Английская Википедия:Cahen's constant
In mathematics, Cahen's constant is defined as the value of an infinite series of unit fractions with alternating signs:
- <math>C = \sum_{i=0}^\infty \frac{(-1)^i}{s_i-1}=\frac11 - \frac12 + \frac16 - \frac1{42} + \frac1{1806} - \cdots\approx 0.643410546288...</math> Шаблон:OEIS
Here <math>(s_i)_{i \geq 0}</math> denotes Sylvester's sequence, which is defined recursively by
- <math>\begin{array}{l}
s_0EducationBot (обсуждение) = 2; \\ s_{i+1} = 1 + \prod_{j=0}^i s_j \text{ for } i \geq 0. \end{array}</math>
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:
- <math>C = \sum\frac{1}{s_{2i}}=\frac12+\frac17+\frac1{1807}+\frac1{10650056950807}+\cdots</math>
This constant is named after Шаблон:Ill (also known for the Cahen–Mellin integral), who was the first to introduce it and prove its irrationality.Шаблон:Sfnp
Continued fraction expansion
The majority of naturally occurring[1] mathematical constants have no known simple patterns in their continued fraction expansions.Шаблон:Sfnp Nevertheless, the complete continued fraction expansion of Cahen's constant <math>C</math> is known: it is <math display=block>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]</math> where the sequence of coefficients Шаблон:Bi is defined by the recurrence relation <math display=block>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}.</math> All the partial quotients of this expansion are squares of integers. Davison and Shallit made use of the continued fraction expansion to prove that <math>C</math> is transcendental.Шаблон:Sfnp
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 <math>n \geq 1</math> that <math>1+a_n a_{n+1} = s_{n-1}</math>. Indeed, we have <math>1+a_1 a_2 = 2 = s_0</math>, and if <math>1+a_n a_{n+1} = s_{n-1}</math> holds for some <math>n \geq 1</math>, then
<math>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_na_{n+1})^2 = s_{n-1} + (s_{n-1}-1)^2 = s_{n-1}^2-s_{n-1}+1 = s_n,</math>where we used the recursion for <math>(a_n)_{n \geq 0}</math> in the first step respectively the recursion for <math>(s_n)_{n \geq 0}</math> in the final step. As a consequence, <math>a_{n+2} = a_n \cdot s_{n-1}</math> holds for every <math>n \geq 1</math>, from which it is easy to conclude that
<math>C = [0;1,1,1,s_0^2, s_1^2, (s_0s_2)^2, (s_1s_3)^2, (s_0s_2s_4)^2,\ldots]</math>.
Best approximation order
Cahen's constant <math>C</math> has best approximation order <math>q^{-3}</math>. That means, there exist constants <math>K_1, K_2 > 0</math> such that the inequality <math> 0 < \Big| C - \frac{p}{q} \Big| < \frac{K_1}{q^3} </math> has infinitely many solutions <math> (p,q) \in \mathbb{Z} \times \mathbb{N} </math>, while the inequality <math> 0 < \Big| C - \frac{p}{q} \Big| < \frac{K_2}{q^3} </math> has at most finitely many solutions <math> (p,q) \in \mathbb{Z} \times \mathbb{N} </math>. This implies (but is not equivalent to) the fact that <math>C</math> has irrationality measure 3, which was first observed by Шаблон:Harvtxt.
To give a proof, denote by <math>(p_n/q_n)_{n \geq 0}</math> the sequence of convergents to Cahen's constant (that means, <math>q_{n-1} = a_n \text{ for every } n \geq 1</math>).[2]
But now it follows from <math>a_{n+2} = a_n \cdot s_{n-1}</math>and the recursion for <math>(s_n)_{n \geq 0}</math> that
- <math>\frac{a_{n+2}}{a_{n+1}^2} = \frac{a_{n} \cdot s_{n-1}}{a_{n-1}^2 \cdot s_{n-2}^2} = \frac{a_n}{a_{n-1}^2} \cdot \frac{s_{n-2}^2 - s_{n-2} + 1}{s_{n-1}^2} = \frac{a_n}{a_{n-1}^2} \cdot \Big( 1 - \frac{1}{s_{n-1}} + \frac{1}{s_{n-1}^2} \Big)</math>
for every <math>n \geq 1</math>. As a consequence, the limits
- <math>\alpha := \lim_{n \to \infty} \frac{q_{2n+1}}{q_{2n}^2} = \prod_{n=0}^\infty \Big( 1 - \frac{1}{s_{2n}} + \frac{1}{s_{2n}^2}\Big)</math> and <math>\beta := \lim_{n \to \infty} \frac{q_{2n+2}}{q_{2n+1}^2} = 2 \cdot \prod_{n=0}^\infty \Big( 1 - \frac{1}{s_{2n+1}} + \frac{1}{s_{2n+1}^2}\Big)</math>
(recall that <math>s_0 = 2</math>) both exist by basic properties of infinite products, which is due to the absolute convergence of <math>\sum_{n=0}^\infty \Big| \frac{1}{s_{n}} - \frac{1}{s_{n}^2} \Big|</math>. Numerically, one can check that <math>0 < \alpha < 1 < \beta < 2</math>. Thus the well-known inequality
- <math>\frac{1}{q_n(q_n + q_{n+1})} \leq \Big| C - \frac{p_n}{q_n} \Big| \leq \frac{1}{q_nq_{n+1}}</math>
yields
- <math>\Big| C - \frac{p_{2n+1}}{q_{2n+1}} \Big| \leq \frac{1}{q_{2n+1}q_{2n+2}} = \frac{1}{q_{2n+1}^3 \cdot
\frac{q_{2n+2}}{q_{2n+1}^2}} < \frac{1}{q_{2n+1}^3}</math> and <math>\Big| C - \frac{p_n}{q_n} \Big| \geq \frac{1}{q_n(q_n + q_{n+1})} > \frac{1}{q_n(q_n + 2q_{n}^2)} \geq \frac{1}{3q_n^3}</math>
for all sufficiently large <math>n</math>. Therefore <math>C</math> has best approximation order 3 (with <math>K_1 = 1 \text{ and } K_2 = 1/3</math>), where we use that any solution <math> (p,q) \in \mathbb{Z} \times \mathbb{N} </math> to
- <math>0 < \Big| C - \frac{p}{q} \Big| < \frac{1}{3q^3}</math>
is necessarily a convergent to Cahen's constant.
Notes
References
External links
- ↑ A number is said to be naturally occurring if it is *not* defined through its decimal or continued fraction expansion. In this sense, e.g., Euler's number <math>e = \lim_{n \to \infty} \Big(1+\frac{1}{n}\Big)^n</math> is naturally occurring.
- ↑ Шаблон:Cite OEIS
