Continued Fractions - Hyperbolic Trigonometric Functions

Hyperbolic Trigonometric Functions

We derive some continued fraction expansions for hyperbolic trigonometric functions.

Theorem: If $a \neq 0, a k + b \neq 0$ for all integers $k$ , and

\begin{aligned} P & = & \sum_{k = 0}^{∞} & \frac{1}{k! a^{k} (a + b) (2 a + b) . . . (a k + b)} \\ Q & = & \sum_{k = 0}^{∞} & \frac{1}{k! a^{k} (a + b) (2 a + b) . . . (a (k + 1) + b)} \end{aligned}

then

\frac{P}{Q} ~ \prod_{k = 1}^{∞} (\begin{matrix} a k + b & 1 \\ 1 & 0 \end{matrix})

Proof: By induction

\prod_{k = 1}^{n} (\begin{matrix} a k + b & 1 \\ 1 & 0 \end{matrix}) = (\begin{matrix} f_{n} & f_{n - 1} \\ g_{n} & g_{n - 1} \end{matrix})

where

\begin{aligned} f_{n} & = & \sum_{k = 0}^{⌊ \frac{n}{2} ⌋} & (\begin{matrix} n - k \\ k \end{matrix}) & \prod_{r = k + 1}^{n - k} (a r + b) \\ g_{n} & = & \sum_{k = 0}^{⌊ \frac{n - 1}{2} ⌋} & (\begin{matrix} n - k - 1 \\ k \end{matrix}) & \prod_{r = k + 2}^{n - k} (a r + b) \end{aligned}

Then define $u_{n, k}$ by

\begin{aligned} \frac{f_{n}}{(a + b) (2 a + b) . . . (a n + b)} & = & \sum_{k = 0}^{⌊ \frac{n}{2} ⌋} & \frac{(n - k) (n - k - 1) . . . (n - 2 k + 1) (a (k + 1) + b) . . . (a (n - k) + b)}{(a + b) (2 a + b) . . . (a n + b) k!} \\ = & \sum_{k = 0}^{⌊ \frac{n}{2} ⌋} & \frac{(1 - \frac{k}{n}) (1 - \frac{k + 1}{n}) . . . (1 - \frac{2 k - 1}{n})}{(1 - \frac{k - 1}{n} + \frac{b}{a n}) . . . (1 + \frac{b}{a n}) (a + b) (2 a + b) . . . (a n + b) a^{k} k!} \\ = & \sum_{k = 0}^{⌊ \frac{n}{2} ⌋} & \frac{u_{n, k}}{(a + b) (2 a + b) . . . (a n + b) a^{k} k!} \end{aligned}

As $n \to ∞$ , $u_{n, k} \to 1$ for fixed $k$ hence $\frac{f_{n}}{(a + b) . . . (a n + b)} \to P$ and similarly, $\frac{g_{n}}{(a + b) . . . (a n + b)} \to Q$ . $▪$

Corollary: For nonzero $x, y \in ℂ$ ,

\prod_{k = 1}^{∞} (\begin{matrix} (2 k - 1) y & x \\ x & 0 \end{matrix}) ~ \coth \frac{x}{y}

and

(\begin{matrix} i & 0 \\ 0 & 1 \end{matrix}) \prod_{k = 1}^{∞} (\begin{matrix} (2 k - 1) y & i x \\ i x & 0 \end{matrix}) ~ \cot \frac{x}{y}

Proof: Use the theorem with $a = 2 y / x, b = - y / x$ to find $\prod_{k = 1}^{∞} (\begin{matrix} a k + b & 1 \\ 1 & 0 \end{matrix}) ~ \frac{\cosh (x / y)}{\sinh (x / y)}$ . Multiplying each matrix by $x$ completes the proof.

Example: Set $x = 1$ and we have

\prod_{k = 1}^{∞} (\begin{matrix} (2 k - 1) y & 1 \\ 1 & 0 \end{matrix}) ~ \coth \frac{1}{y}

In particular, when $y$ is a positive integer, we have $\coth \frac{1}{y} = [y; 3 y,5 y, . . .]$ .

Theorem: If $m, n$ are positive integers then $e^{m / n}$ is irrational.

Proof: Since $\frac{\cosh m / n}{\sinh m / n} = \frac{e^{2 m / n} + 1}{e^{2 m / n} - 1}$ is irrational, it follows $e^{2 m / n}$ and hence $e^{m / n}$ are irrational.

