\documentclass{PJM}


\authorone{Tewodros Amdeberhan}
\addressone{Department of Mathematics,
Tulane University,
New Orleans, LA, 70118}
\countryone{USA}
\emailone{tamdeber@tulane.edu }
\authortwo{Doron Zeilberger}
\addresstwo{Department of Mathematics,
Rutgers University (New Brunswick),
Hill Center - Busch Campus,
110 Frelinghuysen Rd., Piscataway, NJ 08854-8019}
\countrytwo{USA}
\emailtwo{DoronZeil@gmail.com }
\title{An elegant Multi-Integral that implies an even more elegant determinant identity of Dougherty and McCammond}
\headlinetitle{Multi-Integral and determinantal identity}
\begin{document}
\centerline{Communicated  by  Ayman Badawi}
\vskip0.2in
\footnotesize MSC 2010 Classifications: Primary 05A15, 05A05; Secondary 30C15.
\vskip0.1in
Keywords and phrases: Cauchy's alternant formula, Vandermonde and Jacobian matrices and determinants, critical values, multi-integrals, anti-symmetric functions.
\vskip0.1in
\normalsize

 {\bf Abstract} We state and prove an explicit evaluation of a certain multi-variate integral and use it to furnish a new, and shorter, proof of an elegant determinant identity of Michael Dougherty and Jon McCammond that came up in their study of critical values of a complex polynomial. \vspace*{5mm}


\section{Introduction}
Beardon-Carne-Ng [1] investigated invertibility of the Jacobian of the polynomials
$$p(z)=\int_0^z(w-z_1)\cdots(w-z_n)\,dw$$
given by the $n$-by-$n$ matrix
$$\pmb{J}(z)=\left(\frac{\partial}{\partial z_i}p(z_j)\right)_{i,j}^{1,n}.$$
Using Topological arguments and techniques from Several Complex variables, the authors [1] show that every $n$-tuple of complex numbers arises as the critical \it values \rm of some polynomial by proving that the determinant $\det(\pmb{J})\neq0$, as along as $z_1,\dots,z_n$ are non-zero and distinct. More recently, Dougherty and McCammond [2] computed 
$\det(\pmb{J})$, explicitly, and reproved the above-mentioned result of [1] with a different method. 

\smallskip
\noindent
Some nomenclature adopted in the sequel: bold-face lower case letters are reserved for vectors, such as $\pmb{z}=(z_1,\dots,z_n)$ and $\pmb{\hat{z}}^{i}=(z_1,\dots,z_{i-1},z_{i+1},\dots,z_n)$. Given $\pmb{a}=(a_1,\dots,a_n)\in\mathbb{Z}_{\geq0}^n$, denote $\pmb{\bar a}=\sum_{i=1}^{\lceil n/2\rceil}a_{2i-1}$ where $\lceil\cdot\rceil$ is the {\it ceiling function} while $\lfloor\cdot\rfloor$ is the {\em floor function}. 
Let $V_n(\pmb{x})=\prod_{1\leq i<j\leq n}(x_j-x_i)$ be the {\em Vandermonde determinant} of the {\em Vandermonde matrix} $\pmb{M}=(x_j^{i-1})_{i,j}^{1,n}$. We write $\pmb{dx}=dx_1\cdots dx_n$, the dimension $n$ being suppressed when it is clear from the context.


\section{The Main Theorem}
The first purpose of this note is to prove the following elegant identity.

\begin{theorem}  \label{Theorem 2.1}
 Let $z_1, \dots z_n$ be  commuting indeterminates, let $n$ be a positive integer, and let $a_1, \dots, a_n$, and $b$ be non-negative integers. Then
\begin{align*}
&\int_0^{z_n} \dots  \int_0^{z_1} \, \prod_{i=1}^{n} x_i^b \prod_{1\leq j,k\leq n}(x_j -z_k)^{a_k} \prod_{1 \leq i<j \leq n} (x_j -x_i) \, dx_1\cdots dx_n \\
&=
\,(-1)^{\pmb{\bar{a}}} \prod_{1 \leq i<j \leq n} (z_j - z_i)^{a_i+a_j+1} \cdot \prod_{i=1}^{n} z_i^{a_i+b+1} \cdot
\frac{b! \, \prod_{i=1}^{n} a_i!}{(n+b+ \sum_{i=1}^{n} a_i)!}. \quad 
\tag1
%\eqno(1)
\end{align*}
\end{theorem}

