Interesting and unexpected applications of $\pi$

Question

$\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?

Answer 1

$\pi$ 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.

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 $$\lim_{\delta\rightarrow 0^{+}} N(\delta)\sqrt{\delta} = \pi.$$

Answer 2

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! ;-)

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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 people see magic, math sees design. ;-) Hope all this helps shed some light on the subject.

Answer 3

Unexpected appliactions of pi. pi used as stool and pi used as umbrella stand.

Answer 4

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.

Answer 5

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

Answer 6

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:

$$\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt{\pi}.$$

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

Answer 7

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.

Answer 8

Wallis's Product:

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

Answer 9

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

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

Answer 10

I like this more:

$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}\mp\cdots.$$

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.

Answer 11

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 $\pi = 3.1415929$.

Answer 12

Sum of reciprocals:

$$\frac{4}{3}+\frac{2\pi}{9\sqrt{3}}=1+\frac{1}{2}+\frac{1}{6}+\frac{1}{20}+\frac{1}{70}+\frac{1}{252}+\cdots$$

$$2+\frac{4\sqrt{3}\pi}{27}=1+1+\frac{1}{2}+\frac{1}{5}+\frac{1}{14}+\frac{1}{42}+\frac{1}{132}+\cdots$$

Answer 13

$\pi$ 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 $\pi$.

Answer 14

[This is too long for a comment to Vadim's answer.]

Here is an unexpected appearance of $\pi$:

Two real numbers $x$ and $y$ are chosen at random in the interval $]0,1[$ with respect to the uniform distribution. What is the probability that the closest integer to $x/y$ is even?

(Notice that $x/y$ has the form $n+1/2,n\in\Bbb N$ with probability $0$, so that the question is well-defined.)

The closest integer to $x/y$ is:

• $0$ if $2x<y$
• $2n$ if $\frac{2x}{4n+1}<y<\frac{2x}{4n-1}$ (for some $n≥1$)

Therefore, the probability $p$ is the $1/4$ plus the area of the grey shape:

$$ \bigcup\limits_{n≥1} \left\{ (x,y) \in [0,1]^2 \mid \frac{2x}{4n+1}<y<\frac{2x}{4n-1} \right\} $$

$\qquad\qquad\qquad\qquad$

In other words, we get, using the famous Madhava-Gregory-Leibniz series:

$$p = \frac{1}{4} + \left(\frac{1}{3}-\frac{1}{5}\right) + \left(\frac{1}{7}-\frac{1}{9}\right)+\cdots = \frac{5}{4}-\frac{\pi}{4}= \frac{5-\pi}{4}$$

Source: Putnam 1993, problem B-3.

Answer 15

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

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

where $i = \sqrt{-1}$.

Answer 16

I always found it interesting that $e^{-\zeta'(0)} =\sqrt{2\pi}$, which can be interpreted as the regularized product $1\cdot2\cdot3\cdots \infty = \infty!$. See here.

Answer 17

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

Answer 18

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.

Answer 19

Fundamental group

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

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

With

$$[\gamma_1] * [\gamma_2] = [\gamma_1 * \gamma_2]$$ 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.

Answer 20

This is too long 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.

Answer 21

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 $\frac{3}{\pi} \approx 0.955$.

Answer 22

For each $n\in\mathbb N$, let $P_h(n)$ be the number of primitive Pythagorian triples whose hypothenuse is smaller than $n$ and let $P_p(n)$ be the number of primitive Pythagorian triples whose perimeter is smaller than $n$. In 1900, Lehmer proved that

• $\displaystyle P_h(n)\sim\frac n{2\pi}$;
• $\displaystyle P_p(n)\sim\frac{n\log2}{\pi^2}$.

Answer 23

The first time I observed this fact ,I was totally blown away,kindly see my answer here.

Answer 24

This is due to Euler:$$\frac{\pi}{4} = \frac{3}{4}\cdot\frac{5}{4}\cdot\frac{7}{8}\cdot\frac{11}{12}\cdot\cdot\cdot $$

which can written

$$\pi^* = \prod_{n=1}^\infty\ {p_n}^* $$

where $x^*$ denotes $x$ divided by the multiple of $4$ that's nearest to $x$, and $p_n$ is the $n$th odd prime.

