User:AkanoToE/List of formulae involving τ

The following is a list of significant formulae involving the mathematical constant $τ$.

Euclidean geometry

 * $$\tau = \frac Cr $$

where $C$ is the circumference of a circle, $r$ is the radius. More generally,
 * $$\tau=\frac{L}{w/2}$$

where $L$ and $w$ are, respectively, the perimeter and the width of any curve of constant width.


 * $$A = \frac{1}{2}\tau r^2$$

where $A$ is the area of a circle and $r$ is the radius. More generally,


 * $$A = \frac{1}{2}\tau ab$$

where $A$ is the area enclosed by an ellipse with semi-major axis $a$ and semi-minor axis $b$.


 * $$A=2\tau r^2$$

where $A$ is the area between the witch of Agnesi and its asymptotic line; $r$ is the radius of the defining circle.


 * $$A=\frac{\Gamma (1/4)^2}{\sqrt{2\tau}} r^2=\frac{\tau r^2}{2\operatorname{agm}(1,1/\sqrt{2})}$$

where $A$ is the area of a squircle with minor radius $r$, $$\Gamma$$ is the gamma function and $$\operatorname{agm}$$ is the arithmetic–geometric mean.


 * $$A=\frac{(k+1)(k+2)}{2}\tau r^2 = T_{k+1}\tau r^2$$

where $A$ is the area of an epicycloid with the smaller circle of radius $r$ and the larger circle of radius $kr$ ($$k\in\mathbb{N}$$), assuming the initial point lies on the larger circle, and $$T_n$$ is the $n^{th}$ Triangular number.


 * $$A=\frac{(-1)^k+3}{16}\tau a^2$$

where $A$ is the area of a rose with angular frequency $k$ ($$k\in\mathbb{N}$$) and amplitude $a$.


 * $$L=\sqrt{\frac{2}{\tau}}\,\Gamma (1/4)^2c=\frac{\tau c}{\operatorname{agm}(1,1/\sqrt{2})}$$

where $L$ is the perimeter of the lemniscate of Bernoulli with focal distance $c$.


 * $$V = {2 \over 3}\tau r^3$$

where $V$ is the volume of a sphere and $r$ is the radius.


 * $$SA = 2\tau r^2$$

where $SA$ is the surface area of a sphere and $r$ is the radius.


 * $$H = {1 \over 8}\tau^2 r^4$$

where $H$ is the hypervolume of a 3-sphere and $r$ is the radius.


 * $$SV = {1\over 2} \tau^2 r^3$$

where $SV$ is the surface volume of a 3-sphere and $r$ is the radius.

Regular convex polygons
Sum $S$ of internal angles of a regular convex polygon with $n$ sides:
 * $$S=\left({n\over 2}-1\right)\tau$$

Area $A$ of a regular convex polygon with $n$ sides and side length $s$:
 * $$A=\frac{ns^2}{4}\cot\frac{\tau}{2n}$$

Inradius $r$ of a regular convex polygon with $n$ sides and side length $s$:
 * $$r=\frac{s}{2}\cot\frac{\tau}{2n}$$

Circumradius $R$ of a regular convex polygon with $n$ sides and side length $s$:
 * $$R=\frac{s}{2}\csc\frac{\tau}{2n}$$

Physics

 * The cosmological constant:
 * $$\Lambda = {{4\tau G} \over {3c^2}} \rho$$


 * Heisenberg's uncertainty principle:
 * $$ \Delta x\, \Delta p \ge \frac h {2\tau} $$


 * Einstein's field equation of general relativity:
 * $$ R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R + \Lambda g_{\mu\nu} = {4 \tau G \over c^4} T_{\mu\nu} $$


 * Coulomb's law for the electric force in vacuum:
 * $$ F = \frac{|q_1q_2|}{2 \tau \varepsilon_0 r^2}$$


 * Magnetic permeability of free space:
 * $$ \mu_0 \approx 2 \tau \cdot 10^{-7}\,\mathrm{N}/\mathrm{A}^2$$


