I thought it might be fun to ask this question as a way of celebrating Pi Day. One way in which people popularize pi is that they say that even though it's defined in terms of properties of a circle, it shows up in contexts such as the normal distribution which have (apparently) nothing to do with circles, and this is supposed to say something about the mysterious and amazing character of mathematics. So I'm curious what MO have to say about this phenomenon. If I had to venture a guess, it would be something about Fourier analysis, but I know there are many people here who could make such a statement much more precise.

Answers should ideally be given in general terms, but feel free to illustrate your generalities with interesting examples.

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    $e^{\pi i} = -1$ – Steve Huntsman Mar 14 '10 at 17:37
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    Ken Ribet has said that the correct holiday is June 28. – S. Carnahan Mar 14 '10 at 18:18
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    I've heard that 7/22 is the correct holiday. – Ben Weiss Mar 14 '10 at 22:41
  • @Ben Weiss: Really 7/22, as in $7.22$ or $0.318\overline{18}$?...$$ $$ Even if you write July 22 as $22/7$ it's not quite right, though it comes close; see mathoverflow.net/questions/67384 for recent MO discussion of one way to explain "$\pi$ Approximation Day". – Noam D. Elkies Aug 9 '11 at 21:24
  • snorgtees.com/octopi – Will Jagy Aug 10 '11 at 0:15

12 Answers 12

Let me play devil's advocate here: I'm not sure that I agree that the ubiquity of π is so mysterious. After all, how do you ever prove that π appears? You have to relate your situation to some known situation where π already appears, so the mystery is solved almost before it occurs. Of course, if you're just shown the appearance of π without the proof then you may be surprised, but that simply means you haven't yet seen the proof. To take an example, the proof that $\int_{-\infty}^\infty e^{-x^2}dx$ involves π uses the rotational invariance of the normal distribution. But rotations are closely connected with circles and hence with π, so it isn't too surprising that π shows up. To take another example, it is amazing that $\sum n^{-2}=\pi^2/6$, but one nice proof of that uses Parseval's identity, calculating the $\ell_2$ and $L_2$ norms of the Fourier coefficients of a certain function and the function itself, respectively. And Fourier coefficients involve trigonometric functions, so the appearance of π is, once again, not a surprise.

Maybe the right thing to say is that the multiple appearances of π are initially striking and mysterious, but the mystery disappears on closer inspection -- like many mysteries. The statement I would dispute is that there is a general mystery of the kind "Why does π appear so much?" I'd give an answer like "Because circles and rotations appear a lot."

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    I agree, but sometimes the mystery is not within $\pi$ itself but on the appearance of the circle/rotations. Take for example Johanson's theorem of the artic circle of the Aztec diamond, or the Kerov-Vershik "arcsine-law" which describes the shape of large Young tableaux, $\pi$ only appears in the formulas but the true mystery is why do these certain shapes appear. – Gjergji Zaimi Mar 14 '10 at 22:34

In some sense you could say that there are probably not as many sources as you might think: The conjecture of Kontsevich-Zagier says that when $\pi$ appears as some kind of period one should always be able to make some simple transformations of the involved integral to make it visibly equal to the standard definition. This of course is something number theorists/arithmetic algebraic geometers are used to do; try to find that two different $\pi$'s are the same (usually for some kind of ``motivic´´ reason).

This preamble is just to lead up to the amusing fact that wondering whether two different $\pi$ are the same or not is not the exclusive domain of arithmeticians. In his first book in the series of PDO's Hörmander proofs Fourier inversion formula by first proving that it is true up to a constant (using in effect the irreducibility of the Heisenberg representation of the Heisenberg group) and then uses two different methods for determining the constant (which of course involves $\pi$), one by computing the integral $\int_0^\infty e^{-x^2}$ and one by the Cauchy residue formula. He then writes ''The constants $2\pi$ in Cauchy's integral formula and in the inversion formula are therefore 'the same'...''

