I noticed this answer to the question

$\quad$ Continued fraction of a square root

and the comment

enter image description here

So I felt obliged to take this on using the theory of sequences.
Confession: I find it difficult working with continued fractions.

Let $S \gt 0$ be an integer that is not the square of another integer.

Let $a$ be the greatest positive integer satisfying $a^2 \lt S$.

Using the algebra and notation from user Julien Blanchon,

$$ x = \frac{S - a^2}{x + 2a}$$

The positive number $x$ satisfies $x \lt 1$.

Set $x_0$ to any positive number and recursively define

$$ x_{n+1} = \frac{S - a^2}{x_n + 2a}$$

Show that the sequence $a + x_n$ converges to $\sqrt S$.

It would be interesting to see if this can be explained using the theory of continued fractions.

My Work

I checked it out using a Python program and would bet that the claim is true. Showing it is true mathematically is another matter, and I'm hoping to see some short answers; I'm not sure how to proceed to 'deconstruct' the method and avoid an algebraic nightmare.

Also, I suspect that it doesn't really matter what you choose for the value of $a$ - any positive number will work.

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1 Answers1


Marc van Leeuwen is correct for simple continued fractions. The continued fraction corresponding to this recursion

$$ a + \frac{S - a^2}{2a + \dfrac{S - a^2}{2a + \dfrac{S-a^2}{2a+\ldots}}} $$ is not a simple continued fraction.

Let $$ f(x) = \frac{S - a^2}{2a + x} $$ so your recursion is $x_{n+1} = f(x_n)$. This function has two fixed points $-a \pm \sqrt{S}$. Since $f'(-a+\sqrt{S}) = \frac{a-\sqrt{S}}{a+\sqrt{S}}$ has absolute value $< 1$, $-a + \sqrt{S}$ is an attracting fixed point. Thus the iteration starting "sufficiently close" to this fixed point will converge to it. Moreover, it's easy to check that there are no $2$-cycles. The boundaries of the immediate basin of attraction of an attracting fixed point can only be points on a $2$-cycle, repelling fixed points, singular points (where the function goes to $\infty$: here $x = -2a$), or $\pm \infty$. Thus in this case those boundary points are $-2a$ and $+\infty$. We conclude that $x_n$ does converge to $-a + \sqrt{S}$ for any $x_0 > -2a$.

Robert Israel
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