No, really, pi is wrong: The Tau Manifesto

By Michael Hartl

I continue to be impressed with how rich this subject is, and my understanding of $$\pi$$ and $$\tau$$ continues to evolve. On Half Tau Day, 2012, I believed I identified exactly what is wrong with $$\pi$$. My argument hinged on an analysis of the surface area and volume of an $$n$$-dimensional sphere, which (as shown below) makes clear that $$\pi$$ doesn’t have any fundamental geometric significance. My analysis was incomplete, though—a fact brought to my attention in a remarkable message from Tau Manifesto reader Jeff Cornell. As a result, this section is an attempt not only to definitively debunk $$\pi$$, but also to articulate the truth about $$\tau$$, a truth that is deeper and subtler than I had imagined.

Note: This section is more advanced than the rest of the manifesto and can be skipped without loss of continuity. If you find it confusing, I recommend proceeding directly to the conclusion in Section 6.

We start our investigations with the generalization of a circle to arbitrary dimensions. This object, called a hypersphere or an $$n$$-sphere, can be defined as follows. (For convenience, we assume that these spheres are centered on the origin.) A $$0$$-sphere is the empty set, and we define its “interior” to be a point. A $$1$$-sphere is the set of all points satisfying

$x^2 = r^2,$

which consists of the two points $$\pm r$$. Its interior, which satisfies

$x^2 \leq r^2,$

is the line segment from $$-r$$ to $$r$$. A $$2$$-sphere is a circle, which is the set of all points satisfying

$x^2 + y^2 = r^2.$

Its interior, which satisfies,

$x^2 + y^2 \leq r^2,$

is a disk. Similarly, a $$3$$-sphere satisfies

$x^2 + y^2 + z^2 = r^2,$

and its interior is a ball. The generalization to arbitrary $$n$$, although difficult to visualize for $$n > 3$$, is straightforward: an $$n$$-sphere is the set of all points satisfying

$\sum_{i=1}^{n} x_i^2 = r^2.$

The Pi Manifesto (discussed in Section 4.2) includes a formula for the volume of a unit $$n$$-sphere as an argument in favor of $$\pi$$:

$$\label{eq:unit_n_sphere_pi} \frac{\sqrt{\pi}^{n} }{\Gamma(1 + \frac{n}{2})},$$

where the Gamma function is given by Eq. (11). Eq. (14) is a special case of the formula for general radius, which is also typically written in terms of $$\pi$$:

$$\label{eq:n_sphere_pi} V_n(r) = \frac{\pi^{n/2} r^n}{\Gamma(1 + \frac{n}{2})}.$$

Because $$V_n(r) = \int S_n(r)\,dr$$, we have $$S_n(r) = dV_n(r)/dr$$, which means that the surface area can be written as follows:

$$\label{eq:n_sphere_pi_r} S_n(r) = \frac{n \pi^{n/2} r^{n-1}}{\Gamma(1 + \frac{n}{2})}.$$

Rather than simply take these formulas at face value, let’s see if we can untangle them to shed more light on the question of $$\pi$$ vs. $$\tau$$. We begin our analysis by noting that the apparent simplicity of the above formulas is an illusion: although the Gamma function is notationally simple, in fact it is an integral over a semi-infinite domain, which is not a simple idea at all. Fortunately, the Gamma function can be simplified in certain special cases. For example, when $$n$$ is an integer, it is easy to show (using integration by parts) that

$\Gamma(n) = (n-1)(n-2)\ldots 2\cdot 1 = (n-1)!$

Seen this way, $$\Gamma$$ can be interpreted as a generalization of the factorial function to real-valued arguments.

In the $$n$$-dimensional surface area and volume formulas, the argument of $$\Gamma$$ is not necessarily an integer, but rather is $$\left(1 + \frac{n}{2}\right)$$, which is an integer when $$n$$ is even and is a half-integer when $$n$$ is odd. Taking this into account gives the following expression, which is taken from a standard reference, Wolfram MathWorld, and as usual is written in terms of $$\pi$$:

$$\label{eq:surface_area_mathworld} S_n(r) = \begin{cases} \displaystyle \frac{2\pi^{n/2}\,r^{n-1}}{(\frac{1}{2}n - 1)!} & \text{if } n \text{ is even}; \\ \\ \displaystyle \frac{2^{(n+1)/2}\pi^{(n-1)/2}\,r^{n-1}}{(n-2)!!} & \text{if } n \text{ is odd}. \end{cases}$$

