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\title
Jacobian Determinant
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\author
Tewodros Amdeberhan and Doron Zeilberger 
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\affil
Tulane University, Department of Mathematics, New Orleans, LA 70118 \\
Rutgers Univeristy, Department of Mathematics, Piscataway, NJ 08854.
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\noindent
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] reproved this result with a different method.
\bigskip
\noindent
In this short note, we offer a much simpler proof of:
\smallskip
\noindent
\bf Theorem 1. \it We have the explicit determinantal evaluation
$$\det(\pmb{J})=\frac{(-1)^n}{n!}\prod_{j=1}^nz_j\prod_{\Sb j,k=1 \\ j\neq k \endSb}^n(z_k-z_j).$$ \rm

$$ $$ 

$$ $$

\pagebreak

\noindent
\bf Proof of Theorem 1. \rm Let's rewrite the determinant and apply Cauchy's alternant formula (twice) so that
$$\align \det(\pmb{J})&=\det\left(-\int_0^{z_j}\prod_{k=1}^n(w-z_k)\,\frac{dw}{w-z_i}\right)
=\det\left(-\int_0^{z_j}\prod_{k=1}^n(\beta_j-z_k)\,\frac{d\beta_j}{\beta_j-z_i}\right) \\
&=(-1)^n\int_0^{z_1}\cdots\int_0^{z_n}\prod_{j,k}^{1,n}(\beta_j-z_k)\cdot\det\left(\frac1{\beta_j-z_i}\right)_{i,j}^{1,n}\,d\beta_n\cdots d\beta_1 \\
&=(-1)^n\int_0^{z_1}\cdots\int_0^{z_n}\prod_{j,k}^{1,n}(\beta_j-z_k)\left[\frac{\prod_{i<j}^{1,n}(z_i-z_j)(\beta_j-\beta_i)}{\prod_{i,j}^{1,n}(\beta_j-z_i)}\right]\,d\beta_n\cdots d\beta_1 \\
&=(-1)^n\prod_{i<j}^{1,n}(z_i-z_j)\int_0^{z_1}\cdots\int_0^{z_n}\prod_{i<j}^{1,n}(\beta_j-\beta_i)\,d\beta_n\cdots d\beta_1 \\
&=(-1)^n\prod_{i<j}^{1,n}(z_i-z_j)\int_0^{z_1}\cdots\int_0^{z_n}\det\left(\beta_j^{i-1}\right)_{i,j}^{1,n}\,d\beta_n\cdots d\beta_1 \\
&=(-1)^n\prod_{i<j}^{1,n}(z_i-z_j)\cdot\det\left(\int_0^{z_j}\beta_j^{i-1}d\beta_j\right)_{i,j}^{1,n} \\
&=(-1)^n\prod_{i<j}^{1,n}(z_i-z_j)\cdot\det\left(\frac{z_j^i}i\right)_{i,j}^{1,n} \\
&=\frac{(-1)^n}{n!}\prod_{j=1}^nz_j\cdot\prod_{i<j}^{1,n}(z_i-z_j)\cdot\det\left(z_j^{i-1}\right)_{i,j}^{1,n} \\
&=\frac{(-1)^n}{n!}\prod_{j=1}^nz_j\cdot\prod_{i<j}^{1,n}(z_i-z_j)\cdot\prod_{i<j}(z_j-z_i) \\
&=\frac{(-1)^n}{n!}\prod_{j=1}^nz_j\cdot\prod_{\Sb i,j\\ i\neq j\endSb}^{1,n}(z_i-z_j).
\endalign$$
The proof is complete. $\square$

\smallskip
\noindent
\bf Lemma 2. \it If $Q_{i,j}(b,c)=\int_0^{z_{j+c}}w^{i-1+b}\prod_{k=1}^n(w-z_{k+c})^{a_{k+c}-1}\,dw$, then the associated Jacobian has the determinant evaluation
$$\det\left(Q_{i,j}(b,c)\right)_{i,j}^{1,n}
=(-1)^{\epsilon}\frac{b!\prod_{i=1}^n(a_{i+c}-1)!}{(b+\sum_{i=1}^na_{i+c})!}\prod_{j=1}^nz_{j+c}^{b+a_{j+c}}\prod_{i<j}^n(z_{i+c}-z_{j+c})\prod_{\Sb j,k \\ j\neq k\endSb}^{1,n}(z_{k+c}-z_{j+c})^{a_{j+c}-1};$$ 
where $\epsilon$ will be given soon.
\smallskip
\noindent
\bf Proof. \rm Denote $M_n(b,c)=\det(Q_{i,j}(b,c))_{i,j}^{1,n}$. Dodgson's condensation validates 
$$M_{n+1}(b,c)=\frac{M_n(b,c)\cdot M_n(b+1,c+1)-M_n(b+1,c)\cdot M_n(b,c+1)}{M_{n-1}(b+1,c+1)}.$$
So, we need only justify that the expression on the right-hand side of the Lemma does the same. However, this is routine. $\square$.

