There is a famous old theorem by Kronecker that for every positive real $\alpha$ and $\epsilon>0$ there exists a positive integer n such that $\alpha n$ is within $\epsilon$ of an integer.


Recently I found that the same result is true if we replace $\alpha n$ by $\alpha n^2$ or any polinomial p such that $p(0)=0$.


Could this result be generalised to other functions? Particularly I'm curious about sequences $\alpha 2^n$ and $\alpha F\_n$ where by $F\_n$ I denote n-th Fibonacci number.


Does anyone know anything about it?



Kronecker Approximation theorem and Fibonacci numbers
Well, as Gerry has pointed out, this is certainly not true for all $\alpha$. On the other hand, this is true for a.e. $\alpha$. More precisely, the sequence $2^n\alpha$ is equidistributed mod 1 for a.e. $\alpha$.


I believe this result is due to H. Weyl and can be found in Cornfeld, Fomin and Sinai `Ergodic Theory'. (I don't have it with me.) 


The same must be true for the Fibonacci sequence, I'm sure. 


So, what you probably need is for this to be true for all $\alpha$, except some small (countable?) set. After all, $\|2^n\alpha\|<\varepsilon$ is indeed much weaker than equidistibution. 


**Update.** Come to think about it, the answer is as follows: let 
$$
\alpha = \sum\_{k=1}^\infty a\_k2^{-k}
$$
be the binary expansion of $\alpha$. Then the sequence $2^n\alpha\bmod 1$ gets arbitrarily close to 0 if and only if the sequence $(a\_k)$ has unbounded strings of 0s. In particular, any rational $\alpha$ is out of the picture, apart from the binary rationals, of course.


All in all, your set of $\alpha$'s is indeed of full measure, but the exceptional set is of Hausdorff dimension 1, i.e., pretty big.


For the Fibonacci sequence you'll need to replace binary expansion with the $\beta$-expansion, where $\beta=(\sqrt5-1/)2$, with the same conclusion. 


