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Catalan's constant

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Number, approximately 0.916 Not to be confused with Catalan number.

Catalan's constant
RationalityUnknown
SymbolG
Representations
Decimal0.9159655941772190150...

In mathematics, Catalan's constant G, is the alternating sum of the reciprocals of the odd square numbers, being defined by:

G = β ( 2 ) = n = 0 ( 1 ) n ( 2 n + 1 ) 2 = 1 1 2 1 3 2 + 1 5 2 1 7 2 + 1 9 2 , {\displaystyle G=\beta (2)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)^{2}}}={\frac {1}{1^{2}}}-{\frac {1}{3^{2}}}+{\frac {1}{5^{2}}}-{\frac {1}{7^{2}}}+{\frac {1}{9^{2}}}-\cdots ,}

where β is the Dirichlet beta function. Its numerical value is approximately (sequence A006752 in the OEIS)

G = 0.915965594177219015054603514932384110774…

Catalan's constant was named after Eugène Charles Catalan, who found quickly-converging series for its calculation and published a memoir on it in 1865.

Uses

In low-dimensional topology, Catalan's constant is 1/4 of the volume of an ideal hyperbolic octahedron, and therefore 1/4 of the hyperbolic volume of the complement of the Whitehead link. It is 1/8 of the volume of the complement of the Borromean rings.

In combinatorics and statistical mechanics, it arises in connection with counting domino tilings, spanning trees, and Hamiltonian cycles of grid graphs.

In number theory, Catalan's constant appears in a conjectured formula for the asymptotic number of primes of the form n 2 + 1 {\displaystyle n^{2}+1} according to Hardy and Littlewood's Conjecture F. However, it is an unsolved problem (one of Landau's problems) whether there are even infinitely many primes of this form.

Catalan's constant also appears in the calculation of the mass distribution of spiral galaxies.

Properties

Unsolved problem in mathematics: Is Catalan's constant irrational? If so, is it transcendental? (more unsolved problems in mathematics)

It is not known whether G is irrational, let alone transcendental. G has been called "arguably the most basic constant whose irrationality and transcendence (though strongly suspected) remain unproven".

There exist however partial results. It is known that infinitely many of the numbers β(2n) are irrational, where β(s) is the Dirichlet beta function. In particular at least one of β(2), β(4), β(6), β(8), β(10) and β(12) must be irrational, where β(2) is Catalan's constant. These results by Wadim Zudilin and Tanguy Rivoal are related to similar ones given for the odd zeta constants ζ(2n+1).

Catalan's constant is known to be an algebraic period, which follows from some of the double integrals given below.

Series representations

Catalan's constant appears in the evaluation of several rational series including: π 2 16 + G 2 = n = 0 1 ( 4 n + 1 ) 2 . {\displaystyle {\frac {\pi ^{2}}{16}}+{\frac {G}{2}}=\sum _{n=0}^{\infty }{\frac {1}{(4n+1)^{2}}}.} π 2 16 G 2 = n = 0 1 ( 4 n + 3 ) 2 . {\displaystyle {\frac {\pi ^{2}}{16}}-{\frac {G}{2}}=\sum _{n=0}^{\infty }{\frac {1}{(4n+3)^{2}}}.} The following two formulas involve quickly converging series, and are thus appropriate for numerical computation: G = 3 n = 0 1 2 4 n ( 1 2 ( 8 n + 2 ) 2 + 1 2 2 ( 8 n + 3 ) 2 1 2 3 ( 8 n + 5 ) 2 + 1 2 3 ( 8 n + 6 ) 2 1 2 4 ( 8 n + 7 ) 2 + 1 2 ( 8 n + 1 ) 2 ) 2 n = 0 1 2 12 n ( 1 2 4 ( 8 n + 2 ) 2 + 1 2 6 ( 8 n + 3 ) 2 1 2 9 ( 8 n + 5 ) 2 1 2 10 ( 8 n + 6 ) 2 1 2 12 ( 8 n + 7 ) 2 + 1 2 3 ( 8 n + 1 ) 2 ) {\displaystyle {\begin{aligned}G&=3\sum _{n=0}^{\infty }{\frac {1}{2^{4n}}}\left(-{\frac {1}{2(8n+2)^{2}}}+{\frac {1}{2^{2}(8n+3)^{2}}}-{\frac {1}{2^{3}(8n+5)^{2}}}+{\frac {1}{2^{3}(8n+6)^{2}}}-{\frac {1}{2^{4}(8n+7)^{2}}}+{\frac {1}{2(8n+1)^{2}}}\right)\\&\qquad -2\sum _{n=0}^{\infty }{\frac {1}{2^{12n}}}\left({\frac {1}{2^{4}(8n+2)^{2}}}+{\frac {1}{2^{6}(8n+3)^{2}}}-{\frac {1}{2^{9}(8n+5)^{2}}}-{\frac {1}{2^{10}(8n+6)^{2}}}-{\frac {1}{2^{12}(8n+7)^{2}}}+{\frac {1}{2^{3}(8n+1)^{2}}}\right)\end{aligned}}} and G = π 8 log ( 2 + 3 ) + 3 8 n = 0 1 ( 2 n + 1 ) 2 ( 2 n n ) . {\displaystyle G={\frac {\pi }{8}}\log \left(2+{\sqrt {3}}\right)+{\frac {3}{8}}\sum _{n=0}^{\infty }{\frac {1}{(2n+1)^{2}{\binom {2n}{n}}}}.}

