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$\text{What are some interesting cases of $\pi$ appearing in situations that do not seem geometric?}$

Ever since I saw the identity $$\displaystyle \sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

and the generalization of $\zeta (2k)$, my perception of $\pi$ has changed. I used to think of it as rather obscure and purely geometric (applying to circles and such), but it seems that is not the case since it pops up in things like this which have no known geometric connection as far as I know. What are some other cases of $\pi$ popping up in unexpected places, and is there an underlying geometric explanation for its appearance?

In other words, what are some examples of $\pi$ popping up in places we wouldn't expect?

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Is there any known explanation for $\pi$ being there? I wouldn't have expected $\pi$ to pop up when dealing with factorials of all things. –  Soke Feb 25 '14 at 2:39
@user92774 The explanation actually would make a nice addition to my answer below. It's a special case of the central limit theorem. Binomial distributions tend toward the normal distribution for large $n$, which relates the $n!$ in the normalization factor of the binomial distribution to the $\sqrt{2\pi}$ in the normal distribution's normalization factor. –  David H Feb 25 '14 at 3:07
@user92774 The $\sqrt{\pi}$ in the Stirling's formula comes from the Laplace's approximation of a uni-modal function by the Gaussian function which integrates to $\sqrt{\pi}$. So the real question is why it integrates to that. –  Vadim Feb 25 '14 at 3:08
For the fun of it: $\pi=\left[\frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}\right]^{-1}$ –  Z Z Feb 25 '14 at 3:35
@JulienGodawatta Good ol' Ramanujan. –  Soke Feb 25 '14 at 3:47

19 Answers 19

up vote 28 down vote accepted

Too long for a comment:

What are some interesting cases of $\pi$ appearing in situations that are not geometric ?

None! :-) You did well to add “do not seem” in the title! ;-)

All $\zeta(2k)$ are bounded sums of squares, are they not ? And the equation of the circle, $x^2$$+y^2=r^2$, also represents a bounded sum of squares, does it not ? :-) Likewise, if you were to read a proof of why $\displaystyle\int_{-\infty}^\infty e^{-x^2}dx=\sqrt\pi$ , you would see that it also employs the equation of the circle! $\big($Notice the square of x in the exponent ?$\big)$ :-) Similarly for $\displaystyle\int_{-1}^1\sqrt{1-x^2}=\int_0^\infty\frac{dx}{1+x^2}=\frac\pi2$ , both of which can quite easily be traced back to the Pythagorean theorem. The same goes for the Wallis product, whose mathematical connection to the Basel problem is well known, the former being a corollary of the more general infinite product for the sine function, established by the great Leonhard Euler. $\big($Generally, all products of the form $\prod(1\pm a_k)$ are linked to sums of the form $\sum a_k$$\big)$. It is also no mystery that the discrete difference of odd powers of consecutive numbers, as well as its equivalent, the derivative of an odd power, is basically an even power, i.e., a square, so it should come as no surprise if the sign alternating sums $(+/-)$ of the Dirichlet beta function also happen to depend on $\pi$ for odd values of the argument. :-) Euler's formula and his identity are no exception either, since the link between the two constants, e and $\pi$, is also well established, inasmuch as the former is the basis of the natural logarithm, whose derivative describes the hyperbola $y=\dfrac1x$, which can easily be rewritten as $x^2$$-y^2=r^2$, following a rotation of the graphic of $45^\circ$. As for Viete's formula, its geometrical and trigonometrical origins are directly related to the half angle formula known since before the time of Archimedes. Etc. $\big($And the list could go on, and on, and on $\!\ldots\!\big)$ Where men see magic, math sees design. ;-) Hope all this helps shed some light on the subject.

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"Bounded sums of squares"...? –  Jack M Feb 25 '14 at 12:04
This answer clears up so many questions I never knew I had. It's stuff like this I wish my profs had told me years ago. –  David H Feb 25 '14 at 13:31
;-) okay that's a little bit too much –  qwr Feb 27 '14 at 23:54
The number of links makes it seem like this is Wikipedia! –  Sanath Devalapurkar Apr 20 '14 at 19:08