(Here is a derivation from Leibniz's formula, which, ironically, Leibnitz proved geometrically.)

Answer 25

Stirling's Formula : $n!\sim (n/e)^n\sqrt {2\pi n}.$ It is fairly easy to show that $n!\sim (n/e)^n\sqrt {kn}$ for some $k$ but to show $k=2\pi$ is more subtle, and is basically the same as finding the Wallis Product for $\pi.$ Which Wallis did before Newton, that is, before calculus.

Answer 26

Interesting is also the relation of $\pi$ with the Gamma function, like in the reflection formula $$ \Gamma \left( z \right)\,\Gamma \left( {1 - z} \right) = \frac{\pi } {{\sin \left( {\pi \,z} \right)}}\quad \left| {\;\forall z \in {\Bbb C}\;\left( {\backslash {\Bbb Z}} \right)} \right. $$ or, more symmetrically $$ \,\Gamma \left( {\frac{1} {2} + z} \right)\Gamma \left( {\frac{1} {2} - z} \right) = \frac{\pi } {{\cos \left( {\pi \,z} \right)}} $$

Answer 27

This paper gives a polynomial-time approximation algorithm for the Minimum Equivalent Digraph (MEG) problem, with approximation ratio $\pi^2/6$.

The problem is, given a directed graph, to find a min-size subset $S$ of the edges that preserves all reachability relations between pairs of vertices. (That is, for every pair $u, v$ of vertices, if there is a path from $u$ to $v$ in the original graph, then there is such a path that uses only edges in $S$.) The problem is NP-hard. This was the first poly-time algorithm with approximation ratio less than 2.

Answer 28

Dido's problem: find isoperimetric figure with maximum possible area. $$\mathsf{const}=P=\oint \sqrt{x'^2+y'^2}dt$$ $$\mathsf{max}=A=\frac12 \oint (xy'-yx') dt $$

Lagrangian for Euler's equations $0=\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial q'}$ is $$ L= A+\lambda P$$ For $x$ projection it will become $$\frac12 y' - \frac{d}{dt}\left(-\frac{y}{2}+\frac{\lambda x'}{\sqrt{x'^2+y'^2}}\right)=0$$ You can check next but it is pretty obvious from here that it comes to a circle with area of $$A=\frac{P^2}{4\pi}.$$

Answer 29

I find several generalized continued fraction expansions of $\pi$ also quite interesting. It is really astonishing to see how fractions arranged in regular pattern on the RHS gives rise to $\pi$! (which is not only irrational, but transcendental) \begin{align} \displaystyle \pi &={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+{\cfrac {9^{2}}{2+\ddots }}}}}}}}}}}}\\ &={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}\\ &=3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+{\cfrac {7^{2}}{6+{\cfrac {9^{2}}{6+\ddots }}}}}}}}}}\\ &=2+{\cfrac {2}{1+{\cfrac {1}{1/2+{\cfrac {1}{1/3+{\cfrac {1}{1/4+\ddots }}}}}}}}\\ &=2+{\cfrac {2}{1+{\cfrac {1\cdot 2}{1+{\cfrac {2\cdot 3}{1+{\cfrac {3\cdot 4}{1+\ddots }}}}}}}}\\ &=2+{\cfrac {4}{3+{\cfrac {1\cdot 3}{4+{\cfrac {3\cdot 5}{4+{\cfrac {5\cdot 7}{4+\ddots }}}}}}}} \end{align} A particularly interesting that I find is the following: \begin{equation} \pi=3+{\cfrac {1^{3}}{6+{\cfrac {1^{3}+2^{3}}{6+{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}}{6+{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}}{6+{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}+7^{3}+8^{3}}{6+\ddots }}}}}}}}}} \end{equation} Tell me if someone finds a geometrical intuition for these:)

Answer 30

$\pi=2\cdot\int_{0}^{\infty} \frac{\sin t}{t}\,dt =\lim_{x \to \infty}{2\cdot\mathrm {Si}(x)}$