 * Approximate period of a simple pendulum with small amplitude:
 * $$T \approx \tau \sqrt\frac L g $$


 * Exact period of a simple pendulum with amplitude $$\theta_0$$ ($$\operatorname{agm}$$ is the arithmetic–geometric mean):
 * $$T=\frac{\tau}{\operatorname{agm}(1,\cos (\theta_0/2))}\sqrt{\frac{L}{g}}$$


 * Kepler's third law of planetary motion:
 * $$\frac{R^3}{T^2} = \frac{GM}{\tau^2}$$


 * The buckling formula:
 * $$F =\frac{\tau^2EI}{4L^2}$$

Integrals

 * $$2 \int_{-1}^1 \sqrt{1-x^2}\,dx = \pi $$ (integrating two halves $$y(x)=\sqrt{1-x^2}$$ to obtain the area of the unit circle)


 * $$\int_{-\infty}^\infty \operatorname{sech}x \, dx = \pi $$


 * $$\int_{-\infty}^\infty \int_t^\infty e^{-1/2t^2-x^2+xt} \, dx \, dt = \int_{-\infty}^\infty \int_t^\infty e^{-t^2-1/2x^2+xt} \, dx \, dt = \pi$$


 * $$\int_{-1}^1\frac{dx}{\sqrt{1-x^2}} = \pi$$


 * $$\int_{-\infty}^\infty\frac{dx}{1+x^2} = \pi$$ (see also Cauchy distribution)


 * $$\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}$$ (see Gaussian integral).


 * $$\oint\frac{dz} z = 2\pi i$$ (when the path of integration winds once counterclockwise around 0. See also Cauchy's integral formula).


 * $$\int_0^\infty \ln\left(1+\frac{1}{x^2}\right)\, dx=\pi$$


 * $$\int_{-\infty}^\infty \frac{\sin x} x \,dx=\pi $$


 * $$\int_0^1 {x^4(1-x)^4 \over 1+x^2}\,dx = {22 \over 7} - \pi$$ (see also Proof that 22/7 exceeds π).


 * $$\int_0^\infty \frac{x^{\alpha-1}}{x+1}\, dx=\frac{\pi}{\sin \pi\alpha},\quad 0<\alpha<1$$


 * $$\int_0^\infty \frac{dx}{\sqrt{x(x+a)(x+b)}}=\frac{\pi}{\operatorname{agm}(\sqrt{a},\sqrt{b})}$$ (where $$\operatorname{agm}$$ is the arithmetic–geometric mean; see also elliptic integral)

Note that with symmetric integrands $$f(-x)=f(x)$$, formulas of the form $\int_{-a}^af(x)\,dx$ can also be translated to formulas $2\int_{0}^af(x)\,dx$.

Efficient infinite series

 * $$\sum_{k=0}^\infty \frac{k!}{(2k+1)!!} = \sum_{k=0}^\infty\frac{2^k k!^2}{(2k+1)!} = \frac \pi 2 $$ (see also Double factorial)


 * $$12 \sum^\infty_{k=0} \frac{(-1)^k (6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 640320^{3k + 3/2}}=\frac 1 \pi $$ (see Chudnovsky algorithm)


 * $$\frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}=\frac 1 \pi $$ (see Srinivasa Ramanujan, Ramanujan–Sato series)

The following are efficient for calculating arbitrary binary digits of π:


 * $$\sum_{k=0}^\infty \frac{(-1)^k}{4^k}\left(\frac{2}{4k+1}+\frac{2}{4k+2}+\frac{1}{4k+3}\right)=\pi$$


 * $$\sum_{k = 0}^{\infty} \frac{1}{16^k} \left( \frac{4}{8k + 1} - \frac{2}{8k + 4} - \frac{1}{8k + 5} - \frac{1}{8k + 6}\right)=\pi$$ (see Bailey–Borwein–Plouffe formula)