The theoretical foundations for such series are given by Broadhurst, for the first formula, and Ramanujan, for the second formula. The algorithms for fast evaluation of the Catalan constant were constructed by E. Karatsuba. Using these series, calculating Catalan's constant is now about as fast as calculating Apéry's constant, ζ ( 3 ) {\displaystyle \zeta (3)} .

Other quickly converging series, due to Guillera and Pilehrood and employed by the y-cruncher software, include:

G = 1 2 k = 0 ( 8 ) k ( 3 k + 2 ) ( 2 k + 1 ) 3 ( 2 k k ) 3 {\displaystyle G={\frac {1}{2}}\sum _{k=0}^{\infty }{\frac {(-8)^{k}(3k+2)}{(2k+1)^{3}{\binom {2k}{k}}^{3}}}}
G = 1 64 k = 1 256 k ( 580 k 2 184 k + 15 ) k 3 ( 2 k 1 ) ( 6 k 3 k ) ( 6 k 4 k ) ( 4 k 2 k ) {\displaystyle G={\frac {1}{64}}\sum _{k=1}^{\infty }{\frac {256^{k}(580k^{2}-184k+15)}{k^{3}(2k-1){\binom {6k}{3k}}{\binom {6k}{4k}}{\binom {4k}{2k}}}}}
G = 1 1024 k = 1 ( 4096 ) k ( 45136 k 4 57184 k 3 + 21240 k 2 3160 k + 165 ) k 3 ( 2 k 1 ) 3 ( ( 2 k ) ! 6 ( 3 k ) ! 3 k ! 3 ( 6 k ) ! 3 ) {\displaystyle G=-{\frac {1}{1024}}\sum _{k=1}^{\infty }{\frac {(-4096)^{k}(45136k^{4}-57184k^{3}+21240k^{2}-3160k+165)}{k^{3}(2k-1)^{3}}}\left({\frac {(2k)!^{6}(3k)!^{3}}{k!^{3}(6k)!^{3}}}\right)}

All of these series have time complexity O ( n log ( n ) 3 ) {\displaystyle O(n\log(n)^{3})} .

