§15.7 Continued Fractions

If $|\mathop{\mathrm{ph}\/}\nolimits\!\left(1-z\right)|<\pi$, then

 15.7.1 $\frac{\mathop{\mathbf{F}\/}\nolimits\!\left(a,b;c;z\right)}{\mathop{\mathbf{F}% \/}\nolimits\!\left(a,b+1;c+1;z\right)}=t_{0}-\cfrac{u_{1}z}{t_{1}-\cfrac{u_{2% }z}{t_{2}-\cfrac{u_{3}z}{t_{3}-\cdots}}},$

where

 15.7.2 $\displaystyle t_{n}$ $\displaystyle=c+n,$ $\displaystyle u_{2n+1}$ $\displaystyle=(a+n)(c-b+n),$ $\displaystyle u_{2n}$ $\displaystyle=(b+n)(c-a+n).$

If $|z|<1$, then

 15.7.3 $\frac{\mathop{\mathbf{F}\/}\nolimits\!\left(a,b;c;z\right)}{\mathop{\mathbf{F}% \/}\nolimits\!\left(a,b+1;c+1;z\right)}=v_{0}-\cfrac{w_{1}}{v_{1}-\cfrac{w_{2}% }{v_{2}-\cfrac{w_{3}}{v_{3}-\cdots}}},$

where

 15.7.4 $\displaystyle v_{n}$ $\displaystyle=c+n+(b-a+n+1)z,$ $\displaystyle w_{n}$ $\displaystyle=(b+n)(c-a+n)z.$

If $\Re{z}<\tfrac{1}{2}$, then

 15.7.5 $\frac{\mathop{\mathbf{F}\/}\nolimits\!\left(a,b;c;z\right)}{\mathop{\mathbf{F}% \/}\nolimits\!\left(a+1,b+1;c+1;z\right)}={x_{0}+\cfrac{y_{1}}{x_{1}+\cfrac{y_% {2}}{x_{2}+\cfrac{y_{3}}{x_{3}+\cdots}}}},$

where

 15.7.6 $\displaystyle x_{n}$ $\displaystyle=c+n-(a+b+2n+1)z,$ $\displaystyle y_{n}$ $\displaystyle=(a+n)(b+n)z(1-z).$

See also Cuyt et al. (2008, pp. 295–309).