This one’s related to another post on the volume of a sphere in n-dimensions.

The story there goes something like this:  find the next term in the sequence,

The observation is that the sequence represents radians, circumference of a circle, the area of a circle, the surface area of a sphere and the volume of a sphere.  The next term then would be the surface “area” of a sphere in 3-dimensions, which is

followed by its volume, and so on.  Notice it’s pi squared – something I didn’t expect when I did this problem the first time back in 2001.

The conclusion of that post was that the surface area of a sphere in n-1 dimensions is given by,

and its volume is given by,

where Γ(z) is the Gamma Function.

But we’ve glossed over something important.  These terms are the surface areas and volumes for what mathematicians call the n-dimensional 2-sphere, i.e. things that look like

which are the usual geometric objects one encounters in math growing up.  They’re generalizations of the Pythagorean Theorem as to how one should compute distances. 

Notice the powers of 2 everywhere.  That’s why they’re called n-dimensional 2-spheres – n for the number of dimensions, and 2 because we’re generalizing the Pythagorean Theorem. 

There’s no reason we can’t consider more general kinds of spheres, what mathematicians call the n-dimensional p-sphere, i.e. things that look like

They’re just different ways of computing distances from the Pythagorean Theorem, often called L^p norms.  Fun fact – you can compute angles between two points when p=2, but not otherwise!

The purpose of this post:  let’s compute the surface areas and volumes of these n-dimensional p-spheres. 

To get some intuition, I’ve plotted a few of them in 2 dimensions below.  Notice that for p<1, the shapes are not convex (they don’t “round out”), so they’re not strictly speaking ways to measure distances (why?).  As p tends to infinity, we approach the square. 

So nice.  So pretty.  So cool.

This plot establishes some monotonicity: namely that if p>q, then the perimeter (surface area) of the n-dimensional p-sphere should be greater than that of the n-dimensional q-sphere.  Likewise for the area. 

Great.  But what are the surface areas and volumes of these objects?  For n=2, p=2, we know the perimeter is 2 pi r.  What about n=2, p=3, and so on?

Here’s the great inspiration: mimic the proof from the previous post, carrying through powers of p instead of 2.  The one thing we’ll need is way to parametrize the surface of these p-spheres in n-dimensions in the positive orthant. 

Let’s do it in 2-dimensions which will then generalize straightforward.  The parametrization is given by,

Notice the trick: we’re using the standard 2-dimensional polar coordinates, just deforming it using powers of 2/p on just the trig functions.  Since cos^2(t)+sin^2(t)=1, x^p+y^p=r^p.

There are n-dimensional spherical polar coordinates as well, and we can repeat this trick there too.  Let’s call these our deformed n-dimensional spherical coordinates.

Something critical we’ll need is the Jacobian that arises in the original proof when we change coordinates from Euclidean to our deformed n-dimensional spherical coordinates.  We don’t really need to know what it is exactly – in fact, that’s the point since we’re looking for the surface area and we can’t explicitly evaluate the integral.  The only thing we need to know is that it’s of the form

Which is immediate due to how we deformed the trig functions and left the scaling in r alone.

With that in hand, we can proceed as we did in the previous post:

Raising both sides to the n, using Fubini’s theorem, and changing to our deformed n-dimensional spherical coordinates:

Solving for σ, substituting the expression for I_1, and using the well-known property zΓ(z)=Γ(1+z):

This is the surface area of the unit n-dimensional p-sphere.  A radius of r should scale like n-1, so that the surface area of an n-dimensional p-sphere of radius r is: 

Finally, integrating over r and using zΓ(z)=Γ(1+z) again:

Yielding our final formula:

To sanity check, if you substitute p=2, use zΓ(z)=Γ(1+z) again, and use the well-known fact

you get the volume from our original post,

If you’re still not convinced and this looks too good to be true, check out the below plots comparing the estimated volume via Monte Carlo and the numerical values via our formula when r=1.

Note that for fixed values of p, the volume of the ambient space still outpaces that of the p-Sphere like it did in our original post when p=2, so the volumes collapse to zero.  In higher dimensions, the simulated curves are noiser as fewer sampled point falls within the n-dimensional p-sphere.

Volume Found

In our original post, we asked “where did the volume go?”

Seems like for any fixed value of p, as n grows, the volume of our n-dimensional p-spheres still collapse to zero.

To find out where the volume went, we allow p to vary as a function of n.  That way, the p-spheres are allowed to “bulge” out more and more as the dimension grows, so we can hope to capture more volume.  Ideally in the limit of large n, we should capture all of it.

Formalizing this a bit more, how should p(n) behave so that,

To figure this out, let’s start out with p(n) with polynomial growth.  We’ll also need the Weierstrauss’ factorization of the Gamma function:

Actually I lied – we need a slight tweak of the Weierstrauss formula.  Since Γ(1+z) appear in our ratio, we need a infinite factorization of it instead.  This is easy using Γ(1+z) = zΓ(z):

Substituting all this in:

Where α>0, and γ is the Euler–Mascheroni constant.

A couple of observations are immediate:

This result is good, but not great since it only achieves ~ 56% of the volume. 

To do better, we need p(n) to grow a bit faster than n, but not faster than any of its higher powers. 

You should be thinking logarithms.  p(n)=n log(n) to be exact.  You’d get pretty close: 99.08% of the volume. 

p(n)=n li(n) works, where li is the logarithmic integral function.

To summarize these findings:

p(n)=n contains 56% of the volume for large n

p(n) = n log(n) contains 99.08% of the volume for large n.

p(n) = n li(n) contains 100% of the volume for large n.

CONCLUDING REMARKS

We’ve shown that the volume of the p-sphere of radius r described by the algebraic equation,

has volume

The ramifications of this simple, but elegant formula are deep. 

For example, for an positive integer n, we can write n=mp + ℓ, for 0 ≤ ℓ < p.  A bit of algebra on our formula implies that the corresponding volumes are related by,

Which factors the volume of a p-sphere of dimension n into its volume when n=p, and a remainder when the dimension is between 0 and p.

For example, take p=2:

If ℓ=0, we get

While if ℓ = 1 we get,

Which is exactly what happens if we set n=2m+ℓ into our original formula,

For p>2, we have the base case when n=p, and the remaining pieces when 0<n<p.

Next, our formula for the volumes of p-spheres yields geometric interpretation of Γ(r) for r rational.  Wikipedia (correctly) says:

In other words, we don’t know much about Γ(r) for rational number r.  What we do know is that it the numbers appear to be (algebraically) independent of the geometric number we know a lot about, π (p=2).

But what our result shows shows is that while Γ(r) for rational r might appear to be independent of π, it is intimately related to its general p-dimensional variants. 

To see this, note that p=n yields a geometric interpretation of Γ(1/n):

We can then plug this into our formula, and with a little bit of algebra:

This establishes a one-to-one correspondence:

Knowing Γ(ℓ/p) means we know the volumes of p-spheres in dimension ℓ, and conversely, knowing the volumes of p-spheres of dimension ℓ means we know Γ(ℓ/p).

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