Integral identities

As Seán Stewart writes, "There is a rich and seemingly endless source of definite integrals that can be equated to or expressed in terms of Catalan's constant." Some of these expressions include: G = 1 π i 0 π 2 ln ln tan x ln tan x d x G = [ 0 , 1 ] 2 1 1 + x 2 y 2 d x d y G = 0 1 0 1 x 1 1 x 2 y 2 d y d x G = 1 ln t 1 + t 2 d t G = 0 1 ln t 1 + t 2 d t G = 1 2 0 π 2 t sin t d t G = 0 π 4 ln cot t d t G = 1 2 0 π 2 ln ( sec t + tan t ) d t G = 0 1 arccos t 1 + t 2 d t G = 0 1 arcsinh t 1 t 2 d t G = 1 2 0 arctan t t 1 + t 2 d t G = 1 2 0 1 arctanh t 1 t 2 d t G = 0 arccot e t d t G = 1 4 0 π 2 / 4 csc t d t G = 1 16 ( π 2 + 4 1 arccsc 2 t d t ) G = 1 2 0 t cosh t d t G = π 2 1 ( t 4 6 t 2 + 1 ) ln ln t ( 1 + t 2 ) 3 d t G = 1 2 0 arcsin ( sin t ) t d t G = 1 + lim α 1 { 0 α ( 1 + 6 t 2 + t 4 ) arctan t t ( 1 t 2 ) 2 d t + 2 artanh α π α 1 α 2 } G = 1 1 8 R 2 x sin ( 2 x y / π ) ( x 2 + π 2 ) cosh x sinh y d x d y G = 0 0 x 4 ( x y 1 ) ( x + 1 ) 2 y 4 ( y + 1 ) 2 log ( x y ) d x d y {\displaystyle {\begin{aligned}G&=-{\frac {1}{\pi i}}\int _{0}^{\frac {\pi }{2}}\ln \ln \tan x\ln \tan x\,dx\\G&=\iint _{^{2}}\!{\frac {1}{1+x^{2}y^{2}}}\,dx\,dy\\G&=\int _{0}^{1}\int _{0}^{1-x}{\frac {1}{1-x^{2}-y^{2}}}\,dy\,dx\\G&=\int _{1}^{\infty }{\frac {\ln t}{1+t^{2}}}\,dt\\G&=-\int _{0}^{1}{\frac {\ln t}{1+t^{2}}}\,dt\\G&={\frac {1}{2}}\int _{0}^{\frac {\pi }{2}}{\frac {t}{\sin t}}\,dt\\G&=\int _{0}^{\frac {\pi }{4}}\ln \cot t\,dt\\G&={\frac {1}{2}}\int _{0}^{\frac {\pi }{2}}\ln \left(\sec t+\tan t\right)\,dt\\G&=\int _{0}^{1}{\frac {\arccos t}{\sqrt {1+t^{2}}}}\,dt\\G&=\int _{0}^{1}{\frac {\operatorname {arcsinh} t}{\sqrt {1-t^{2}}}}\,dt\\G&={\frac {1}{2}}\int _{0}^{\infty }{\frac {\operatorname {arctan} t}{t{\sqrt {1+t^{2}}}}}\,dt\\G&={\frac {1}{2}}\int _{0}^{1}{\frac {\operatorname {arctanh} t}{\sqrt {1-t^{2}}}}\,dt\\G&=\int _{0}^{\infty }\operatorname {arccot} e^{t}\,dt\\G&={\frac {1}{4}}\int _{0}^{{\pi ^{2}}/{4}}\csc {\sqrt {t}}\,dt\\G&={\frac {1}{16}}\left(\pi ^{2}+4\int _{1}^{\infty }\operatorname {arccsc} ^{2}t\,dt\right)\\G&={\frac {1}{2}}\int _{0}^{\infty }{\frac {t}{\cosh t}}\,dt\\G&={\frac {\pi }{2}}\int _{1}^{\infty }{\frac {\left(t^{4}-6t^{2}+1\right)\ln \ln t}{\left(1+t^{2}\right)^{3}}}\,dt\\G&={\frac {1}{2}}\int _{0}^{\infty }{\frac {\arcsin \left(\sin t\right)}{t}}\,dt\\G&=1+\lim _{\alpha \to {1^{-}}}\!\left\{\int _{0}^{\alpha }\!{\frac {\left(1+6t^{2}+t^{4}\right)\arctan {t}}{t\left(1-t^{2}\right)^{2}}}\,dt+2\operatorname {artanh} {\alpha }-{\frac {\pi \alpha }{1-\alpha ^{2}}}\right\}\\G&=1-{\frac {1}{8}}\iint _{\mathbb {R} ^{2}}\!\!{\frac {x\sin \left(2xy/\pi \right)}{\,\left(x^{2}+\pi ^{2}\right)\cosh x\sinh y\,}}\,dx\,dy\\G&=\int _{0}^{\infty }\int _{0}^{\infty }{\frac {{\sqrt{x}}\left({\sqrt {x}}{\sqrt {y}}-1\right)}{(x+1)^{2}{\sqrt{y}}(y+1)^{2}\log(xy)}}dxdy\end{aligned}}}

where the last three formulas are related to Malmsten's integrals.

If K(k) is the complete elliptic integral of the first kind, as a function of the elliptic modulus k, then G = 1 2 0 1 K ( k ) d k {\displaystyle G={\tfrac {1}{2}}\int _{0}^{1}\mathrm {K} (k)\,dk}

If E(k) is the complete elliptic integral of the second kind, as a function of the elliptic modulus k, then G = 1 2 + 0 1 E ( k ) d k {\displaystyle G=-{\tfrac {1}{2}}+\int _{0}^{1}\mathrm {E} (k)\,dk}

With the gamma function Γ(x + 1) = x! G = π 4 0 1 Γ ( 1 + x 2 ) Γ ( 1 x 2 ) d x = π 2 0 1 2 Γ ( 1 + y ) Γ ( 1 y ) d y {\displaystyle {\begin{aligned}G&={\frac {\pi }{4}}\int _{0}^{1}\Gamma \left(1+{\frac {x}{2}}\right)\Gamma \left(1-{\frac {x}{2}}\right)\,dx\\&={\frac {\pi }{2}}\int _{0}^{\frac {1}{2}}\Gamma (1+y)\Gamma (1-y)\,dy\end{aligned}}}

The integral G = Ti 2 ( 1 ) = 0 1 arctan t t d t {\displaystyle G=\operatorname {Ti} _{2}(1)=\int _{0}^{1}{\frac {\arctan t}{t}}\,dt} is a known special function, called the inverse tangent integral, and was extensively studied by Srinivasa Ramanujan.

