Hyperbolic Trigonometric Functions

We derive some continued fraction expansions for hyperbolic trigonometric functions.

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

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

then

\[ \frac{P}{Q} ~ \prod_{k=1}^\infty \begin{pmatrix} a k + b & 1 \\ 1 & 0 \end{pmatrix} \]

Proof: By induction

\[ \prod_{k=1}^n \begin{pmatrix} a k + b & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} f_n & f_{n-1} \\ g_n & g_{n-1} \end{pmatrix} \]

where

\[ \begin{aligned} f_n &=& \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} & \begin{pmatrix} n-k \\ k \end{pmatrix} & \prod_{r=k+1}^{n-k} (a r + b) \\ g_n &=& \sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} & \begin{pmatrix} n-k-1 \\ k \end{pmatrix} & \prod_{r=k+2}^{n-k} (a r + b) \end{aligned} \]

Then define $u_{n,k}$ by

\[ \begin{aligned} \frac{f_n}{ (a+b)(2a+b)...(a n + b) } &=& \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} & \frac { (n-k)(n-k-1)...(n-2k+1)(a(k+1)+b)...(a(n-k)+b) }{ (a+b)(2a+b)...(a n+b)k! } \\ &=& \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} & \frac { (1-\frac{k}{n})(1-\frac{k+1}{n})...(1-\frac{2k-1}{n}) }{ (1-\frac{k-1}{n}+\frac{b}{a n})...(1+\frac{b}{a n}) (a+b)(2a+b)...(a n+b)a^k k! } \\ &=& \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} & \frac { u_{n,k} }{ (a+b)(2a+b)...(a n+b)a^k k! } \end{aligned} \]

As $n \rightarrow\infty$, $u_{n,k}\rightarrow 1$ for fixed $k$ hence $\frac{f_n}{(a+b)...(a n+b)} \rightarrow P$ and similarly, $\frac{g_n}{(a+b)...(a n+b)} \rightarrow Q$.∎

Corollary: For nonzero $x, y \in \mathbb{C}$,

\[ \prod_{k=1}^\infty \begin{pmatrix} (2k-1)y & x\\ x & 0 \end{pmatrix} ~ \coth \frac{x}{y} \]

and

\[ \begin{pmatrix} i&0\\0&1 \end{pmatrix} \prod_{k=1}^\infty \begin{pmatrix} (2k-1)y & i x\\ i x & 0 \end{pmatrix} ~ \cot \frac{x}{y} \]

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

Example: Set $x = 1$ and we have

\[ \prod_{k=1}^\infty \begin{pmatrix} (2k-1)y & 1\\ 1 & 0 \end{pmatrix} ~ \coth \frac{1}{y} \]

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

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

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

Theorem: If $x, y$ are nonzero then

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

Proof: By the following identities:

\[ \begin{aligned} \begin{pmatrix} -i & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & i b \\ i b & 0 \end{pmatrix} \begin{pmatrix} i & 0 \\ 0 & 1 \end{pmatrix} &=& - \begin{pmatrix} a & b \\ b & 0 \end{pmatrix} \\ \begin{pmatrix} -i & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & i b \\ i b & 0 \end{pmatrix} \begin{pmatrix} - i & 0 \\ 0 & 1 \end{pmatrix} &=& \begin{pmatrix} - a & b \\ b & 0 \end{pmatrix} \\ \begin{pmatrix} i & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} - i & 0 \\ 0 & 1 \end{pmatrix} &=& \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{aligned} \]

we have

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

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

\[ \frac{cos(x/y)}{sin(x/y)} ~ \begin{pmatrix} y-x & x \\ x & 0 \end{pmatrix} \prod_{k=2}^\infty \left\{ \begin{pmatrix}1&1\\1&0\end{pmatrix} \begin{pmatrix} (2k-1)y-2x & x \\ x & 0 \end{pmatrix} \right\} \]

Proof: We have the following identities:

1.

\[ \begin{pmatrix} a & b \\ b & 0 \end{pmatrix} = \begin{pmatrix} a-b & b \\ b & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \]

2.

\[ \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} -a & b \\ b & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix} = - \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} a-2b & b \\ b & 0 \end{pmatrix} \]

3.

\[ \begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} a & b \\ b & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} a-2b & b \\ b & 0 \end{pmatrix} \]

4.

\[ \begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

5.

\[ \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

which gives

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

and similarly

\[ \begin{aligned} & & \prod_{k=1}^{2n} \begin{pmatrix} (-1)^{k+1}(2k-1)y-2x & x \\ x & 0 \end{pmatrix} \\ & = & \begin{pmatrix} y-x & x \\ x & 0 \end{pmatrix} \prod_{k=2}^{2n} \left\{ \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} (2k-1)y-2x & x \\ x & 0 \end{pmatrix} \right\} \begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix} ∎ \end{aligned} \]