Integrating with respect to $$r$$ then gives

$$\label{eq:volume_mathworld} V_n(r) = \begin{cases} \displaystyle \frac{\pi^{n/2}\,r^n}{(\frac{n}{2})!} & \text{if } n \text{ is even}; \\ \\ \displaystyle \frac{2^{(n+1)/2}\pi^{(n-1)/2}\,r^n}{n!!} & \text{if } n \text{ is odd}. \end{cases}$$

Let’s examine Eq. (18) in more detail. Notice first that MathWorld uses the double factorial function $$n!!$$—but, strangely, it uses it only in the odd case. (This is a hint of things to come.) The double factorial function, although rarely encountered in mathematics, is elementary: it’s like the normal factorial function, but involves subtracting $$2$$ at a time instead of $$1$$, so that, e.g., $$5!! = 5 \cdot 3 \cdot 1$$ and $$6!! = 6 \cdot 4 \cdot 2$$. In general, we have

$$\label{eq:double_factorial} n!! = \begin{cases} n(n-2)(n-4)\ldots6\cdot4\cdot2 & \text{if } n \text{ is even}; \\ \\ n(n-2)(n-4)\ldots5\cdot3\cdot1 & \text{if } n \text{ is odd}. \end{cases}$$

(By definition, $$0!! = 1!! = 1$$.) Note that Eq. (19) naturally divides into even and odd cases, making MathWorld’s decision to use it only in the odd case still more mysterious.

To solve this mystery, we’ll start by taking a closer look at the formula for odd $$n$$ in Eq. (18):

$\frac{2^{(n+1)/2}\pi^{(n-1)/2}\,r^n}{n!!}$

Upon examining the expression

$2^{(n+1)/2}\pi^{(n-1)/2},$

we notice that it can be rewritten as

$2(2\pi)^{(n-1)/2},$

and here we recognize our old friend $$2\pi$$.

Now let’s look at the even case in Eq. (18). We noted above how strange it is to use the ordinary factorial in the even case but the double factorial in the odd case. Indeed, because the double factorial is already defined piecewise, if we unified the formulas by using $$n!!$$ in both cases we could pull it out as a common factor:

$V_n(r) = \frac{1}{n!!}\times \begin{cases} \ldots & \text{if } n \text{ is even}; \\ \\ \ldots & \text{if } n \text{ is odd}. \end{cases}$

So, is there any connection between the factorial and the double factorial? Yes—when $$n$$ is even, the two are related by the following identity:

$\left(\frac{n}{2}\right)! = \frac{n!!}{2^{n/2}}.$

(This is easy to verify using mathematical induction.) Substituting this into the volume formula for even $$n$$ then yields

$\frac{2^{n/2}\pi^{n/2}\,r^n}{n!!},$

which bears a striking resemblance to

$\frac{(2\pi)^{n/2}\,r^n}{n!!},$

and again we find a factor of $$2\pi$$.

Putting these results together, we see that Eq. (18) can be rewritten as

$$\label{eq:volume_2pi} V_n(r) = \begin{cases} \displaystyle \frac{(2\pi)^{n/2}\,r^n}{n!!} & \text{if } n \text{ is even}; \\ \\ \displaystyle \frac{2(2\pi)^{(n-1)/2}\,r^n}{n!!} & \text{if } n \text{ is odd} \end{cases}$$

and Eq. (17) can be rewritten as

$$\label{eq:surface_area_2pi} S_n(r) = \begin{cases} \displaystyle \frac{(2\pi)^{n/2}\,r^{n-1}}{(n-2)!!} & \text{if } n \text{ is even}; \\ \\ \displaystyle \frac{2(2\pi)^{(n-1)/2}\,r^{n-1}}{(n-2)!!} & \text{if } n \text{ is odd.} \end{cases}$$

Making the substitution $$\tau=2\pi$$ in Eq. (21) then yields

$S_n(r) = \begin{cases} \displaystyle \frac{\tau^{n/2}\,r^{n-1}}{(n-2)!!} & \text{if } n \text{ is even}; \\ \\ \displaystyle \frac{2\tau^{(n-1)/2}\,r^{n-1}}{(n-2)!!} & \text{if } n \text{ is odd.} \end{cases}$

To unify the formulas further, we can use the floor function $$\lfloor x \rfloor$$, which is simply the largest integer less than or equal to $$x$$ (equivalent to chopping off the fractional part, so that, e.g., $$\lfloor 3.7 \rfloor = \lfloor 3.2 \rfloor = 3$$). This gives

