How to express an irrational number as generalized continued fraction?

Question

With simple continued fraction, i.e. $$a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 \ldots}}}$$ I can use this formula: $$a_k = \lfloor \alpha_k \rfloor$$ $$\alpha_{k+1} = \dfrac{1}{\alpha_k - a_k}$$

I wonder is there a formula to express the "generalized continued fraction" of the form: $$a_0 + \cfrac{b_0}{a_1 + \cfrac{b_1}{a_2 + \cfrac{b_2}{a_3 \ldots}}}$$ ?

Thank you,

Answer 1

You want $\alpha_k = a_k + \frac{b_k}{\alpha_{k+1}}$ so $\alpha_{k+1} = \frac{b_k}{\alpha_k - a_k}$.

Answer 2

The simplest way: take all $b_k=1$ and use your previous formula.

Answer 3

@Chan, Here is an example for $\pi$:

$$ \pi=\textstyle \cfrac{4}{1+\textstyle \frac{1^2}{2+\textstyle \frac{3^2}{2+\textstyle \frac{5^2}{2+\textstyle \frac{7^2}{2+\textstyle \frac{9^2}{2+\ddots}}}}}} =3+\textstyle \frac{1^2}{6+\textstyle \frac{3^2}{6+\textstyle \frac{5^2}{6+\textstyle \frac{7^2}{6+\textstyle \frac{9^2}{6+\ddots}}}}} =\textstyle \cfrac{4}{1+\textstyle \frac{1^2}{3+\textstyle \frac{2^2}{5+\textstyle \frac{3^2}{7+\textstyle \frac{4^2}{9+\ddots}}}}} $$

Each continued fraction converges to $\pi$, but at greatly different rates.

The first is horrifically slow, requiring roughly $3\times 10^n$ terms for n decimal digits.

The second starts off nicely but later requires nearly 50 terms for five decimal digits, 120 for six.

The third is the best, requiring just four terms for each three decimal digits.