$\pi$ and the Mandelbrot set

Suppose we iterate the function $f(z)=z^2+c$ starting at $z_0=0$. For example, if $c=1/4$, the first few terms are \begin{align} z_0&=0 \\ z_1&=0^2+1/4=1/4\\ z_2&=(1/4)^2+1/4=5/16 \end{align} It can be shown that the sequence converges slowly up to $1/2$. On the other hand, if $c=1/4+\delta$, where $\delta>0$ (no matter how small), then the sequence diverges to $\infty$. This corresponds to the fact that $c=1/4$ is on the boundary of the Mandelbrot set.

enter image description here

We now ask the following: given $\delta>0$, how many iterates $N$ does it take until $z_N>2$? Here are the answers for several choices of $\delta$:

\begin{array}{c|c} \delta & \text{number of iterates until escape} \\ \hline 0.01 & 31 \\ \hline 0.0001& 313 \\ \hline 0.000001&3141\\ \hline 0.00000001&31415 \end{array}

In fact, if $N(\delta)$ represents the number of iterates until the iterate value exceeds two, then it can be proved that $$\lim_{\delta\rightarrow 0^{+}} N(\delta)\sqrt{\delta} = \pi.$$

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That is amazing! –  Jair Taylor Feb 25 '14 at 6:59
There seems to be an error in the third row of the tabulated values: When $\delta=10^{-6}$, we have $31415\sqrt\delta=31.415\approx 10\pi$. –  John Bentin Feb 25 '14 at 8:35
@JohnBentin Thanks! –  Mark McClure Feb 25 '14 at 11:28

When I first encountered the normal distribution in my high school statistics class, I was shocked to discover pi in the normalization of the Gaussian integral:


The statistical of analysis of data is about as far removed from purely geometric situations as I can think of.

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At some deep level this is how you get $\sqrt{\pi}$ in Stirling's formula. –  Vadim Feb 25 '14 at 3:11
However, the Gaussian is quite geometric, and the Gaussian integral is not a result of statistical analysis of data, but rather statistical analysis of data piggybacks on the geometric properties of the Gaussian. –  Arkamis Feb 25 '14 at 21:46

I don't know if this is what you are looking for, but the formula for $\pi$ discovered (somehow) by Ramanujan sometime around $1910$ is given by,

$$\frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum_{k\geq0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}.$$

If there is a geometric interpretation for this, I would like to know.

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Is there any interpretation for this? How does one derive this? –  Soke Feb 28 '14 at 23:13
@user92774: It's based on modular forms and the monster group. See here. –  Lucian Mar 1 '14 at 14:31

The probability that two positive integers are coprime is $\frac{6}{\pi^2}$.

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This is based on the Euler product formula for the Riemann $\zeta$ function. –  Lucian Mar 1 '14 at 14:24
The result and the proof still do not seem geometric. –  user37238 Mar 1 '14 at 15:26
That $\zeta(2)=\dfrac{\pi^2}6$ follows from rewriting the trigonometric sine function as an infinite product. On the other hand, the $\zeta$ function itself can be easily rewritten as a product over primes, as shown. Hence, the probability that k positive integers chosen at random are relatively prime is $\dfrac1{\zeta(k)}$ , which, for even values of the argument, is a rational multiple of a power of $\pi$. –  Lucian Mar 1 '14 at 15:43
@Lucian Thank you for the generalization. There is something that I don't understand : are you saying that the result I gave can been prove geometrically? –  user37238 Mar 4 '14 at 12:25
It ultimately lies on a geometric and trigonometric foundation. (The only result in this entire thread that -to my knowledge- does not possess a geometric or trigonometric interpretation, is Ramanujan's $\pi$ formula, all others being ultimately linked or connected with these two closely-related fields). –  Lucian Mar 4 '14 at 13:01

I think "seem" is an opinion, but I've always found BBP type formulas interesting:

$$\pi = \sum_{k=0}^\infty \frac{1}{16^k} \left( \frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} - \frac{1}{8k+6}\right)$$

This formula can find arbitrary digits of pi without calculating the previous.

