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Vectors/Cartesian products

  1. Jul 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Let's say we have a notation:
    gif.gif

    Does it mean that x is an L-tuple? Also, if it does, does it have to have L members?
    If it x isn't an L-tuple, does it mean that x belongs to an L-tuple? That is, x is inside of an L-tuple: (1, x, ... L)? And again, does it mean that that L-tuple has to have exactly L members?


    2. Relevant equations
    gif.gif


    3. The attempt at a solution
    Googling
     
    Last edited: Jul 6, 2012
  2. jcsd
  3. Jul 6, 2012 #2

    micromass

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    Yes.

    Yes.

    For example, typical members of [itex]\mathbb{R}^6[/itex] are x=(1,2,3,6,4,2). An element like (1,2,3) is not an element of [itex]\mathbb{R}^6[/itex].
     
  4. Jul 6, 2012 #3

    tiny-tim

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    Hi Cinitiator! :smile:
    Yes. :smile:
    Yes!!

    A Cartesian product space, of L individual spaces, only has elements that are L-tuples.

    If the individual spaces are all different, that's fairly obvious … it's the same even if the individual spaces are identical. :wink:
     
  5. Jul 6, 2012 #4
    Hi Cinitiator,

    Does it mean that x is an L-tuple?

    No, x [itex]\in[/itex] ℝL+ usually means that x is a column vector with L rows with real entries where the subscript + requires that all its entries are positive.

    http://en.wikipedia.org/wiki/Column_vector

    A tuple is also a kind of array but such that the listed objects are ordered horizontally. It is usually notated like this:
    L-Tupel: (n1,n2,...,nL)

    http://en.wikipedia.org/wiki/Tuple

    Also, if it does, does it have to have L members?

    Yes, exactly L entries (or alternatevely members)
     
    Last edited: Jul 6, 2012
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