$S_n(r) = \begin{cases} \displaystyle \frac{\tau^{\left\lfloor \frac{n}{2} \right\rfloor}\,r^{n-1}}{(n-2)!!} & \text{if } n \text{ is even}; \\ \\ \displaystyle \frac{2\tau^{\left\lfloor \frac{n}{2} \right\rfloor}\,r^{n-1}}{(n-2)!!} & \text{if } n \text{ is odd}, \end{cases}$

which allows us to write the formula as follows:

$$\label{eq:surface_area_tau} S_n(r) = \frac{\tau^{\left\lfloor \frac{n}{2} \right\rfloor}\,r^{n-1}}{(n-2)!!}\times \begin{cases} 1 & \text{if } n \text{ is even}; \\ \\ 2 & \text{if } n \text{ is odd}. \end{cases}$$

Integrating Eq. (22) with respect to $$r$$ then yields

$$\label{eq:volume_tau} V_n(r) = \frac{\tau^{\left\lfloor \frac{n}{2} \right\rfloor}\,r^n}{n!!}\times \begin{cases} 1 & \text{if } n \text{ is even}; \\ \\ 2 & \text{if } n \text{ is odd}. \end{cases}$$

The formulas in Eq. (22) and Eq. (23) represent a major improvement over the original formulations (Eq. (17) and Eq. (18)) in terms of $$\pi$$. But in fact an additional simplification is possible, using the measure of a right angle:

$$\label{eq:lambda} \lambda = \frac{\tau}{4}.$$

As we’ll see in Section 5.2, Eq. (24) can be more naturally rewritten in terms of the symmetries of the circle:

$$\label{eq:tau_lambda} \tau = 2^2 \lambda,$$

where the factor of $$2^2$$ comes from the $$2^2$$ congruent circular arcs (one in each quadrant) in two-dimensional space.

The biggest advantage of $$\lambda$$ is that it completely unifies the even and odd cases in Eq. (22) and Eq. (23), each of which has a factor of $$\tau^{\left\lfloor \frac{n}{2} \right\rfloor}$$. Making the substitution in Eq. (25) then gives

$\tau^{\left\lfloor \frac{n}{2} \right\rfloor} = (2^2\lambda)^{\left\lfloor \frac{n}{2} \right\rfloor} = 2^{2\left\lfloor \frac{n}{2} \right\rfloor} \lambda^{\left\lfloor \frac{n}{2} \right\rfloor} = \lambda^{\left\lfloor \frac{n}{2} \right\rfloor}\times \begin{cases} 2^n & \text{if } n \text{ is even}; \\ \\ 2^{n-1} & \text{if } n \text{ is odd}. \end{cases}$

This means that we can rewrite the product

$\tau^{\left\lfloor \frac{n}{2} \right\rfloor}\times \begin{cases} 1 & \text{if } n \text{ is even}; \\ \\ 2 & \text{if } n \text{ is odd}. \end{cases}$

as

$$\label{eq:prefactor} \lambda^{\left\lfloor \frac{n}{2} \right\rfloor} \times \begin{cases} 2^n & \text{if } n \text{ is even}; \\ \\ 2^{n-1} & \text{if } n \text{ is odd}. \end{cases} \times \begin{cases} 1 & \text{if } n \text{ is even}; \\ \\ 2 & \text{if } n \text{ is odd}. \end{cases} = 2^n\,\lambda^{\left\lfloor \frac{n}{2} \right\rfloor},$$

which eliminates the explicit dependence on parity. Applying Eq. (26) to Eq. (22) and Eq. (23) then gives

$$\label{eq:surface_area_lambda} S_n(r) = \frac{2^n\,\lambda^{\left\lfloor \frac{n}{2} \right\rfloor}\,r^{n-1}}{(n-2)!!}$$

and

$$\label{eq:volume_lambda} V_n(r) = \frac{2^n\,\lambda^{\left\lfloor \frac{n}{2} \right\rfloor}\,r^n}{n!!}.$$

The simplification in Eq. (27) and Eq. (28) appears to come at the cost of a factor of $$2^n$$, but even this has a clear geometric meaning: a sphere in $$n$$ dimensions divides naturally into $$2^n$$ congruent pieces, corresponding to the $$2^n$$ families of solutions to $$\sum_{i=1}^{n} x_i^2 = r^2$$ (one for each choice of $$\pm x_i$$). In two dimensions, these are the circular arcs in each of the four quadrants; in three dimensions, they are the sectors of the sphere in each octant; and so on in higher dimensions. In other words, we can exploit the symmetry of the sphere by calculating the surface area or volume of one piece—typically the principal part where $$x_i > 0$$ for every $$i$$—and then find the full value by multiplying by $$2^n$$.

