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
\[
\frac{P}{Q} \sim \prod_{k=1}^\infty \begin{pmatrix}
a k + b & 1 \\
1 & 0
\end{pmatrix}
\]
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 \(x, y \in \mathbb{C}\),
\[
\prod_{k=1}^\infty \begin{pmatrix}
(2k-1)y & x\\
x & 0
\end{pmatrix} \sim \coth \frac{x}{y}
\]
and
\[
\begin{pmatrix} i&0\\0&1 \end{pmatrix}
\prod_{k=1}^\infty \begin{pmatrix}
(2k-1)y & i x\\
i x & 0
\end{pmatrix} \sim \cot \frac{x}{y}
\]
Proof: Use the theorem with \(a = 2y / x, b = -y /x\) to find \( \prod_{k=1}^\infty \left(\begin{smallmatrix} a k+b&1\\1&0 \end{smallmatrix}\right) \sim \frac{\cosh(x/y)}{\sinh(x/y)} \). Multiplying each matrix by \(x\) completes the proof.
Example: Set \(x = 1\) and we have
\[
\prod_{k=1}^\infty \begin{pmatrix}
(2k-1)y & 1\\
1 & 0
\end{pmatrix} \sim \coth \frac{1}{y}
\]
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
\[\frac{\cos(x/y)}{\sin(x/y)} \sim
\prod_{k=1}^\infty
\begin{pmatrix}
(-1)^{k+1} (2k - 1)y & x \\
x & 0
\end{pmatrix}
\]
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
\[
\begin{aligned}
&
\begin{pmatrix} -i & 0 \\ 0 & 1 \end{pmatrix}
\left\{
\prod_{k=1}^{2n}
\begin{pmatrix} (2k-1)y & i x \\ i x & 0 \end{pmatrix}
\right\}
\begin{pmatrix} i & 0 \\ 0 & 1 \end{pmatrix}
\\
=&
\begin{pmatrix} -i & 0 \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} y & i x \\ i x & 0 \end{pmatrix}
\begin{pmatrix} - i & 0 \\ 0 & 1 \end{pmatrix}
\times
\begin{pmatrix} -i & 0 \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} 3y & i x \\ i x & 0 \end{pmatrix}
\begin{pmatrix} - i & 0 \\ 0 & 1 \end{pmatrix}
\times ...
\\
\sim &
\begin{pmatrix} y & x \\ x & 0 \end{pmatrix}
\begin{pmatrix} -3y & x \\ x & 0 \end{pmatrix}
\begin{pmatrix} 5y & x \\ x & 0 \end{pmatrix}
...
\\
=&
\prod_{k=1}^{2n}
\begin{pmatrix} (-1)^{k+1}(2k-1)y & x \\ x & 0 \end{pmatrix}
\end{aligned}
\]
Theorem: If \(x,y\) are positive integers then
\[
\frac{cos(x/y)}{sin(x/y)} \sim
\begin{pmatrix} y-x & x \\ x & 0 \end{pmatrix}
\prod_{k=2}^\infty \left\{
\begin{pmatrix}1&1\\1&0\end{pmatrix}
\begin{pmatrix} (2k-1)y-2x & x \\ x & 0 \end{pmatrix}
\right\}
\]
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
\[
\begin{aligned}
& &
\prod_{k=1}^{2n+1} \begin{pmatrix}
(-1)^{k+1}(2k-1)y & x \\ x & 0 \end{pmatrix} \\
&=&
\begin{pmatrix} y & x \\ x & 0 \end{pmatrix}
\begin{pmatrix} -3y & x \\ x & 0 \end{pmatrix}
\begin{pmatrix} 5y & x \\ x & 0 \end{pmatrix} ... \\
&=&
\begin{pmatrix} y-x & x \\ x & 0 \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}
\begin{pmatrix} -3y & x \\ x & 0 \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix} \times
\begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix}
\begin{pmatrix} 5y & x \\ x & 0 \end{pmatrix} ... \\
&\sim&
\begin{pmatrix} y-x & x \\ x & 0 \end{pmatrix}
\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}
\begin{pmatrix} -3y-2x & x \\ x & 0 \end{pmatrix} \times
\begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix}
\begin{pmatrix} 5y & x \\ x & 0 \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}
\begin{pmatrix} -7y & x \\ x & 0 \end{pmatrix} ... \\
&\sim&
\begin{pmatrix} y-x & x \\ x & 0 \end{pmatrix}
\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}
\begin{pmatrix} -3y-2x & x \\ x & 0 \end{pmatrix}
\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}
\begin{pmatrix} 5y-2x & x \\ x & 0 \end{pmatrix} \times
\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}
\begin{pmatrix} -7y & x \\ x & 0 \end{pmatrix} ... \\
&\sim&
\begin{pmatrix} y-x & x \\ x & 0 \end{pmatrix}
\prod_{k=2}^{2n+1} \left\{
\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}
\begin{pmatrix} (2k-1)y-2x & x \\ x & 0 \end{pmatrix} \right\}
\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}
\end{aligned}
\]
and similarly
\[
\begin{aligned}
& &
\prod_{k=1}^{2n}
\begin{pmatrix} (-1)^{k+1}(2k-1)y-2x & x \\ x & 0 \end{pmatrix} \\
& = &
\begin{pmatrix} y-x & x \\ x & 0 \end{pmatrix}
\prod_{k=2}^{2n} \left\{
\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}
\begin{pmatrix} (2k-1)y-2x & x \\ x & 0 \end{pmatrix} \right\}
\begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix}
∎
\end{aligned}
\]