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)

I would be very grateful if someone could check for mistakes in the changes that I've made although I've looked really carefully and checked results with MATLAB. One can never be too certain--Riddler (talk) 21:55, 11 May 2015 (UTC)


 * Okay, thx -- I'll go over it in due course and give it a rewrite so as to put it back into its original form (as found in Knuth). --prime mover (talk) 22:03, 11 May 2015 (UTC)