To my knowledge, Eq. (27) and Eq. (28) are the simplest possible formulations of the spherical surface area and volume formulas (and indeed are the only forms I have ever consistently been able to memorize). Consider the volume formula in particular: unlike the faux simplicity of Eq. (15), Eq. (28) involves no fancy integrals—just the slightly exotic but nevertheless elementary floor and double-factorial functions. The volume of a unit $$n$$-sphere is just the volume of each symmetric piece, $$\lambda^{\left\lfloor \frac{n}{2} \right\rfloor}/n!!$$, multiplied by the number of pieces, $$2^n$$.

We’ve now seen, via Eq. (27) and Eq. (28), that the surface area and volume formulas are simplest in terms of the right angle $$\lambda$$. Nevertheless, we’re still not done with $$\tau$$.

As seen in Eq. (28), the volume formula divides naturally into two families, corresponding to even- and odd-dimensional spaces, respectively. This means that the four-dimensional volume, $$V_4$$, is related simply to $$V_2$$ but not to $$V_3$$, while $$V_3$$ is related to $$V_1$$ but not to $$V_2$$. How exactly are they related?

We can find the answer by deriving the recurrence relations between dimensions. In particular, let’s divide the volume of an $$n$$-dimensional sphere by the volume for an $$(n-2)$$-dimensional sphere:

$$\label{eq:volume_recurrence} \frac{V_n(r)}{V_{n-2}(r)} = \frac{2^n}{2^{n-2}} \frac{\lambda^{\left\lfloor \frac{n}{2} \right\rfloor}}{\lambda^{\left\lfloor \frac{n-2}{2} \right\rfloor}} \frac{(n-2)!!}{n!!} \frac{r^{n}}{r^{n-2}} = \frac{2^2\lambda}{n}\,r^2.$$

We see from Eq. (29) that we can obtain the volume of an $$n$$-sphere simply by multiplying the formula for an $$(n-2)$$-sphere by $$r^2$$ (a factor required by dimensional analysis), dividing by $$n$$, and multiplying by the “recurrence constant” $$2^2\lambda$$.

Similarly, for the surface area we have

$$\label{eq:surface_area_recurrence} \frac{S_n(r)}{S_{n-2}(r)} = \frac{2^n}{2^{n-2}} \frac{\lambda^{\left\lfloor \frac{n}{2} \right\rfloor}}{\lambda^{\left\lfloor \frac{n-2}{2} \right\rfloor}} \frac{(n-2-2)!!}{(n-2)!!} \frac{r^{n}}{r^{n-2}} = \frac{2^2\lambda}{n-2}\,r^2,$$

with the same recurrence constant $$2^2\lambda$$.

Thus, in both Eq. (29) and Eq. (30), the constant relating the different dimensions is not $$\lambda$$ itself but rather the combination $$2^2\lambda$$. Comparing with Eq. (25), we see this is none other than $$\tau$$! Indeed, an alternate derivation of the volume recurrence by direct calculation (which uses $$R$$ where we write $$r$$) concludes with the integral

\begin{align} \label{eq:integral_recurrence} V_n(R) &= \int_0^{2\pi} \int_0^R V_{n-2}\left(\sqrt{R^2 - r^2}\right) \,r\,dr\,d\theta \nonumber \\ &= 2\pi V_{n-2}(R) \cdot \left[-\frac{R^2}{n}\left(1 - \left(\frac{r}{R}\right)^2\right)^\frac{n}{2}\right]_{r=0}^{r=R} \\ &= \frac{2\pi R^2}{n} V_{n-2}(R), \nonumber \end{align}

thus showing that the identification of $$\tau$$ as the “recurrence constant” isn’t a coincidence—the recurrence constant and the circle constant really are one and the same:

$\tau = \mbox{circle constant} = \mbox{recurrence constant} = 2^2\lambda.$

As a result, it is $$\tau$$, not $$\lambda$$, that provides the common thread tying together the two families of even and odd solutions, as illustrated by Joseph Lindenberg in Tau Before It Was Cool (Figure 16).

Surface area and volume recurrences.

When discussing general $$n$$-dimensional spheres, for convenience we’ll write the surface area and volume formulas in terms of $$\lambda$$ as in Eq. (27) and Eq. (28), but for any given $$n$$ we’ll express the results in terms of the recurrence constant $$\tau$$.

Equipped with the tools developed in Section 5.1, we’re now ready to get to the bottom of $$\pi$$ and $$\tau$$. To complete the excavation, we’ll use Eq. (27) and Eq. (28) to define two families of constants, and then use the definition of $$\pi$$ (Eq. (1)) to define a third, thereby revealing exactly what is wrong with $$\pi$$.

