Continued fraction manipulation 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
 A: 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.