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I like that one too! I believe, though, that it can be used to find hexadecimal digits of pi without calculating the all previous. The relationship between base 16 and base 2 allows you to do the same with binary but I don't believe it works in base 10. –  Mark McClure Feb 25 '14 at 12:02
@MarkMcClure Yes, they are hex digits. A bonus is that it's only practical to use these formulas with computers which can work with binary very efficiently. –  qwr Feb 26 '14 at 2:09
This is called a spigot algorithm. IIRC there was a version for base 10 but I can't find it now. –  DanielV Feb 26 '14 at 6:15

Wallis's Product:

$$\frac{\pi}{2} = \frac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdot8\cdot8\cdot\ldots}{1\cdot 3\cdot3\cdot5\cdot5\cdot7\cdot7\cdot9\cdot\ldots}.$$

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This one has a very cute derivation –  Soke Feb 28 '14 at 23:04

I like this more:


I think it is the simplest form that may have some geometric interpretation. Actually, once I discovered this expression in high school, I spent some time thinking about what the geometric explanation for this might be, but don't think came up with something nice.

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Oh yeah, that's a fun one. When I'm really bored in Calc class (I already know all of the material but have to take the class anyway...) I would compute $\pi$ using this formula. The incredibly slow convergence of this formula made my unproductive Calc days even more unproductive! (I almost managed to calculate it to three decimal points of accuracy one day...) –  Soke Feb 25 '14 at 3:08
Yes, it is slow. But if you are looking for something that might have a geometric explanation, this is the closest I can think of. –  Vadim Feb 25 '14 at 3:10
isn't this the just the series expansion of $\arctan 1$? –  ratchet freak Feb 26 '14 at 13:17
@ratchetfreak Yes, but I do not know a geometric explanation for the series expansion of $\arctan$. –  Vadim Feb 26 '14 at 17:42

You have for example $$ \int_{-\infty}^\infty \dfrac1{1+x^2}\,dx=\pi. $$

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Not fair - that has a rather obvious geometric interpretation. –  nbubis Feb 25 '14 at 3:23
I have no idea what that would be. –  Martin Argerami Feb 25 '14 at 3:30
It's the anti-derivative of $\arctan$, which is equal to the sum of the angles of two infinitely long triangles. –  nbubis Feb 25 '14 at 3:45
It's the derivative, not the anti-derivative, but yes, the geometric / trigonometric relation to $\pi$ is quite evident. –  Soke Feb 25 '14 at 3:48
I drew your example on a large poster and hung it on a wall during a math club event. :) –  neofoxmulder Feb 25 '14 at 6:39

$e^{\sqrt{163}\pi} = 262537412640768743.9999999999992\ldots$

This does not seem geometric, though it is in several ways.

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whats so interesting about 262537412640768743.9999999999992... –  Joao Oct 15 '14 at 6:08
That it is very close to an integer. In fact we have $262537412640768743.9999999999992 = (640319.9999999999999999999999993\ldots)^3 + 744$ but this is of course not easy to see from the number alone, whereas the other fact is. Likewise, we have $e^{\pi\sqrt{67}} = (5279.99999999999998\ldots)^3 + 744$, $e^{\pi\sqrt{43}} = (959.99999999991\ldots)^3 + 744$ and a few more like that, in descending order of spectacularity. –  doetoe Oct 15 '14 at 7:37
+1. Is there any reason or pattern to these numbers? –  Joao Oct 16 '14 at 3:02
@Joao Yes there is. It has to do with the fact that $\Bbb Q(\sqrt{-163})$, $\Bbb Q(\sqrt{-67})$, $\Bbb Q(\sqrt{-43})$ have unique factorization. The values $e^{\sqrt{163}\pi}$ etc. (essentially) are the most significant term of the Fourier coefficients of the j-invariant as a complex function on the complex upper half plane evaluated in $\tau = \frac12 + \frac12\sqrt{-163}$. Writing $q = e^{2\pi i\tau}$ the Fourier series of $j$ starts like $q^{-1} + 744 + 196884q + \cdots$, and for theoretical reasons this has to be an integer for these values. –  doetoe Oct 16 '14 at 7:35
$q$ is very small for the $\tau$ I mentioned, namely $q = -e^{-\sqrt{163}\pi}$, so that $q^{-1}$ should be close to an integer. Note that much more can be said. –  doetoe Oct 16 '14 at 7:40

Sum of reciprocals:



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for the first $\frac{1}{{2n\choose n}}$ the Central Binomial Coefficients an for the second $\frac{n+1}{{2n\choose n}}$ the Catalan Numbers, reciprocals both –  janmarqz Feb 25 '14 at 4:16
even more interesting are the relation with generating functions, see zum Beispiel: juanmarqz.wordpress.com/about/sum-of-reciprocals –  janmarqz Jul 2 '14 at 18:16

Fundamental group

Let $X$ be a topological space and $x \in X$. Then

$$\pi_1(X,x) := \{[\gamma] | \gamma \text{ is a path with } \gamma(0) = x = \gamma(1)\}$$


$$[\gamma_1] * [\gamma_2] = [\gamma_1 * \gamma_2]$$ is $\pi_1(X,x)$ a group and called fundamental group of $X$ in the point $x$.