First, we’ll define a family of “surface area constants” $$\tau_n$$ by dividing Eq. (27) by $$r^{n-1}$$, the power of $$r$$ needed to yield a dimensionless constant:

$$\label{eq:surface_area_constants} \tau_n \equiv \frac{S_n(r)}{r^{n-1}} = \frac{2^n\,\lambda^{\left\lfloor \frac{n}{2} \right\rfloor}}{(n-2)!!}$$

Second, we’ll define a family of “volume constants” $$\sigma_n$$ by dividing the volume formula Eq. (28) by $$r^n$$:

$$\label{eq:volume_constants} \sigma_n \equiv \frac{V_n(r)}{r^n} = \frac{2^n\,\lambda^{\left\lfloor \frac{n}{2} \right\rfloor}}{n!!}.$$

With the two families of constants defined in Eq. (32) and Eq. (33), we can write the surface area and volume formulas (Eq. (27) and Eq. (28)) compactly as follows:

$S_n(r) = \tau_n\,r^{n-1}$

and

$V_n(r) = \sigma_n\,r^n.$

Because of the relation $$V_n(r) = \int S_n(r)\,dr$$, we have the simple relationship

$\sigma_n = \frac{\tau_n}{n}.$

Let us make some observations about these two families of constants. The family $$\tau_n$$ has an important geometric meaning: by setting $$r=1$$ in Eq. (32), we see that each $$\tau_n$$ is the surface area of a unit $$n$$-sphere, which is also the angle measure of a full $$n$$-sphere. In particular, by writing $$s_n(r)$$ as the $$n$$-dimensional “arclength” equal to a fraction $$f$$ of the full surface area $$S_n(r)$$, we have

$\theta_n \equiv \frac{s_n(r)}{r^{n-1}} = \frac{f S_n(r)}{r^{n-1}} = f\left(\frac{S_n(r)}{r^{n-1}}\right) = f\tau_n.$

Here $$\theta_n$$ is simply the $$n$$-dimensional generalization of radian angle measure, and we see that $$\tau_n$$ is the generalization of “one turn” to $$n$$ dimensions, which explains why the 2-sphere (circle) constant $$\tau_2 = 2^2\lambda = \tau$$ leads naturally to the diagram shown in Figure 10. Furthermore, we learned in Section 5.1 that $$\tau_2$$ is also the “recurrence constant” for $$n$$-sphere surface areas and volumes.

Meanwhile, the $$\sigma_n$$ are the volumes of unit $$n$$-spheres. In particular, $$\sigma_2$$ is the area of a unit disk:

$\sigma_2 = \frac{\tau_2}{2} = \frac{\tau}{2}.$

This shows that $$\sigma_2 = \tau/2 = 3.14159\ldots$$ does have an independent geometric significance. Note, however, that it has nothing to do with circumferences or diameters. In other words, $$\pi = C/D$$ is not a member of the family $$\sigma_n$$.

So, to which family of constants does $$\pi$$ naturally belong? Let’s rewrite Eq. (1) in terms more appropriate for generalization to higher dimensions:

$\pi = \frac{C}{D} = \frac{S_2}{D^{2-1}}.$

We thus see that $$\pi$$ is naturally associated with surface areas divided by the power of the diameter necessary to yield a dimensionless constant. This suggests introducing a third family of constants $$\pi_n$$:

$$\label{eq:diameter_constants} \pi_n \equiv \frac{S_n(r)}{D^{n-1}}.$$

We can express this in terms of the family $$\tau_n$$ by substituting $$D = 2r$$ in Eq. (34) and applying Eq. (32):

$\pi_n = \frac{S_n(r)}{D^{n-1}} = \frac{S_n(r)}{(2r)^{n-1}} = \frac{S_n(r)}{2^{n-1}r^{n-1}} = \frac{\tau_n}{2^{n-1}}.$

We are now finally in a position to understand exactly what is wrong with $$\pi$$. The principal geometric significance of $$3.14159\ldots$$ is that it is the area of a unit disk. But this number comes from evaluating $$\sigma_n = \tau_n/n$$ when $$n=2$$:

$\sigma_2 = \frac{\tau_2}{2} = \frac{\tau}{2}.$

It’s true that this equals $$\pi_2$$:

$\pi_2 = \pi = \frac{\tau_2}{2^{2-1}} = \frac{\tau}{2}.$

But this equality is a coincidence: it occurs only because $$2^{n-1}$$ happens to equal $$n$$ when $$n=2$$ (that is, $$2^{2-1} = 2$$). In all higher dimensions, $$n$$ and $$2^{n-1}$$ are distinct. In other words, the geometric significance of $$\pi$$ is the result of a mathematical pun.