Relation to special functions

G appears in values of the second polygamma function, also called the trigamma function, at fractional arguments:

ψ 1 ( 1 4 ) = π 2 + 8 G ψ 1 ( 3 4 ) = π 2 8 G . {\displaystyle {\begin{aligned}\psi _{1}\left({\tfrac {1}{4}}\right)&=\pi ^{2}+8G\\\psi _{1}\left({\tfrac {3}{4}}\right)&=\pi ^{2}-8G.\end{aligned}}}

Simon Plouffe gives an infinite collection of identities between the trigamma function, π and Catalan's constant; these are expressible as paths on a graph.

Catalan's constant occurs frequently in relation to the Clausen function, the inverse tangent integral, the inverse sine integral, the Barnes G-function, as well as integrals and series summable in terms of the aforementioned functions.

As a particular example, by first expressing the inverse tangent integral in its closed form – in terms of Clausen functions – and then expressing those Clausen functions in terms of the Barnes G-function, the following expression is obtained (see Clausen function for more):

G = 4 π log ( G ( 3 8 ) G ( 7 8 ) G ( 1 8 ) G ( 5 8 ) ) + 4 π log ( Γ ( 3 8 ) Γ ( 1 8 ) ) + π 2 log ( 1 + 2 2 ( 2 2 ) ) . {\displaystyle G=4\pi \log \left({\frac {G\left({\frac {3}{8}}\right)G\left({\frac {7}{8}}\right)}{G\left({\frac {1}{8}}\right)G\left({\frac {5}{8}}\right)}}\right)+4\pi \log \left({\frac {\Gamma \left({\frac {3}{8}}\right)}{\Gamma \left({\frac {1}{8}}\right)}}\right)+{\frac {\pi }{2}}\log \left({\frac {1+{\sqrt {2}}}{2\left(2-{\sqrt {2}}\right)}}\right).}

If one defines the Lerch transcendent Φ(z,s,α) by Φ ( z , s , α ) = n = 0 z n ( n + α ) s , {\displaystyle \Phi (z,s,\alpha )=\sum _{n=0}^{\infty }{\frac {z^{n}}{(n+\alpha )^{s}}},} then G = 1 4 Φ ( 1 , 2 , 1 2 ) . {\displaystyle G={\tfrac {1}{4}}\Phi \left(-1,2,{\tfrac {1}{2}}\right).}

Continued fraction

G can be expressed in the following form:

G = 1 1 + 1 4 8 + 3 4 16 + 5 4 24 + 7 4 32 + 9 4 40 + {\displaystyle G={\cfrac {1}{1+{\cfrac {1^{4}}{8+{\cfrac {3^{4}}{16+{\cfrac {5^{4}}{24+{\cfrac {7^{4}}{32+{\cfrac {9^{4}}{40+\ddots }}}}}}}}}}}}}

The simple continued fraction is given by:

G = 1 1 + 1 10 + 1 1 + 1 8 + 1 1 + 1 88 + {\displaystyle G={\cfrac {1}{1+{\cfrac {1}{10+{\cfrac {1}{1+{\cfrac {1}{8+{\cfrac {1}{1+{\cfrac {1}{88+\ddots }}}}}}}}}}}}}

This continued fraction would have infinite terms if and only if G {\displaystyle G} is irrational, which is still unresolved.

Known digits

The number of known digits of Catalan's constant G has increased dramatically during the last decades. This is due both to the increase of performance of computers as well as to algorithmic improvements.

Number of known decimal digits of Catalan's constant G
Date Decimal digits Computation performed by
1832 16 Thomas Clausen
1858 19 Carl Johan Danielsson Hill
1864 14 Eugène Charles Catalan
1877 20 James W. L. Glaisher
1913 32 James W. L. Glaisher
1990 20000 Greg J. Fee
1996 50000 Greg J. Fee
August 14, 1996 100000 Greg J. Fee & Simon Plouffe
September 29, 1996 300000 Thomas Papanikolaou
1996 1500000 Thomas Papanikolaou
1997 3379957 Patrick Demichel
January 4, 1998 12500000 Xavier Gourdon
2001 100000500 Xavier Gourdon & Pascal Sebah
2002 201000000 Xavier Gourdon & Pascal Sebah
October 2006 5000000000 Shigeru Kondo & Steve Pagliarulo
August 2008 10000000000 Shigeru Kondo & Steve Pagliarulo
January 31, 2009 15510000000 Alexander J. Yee & Raymond Chan
April 16, 2009 31026000000 Alexander J. Yee & Raymond Chan
June 7, 2015 200000001100 Robert J. Setti
April 12, 2016 250000000000 Ron Watkins
February 16, 2019 300000000000 Tizian Hanselmann
March 29, 2019 500000000000 Mike A & Ian Cutress
July 16, 2019 600000000100 Seungmin Kim
September 6, 2020 1000000001337 Andrew Sun
March 9, 2022 1200000000100 Seungmin Kim

See also

References

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Further reading

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