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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,

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    @Peter Bala: I guess there isn't a unique one. Thanks for the answer.2011-08-09

3 Answers 3

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

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@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.

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    @vonbrand I think this is a nice way of showing that such expansions can't possibly be unique in the way the OP was hoping.2013-03-05
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The simplest way: take all $b_k=1$ and use your previous formula.

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    Thank you. But then, how would I generate $b_k$ again if initially I take them all to 1?2011-08-07