This is a temporary page for a quick study on Möbius transformations and continued fractions.
C F 3 ( x ) = a 0 + b 0 a 1 + b 1 a 2 + b 1 a 3 + b 3 x = A 3 + B 3 x C 3 + D 3 x {\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 C F n {\displaystyle \mathrm {CF} _{n}} :
C F n ( x ) = C F n − 1 ( a n + b n x ) {\displaystyle \mathrm {CF} _{n}(x)=\mathrm {CF} _{n-1}(a_{n}+{\frac {b_{n}}{x}})}
It should lead to a recursive formula for A n , B n , C n , D n {\displaystyle A_{n},\,B_{n},\,C_{n},\,D_{n}} :
A n + B n x C n + D n x = A n − 1 + B n − 1 ( a n + b n x ) C n − 1 + D n − 1 ( a n + b n x ) = x A n − 1 + B n − 1 ( x a n + b n ) x C n − 1 + D n − 1 ( x a n + b n ) = 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 {\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:
A n = B n − 1 b n , C n = D n − 1 b n , B n = A n − 1 + B n − 1 a n , D n = C n − 1 + D n − 1 a n {\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)
:
:
