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Inversion of infinite continued fractions

  • Thread starter denijane
  • Start date
  • #1
3
0
Hello to everyone!
I need to invert the following infinite continued fraction:
[tex]0=\beta_{0}-\frac{\alpha_{0}\gamma_{1}}{\beta_{1}-}\frac{\alpha_{1}\gamma_{2}}{\beta_{2}-}\ldots, n=1..\infty
[/tex]
to something starting with the [tex]\beta_{n}[/tex] term and going down to n=1 (where n is the term the fractions will be cut to).

I know there is the following identity for finite continued fractions:
[tex]
\frac{[a_{0},\ldots, a_{n}]}{[a_{0},\ldots, a_{n-1}]}=\frac{[a_{n},\ldots, a_{0}]}{[a_{n-1},\ldots, a_{0}]}
[/tex] http://mathworld.wolfram.com/ContinuedFraction.html" [Broken]

But I'm not sure if it remains true for truncated infinite continued fractions.
Any help or resources will be appreciated.
 
Last edited by a moderator:

Answers and Replies

  • #2
11
0
If
[tex]x = [a_{0} ; a_{1}, a_{2},...][/tex]

[tex]1/x = [ 0 ; a_{0}, a_{1}, a_{2}, ... ] \ when \ a_{0} \neq 0 [/tex]
[tex] \ \ \ \ \ \ = [ a_{1}; a_{2}, a_{3},... ] \ when \ a_{0} = 0 [/tex]
 

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