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Vectors and derivasion

  1. Jan 30, 2004 #1
    What is a vector?
    What is derivation?

    I have allready checked on mathworld.wolfram.com, but there were too many foreign words I could not find translations/clarifications for.

    Please help!
     
  2. jcsd
  3. Jan 30, 2004 #2

    matt grime

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    Please could you provide a translation of your sig to English. At first glance I found it slightly offensive, because I cannot understand all of the second line.
     
  4. Jan 30, 2004 #3
    vestu heil ok sæl is an Old Norse solemn greeting. It is best translated as "Good health and soul/life". Was there anything else?
     
  5. Jan 31, 2004 #4
    Will no one answer my questions?
     
  6. Jan 31, 2004 #5

    selfAdjoint

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    A vector is a member of a vector space.

    A vector space is a set of 'things' (vectors) which obey the linear laws of vector addition: for every pair of vectors u and v there is a vector u+v such that v+u = u+v and if a is a constant then a(u+v) = au+av, and if a and b are constants then (a+b)u = au + bu. The constants are from some algebraic structure, often a field, and the vector space is said to be "over" that field. In many physics applications the field is R the reals or C the complex numbers. There is also the inner product <,> of vectors. This maps the vector space into the field, thus <u,v> = c. If it happens that <u,v> = 0 then the vectors are said to be orthogonal.

    Derivation.

    Let U be an open set in some Euclidean space. The symbol [tex]C^{\infty}(U)[/tex] means the set of diffeomorphisms of the set U onto itself. See my post What is a Diffeomorphism on the Strings and LQG forum. So now you have this set of functions. And now imagine a function from this set [tex]C^{\infty}(U)[/tex] to the real numbers, call the function Z. [tex]Z:C^{\infty}(U) \rightarrow \mathbf {R}[/tex].

    The function Z will be called a derivation if
    For all [tex]f,g \in C^{\infty}(U)[/tex] and real number a and b,
    [tex]Z(af+bg) = aZ(f)+bZ(g)[/tex], and
    [tex]Z(f \cdot g) = f \cdot Zg + Zf \cdot g[/tex],
    where the product [tex]f \cdot Zg[/tex] is defined to be
    [tex](f \cdot Zg)(x) = f(x)Zg(x)[/tex]

    So you see this function Z acts like a derivative in the way it acts on sums and products.

    <edited to fix incorrect range for Z>
     
    Last edited: Jan 31, 2004
  7. Jan 31, 2004 #6

    matt grime

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    So it bears no relation to the first line then? Well, seeing as aryan is merely an adjective pertaining to set of indo-european langauges, I really can't see how there is any other way to read that (the first line) other than as a standard, ignorant neo-facist slogan. I wouldn't be surprised at a lack of response. 'White pride' would be considered potentially, and perhaps unfairly (eg black pride and gay pride), insulting, because of modern PC attitudes. However including the inaccurate adjective implies that you are are a neo-nazi, at least in this country. As someone who lists an interest in languages, surely you are aware of the misuse of the word 'aryan'?That is why I didn't bother answering your question. I may be being unfair, of course.
     
    Last edited: Jan 31, 2004
  8. Feb 1, 2004 #7
    Well noticed matt grime.

    I am not a neo-nazi. I wanted to see if there was any reactions to this slogan. I love Jews, I love Afro-Americans and all coloured people. I will remove the slogan but keep the Old Norse greeting as I love Old Norse and Vikings!

    And thank you selfAdjoint!
     
  9. Feb 2, 2004 #8

    uart

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    I personally find nothing at all offensive in the term "White Pride", I say go for it. I just hate the whole political correctness straight-jacket and like to be politically incorrect as often as possible :).

    I agree with Matt however about the word "aryan" being just totally taken over by Neo-Nazi's, almost as some kind of catch-cry. So I'd avoid that one, but have no worries about the rest.
     
  10. Feb 2, 2004 #9

    turin

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    self adjoint,
    I didn't see any mention of a null vector or inverse vector. Did you just feel that it wasn't important to display all of the axioms, or are those two actually not axioms?
     
  11. Feb 2, 2004 #10

    selfAdjoint

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    I wanted to say the vectors form a group under vector addition, but I thought that would make it even harder to understand than it was. I just left them out.
     
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