Gauss's constant

Lemniscate of Bernoulli

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 π for the circle. It is approximately equal to 2.62205755.

Gauss's constant, denoted by G, is equal to ϖ /π ≈ 0.8346268.

John Todd named two more lemniscate constants, the first lemniscate constant L = ϖ/2 ≈ 1.3110287771 and the second lemniscate constant M = π/(2ϖ) ≈ 0.5990701173.

Sometimes the quantities 2ϖ or L are referred to as the lemniscate constant. Throughout this article, we strictly follow Gauss' definition for the lemniscate constant.

History

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

The lemniscate constant and first lemniscate constant $L$ were proven transcendental by Theodor Schneider in 1937 and the second lemniscate constant $M$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 $\mathbb{\mathbb{Q}}$, which implies that $L$ and $M$ are algebraically independent as well. But the set $\{\pi ,M(1,1/\sqrt{2}),M^{\prime }(1,1/\sqrt{2})\}$ (where the prime denotes the derivative with respect to the second variable) is not algebraically independent over $\mathbb{\mathbb{Q}}$. In fact,

$$\pi =2\sqrt{2}\frac{M^3(1,1/\sqrt{2})}{M^{\prime }(1,1/\sqrt{2})}=\frac{1}{G^3{M}^{\prime }(1,1/\sqrt{2})}.$$

Forms

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

$$\begin{array}{rl}\varpi & =2\underset{0}{\overset{1}{\int }}\frac{\mathrm{d}t}{\sqrt{1-t^4}}=\sqrt{2}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\mathrm{d}t}{\sqrt{1+t^4}}=\underset{0}{\overset{1}{\int }}\frac{\mathrm{d}t}{\sqrt{t-t^3}}\\ & =4\underset{0}{\overset{\mathrm{\infty }}{\int }}(\sqrt[4]{1+t^4}-t)\mathrm{d}t=2\sqrt{2}\underset{0}{\overset{1}{\int }}\sqrt[4]{1-t^4}\mathrm{d}t=3\underset{0}{\overset{1}{\int }}\sqrt{1-t^4}\mathrm{d}t\\ & =2K(i)=\frac{1}{2}\mathrm{B}(\frac{1}{4},\frac{1}{2})=\frac{\mathrm{\Gamma }(1/4{)}^2}{2\sqrt{2\pi }}=\sqrt{\pi }{e}^{{\beta }^{\prime }(0)}=\frac{2-\sqrt{2}}{4}\frac{\zeta (3/4{)}^2}{\zeta (1/4{)}^2}\\ & =2.62205755429211981046483958989111941\dots ,\end{array}$$

where K is the complete elliptic integral of the first kind with modulus k, Β is the beta function, Γ is the gamma function, β' is the derivative of the Dirichlet beta function and ζ is the Riemann zeta function.

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

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

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 $\mathrm{M}(1,\sqrt{2})$ published in 1800:

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

Gauss's constant is equal to

$$G=\frac{1}{2\pi }\mathrm{B}(\frac{1}{4},\frac{1}{2})$$

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 ⁠1/4⁠:

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

John Todd's lemniscate constants may be given in terms of the beta function B: $$\begin{array}{rl}L& =\frac{1}{2}\pi G=\frac{1}{2}\varpi =\frac{1}{4}\mathrm{B}(\frac{1}{4},\frac{1}{2}),\\ M& =\frac{1}{2G}=\frac{1}{4}\mathrm{B}(\frac{1}{2},\frac{3}{4}).\end{array}$$

Series

Viète's formula for π can be written:

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

An analogous formula for ϖ is:

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

The Wallis product for π is:

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

An analogous formula for ϖ is:

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

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

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

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

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

The Machin formula for π is $\frac{1}{4}\pi =4\arctan \frac{1}{5}-\arctan \frac{1}{239},$ and several similar formulas for π 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 for ϖ, including the following found by Gauss: $\frac{1}{2}\varpi =2\mathrm{arcsl}\frac{1}{2}+\mathrm{arcsl}\frac{7}{23}$, where $\mathrm{arcsl}$ is the lemniscate arcsine.

The lemniscate constant can be rapidly computed by the series

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

where $p_n=(3n^2-n)/2$ (for $n\ge 1$, these are the pentagonal numbers)

In a spirit similar to that of the Basel problem,

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

where $\mathbb{\mathbb{Z}}[i]$ are the Gaussian integers and $G_4(\tau )$ is the Eisenstein series of weight $4$.

Gauss's constant is given by the rapidly converging series

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

The constant is also given by the infinite product

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

An analog of the Wallis product is:

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

Gauss' constant as a continued fraction is [0, 1, 5, 21, 3, 4, 14, ...]. (sequence A053002 in the OEIS)

Integrals

A geometric representation of $\varpi /2$ and $\varpi /\sqrt{2}$

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

Gauss's constant appears in the evaluation of the integrals

$$\frac{1}{G}=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin (x)}dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\cos (x)}dx$$

$$G=\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{\sqrt{\cosh (\pi x)}}$$

The first and second lemniscate constants are defined by integrals:

$$L=\underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{1-x^4}}$$

$$M=\underset{0}{\overset{1}{\int }}\frac{x^{2}dx}{\sqrt{1-x^4}}$$

Circumference of an ellipse

Gauss's constant satisfies the equation

$$\frac{1}{G}=2\underset{0}{\overset{1}{\int }}\frac{x^{2}dx}{\sqrt{1-x^4}}$$

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

$$\text{<mrow data-mjx-texclass="ORD">arc</mrow>}\text{<mrow data-mjx-texclass="ORD">length</mrow>}\cdot \text{<mrow data-mjx-texclass="ORD">height</mrow>}=L\cdot M=\underset{0}{\overset{1}{\int }}\frac{\mathrm{d}x}{\sqrt{1-x^4}}\cdot \underset{0}{\overset{1}{\int }}\frac{x^2\mathrm{d}x}{\sqrt{1-x^4}}=\frac{\varpi }{2}\cdot \frac{\pi }{2\varpi }=\frac{\pi }{4}$$

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

$$\frac{C}{2}=\underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{1-x^4}}+\underset{0}{\overset{1}{\int }}\frac{x^{2}dx}{\sqrt{1-x^4}}$$

Hence the full circumference is

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

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

$$C=\underset{0}{\overset{\pi }{\int }}\sqrt{1+{\cos }^2(x)}dx$$

See also

• Lemniscate elliptic function

References

• Weisstein, Eric W.. "Gauss's Constant". http://mathworld.wolfram.com/GausssConstant.html. 
• Sequences A014549, A053002, and A062539 in OEIS
• Cox, David A. (January 1984). "The Arithmetic-Geometric Mean of Gauss". L'Enseignement Mathématique 30 (2): 275-330. doi:10.5169/seals-53831. https://webspace.science.uu.nl/~wepst101/elliptic/cox_agm.pdf. Retrieved 25 June 2022. 

External links

• "Gauss's constant and where it occurs" (in en-US). 2021-10-17. https://www.johndcook.com/blog/2021/10/17/gauss-constant/.