4
$\begingroup$

Assuming I've been given a number in the form of a continued fraction. Is there a general way to write the square root of that number as continued question?

For example, consider $$1+\sqrt{2} = 2+\frac1{2+\frac1{2+\frac1{2+\dots}}} = [2;2,2,2,2,2,\dots]$$ Its square root has, according to Mathematica, the form $$\sqrt{1+\sqrt{2}} = [1; 1, 1, 4, 6, 1, 2, 2, 2, 1, \dots]$$ I don't see a pattern here. Is there one?

$\endgroup$
  • 1
    $\begingroup$ There is no known pattern. If you find one, you will be famous! $\endgroup$ – GEdgar Oct 31 '15 at 13:29
0
$\begingroup$

According wise words of Alan Baker, the continued fractions algorithm establishes a bijective correspondence between irrational numbers $\theta$ and infinite sets of positive integers $n_0, n_1,n_2....$.

Hence for all sequence $\mathcal S$ of positive integers with no apparent pattern, there is a irrational $\theta$ whose expansion in continued fractions is $\mathcal S$. It could be maybe the case, I mean no apparent pattern, for $\sqrt{1+\sqrt{2}}$. How do you know whether or not there is a viewable pattern?

$\endgroup$
  • $\begingroup$ That doesn't answer my question. Note that the question is in the first paragraph; there's a reason the second paragraph begins with "For example". $\endgroup$ – celtschk Oct 18 '15 at 12:28
  • $\begingroup$ I understand and your question it seems to me plausible, I like it. However $ 1+\sqrt 2$ and $\sqrt{1+\sqrt 2}$ are quite different numbers. More, a rational $r$ has a finite continued fraction but $\sqrt r$ has an infinite one in general. $\endgroup$ – Piquito Oct 18 '15 at 12:42
  • $\begingroup$ I think you have to change "question" by "fraction". $\endgroup$ – Piquito Oct 18 '15 at 13:12
0
$\begingroup$

There will be no 'non-generalized' continued fraction with a repetitive pattern. These will always lead to values of the form $\frac{P+\sqrt{D}}{Q}$. There could be a generalized one though, having a repetitive or otherwise clear pattern. Generalized meaning that not only the $a_j$ can be any natural number from 1, but also the quotient part added to it may have a numerator unequal to 1. See for example https://en.wikipedia.org/wiki/Continued_fraction for generalized continued fractions for $\pi$ having a regular pattern.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.