What is the dimension of the volume


26. A square has about 63.7% of the area of ​​its circumference, but the corresponding inscribed circle about 78.5% of the area of ​​the square. A square therefore hugs its inscribed circle more closely than its periphery. If one compares the volume of a cube with the volume of its surrounding sphere and that of its incugel, one finds this effect even stronger. If, on the other hand, one looks at hypercubes in ever higher dimensions, one finds a dimension in which this effect becomes maximum, then decreases again and finally even reverses. From which dimension does a hypercube cling more closely to its surrounding sphere than to its inc sphere?

The volume V of an n-dimensional sphere with the radius r is (Γ = gamma function): V = πn / 2 · Rn / Γ (n / 2 + 1)

For even-numbered n we have wegen (n) = (n - 1)! the equation: V = πn / 2 · Rn / (n / 2)!

If you define that (1/2)! = Γ (3/2) and uses that Γ (3/2) = √π / 2 and (n + 1)! = (n + 1) n! is
then the above equation and the following equations can also be used for all integer n.


If a hypersphere crosses the sphere (radius r = ru) of an n-dimensional cube with side a, then:

a = 2 * ru / √n

The volume for an n-dimensional cube with side a is: Vw = an = 2n · Run / nn / 2

The in- sphere of a hypercube, in turn, has a radius ri from: ri = ru / √n

The volume of this sphere is thus: Vi = πn / 2 · Run / (nn / 2 · (N / 2)!)


The ratio of the volume ratios is then: (Vw / Vu) / (Vi / Vw) = 22n / (πn · Nn / 2) · (N / 2)!2

If this ratio is less than one, the hyper-cube hugs its sphere more closely; if it is greater than one, it hugs its surrounding sphere more closely. For ru = 1, the following table contains the values ​​for the dimension, for the volumes of incugel, hypercube and umkugel, and for the ratio of the volume ratios mentioned above:

dimensionvolumevolumevolume(Vw / Vu) /
nVi theVw theVu the(Vi / Vw)
IncugelCubeUmkugel
 
  12,0000000002,0000002,0000  1,0000
  21,5707963252,0000003,1416  0,8106
  30,8061330501,5396014,1888  0,7020
  40,3084251371,0000004,9348  0,6570
  50,0941615200,5724335,2638  0,6611
  60,0239245960,2962965,1677  0,7101
  70,0052063960,1410484,7248  0,8088
  80,0009908970,0625004,0587  0,9713
  90,0001675820,0260123,2985  1,2241
100,0000255020,0102402,5502  1,6124
110,0000035270,0038341,8841  2,2120
120,0000004470,0013721,3353  3,1514



According to Stirling's formula, the ratio of the volume ratios already applies to a good approximation for small n:

(Vw / Vu) / (Vi / Vw) = π n (2 √n / π / e)n

You can see from the right column of the table that in two-dimensional to eight-dimensional space the hypercube is closer to its in-sphere. This effect is greatest in four-dimensional space. From the ninth dimension onwards, the hypercube clings more closely to its sphere. The approximation formula shows that this snuggling becomes stronger with increasing n.

What's amazing about this math puzzle? Many intuitively believe that a certain relationship between a body and its incugel and umkugel cannot simply turn into its opposite from a certain dimension. The following considerations show that this intuition is not entirely wrong, even if it lures us on the wrong track in the case of the dice.



In addition to the hypercube, there is also the tetrahedron (simplex) and the octahedron (cross polytope) in all n-dimensional spaces. The same considerations as above can be made for these bodies. For this you need the corresponding formulas again:

Volume of the n-dimensional sphere: Vu = πn / 2 · Run / (n / 2)!

Side a of the n-dimensional tetrahedron: a = 21/2 (N + 1)1/2 · Ru / √n

Volume of the n-dimensional tetrahedron with side a: Vt = (n + 1)1/2 · An / 2n / 2 / n! = (n + 1)(n + 1) / 2 · Run / nn / 2 / n!

Volume of the n-dimensional sphere: Vi = πn / 2 · Run / (nn · (N / 2)!)

The ratio of the volume ratios is then: (Vt / Vu) / (Vi / Vt) = (n + 1)n + 1 / πn · (N / 2)!2 / n!2

This results in the following table (ru = 1):

dimensionvolumevolumevolume(Vt / Vu) /
nVi theVt theVu the(Vi / Vt)
IncugelTetrahedronUmkugel
 
  12,0000000000002,0000000002,00001,0000
  20,7853981625001,2990381063,14160,6839
  30,1551403777780,5132002394,18880,4053
  40,0192765710520,1455773424,93480,2228
  50,0016844124800,0321993795,26380,1169
  60,0001107620190,0058352155,16770,0595
  70,0000057371210,0008955434,72480,0296
  80,0000002419180,0001191824,05870,0145
  90,0000000085140,0000140013,29850,0070
100,0000000002550,0000014722,55020,0033
110,0000000000070,0000001401,88410,0016
120,0000000000000,0000000121,33530,0007



According to Stirling's formula, the ratio of the volume ratios already applies to a good approximation for small n:

(Vt / Vu) / (Vi / Vt) = (n + 1) / 2 (1 + 1 / n)n (E / 2π)n

It can be seen that the ratio becomes smaller and smaller for increasing n. The n-dimensional tetrahedron hugs its inc sphere more and more and does not show the amazing behavior of the hypercube.



Finally, the corresponding formulas for the n-dimensional octahedron with its incugel and umkugel:

Volume of the n-dimensional sphere: Vu = πn / 2 · Run / (n / 2)!

Side a of the n-dimensional octahedron: a = 21/2 · Ru

Volume of the n-dimensional octahedron with side a: VO = 2n / 2 · An / n! = 2n · Run / n!

Volume of the n-dimensional sphere: Vi = πn / 2 · Run / (nn / 2 · (N / 2)!)

The ratio of the volume ratios is then: (VO / Vu) / (Vi / VO) = 22n / πn · Nn / 2 · (N / 2)!2 / n!2

The last table follows from this (ru = 1):

dimensionvolumevolumevolume(VO / Vu) /
nVi theVO theVu the(Vi / VO)
IncugelOctahedronUmkugel
 
  12,0000000002,0000002,00001,0000
  21,5707963252,0000003,14160,8106
  30,8061330501,3333334,18880,5265
  40,3084251370,6666674,93480,2920
  50,0941615200,2666675,26380,1435
  60,0239245960,0888895,16770,0639
  70,0052063960,0253974,72480,0262
  80,0009908970,0063494,05870,0100
  90,0001675820,0014113,29850,0036
100,0000255020,0002822,55020,0012
110,0000035270,0000511,88410,0004
120,0000004470,0000091,33530,0001



According to Stirling's formula, the ratio of the volume ratios already applies to a good approximation for small n:

(VO / Vu) / (Vi / VO) = 1/2 (2e / (π √n))n

It can be seen that here too the ratio for increasing n becomes smaller and smaller. The n-dimensional octahedron hugs its inc sphere more and more and thus does not show the amazing behavior of the hypercube either.



By the way, a hypersphere with the radius r has the surface O = 2 · π in the nth dimensionn / 2 · Rn - 1 / (n / 2 - 1) !.

And a hypercube with side a has surface Ow = 2n an - 1.


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