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Consider following partial Vandermonde type, circulant matrix $\begin{bmatrix} x_1 & x_2 & 0 & \dots & 0 & x_n\\ x_1^2 & x_2^2 & x_3^2 & \dots & 0 & 0\\ \vdots &\vdots &\vdots &\ddots &\vdots &\vdots\\ x_1^n & 0 & 0 & \dots & x_{n-1}^n & x_n^n\\ \end{bmatrix}$.

Is there a closed form expression for this determinant like Vandermonde determinant expression?

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    $\begingroup$ I think your typesetting of the matrix (which seems not to be circulant) is off--it appears you have two more columns than rows. And can you be more clear about which entries are zero? $\endgroup$
    – Kimball
    Commented Jun 7, 2015 at 9:40
  • $\begingroup$ no $x_1$ to $x_n$ variables $1$ to $n$ exponents. $\endgroup$
    – Turbo
    Commented Jun 7, 2015 at 9:41
  • $\begingroup$ Every row/column has $3$ non-zero entries. $\endgroup$
    – Turbo
    Commented Jun 7, 2015 at 9:43
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    $\begingroup$ so it's not circulant, but a termwise product of the vandermonde matrix $V(x_1,\ldots,x_n)$ with the circulant matrix $circ(1,1,0,\ldots,0,1)$. $\endgroup$
    – kodlu
    Commented Jun 7, 2015 at 10:09
  • $\begingroup$ yes thats what I am meaning here. $\endgroup$
    – Turbo
    Commented Jun 7, 2015 at 18:29

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The "closest" form you can expect is the well known determinant formula for tridiagonal matrices, which in your case can be written as $$ \Delta = \det \left[ \begin{pmatrix} -x_n^{-n} & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & -x_1^{n} \end{pmatrix} \mathbf T_n \mathbf T_{n-1} \cdots \mathbf T_2 \mathbf T_1 \right] \,\,\prod_{k=1}^n (-x_k^{k+1})\,, $$ with transfer matrix $$ \mathbf T_k = \begin{pmatrix} -x_k^{-1} & -x_k^{-2} \\ 1 & 0 \end{pmatrix}. $$

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