\section{An Example}
We offer an illustrative example for Theorem 2.1. Choose $n=2, b=2022, a_1=1, a_2=2$. Then Eq. (1) reads 
$$\int_0^{z_2}\int_0^{z_1}(x_1x_2)^{2022}(x_1-z_1)(x_2-z_1)(x_1-z_2)^2(x_2-z_2)^2(x_2-x_1) \,dx_1dx_2 
=\frac{-\,(z_2-z_1)^4z_1^{2024}z_2^{2025}}{17025293698597800}.$$

\section{An interesting consequence}
The second purpose our work here is to deduce from Theorem 2.1 (and thereby give a shorter proof) of the following even more elegant identity, discovered, and first proved in [2].

\begin{theorem}  ({\bf Dougherty and McCammond}) Let
$$
p(Z) := \, \int_{0}^{Z} \, \prod_{i=1}^{n} (w-z_i)^{a_i} \, dw, \quad
$$
and let ${\bf J}(z_1, \dots, z_n)$ be the $n \times n$ matrix whose $(i,j)$-entry is
${\bf J}(z_1, \dots, z_n)_{i,j} \, := \, \frac{\partial}{\partial z_i} p(z_j),$
then
\begin{align*}
\det \, {\bf J}(z_1, \dots, z_n) \, = \, 
\frac{\prod_{i=1}^{n} a_i!} { (\sum_{i=1}^{n} a_i)!}
\cdot \prod_{i=1}^{n} (-z_j)^{a_j}  \cdot \prod_{\substack {1 \leq i,j\leq n \\ i \neq j}} (z_i - z_j)^{a_j}. \quad 
\tag2
\end{align*}
\end{theorem}
\noindent
{\bf Proof that Theorem 2.1 $\implies$ Theorem 4.1:} Let's rewrite the determinant and apply {\em Cauchy's alternant formula} so that
\begin{align*}
\det ({\bf J}) 
&= \det \left ( -a_i \int_0^{z_j} \prod_{k=1}^{n} (w-z_k)^{a_k} \frac{dw}{w-z_i} \right ) \, = \, \det \left ( -a_i \int_0^{z_j} \prod_{k=1}^{n} (x_j-z_k)^{a_k} \frac{d x_j}{x_j-z_i} \right ) \\
&=\prod_{i=1}^{n} (-a_i) \int_0^{z_n} \dots \int_0^{z_1} \, \prod_{1 \leq j,k \leq n} (x_j -z_k)^{a_k} \cdot \det \left ( \frac{1}{x_j-z_i} \right)_{i,j}^{1,n} \, \pmb{dx} \\
&=\, \prod_{i=1}^{n} (-a_i) \int_0^{z_n} \dots \int_0^{z_1} \, \prod_{1 \leq j,k \leq n} (x_j -z_k)^{a_k} \cdot
\left [ \frac{ \prod_{1 \leq i < j \leq n} (z_i - z_j) (x_j -x_i)}{ \prod_{1 \leq i,j \leq n} (x_j -z_i)} \right ] \pmb{dx} \\
&=\prod_{i=1}^{n} (-a_i)  \cdot \prod_{1 \leq i < j \leq n} (z_i - z_j)
\cdot \int_0^{z_n} \dots \int_0^{z_1}  \prod_{1 \leq j,k \leq n} (x_j -z_k)^{a_k-1}  \cdot V_n(\pmb{x})\,\, \pmb{dx}.
\end{align*}
But by Theorem 2.1, with $b=0$ and $(a_1, \dots, a_n)$ replaced by $(a_1-1, \dots a_n-1)$, this equals
\begin{align*}
\det({\bf J})
&=(-1)^{\pmb{\bar{a}}-\lceil n/2\rceil}
 \prod_{i=1}^{n} (-a_i)  \prod_{1 \leq i < j \leq n} (z_i - z_j)
\prod_{1 \leq i<j \leq n} (z_j - z_i)^{a_i+a_j-1}  \prod_{i=1}^{n} z_i^{a_i} 
\frac{ \prod_{i=1}^{n} (a_i-1)!}{(\sum_{i=1}^{n} a_i)!} \\
&=\frac{\prod_{i=1}^{n} a_i!} { (\sum_{i=1}^{n} a_i)!}\,\,
\prod_{i=1}^{n} (-z_j)^{a_j}  \prod_{\substack {1 \leq i,j \leq n \\ i \neq j}} (z_i - z_j)^{a_j}. \quad \Box
\end{align*}

\section{Proof of Theorem 2.1:} Proceed by induction on $n$ and $b$. When $n=1$ and $b=0$, whose proof is left to
the reader's five-year-old, the claim is saying $\int_0^{z_1} (x_1-z_1)^{a_1}\, dx_1 = (-1)^{a_1} \frac{z_1^{a_1+1}}{a_1+1}$.