 * $$\frac{1}{2^6} \sum_{k=0}^{\infty} \frac{{(-1)}^k}{2^{10k}} \left( - \frac{2^5}{4k+1} - \frac{1}{4k+3} + \frac{2^8}{10k+1} - \frac{2^6}{10k+3} - \frac{2^2}{10k+5} - \frac{2^2}{10k+7} + \frac{1}{10k+9} \right)=\pi$$

Plouffe's series for calculating arbitrary decimal digits of π:


 * $$\sum_{k=1}^\infty k\frac{2^kk!^2}{(2k)!}=\pi +3$$

Other infinite series

 * $$\zeta(2) = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6}$$ (see also Basel problem and Riemann zeta function)


 * $$\zeta(4)= \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots = \frac{\pi^4}{90}$$


 * $$\zeta(2n) = \sum_{k=1}^{\infty} \frac{1}{k^{2n}}\, = \frac{1}{1^{2n}} + \frac{1}{2^{2n}} + \frac{1}{3^{2n}} + \frac{1}{4^{2n}} + \cdots = (-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}$$, where B2n is a Bernoulli number.


 * $$\sum_{n=1}^\infty \frac{3^n - 1}{4^n}\, \zeta(n+1) = \pi$$


 * $$\sum_{n=2}^\infty \frac{2(3/2)^n-3}{n}(\zeta (n)-1)=\ln \pi$$


 * $$\sum_{n=1}^\infty \zeta (2n)\frac{x^{2n}}{n}=\ln\frac{\pi x}{\sin \pi x},\quad 0<|x|<1$$


 * $$\sum_{n=0}^\infty \frac{(-1)^{n}}{2n+1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots = \arctan{1} = \frac{\pi}{4}$$ (see Leibniz formula for pi)


 * $$\sum_{n=0}^\infty \frac{(-1)^{(n^2-n)/2}}{2n+1}=1+\frac13-\frac15-\frac17+\frac19+\frac{1}{11}-\cdots=\frac{\pi}{2\sqrt{2}}$$ (Newton, Second Letter to Oldenburg, 1676)


 * $$\sum_{n=0}^\infty \frac{(-1)^n}{3^n(2n+1)}=1-\frac{1}{3^1\cdot 3}+\frac{1}{3^2\cdot 5}-\frac{1}{3^3\cdot 7}+\frac{1}{3^4\cdot 9}-\cdots =\sqrt{3}\arctan\frac{1}{\sqrt{3}}=\frac{\pi}{2\sqrt{3}}$$ (Madhava series)


 * $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}=\frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \cdots=\frac{\pi^2}{12}$$


 * $$\sum_{n=1}^\infty \frac1{(2n)^2} = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \frac{1}{8^2} + \cdots = \frac{\pi^2}{24}$$


 * $$\sum_{n=0}^\infty \left( \frac{(-1)^{n}}{2n+1} \right)^2 = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots = \frac{\pi^2}{8}$$


 * $$\sum_{n=0}^\infty \left( \frac{(-1)^{n}}{2n+1} \right)^3 = \frac{1}{1^3} - \frac{1}{3^3} + \frac{1}{5^3} - \frac{1}{7^3} + \cdots = \frac{\pi^3}{32}$$


 * $$\sum_{n=0}^\infty \left( \frac{(-1)^{n}}{2n+1} \right)^4 = \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \frac{1}{7^4} + \cdots = \frac{\pi^4}{96}$$


 * $$\sum_{n=0}^\infty \left( \frac{(-1)^{n}}{2n+1} \right)^5 = \frac{1}{1^5} - \frac{1}{3^5} + \frac{1}{5^5} - \frac{1}{7^5} + \cdots = \frac{5\pi^5}{1536}$$


 * $$\sum_{n=0}^\infty \left( \frac{(-1)^{n}}{2n+1} \right)^6 = \frac{1}{1^6} + \frac{1}{3^6} + \frac{1}{5^6} + \frac{1}{7^6} + \cdots = \frac{\pi^6}{960}$$


