I would like to find the continued fraction expansion of the roots of:
$$x^2 - \frac{53793390359}{1088391168}x + \frac{823543}{12230590464} = 0$$
Eq 1.6 from [1]
What makes this problem so difficult is obtaining numerical accuracy. However, we have some starting points:
- since this is a quadratic the digits are ultimately periodic
- the continued fraction of the middle coefficient is $$ [49, 2, 2, 1, 4, 1, 1, 6, 2, 1, 6, 1, 9, 1, 454, 3, 1, 4]$$
- the constant term is of order $10^{-6}$.
Since this is a quadratic, I am hoping an exact answer is possible. Although the number $454$ may already rule out the possibility of a "small divisors" explanation of the equation here.
Work I have no idea what to do here. My first guess is to approximate this equation as:
$$ x^2 - 49x + 0 = x(x-49) = 0 $$
so the roots are $x = 0, 49$. If I wanted to get more accuracy I could try dividing by $x$ and hoping for the best:
\begin{eqnarray} x &=& \underbrace{\frac{53793390359}{1088391168}}_A + \underbrace{\frac{823543}{12230590464}}_{1/B}\frac{1}{x} \\\\ &=& A + \frac{1}{B\, x} = A + \frac{1}{AB + \frac{1}{x}}\end{eqnarray}
The first two digits are... $x = [\lfloor A \rfloor, \lfloor AB \rfloor,...] = [49,734015,... ]$ so $x = \mathbf{49} $ is a really good guess!


I computed the continued fraction out many terms, but didn't see a pattern.
Here's the Maple code that I used (following the VAS algorithm) to generate the continued fraction for the root near 49:
Here are the first 101 terms of the continued fraction:
For the smaller root: $$[0, 2, 2, 1, 4, 1, 1, 6, 1, 2, 2, 6, 1, 2, 1, 17, 1, 3, 3, 6, 4, 3, 1, 1, 2, 2, 4, 14, 35, 1, 1, 247, 1, 1, 4, 2, 15, 1, 1, 1, 16, 42, 1, 6, 3, 1, 5, 1, 4, 1, 1, 1, 4, 16, 3, 17, 1, 1, 3, 1, 1, 4, 1, 1, 11, 1, 5, 1, 71, 1, 1, 1, 1, 7, 61, 2, 1, 1, 4, 4, 62, 1, 5, 1, 1, 2, 3, 2, 3, 45, 1, 1, 3, 6, 1, 3, 1, 47, 1, 27, 1]$$
For the larger root: $$[49, 2, 2, 1, 4, 1, 1, 6, 1, 2, 2, 6, 1, 2, 1, 17, 1, 3, 3, 6, 4, 3, 1, 1, 2, 2, 4, 14, 35, 1, 1, 247, 1, 1, 4, 2, 15, 1, 1, 1, 16, 42, 1, 6, 3, 1, 5, 1, 4, 1, 1, 1, 4, 16, 3, 17, 1, 1, 3, 1, 1, 4, 1, 1, 11, 1, 5, 1, 71, 1, 1, 1, 1, 7, 61, 2, 1, 1, 4, 4, 62, 1, 5, 1, 1, 2, 3, 2, 3, 45, 1, 1, 3, 6, 1, 3, 1, 47, 1, 27, 1]$$
By the way, the continued fraction expansion does not repeat before the 10,000th term.