Re: derivatives and determinant

In article <0hja125fj79lj9phc5va55lugestq93ins@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:
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).

Let g_{ij} = product_{k <> j, k <> i} (x_i - x_k).
It looks to me like Av = 0 where
v_i = g_{ni}/g_{in} for i < n, v_n = -1.
I haven't proven it in general, though I've verified it
symbolically with Maple for n <= 7, and using random values for the
x[i] for n <= 20.

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

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