Lemniscate constant

In mathematics, the lemniscate constant $ϖ$  is a transcendental mathematical constant that is the ratio of the perimeter of Bernoulli's lemniscate to its diameter, analogous to the definition of $\pi$ for the circle. Equivalently, the perimeter of the lemniscate $$(x^2+y^2)^2=x^2-y^2$$ is $2ϖ$. The lemniscate constant is closely related to the lemniscate elliptic functions and approximately equal to 2.62205755. The symbol $ϖ$ is a cursive variant of $π$; see Pi § Variant pi.

Gauss's constant, denoted by G, is equal to $ϖ⁄\pi ≈ 0.8346268$.

John Todd named two more lemniscate constants, the first lemniscate constant $A = ϖ/2 ≈ 1.3110287771$ and the second lemniscate constant $B = \pi/(2ϖ) ≈ 0.5990701173$.

Sometimes the quantities $2ϖ$ or $A$ are referred to as the lemniscate constant.

History
Gauss's constant $$G$$ is named after Carl Friedrich Gauss, who calculated it via the arithmetic–geometric mean as $$\tfrac{1}{M\left(1,\sqrt{2}\right)}$$. By 1799, Gauss had two proofs of the theorem that $$M\left(1,\sqrt{2}\right)=\tfrac{\pi}{\varpi}$$ where $$\varpi$$ is the lemniscate constant.

The lemniscate constant $$\varpi$$ and first lemniscate constant $$A$$ were proven transcendental by Theodor Schneider in 1937 and the second lemniscate constant $$B$$ and Gauss's constant $$G$$ were proven transcendental by Theodor Schneider in 1941. In 1975, Gregory Chudnovsky proved that the set $$\{\pi,\varpi\}$$ is algebraically independent over $$\mathbb{Q}$$, which implies that $$A$$ and $$B$$ are algebraically independent as well. But the set $$\left\{\pi,M\left(1,\tfrac{1}{\sqrt{2}}\right),M'\left(1,\tfrac{1}{\sqrt{2}}\right)\right\}$$ (where the prime denotes the derivative with respect to the second variable) is not algebraically independent over $$\mathbb{Q}$$. In fact,

$$\pi=2\sqrt{2}\frac{M^3\left(1,\frac{1}{\sqrt{2}}\right)}{M'\left(1,\frac{1}{\sqrt{2}}\right)}=\frac{1}{G^3 M'\left(1,\frac{1}{\sqrt{2}}\right)}.$$

Forms
Usually, $$\varpi$$ is defined by the first equality below.

$$\begin{aligned} \varpi &= 2\int_0^1\frac{\mathrm{d}t}{\sqrt{1-t^4}} = \sqrt2\int_0^\infty\frac{\mathrm{d}t}{\sqrt{1+t^4}} = \int_0^1\frac{\mathrm dt}{\sqrt{t-t^3}} = \int_1^\infty \frac{\mathrm dt}{\sqrt{t^3-t}}\\[6mu] &= 4\int_0^\infty\left(\sqrt[4]{1+t^{4}}-t\right)\,\mathrm{d}t = 2\sqrt2\int_0^1 \sqrt[4]{1-t^{4}}\mathop{\mathrm{d}t} =3\int_0^1 \sqrt{1-t^4}\,\mathrm dt\\[2mu] &= 2K(i) = \tfrac{1}{2}\Beta\left( \tfrac14, \tfrac12\right) = \frac{\Gamma \left(\frac14\right)^2}{2\sqrt{2\pi}} = \frac{2-\sqrt{2}}{4}\frac{\zeta\left(\frac34\right)^2}{\zeta\left(\frac14\right)^2}\\[5mu] &= 2.62205\;75542\;92119\;81046\;48395\;89891\;11941\ldots, \end{aligned}$$

where $B$ is the complete elliptic integral of the first kind with modulus $G$, $&Beta;$ is the beta function, $&Gamma;$ is the gamma function and $K$ is the Riemann zeta function.

The lemniscate constant can also be computed by the arithmetic–geometric mean $$M$$,

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

Moreover, $$e^{\beta'(0)}=\frac{\varpi}{\sqrt{\pi}}$$

which is analogous to

$$e^{\zeta'(0)}=\frac{1}{\sqrt{2\pi}}$$

where $$\beta$$ is the Dirichlet beta function and $$\zeta$$ is the Riemann zeta function.