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    Re Hörmander story: Harish-Chandra was known to start his papers/lectures with: "Let us fix a square root of $-1$ and call it $i$". Varadarajan, who admired Harish-Chandra, was doing the same, until he was once interrupted with: "Is your choice of the square root the same as Harish-Chandra's?" – Victor Protsak Jun 30 '10 at 4:42

As a counterpoint to gowers's devil's advocacy, I'd mention that some formulas for $\pi$ have been discovered experimentally, and in some cases we still don't know how to prove them. For example, in the paper "About a New Kind of Ramanujan-type Series" by Jesús Guillera (Experimental Mathematics 12 (2003), 507–510), the following conjecture by Gourevitch is stated: $$\sum_{n=0}^\infty \frac{1+14n+76n^2+168n^3}{2^{20n}}\binom{2n}{n}^7 = \frac{32}{\pi^3}.$$ As far as I know, this is still unproved. In principle this kind of formula should be WZ-able, but it seems to be just out of reach of current computers. And probably there ultimately does exist some "motivic explanation" as Torsten Ekedahl said, but since we don't currently know of one, I think that one answer to Qiaochu's question is, "experimental observation."

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    To apply the WZ method (or Zeilberger's algorithm) you need another parameter. You would have to find an identity with a free parameter that gives this identity as a limit or special case, and as far as I know, there's no automatic way to do this. – Ira Gessel May 20 '14 at 0:22

$\pi$ shows up in at least two different ways related to factorials.

The $\pi$ in Stirling's Formula $n! \approx (\frac{n}{e})^n \times\sqrt{2 \pi n}$ comes down to Wallis's Formula

$$\frac {\pi}{2} = \prod_{n=1}^\infty \frac{2n\times 2n}{2n-1 \times 2n+1}$$

which follows from the infinite product for sine

$$\sin x = x \prod_{n=1}^\infty \bigg( 1 - \frac {x^2}{\pi^2 n^2}\bigg)$$

evaluated at $x=\frac\pi2$. I guess that comes down to the circle after all.

Also, $(-1/2)! = \Gamma(1/2) = \sqrt \pi.$ That can be seen from Euler's reflection formula,

$$\Gamma(x)\Gamma(1-x) = \frac {\pi}{\sin \pi x}$$

As for the normal distribution, you can characterize it as the unique distribution with the following properties:

Let $X_1, X_2, \cdots X_n$ be independent identically distributed normal random variables. Then the joint distribution of the vector $X=(X_1, X_2, \cdots X_n)$ is the same as that of $AX$ where $A$ is any orthogonal matrix. So the normal distribution is intimately related to the geometry of real inner product spaces.

The $\pi$ comes from the fact that you can integrate such a distribution by first integrating over a sphere and then integrating over $[0,\infty]$. Because the distribution is orthogonally invariant, you pick up a constant corresponding to the area of the sphere. For $n=2$ you get the circle, and this is the usual calculation for computing the normalization constant for the normal distribution.

So then the mystery becomes: given that the normal distribution is so closely tied to inner product spaces, why does it show up all the time? The central limit theorem tells us that all that really matters in large scale limits are the first and second moments. The first moment can always be eliminated by re-centering. So all that matters is the second moment. But the second moment comes from the covariance, which is an inner product! (technically, only once you restrict to re-centered random variables, but we are doing that)

I'd venture a guess that most, if not all, appearances of $\pi$ in statistics boil down to this fact that covariance is an inner product, and the fact that spheres, which are the norm-level sets for inner product spaces, have areas related to $\pi$

The appearance of $\pi$ in Stirling's formula still astounds me.

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    While we're on this subject, what's e doing there? – Michael Lugo Mar 14 '10 at 17:59
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    The appearance of e is much less mysterious. Knowing only that (1 + 1/n)^n \le e for all n you can already deduce that n! \ge (n/e)^n. What's really mysterious is the extra work you have to do to get that last factor in. – Qiaochu Yuan Mar 14 '10 at 18:08
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    The presence of $e$ is easily explained by the geometric mean of $[0,1]$ which is $1/e$. The $\sqrt n$ factor comes from the difference between the trapezoid rule and right endpoint rule for integrating $\log x$ on $[0,1]$. – Douglas Zare Mar 14 '10 at 18:21
  • That should have been the left endpoint rule on $[1/n,1]$ with subintervals of width $1/n$, or you can also use $[1,n]$ with subintervals of width $1$. – Douglas Zare Mar 14 '10 at 19:26
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    pi naturally arises as a normalisation constant in any integral involving an exponential (see: Laplace's method, or the principle of stationary phase), thanks ultimately to the identity $\int_R e^{-\pi x^2}\ dx = 1$ (which is not a bad definition of pi, really). The factorial is a special case of the Gamma function, which is an integral involving an exponential. – Terry Tao Mar 15 '10 at 5:11