 * $$\sum_{n=0}^\infty \binom{\frac{1}{2}}{n}\frac{(-1)^n}{2n+1} = 1 - \frac{1}{6} - \frac{1}{40}-\cdots = \frac{\pi}{4}$$


 * $$\sum_{n=0}^\infty \frac{1}{(4n+1)(4n+3)} = \frac{1}{1\cdot 3}+\frac{1}{5\cdot 7} +\frac{1}{9\cdot 11} +\cdots=\frac{\pi}{8}$$


 * $$ \sum_{n=1}^\infty (-1)^{(n^2+n)/2+1}\left|G_{\left((-1)^{n+1}+6n-3\right)/4}\right|=|G_1|+|G_2|-|G_4|-|G_5|+|G_7|+|G_8|-|G_{10}|-|G_{11}|+\cdots =\frac{\sqrt{3}}{\pi} $$ (see Gregory coefficients)


 * $$ \sum_{n=0}^\infty \frac{(1/2)_n^2}{2^n n!^2}\sum_{n=0}^\infty \frac{n(1/2)_n^2}{2^n n!^2}=\frac{1}{\pi}$$ (where $$(x)_n$$ is the rising factorial)


 * $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)(2n+1)}=\pi -3$$ (Nilakantha series)


 * $$\sum_{n=1}^\infty \frac{F_{2n}}{n^2 \binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5}$$ (where $$F_n$$ is the n-th Fibonacci number)


 * $$ \pi = \sum_{n=1}^\infty \frac{(-1)^{\epsilon (n)}}{n}=1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} - \frac{1}{10} + \frac{1}{11} + \frac{1}{12} - \frac{1}{13} + \cdots $$  (where $$\epsilon (n)$$ is the number of prime factors of the form $$p\equiv 1\,(\mathrm{mod}\,4)$$ of $$n$$)


 * $$\frac{\pi}{2}=\sum_{n=1}^\infty \frac{(-1)^{\varepsilon (n)}}{n}=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\cdots$$  (where $$\varepsilon (n)$$ is the number of prime factors of the form $$p\equiv 3\, (\mathrm{mod}\, 4)$$ of $$n$$)


 * $$\pi=\sum_{n=-\infty}^\infty \frac{(-1)^n}{n+1/2}$$


 * $$\pi^2=\sum_{n=-\infty}^\infty \frac{1}{(n+1/2)^2}$$

The last two formulas are special cases of


 * $$\begin{align}\frac{\pi}{\sin\pi x}&=\sum_{n=-\infty}^\infty \frac{(-1)^n}{n+x}\\

\left(\frac{\pi}{\sin \pi x}\right)^2&=\sum_{n=-\infty}^\infty \frac{1}{(n+x)^2}\end{align}$$

which generate infinitely many analogous formulas for $$\pi$$ when $$x\in\mathbb{Q}\setminus\mathbb{Z}.$$

Some formulas relating π and harmonic numbers are given here. Further infinite series involving π are:

where $$(x)_n $$ is the Pochhammer symbol for the rising factorial. See also Ramanujan–Sato series.

Machin-like formulae

 * $$\frac{\pi}{4} = \arctan 1$$


 * $$\frac{\pi}{4} = \arctan\frac{1}{2} + \arctan\frac{1}{3}$$


 * $$\frac{\pi}{4} = 2 \arctan\frac{1}{2} - \arctan\frac{1}{7}$$


 * $$\frac{\pi}{4} = 2 \arctan\frac{1}{3} + \arctan\frac{1}{7}$$


 * $$\frac{\pi}{4} = 4 \arctan\frac{1}{5} - \arctan\frac{1}{239} $$ (the original Machin's formula)


 * $$\frac{\pi}{4} = 5 \arctan\frac{1}{7} + 2 \arctan\frac{3}{79}$$


 * $$\frac{\pi}{4} = 6 \arctan\frac{1}{8} + 2 \arctan\frac{1}{57} + \arctan\frac{1}{239}$$


