- #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.

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: