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Let $x_1,...,x_n $ are distinct real numbers.

Is it a formula for the Vandermonde type determinant $V(x_1, \cdots,x_n)$ whose last column is $x_1^k,\ \cdots,\ x_n^k$, where $k \geq n$, instead of $x_1^{n-1},\ \cdots,\ x_n^{n-1}$?


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All the other columns are the same as usual? – EuYu Nov 16 '12 at 15:01
Yes, the first $n-1$ column are the same. – Richard Nov 16 '12 at 15:01
up vote 8 down vote accepted

Sure, at least you can find such a formula for any fixed $k \geqslant n$. Not sure about a general formula for an unknown $k$ though.

Here is a hint: if all the $x_i$ are distinct, vector $(x_1^k,\ldots,x_n^k)$ is a linear combination of vectors $(x_1^i,\ldots,x_n^i)$, $i=0,\,1,\ldots,n-1$. If $$ (x_1^k,\ldots,x_n^k) = \sum_{i=0}^{n-1} \lambda_i (x_1^i,\ldots,x_n^i), $$ then your determinant is simply equal to $\lambda_{n-1}$ times the usual Vandermonde determinant. So all you need to do is find $\lambda_{n-1}$.

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To continue Dan's answer, we want $$ x_i^k = \sum_{j=0}^{n-1} \lambda_j x_i^j, \qquad 1 \le i \le n $$ That is the polynomial $p(x) := \sum_{j=0}^{n-1} \lambda_j x^j$ interpolates $x^k$ at $x_0, \ldots, x_{n-1}$. Lagrange interpolation gives $$ p(x) = \sum_{j=0}^{n-1} x_j^k \cdot \prod_{\ell \ne j} \frac{x-x_\ell}{x_j - x_\ell} $$ $\lambda_{n-1}$ is the coefficient of $x^{n-1}$, so it is $$ \lambda_{n-1} = \sum_{j=0}^{n-1} x_j^k \prod_{\ell\ne j} \frac 1{x_j - x_\ell}. $$

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Perform Laplace expansion along the last column. As the deletion of the $\ell$-th row and the last column gives a $(n-1)\times(n-1)$ Vandermonde matrix (in the original flavour), we get $$ \sum_{\ell=1}^n (-1)^{\ell+n} x_\ell^k\prod_{i<j\,\textrm{ and }\,i,j\not=\ell}(x_j-x_i). $$

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Actually there is a general formula but not an easy way to describe it.

Let me first "define" the polynomials $f_m(x_1,x_2,\ldots,x_k)$. I will give some examples.

  • $f_0(x_1,x_2,x_3)=1.$
  • $f_1(x_1,x_2,x_3)=x_1+x_2+x_3.$
  • $f_2(x_1,x_2,x_3)=x_1^2+x_2^2+x_3^2+x_1x_2+x_1x_3+x_2x_3.$
  • $f_3(x_1,x_2,x_3)=x_1^3+x_2^3+x_3^3+x_1^2x_2+x_1^2x_3+x_2^2x_1+x_2^2x_3+x_3^2x_1+x_3^2x_2+x_1x_2x_3.$
  • $f_3(x_1,x_2,x_3,x_4)=x_1^3+x_1^2x_2+x_1^2x_3+x_1^2x_4+x_1x_2^2+x_1x_2x_3+x_1x_2x_4+x_1x_3^2+x_1x_3x_4+x_1x_4^2+x_2^3+x_2^2x_3+x_2^2x_4+x_2x_3^2+x_2x_3x_4+x_2x_4^2+x_3^3+x_3^2x_4+x_3x_4^2+x_4^3 .$
  • $\ldots$

Let $V(x_1,x_2,\ldots,x_n,k)$ denote the Vandermonde type determinant whose last column is $x_1^k,x_2^k,\ldots,x_n^k$ where $k\geq n-1$. So $V(x_1,x_2,\ldots,x_n,n-1)$ is the usual Vandermonde. Then $$V(x_1,x_2,\ldots,x_n,k)=V(x_1,x_2,\ldots,x_n,n-1)\cdot f_{k-n+1}(x_1,x_2,\ldots,x_n).$$

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