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 March 11th, 2017, 04:52 PM #1 Member   Joined: Mar 2017 From: venezuela Posts: 36 Thanks: 3 Prove this New Determinant Property (New High-Order numerical methods) I suspect this property does not appear in the math literature, so please kindly prove it: Given the square matrix $A$ with $k=n+1$ rows and columns:% $A=% \begin{pmatrix} x_{1} & x^{n/k} & x^{(n-1)/k} & x^{(n-2)/k} & x^{(n-3)/k} & \cdots & {x}^{3/k} & {x}^{2/k} & {x}^{1/k}\\ 0 & {x}_{2} & x^{n/k} & x^{(n-1)/k} & x^{(n-2)/k} & \cdots & {x}^{4/k} & {x}^{3/k} & x^{2/k}\\ 0 & 0 & x_{3} & x^{n/k} & x^{(n-1)/k} & \cdots & {x}^{5/k} & {x}^{4/k} & {x}^{3/k}\\ 0 & 0 & 0 & {x}_{4} & x^{n/k} & \cdots & {x}^{6/k} & {x}^{5/k} & {x}^{4/k}\\ 0 & 0 & 0 & 0 & {x}_{5} & \cdots & {x}^{7/k} & {x}^{6/k} & {x}^{5/k}\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & 0 & \cdots & {x}_{n-1} & x^{n/k} & {x}^{(n-1)/k}\\ 0 & 0 & 0 & 0 & 0 & \cdots & 0 & {x}_{n} & {x}^{n/k}\\ x^{n/k} & x^{(n-1)/k} & x^{(n-2)/k} & x^{(n-3)/k} & x^{(n-4)/k} & \cdots & {x}^{2/k} & {x}^{1/k} & 1 \end{pmatrix}$ $\$ $\left\vert A\right\vert =\ \ (-1)^{n}\ \left( x-x_{1}\right) \left( x-x_{2}\right) \left( x-x_{3}\right) \cdots\left( x-x_{n}\right)$ Best regards, www.domingogomezmorin.wordpress.com .
 March 12th, 2017, 04:52 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 That looks like a straight forward "expansion by minors". Have you tried proof by induction on n? If n= 1, so k= 2, then this is the determinant $\left|\begin{array}{cc}x_1=&x^{1/2} \\ x^{1/2}=&1\end{array}\right|= x_1- x$= -(x- x_1)[/tex] so the theorem is true for n= 1. Now try doing it for n= 2 so k= 3. Do you see how this determinant is related to the first matrix?
 March 14th, 2017, 07:38 AM #3 Member   Joined: Mar 2017 From: venezuela Posts: 36 Thanks: 3 Many thanks, indeed. I already did a proof, however, I thought this matrix could be of interest to some people and some other ideas could arise from this. This matrix is based on new high-order root-approximating algorithms based on the Arithmonic Mean (Particular case of the Generalized Mediant). Domingo Gomez Morin
March 14th, 2017, 08:14 AM   #4
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 Originally Posted by arithmo Many thanks, indeed. I already did a proof, however, I thought this matrix could be of interest to some people and some other ideas could arise from this. This matrix is based on new high-order root-approximating algorithms based on the Arithmonic Mean (Particular case of the Generalized Mediant). Domingo Gomez Morin
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