Interesting and unexpected applications of $pi$

I know I'm six and a half years late, but you would enjoy this 3Blue1Brown playlist: https://www.youtube.com/watch?v=d-o3eB9sfls&list=PLZHQObOWTQDMVQcT3414TcPMeEYf_VtPM

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 $$A=frac{P^2}{4pi}.$$

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.

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 $$ ,Gamma left( {frac{1} {2} + z} right)Gamma left( {frac{1} {2} - z} right) = frac{pi } {{cos left( {pi ,z} right)}} $$

For each $ninmathbb N$, 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

• $displaystyle P_h(n)simfrac n{2pi}$;
• $displaystyle P_p(n)simfrac{nlog2}{pi^2}$.

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

$$pi=2cdotint_{0}^{infty} frac{sin t}{t},dt =lim_{x to infty}{2cdotmathrm {Si}(x)}$$ enter image description here

[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 $n+1/2,ninBbb N$ 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:

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

$qquadqquadqquadqquad$

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.

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

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.)

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

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

$pi$ 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, $$M=m^2left( begin{array}{cccccc} -2 & 1 & 0 & cdots & cdots & 0 \ 1 & -2 & 1 & cdots & cdots & 0 \ 0 & 1 & -2 & 1 & cdots & 0 \ vdots & cdots & vdots & vdots & cdots & vdots \ 0 & cdots & cdots & 1 & -2 & 1 \ 0 & cdots & cdots & 0 & 1 & -2 \ end{array} right). $$ 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 $mrightarrowinfty$ of this process yields exactly $pi$.

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

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 $pi = 3.1415929$.

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.

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

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

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

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.

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 $i = sqrt{-1}$.

I always found it interesting that $e^{-zeta'(0)} =sqrt{2pi}$, which can be interpreted as the regularized product $1cdot2cdot3cdots infty = infty!$. See here.

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

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,

$$frac{1}{pi} = frac{2sqrt{2}}{9801} sum_{kgeq0} frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}.$$

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

Sum of reciprocals:

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

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

Wallis's Product:

$$frac{pi}{2} = frac{2cdot2cdot4cdot4cdot6cdot6cdot8cdot8cdotldots}{1cdot 3cdot3cdot5cdot5cdot7cdot7cdot9cdotldots}.$$

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.

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.

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

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.

I like this more:

$$frac{pi}{4}=1-frac{1}{3}+frac{1}{5}-frac{1}{7}+frac{1}{9}mpcdots.$$

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.