\newpage

\bigskip
\noindent
Let's denote the statement of Theorem 2.1 by ${\bf A}(n,b)$.

\bigskip
\noindent
{\bf Proof that ${\bf A}(n,b) \implies {\bf A}(n,b+1)$:}

\smallskip
\noindent
Let $\pmb{a}=(a_1,\dots,a_n)$. We claim that {\it both} sides of Eq. (1), let's call them $L(\pmb{a};b)$ and  $R(\pmb{a};b)$ respectively,
satisfy the recurrence
$$
X(\pmb{a}; b+1) = 
\sum_{i=1}^{n} \left ( \prod_{\substack {j=1 \\ j \neq i}}^{n} \frac{z_j}{z_j-z_i} \right ) \cdot
X(a_1, \dots, 1+a_i, \dots, a_n;b) 
+ \left ( \prod_{j=1}^{n} z_j \right) \cdot X(\pmb{a}; b). \quad
\eqno(3)
$$
In other words, if you replace $X$ by either $L$ or $R$ you get a true statement. Regarding the left-hand side of (1), in fact, this identity is already true if you replace $X$ by the {\bf integrand} of the left side of $(1)$,
since there are no $x_i$'s in sight, it is still true when you integrate with respect to $x_1, \dots, x_n$. We leave both checks as pleasant exercises for the curious reader. $\Box$

\bigskip
\noindent
{\bf Proof that ${\bf A}(n-1,b)$ for all $b \implies {\bf A}(n,0)$:}