\pagebreak

\bigskip
\noindent
There is some generalization to Theorem 1 that appeared in [2], which is stated as:
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\noindent
\bf Theorem 3. \it If $a_1,\dots,a_n$ are positive integers and $p(z)=\int_0^z(w-z_1)^{a_1}\cdots(w-z_n)^{a_n}\,dw$ then the corresponding Jacobian has the determinant
$$\det(\pmb{J})=\frac1{\binom{a_1+\cdots+a_n}{a_1,\dots,a_n}}\prod_{j=1}^n(-z_j)^{a_j}\prod_{\Sb j,k \\ j\neq k\endSb}^{1,n}(z_k-z_j)^{a_j}.$$ \rm

\smallskip
\noindent
\bf Proof. \rm Let's rewrite the determinant and apply Cauchy's alternant formula so that
$$\align \det(\pmb{J})&=\det\left(-a_j\int_0^{z_j}\prod_{k=1}^n(w-z_k)^{a_j}\,\frac{dw}{w-z_i}\right)
=\det\left(-a_j\int_0^{z_j}\prod_{k=1}^n(\beta_j-z_k)^{a_j}\,\frac{d\beta_j}{\beta_j-z_i}\right) \\
&=\prod_{j=1}^n(-a_j)\int_0^{z_1}\cdots\int_0^{z_n}\prod_{j,k}^{1,n}(\beta_j-z_k)^{a_j}\cdot\det\left(\frac1{\beta_j-z_i}\right)_{i,j}^{1,n}\,d\beta_n\cdots d\beta_1 \\
&=\prod_{j=1}^n(-a_j)\int_0^{z_1}\cdots\int_0^{z_n}\prod_{j,k}^{1,n}(\beta_j-z_k)^{a_j}\left[\frac{\prod_{i<j}^{1,n}(z_i-z_j)(\beta_j-\beta_i)}{\prod_{i,j}^{1,n}(\beta_j-z_i)}\right]\,d\beta_n\cdots d\beta_1 \\
&=\prod_{j=1}^n(-a_j)\prod_{i<j}^{1,n}(z_i-z_j)\int_0^{z_1}\cdots\int_0^{z_n}\prod_{j,k}^{1,n}(\beta_j-z_k)^{a_j-1}\prod_{i<j}^{1,n}(\beta_j-\beta_i)\,d\beta_n\cdots d\beta_1 \\
&=\prod_{j=1}^n(-a_j)\prod_{i<j}^{1,n}(z_i-z_j)\int_0^{z_1}\cdots\int_0^{z_n}\prod_{j,k}^{1,n}(\beta_j-z_k)^{a_j-1}
\det\left(\beta_j^{i-1}\right)\,d\beta_n\cdots d\beta_1 \\
&=\prod_{j=1}^n(-a_j)\prod_{i<j}^{1,n}(z_i-z_j)\int_0^{z_1}\cdots\int_0^{z_n}\det\left(\beta_j^{i-1}\prod_{k=1}^n(\beta_j-z_k)^{a_j-1}\right)_{i,j}^{1,n}\,d\beta_n\cdots d\beta_1 \\
&=\prod_{j=1}^n(-a_j)\prod_{i<j}^{1,n}(z_i-z_j)\det\left(\int_0^{z_j}w^{i-1}\prod_{k=1}^n(w-z_k)^{a_j-1}dw\right)_{i,j}^{1,n}.
\endalign$$
Now, apply Lemma 2 with $b=c=0$ and rearrange the terms to complete the proof. $\square$


\Refs
\widestnumber\key{10}

\ref \key 1 \by A F Beardon, T K Carne, T W Ng \paper The critical values of a polynomial \jour Constr. Approx. \vol 18 \#3 \yr 2002 \pages 343-354
\endref

\ref \key 2 \by M Dougherty, J McCammond \paper Critical Points, Critical Values, and a Determinant Identity for Complex Polynomials
\jour \tt{arXiv:1908.10477}
\endref

\endRefs
















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