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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?

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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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

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