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I have the following continued fraction

$$ \frac{1}{a_1x+}\;\;\frac{1}{b_1+}\;\;\frac{1}{a_2x+}\;\;\frac{1}{b_2} $$

The paper I am reading then converts this to the following continued z-fraction but does not show any work

$$ \frac{\frac{1}{a_1x}}{1+}\;\;\frac{\frac{1}{b_1a_1x}}{1+}\;\;\frac{\frac{1}{a_2x}}{1+}\;\;\frac{\frac{1}{a_2b_2x}}{1+0} $$

Is this relationship correct? When I try a variety of numbers, the relationship seems to hold for large values of $x$ but becomes increasingly inaccurate as $x$ becomes smaller ($x<1$).

Any help is greatly appreciated.

Thanks

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  • $\begingroup$ Could you please provide a reference for this paper, preferably with a DOI link or a link to a freely accessible version, if available? $\endgroup$ – A.P. Mar 24 '15 at 22:24
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It works like this:

$$\frac{1}{a_1x+\frac{1}{b_1+\frac{1}{a_2x+\frac{1}{b_2}}}}=\frac{\frac{1}{a_1x}}{1+\frac{\frac{1}{a_1x}}{b_1+\frac{1}{a_2x+\frac{1}{b_2}}}}$$

because we multiply the top and bottom of the outermost fraction by $\frac{1}{a_1x}$. Now, isolate the fraction from the denominator above:

$$\frac{\frac{1}{a_1x}}{b_1+\frac{1}{a_2x+\frac{1}{b_2}}}=\frac{\frac{1}{b_1a_1x}}{1+\frac{\frac{1}{b_1}}{a_2x+\frac{1}{b_2}}}$$

because we multiply the top and bottom by $\frac{1}{b_1}$. Moving along, we're going to multiply the top and bottom of that fraction in the denominator we just looked at by $\frac{1}{a_2x}$:

$$\frac{\frac{1}{b_1}}{a_2x+\frac{1}{b_2}}=\frac{\frac{1}{b_1a_2x}}{1+\frac{\frac{1}{a_2x}}{b_2}}$$

Finally, we take the fraction in that last denominator, and multiply it, top and bottom, by $\frac{1}{b_2}$:

$$\frac{\frac{1}{a_2x}}{b_2}=\frac{\frac{1}{a_2b_2x}}{1}=\frac{\frac{1}{a_2b_2x}}{1+0}$$

In the notation from your question, this calculation is giving us:

$$ \frac{\frac{1}{a_1x}}{1+}\;\;\frac{\frac{1}{b_1a_1x}}{1+}\;\;\frac{\frac{1}{b_1a_2x}}{1+}\;\;\frac{\frac{1}{a_2b_2x}}{1+0} $$

which differs from what you posted in having a $b_1$ term in the denominator of the third numerator.

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