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 $x_1^{n-1},\ \cdots,\ x_n^{n-1}$?
Thanks
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 $x_1^{n-1},\ \cdots,\ x_n^{n-1}$?
Thanks
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 $\lambda_{n-1}$.
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}. $
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
Actually there is a general formula but not an easy way to describe it.
Let me first "define" the polynomials $f_m(x_1,x_2,\ldots,x_k)$ through some examples.
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 $V(x_1,x_2,\ldots,x_n,k)=V(x_1,x_2,\ldots,x_n,n-1)\cdot f_{k-n+1}(x_1,x_2,\ldots,x_n).$