Inmathematics, thelemniscate constantϖ is atranscendental mathematical constant that is the ratio of theperimeter ofBernoulli's lemniscate to itsdiameter, analogous to the definition ofπ for the circle.^[1] Equivalently, the perimeter of the lemniscate${\displaystyle (x^2+y^2{)}^2=x^2-y^2}$ is2ϖ. The lemniscate constant is closely related to thelemniscate elliptic functions and approximately equal to 2.62205755.^[2] It also appears in evaluation of thegamma andbeta function at certain rational values. The symbolϖ is acursive variant ofπ known asvariant pi represented in Unicode by the characterU+03D6 ϖGREEK PI SYMBOL.

Sometimes the quantities2ϖ orϖ/2 are referred to asthe lemniscate constant.^[3][4]

As of 2024 over 1.2 trillion digits of this constant have been calculated.^[5]

Gauss's constant, denoted byG, is equal toϖ /π ≈ 0.8346268^[6] and named afterCarl Friedrich Gauss, who calculated it via thearithmetic–geometric mean as${\displaystyle 1/M(1,\sqrt{2})}$.^[7] By 1799, Gauss had two proofs of the theorem that${\displaystyle M(1,\sqrt{2})=\pi /\varpi }$ where${\displaystyle \varpi }$ is the lemniscate constant.^[8]

John Todd named two more lemniscate constants, thefirst lemniscate constantA =ϖ/2 ≈ 1.3110287771 and thesecond lemniscate constantB =π/(2ϖ) ≈ 0.5990701173.^[9][10][11]

The lemniscate constant${\displaystyle \varpi }$ and Todd's first lemniscate constant${\displaystyle A}$ were proventranscendental byCarl Ludwig Siegel in 1932 and later byTheodor Schneider in 1937 and Todd's second lemniscate constant${\displaystyle B}$ and Gauss's constant${\displaystyle G}$ were proven transcendental by Theodor Schneider in 1941.^[9][12][13] In 1975,Gregory Chudnovsky proved that the set${\displaystyle \{\pi ,\varpi \}}$ isalgebraically independent over${\displaystyle \mathbb{Q}}$, which implies that${\displaystyle A}$ and${\displaystyle B}$ are algebraically independent as well.^[14][15] But the set${\displaystyle \{\pi ,M(1,1/\sqrt{2}),M^{\prime }(1,1/\sqrt{2})\}}$ (where the prime denotes thederivative with respect to the second variable) is not algebraically independent over${\displaystyle \mathbb{Q}}$.^[16] In 1996,Yuri Nesterenko proved that the set${\displaystyle \{\pi ,\varpi ,e^{\pi }\}}$ is algebraically independent over${\displaystyle \mathbb{Q}}$.^[17]

Usually,${\displaystyle \varpi }$ is defined by the first equality below, but it has many equivalent forms:^[18]

whereK is thecomplete elliptic integral of the first kind with modulusk,Β is thebeta function,Γ is thegamma function andζ is theRiemann zeta function.

The lemniscate constant can also be computed by thearithmetic–geometric mean${\displaystyle M}$,

$${\displaystyle \varpi =\frac{\pi }{M(1,\sqrt{2})}.}$$

Gauss's constant is typically defined as thereciprocal of thearithmetic–geometric mean of 1 and thesquare root of 2, after his calculation of${\displaystyle M(1,\sqrt{2})}$ published in 1800:^[19]$${\displaystyle G=\frac{1}{M(1,\sqrt{2})}}$$John Todd's lemniscate constants may be given in terms of thebeta function B:

$${\displaystyle {\beta }^{\prime }(0)=\log \frac{\varpi }{\sqrt{\pi }}}$$

which is analogous to

$${\displaystyle {\zeta }^{\prime }(0)=\log \frac{1}{\sqrt{2\pi }}}$$

where${\displaystyle \beta }$ is theDirichlet beta function and${\displaystyle \zeta }$ is theRiemann zeta function.^[20]

Analogously to theLeibniz formula for π,$${\displaystyle \beta (1)=\sum _{n=1}^{\mathrm{\infty }}\frac{\chi (n)}{n}=\frac{\pi }{4},}$$we have^{[21][22][23][24][25]}$${\displaystyle L(E,1)=\sum _{n=1}^{\mathrm{\infty }}\frac{\nu (n)}{n}=\frac{\varpi }{4}}$$where${\displaystyle L}$ is theL-function of theelliptic curve${\displaystyle E:y^2=x^3-x}$ over${\displaystyle \mathbb{Q}}$; this means that${\displaystyle \nu }$ is themultiplicative function given bywhere${\displaystyle {\mathcal{N}}_p}$ is the number of solutions of the congruence$${\displaystyle a^3-a\equiv b^2(\mathrm{mod}p),p\in \mathbb{P}}$$in variables${\displaystyle a,b}$ that are non-negative integers (${\displaystyle \mathbb{P}}$ is the set of all primes).Equivalently,${\displaystyle \nu }$ is given by$${\displaystyle F(\tau )=\eta (4\tau {)}^2\eta (8\tau {)}^2=\sum _{n=1}^{\mathrm{\infty }}\nu (n)q^n,q=e^{2\pi i\tau }}$$where${\displaystyle \tau \in \mathbb{C}}$ such that${\displaystyle \mathrm{\Im }\tau >0}$ and${\displaystyle \eta }$ is theeta function.^[26][27][28]The above result can be equivalently written as$${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{\nu (n)}{n}{e}^{-2\pi n/\sqrt{32}}=\frac{\varpi }{8}}$$(the number${\displaystyle 32}$ is theconductor of${\displaystyle E}$) and also tells us that theBSD conjecture is true for the above${\displaystyle E}$.^[29]The first few values of${\displaystyle \nu }$ are given by the following table; if${\displaystyle 1\le n\le 113}$ such that${\displaystyle n}$ doesn't appear in the table, then${\displaystyle \nu (n)=0}$:

Let${\displaystyle \mathrm{\Delta }}$ be the minimal weight level${\displaystyle 1}$ new form. Then^[30]$${\displaystyle \mathrm{\Delta }(i)=\frac{1}{64}{\left(\frac{\varpi }{\pi }\right)}^{12}.}$$The${\displaystyle q}$-coefficient of${\displaystyle \mathrm{\Delta }}$ is theRamanujan tau function.

Viète's formula forπ can be written:

$${\displaystyle \frac{2}{\pi }=\sqrt{\frac{1}{2}}\cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}\cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}\cdots }$$

An analogous formula forϖ is:^[31]

$${\displaystyle \frac{2}{\varpi }=\sqrt{\frac{1}{2}}\cdot \sqrt{\frac{1}{2}+\frac{1}{2}/\sqrt{\frac{1}{2}}}\cdot \sqrt{\frac{1}{2}+\frac{1}{2}/\sqrt{\frac{1}{2}+\frac{1}{2}/\sqrt{\frac{1}{2}}}}\cdots }$$

TheWallis product forπ is:

$${\displaystyle \frac{\pi }{2}=\prod _{n=1}^{\mathrm{\infty }}{\left(1+\frac{1}{n}\right)}^{(-1{)}^{n+1}}=\prod _{n=1}^{\mathrm{\infty }}\left(\frac{2n}{2n-1}\cdot \frac{2n}{2n+1}\right)=(\frac{2}{1}\cdot \frac{2}{3})(\frac{4}{3}\cdot \frac{4}{5})(\frac{6}{5}\cdot \frac{6}{7})\cdots }$$

An analogous formula forϖ is:^[32]

$${\displaystyle \frac{\varpi }{2}=\prod _{n=1}^{\mathrm{\infty }}{\left(1+\frac{1}{2n}\right)}^{(-1{)}^{n+1}}=\prod _{n=1}^{\mathrm{\infty }}\left(\frac{4n-1}{4n-2}\cdot \frac{4n}{4n+1}\right)=(\frac{3}{2}\cdot \frac{4}{5})(\frac{7}{6}\cdot \frac{8}{9})(\frac{11}{10}\cdot \frac{12}{13})\cdots }$$

A related result for Gauss's constant (${\displaystyle G=\varpi /\pi }$) is:^[33]

$${\displaystyle \frac{\varpi }{\pi }=\prod _{n=1}^{\mathrm{\infty }}\left(\frac{4n-1}{4n}\cdot \frac{4n+2}{4n+1}\right)=(\frac{3}{4}\cdot \frac{6}{5})(\frac{7}{8}\cdot \frac{10}{9})(\frac{11}{12}\cdot \frac{14}{13})\cdots }$$

An infinite series discovered by Gauss is:^[34]

$${\displaystyle \frac{\varpi }{\pi }=\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n\prod _{k=1}^n\frac{(2k-1{)}^2}{(2k{)}^2}=1-\frac{1^2}{2^2}+\frac{1^2\cdot 3^2}{2^2\cdot 4^2}-\frac{1^2\cdot 3^2\cdot 5^2}{2^2\cdot 4^2\cdot 6^2}+\cdots }$$