\smallskip
\noindent
Fix $a_1, \dots, a_n$. Notations: $\pmb{a}=(a_1,\dots,a_n), \pmb{z}=(z_1,\dots,z_n), V_n(\pmb{x})=\prod_{1 \leq i<j \leq n} (x_j -x_i)$ and
$\pmb{\hat{x}}^i=(x_1,\dots,x_{i-1},x_{i+1},\dots,x_n)$. Define the multi-variable polynomial
$$F(\pmb{x},\pmb{z}):= \prod_{1\leq j,k\leq n} (x_j-z_k)^{a_k}. $$
We claim (check! this reduces to  the Laplace expansion $V_n(\pmb{x})=\sum_{i=1}^n(-1)^iV_{n-1}(\pmb{\hat{x}}^i)x_i^{n-1}$) that
\begin{align*}
\left(n+\sum_{i=1}^{n} a_i\right) \,F(\pmb{x},\pmb{z}) V_n(\pmb{x})
=\sum_{i=1}^{n} (-1)^i V_{n-1}(\pmb{\hat{x}}^i) \cdot \frac{\partial}{\partial x_i} \left [ \prod_{j=1}^{n} (x_i-z_j) \cdot F(\pmb{x},\pmb{z}) \right ]. \quad
\tag4
\end{align*}
Applying  $\int_0^{z_n} \dots \int_0^{z_1} (\cdots)dx_1 \cdots dx_n$, we get
\begin{align*}
&\left(n+\sum_{i=1}^{n} a_i\right)\,\int_0^{z_n} \dots \int_0^{z_1} F(\pmb{x},\pmb{z}) V_n(\pmb{x}) \,\pmb{dx} \\
& = \sum_{i=1}^{n} (-1)^i \,\int_0^{z_n} \dots \int_0^{z_1}V_{n-1}(\pmb{\hat{x}}^i) \cdot \frac{\partial}{\partial x_i} \left [ \prod_{j=1}^{n} (x_i-z_j)
 \cdot F(\pmb{x},\pmb{z}) \right ] \pmb{dx} \\
& =  
\sum_{i=1}^{n} (-1)^i \, 
\int_0^{z_n} \dots \int_0^{z_{i+1}} \int_0^{z_{i-1}} \dots \int_0^{z_1} V_{n-1}(\pmb{\hat{x}}^i) \,\, \pmb{d\hat{x}^i} 
 \times \int_{0}^{z_i} \frac{\partial}{\partial x_i} \left [ \prod_{j=1}^{n} (x_i-z_j) \cdot F(\pmb{x},\pmb{z}) \right ] dx_i. 
\end{align*}
By the {\it Fundamental Theorem of Calculus}, we have
\begin{align*}
\int_{0}^{z_i} \frac{\partial}{\partial x_i} \left [ \prod_{j=1}^{n} (x_i-z_j) \cdot F(\pmb{x},\pmb{z}) \right ] dx_i 
&= \prod_{j=1}^{n} (x_i-z_j) \cdot F(\pmb{x},\pmb{z}) \bigl \vert_{x_i=0}^{x_i=z_i} \\
&= \prod_{j=1}^{n} (x_i-z_j)^{a_{j}+1}  \prod_{j=1}^{n}\prod_{\substack {1 \leq i'\leq n \\ i' \neq i}} (x_{i'}-z_j)^{a_{j}} \bigl \vert_{x_i=0}^{x_i=z_i} \\
&= - \prod_{j=1}^{n} (-z_j)^{a_{j}+1}  \prod_{j=1}^{n}\prod_{\substack {1 \leq i'\leq n \\  i' \neq i}} (x_{i'}-z_j)^{a_{j}}.  
\end{align*}
Going back, we have that the left side of Eq. (1), when $b=0$, is
\begin{align*}
&\frac{-1}{n+\sum_{i=1}^{n} a_i} \cdot
\prod_{j=1}^{n} (-z_j)^{a_{j}+1}  \cdot
\sum_{i=1}^{n} (-1)^i \,\int_0^{z_n} \dots  \int_{0}^{z_{i+1}}  \int_{0}^{z_{i-1}}  \dots \int_0^{z_1}
V_{n-1}(\pmb{\hat{x}}^i)  \\
& \qquad \qquad \qquad \qquad \qquad \times \prod_{j=1}^{n}\prod_{\substack {1 \leq i'\leq n \\ i' \neq i}} (x_{i'}-z_j)^{a_{j}} \pmb{d\hat{x}}^i \\
&= \frac{-1}{n+\sum_{i=1}^{n} a_i} \cdot \prod_{j=1}^{n} (-z_j)^{a_{j}+1}  \cdot
 \sum_{i=1}^{n} (-1)^i \,\int_0^{z_n} \dots  \int_{0}^{z_{i+1}}  \int_{0}^{z_{i-1}} \dots \int_0^{z_1}V_{n-1}(y_1, \dots, y_{n-1})  \\
& \qquad \qquad \qquad \qquad \qquad \times \prod_{j=1}^{n}\prod_{i=1}^{n-1} (y_i-z_j)^{a_{j}} dy_1\cdots dy_{n-1}.
\end{align*}
We now claim that
\begin{align*}
& \sum_{i=1}^{n} (-1)^{i-1} \,\int_0^{z_n} \dots  \int_{0}^{z_{i+1}}  \int_{0}^{z_{i-1}} \dots \int_0^{z_1}
V_{n-1}(y_1, \dots, y_{n-1}) \cdot
\prod_{j=1}^{n}\prod_{i=1}^{n-1} (y_i-z_j)^{a_{j}} dy_1 \cdots  dy_{n-1} \\
& \qquad \qquad =\int_{z_1}^{z_n} \dots \int_{z_1}^{z_2} V_{n-1}(y_1, \dots , y_{n-1}) \cdot \prod_{j=1}^{n}\prod_{i=1}^{n-1} (y_i-z_j)^{a_{j}} dy_1 \cdots  dy_{n-1}. 
\tag5
\end{align*}
In order to prove this, notice that each of the integrands on the left-hand side, and the integrand on the right-hand side, are {\it anti-symmetric} in their arguments. Hence, for any given permutation of the integration variables, the effect is to multiple it by the sign of that permutation. Calling the common integrand $f(y_1,\dots, y_{n-1})$ and denoting $A_n(i)=Per(1,\dots, i-1, i+1, \dots n)$, we claim that
\begin{align*}
& \sum_{i=1}^{n} (-1)^{i-1} 
\sum_{\pi \in A_n(i)} \, sgn(\pi) \, \int_0^{z_{\pi(n)}} \dots  \int_{0}^{z_{\pi(i+1)}} \int_{0}^{z_{\pi(i-1)}} \cdots \int_0^{z_{\pi(1)}}
f(y_1, \dots, y_{n-1}) dy_1\cdots dy_{n-1} \\
& \qquad \qquad =\sum_{\pi \in A_n(1)}  sgn(\pi) \, \int_{z_1}^{z_{\pi(n)}} \dots \int_{z_1}^{z_{\pi(2)}} f(y_1, \dots, y_{n-1}) dy_1 \dots dy_{n-1}.
\tag6
\end{align*}
Since both sides of Eq. $(6)$ are $(n-1)!$ times the respective sides of Eq. $(5)$, if we can prove $(6)$, then $(5)$ would follow.

