Cheat Sheet
From brute force matrix calculations we can prove:
\[
\begin{aligned}
\coth (1/n) &=& [n; 3n, 5n,...] \\
\tan(1) &=& [1; 1, 1, 3, 1, 5, 1, 7, 1, ...] & \\
\tan(1/n) &=& [0; n-1, 1, 3n - 2, 1, 5n -2, 1, 7n -2, 1, ...] & (n \gt 1) \\
\end{aligned}
\]
Invert (insert or remove a 0) to derive expansions for \(\tanh\) and \(\cot\). We can compute the other trigonometric and hyperbolic trigonometric functions by solving quadratic equations involving continued fractions via the \(\tan x/2\) and \(\tanh x/2\) identities.
From the continued fraction of Gauss we have:
\[
\begin{aligned}
\arctan z &=& \frac{z}{1 + \frac{z^2}{3 + \frac{(2 z)^2}{5 + \frac{(3 z)^2}{...}}}} \\
\tanh z &=& \frac{z}{1 + \frac{z^2}{3 + \frac{z^2}{5 + \frac{z^2}{...}}}} \\
\tan z &=& \frac{z}{1 - \frac{z^2}{3 - \frac{z^2}{5 - \frac{z^2}{...}}}}
\end{aligned}
\]
The inverse tangent is useful for computing other inverse trigonometric functions. Its expansion also gives an expansion for \(\pi\) by setting \(z = 1\). Also,
\[
\pi =
3 + \frac{1}{6 + \frac{9}{6 + \frac{25}{6 + ...}}}
\]
but this converges far too slowly for practical purposes.
Euler’s continued fraction formula can show:
\[
\begin{aligned}
\log(1+x) &=& \frac{x}{1 + \frac{1^2 x}{2-x +\frac{2^2 x}{3-2x + ...}}} \\
e^x &=& 1+\frac{x}{1-\frac{x}{x+2-\frac{2x}{x+3-\frac{3x}{x+4-...}}}}
\end{aligned}
\]
Other well-known expansions [scoured from web searches; I wish I knew how these were derived]:
\[
\begin{aligned}
e &=& [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, ...] & \\
\exp(1/n) &=& [1; n - 1, 1, 1, 3n - 1, 1 ,1 , 5n - 1, 1 ,1, ...] & (n \gt 1) \\
\exp(2m/n) &=& 1 + \frac{2m}{(n-m)+\frac{m^2}{3n + \frac{m^2}{5n +\frac{m^2}{7n+...}}}} \\
\exp(z) &=& 1+\frac{z}{1-\frac{z}{2+\frac{z}{3-\frac{z}{2+\frac{z}{5-\frac{z}{2+...}}}}}} \\
\end{aligned}
\]