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Thread: I suspect this property does not appear in the math literature.

  1. #1
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    I suspect this property does not appear in the math literature.

    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:%

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


    $\ $

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


    This is based on new high-order numerical methods that can be taught to young students and even children


    Best regards,



    .
    Last edited by topsquark; Mar 12th 2017 at 12:22 PM.
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    Re: I suspect this property does not appear in the math literature.

    Domingo Gomez Morin is the name of the author of this post
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    Re: I suspect this property does not appear in the math literature.

    misread matrix.
    Last edited by HallsofIvy; Mar 12th 2017 at 03:35 PM.
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    Re: I suspect this property does not appear in the math literature.

    Quote Originally Posted by HallsofIvy View Post
    misread matrix.
    What does that mean?
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    Forum Admin topsquark's Avatar
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    Re: I suspect this property does not appear in the math literature.

    Quote Originally Posted by arithmo View Post
    What does that mean?
    He's saying he made a mistake.

    -Dan
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    Re: I suspect this property does not appear in the math literature.

    Hey arithmo.

    Even regardless of a mistake or not I'd encourage you to think about how row operations change a determinant and to combine that with the fact that the determinant of a triangular matrix is the product of the diagonals.

    If you can combine that then you can get something a bit more conclusive.
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    Re: I suspect this property does not appear in the math literature.

    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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  8. #8
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    Re: I suspect this property does not appear in the math literature.

    Quote Originally Posted by arithmo View Post
    What does that mean?
    It means that I wrote out an answer but then realized that the given matrix was not exactly what I had thought it was and that my answer did not apply. I would have preferred to remove the post so it did not even appear (and no one would know I had made a mistake!) but was not able to.
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    Re: I suspect this property does not appear in the math literature.

    Quote Originally Posted by HallsofIvy View Post
    It means that I wrote out an answer but then realized that the given matrix was not exactly what I had thought it was and that my answer did not apply. I would have preferred to remove the post so it did not even appear (and no one would know I had made a mistake!) but was not able to.
    No matter your response was an error you pointed out a very important issue:
    ***************
    There's a good reason why that "does not appear in the math literature"- its *not* true! That's an "upper triangular" matrix. If you "expand by the first column", repeatedly, you can show that the determinant of any triangular determinant is just the product of the numbers on the main diagonal. The determinant does not involve "x" at all. it is x_1(x_2)(x_3)\cdot\cdot\\cdot x_n
    ***************


    You said that it does not appear in the literature because it was false.

    So, now that you realized it is true and so simple, and considering your True-False argument for analyzing math objects then:

    Why this matrix appear does not appear in the math literature?

    Notice that the determinant yields the general algebraic equation, and the matrix is based on new extremely simple high-order root-approximating methods that
    also do not appear in the math literature and could be developed by any inexperienced young student, even by children.

    Regards,
    Domingo Gomez Morin
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  10. #10
    Forum Admin topsquark's Avatar
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    Re: I suspect this property does not appear in the math literature.

    Quote Originally Posted by arithmo View Post
    No matter your response was an error you pointed out a very important issue:
    ***************
    There's a good reason why that "does not appear in the math literature"- its *not* true! That's an "upper triangular" matrix. If you "expand by the first column", repeatedly, you can show that the determinant of any triangular determinant is just the product of the numbers on the main diagonal. The determinant does not involve "x" at all. it is x_1(x_2)(x_3)\cdot\cdot\\cdot x_n
    ***************


    You said that it does not appear in the literature because it was false.

    So, now that you realized it is true and so simple, and considering your True-False argument for analyzing math objects then:

    Why this matrix appear does not appear in the math literature?

    Notice that the determinant yields the general algebraic equation, and the matrix is based on new extremely simple high-order root-approximating methods that
    also do not appear in the math literature and could be developed by any inexperienced young student, even by children.

    Regards,
    Domingo Gomez Morin
    First: Please educate us on how "children" would even understand the meaning of a matrix, much less how they would understand a determinant. Just what age range are you calling children?

    Also, the determinant
    \left | \begin{matrix} x_1 & x_1 & x_1 ^2 \\ x_2 ^2 & x_2 ^2 & x_2 ^3 \\ x_3 ^3 & x_3 ^3 & x_3 ^4 \end{matrix} \right |
    does not appear in the literature. I can think of no use for such a matrix so who cares? What use does your determinant have?

    -Dan
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    Re: I suspect this property does not appear in the math literature.

    Quote Originally Posted by topsquark View Post
    First: Please educate us on how "children" would even understand the meaning of a matrix, much less how they would understand a determinant. Just what age range are you calling children?

    Also, the determinant
    \left | \begin{matrix} x_1 & x_1 & x_1 ^2 \\ x_2 ^2 & x_2 ^2 & x_2 ^3 \\ x_3 ^3 & x_3 ^3 & x_3 ^4 \end{matrix} \right |
    does not appear in the literature. I can think of no use for such a matrix so who cares? What use does your determinant have?

    -Dan
    If you don't know anything on this matter then that's your problem not mine.
    I clearly said:
    "Notice that the determinant yields the general algebraic equation, and the matrix is based on new extremely simple high-order root-approximating methods that
    also do not appear in the math literature and could be developed by any inexperienced young student, even by children."

    Thus, I never said as you falsely claimed that the matrix and the determinant could be developed by children, but the high-order extremely simple root-solving methods that lead to that matrix.


    Actually, one never want to be rude, but frankly things like this from "math-helpers" deserves some response:
    ***************
    There's a good reason why that "does not appear in the math literature"- its *not* true! That's an "upper triangular" matrix. If you "expand by the first column", repeatedly, you can show that the determinant of any triangular determinant is just the product of the numbers on the main diagonal. The determinant does not involve "x" at all. it is x_1(x_2)(x_3)\cdot\cdot\\cdot x_n
    ***************


    and your response is even worse:
    " I can think of no use for such a matrix so who cares?"

    It is clear you crankly think of yourself as the Center, the wisdom.
    Who cares about any single bit you could say to anyone.
    You are far beyond any help from anyone, find a doctor.
    Last edited by arithmo; Mar 15th 2017 at 07:58 PM.
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    Re: I suspect this property does not appear in the math literature.

    I think we have a verified troll sighting.
    Last edited by JeffM; Mar 15th 2017 at 08:01 PM.
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  13. #13
    Forum Admin topsquark's Avatar
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    Re: I suspect this property does not appear in the math literature.

    Quote Originally Posted by arithmo View Post
    It is clear you crankly think of yourself as the Center, the wisdom.
    Of course I do. I'm a Physicist.

    Quote Originally Posted by arithmo View Post
    Who cares about any single bit you could say to anyone.
    This "question" is on an open Forum. If you don't want discussion or feedback of some kind you should post it elsewhere.

    Quote Originally Posted by arithmo View Post
    You are far beyond any help from anyone, find a doctor.
    I already have one and I have the pills to prove it.

    Seeing as you are apparently not seeking to answer any questions we may have on this topic, I think this thread has served your purpose...to simply post the determinant and not explain anything. This being the case I am closing the thread.

    -Dan
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