# Interesting and unexpected applications of $\pi$

$\text{What are some interesting cases of$\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?

• @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. Feb 25, 2014 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. Feb 25, 2014 at 3:08
• For the fun of it: $\pi=\left[\frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}\right]^{-1}$
– ocg
Feb 25, 2014 at 3:35
• @JulienGodawatta Good ol' Ramanujan.
– MT_
Feb 25, 2014 at 3:47
• I wonder why this interesting question "where does $\pi$ pop up unexpectedly" has been misinterpreted as "what are fancy formulas for $\pi$" (except for two answers). Mar 1, 2014 at 12:43

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

• That is amazing! Feb 25, 2014 at 6:59
• There seems to be an error in the third row of the tabulated values: When $\delta=10^{-6}$, we have $31415\sqrt\delta=31.415\approx 10\pi$. Feb 25, 2014 at 8:35
• @JohnBentin Thanks! Feb 25, 2014 at 11:28

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

• "Bounded sums of squares"...? Feb 25, 2014 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. Feb 25, 2014 at 13:31
• ;-) okay that's a little bit too much
– qwr
Feb 27, 2014 at 23:54
• The number of links makes it seem like this is Wikipedia!
– user122283
Apr 20, 2014 at 19:08
• This is very nice. Aug 3, 2016 at 18:54

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

• Indeed, I didn't expect those. Aug 3, 2016 at 19:01

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.

• Is there any interpretation for this? How does one derive this?
– MT_
Feb 28, 2014 at 23:13
• @user92774: It's based on modular forms and the monster group. See here. Mar 1, 2014 at 14:31

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

• This is based on the Euler product formula for the Riemann $\zeta$ function. Mar 1, 2014 at 14:24
• The result and the proof still do not seem geometric. Mar 1, 2014 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 $\pi$. Mar 1, 2014 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? Mar 4, 2014 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). Mar 4, 2014 at 13:01

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.

• At some deep level this is how you get $\sqrt{\pi}$ in Stirling's formula. Feb 25, 2014 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. Feb 25, 2014 at 21:46
• @Emily "However, the Gaussian is quite geometric": can you substantiate this ?
– user65203
Feb 2, 2017 at 15:40
• Consider computing its square (for which the answer is $\pi$), and the circle becomes clear by looking at the exponent. If it's not obvious, convert to polar coordinates. Apr 2, 2019 at 7:44

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.

• 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. Feb 25, 2014 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, 2014 at 2:09
• This is called a spigot algorithm. IIRC there was a version for base 10 but I can't find it now. Feb 26, 2014 at 6:15
• @DanielV, could it be that you are thinking of jstor.org/stable/2975006 ? Aug 2, 2015 at 21:29
• @LSpice Great article! Here is the non-paywall version: dept.cs.williams.edu/~heeringa/classes/cs135/s15/readings/…
– qwr
Aug 13, 2015 at 1:56

Wallis's Product:

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

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

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

• whats so interesting about 262537412640768743.9999999999992...
– Joao
Oct 15, 2014 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. Oct 15, 2014 at 7:37
• +1. Is there any reason or pattern to these numbers?
– Joao
Oct 16, 2014 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. Oct 16, 2014 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. Oct 16, 2014 at 7:40

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.

• 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...)
– MT_
Feb 25, 2014 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. Feb 25, 2014 at 3:10
• isn't this the just the series expansion of $\arctan 1$? Feb 26, 2014 at 13:17
• @ratchetfreak Yes, but I do not know a geometric explanation for the series expansion of $\arctan$. Feb 26, 2014 at 17:42
• Geometric interpretation: math.stackexchange.com/questions/682532/… May 12, 2017 at 0:31

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 $\pi = 3.1415929$.

• However, the geometry is not so distant, as the orientation of the needle can be somehow identified with "angle". Jun 27, 2014 at 11:32
• Note that Lazzarini's claim is dubious. Jan 15, 2021 at 14:04
• It looks like exactly 2169.6 out of his 3408 tosses had the needle cross the line! Sep 30, 2021 at 21:19

Sum of reciprocals:

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

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

• 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 Feb 25, 2014 at 4:16
• even more interesting are the relation with generating functions, see zum Beispiel: juanmarqz.wordpress.com/about/sum-of-reciprocals Jul 2, 2014 at 18:16
• In general, $~\displaystyle\sum_{n=1}^\infty\dfrac{(2x)^{2n}}{\displaystyle{2n\choose n}~n^2}~=~2\arcsin^2x.~$ By repeatedly differentiating with regard to $x,~$ we can deduce the general formulas behind the two provided identities. Feb 2, 2017 at 18:14
• @Lucian, just like that? well if you can read spanish you can find other kind of details at researchgate.net/profile/Jan_Marqz_Bobadilla/project/BSc-theses/… Feb 2, 2017 at 20:35

