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Hypersphere to Hypercube volume ratio

  1. Jan 14, 2008 #1
    I have a problem and i would like some expert advise. I want proof is to whether i am correct or wrong.....not just 'You're wrong'. Please either paste your proof or send me an email...at edited to remove e-mail address

    i will explain the problem....i wanted to see if there was a relationship between the area of a square with the same dimensions as a circles diametre. and the area of a circle. for this example lets say that r = 2. then the square would be (2r)^2 and the circle would be PI*r^2.

    i divided the area of the square by the area of the circle and i recieved 1.273239545. It is always the same no matter what the radius....i then multiplied this number by PI. It returned 4. No excess decimals just 4. This intrigued me so i then did the same for 3D sphere and cube with same diametre as measurements. This returned 1.909859317. once again i multiplied it against Pi and it returned 6 without any decimals. This intrigued me once again. So i decided to do it for a 1D version. It gave me 0.6366197724 which when multiplied against Pi gave me 2. so i saw a pattern. for each dimension increased it also increased by 2. So i guessed that it would be 8 for a 4D version. 8/Pi = 2.546479089. When i knew the number i decided that i would use some basic algebra and figure out the equation for a 4d hypersphere. I didn't know the equation so i had to figure it out. I then continued to do step by step and i came up with an equation which recently i noticed that it was different then the one that is well known. But the one that is publicized doesnt work with the relationship of the sphere and cube. I tested the way and it managed to get the correct equations for 1D, 2D, and 3D so i think that it works for 4D. I want someone or some people to help me find any errors in the forumal.....please state if it doesnt work but once again please have proof. here is a file with the formula in it attached to this article. i would just ;like to know if i am wrong...plus if anyone knows where i went wrong or if it is just a weird coincidence but nothing bigger or better.

    thanks:

    LOck-doWN(Kyle Derby MacInnis)
     

    Attached Files:

    Last edited by a moderator: Jan 15, 2008
  2. jcsd
  3. Jan 14, 2008 #2
    If the .doc files doesnt work i will send you the original file....i used open office and i dont know if it exported properly....so if it doesnt work just ask me for it and i will give you the .odf file
     
  4. Jan 14, 2008 #3
    The attachment is "pending approval", but while it comes out, I have a question: how did you figured the number 2/Pi for the 1-D case? It looks to me that, in 1-D, the 'volumes' are line segments; and both the 'square' and 'circle' would coincide, both being the locus of all points not farther than r from the center. Thus the 1-D ratio should have been 1. With no Pi factor. What was your reasoning for the 1-D case?

    P.S.: On this page,
    http://mathworld.wolfram.com/Ball.html
    the authors compare the volumes of hyper-balls to hyper-cubes; there is a table in mid-page doing so. On that table, the radius is fixed as =1 (no problem, if you are interested in the ratio), and the center column gives a ratio, using a inverted definition with respect to yours (i.e. ball/cube instead of cube/ball, showing Pi/4 or Pi/6 instead of 4/Pi or 6/Pi).
     
    Last edited: Jan 14, 2008
  5. Jan 14, 2008 #4
    for the 1-d it would be just 2r or d for the length of the 1d square and for the circle it would be just Pi*r should you turn a circle on its side....and i will check on the Pi/4 thing thanks for t he link.
     
  6. Jan 14, 2008 #5
    i noticed why....for the 4/Pi Pi/4 accurence is because on the site they are dividing a sphere by a cube....i was dividing a cube by a sphere that is the difference.

    also if it is a serious mistake on any of my stuff i am sorry....i have yet to take calculus....i am only 15.
     
    Last edited: Jan 14, 2008
  7. Jan 16, 2008 #6
    This is not surprising.

    The area of your square is (2r)^2 = 4r^2 no matter what radius you choose. The area of your cricle is pi*r^2.

    When you divide the two numbers you are dividing (4r^2) / (pi*r^2) The r^2 cancels out and you are left with 4/pi... With 3-d you are doing the same thing, dividing the formulas of volume of a cube and volume of a sphere.

    Vcuber = (2r)^3 = 8r^3
    Vsphere = (4/3)pi*r^3

    when you divide the two you the r^3 cancel out and you get 8/((4/3)pi) = 6/pi

    it appears what you are looking for is that Vcube(n)/VSphere (n), call it R(n), where the side of your cube is 2* radius of a sphere(or the equivalent of a cube or sphere in n dimension), and n is the dimension. Your conjecture is that R(n) = 2n/pi. This is not true, and the counter example is d4. where the volumne of the 4-d sphere is calculated by .5*pi^2*r^4 and the cube or Tesseract would have "volume" (2r)^4 = 16r^4.

    In this case, when you divide you would be left with 32/p^2. So when you multiply by pi you would not get 8/pi, you would get 32/pi^2.
     
    Last edited: Jan 16, 2008
  8. Jan 16, 2008 #7

    HallsofIvy

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    This gives formulas for the n-volume of an n-sphere:
    http://en.wikipedia.org/wiki/Hypersphere#n-spherical_volume
    In general, the ratio you are seeking is, for dimension n,
    [tex]\frac{2^n\Gamma (\frac{n}{2}+1)}{\pi^\frac{n}{2}}[/tex]

    That reduces to the values you give: [itex]4/\pi[/itex] for n= 2 and [itex]6/\pi[/itex] for n= 3. It also reduces to the [itex]32/\pi^2[/itex] Diffy gives for n= 4.
    The reason why both n= 2 and n= 3 have only [itex]\pi[/itex] while the higher dimensions have higher powers of [itex]\pi[/itex] is that "gamma function" in the denominator. For n= 3, that will give a [itex]\sqrt{\pi}[/itex] in the denominator which cancels part of the [itex]\pi^{3/2}[/itex] in the numerator.
     
  9. Jan 16, 2008 #8
    Lock-down,
    part of your question was, if the formulas in the document are wrong, please explain why. I gave a look, but it is hard to point a reason as for why they are wrong, when in the document there was given no reason as for why they are right. In the document, the volume of a 4-D sphere is given as 2 * Pi * r^4, with no explanation. Unfortunately, you'll need to wait until you learn calculus, in order to be able to deduce the right formula yourself.

    In the meantime, if you can write computer programs, there is an experiment that you can try. (This experiment proves nothing, but it can serve as an illustration.) I will describe it for the 2-D case, but it is easy to do the same for any dimension (although, the higher the dimension, the longer the calculation time will be).

    For 2-D, suppose you pick 2 random numbers between 0 and 1 (such as X1=0.2548623537 and X2=0.8776532642). These two numbers represent the coordinates of a point inside the 2-D "cube" (square), which has a side of length 1 and is centered at coordinates (0.5, 0.5). Now suppose you also have a 2-D sphere (a circle) centered at (0.5, 0.5), with radius 0.5. Your random point will be inside the sphere if X1^2 + X2^2 <= 1, and outside the sphere if > 1.

    So generate a million random points, and count how many fall inside the sphere. The ratio (total_points/points_inside) gives you an approximation of your magic number. You can try for higher dimensions and see if it goes close or far from your predictions.
     
  10. Jan 17, 2008 #9
    thanks i will do that dodo and i will check to see if it is near my numbers.
     
  11. Jan 17, 2008 #10
    Sorry, there is a mistake here - the proper test is
    (X1 - 0.5)^2 + (X1 - 0.5)^2 <= 1 ... or > 1,
    since the 'sphere' is centered at (0.5, 0.5).
     
  12. Jan 17, 2008 #11

    CRGreathouse

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    I remember how much fun it was for me to derive that equation for the first time. It was hard! I'm not very geometrically inclined.
     
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