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Change of variables theorem

  1. Feb 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Let S = {x∈R^n : x_i ≥ 0 for all i, x_1 + 2x_2 + 3x_3 + ... + nx_n ≤ n}. Find the n-dimensional volume of S.

    2. Relevant equations

    I'm 95% sure that I'm supposed to use the change of variables theorem here.

    3. The attempt at a solution

    So far, I have calculated the values for n from 1 to 6 using iterated integrals (calculated with mathematica). They are 1, 1, 3/4, 4/9, 125/576, and 9/100, respectively. I was hoping to get an easy pattern that I could use to lead me toward the general solution, but it looks like that won't be the case.

    This is only the second COVT problem that we've been assigned, so I'm not quite comfortable with the theorem yet. I don't really have any idea what sort of change of variables would make this easy to deal with. Does anyone have any idea?

    Edit: I just figured out the pattern based on some intuition and a lucky guess. The volume of S is (or seems to be, at least) n^n/(n!)^2. I have no idea how I could possibly get this by integration, however.
    Last edited: Feb 19, 2009
  2. jcsd
  3. Feb 19, 2009 #2


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    Hi Frillth! :smile:

    Hint: what is the volume of {x∈R^n : x_i ≥ 0 for all i, x_1 + x_2 + x_3 + ... + x_n ≤ n}? :wink:
  4. Feb 19, 2009 #3
    The volume of that should be (I think):
    Is there a simple way to evaluate this, or am I doing this completely wrong?

    By the way, sorry if that's hard to read. I don't know latex, so I just used Mathematica.

    Edit: Ah, I just realized how to compute that integral. It comes out to n^n/n!, right?
    Edit 2: So for my change of variables to, say, u, I want u_1 = x_1, u_2 = 2x_2, u_3 = 3x_3 ... u_n = nx_n. Then the determinant of the derivative matrix for this function would be 1/n!. So I take the integral of the surface described by u_1 + u_2 + ... u_n = n, which I already showed was n^n/n!, and multiply by 1/n! from the determinant to get volume = n^n/(n!)^2. Is that correct?
    Edit 3: All right, I believe I have a rigorous proof now. Thank you very much for your hint, tiny-tim.
    Last edited: Feb 19, 2009
  5. Feb 19, 2009 #4


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    Hi Frillth! :smile:

    (for LaTeX integrals, type \int_{a}^{b} and for three dots type \cdots :wink:)

    That was fun! :smile:

    Which way did you get the n^n/n!?

    I found two ways (I did it for a "pyramid" of side 1, so I just got 1/n!) …

    one was ordinary integrating, by induction on n,

    and the other was by comparing

    {x∈R^n : x_i ≥ 0 for all i, x_1 + x_2 + x_3 + ... + x_n ≤ 1}


    {x∈R^n : x_i ≥ 0 for all i, x_1 + x_2 + x_3 + ... + x_n ≤ 1

    and x_1 ≤ x_2 ≤ x_3 ≤ ... ≤ x_n} :wink:

    As a result of more geometric juggling, I also managed to prove:

    [tex]\sum_{0\,\leq\,m\,< n/2}\,(-1)^m(n - 2m)^n\,^nC_m\ =\ 2^{n-1}\,n![/tex]

    … see https://www.physicsforums.com/showthread.php?p=2083077". :smile:
    Last edited by a moderator: Apr 24, 2017
  6. Feb 19, 2009 #5
    I did it by induction on n. How would you approach it with your second method?
  7. Feb 20, 2009 #6


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    Hi Frillth! :smile:
    waaah! :cry: i've just spent ages trying to re-create it because i couldn't read my own handwriting! :redface:

    that second set should have been …

    {y∈R^n : 0≤ y_1 ≤ y_2 ≤ y_3 ≤ ... ≤ y_n ≤ 1}

    with y_m = x_i + … + x_m :wink:

    then the Jacobian is 1, and the original integral equals

    [tex]\int_0^1\int_0^{y_n}\cdots\int_0^{y_2}dy_n\cdots dy_1[/tex]

    = 1/n! :rolleyes:

    neat, huh? :wink:

    btw, this technique, of ordering the variables, is used a lot in quantum field theory, where each y is a time, and the integral is called a time-ordered integral

    now it's your turn …

    i've only just noticed that there's a direct proof which is (obviously :rolleyes:) what they intended you to use …

    can you see a similar method, using a change of variables with a Jacobian of n!, which directly solves for the original {x∈R^n : x_i ≥ 0 for all i, x_1 + 2x_2 + 3x_3 + ... + nx_n ≤ n}? :smile:
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