Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
I guess you are assuming that $a$ is not in $(-1,1)$, as that's the boring case... –  J. M. Nov 18 '11 at 6:09
    
We don't have to assume that, it just doesn't help answer the question; if $a$ is in (-1,1), the fractional part will be 0, which we already know is possible. I'm curious as to what other possible values we can get. –  Rebecca Saramosa Nov 18 '11 at 6:18
    
Some 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: go.helms-net.de/math/collatz/aboutloop/… –  Gottfried Helms Nov 18 '11 at 6:52
add comment

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 $1$.

share|improve this answer
    
Or 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 $a_0=2, a_1=6$. –  J. M. Nov 18 '11 at 13:23
add comment

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 $x=\frac32$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.