Is $$1+\cfrac{1}{1+\cfrac{1}{2+\cdots}} $$ or$\{1,1,2,3,4,5,\cdots,i,\cdots \} , i\in \mathbb{N}$ the simple continued fraction algebraic or transcendental?

Any reference is appreciated

EDIT and replacing $\{1,1,2,3,4,5,\cdots,i,\cdots \}$ with $\{1,2,2^2,2^3,2^4,2^5,\cdots,2^i,\cdots \} $ $i\in \mathbb{N}$ or any other patternful sequences that are upper unbounded ,can we get an algebraic number?


I believe this value is known to be transcendental; its value is known explicitly to be $1+\frac{I_1(2)}{I_0(2)}$, where $I_0()$ and $I_1()$ are the modified Bessel functions. For more details, see http://mathworld.wolfram.com/ContinuedFractionConstant.html . This continued fraction (and continued fractions with arithmetic progressions of coefficients in general) are related to certain Riccati equations, but unfortunately a bit of search on those terms doesn't really turn up any good lightweight accounts of the link between the two; you'll have to hunt about a bit more on your own for that.

  • $\begingroup$ Steven, as to the OP's earlier question today, are there any known real algebraic numbers with unbounded "elements," as Khinchin calls them? These would be the {1,1,2,3,4,5...} above $\endgroup$ – Will Jagy Jul 30 '14 at 23:58
  • $\begingroup$ yes,it is relevant to my other question about upper bound of continued fraction of algebraic number $\endgroup$ – XL _at_China Jul 31 '14 at 0:00
  • $\begingroup$ @WillJagy As noted there, the expectation would be that almost all algebraic numbers should have unbounded CF coefficients, but just like with normal numbers I unfortunately don't know of any specific examples. I wonder if something can be said about the continued fraction expansion of Pisot numbers; hrm... $\endgroup$ – Steven Stadnicki Jul 31 '14 at 0:04
  • 2
    $\begingroup$ @XL_at_China, appears we do not know, but there are many books on continued fractions en.wikipedia.org/wiki/Oskar_Perron $\endgroup$ – Will Jagy Jul 31 '14 at 0:07
  • 1
    $\begingroup$ @XL_at_China, after searching online, seems an open problem. Repeat mentions for cube root of 2, people think unbounded but no proof. Also, bounded but not periodic gives something completely different; see The Markoff and Lagrange Spectra by Cusick and Flahive $\endgroup$ – Will Jagy Jul 31 '14 at 0:22

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.