 * $$\frac{\pi}{4} = 12 \arctan\frac{1}{49} + 32 \arctan\frac{1}{57} - 5 \arctan\frac{1}{239} + 12 \arctan\frac{1}{110443}$$


 * $$\frac{\pi}{4} = 44 \arctan\frac{1}{57} + 7 \arctan\frac{1}{239} - 12 \arctan\frac{1}{682} + 24 \arctan\frac{1}{12943}$$

Infinite products

 * $$\frac{\pi}{4} = \left(\prod_{p\equiv 1\pmod 4}\frac{p}{p-1}\right)\cdot\left( \prod_{p\equiv 3\pmod 4}\frac{p}{p+1}\right)=\frac{3}{4} \cdot \frac{5}{4} \cdot \frac{7}{8} \cdot \frac{11}{12} \cdot \frac{13}{12} \cdots,$$ (Euler)
 * where the numerators are the odd primes; each denominator is the multiple of four nearest to the numerator.


 * $$\frac{\sqrt{3}\pi}{6}=\left(\displaystyle\prod_{p \equiv 1 \pmod{6} \atop p \in \mathbb{P} } \frac{p}{p-1}\right) \cdot \left(\displaystyle\prod_{p \equiv 5 \pmod{6} \atop p \in \mathbb{P} } \frac{p}{p+1}\right)=\frac{5}{6} \cdot \frac{7}{6} \cdot \frac{11}{12} \cdot \frac{13}{12} \cdot \frac{17}{18} \cdots ,$$


 * $$\frac{\pi}{2}=\prod_{n=1}^{\infty} \frac{(2n)(2n)}{(2n-1)(2n+1)} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots$$ (see also Wallis product)


 * $$\frac{\pi}{2}=\prod_{n=1}^\infty\left(1+\frac{1}{n}\right)^{(-1)^{n+1}}=\left(1+\frac{1}{1}\right)^{+1}\left(1+\frac{1}{2}\right)^{-1}\left(1+\frac{1}{3}\right)^{+1}\cdots$$ (another form of Wallis product)

Viète's formula:
 * $$\frac{2}{\pi}=\frac{\sqrt2}2 \cdot \frac{\sqrt{2+\sqrt2}}2 \cdot \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2 \cdot \cdots$$

A double infinite product formula involving the Thue–Morse sequence:
 * $$\frac{\pi}{2}=\prod_{m\geq1} \prod_{n\geq1} \left( \frac{(4 m^2 + n - 2) (4 m^2 + 2 n - 1)^2}{4 (2 m^2 + n - 1) (4 m^2 + n - 1) (2 m^2 + n)} \right) ^{\epsilon_n},$$
 * where $$\epsilon_n = (-1)^{t_n}$$ and $$t_n$$ is the Thue–Morse sequence.

Arctangent formulas

 * $$ \frac{\pi}{2^{k+1}}=\arctan \frac{\sqrt{2-a_{k-1}}}{a_k}, \qquad\qquad k\geq 2 $$
 * $$ \frac{\pi}{4}=\sum_{k\geq 2}\arctan \frac{\sqrt{2-a_{k-1}}}{a_k}, $$

where $$ a_k=\sqrt{2+a_{k-1}} $$ such that $$ a_1=\sqrt{2} $$.


 * $$\frac{\pi}{2} = \sum_{k=0}^\infty \arctan\frac{1}{F_{2k+1}} = \arctan\frac{1}{1} + \arctan\frac{1}{2} + \arctan\frac{1}{5} + \arctan\frac{1}{13} + \cdots $$

where $$F_k$$ is the k-th Fibonacci number.


