Wikipedia:Reference desk/Archives/Mathematics/2013 July 2

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July 2

From Unit sphere

The surface areas and the volumes for some values of $n$ are as follows:

$n$	$A_{n}$ (surface area)		$V_{n}$ (volume)
0	$0(1/0!)\pi ^{0}$	0.000	$(1/0!)\pi ^{0}$	1.000
1	$1(2^{1}/1!!)\pi ^{0}$	2.000	$(2^{1}/1!!)\pi ^{0}$	2.000
2	$2(1/1!)\pi ^{1}=2\pi$	6.283	$(1/1!)\pi ^{1}=\pi$	3.141
3	$3(2^{2}/3!!)\pi ^{1}=4\pi$	12.57	$(2^{2}/3!!)\pi ^{1}=(4/3)\pi$	4.189
4	$4(1/2!)\pi ^{2}=2\pi ^{2}$	19.74	$(1/2!)\pi ^{2}=(1/2)\pi ^{2}$	4.935
5	$5(2^{3}/5!!)\pi ^{2}=(8/3)\pi ^{2}$	26.32	$(2^{3}/5!!)\pi ^{2}=(8/15)\pi ^{2}$	5.264
6	$6(1/3!)\pi ^{3}=\pi ^{3}$	31.01	$(1/3!)\pi ^{3}=(1/6)\pi ^{3}$	5.168
7	$7(2^{4}/7!!)\pi ^{3}=(16/15)\pi ^{3}$	33.07	$(2^{4}/7!!)\pi ^{3}=(16/105)\pi ^{3}$	4.725
8	$8(1/4!)\pi ^{4}=(1/3)\pi ^{4}$	32.47	$(1/4!)\pi ^{4}=(1/24)\pi ^{4}$	4.059
9	$9(2^{5}/9!!)\pi ^{4}=(32/105)\pi ^{4}$	29.69	$(2^{5}/9!!)\pi ^{4}=(32/945)\pi ^{4}$	3.299
10	$10(1/5!)\pi ^{5}=(1/12)\pi ^{5}$	25.50	$(1/5!)\pi ^{5}=(1/120)\pi ^{5}$	2.550

where the decimal expanded values for n ≥ 2 are rounded to the displayed precision.

Why?--Gilderien Chat|List of good deeds 22:09, 2 July 2013 (UTC)

This is explained at length in the article n-sphere. Sławomir Biały (talk) 22:17, 2 July 2013 (UTC)

(edit conflict)Just to clarify, I would expect the n-dimensional hypervolumes to increase without bound as n→∞ but the table implies that does not happen and the hypervolumes decrease after the fifth dimension.--Gilderien Chat|List of good deeds 22:18, 2 July 2013 (UTC)

Put simply, every two steps the volume increases by $2\pi /n$. When n>6.28..., this quantity is less than one (therefore V(7)<V(5)). As for a simple, understandable geometric explaination for this, my n-dimensional thinking isn't that good! MChesterMC (talk) 08:33, 3 July 2013 (UTC)

As the dimension goes up you have more corners on a hypercube so more of the volume is there rather than in the hypersphere around the centre. Well in fact the hypersphere only touches one point on each outside face so even the space between the corners isn't in the hypersphere. Dmcq (talk) 10:02, 3 July 2013 (UTC)

...and not only does the number of corners increase as the dimension increases (2ⁿ corners in n dimensions) but they also get further away from the centre, as the longest diagonal of an n dimensional unit hypercube is ${\sqrt {n}}$ units long. So the proportion of a unit hypercube that lies within its inscribed hypersphere decreases with increasing dimension because there are more and bigger corners that lie outside of the hypersphere. Gandalf61 (talk) 10:30, 3 July 2013 (UTC)

Remember, however, that the unit ball has a unit radius, not the diameter, so your hypercubes' volume grows exponentially with the number of dimensions (a hypercube described on the n–dimensional ball has volume 2ⁿ), and even so the hypervolume of the ball decreases. --CiaPan (talk) 10:37, 3 July 2013 (UTC)

No, the expectation is wrong; I even saw the corresponding graph at some math forum. Actually, whether “volumes” (for different n) decrease or increase depends on the system of measurement. Were the standard simplex the unit of n-volume instead of hypercube, they would increase. Incnis Mrsi (talk) 11:09, 3 July 2013 (UTC)