But that's perhaps a little bit boring as it is only the use of a symbol "$\pi$" and not the constant $3.141...$

Theorem of Gauß-Bonnet

Although the theorem of Gauß-Bonnet is clearly geometrical, it does not seem to be related to circles, so I think it's quite surprising to so $\pi$ here.

Let $S \subseteq \mathbb{R}^3$ be a compact, orientable regular surface. Then: $$\int_S K(s) \mathrm{d}A = 2 \pi \chi(S)$$ where $\chi$ is the Euler-characteristic of $S$ and $K$ is the Gaussian curvature.

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The mentioning of the fundamental group is a joke, right? –  Martin Brandenburg Mar 1 '14 at 12:41
@MartinBrandenburg: user92774 never said he was talking about the constant defined as the ratio of a circle's circumference to its diameter :-) I thought about adding more from this list, but that would be too much. Sometimes I have the feeling that people mix up the representant ($\pi$) with what they represent (the ratio of a circle's circumference to its diameter). That's why I've added the fundamental group. –  moose Mar 1 '14 at 13:21
I wouldn't say that the fundamental group "do not seem geometric", as $\pi_1$ measures the loops/circles. For the Gauss-Bonet, +1 :) –  Peter Franek Jun 27 '14 at 11:35
$+1$ for $\pi_1$ –  miracle173 Sep 20 '14 at 2:49

The "Buffon's needle experiment" says that if a needle of length $l$ is tossed on a paper ruled with lines with $d$ distance apart and equidistant from each other and also $l<d$, then the probability of the needle crossing one of the ruled line is

$${P=\large \frac{2l}{\pi d}}$$

Consequently, if $l=d$, then $\pi$ can be calculated as

$\pi=\Large \frac{2}{P}$, where $P=\Large \frac{\text{number of tosses when the needle crosses on of the lines}}{\text{Total number of tosses}}$

In 1901 the Italian mathematician Mario Lazzarini tried this with 3,408 tosses of the needle and got $\pi = 3.1415929$.

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However, the geometry is not so distant, as the orientation of the needle can be somehow identified with "angle". –  Peter Franek Jun 27 '14 at 11:32

Euler's Formula has already been mentioned but i can't help myself and must give the principle value of

$$\huge{i^i \ = \ (\frac{1}{ \ \ \sqrt{e} \ \ })^{ \pi}} $$

Where $i = \sqrt{-1} $

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This still kind of follows from $e^{\frac{1}{2} i \pi} = i$ which has an obvious geometric interpretation. –  Thomas Feb 25 '14 at 14:29
An imaginary number raised to an imaginary power can be a real number and $ \pi $ shows up. I would think that is a very strange place to find $ \pi $ :) –  neofoxmulder Mar 1 '14 at 12:54

How about $$ e^{i \pi} = -1? $$

I'm not sure what you mean by "geometric". If you mean ratios of circumference to diameter and such, then I think this might fit your criterion. :) Nevertheless, this is such a beautiful formula that I felt it was worth mentioning.

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Well, to me there is a clear connection with complex numbers and geometry: $e^{ix}$ is the graph of a circle of radius $1$ on the complex plane. –  Soke Feb 25 '14 at 2:42
Sure, the image of the set $[0,2\pi),$ under the map $e^{ix},$ is the unit circle. The mentioned identity says that the $\pi \mapsto -1.$ But it's still not so much the "circumference to diameter" business. –  Raghav Feb 25 '14 at 2:47

I always found the "fact" (this is actually a regularized product) that $\infty !=\sqrt{2\pi}$ interesting. See here.

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This is too short for a comment, but this regards the $\frac{\pi}{4}$ series mentioned earlier.

Consider a unit square and a quarter of the unit circle. Cut the square down the diagonal; half of the arc of that circle will equal $\frac{\pi}{4}$. Split $AB$ into $n$ equal pieces. Then it can be shown that $$\lim_{n \to \infty} \displaystyle \sum_{r = 1}^{n} \frac{\frac1n}{1 + (\frac rn)^2} = \frac{\pi}{4}$$

A full explanation is beautifully done in a comment here.

