5
$\begingroup$

The simple continued-fraction-expansion for the transcendental number $e$ is known to be unbounded. What about bounded continued fractions ?

Do we know any transcendental number for which it is proven that the simple continued-fraction-expansion is bounded ?

It is conjectured that the simple continued-fraction-expansion of the algebraic numbers with minimal polynomial degree greater than $2$ are unbounded.

If this would be true, every bounded non-periodic infinite simple continued-fraction-expansion would correspond with a transcendental number.

But to my knowledge, it was not proven for a single algebraic number with minimal polynomial degree greater than $2$, that its simple continued-fraction-expansion is unbounded.

$\endgroup$
9
$\begingroup$

Do we know any transcendental number for which it is proven that the simple continued-fraction-expansion is bounded?

It has been proven that $\sum^\infty_{n=0}2^{-2^n}$ is transcendental, and in fact $\operatorname{\kappa}\left(b\right)=\sum^\infty_{n=0}b^{2^n}$ is transcendental for any algebraic $b\in\left(0,1\right)$.

So let's do the exact same sum but in base 10 instead of base 2 so we can visualise it better:

$\begin{align} \operatorname{\kappa}\left(10\right) &= \sum^\infty_{n=0}10^{-2^{n}} \\ &= 10^{-1}+10^{-2}+10^{-4}+10^{-8}+10^{-16}+10^{-32}+10^{-64}+\ldots \\ &= 0.\mathbf{1}\mathbf{1}0\mathbf{1}000\mathbf{1}0000000\mathbf{1}000000000000000\mathbf{1}0000000000000000000000000000000\mathbf{1}\ldots \end{align}$

Since $\frac{1}{10}$ is algebraic, $\operatorname{\kappa}\left(10\right)$ is transcendental. Its canonical continued fraction is:

$\left[0; 9, 12, 10, 10, 8, 10, 12, 10, 8, 12, 10, 8, 10, 10, 12, 10, 8, 12, 10, 10, 8, 10, 12, 8, 10, 12, 10, 8, 10, 10, 12, 10, 8, 12, 10, 10, 8, 10, 12, 10, 8, 12, 10, 8, 10, 10, 12, 8, 10, 12, 10, 10, 8, 10, 12, 8, 10, 12, 10, 8, 10, 10, 12, 10, 8, 12, 10, 10, 8, 10, 12, 10, 8, 12, 10, 8, 10, 10, 12, 10, 8, 12, 10, 10, 8, 10, \ldots\right]$

This constant has a canonical continued fraction whose terms are (with the exception of the first two terms which are 0 and 9, respectively) nothing other than 8, 10, or 12.

The 12's occur at positions congruent to 2 or 7 mod 8. And the positions of the eights (5, 9, 12, 17, 21, 24, ...) are all congruent to 1 or 0 mod 4. But in a particular order.

If we write this as $\left[a_0;a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8,\ldots\right]$:

$\forall~n\in\mathbb{Z}_{\geqslant 0},~a_n=\begin{cases} 0 & n=0 \\ 8 & n\in\left\{\frac{8m+\left(\frac{-1}{m-1}\right)+1}{2}~:~m\in\mathbb{Z}^{+}\right\} \\ 9 & n=1 \\ 10 & \text{otherwise} \\ 12 & n\equiv 2\left(\operatorname{mod}8\right)\text{or}~7\left(\operatorname{mod}8\right) \end{cases}$

where $\left(\frac{n}{m}\right)$ is the Jacobi symbol.

So there we have it. A transcendental number whose continued fraction has bounded terms.

In fact, it is conjectured that all continued fractions that are infinite, bounded in their terms, and not eventually periodic produce transcendental numbers. If this is indeed true, then it would also mean that the terms in the continued fractions for any algebraic number of degree 3 or higher are unbounded.

$\endgroup$
3
$\begingroup$

Yes, but the transcendentals that this answer describes are quite unnatural.

Fix any noncomputable bounded sequence of positive integers $\alpha$, and let $r_\alpha$ be the real number whose continued fraction expansion is given by $\alpha$. Then - since the continued fraction expansion of a computable real is computable, and every algebraic real is computable - $r_\alpha$ is transcendental. Note that the nontrivial part here is proving transcendentality!

$\endgroup$

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.