View source for User:Grondilu/Möbius transformations

This is a temporary page for a quick study on [[wp:Möbius transformations|Möbius transformations]] and [[continued fractions]].


<math>\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_3x}{C_3+D_3x}</math>


There is a recursive expression for <math>\mathrm{CF}_n</math>:

<math>\mathrm{CF}_n(x) = \mathrm{CF}_{n-1}(a_n + \frac{b_n}{x})</math>

It should lead to a recursive formula for <math>A_n,\, B_n,\, C_n,\, D_n</math>:

<math>\frac{A_n+B_nx}{C_n+D_nx} = \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{x A_{n-1} + B_{n-1} (x a_n + b_n)}{x C_{n-1} + D_{n-1} (x a_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  }</math>

And Thus:

<math>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</math>

--[[User:Grondilu|Grondilu]] 15:33, 3 January 2013 (UTC)