Gauss's constant is typically defined as the reciprocal of the arithmetic–geometric mean of 1 and the square root of 2, after his calculation of $$M\left(1, \sqrt{2}\right)$$ published in 1800:

$$G = \frac{1}{M(1, \sqrt{2})}$$

Gauss's constant is equal to

$$ G = \frac{1}{2\pi}\Beta\left( \tfrac14, \tfrac12\right)$$

where Β denotes the beta function. A formula for G in terms of Jacobi theta functions is given by

$$G = \vartheta_{01}^2\left(e^{-\pi}\right) $$

Gauss's constant may be computed from the gamma function at argument $k$:

$$ G = \frac{\Gamma\left( \tfrac{1}{4}\right){}^2}{2\sqrt{ 2\pi^3}} $$

John Todd's lemniscate constants may be given in terms of the beta function B: $$\begin{aligned} A &= \tfrac12\pi G = \tfrac12\varpi = \tfrac14 \Beta \left(\tfrac14,\tfrac12\right), \\[3mu] B &= \frac{1}{2G} =\tfrac14\Beta \left(\tfrac12,\tfrac34\right). \end{aligned}$$

Series
Viète's formula for $ζ$ can be written:

$$ \frac2\pi = \sqrt\frac12 \cdot \sqrt{\frac12 + \frac12\sqrt\frac12} \cdot \sqrt{\frac12 + \frac12\sqrt{\frac12 + \frac12\sqrt\frac12}} \cdots $$

An analogous formula for $1⁄4$ is:

$$ \frac2\varpi = \sqrt\frac12 \cdot \sqrt{\frac12 + \frac{\frac12 }{ \sqrt\frac12} } \cdot \sqrt{\frac12 +  \frac{\frac12 }{ \sqrt{\frac12 +  \frac{\frac12 }{ \sqrt\frac12}}}} \cdots $$

The Wallis product for $π$ is:

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

An analogous formula for $ϖ$ is:

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

A related result for Gauss's constant ($$G=\tfrac{\varpi}{\pi}$$) is:

$$ G = \prod_{n=1}^{\infty} \left(\frac{4n-1}{4n} \cdot \frac{4n+2}{4n+1}\right) = \left(\frac{3}{4} \cdot \frac{6}{5}\right) \left(\frac{7}{8} \cdot \frac{10}{9}\right) \left(\frac{11}{12} \cdot \frac{14}{13}\right) \cdots $$

An infinite series of Gauss's constant discovered by Gauss is:

$$ G = \sum_{n=0}^\infty (-1)^n \prod_{k=1}^n \frac{(2k-1)^2}{(2k)^2} = 1 - \frac{1^2}{2^2} + \frac{1^2\cdot3^2}{2^2\cdot4^2} - \frac{1^2\cdot3^2\cdot5^2}{2^2\cdot4^2\cdot6^2} + \cdots $$

The Machin formula for $π$ is $\tfrac14\pi = 4 \arctan \tfrac15 - \arctan \tfrac1{239},$ and several similar formulas for $ϖ$ can be developed using trigonometric angle sum identities, e.g. Euler's formula $\tfrac14\pi = \arctan\tfrac12 + \arctan\tfrac13$. Analogous formulas can be developed for $π$, including the following found by Gauss: $$\tfrac12\varpi = 2 \operatorname{arcsl} \tfrac12 + \operatorname{arcsl} \tfrac7{23}$$, where $$\operatorname{arcsl}$$ is the lemniscate arcsine.

The lemniscate constant can be rapidly computed by the series


 * $$\varpi=\frac{1}{\sqrt 2}\pi\left(\sum_{n\in\mathbb{Z}}e^{-\pi n^2}\right)^2=\sqrt[4]{2} \pi e^{-\frac{\pi}{12}} \left(\sum_{n\in\mathbb{Z}}(-1)^n e^{-\pi p_n}\right)^2$$

where $$p_n=\tfrac{3n^2-n}{2}$$ (these are the generalized pentagonal numbers). Also


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

In a spirit similar to that of the Basel problem,
 * $$\sum_{z\in\mathbb{Z}[i]\setminus\{0\}}\frac{1}{z^4}=G_4(i)=\frac{\varpi ^4}{15}$$

where $$\mathbb{Z}[i]$$ are the Gaussian integers and $$G_4$$ is the Eisenstein series of weight 4 (see Lemniscate elliptic functions § Hurwitz numbers for a more general result).

A related result is
 * $$\sum_{n=1}^\infty \sigma_3(n)e^{-2\pi n}=\frac{\varpi^4}{80 \pi^4}-\frac{1}{240}$$

where $$\sigma_3$$ is the sum of positive divisors function.

In 1842, Malmsten found
 * $$\sum_{n=1}^\infty (-1)^{n+1}\frac{\log (2n+1)}{2n+1}=\frac{\pi}{4}\left(\gamma+2\log\frac{\pi}{\varpi\sqrt{2}}\right)$$

where $$\gamma$$ is Euler's constant.

Gauss's constant is given by the rapidly converging series

$$G = \sqrt[4]{32}e^{-\frac{\pi}{3}}\left (\sum_{n = -\infty}^\infty (-1)^n e^{-2n\pi(3n+1)} \right )^2.$$

The constant is also given by the infinite product


 * $$G = \prod_{m = 1}^\infty \tanh^2 \left( \frac{\pi m}{2}\right).$$

Continued fractions
A (generalized) continued fraction for $π$ is $$\frac\pi2=1 + \cfrac{1}{1 + \cfrac{1\cdot 2}{1 + \cfrac{2\cdot 3}{1 + \cfrac{3\cdot 4}{1+\ddots}}}}$$ An analogous formula for $ϖ$ is $$\frac\varpi2= 1 + \cfrac{1}{2 + \cfrac{2\cdot 3}{2 + \cfrac{4\cdot 5}{2 + \cfrac{6\cdot 7}{2+\ddots}}}}$$