So yes, geometric explanations exist. Circles appear everywhere in mathematics, it's almost scary.

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This is the Riemann sum for $\displaystyle\int_0^1\dfrac{dx}{1+x^2}=[\arctan x]_0^1=\dfrac\pi4$. –  Lucian Mar 1 '14 at 14:27
do you mean too long for a comment? (see your first sentence) –  Joao Apr 30 '14 at 8:31

In statistics, if hypothesis test A requires sample size $n_A$ to attain a certain power $\beta$, and hypothesis test B requires sample size $n_B$, we say that the relative efficiency of test A with respect to test B is $\frac{n_B}{n_A}$. Effectively this tells us how efficiently the tests make use of the information in the samples to draw inferences about the population. If Test A requires only half the sample size that Test B did, then we say it has twice the efficiency. The asymptotic relative efficiency considers how the relative efficiency behaves for increasingly large sample sizes.

Two frequently-used hypothesis tests are the Student's t-test for independent samples (which assumes data are drawn from a normal distribution), and the Wilcoxon-Mann-Whitney U test (which is a non-parametric test: it does not assume a normal distribution). If the data are actually drawn from an exponential distribution, the U test "wins": compared to the t-test it has an ARE of 3, i.e. it is 3 times as efficient.

But if the data are drawn from a normally distributed population, then Student's t-test is on home turf and its power benefits from the data fulfilling its assumption of normality. The U test is now at a slight disadvantage. Compared to the t-test, its ARE drops to $\frac{3}{\pi} \approx 0.955$.

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This is David H's answer in disguise: the ARE is algebraically related to (a) $e^0=1$ and (b) $\int_\mathbb{R}e^{-x^2}dx=\Gamma(1/2)=\sqrt{\pi}$. The appearance of $\pi$ is due to the duplication formula for the Gamma function, $\Gamma(z)\Gamma(1-z)=\pi\csc(\pi z)$, exhibiting $\Gamma$ as a kind of "square root" of a trig function. This is because $\Gamma(z)=\Gamma(z+1)/z$ implies $\Gamma$ has (simple) poles at $0,-1,-2,\ldots$, whence $\Gamma(z)\Gamma(1-z)$ has poles at $\mathbb Z$, strongly suggesting periodicity--which is why $\pi$ should show up! –  whuber Dec 29 '14 at 16:56

$\pi$ as a matrix eigenvalue

For a given positive integer $m$, let us form an $(m-1)\times(m-1)$ dimensional matrix $M$. The terms on the diagonal of $M$ will be $-2m^2$, the terms on the off diagonals of $M$ will be $m^2$ and all other terms will be zero. Thus, $$M=m^2\left( \begin{array}{cccccc} -2 & 1 & 0 & \cdots & \cdots & 0 \\ 1 & -2 & 1 & \cdots & \cdots & 0 \\ 0 & 1 & -2 & 1 & \cdots & 0 \\ \vdots & \cdots & \vdots & \vdots & \cdots & \vdots \\ 0 & \cdots & \cdots & 1 & -2 & 1 \\ 0 & \cdots & \cdots & 0 & 1 & -2 \\ \end{array} \right). $$ For $m=10$, the matrix is $$M = 100 \left( \begin{array}{ccccccccc} -2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & -2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & -2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & -2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & -2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & -2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & -2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 \\ \end{array} \right). $$ If we numerically estimate the eigenvalues of this $M$ (for $m=10$), we get $$ -390.211, -361.803, -317.557, -261.803, -200., -138.197, -82.4429, -38.1966, -9.7887. $$ Of note, they are all negative. If we take the square root of the absolute value of the largest eigenvalue (the last in the above list) we get $3.12869$.

If we perform this exact same process for $m=1000$, we find that the square root of the absolute value of largest eigenvalue is $3.14159$.

It can be proved that the limit as $m\rightarrow\infty$ of this process yields exactly $\pi$.

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This seems pretty geometric to me, given that these matrices are approximations of $f \mapsto f''$... –  Ian Jan 28 at 17:33
@Ian Of course, that is exactly the origin. Does the statement seem geometric as is? Without that knowledge? :) –  Mark McClure Jan 28 at 17:36

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