Inversion of infinite continued fractions

AI Thread Summary
The discussion focuses on inverting an infinite continued fraction represented by a specific formula involving terms \(\beta\), \(\alpha\), and \(\gamma\). The user seeks to transform this infinite fraction into a form that begins with the \(\beta_n\) term and proceeds down to \(n=1\). They reference an identity for finite continued fractions and express uncertainty about its applicability to truncated infinite continued fractions. The user requests assistance or resources to clarify this inversion process. The conversation highlights the complexities of manipulating continued fractions and the need for further exploration of their properties.
denijane
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Hello to everyone!
I need to invert the following infinite continued fraction:
0=\beta_{0}-\frac{\alpha_{0}\gamma_{1}}{\beta_{1}-}\frac{\alpha_{1}\gamma_{2}}{\beta_{2}-}\ldots, n=1..\infty<br />
to something starting with the \beta_{n} 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:
<br /> \frac{[a_{0},\ldots, a_{n}]}{[a_{0},\ldots, a_{n-1}]}=\frac{[a_{n},\ldots, a_{0}]}{[a_{n-1},\ldots, a_{0}]}<br /> http://mathworld.wolfram.com/ContinuedFraction.html"

But I'm not sure if it remains true for truncated infinite continued fractions.
Any help or resources will be appreciated.
 
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If
x = [a_{0} ; a_{1}, a_{2},...]

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