\smallskip
\noindent
But {\bf surprise!}, Eq. $(6)$ is valid for {\it any} integrand! It is just a relation between {\it regions} in $\mathbb{R}^{n-1}$ that is equivalent to an easy symmetric function identity,
that we also leave as a pleasant exercise to the reader. Now make the change of variables $(y_1, \dots, y_{n-1}) \rightarrow (y_1-z_1, \dots, y_{n-1}-z_1) $, thereby making
it a case of ${\bf A}(n-1,b)$ with $b=a_1$; and $a_1, \dots, a_{n-1}$ replaced by $a_2, \dots, a_n$, respectively; and
 $z_1, \dots, z_{n-1}$ replaced by $z_2-z_1, \dots, z_n-z_1$, respectively. Plugging it in and simplifying, completes
the induction. $\Box$ $\Box$

\bigskip
\noindent
{\bf Remark:} Readers that prefer not do the `exercises' can convince themselves of all the claims, empirically, by playing with the Maple package {\tt CritVal.txt} available
from the front of this paper

{\tt https://sites.math.rutgers.edu/\~{}zeilberg/mamarim/mamarimhtml/crit.html} 

\newpage

\bigskip
\noindent
\section{Appendix: proof of the "exercises"}

\smallskip
\noindent
{\bf Proof of Equation (3):}

\bigskip
\noindent
Let $\pmb{e}_i=(0,\dots,0,1,0,\dots,0)$ be the $i$-th unit vector. So, the recurrence (3) reads as
$$X(\pmb{a};b+1)=\sum_{i=1}^nX(\pmb{a}+\pmb{e}_i;b)\cdot \prod_{\substack {j=1 \\ j\neq i}}^n\frac{z_j}{z_j-z_i}+X(\pmb{a};b)\cdot \prod_{i=1}^nz_i.
$$
The integrand on the left-hand side of (1) satisfies this recurrence  because it reduces to (check!)
\begin{align*}
\prod_{i=1}^n\frac{x_i}{z_i}&=1+\sum_{i=1}^n\left(\frac{x_i-z_i}{z_i}\right)\cdot \prod_{\substack {j=1 \\ j\neq i}}^n\frac{x_j-z_i}{z_j-z_i}. \tag7
\end{align*}
Induct on $n$. The case $n=1$ is obvious. Let $w=x_1$ and treat (7) as a {\em linear} equation in $w$:
\begin{align*}
\frac{w}{z_1}\cdot \prod_{i=2}^n\frac{x_i}{z_i}=1+\frac{w-z_1}{z_1}\prod_{j=2}^n\frac{x_j-z_1}{z_j-z_1}+\sum_{i=2}^n\frac{x_i-z_i}{z_i}\cdot
\frac{w-z_i}{z_1-z_i}\cdot\prod_{\substack {j=2 \\ j\neq i}}^n\frac{x_j-z_i}{z_j-z_{i}}.
\end{align*}
If $w=z_1$, equality holds by the induction assumption. Putting $w=0$, turns the task into 
\begin{align*}
\prod_{i=2}^n\frac{x_i-z_1}{z_i-z_1}=1+\sum_{i=2}^n\frac{x_i-z_i}{z_i-z_1}\cdot \prod_{\substack {j=2 \\ j\neq i}}^n\frac{x_j-z_i}{z_j-z_{i}}
\end{align*}
which is {\em again} the inductive step with $x_i\rightarrow x_i-z_1$ and $z_i\rightarrow z_i-z_1$. So, the first claim holds.