Define Brouncker's continued fraction by $$b(s)=s + \cfrac{1^2}{2s + \cfrac{3^2}{2s + \cfrac{5^2}{2s+\ddots}}},\quad s>0.$$ Let $$n\ge 0$$ except for the first equality where $$n\ge 1$$. Then $$\begin{align}b(4n)&=(4n+1)\prod_{k=1}^n \frac{(4k-1)^2}{(4k-3)(4k+1)}\frac{\pi}{\varpi^2}\\ b(4n+1)&=(2n+1)\prod_{k=1}^n \frac{(2k)^2}{(2k-1)(2k+1)}\frac{4}{\pi}\\ b(4n+2)&=(4n+1)\prod_{k=1}^n \frac{(4k-3)(4k+1)}{(4k-1)^2}\frac{\varpi^2}{\pi}\\ b(4n+3)&=(2n+1)\prod_{k=1}^n \frac{(2k-1)(2k+1)}{(2k)^2}\,\pi.\end{align}$$ For example, $$\begin{align}b(1)&=\frac{4}{\pi}\\ b(2)&=\frac{\varpi^2}{\pi}\\ b(3)&=\pi\\ b(4)&=\frac{9\pi}{\varpi^2}.\end{align}$$

===Simple continued fractions === $$\begin{align}\varpi&=[2,1,1,1,1,1,4,1,2,\ldots]\\ 2\varpi&=[5,4,10,2,1,2,3,29,\ldots]\\ \frac{\varpi}{2}&=[1,3,4,1,1,1,5,2,\ldots]\\ G&=[0,1,5,21,3,4,14,\ldots]\end{align}$$

Integrals
$π$ is related to the area under the curve $$x^4 + y^4 = 1$$. Defining $$\pi_n \mathrel{:=} \Beta\left(\tfrac1n, \tfrac1n \right)$$, twice the area in the positive quadrant under the curve $$x^n + y^n = 1$$ is $$2 \int_0^1 \sqrt[n]{1 - x^n}\mathop{\mathrm{d}x} = \tfrac1n \pi_n.$$ In the quartic case, $$\tfrac14 \pi_4 = \tfrac1\sqrt{2} \varpi.$$

In 1842, Malmsten discovered that

$$\int_0^1 \frac{\log (-\log x)}{1+x^2}\, dx=\frac{\pi}{2}\log\frac{\pi}{\varpi\sqrt{2}}.$$

Furthermore, $$\int_0^\infty \frac{\tanh x}{x}e^{-x}\, dx=\log\frac{\varpi^2}{\pi}$$

and

$$\int_0^\infty e^{-x^4}\, dx=\frac{\sqrt{2\varpi\sqrt{2\pi}}}{4},\quad\text{analogous to}\,\int_0^\infty e^{-x^2}\, dx=\frac{\sqrt{\pi}}{2},$$ a form of Gaussian integral.

Gauss's constant appears in the evaluation of the integrals

$$ {\frac{1}{G}} = \int_0^{\frac{\pi}{2}}\sqrt{\sin(x)}\,dx=\int_0^{\frac{\pi}{2}}\sqrt{\cos(x)}\,dx $$

$$ G = \int_0^{\infty}{\frac{dx}{\sqrt{\cosh(\pi x)}}} $$

The first and second lemniscate constants are defined by integrals:

$$A = \int_0^1\frac{dx}{\sqrt{1 - x^4}}$$

$$B = \int_0^1\frac{x^2\, dx}{\sqrt{1 - x^4}} $$

Circumference of an ellipse
Gauss's constant satisfies the equation

$$\frac{1}{G} = 2 \int_0^1\frac{x^2\, dx}{\sqrt{1 - x^4}} $$

Euler discovered in 1738 that for the rectangular elastica (first and second lemniscate constants)

$$\textrm{arc}\ \textrm{length}\cdot\textrm{height} = A \cdot B = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1 - x^4}} \cdot \int_0^1 \frac{x^2 \mathop{\mathrm{d}x}}{\sqrt{1 - x^4}} = \frac\varpi2 \cdot \frac\pi{2\varpi} = \frac\pi4$$

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

$$ \frac{C}{2} = \int_0^1\frac{dx}{\sqrt{1 - x^4}} + \int_0^1\frac{x^2\,dx}{\sqrt{1 - x^4}}$$

Hence the full circumference is

$$ C = \frac{1}{G} + G \pi \approx 3.820197789\ldots $$

This is also the arc length of the sine curve on half a period:

$$ C = \int_0^\pi \sqrt{1+\cos^2(x)}\,dx $$

Other limits
Analogously to $$2\pi=\lim_{n\to\infty}\left|\frac{(2n)!}{\mathrm{B}_{2n}}\right|^{\frac{1}{2n}}$$ where $$\mathrm{B}_n$$ are Bernoulli numbers, we have $$2\varpi=\lim_{n\to\infty}\left(\frac{(4n)!}{\mathrm{H}_{4n}}\right)^{\frac{1}{4n}}$$ where $$\mathrm{H}_n$$ are Hurwitz numbers.