Party in the 5th dimension

[marginbuu]

The 5th dimension is where RPG characters store all their inventory.

[vahidmirkhani]

Although our dimension's sphere isn't the largest, I still feel very proud and supportive for our sphere.

[NXaiUL]

That's totally unexpected. My mind is blown.

[Gustav.J]

Sometimes it seems as if mathematicians are just being paid to have fun with numbers.

[tyzonemusic]

Party rock is in the 5-dimensional sphere tonight

[raphaelthorp]

To be more clear about the 0D sphere (formally a 0-ball), and the term “volume” in general: you’re looking at the counting measure of the given dimensional space, not strictly the “3D Volume”, so in a 2D space you’re computing area, and in a 1D space you’re computing length. In a “1-ball” of radius 1, there are 2 units of length (one on each side of the center) so it’s “volume” is 2. In a 0-ball, the counting measure is itself a point, and within any 0-ball, there is only 1 point, so the “volume” is 1.

[compiling]

Since we are talking about hyperspheres of radius 1, they are contained in hypercubes of sidelength 2. So there are 2 competing exponential factors involved with how the size changes as you increase the number of dimensions - the size of the containing hypercube doubles, but you also double the number of corners which greatly reduces the proportion of the hypercube that the hypersphere takes up.

[somewhat]

Wait why? I need an explanation. I can’t comprehend beyond 3 dimensions 😭

[YoungGandalf]

Infinite dimensions. It's kind of hard to wrap my head around that thought.

[infernothegreat]

I’ve been scrolling through these Multi dimensional shorts for half an hour and my brain is both enriched and broken. I think i went through the 4th dimension!

[TechnoSan]

really appreciate your work on studying about higher dimensions
we need to push human limits to more than 3 dimension

[FiberOptikAssassin]

If you ever get into ASMR as a side gig, don't change what you're doing naturally. I don't understand a word you're saying, but it's relaxing.

[MichaelRothwell]

This is a beautiful video, and it's great to see the recursive formula V(n)=2πr²/n V(n-2) that allows one to calculate the volume of a hypersphere of any dimension (please see my proof given as Proof 2 below).

However, when comparing volumes of hyperspheres across different dimensions, I think that it makes more sense to consider the proportion of the volume of a hypercube occupied by the inscribed hypersphere.

Considering that a hypersphere of radius r sits snugly inside a hypercube of side 2r, this proportion is actually the same as the volume of a hypersphere with radius r=½, i.e. the volume of a hypersphere inscribed in a unit hypercube (as krⁿ/(2r)ⁿ=k(½)ⁿ where V(n)=krⁿ).

Taking r=½, then, starting at n=1, we get:

For n=1, V=2r=2×½=1.
For n=2, V=πr²=π×(½)²=π/4≈0.7854.
For n=3, V=4/3 πr³=4/3 π×(½)³=π/6≈0.5236.
For n=4,

It looks like these values are always strictly decreasing, starting from the maximum possible value of 1 at n=1, which is what one would expect (though if you include dimension zero, as in the video, you get V(0)=V(1)=1).

What is easily proved is that the values are decreasing separately for odd dimensions and even dimensions, as, for r=½,
V(n)=2π(¼)/n V(n-2)=π/(2n) V(n-2),
and for n≥2, 2n≥4, so π/(2n)<1.

EDIT
I have added below two proofs:

Proof 1: the hypervolume of n-dimensional hyperspheres of radius ½ is strictly decreasing for n≥1.

Proof 2: of the recursive volume formula for hyperspheres: V(n)=2πr²/n V(n-2), where V(n) is the volume of an n-dimensional hypersphere of radius r.

PROOF 1
Here is a proof that the hypervolume of n-dimensional hyperspheres of radius ½ is strictly decreasing for n≥1.

Using the values V(1)=2r and V(2)=πr² and the recursive rule V(n)=2πr²/n V(n-2), we have

=2ⁿ⁺¹πⁿr²ⁿ⁺¹/(2n+1)!!
where (2n+1)!!=1×3×5×...×(2n+1), the double factorial for positive odd numbers,
and

=(2πr²)ⁿ/(2n)!!
where (2n)!!=2×4×6×...×2n, the double factorial for positive even numbers.