\smallskip
\noindent
Now, we focus on the right-hand side of (1) and show it, too, fulfils (3). This tantamount
\begin{align*}
\frac{1+b}{n+b+1+\sum_{i=1}^na_i}&=1+\sum_{i=1}^n(-1)^{\delta_i}\frac{1+a_i}{n+b+1+\sum_{i=1}^na_i}\cdot (-1)^{i-1}
\end{align*}
where $\delta_i=1$ if $i$ is odd; $\delta_i=0$, otherwise. The justification is immediate. $\Box$

\bigskip
\noindent
{\bf Proof of Equation (4):}

\bigskip
\noindent
After computing the derivatives and cancelling out the term $F(\pmb{x},\pmb{z})$, the claim boils down to
\begin{align*}\left(n+\sum_{i=1}^na_i\right) V_n(\pmb{x})
&=\sum_{i=1}^n(-1)^i\,V_{n-1}(\pmb{\hat{x}}^i)\cdot \sum_{j=1}^n(1+a_j)\prod_{\substack {k=1 \\ k\neq j}}^n(x_i-z_k) \\
&=\sum_{j=1}^n(1+a_j)\,\,\sum_{i=1}^n(-1)^i\,V_{n-1}(\pmb{\hat{x}}^i)\cdot \prod_{\substack {k=1 \\ k\neq j}}^n(x_i-z_k). \tag8
\end{align*}
If we consider the sum
$\sum_{i=1}^n(-1)^i\,V_{n-1}(\pmb{\hat{x}}^i)\cdot x_i^m$, it is recognized as the determinant of the Vandermonde matrix $\pmb{M}$
where the last row is replaced by the vector $(x_1^m,\dots,x_n^m)$. This, however, is well-known to be $V_n(\pmb{x})\cdot s_{m-n+1}$ where $s_{\mu}$ is 
the {\em Schur polynomial}. Notice that $\prod_{\substack {k=1 \\ k\neq j}}^n(x_i-z_k)$ is a polynomial of degree $n-1$, in $x_i$. But, $s_{m-n+1}=1$ if $m=n-1$; and $s_{m-n+1}=0$ for $m<n-1$. Thus, equation (8) becomes trivial $(n+\sum_{i=1}^na_i)\,V_n(\pmb{x})=\sum_{j=1}^n(1+a_j)\,V_n(\pmb{x})$. $\Box$

\newpage

\bigskip
\noindent
{\bf Proof of Equation (5):}

\bigskip
\noindent
Eq. (5) has been elucidated, neatly, using a symmetrizing process as depicted by Eq. (6). Here, we offer yet another verification. To this end, start with the right-hand side of Eq. (5) by replacing each integral with $\int_{z_1}^{z_i}=\int_0^{z_i}-\int_0^{z_1}$ and expand the product $(\int_0^{z_n}-\int_0^{z_1})\cdots (\int_0^{z_2}-\int_0^{z_1})$. Based on the fact that the integrand is an anti-symmetric, any multi-integral involving repeated evaluation $\int_0^{z_1}\int_0^{z_1}$ vanishes while out-of-order pairs, such as $\int_0^{z_3}\int_0^{z_1}\int_0^{z_2}$, changes sign where {\em reordered}, that is, $-\int_0^{z_3}\int_0^{z_2}\int_0^{z_1}$. On account of this, we find the $(n-1)$-tuple multi-integrals
\begin{align*}
\int_{z_1}^{z_n}\cdots \int_{z_1}^{z_2}
&=\left(\int_0^{z_n}-\int_0^{z_1}\right)\cdots \left(\int_0^{z_2}-\int_0^{z_1}\right) \\
&=\int_0^{z_n}\cdots \int_0^{z_1}\,\, - \,\, \sum_{i=2}^n\int_0^{z_n}\cdots \int_0^{z_{i-1}}\int_0^{z_1}\int_0^{z_{i+1}}\cdots \int_0^{z_2} \\
&=\sum_{i=1}^n(-1)^{i-1}\int_0^{z_n}\cdots \int_0^{z_{i-1}} \int_0^{z_{i+1}} \cdots \int_0^{z_1}.
\end{align*}
The proof follows. $\Box$

\bigskip
\noindent
{\bf Acknowledgment:}

\smallskip
\noindent
We wish to thank the referee and Peter Larcombe for insightful comments on an earlier draft.




\begin{thebibliography}{99}
\bibitem{BCN} A. F. Beardon, T. K. Carne, T. W. Ng, The critical values of a polynomial {\em Constr. Approx.}, {\bf18} \#3 343-354 (2002).

\bibitem{ZS} M. Dougherty, J. M. McCammond, Critical points, critical values, and a determinant identity for complex polynomials, {\em Proc. Amer. Math. Soc.}, 
{\bf 148} 5722-5280 (2020).


\end{thebibliography}
\end{document}

