In mathematics, Cahen's constant is defined as the value of an infinite series of unit fractions with alternating signs:
Here ( s i ) i ≥ ≥ --> 0 {\displaystyle (s_{i})_{i\geq 0}} denotes Sylvester's sequence, which is defined recursively by
Combining these fractions in pairs leads to an alternative expansion of Cahen's constant as a series of positive unit fractions formed from the terms in even positions of Sylvester's sequence. This series for Cahen's constant forms its greedy Egyptian expansion:
This constant is named after Eugène Cahen [fr] (also known for the Cahen–Mellin integral), who was the first to introduce it and prove its irrationality.[1]
The majority of naturally occurring[2] mathematical constants have no known simple patterns in their continued fraction expansions.[3] Nevertheless, the complete continued fraction expansion of Cahen's constant C {\displaystyle C} is known: it is C = [ a 0 2 ; a 1 2 , a 2 2 , a 3 2 , a 4 2 , … … --> ] = [ 0 ; 1 , 1 , 1 , 4 , 9 , 196 , 16641 , … … --> ] {\displaystyle C=\left[a_{0}^{2};a_{1}^{2},a_{2}^{2},a_{3}^{2},a_{4}^{2},\ldots \right]=[0;1,1,1,4,9,196,16641,\ldots ]} where the sequence of coefficients
is defined by the recurrence relation a 0 = 0 , a 1 = 1 , a n + 2 = a n ( 1 + a n a n + 1 ) ∀ ∀ --> n ∈ ∈ --> Z ⩾ ⩾ --> 0 . {\displaystyle a_{0}=0,~a_{1}=1,~a_{n+2}=a_{n}\left(1+a_{n}a_{n+1}\right)~\forall ~n\in \mathbb {Z} _{\geqslant 0}.} All the partial quotients of this expansion are squares of integers. Davison and Shallit made use of the continued fraction expansion to prove that C {\displaystyle C} is transcendental.[4]
Alternatively, one may express the partial quotients in the continued fraction expansion of Cahen's constant through the terms of Sylvester's sequence: To see this, we prove by induction on n ≥ ≥ --> 1 {\displaystyle n\geq 1} that 1 + a n a n + 1 = s n − − --> 1 {\displaystyle 1+a_{n}a_{n+1}=s_{n-1}} . Indeed, we have 1 + a 1 a 2 = 2 = s 0 {\displaystyle 1+a_{1}a_{2}=2=s_{0}} , and if 1 + a n a n + 1 = s n − − --> 1 {\displaystyle 1+a_{n}a_{n+1}=s_{n-1}} holds for some n ≥ ≥ --> 1 {\displaystyle n\geq 1} , then
1 + a n + 1 a n + 2 = 1 + a n + 1 ⋅ ⋅ --> a n ( 1 + a n a n + 1 ) = 1 + a n a n + 1 + ( a n a n + 1 ) 2 = s n − − --> 1 + ( s n − − --> 1 − − --> 1 ) 2 = s n − − --> 1 2 − − --> s n − − --> 1 + 1 = s n , {\displaystyle 1+a_{n+1}a_{n+2}=1+a_{n+1}\cdot a_{n}(1+a_{n}a_{n+1})=1+a_{n}a_{n+1}+(a_{n}a_{n+1})^{2}=s_{n-1}+(s_{n-1}-1)^{2}=s_{n-1}^{2}-s_{n-1}+1=s_{n},} where we used the recursion for ( a n ) n ≥ ≥ --> 0 {\displaystyle (a_{n})_{n\geq 0}} in the first step respectively the recursion for ( s n ) n ≥ ≥ --> 0 {\displaystyle (s_{n})_{n\geq 0}} in the final step. As a consequence, a n + 2 = a n ⋅ ⋅ --> s n − − --> 1 {\displaystyle a_{n+2}=a_{n}\cdot s_{n-1}} holds for every n ≥ ≥ --> 1 {\displaystyle n\geq 1} , from which it is easy to conclude that
C = [ 0 ; 1 , 1 , 1 , s 0 2 , s 1 2 , ( s 0 s 2 ) 2 , ( s 1 s 3 ) 2 , ( s 0 s 2 s 4 ) 2 , … … --> ] {\displaystyle C=[0;1,1,1,s_{0}^{2},s_{1}^{2},(s_{0}s_{2})^{2},(s_{1}s_{3})^{2},(s_{0}s_{2}s_{4})^{2},\ldots ]} .
Cahen's constant C {\displaystyle C} has best approximation order q − − --> 3 {\displaystyle q^{-3}} . That means, there exist constants K 1 , K 2 > 0 {\displaystyle K_{1},K_{2}>0} such that the inequality 0 < | C − − --> p q | < K 1 q 3 {\displaystyle 0<{\Big |}C-{\frac {p}{q}}{\Big |}<{\frac {K_{1}}{q^{3}}}} has infinitely many solutions ( p , q ) ∈ ∈ --> Z × × --> N {\displaystyle (p,q)\in \mathbb {Z} \times \mathbb {N} } , while the inequality 0 < | C − − --> p q | < K 2 q 3 {\displaystyle 0<{\Big |}C-{\frac {p}{q}}{\Big |}<{\frac {K_{2}}{q^{3}}}} has at most finitely many solutions ( p , q ) ∈ ∈ --> Z × × --> N {\displaystyle (p,q)\in \mathbb {Z} \times \mathbb {N} } . This implies (but is not equivalent to) the fact that C {\displaystyle C} has irrationality measure 3, which was first observed by Duverney & Shiokawa (2020).
To give a proof, denote by ( p n / q n ) n ≥ ≥ --> 0 {\displaystyle (p_{n}/q_{n})_{n\geq 0}} the sequence of convergents to Cahen's constant (that means, q n − − --> 1 = a n for every n ≥ ≥ --> 1 {\displaystyle q_{n-1}=a_{n}{\text{ for every }}n\geq 1} ).[5]
But now it follows from a n + 2 = a n ⋅ ⋅ --> s n − − --> 1 {\displaystyle a_{n+2}=a_{n}\cdot s_{n-1}} and the recursion for ( s n ) n ≥ ≥ --> 0 {\displaystyle (s_{n})_{n\geq 0}} that
for every n ≥ ≥ --> 1 {\displaystyle n\geq 1} . As a consequence, the limits
(recall that s 0 = 2 {\displaystyle s_{0}=2} ) both exist by basic properties of infinite products, which is due to the absolute convergence of ∑ ∑ --> n = 0 ∞ ∞ --> | 1 s n − − --> 1 s n 2 | {\displaystyle \sum _{n=0}^{\infty }{\Big |}{\frac {1}{s_{n}}}-{\frac {1}{s_{n}^{2}}}{\Big |}} . Numerically, one can check that 0 < α α --> < 1 < β β --> < 2 {\displaystyle 0<\alpha <1<\beta <2} . Thus the well-known inequality
yields
for all sufficiently large n {\displaystyle n} . Therefore C {\displaystyle C} has best approximation order 3 (with K 1 = 1 and K 2 = 1 / 3 {\displaystyle K_{1}=1{\text{ and }}K_{2}=1/3} ), where we use that any solution ( p , q ) ∈ ∈ --> Z × × --> N {\displaystyle (p,q)\in \mathbb {Z} \times \mathbb {N} } to
is necessarily a convergent to Cahen's constant.
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