Re: derivatives and determinant

On Mon, 13 Mar 2006 05:23:52 EST, eugene <jane1806@xxxxxxx> wrote:

It was a typo in my first message
Let f(x)=(x-x_1)(x-x_2)...(x-x_n)where the numbers x_1,x_2,...,x_n-pairwise distinct and a_{ii}=f''(x_i), a_{ij}=(f'(x_i)-f'(x_j))/(x_i-x_j). i \neq j.
1<=i,j<=n.

Now, A mustn't be always diagonal..

And with that correction, det(A)=0 now appears to be true (but I don't
have time to look at it right now).

quasi
.