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March 11th, 2017, 05:52 PM   #1
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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:%

\[
A=%
\begin{pmatrix}
x_{1} & x^{n/k} & x^{(n-1)/k} & x^{(n-2)/k} & x^{(n-3)/k} & \cdots & {x}^{3/k}
& {x}^{2/k} & {x}^{1/k}\\
0 & {x}_{2} & x^{n/k} & x^{(n-1)/k} & x^{(n-2)/k} & \cdots & {x}^{4/k} &
{x}^{3/k} & x^{2/k}\\
0 & 0 & x_{3} & x^{n/k} & x^{(n-1)/k} & \cdots & {x}^{5/k} & {x}^{4/k} &
{x}^{3/k}\\
0 & 0 & 0 & {x}_{4} & x^{n/k} & \cdots & {x}^{6/k} & {x}^{5/k} & {x}^{4/k}\\
0 & 0 & 0 & 0 & {x}_{5} & \cdots & {x}^{7/k} & {x}^{6/k} & {x}^{5/k}\\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots &
\vdots\\
0 & 0 & 0 & 0 & 0 & \cdots & {x}_{n-1} & x^{n/k} & {x}^{(n-1)/k}\\
0 & 0 & 0 & 0 & 0 & \cdots & 0 & {x}_{n} & {x}^{n/k}\\
x^{n/k} & x^{(n-1)/k} & x^{(n-2)/k} & x^{(n-3)/k} & x^{(n-4)/k} & \cdots &
{x}^{2/k} & {x}^{1/k} & 1
\end{pmatrix}
\]


$\ $

\[
\left\vert A\right\vert =\ \ (-1)^{n}\ \left( x-x_{1}\right) \left(
x-x_{2}\right) \left( x-x_{3}\right) \cdots\left( x-x_{n}\right)
\]



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www.domingogomezmorin.wordpress.com

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March 12th, 2017, 04:52 AM   #2
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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?
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March 14th, 2017, 07:38 AM   #3
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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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March 14th, 2017, 08:14 AM   #4
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Quote:
Originally Posted by arithmo View Post
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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One post is more than enough.
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