ is some nonzero coefficient. This means that for small variations, these roots will move at much higher velocities than the simple roots, and they will constitute the major contribution to the variation of $\mathrm{Conv}(\Pi_{\omega})$; also, they spread out evenly, and (at least if the multiplicity is greater or equal to $3$) they will tend to increase the convex hull around $z_0$. Thus it seems not too unreasonable to conjecture that the convex hull $\mathrm{Conv}(\Pi_{\omega})$ has what one can only describe as critical points at the $\omega_0$ that produce roots with multiplicities. I'm fairly certain there is a sort of calculus on convex sets that would allow one to make this statement precise, but I don't know see what it could be.

Back to $X^3-1$: explicit calculations suggest that up to second order, the double root $1$ of $X^4-4X+3-h$ for $|h|<<1$ splits in half nicely (here $\omega=-3+h$), and the convex hull will continue to contain the aforementioned hexagon.

$\mathrm{QUESTION (Conjecture)}$: is it true that $\mathrm{Hull}(P)=\bigcap_{\omega\in\mathrm{MR}}\mathrm{Conv}(\Pi_{\omega})$, where $\mathrm{MR}$ is the set of all $\omega_0$ such that $\Pi_{\omega_0}$ has a multiple root, i.e., the set of all $\Pi(\alpha_i)$ where the $\alpha_i$ are the roots of $P$?

Are you aware of a solution? Is this a classical problem? Is anybody brave enough to make a computer program that would compute some intersections of convex hulls obtained from the roots to see if my conjecture is valid?