Timeline for Kronecker Approximation theorem and Fibonacci numbers
Current License: CC BY-SA 2.5
8 events
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| Mar 13, 2011 at 1:17 | comment | added | Nikita Sidorov | I agree that the case of the Fibonacci sequence is different. What we have here is the condition $\|\xi\tau^{n_k}\|\to0$ as $k\to\infty$ for some subsequence $n_k$ (with $\tau=(1+\sqrt5)/2$). If we had just $n_k=k$, then we would know that $\xi\in\mathbf Q(\tau)$, whence $\xi$ would have an eventually periodic greedy $\tau$-expansion (combined Pisot + K. Schmidt). I believe we must have $n_k=bk$ for some $b\ge1$ but don't know how to prove it. Perhaps, re-reading Cassels `Introduction in Diophantine Approximations' (chapter on Pisot-Vijayaraghavan numbers) might help... | |
| Mar 11, 2011 at 16:19 | comment | added | Ostap Chervak | Actually we must forbid not only pairs of consecutive 1s but also infinite sequences of the form 010101... e.g. Let β be a golden ratio, then $\sum \beta^{2k−1}={\beta\over{\beta^2−1}}=1=\beta^0$.As you see there are no consecutive 1s in any of two expansions, yet they are equal. Though now I see that nonuniqueness isn't very bad there, sorry for misconseption. Though for β-expansion there exist a number $\alpha=\sum\limits_{k=1}^{\infty}\beta^{3k-1} ={\beta\over{\beta^3- 1}}= {\beta\over 2}$ with no unbounded string of 0s but $\alpha\beta^{3k-1}$ gets close to integer($F_{3n}$ are even). | |
| Mar 11, 2011 at 15:00 | comment | added | Nikita Sidorov | Ostap, to make them unique, one usually forbids two consecutive 1's. Then they exhibit pretty much the same ergodic and arithmetic properties as the binary expansions. (See, e.g., my survey paper maths.manchester.ac.uk/~nikita/ad.pdf). The Fibonacci numbers are closely related to the powers of the golden ratio, of course - you just need to divide them by $sqrt5$, I guess. | |
| Mar 11, 2011 at 14:46 | comment | added | Ostap Chervak | Thanks for an answer, though in the case of Fibonacci sequence this trick works not so good as in binary case. Just as β-expansion is not unique it is not so obvious that there exists reals with uniformly bounded number of consecutive 0s. At least typical example $0.10(10)_\beta=10_\beta$ shows it. Though it looks like looking on expansions $\sum\limits_{k=1}^{\infty} a_k \beta^k$ with maximum value of $\sum a_k 2^{-k}$ must work. And a.e. part is certainly true. Thanks! | |
| Mar 11, 2011 at 13:49 | vote | accept | Ostap Chervak | ||
| Mar 11, 2011 at 4:59 | comment | added | Gerry Myerson | @Nikita, yes, Weyl proved that if $a_1,a_2,\dots$ is any sequence of distinct integers then the sequence $a_1x,a_2x,\dots$ is uniformly distributed modulo 1 for almost all real $x$. This of course includes the Fibonaccis. Hardy and Littlewood had already studied the case $a_n=b^n$, $b$ an integer, $b\ge2$. | |
| Mar 11, 2011 at 1:13 | history | edited | Nikita Sidorov | CC BY-SA 2.5 |
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| Mar 11, 2011 at 0:35 | history | answered | Nikita Sidorov | CC BY-SA 2.5 |