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Visualizing R^4

  1. Jan 31, 2008 #1
    I'm having trouble visualizing [tex]\ R^{4}[/itex](a domain of reals in four dimensions).

    1. Describe a procedure in given 3 vectors, finds a fourth vector perpendicular to those three. Explain why we can use it in analogous fashion to the normal vector to a plane in [tex]\ R^{3}[/itex].

    Here, I'm thinking taking the normal vector of each vector and then adding the three normals would be sufficient, but I am not sure how this is analogous to the normal vector to a plane in the xyz system.

    2. How many vertices does the unit cube have in [tex]\ R^{n}[/itex] have? What is the furthest distance from the origin that one can be on the unit cube in [tex]\ R^{n}[/itex]? What is the average distance of the vertices to the origin?

    First part I have [tex]\ 2^{n}[/itex], second part a square root of [ a summation from i = 1 to i = n of [tex]\ n^{2}[/itex]]

    But the last part stumps me...average distance? and writing a formula for this is a bit confusing too.
     
    Last edited: Jan 31, 2008
  2. jcsd
  3. Feb 1, 2008 #2
    for the first you should look at the gram-schmidt proces:

    http://en.wikipedia.org/wiki/Gram-Schmidt_process

    for the second, you are right about the [tex]2^n[/tex]. In general the distance from a vertice to the origin is given by

    [tex] \frac{\sqrt{1^2+1^2+\dots+1^2}}{2} = \frac{\sqrt{n}}{2} [/tex]

    so the average must be

    [tex] lim_{n\rightarrow \infty}\frac{1}{n}\sum_i^{n} \frac{\sqrt{i}}{2} = lim_{n\rightarrow \infty} \frac{1}{n}[\frac{\sqrt{1}+\sqrt{2}+\dots+\sqrt{n}}{2}] = lim_{n\rightarrow \infty} \frac{\sqrt{1}}{2n}+\frac{\sqrt{2}}{2n}+\dots +\frac{\sqrt{n}}{2n} = \infty [/tex]

    edited from:

    [tex] lim_{n\rightarrow \infty}\frac{1}{n}\sum_i^{n} \frac{\sqrt{i}}{2} = lim_{n\rightarrow \infty} \frac{1}{n}[\frac{\sqrt{1}+\sqrt{2}+\dots+\sqrt{n}}{2}] = lim_{n\rightarrow \infty} \frac{\sqrt{1}}{2n}+\frac{\sqrt{2}}{2n}+\dots +\frac{\sqrt{n}}{2n} = 0[/tex]

    if the average is over dimension.
     
    Last edited: Feb 1, 2008
  4. Feb 1, 2008 #3
    ups, don't think the limit converge
     
  5. Feb 1, 2008 #4
    then i'm not sure what average you should calculate
     
  6. Feb 1, 2008 #5

    HallsofIvy

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    There is no such thing as "the" normal vector to a vector- any vector in the 3 dimensional "plane" normal to that vector. And even in 3 dimensions, adding normal vectors to two given vectors will not in general give you a vector normal to both. In 3 dimensions you have to take the cross product of the two given vectors. Can you thknk of a procedure analagous to that?

    Assuming by "unit cube" they mean an n-dimensional "cube" with center at the origin and each side of length 1, then you have, NOT a " summation from i = 1 to i = n of n^2" but only a summation from 1 to n of 1 (the square of the side length= 1)- and that is just n. The diagonal distance between opposite vertices is [itex]\sqrt{n}[/itex] and if the origin is at the center of the "cube" the distance you want is [itex]\sqrt{n}/2[/itex] (I see that mrandersdk has already given that). Now I guess you could divide that by n and sum to get an average but that does not converge!
     
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