# Continued Fraction Recursion: Loose Ends

Mitch Harris maharri at gmail.com
Tue Feb 27 20:34:34 CET 2007

```On 2/27/07, Leroy Quet <qq-quet at mindspring.com> wrote:
> I wrote in part:
>
> >So, if x = the root of x^3 - x -1 = 0,* (I am assuming that b(n) -> x.)...
> >
> >*(What is this root? I have no access to a computer algebra program, and
> >I do not know how to solve the cubic off the top of my head. I am
> >embarrassed that I don't know the root to even this simple cubic.)
>
> Woops...
> Thanks to Hugo for getting the numerical real root, but....
>
> I made a mistake. I REALLY want x to be a real root of:
>
> x^3 - x^2 -1 = 0.

the real root to x^3 - x -1 = 0  is

(1/(3 * 2^(1/3))) * (cbrt(27 + 3sqrt(69)) + cbrt(27 - 3sqrt(69)))

the real root to x^3 - x^2 -1 = 0  is

(1/(3 * 2^(1/3))) * (2^(1/3) + cbrt(29 + 3sqrt(93)) + cbrt(29 - 3sqrt(93)))

(uh... mathematica helped. I remember being annoyed as a youth at not
being taught in school things like the solution to the general cubic
or spherical trigonometry. Now I am not so annoyed. I wouldn't
consider not being able to do them a disgrace.)

Mitch

> I made a mistake. I REALLY want x to be a real root of:
>
> x^3 - x^2 -1 = 0.

A closed form for the real root x, which I didn't try very
hard to simplify:

The cubic can be solved by "hand" by making the substitutions
x = y + 1/3 and then y = z + 1/(9*z).  Which leaves a sextic
in z of the form A*z^6 + B*z^3 + C == 0.  Solve for z and
then work backward.

The numeric value of x is:

```