Talk:Inverse of Vandermonde Matrix

Does anyone know if Knuth give details of the proof? Then line:

Identifying the $k^\text{th}$ order coefficient in these two polynomials yields:
 * $b_{kj} = (-1)^{n-k-1} \left({\dfrac{\displaystyle \sum_{\stackrel{0 \mathop \le m_0 \mathop < \ldots \mathop < m_{n-k} \mathop \le n} {m_0, \ldots, m_{n-k} \mathop \ne j} } x_{m_0} \cdots x_{m_{n-k}} } {\displaystyle \prod_{\stackrel {0 \mathop \le m \mathop \le n} {m \mathop \ne j} } \left({x_j - x_m}\right)}}\right)$

isn't correct (I think); the indices are one out of sync, and I can't think of a tidy way of correcting it without defining the empty sum to be $1$, not $0$. --Linus44 (talk) 23:20, 12 October 2012 (UTC)


 * Actually it's fine. I'm going to stop doing maths at half 1 now. --Linus44 (talk) 23:22, 12 October 2012 (UTC)

The matrix $V_n$ is $n \times n$ and the matrix $W_n$ is $(n+1) \times (n+1)$... You can't write $V_n = diag(x_i) W_n$ with $i = 1, ..., n $. --Ostrogradsky (talk) 14:16, 12 March 2015 (UTC)


 * Good call. I'll put a maintenance tag on it as that needs to be addressed. Thanks. --prime mover (talk) 17:04, 12 March 2015 (UTC)