Return to Revisions

Grouping the terms of $F(z)$ by the height reached, we get $$F(z) = \frac{1}{(1 - z\gamma_0)} + \frac{z^2 \beta_1}{(1 - z\gamma_0)^2 (1 - z\gamma_1)} + \frac{z^4 \beta_1 \beta_2}{(1 - z\gamma_0)^2 (1 - z\gamma_1)^2 (1 - z\gamma_2)} + \cdots \\ $$ This has the form of Euler's continued fraction $$a_0 + a_0a_1 + a_0a_1a_2 + \cdots = \cfrac{a_0}{1 - \cfrac{a_1}{1 + a_1 - \cfrac{a_2}{1 + a_2 - \ddots}}}$$ with $$a_0 = \frac{1}{1 - z\gamma_0} \\ a_1 = \frac{z^2 \beta_1}{(1 - z\gamma_0)(1 - z \gamma_1)} \\ a_2 = \frac{z^2 \beta_2}{(1 - z\gamma_1)(1 - z \gamma_2)} \\ \vdots $$ It is perhaps more natural to drop the denominators to the next level: i.e. instead of $$\cfrac{\frac{1}{1 - z\gamma_0}}{1 - \cfrac{\frac{z^2 \beta_1}{(1 - z\gamma_0)(1 - z \gamma_1)}}{1 + \frac{z^2 \beta_1}{(1 - z\gamma_0)(1 - z \gamma_1)} - \cfrac{\frac{z^2 \beta_2}{(1 - z\gamma_1)(1 - z \gamma_2)}}{1 + \frac{z^2 \beta_2}{(1 - z\gamma_1)(1 - z \gamma_2)} - \ddots}}}$$ we could write $$\cfrac{1}{(1 - z\gamma_0) - \cfrac{(1 - z\gamma_0)z^2 \beta_1}{(1 - z\gamma_0)(1 - z \gamma_1) + z^2 \beta_1 - \cfrac{(1 - z\gamma_0)z^2 \beta_2}{(1 - z \gamma_2) + \frac{ z^2 \beta_2}{(1 - z\gamma_1)} - \ddots}}}$$ although, as can readily be seen, this is not a strict panacea.