|
||||
|
Title: Fractional Parts of Powers of Rationals Post by Barukh on Feb 26th, 2004, 1:10am Prove: if r[smiley=in.gif][smiley=bbq.gif], r > 1 and lim {rn} [smiley=to.gif] 0 when n [smiley=to.gif][infty], then r[smiley=in.gif][smiley=bbz.gif]. In words: if a rational number r > 1 has a property that the sequence of the fractional parts of its powers converges to 0, then r is an integer. |
||||
|
Title: Re: Fractional Parts of Powers of Rationals Post by towr on Feb 26th, 2004, 1:43am Doesn't lim frac((1/i)n) [to] 0 when n [to][infty] hold for every i in [bbz]\{-1,0,1} I mean, seeing as frac((1/i)n) = 1/in and lim 1/in [to] 0 when n [to][infty] and considering (1/i)n [notin] [bbz] for every i in [bbz]\{-1,0,1} that seems to be contradictory to your cliam.. Unless I misunderstand what you're getting at.. |
||||
|
Title: Re: Fractional Parts of Powers of Rationals Post by Barukh on Feb 26th, 2004, 3:56am towr, you are right... I forgot to specify that the number is greater than 1. I already modified the post. Sorry for inconvinience. :( |
||||
|
Title: Re: Fractional Parts of Powers of Rationals Post by Eigenray on Feb 26th, 2004, 4:35am Write r = k + [alpha], where [alpha] = {r} = p/q [in] [bbq], with 0 [le] p < q relatively prime. Suppose lim {rn} = 0. Take [epsilon] < 1/(qr) < 1/q. Then there's some N such that for all n > N, we have {rn} < [epsilon], so for some integer m we have: m [le] rn < m + [epsilon] mk + mp/q [le] rn+1 < mk + mp/q + [epsilon]r < mk + (mp+1)/q Now, write mp/q = t + x/q, where 0 [le] x < q. If x [ne] 0, then 1/q [le] x/q [le] 1-1/q, so: mk + t + 1/q [le] rn+1 < mk + t + 1, which would make 1/q [le] {rn+1} < [epsilon], which is a contradiction. Therfore x = 0, and therefore q | m. We therefore have: m' = mk+mp/q [le] rn+1 < m'+[epsilon], and we can repeat the argument to get q | m', and since q | m, we have q | mp/q, so q2 | m. But then m'' = m'k + m'p/q [le] rn+2 < m''+[epsilon], and again, q | m'', so q2 | m', and q3 | m. In this way, qs | m for all s. Now, m [ne] 0 because r > 1, so q can only equal 1. Therefore [alpha] = 0, and r = k. |
||||
|
Title: Re: Fractional Parts of Powers of Rationals Post by Eigenray on Feb 26th, 2004, 6:41am Is there an x [notin] [bbq], x>1, such that lim {xn} = 0? This comes close: Claim: If x - 1/x [in] [bbz], then xn + (-1/x)n [in] [bbz] for all n [in] [bbn] Proof: Induction: xn+1 + (-1/x)n+1 = (xn + (-1/x)n)(x - 1/x) + (xn-1 + (-1/x)n-1) [in] [bbz]. Now, take x>0 such that x - 1/x = 1; i.e., x = [phi] = (1 + [sqrt]5)/2. Then {x2n} = {mn - 1/x2n} = 1-x2n [to] 1 as n[to][infty]. And {x2n+1} = {mn + 1/x2n+1} = 1/x2n+1 [to] 0 as n[to][infty]. So we can pick y=x2 and have {yn}[to]1. I can't come up with anything to get {yn}[to]0 though. |
||||
|
Title: Re: Fractional Parts of Powers of Rationals Post by Barukh on Feb 26th, 2004, 8:46am Very nice, Eigenray! It think you did a typo here: on 02/26/04 at 04:35:14, Eigenray wrote:
It should be 1 [le] x [le] q-1, and the next step follows based on this. |
||||
|
Title: Re: Fractional Parts of Powers of Rationals Post by Eigenray on Feb 26th, 2004, 3:38pm Thanks. Fixed. Any idea how to construct x [in] [bbr]\[bbz], x > 1, such that limn[to][infty]{xn}=0? I'd be pretty surprised if it didn't exist. |
||||
|
Title: Re: Fractional Parts of Powers of Rationals Post by Barukh on Feb 27th, 2004, 3:54am on 02/26/04 at 15:38:45, Eigenray wrote:
Maybe, the following link (http://mathworld.wolfram.com/PowerFractionalParts.html) will be of some help... |
||||
|
Title: Re: Fractional Parts of Powers of Rationals Post by Barukh on Feb 27th, 2004, 7:16am on 02/26/04 at 06:41:47, Eigenray wrote:
I think this may be extended as follows: let y be a root of the quadratic y2-by+c = 0, b > 0, 0 < c < b-1. Then {yn}[to]1. For instance, in your example, b = 3, c = 1. Quote:
I started to doubt it’s possible… :-/ |
||||
|
Powered by YaBB 1 Gold - SP 1.4! Forum software copyright © 2000-2004 Yet another Bulletin Board |