For r=½,

=(π/2)ⁿ/(2n+1)!!
and
V(2n)=(2π(½)²)ⁿ/(2n)!!
=(π/2)ⁿ/(2n)!!

So V(2n+1)/V(2n)=(2n)!!/(2n+1)!!
=(2/3)(4/5)...(2n/(2n+1))<1 (as n≥1)

and V(2n)/V(2n-1)=(π/2)ⁿ/(2n)!! / [(π/2)ⁿ⁻¹/(2n-1)!!]
=(π/2)ⁿ/(2n)!! × (2n-1)!!/(π/2)ⁿ⁻¹
=π/2 (2n-1)!!/(2n)!!

=(π/4)(3/4)...(2n-1)/(2n)<1 (as π<4 and n≥1)

Hence V(2n+1)<V(2n) and V(2n)<V(2n-1) (both for n≥1), so V(n+1)<V(n) for all n≥1.


PROOF 2
Here is a proof of the recursive volume formula V(n)=2πr²/n V(n-2).

Let's start with a derivation of the volume of a sphere.

Suppose a sphere of radius r centred at the origin is split, perpendicular to the x-axis, into thin disc-like slices, each slice having as one surface area (at x-coordinate x) the area of a disc of radius R (where R²=r²-x²) and thickness δx. The volume of each of these slices is approximately πR²δx.

So V(3)≈∑πR²δx
As the thickness δx tends to zero, the error in this approximation tends to zero, so we get
V(3)=∫₋ᵣʳ πR²dx, where R²=r²-x²,
=π∫₋ᵣʳ(r²-x²)dx
=π[r²x-1/3 x³]₋ᵣʳ
=π[(r³-1/3 r³)-(-r³-1/3(-r³))]
=π[2/3 r³-(-2/3r³)]
=4/3 πr³

Note that for each value of x satisfying 0<x≤r, x and -x give the same value of R, allowing us to simplify the calculation as follows:
V(3)=∫₀ʳ 2×πR²dx, where R²=r²-x²,
=2π∫₀ʳ(r²-x²)dx
=2π[r²x-1/3 x³]₀ʳ
=2π[(r³-1/3 r³)-0]
=2π(2/3 r³)
=4/3 πr³

Note also that for 0<x≤r the two points -x and x that give the same value of R are the boundary points of the line segment [-x, x] which can be thought of as the one-dimensional "sphere" of "radius" x centred at the origin. This perspective will allows us to generalise the method to higher dimensions.

Now let's derive the volume of a 4D hypersphere.

Consider a 4D hypersphere of radius r centred at the origin, and label the coordinate axes x, y, z, w.

Consider the 2-dimensional disc formed by the part of the 4D hypersphere in the xy-plane (i.e. where z=w=0). Then, at each point (a, b, 0, 0) on the disc there will be a 2D cross-section of the 4D hypersphere with the points (a, b, z, w), where a and b are fixed, such that z²+w²=r²-a²-b², so this 2D cross-section is a circle of radius R=√(r²-a²-b²), and this radius will be the same for all points (a, b, 0, 0) on the original 2D disc belonging to a circle centre O of radius x (0≤x≤r), say, i.e. such that a²+b²=x², so that R=√(r²-x²).

So the volume of the 4D hypersphere is
V(4)=∫₀ʳ (2πx)(πR²)dx, where R²=r²-x², obtained by integrating the product of the circumference of the circle radius x (2πx) and the area of the corresponding 2D cross-section (πR²), for x from 0 to r.
So V(4)=2π²∫₀ʳ x(r²-x²)dx
=2π²∫₀ʳ (r²x-x³)dx
=2π²[½r²x²-¼x⁴]₀ʳ
=2π²[(½r⁴-¼r⁴)-0]
=2π²(¼r⁴)
=π²r⁴/2

We can derive the volume of a 5D hypersphere in a similar way.

