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Thread: Cross product vs Tensor product

  1. #21
    Moderator grapes's Avatar
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by BuleriaChk View Post
    The subject got me to thinking this morning, however.

    The reason Fermat's expression can be related to the Binomial Theorem is becauset the integers a and b are independent. That is, with the exception of Pythagorean triples, is not an integer.
    Pythagorean triples are not exceptions
    That means that with the exception of Pythagorean triples, the coefficients each vector are a ring without division, which is why fractions consisting only of individual number lines as coefficients of vectors are independent.

    E.g. For fractions such as z = y/x, x and y are generally independent (which it is why z is the slope of a straight line in two dimensions), but not fractions such as z = f(x,y)/g(x,y). So they are also independent for , which is why they can form the bases of linear systems consisting of polynomials with constant coefficients.

  2. #22
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by grapes View Post
    Pythagorean triples are not exceptions
    Good for my proof, though, since is not an integer for .

    Think real hard, I mean really, really hard about why.



    (I can't believe the idea that Fermat's theorem hasn't been proved long ago has been a practical joke in math departments for several centuries ..)
    Last edited by BuleriaChk; 12-23-2016 at 11:40 AM.
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  3. #23
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by BuleriaChk View Post
    correct, my bad. Good for my proof, though, since is not an integer.
    Yw .

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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by BuleriaChk View Post
    Ok, they're abstract algebra. I'm not sure which is worse, number theory or abstract algebra

    Finally, modules do not necessarily have a BASIS. (From the Wiki article) Those that don't are maybe affine vectors as the scalar coefficient of another vector? (I'm not even sure they have a metric, but I'm going to avoid researching it to find out).

    In any case I have zero interest in them personally, since my proof of Fermat's theorem does not depend on them, and plain ol' functions as coefficients of vectors are fine with me.

    "Twas brillig, and the slithy toves,
    Did gyre and gimble in the wabe...

    But did the momraths really outgrabe?..

    Anyway, I am outta here.
    Good luck in finding someone who is interested in that sort of thing....
    (or who knows enough to at least read the Wiki article on it...
    Both topics are equally superior to you.

    You have no interest in anything as you are too stupid to understand anything, including your obsession about dimensions is unfounded.

  5. #25
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by emperorzelos View Post
    Both topics are equally superior to you.

    You have no interest in anything as you are too stupid to understand anything, including your obsession about dimensions is unfounded.
    So "superior" is anything you want to thnk about instead of anything I want to think about..
    I think not... (that is intellectual bullshit)...
    (It is true that I like some people a whole lot better if I don't know them very well.)
    (and you are hardly a shining light either socially or intellectually.)
    ---------------------------------

    That said, the concept of z = x/y where x and y are function variables as coefficients of a tensor is the subject of Einstein's application of differential geometry to linear algebera in the form of tensor analysis, where z = x/t where x and t are the "space-time" variables and and c is an invariant for STR. This means that , where

    ( where for x = r in radial coordinates.

    For STR, and for x=ct (inertial frames).

    -------------------------------

    Note that the area of the unit circle can be thought of as the cross product beteween the quotient of the circumference and the circumference, since

    ,

    which represents the gradient to the unit circle in the plane , and so a third dimension.

    Hold on, more editing coming; this is starting to get interesting. Back in a while.
    -------------------------------
    For GTR, the question arises:

    Suppose

    However, for mass in STR, so

    Suppose where the components of the fraction are now variables... ?

    This brings up some interesting questions, which are too long for this thread..... (But ultimately are related to my proof of Fermat's theorem in terms of the relativistic unit circle).
    Last edited by BuleriaChk; 12-23-2016 at 07:13 PM.
    _______________________________________
    "Flamenco Chuck" Keyser
    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
    Ignore List -The Peanut Gallery.

  6. #26
    Moderator grapes's Avatar
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by BuleriaChk View Post
    So "superior" is anything you want to thnk about instead of anything I want to think about..
    I think not... (that is intellectual bullshit)...
    What they meant was, you haven't engaged those subjects yet, on your intellectual journey
    (It is true that I like some people a whole lot better if I don't know them very well.)
    (and you are hardly a shining light either socially or intellectually.)
    ---------------------------------

    That said, the concept of z = x/y where x and y are function variables as coefficients of a tensor is the subject of Einstein's application of differential geometry to linear algebera in the form of tensor analysis, where z = x/t where x and t are the "space-time" variables and and c is an invariant for STR. This means that , where

    ( where for x = r in radial coordinates.