TheMachin formula forπ is$\frac{1}{4}\pi =4\arctan \frac{1}{5}-\arctan \frac{1}{239},$ and several similar formulas forπ can be developed using trigonometric angle sum identities, e.g. Euler's formula$\frac{1}{4}\pi =\arctan \frac{1}{2}+\arctan \frac{1}{3}$. Analogous formulas can be developed forϖ, including the following found by Gauss:${\displaystyle \frac{1}{2}\varpi =2\mathrm{arcsl}\frac{1}{2}+\mathrm{arcsl}\frac{7}{23}}$, where${\displaystyle \mathrm{arcsl}}$ is thelemniscate arcsine.^[35]

The lemniscate constant can be rapidly computed by the series^[36][37]

${\displaystyle \varpi =2^{-1/2}\pi (\sum _{n\in \mathbb{Z}}{e}^{-\pi n^2}{)}^2=2^{1/4}\pi e^{-\pi /12}(\sum _{n\in \mathbb{Z}}(-1{)}^n{e}^{-\pi p_n}{)}^2}$

where${\displaystyle p_n=\frac{1}{2}(3n^2-n)}$ (these are thegeneralized pentagonal numbers). Also^[38]

${\displaystyle \sum _{m,n\in \mathbb{Z}}{e}^{-2\pi (m^2+mn+n^2)}=\sqrt{1+\sqrt{3}}{\displaystyle \frac{\varpi }{{12}^{1/8}\pi }}.}$

In a spirit similar to that of theBasel problem,

${\displaystyle \sum _{z\in \mathbb{Z}[i]\setminus \{0\}}\frac{1}{z^4}=G_4(i)=\frac{{\varpi }^4}{15}}$

where${\displaystyle \mathbb{Z}[i]}$ are theGaussian integers and${\displaystyle G_4}$ is theEisenstein series of weight⁠${\displaystyle 4}$⁠ (seeLemniscate elliptic functions § Hurwitz numbers for a more general result).^[39]

A related result is

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{\sigma }_3(n)e^{-2\pi n}=\frac{{\varpi }^4}{80{\pi }^4}-\frac{1}{240}}$

where${\displaystyle {\sigma }_3}$ is thesum of positive divisors function.^[40]

In 1842,Malmsten found

${\displaystyle {\beta }^{\prime }(1)=\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{\log (2n+1)}{2n+1}=\frac{\pi }{4}\left(\gamma +2\log \frac{\pi }{\varpi \sqrt{2}}\right)}$

where${\displaystyle \gamma }$ isEuler's constant and${\displaystyle \beta (s)}$ is the Dirichlet-Beta function.

The lemniscate constant is given by the rapidly converging series

$${\displaystyle \varpi =\pi \sqrt[4]{32}{e}^{-\frac{\pi }{3}}(\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}(-1{)}^n{e}^{-2n\pi (3n+1)}{)}^2.}$$

The constant is also given by theinfinite product

${\displaystyle \varpi =\pi \prod _{m=1}^{\mathrm{\infty }}{\tanh }^2\left(\frac{\pi m}{2}\right).}$

Also^[41]

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{{6635520}^n}\frac{(4n)!}{n{!}^4}=\frac{24}{5^{7/4}}\frac{{\varpi }^2}{{\pi }^2}.}$

A (generalized)continued fraction forπ is$${\displaystyle \frac{\pi }{2}=1+\frac{1}{1+\frac{1\cdot 2}{1+\frac{2\cdot 3}{1+\frac{3\cdot 4}{1+\ddots }}}}}$$An analogous formula forϖ is^[10]$${\displaystyle \frac{\varpi }{2}=1+\frac{1}{2+\frac{2\cdot 3}{2+\frac{4\cdot 5}{2+\frac{6\cdot 7}{2+\ddots }}}}}$$

DefineBrouncker's continued fraction by^[42]$${\displaystyle b(s)=s+\frac{1^2}{2s+\frac{3^2}{2s+\frac{5^2}{2s+\ddots }}},s>0.}$$Let${\displaystyle n\ge 0}$ except for the first equality where${\displaystyle n\ge 1}$. Then^[43][44]For example,

In fact, the values of${\displaystyle b(1)}$ and${\displaystyle b(2)}$, coupled with the functional equation$${\displaystyle b(s+2)=\frac{(s+1{)}^2}{b(s)},}$$determine the values of${\displaystyle b(n)}$ for all${\displaystyle n}$.