Consider a 5D hypersphere of radius r centred at the origin, and label the coordinate axes x, y, z, w, v.

Consider the 3-dimensional sphere formed by the part of the 5D hypersphere in the xyz-space (i.e. where w=v=0). Then, at each point (a, b, c, 0, 0) in the sphere there will be a 2D cross-section of the 5D hypersphere with the points (a, b, c, w, v) such that w²+v²=r²-a²-b²-c², so this 2D cross-section is a circle of radius R=√(r²-a²-b²-c²), and this radius will be the same for all points (a, b, c, 0, 0) in the 3D sphere belonging to a spherical surface centre O of radius x (0≤x≤r), say, i.e. such that a²+b²+c²=x², so that R=√(r²-x²).

So the volume of the 5D hypersphere is
V(5)=∫₀ʳ (4πx²)(πR²)dx, where R²=r²-x², obtained by integrating the product of the area of the surface of the 3D sphere of radius x (4πx²) and the volume of the corresponding 2D cross-section (πR²), for x from 0 to r.
So V(5)=4π²∫₀ʳ x²(r²-x²)dx
=4π²∫₀ʳ (r²x²-x⁴)dx
=4π²[1/3 r²x³-1/5 x⁵]₀ʳ
=4π²[(1/3r⁵-1/5 r⁵)-0]
=4π²(2/15 r⁵)
=8π²r⁵/15

Now let's turn to the general case, and find the volume of an (n+2)-dimensional hypersphere in terms of the volume of an n-dimensional a hypersphere

Now the volume of an n-dimensional hypersphere of radius r is krⁿ, for some constant k.

To find its (n-1)-dimensional "surface area" S consider the difference δV between the volume of an n-dimensional hypersphere of radius r and an n-dimensional hypersphere of radius r+δr, both centre O (for small δr), which is a "skin" of inner surface area S and thickness δr, so δV≈Sδr and δV/δr≈S, and in the limit as δr tends to zero we get dV/dr=S.
So S=nkrⁿ⁻¹.

Applying the same method as previously to find the volume of an (n+2)-dimensional hypersphere we get
V(n+2)=∫₀ʳ (nkxⁿ⁻¹)(πR²)dx, where R²=r²-x², obtained by integrating the product of the area of the surface of the (n-1)-dimensional hypersphere of radius x (nkxⁿ⁻¹) and the volume of the corresponding 2D cross-section (πR²), for x from 0 to r.

So V(n+2)=nkπ∫₀ʳ xⁿ⁻¹(r²-x²)dx
=nkπ∫₀ʳ (r²xⁿ⁻¹-xⁿ⁺¹)dx
=nkπ[1/n r²xⁿ-1/(n+2) xⁿ⁺²]₀ʳ
=nkπ[(1/n rⁿ⁺²-1/(n+2) rⁿ⁺²)-0]
=nkπ(2/[n(n+2)]rⁿ⁺²)
=2kπ/(n+2) rⁿ⁺²
=2πr²/(n+2) krⁿ
V(n+2)=2πr²/(n+2) V(n)

and V(n)=2πr²/n V(n-2)

[FiveTrackTape]

Up, up and away! Party song (1967) for the fifth dimension by The Fifth Dimension.

[markjulius]

Toby, listening to you speak about all these wonderful topics is on my bucket list. I really hope I can see you in person one day. Not only are you smart, but your voice is soothing and fun to listen to.

[jaybestnz]

How someone could think to work this out is incredible to me.

[StabbyMcStabStab]

Teaching out in nature, in a nice dress, looking all pretty, talking all quiet. This is the true evolution of education!

[iEuno]

Nice thank you miss, much appreciated and it can be useful for arrays and calculations with those, including how to integrate those calculations with dimensions.

[Timelesone]

This showcases more than meet the eye. The understanding of the sphere, from my experience, tends to provide space for safe passage and is where thought thing are (simplifying things a lot).

[MaatthewHiggs]

There’s a good party happening somewhere.