Continued Fractions - Cheat Sheet

Cheat Sheet

From brute force matrix calculations we can prove:

\begin{aligned} \coth (1 / n) & = & [n; 3 n, 5 n, . . .] \\ \tan (1) & = & [1; 1, 1, 3, 1, 5, 1, 7, 1, . . .] \\ \tan (1 / n) & = & [0; n - 1, 1, 3 n - 2, 1, 5 n - 2, 1, 7 n - 2, 1, . . .] & (n > 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 $π$ by setting $z = 1$ . Also,

π = 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 - 2 x + . . .}}} \\ e^{x} & = & 1 + \frac{x}{1 - \frac{x}{x + 2 - \frac{2 x}{x + 3 - \frac{3 x}{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, 3 n - 1, 1,1, 5 n - 1, 1,1, . . .] & (n > 1) \\ \exp (2 m / n) & = & 1 + \frac{2 m}{(n - m) + \frac{m^{2}}{3 n + \frac{m^{2}}{5 n + \frac{m^{2}}{7 n + . . .}}}} \\ \exp (z) & = & 1 + \frac{z}{1 - \frac{z}{2 + \frac{z}{3 - \frac{z}{2 + \frac{z}{5 - \frac{z}{2 + . . .}}}}}} \end{aligned}