# User:Grondilu/Möbius transformations

This is a temporary page for a quick study on Möbius transformations and continued fractions.

${\displaystyle \mathrm {CF} _{3}(x)=a_{0}+{\cfrac {b_{0}}{a_{1}+{\cfrac {b_{1}}{a_{2}+{\cfrac {b_{1}}{a_{3}+{\cfrac {b_{3}}{x}}}}}}}}={\frac {A_{3}+B_{3}x}{C_{3}+D_{3}x}}}$

There is a recursive expression for ${\displaystyle \mathrm {CF} _{n}}$:

${\displaystyle \mathrm {CF} _{n}(x)=\mathrm {CF} _{n-1}(a_{n}+{\frac {b_{n}}{x}})}$

It should lead to a recursive formula for ${\displaystyle A_{n},\,B_{n},\,C_{n},\,D_{n}}$:

${\displaystyle {\frac {A_{n}+B_{n}x}{C_{n}+D_{n}x}}={\frac {A_{n-1}+B_{n-1}(a_{n}+{\frac {b_{n}}{x}})}{C_{n-1}+D_{n-1}(a_{n}+{\frac {b_{n}}{x}})}}={\frac {xA_{n-1}+B_{n-1}(xa_{n}+b_{n})}{xC_{n-1}+D_{n-1}(xa_{n}+b_{n})}}={\frac {B_{n-1}b_{n}+(A_{n-1}+B_{n-1}a_{n})x}{D_{n-1}b_{n}+(C_{n-1}+D_{n-1}a_{n})x}}}$

And Thus:

${\displaystyle A_{n}=B_{n-1}b_{n},\,\quad C_{n}=D_{n-1}b_{n},\quad B_{n}=A_{n-1}+B_{n-1}a_{n},\quad D_{n}=C_{n-1}+D_{n-1}a_{n}}$

--Grondilu 15:33, 3 January 2013 (UTC)

Of course:

${\displaystyle \mathrm {CF} _{0}(x)=a_{0}+{\frac {b_{0}}{x}}={\frac {b_{0}+a_{0}x}{x}}}$

Thus:

${\displaystyle A_{0}=b_{0},\quad B_{0}=a_{0}\,\quad C_{0}=0,\quad D_{0}=1}$

--Grondilu 01:49, 4 January 2013 (UTC)