Simple continued fractions for the lemniscate constant and related constants include^[45][46]

The lemniscate constantϖ is related to the area under the curve${\displaystyle x^4+y^4=1}$. Defining${\displaystyle {\pi }_n:=\mathrm{B}(\frac{1}{n},\frac{1}{n})}$, twice the area in the positive quadrant under the curve${\displaystyle x^n+y^n=1}$ is$2{\int }_0^1\sqrt[n]{1-x^n}\mathrm{d}x=\frac{1}{n}{\pi }_n.$ In the quartic case,${\displaystyle \frac{1}{4}{\pi }_4=\frac{1}{\sqrt{2}}\varpi .}$

In 1842, Malmsten discovered that^[47]

$${\displaystyle {\int }_0^1\frac{\log (-\log x)}{1+x^2}dx=\frac{\pi }{2}\log \frac{\pi }{\varpi \sqrt{2}}.}$$

Furthermore,$${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{\tanh x}{x}{e}^{-x}dx=\log \frac{{\varpi }^2}{\pi }}$$

and^[48]

$${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-x^4}dx=\frac{\sqrt{2\varpi \sqrt{2\pi }}}{4},\text{analogous to}{\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx=\frac{\sqrt{\pi }}{2},}$$a form ofGaussian integral.

The lemniscate constant appears in the evaluation of the integrals

$${\displaystyle \frac{\pi }{\varpi }={\int }_0^{\frac{\pi }{2}}\sqrt{\sin (x)}dx={\int }_0^{\frac{\pi }{2}}\sqrt{\cos (x)}dx}$$

$${\displaystyle \frac{\varpi }{\pi }={\int }_0^{\mathrm{\infty }}\frac{dx}{\sqrt{\cosh (\pi x)}}}$$

John Todd's lemniscate constants are defined by integrals:^[9]

$${\displaystyle A={\int }_0^1\frac{dx}{\sqrt{1-x^4}}}$$

$${\displaystyle B={\int }_0^1\frac{x^{2}dx}{\sqrt{1-x^4}}}$$

The lemniscate constant satisfies the equation^[49]

$${\displaystyle \frac{\pi }{\varpi }=2{\int }_0^1\frac{x^{2}dx}{\sqrt{1-x^4}}}$$

Euler discovered in 1738 that for the rectangular elastica (first and second lemniscate constants)^[50][49]

$${\displaystyle \text{arc}\text{length}\cdot \text{height}=A\cdot B={\int }_0^1\frac{\mathrm{d}x}{\sqrt{1-x^4}}\cdot {\int }_0^1\frac{x^2\mathrm{d}x}{\sqrt{1-x^4}}=\frac{\varpi }{2}\cdot \frac{\pi }{2\varpi }=\frac{\pi }{4}}$$

Now considering the circumference${\displaystyle C}$ of the ellipse with axes${\displaystyle \sqrt{2}}$ and${\displaystyle 1}$, satisfying${\displaystyle 2x^2+4y^2=1}$, Stirling noted that^[51]

$${\displaystyle \frac{C}{2}={\int }_0^1\frac{dx}{\sqrt{1-x^4}}+{\int }_0^1\frac{x^{2}dx}{\sqrt{1-x^4}}}$$

Hence the full circumference is

$${\displaystyle C=\frac{\pi }{\varpi }+\varpi =3.820197789\dots }$$

This is also the arc length of thesine curve on half a period:^[52]

$${\displaystyle C={\int }_0^{\pi }\sqrt{1+{\cos }^2(x)}dx}$$

Analogously to$${\displaystyle 2\pi =\underset{n\to \mathrm{\infty }}{lim}{\left|\frac{(2n)!}{{\mathrm{B}}_{2n}}\right|}^{\frac{1}{2n}}}$$where${\displaystyle {\mathrm{B}}_n}$ areBernoulli numbers, we have$${\displaystyle 2\varpi =\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{(4n)!}{{\mathrm{H}}_{4n}}\right)}^{\frac{1}{4n}}}$$where${\displaystyle {\mathrm{H}}_n}$ areHurwitz numbers.

• Weisstein, Eric W."Lemniscate Constant".MathWorld.
• Sequences A014549, A053002, andA062539 inOEIS
• Cox, David A. (January 1984)."The Arithmetic-Geometric Mean of Gauss"(PDF).L'Enseignement Mathématique.30 (2):275–330.doi:10.5169/seals-53831. Retrieved25 June 2022.
• Finch, Steven R. (18 August 2003).Mathematical Constants. Cambridge University Press. pp. 420–422.ISBN 978-0-521-81805-6.

• "Gauss's constant and where it occurs".www.johndcook.com. 2021-10-17.

• Mathematical constants
• Real transcendental numbers

• CS1 Latin-language sources (la)
• CS1 German-language sources (de)
• Articles with short description
• Short description is different from Wikidata
• Pages that use a deprecated format of the math tags