## $\pi$ 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 $\pi$.

• This seems pretty geometric to me, given that these matrices are approximations of $f \mapsto f''$...
– Ian
Jan 28, 2015 at 17:33
• @Ian Of course, that is exactly the origin. Does the statement seem geometric as is? Without that knowledge? :) Jan 28, 2015 at 17:36
• "If we numerically estimate the eigenvalues of this $M$" - You don't need to numerically estimate them. $M$ is a Toeplitz Tri-Diagonal matrix, whose eigenvalues can be found exactly: $$\lambda_k = -2 + 2\cos(k\pi/m)$$ in this case, where $1 \le k < m$. Aug 20, 2015 at 14:48
• @wltrup I think you forgot to multiply by $m^2$, but yeah. Aug 20, 2015 at 15:18
• @MarkMcClure Ha.. yes. I didn't see the $m^2$ in front. Thanks for pointing that out. Aug 20, 2015 at 15:19

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.

• (Another unexpected appearance of $\pi$, but much more well-known that the above one, is : the probability of a random integer to be square-free is $6 / \pi^2$) May 28, 2018 at 21:11
• Another example: what is the average distance between two random points in the unit cube? If you think that this is unrelated to $\pi$ and to logarithms, see here. Nov 29, 2018 at 17:34
• Another example: see this interesting collision problem explained by 3Blue1Brown (youtube.com/watch?v=HEfHFsfGXjs), inspired by a paper of Gregory Galperin. Feb 14 at 8:48

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

• This still kind of follows from $e^{\frac{1}{2} i \pi} = i$ which has an obvious geometric interpretation. Feb 25, 2014 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 $\pi$ :) Mar 1, 2014 at 12:54

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.

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

• Not fair - that has a rather obvious geometric interpretation. Feb 25, 2014 at 3:23
• I have no idea what that would be. Feb 25, 2014 at 3:30
• I drew your example on a large poster and hung it on a wall during a math club event. :) Feb 25, 2014 at 6:39
• The really interesting break from geometry here occurs at the point where you prove that the anti-derivative of $\arctan$ can be expressed without any trig functions. This is just a corollary. Feb 25, 2014 at 12:03
• The equivalence of areas can be shown without trig and calculus, by mere triangles similarity in the greek style. If this helps: i.imgur.com/amN1Exb.png Mar 20, 2014 at 15:04

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.

• 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.
– MT_
Feb 25, 2014 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. Feb 25, 2014 at 2:47
• Well, it sort of is; the statement is that, if you want to trace out a circle of radius $r = 1$ (i.e., $d = 2$), then you need to consider $x$-values from $0$ to $2\pi$ to do it (i.e., $C = 2\pi$). The mere numerical coincidence doesn't prove it—one needs also to know that $x \mapsto e^{i x}$ traces out the unit circle at unit speed—but the connexion is surely there. Aug 2, 2015 at 21:43

## 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.

• The mentioning of the fundamental group is a joke, right? Mar 1, 2014 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. Mar 1, 2014 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 :) Jun 27, 2014 at 11:35
• $+1$ for $\pi_1$ Sep 20, 2014 at 2:49
• @MartinThoma Hey, this is OP (I've since changed my name). Just learned about the fundamental group and couldn't help but think about this post! :) That said I don't think people "mix up the representant with what they represent" -- isn't the point of notation so that I can write $\pi$ and not "the ratio of the circumference of a circle and the diameter of a circle of same radius", which is pretty reminiscent of ancient math textbooks...
– MT_
Nov 1, 2016 at 0:10

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.

• This is the Riemann sum for $\displaystyle\int_0^1\dfrac{dx}{1+x^2}=[\arctan x]_0^1=\dfrac\pi4$. Mar 1, 2014 at 14:27
• do you mean too long for a comment? (see your first sentence)
– Joao
Apr 30, 2014 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 $\frac{3}{\pi} \approx 0.955$.

• 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! Dec 29, 2014 at 16:56

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

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

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

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.)

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.

• At least before the full generality of the Fundamental Theorem of Calculus was known. Aug 4, 2019 at 12:54

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

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.

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

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: $$$$\pi=3+{\cfrac {1^{3}}{6+{\cfrac {1^{3}+2^{3}}{6+{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}}{6+{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}}{6+{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}+7^{3}+8^{3}}{6+\ddots }}}}}}}}}}$$$$ Tell me if someone finds a geometrical intuition for these:)

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