 * $$ \pi =\arctan a+\arctan b+\arctan c$$

whenever $$a+b+c=abc$$ and $$a$$, $$b$$, $$c$$ are positive real numbers (see List of trigonometric identities). A special case is
 * $$\pi =\arctan 1+\arctan 2+\arctan 3.$$

Complex exponential formulas

 * $$e^{i \tau} = 1$$ (Euler's identity)

The following equivalences are true for any complex $$z$$:


 * $$e^z\in\mathbb{R}\leftrightarrow\Im z\in {1 \over 2} \tau\mathbb{Z}$$
 * $$e^z=1\leftrightarrow z\in \tau i\mathbb{Z}$$

Continued fractions


\tau= {6 + \cfrac{1^2}{3 + \cfrac{3^2}{12 + \cfrac{5^2}{3 + \cfrac{7^2}{12 + \ddots\,}}}}} $$



\tau = \cfrac{8}{1 + \cfrac{1^2}{3 + \cfrac{2^2}{5 + \cfrac{3^2}{7 + \cfrac{4^2}{9 + \ddots}}}}} $$



\tau = \cfrac{8}{1 + \cfrac{1^2}{2 + \cfrac{3^2}{2 + \cfrac{5^2}{2 + \cfrac{7^2}{2 + \ddots}}}}}$$



\tau = {6 + \cfrac{2^2}{12 + \cfrac{6^2}{12 + \cfrac{10^2}{12+ \cfrac{14^2}{12 + \cfrac{18^2}{12 + \ddots}}}}}} $$ For more on the third identity, see Euler's continued fraction formula.

(See also Continued fraction and Generalized continued fraction.)

Iterative algorithms

 * $$ a_0=1,\, a_{n+1}=\left(1+\frac{1}{2n+1}\right)a_n,\, \pi=\lim_{n\to\infty}\frac{a_n^2}{n}$$


 * $$a_1=0,\, a_{n+1}=\sqrt{2+a_n},\, \pi =\lim_{n\to\infty} 2^n\sqrt{2-a_n}$$ (closely related to Viète's formula)
 * $$\omega(i_n,i_{n-1},\dots,i_{1})=2+i_{n} \sqrt{2+i_{n-1} \sqrt{2+\cdots+i_{1} \sqrt{2}}}=\omega(b_n,b_{n-1},\dots,b_{1}),\, i_{k} \in\{-1,1\}, \, b_k=\begin{cases}

0& \text{if } i_k=1\\ 1& \text{if } i_k=-1 \end{cases}, \, \pi={\displaystyle\lim _{n \rightarrow \infty} \frac{2^{n+1}}{2 h+1} \sqrt{\omega\left(\underbrace{10 \ldots 0}_{n-m} g_{m, h+1}\right)}} $$ (where $$g_{m, h+1} $$ is the h+1-th entry of m-bit Gray code, $$h \in \left\{0,1, \ldots, 2^{m}-1\right\} $$)
 * $$a_1=1,\, a_{n+1}=a_n+\sin a_n,\, \pi =\lim_{n\to\infty}a_n$$ (cubic convergence)


 * $$a_0=2\sqrt{3},\, b_0=3,\, a_{n+1}=\operatorname{hm}(a_n,b_n),\, b_{n+1}=\operatorname{gm}(a_{n+1},b_n),\, \pi =\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n$$ (Archimedes' algorithm, see also harmonic mean and geometric mean)

For more iterative algorithms, see the Gauss–Legendre algorithm and Borwein's algorithm.

Asymptotics

 * $$\binom{2n}{n}\sim \frac{4^{n}}{\sqrt{\pi n}}$$ (asymptotic growth rate of the central binomial coefficients)


 * $$C_n\sim \frac{4^{n}}{\sqrt{\pi n^3}}$$ (asymptotic growth rate of the Catalan numbers)


 * $$n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$$ (Stirling's approximation)


 * $$\sum_{k=1}^{n} \varphi (k) \sim \frac{3n^2}{\pi^2}$$ (where $$\varphi$$ is Euler's totient function)


 * $$\sum_{k=1}^{n} \frac {\varphi (k)} {k} \sim \frac{6n}{\pi^2}$$

Miscellaneous

 * $$\Gamma (s)\Gamma (1-s)=\frac{\pi}{\sin \pi s}$$ (Euler's reflection formula, see Gamma function)