    For STR, and for x=ct (inertial frames).

    Note that the area A of a circle can be thought of as the cross product beteween the diameter and the circumference, since
    Is D the circumference? Where did the 2 in that "2r" come from?
    where at the point where the diametter intersects the circumference (so that and are connected ( at an "origin", i.e., not affine)).

    For GTR, the question arises:

    Suppose

    However, for mass in STR, so

    Suppose where the components of the fraction are now variables... ?

    This brings up some interesting questions, which are too long for this thread..... (But ultimately are related to my proof of Fermat's theorem in terms of the relativistic unit circle).

  7. #27
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by grapes View Post
    What they meant was, you haven't engaged those subjects yet, on your intellectual journey

    Is D the circumference? Where did the 2 in that "2r" come from?


    It comes from D=2r, but the equation is wrong (tex dyslexia) I'll fix it shortly...

    Thank you .

    This is the correct version, I think (fixed/edited in previous post #25)


    but hold on, more editing coming... (this is getting interesting)

    (The vector is unrelated to the scalar r, which was my point; it could just as easily have been designated

    (Phrase deleted for now, will be back)
    Last edited by BuleriaChk; 12-23-2016 at 07:04 PM.
    _______________________________________
    "Flamenco Chuck" Keyser
    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
    Ignore List -The Peanut Gallery.

  8. #28
    Moderator grapes's Avatar
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by BuleriaChk View Post
    It comes from D=2r, but the equation is wrong (tex dyslexia) I'll fix it shortly...

    Thank you .

    This is the correct version, I think (fixed/edited in previous post)

    The 2 is still extraneous. With it, that expression doesn't equal what is before it, nor does it equal what comes after. Just get rid of the "2"
    but hold on, more editing coming... (this is getting interesting)

    (The vector is unrelated to the scalar r, which was my point; it could just as easily have been designated

    Since it is unrelated (i.e., independent), it is orthogonal to the area A, and thus its magnitude represents the scalar gradient at that point (on the circumference, but since it is a vectors, can be translated to the center of the circle - the concept of "parallel transport").

  9. #29
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by grapes View Post
    The 2 is still extraneous. With it, that expression doesn't equal what is before it, nor does it equal what comes after. Just get rid of the "2"
    Still editing (I think it is fixed for now in post #25, but a lot more coming...

    Will return with 2nd phrase (temporarily deleted) shortly...
    Last edited by BuleriaChk; 12-23-2016 at 07:05 PM.
    _______________________________________
    "Flamenco Chuck" Keyser
    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
    Ignore List -The Peanut Gallery.

  10. #30
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by BuleriaChk View Post
    So "superior" is anything you want to thnk about instead of anything I want to think about..
    I think not... (that is intellectual bullshit)...
    (It is true that I like some people a whole lot better if I don't know them very well.)
    (and you are hardly a shining light either socially or intellectually.)
    ---------------------------------

    That said, the concept of z = x/y where x and y are function variables as coefficients of a tensor is the subject of Einstein's application of differential geometry to linear algebera in the form of tensor analysis, where z = x/t where x and t are the "space-time" variables and and c is an invariant for STR. This means that , where

    ( where for x = r in radial coordinates.

    For STR, and for x=ct (inertial frames).

    Note that the unit area of a circle can be thought of as the cross product beteween the quotient of the circumference and the circumference, since



    Hold on, more editing coming; this is starting to get interesting. Back in a while.


    For GTR, the question arises:

    Suppose

    However, for mass in STR, so

    Suppose where the components of the fraction are now variables... ?

    This brings up some interesting questions, which are too long for this thread..... (But ultimately are related to my proof of Fermat's theorem in terms of the relativistic unit circle).
    Anything you think is bullshit, as we have seen here.

    I am considerbly superior intellectully to you and even after being corrected on your abuse of notation you are still using wrong, it is laughable.

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