Theorem: If $x, y$ are nonzero then

\frac{\cos (x / y)}{\sin (x / y)} ~ \prod_{k = 1}^{∞} (\begin{matrix} (- 1)^{k + 1} (2 k - 1) y & x \\ x & 0 \end{matrix})

Proof: By the following identities:

\begin{aligned} (\begin{matrix} - i & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix} a & i b \\ i b & 0 \end{matrix}) (\begin{matrix} i & 0 \\ 0 & 1 \end{matrix}) & = & - (\begin{matrix} a & b \\ b & 0 \end{matrix}) \\ (\begin{matrix} - i & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix} a & i b \\ i b & 0 \end{matrix}) (\begin{matrix} - i & 0 \\ 0 & 1 \end{matrix}) & = & (\begin{matrix} - a & b \\ b & 0 \end{matrix}) \\ (\begin{matrix} i & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix} - i & 0 \\ 0 & 1 \end{matrix}) & = & (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) \end{aligned}

we have

\begin{aligned} (\begin{matrix} - i & 0 \\ 0 & 1 \end{matrix}) {\prod_{k = 1}^{2 n} (\begin{matrix} (2 k - 1) y & i x \\ i x & 0 \end{matrix})} (\begin{matrix} i & 0 \\ 0 & 1 \end{matrix}) \\ = & (\begin{matrix} - i & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix} y & i x \\ i x & 0 \end{matrix}) (\begin{matrix} - i & 0 \\ 0 & 1 \end{matrix}) \times (\begin{matrix} - i & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix} 3 y & i x \\ i x & 0 \end{matrix}) (\begin{matrix} - i & 0 \\ 0 & 1 \end{matrix}) \times . . . \\ ~ & (\begin{matrix} y & x \\ x & 0 \end{matrix}) (\begin{matrix} - 3 y & x \\ x & 0 \end{matrix}) (\begin{matrix} 5 y & x \\ x & 0 \end{matrix}) . . . \\ = & \prod_{k = 1}^{2 n} (\begin{matrix} (- 1)^{k + 1} (2 k - 1) y & x \\ x & 0 \end{matrix}) \end{aligned}

Theorem: If $x, y$ are positive integers then

\frac{\cos (x / y)}{\sin (x / y)} ~ (\begin{matrix} y - x & x \\ x & 0 \end{matrix}) \prod_{k = 2}^{∞} {(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} (2 k - 1) y - 2 x & x \\ x & 0 \end{matrix})}

Proof: We have the following identities:

(\begin{matrix} a & b \\ b & 0 \end{matrix}) = (\begin{matrix} a - b & b \\ b & 0 \end{matrix}) (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix})

(\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}) (\begin{matrix} - a & b \\ b & 0 \end{matrix}) (\begin{matrix} 1 & 0 \\ 1 & - 1 \end{matrix}) = - (\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} a - 2 b & b \\ b & 0 \end{matrix})

(\begin{matrix} 1 & 0 \\ 1 & - 1 \end{matrix}) (\begin{matrix} a & b \\ b & 0 \end{matrix}) (\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix}) = (\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} a - 2 b & b \\ b & 0 \end{matrix})

(\begin{matrix} 1 & 0 \\ 1 & - 1 \end{matrix}) (\begin{matrix} 1 & 0 \\ 1 & - 1 \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

(\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix}) (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

which gives

\begin{aligned} \prod_{k = 1}^{2 n + 1} (\begin{matrix} (- 1)^{k + 1} (2 k - 1) y & x \\ x & 0 \end{matrix}) \\ = & (\begin{matrix} y & x \\ x & 0 \end{matrix}) (\begin{matrix} - 3 y & x \\ x & 0 \end{matrix}) (\begin{matrix} 5 y & x \\ x & 0 \end{matrix}) . . . \\ = & (\begin{matrix} y - x & x \\ x & 0 \end{matrix}) (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}) (\begin{matrix} - 3 y & x \\ x & 0 \end{matrix}) (\begin{matrix} 1 & 0 \\ 1 & - 1 \end{matrix}) \times (\begin{matrix} 1 & 0 \\ 1 & - 1 \end{matrix}) (\begin{matrix} 5 y & x \\ x & 0 \end{matrix}) . . . \\ ~ & (\begin{matrix} y - x & x \\ x & 0 \end{matrix}) (\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} - 3 y - 2 x & x \\ x & 0 \end{matrix}) \times (\begin{matrix} 1 & 0 \\ 1 & - 1 \end{matrix}) (\begin{matrix} 5 y & x \\ x & 0 \end{matrix}) (\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix}) (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}) (\begin{matrix} - 7 y & x \\ x & 0 \end{matrix}) . . . \\ ~ & (\begin{matrix} y - x & x \\ x & 0 \end{matrix}) (\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} - 3 y - 2 x & x \\ x & 0 \end{matrix}) (\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} 5 y - 2 x & x \\ x & 0 \end{matrix}) \times (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}) (\begin{matrix} - 7 y & x \\ x & 0 \end{matrix}) . . . \\ ~ & (\begin{matrix} y - x & x \\ x & 0 \end{matrix}) \prod_{k = 2}^{2 n + 1} {(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} (2 k - 1) y - 2 x & x \\ x & 0 \end{matrix})} (\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix}) \end{aligned}

and similarly

\begin{aligned} \prod_{k = 1}^{2 n} (\begin{matrix} (- 1)^{k + 1} (2 k - 1) y - 2 x & x \\ x & 0 \end{matrix}) \\ = & (\begin{matrix} y - x & x \\ x & 0 \end{matrix}) \prod_{k = 2}^{2 n} {(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} (2 k - 1) y - 2 x & x \\ x & 0 \end{matrix})} (\begin{matrix} 1 & 0 \\ 1 & - 1 \end{matrix}) ▪ \end{aligned}