 * $$\pi^{-s/2}\Gamma \left(\frac{s}{2}\right)\zeta (s)=\pi^{-(1-s)/2}\Gamma\left(\frac{1-s}{2}\right)\zeta (1-s)$$ (the functional equation of the Riemann zeta function)


 * $$e^{-\zeta'(0)}=\sqrt{2\pi}$$


 * $$e^{\zeta'(0,1/2)-\zeta'(0,1)}=\sqrt{\pi}$$ (where $$\zeta (s,a)$$ is the Hurwitz zeta function and the derivative is taken with respect to the first variable)


 * $$\pi =\Beta (1/2,1/2)=\Gamma (1/2)^2$$ (see also Beta function)


 * $$\pi = \frac{\Gamma (3/4)^4}{\operatorname{agm}(1,1/\sqrt{2})^2}=\frac{\Gamma\left({1/4}\right)^{4/3} \operatorname{agm}(1, \sqrt{2})^{2/3}}{2}$$ (where agm is the arithmetic–geometric mean)


 * $$\pi = \operatorname{agm}\left(\theta_2^2(1/e),\theta_3^2(1/e)\right)$$ (where $$\theta_2$$ and $$\theta_3$$ are the Jacobi theta functions )


 * $$\pi=-\frac{\operatorname{K}(k)}{\operatorname{K}\left(\sqrt{1-k^2}\right)}\ln q,\quad k=\frac{\theta_2^2(q)}{\theta_3^2(q)}$$ (where $$q\in (0,1)$$ and $$\operatorname{K}(k)$$ is the complete elliptic integral of the first kind with modulus $$k$$; reflecting the nome-modulus inversion problem)


 * $$\pi =-\frac{\operatorname{agm}\left(1,\sqrt{1-k'^2}\right)}{\operatorname{agm}(1,k')}\ln q,\quad k'=\frac{\theta_4^2(q)}{\theta_3^2(q)}$$ (where $$q\in (0,1)$$)


 * $$\operatorname{agm}(1,\sqrt{2})=\frac{\pi}{\varpi}$$ (due to Gauss, $$\varpi$$ is the lemniscate constant)


 * $$i\pi=\operatorname{Log}(-1)=\lim_{n\to\infty}n\left((-1)^{1/n}-1\right)$$ (where $$\operatorname{Log}$$ is the principal value of the complex logarithm)


 * $$1-\frac{\pi^2}{12}=\lim_{n\rightarrow \infty}\frac{1}{n^2} \sum_{k=1}^n (n\bmod k)$$ (where $ n\bmod k $ is the remainder upon division of n by k)


 * $$\pi = \lim_{r \to \infty} \frac{1}{r^2} \sum_{x=-r}^{r} \; \sum_{y=-r}^{r} \begin{cases}

1 & \text{if } \sqrt{x^2+y^2} \le r \\ 0 & \text{if } \sqrt{x^2+y^2} > r \end{cases} $$ (summing a circle's area)


 * $$ \pi = \lim_{n \rightarrow \infty} \frac{4}{n^2} \sum_{k=1}^n \sqrt{n^2 - k^2} $$ (Riemann sum to evaluate the area of the unit circle)


 * $$ \pi = \lim_{n\to\infty}\frac{2^{4n}n!^4}{n(2n)!^2}=\lim_{n \rightarrow \infty} \frac{2^{4n}}{n {2n\choose n}^2} = \lim_{n \rightarrow \infty} \frac{1}{n}\left(\frac{(2n)!!}{(2n-1)!!}\right)^2$$ (by combining Stirling's approximation with Wallis product)


 * $$\pi=\lim_{n\to\infty}\frac{1}{n}\ln\frac{16}{\lambda (ni)}$$ (where $$\lambda$$ is the modular lambda function)


 * $$\pi=\lim_{n\to\infty}\frac{24}{\sqrt{n}}\ln \left(2^{1/4} G_n\right)=\lim_{n\to\infty}\frac{24}{\sqrt{n}}\ln \left(2^{1/4}g_n\right)$$ (where $$G_n$$ and $$g_n$$ are Ramanujan's class invariants)