Continued fractions for rationals terminate, for transcendentals like pi, they do not terminate and for irrationals (but non transcendentals) they repeat -- is this correct?

  • $\begingroup$ Are you referring to en.wikipedia.org/wiki/… ? $\endgroup$ – user58512 Feb 8 '13 at 14:23
  • 1
    $\begingroup$ you should ask the bit about convolution as a separate question $\endgroup$ – user58512 Feb 8 '13 at 15:23
  • 1
    $\begingroup$ Well, okay, since I liked your answer for the first part so much but you didn't talk about the second part I will take your suggestion and ask again. Also that wiki link is fascinating but not exactly what I am asking. $\endgroup$ – Cris Stringfellow Feb 8 '13 at 17:36

We know all simple continued fractions converge so if a continued fraction was periodic (period 4 for illustration) with value $x$ we could write $$x = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2+ \cfrac{1}{a_3 + \cfrac{1}{x}}}}.$$

Each part of a continued fraction maps $\frac{p}{q} \mapsto a_i + \frac{1}{(\tfrac{p}{q})} = \frac{q + a_i p}{p}$ so apply this iteration to get $$x = \frac{1 + a_2 a_3 + a_0 a_3 + a_0 a_1 + a_0 a_1 a_2 a_3 + (a_2 + a_0 (a_1 a_2 + 1)) x}{a_3 + a_1 + a_1 a_2 a_3 + (a_1 a_2 + 1) x}$$ it is instructive to perform this iteration by hand. The key thing to notice is we have a form $\frac{A + xB}{C + xD}$ where $A,B,C,D$ are integers.

Theorem The continued fraction transform preserves the form $\frac{A + xB}{C + xD}$. proof: $$\frac{A + xB}{C + xD} \mapsto \frac{(C + a_i A) + x (a_i B + D)}{A + xB}.$$

From this we can deduce that all periodic continued fractions are quadratic irrationals: if $x = \frac{A + xB}{C + xD}$ then $x^2 D + x (C - B) - A = 0$ therefore (by the quadratic formula) $x$ is a quadratic irrational.

Yes, A continued fraction is finite if and only if the number is rational. This follows from the fact that a continued fraction gives infinitely many excellent rational approximations of the number and rationals only have finitely many such approximations (See Dirichlet's theorem for more on this).

The periodic continued fractions are exactly the quadratic irrationals: numbers of the form $\frac{A + \sqrt{B}}{C}$. The converse (that every quadratic irrational has a periodic continued fraction) is more involved.

So the pattern of cubic irrationals - which are not transcendental - doesn't have any periodic pattern (and I am not sure if any patterns are known at all).


That is not correct. Continued fractions for quadratic irrationals repeat. But for higher-degree irrationals, such as $\sqrt[3]2$, the continued fraction does not repeat.

I am not aware of any research into the particular convolution operation you mention, but that doesn't mean there is none.

  • $\begingroup$ Okay I see the proof below. Thanks. $\endgroup$ – Cris Stringfellow Feb 8 '13 at 17:35

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.