My friend and I were talking earlier today and he posed the following problem, that he does not know the answer to: Take a real number, $a$, and look at consecutive powers of $a$: $a,a^{2},a^{3},...$ and look at the fractional part of the powers, i.e. $a^k - \lfloor a^k \rfloor$. What values can the fractional part of the powers converge to, if any? Obviously $0$ is one fractional part, but he believes that all rationals can be. He then extended his answers to all numbers that are not transcendental. Just thought it was a nice question and I'm curious as to what the answer is myself.
Fractional Powers
-
0Some years ago I considered$a$similar question in the context of the collatz-problem. For some rational *a=b/c* I expressed the powers as irregular fractions and plotted the fractional part against the denominator and got some nice pictures. For the collatz-problem it was relevant, whether the dots for *1.5^N* are all below the main diagonal. But I found some nice structures for some common irrational numbers like golden ratio and pi so perhaps you'll find it interesting, too. Here it is: http://go.helms-net.de/math/collatz/aboutloop/GraphsOn3N_2NApproximations.htm – 2011-11-18
2 Answers
In case you had not observed it, I should mention that you can get $1$ as the limit. For example, let $a=3+2\sqrt{2}$. It is not hard to verify that $(3+2\sqrt{2})^n +(3-2\sqrt{2})^n$ is an integer for every non-negative integer $n$. One way is to expand each power using the Binomial Theorem, and observe that the terms that involve odd powers of $\sqrt{2}$ cancel.
For large $n$, $(3-2\sqrt{2})^n$ is positive and close to $0$. It follows that $(3+2\sqrt{2})^n$ is slightly below an integer, and therefore $(3+2\sqrt{2})^n -\lfloor(3+2\sqrt{2})^n\rfloor$ is nearly equal to $1$.
-
0Or note that $a_n=(3+2\sqrt{2})^n +(3-2\sqrt{2})^n$ satisfies the recursion relation $a_{n+1}=6a_n-a_{n-1}$, with initial conditions $a_0=2, a_1=6$. – 2011-11-18
It's pretty well-studied. See this classic by Hardy and Littlewood, and this series of papers by Vijayaraghavan. In particular, it is known that the sequence $\{x^k\},\quad x > 1$ is equidistributed for most numbers, with a few exceptions like $x=1+\sqrt 2$ and $x=\phi$. See the MathWorld link for the interesting behavior when $x=\frac32$.