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March 11th, 2017, 04:52 PM  #1 
Member Joined: Mar 2017 From: venezuela Posts: 34 Thanks: 3  Prove this New Determinant Property (New HighOrder 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:% \[ A=% \begin{pmatrix} x_{1} & x^{n/k} & x^{(n1)/k} & x^{(n2)/k} & x^{(n3)/k} & \cdots & {x}^{3/k} & {x}^{2/k} & {x}^{1/k}\\ 0 & {x}_{2} & x^{n/k} & x^{(n1)/k} & x^{(n2)/k} & \cdots & {x}^{4/k} & {x}^{3/k} & x^{2/k}\\ 0 & 0 & x_{3} & x^{n/k} & x^{(n1)/k} & \cdots & {x}^{5/k} & {x}^{4/k} & {x}^{3/k}\\ 0 & 0 & 0 & {x}_{4} & x^{n/k} & \cdots & {x}^{6/k} & {x}^{5/k} & {x}^{4/k}\\ 0 & 0 & 0 & 0 & {x}_{5} & \cdots & {x}^{7/k} & {x}^{6/k} & {x}^{5/k}\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & 0 & \cdots & {x}_{n1} & x^{n/k} & {x}^{(n1)/k}\\ 0 & 0 & 0 & 0 & 0 & \cdots & 0 & {x}_{n} & {x}^{n/k}\\ x^{n/k} & x^{(n1)/k} & x^{(n2)/k} & x^{(n3)/k} & x^{(n4)/k} & \cdots & {x}^{2/k} & {x}^{1/k} & 1 \end{pmatrix} \] $\ $ \[ \left\vert A\right\vert =\ \ (1)^{n}\ \left( xx_{1}\right) \left( xx_{2}\right) \left( xx_{3}\right) \cdots\left( xx_{n}\right) \] Best regards, www.domingogomezmorin.wordpress.com . 
March 12th, 2017, 04:52 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,170 Thanks: 867 
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 = (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: 34 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 highorder rootapproximating algorithms based on the Arithmonic Mean (Particular case of the Generalized Mediant). Domingo Gomez Morin 
March 14th, 2017, 08:14 AM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 1,944 Thanks: 1011  Quote:
One post is more than enough.  

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