Of course $\pi$ shows up in the circumference of a circle of radius $1$, but why does it come up in the surface area of a $2$-sphere of radius $1$? One relation between the two is the remarkable fact observed by Archimedes that the horizontal projection from a sphere to a circumscribing cylinder preserves area.

I'd like to see a similar explanation for the higher powers of $\pi$ in the surface measures of higher dimensional spheres.

  • In the same way: horizontal projection from a sphere to a circumscribing cylinder preserves surface measure in any number of dimensions >= 2, so long as "circumscribing cylinder" means "Those points which would land on the sphere if all but their first two coordinates were changed to zero" and "horizontal projection" means "Scaling up the first two coordinates as needed while preserving the rest". This, along with that the interior content of a sphere in n-dimensional space is 1/n * its radius * its surface content, keeps $pi$ coming up in sphere surface and interior content in any dimension. – Sridhar Ramesh Aug 9 '11 at 22:04

One surprising place where $\pi$ appears, though I don't know whether you could call it "fundamental", is Buffon's needle problem, which I've just been writing about.

I remember this from an old blog post of yours. There is a way to prove the theorem that primes of the form $4k+1$ are sum of two squares, and the theorem that every natural number is the sum of four squares that basically says "because $\pi>2$ for the first one, and because $\pi^2>8$ for the second one". The proofs rely on Minkowski's theorem which is a great source for $\pi$ in number theory :-), but it does sound very surprising at first.

On a different note, from analytic number theory: the average order of Euler's function $\varphi$: $$\sum_{n\le x}\varphi(n)$$ is equal to $\frac{3x^2}{\pi^2}$ "on average"

Gjergji Zaimi's note on the average value of the Euler phi-function reminds me of some other appearances in Number Theory that don't have any obvious circles in them. The probability that two randomly selected positive integers are relatively prime is $6/\pi^2$ (where "randomly selected positive integers" is shorthand for, fix $n$, choose two integers uniformly and independently from $[1,n]$, then take a limit as $n$ goes to infinity). And the density of the squarefree integers (integers divisible by no square number other than 1) is also $6/\pi^2$.

In a horse race, a trifecta is a choice of horses finishing first, second, and third, in order. If the horses are numbered from 1 to $n$, one can ask for the number $S(n)$ of trifectas in geometric progression (e.g., 4-6-9 is a trifecta in geometric progression when interpreted as horse number 4 finishing 1st, horse number 6 finishing 2nd, etc.). I proved that $S(n)=(6/\pi^2)n\log n+O(n)$, and I attributed the appearance of $\pi$ in the formula to the circular portion of the race track.

  • ;) Very nice... – username Mar 7 '14 at 20:05

Here is another devil's advocacy: We (mathematicians) are not familiar with many transcendental numbers such as $\pi$ or $e$. When in a formula appears some transcendental number that can be expressed as a function of $\pi$, we are happy with that and we have our "solution" for the problem. But if it is not related to $\pi$ (or $e$, or etc...), we just think "OK, we don't have a solution yet."

Another way of saying pretty much the same thing is the fact that since we know very well $\pi$ and its properties, as soon as it is "in the neighborhood" of our problem, we see it, but if it is not here (and neither $e$ or another famous constant), we do not know where exactly to look to find an acceptable solution.

My 2-cent guess is "the better we know a number (through formulas for instance), the more often it will appear in new formula in the future."

Have a look at Pi, a source book by Berggren, Borwein, and Borwein.

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