Cahen's constant

Template:Short description In mathematics, Cahen's constant is defined as the value of an infinite series of unit fractions with alternating signs:

C = \sum_{i = 0}^{\infty} \frac{(- 1)^{i}}{s_{i} - 1} = \frac{1}{1} - \frac{1}{2} + \frac{1}{6} - \frac{1}{42} + \frac{1}{1806} - \dots \approx 0.643410546288 . . .

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Here $(s_{i})_{i \geq 0}$ denotes Sylvester's sequence, which is defined recursively by

\begin{matrix} s_{0} = 2; \\ s_{i + 1} = 1 + \prod_{j = 0}^{i} s_{j} for i \geq 0 . \end{matrix}

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:

C = \sum \frac{1}{s_{2 i}} = \frac{1}{2} + \frac{1}{7} + \frac{1}{1807} + \frac{1}{10650056950807} + \dots

This constant is named after Template:Ill (also known for the Cahen–Mellin integral), who was the first to introduce it and prove its irrationality.Template:Sfnp

Continued fraction expansion

The majority of naturally occurring^[1] mathematical constants have no known simple patterns in their continued fraction expansions.Template:Sfnp Nevertheless, the complete continued fraction expansion of Cahen's constant $C$ is known: it is $C = [a_{0}^{2}; a_{1}^{2}, a_{2}^{2}, a_{3}^{2}, a_{4}^{2}, \dots] = [0; 1, 1, 1, 4, 9, 196, 16641, \dots]$ where the sequence of coefficients Template:Bi is defined by the recurrence relation $a_{0} = 0, a_{1} = 1, a_{n + 2} = a_{n} (1 + a_{n} a_{n + 1}) \forall n \in ℤ_{⩾ 0} .$ All the partial quotients of this expansion are squares of integers. Davison and Shallit made use of the continued fraction expansion to prove that $C$ is transcendental.Template: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 $n \geq 1$ that $1 + a_{n} a_{n + 1} = s_{n - 1}$ . Indeed, we have $1 + a_{1} a_{2} = 2 = s_{0}$ , and if $1 + a_{n} a_{n + 1} = s_{n - 1}$ holds for some $n \geq 1$ , then

$1 + a_{n + 1} a_{n + 2} = 1 + a_{n + 1} \cdot a_{n} (1 + a_{n} a_{n + 1}) = 1 + a_{n} a_{n + 1} + (a_{n} a_{n + 1})^{2} = s_{n - 1} + (s_{n - 1} - 1)^{2} = s_{n - 1}^{2} - s_{n - 1} + 1 = s_{n},$ where we used the recursion for $(a_{n})_{n \geq 0}$ in the first step respectively the recursion for $(s_{n})_{n \geq 0}$ in the final step. As a consequence, $a_{n + 2} = a_{n} \cdot s_{n - 1}$ holds for every $n \geq 1$ , from which it is easy to conclude that

$C = [0; 1, 1, 1, s_{0}^{2}, s_{1}^{2}, (s_{0} s_{2})^{2}, (s_{1} s_{3})^{2}, (s_{0} s_{2} s_{4})^{2}, \dots]$ .

Best approximation order

Cahen's constant $C$ has best approximation order $q^{- 3}$ . That means, there exist constants $K_{1}, K_{2} > 0$ such that the inequality $0 < | C - \frac{p}{q} | < \frac{K_{1}}{q^{3}}$ has infinitely many solutions $(p, q) \in ℤ \times ℕ$ , while the inequality $0 < | C - \frac{p}{q} | < \frac{K_{2}}{q^{3}}$ has at most finitely many solutions $(p, q) \in ℤ \times ℕ$ . This implies (but is not equivalent to) the fact that $C$ has irrationality measure 3, which was first observed by Template:Harvtxt.

To give a proof, denote by $(p_{n} / q_{n})_{n \geq 0}$ the sequence of convergents to Cahen's constant (that means, $q_{n - 1} = a_{n} for every n \geq 1$ ).^[2]

But now it follows from $a_{n + 2} = a_{n} \cdot s_{n - 1}$ and the recursion for $(s_{n})_{n \geq 0}$ that

\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 (1 - \frac{1}{s_{n - 1}} + \frac{1}{s_{n - 1}^{2}})

for every $n \geq 1$ . As a consequence, the limits

α : = \lim_{n \to \infty} \frac{q_{2 n + 1}}{q_{2 n}^{2}} = \prod_{n = 0}^{\infty} (1 - \frac{1}{s_{2 n}} + \frac{1}{s_{2 n}^{2}})

and

β : = \lim_{n \to \infty} \frac{q_{2 n + 2}}{q_{2 n + 1}^{2}} = 2 \cdot \prod_{n = 0}^{\infty} (1 - \frac{1}{s_{2 n + 1}} + \frac{1}{s_{2 n + 1}^{2}})

(recall that $s_{0} = 2$ ) both exist by basic properties of infinite products, which is due to the absolute convergence of $\sum_{n = 0}^{\infty} | \frac{1}{s_{n}} - \frac{1}{s_{n}^{2}} |$ . Numerically, one can check that $0 < α < 1 < β < 2$ . Thus the well-known inequality

\frac{1}{q_{n} (q_{n} + q_{n + 1})} \leq | C - \frac{p_{n}}{q_{n}} | \leq \frac{1}{q_{n} q_{n + 1}}

yields

| C - \frac{p_{2 n + 1}}{q_{2 n + 1}} | \leq \frac{1}{q_{2 n + 1} q_{2 n + 2}} = \frac{1}{q_{2 n + 1}^{3} \cdot \frac{q_{2 n + 2}}{q_{2 n + 1}^{2}}} < \frac{1}{q_{2 n + 1}^{3}}

and

| C - \frac{p_{n}}{q_{n}} | \geq \frac{1}{q_{n} (q_{n} + q_{n + 1})} > \frac{1}{q_{n} (q_{n} + 2 q_{n}^{2})} \geq \frac{1}{3 q_{n}^{3}}

for all sufficiently large $n$ . Therefore $C$ has best approximation order 3 (with $K_{1} = 1 and K_{2} = 1 / 3$ ), where we use that any solution $(p, q) \in ℤ \times ℕ$ to

0 < | C - \frac{p}{q} | < \frac{1}{3 q^{3}}

is necessarily a convergent to Cahen's constant.

Notes

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References

External links

Template:Irrational number

↑ 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 $e = \lim_{n \to \infty} (1 + \frac{1}{n})^{n}$ is naturally occurring.
↑ Template:Cite OEIS

[1] 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 $e = \lim_{n \to \infty} (1 + \frac{1}{n})^{n}$ is naturally occurring.

[2] Template:Cite OEIS

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Cahen's constant

Contents

Continued fraction expansion

Best approximation order

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Cahen's constant

Continued fraction expansion

Best approximation order

Notes

References

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