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N-dimensional trianglehedron?

  1. Jan 20, 2005 #1

    T@P

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    if you take two unit orthogonal vectors in R2, the triangle they form has area 1/2. take 3 in R3, the volume is 1/6. i claim (and half proved) that the voluarea of an n dimensional bunch of orthognal vectors would give you 1/n!. can anyone prove it fully?

    (go easy on the linear algebra, im infantile when it comes to that)
     
  2. jcsd
  3. Jan 20, 2005 #2

    NateTG

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    I take it you mean the n-volume of the convex hull formed by the origin and the usual unit vectors in [itex]\Re^n[/itex] under the usual metric.

    Let's instead of looking at just the unit length version, look at [itex]v_n(r)[/itex] the 'trianglehedron' with legs of length [itex]r[/itex].

    Now, let's prove by induction that
    [tex]v_n(r)=\frac{1}{n!} r^n[/tex]
    Then it's easy to see that
    [tex]v_1(r)=\frac{1}{n!} r^1[/tex]
    And that for the n+1 case, we can integrate by slicing to get:
    [tex]v_{n+1}(r)=\int_0^rv_n(x)dx[/tex]
    but from induction we have:
    [tex]v_n(x)=\frac{1}{n!}x^n[/tex]
    so
    [tex]v_{n+1}(r)=\int_0^r\frac{1}{n!}x^n dx[/tex]
    which readily works out to
    [tex]v_{n+1}(r)=\frac{1}{n!} \times \frac{1}{n+1} r^{n+1}=\frac{1}{(n+1)!}r^{n+1}[/tex]

    Sepecifically,
    [tex]v_n(1)=\frac{1}{n!}[/tex]
     
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