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Let $c$ be an irrational real number. Let $\{\cdot\}$ be the fractional part operator. I would like to get some sense of how in-the-dark we are about the distribution of values of $\{cn!\}$, for familiar values of $c$. This is related to a previous post which (essentially) asks the question "Does $n!/(2\pi)$ tend to a limit mod 1?"

Here is the question: Can anyone give a value of $c$ which is either algebraic, or a familiar transcendental, or defined in some reasonably simple way using the elementary functions of calculus, such that

  1. $\{cn!\}<1/2$ infinitely often, or
  2. $\{cn!\}$ tends to a limit, or
  3. The values of $\{cn!\}$ are dense in the interval $[0,1]$.

What do I know that we know about all this? First of all, it is a theorem of P. Diaconis (The Annals of Probability 1977, v5) that $\log(n!)$ is uniformly distibuted mod 1. This has the consequence that any sequence of leading (most significant) digits appears infinitely often. This is probably not going to be of any direct help, but it seems like it deserves to be mentioned.

Secondly, and importantly, it is known that for any lacunary sequence of positive integers $a_n$ (meaning that there is a fixed $\rho>1$ such that the inequality $a_{n+1}>\rho a_n$ holds for all large enough $n$) there are real numbers $c$ such that the sequence $\{ca_n\}$ is bounded away from 0 mod 1, and in fact the set of such real numbers has Hausdorff dimension 1. This is trivial to prove for $\rho > 2$, and in fact in this case we can easily choose $c$ (nonconstuctively!) to get any of the behaviors described in Items 1,2 and 3.

For $\rho$ near 1 the above statement about lacunary sequences was an Erdos problem, first solved by B. de Mathan (Acta Math. Hungar. 1980 v36). There is an exposition by Katznelson here.

Since the sequence $n!$ is lacunary (with $\rho$ as large as one wants) we already know that the behaviors described in Items 1-3 occur in abundance. The question is whether we know any specific examples.

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3 Answers

Isn't it well known that $\{e n!\}\rightarrow0$ as $n\rightarrow+\infty$ ? This is an immediate consequence of the formula $$e=\sum_0^\infty \frac{1}{k!}.$$ To prove the limit, we just remark that the remainder (for $k\ge n+1$) is bounded by $\frac{1}{n!n}$.

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Thanks. I wish I had thought of that! Any ideas for $c$ algebraic, or values that are dense mod 1? –  SJR Nov 11 '10 at 9:52
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I think conventional wisdom holds that if $c$ is a real algebraic irrational then $cn!$ is uniformly distributed modulo 1. I think conventional wisdom further holds that no one has a clue as to how to prove this. –  Gerry Myerson Nov 11 '10 at 22:57
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I'll have to look this up in Kuipers and Niederreiter to make sure I have it right, but I think that for any increasing sequence $a_1,a_2,\dots$ of integers the sequence $a_1x,a_2x,\dots$ is uniformly distributed modulo 1 for almost all $x$ (meaning all $x$ except a set of measure zero). Actually finding such an $x$ is a different story.

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Yes, This is Theorem 4.1 in Kuipers and Niederreiter. As you say the problems is to find an example. Maybe it is possible, to use some expression in $e$ to get an example that is dense mod 1? As for algebraic numbers $c$, could it be that we have no example where $cn!<1/2$ mod 1 infinitely often?! –  SJR Nov 11 '10 at 10:34
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Maybe this works. Let $s_1,s_2,\dots$ be u.d. in $[0,1)$. Let $c=\sum[ks_k]/k!$. Then $\lbrace cn!\rbrace=t_{n+1}$ where $t_{n+1}-s_{n+1}$ goes to zero, so $\lbrace cn!\rbrace$ is u.d. in $[0,1)$. –  Gerry Myerson Nov 11 '10 at 11:09
    
I like this - The problem is can one choose the $s_k$ so that $c$ is something familiar? And I wonder what are ALL POSSIBLE $c$'s defined in this way? –  SJR Nov 11 '10 at 11:38
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See David Speyer's thread

http://tea.mathoverflow.net/discussion/741/does-lim-cosn-exist/

which is about a closed question involving limit points of $\cos (n!)$ which is to say values of $n! \pmod {2 \pi},$ or in turn values of $$ \frac{n!}{2 \pi} \pmod 1. $$ David's conclusion, evidently posted on math.stackexchange, is that we do not know enough about $\pi$ to answer this.

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OP mentioned this thread, but thanks for posting the link. –  Gerry Myerson Nov 12 '10 at 4:27
    
You are correct, oh gerriferous one. I think Anton made it a link in the question as well. –  Will Jagy Nov 12 '10 at 4:54
    
Notice that by Denis's reply if $e+1/(2\pi)$ is rational, then $n!/(2\pi)$ tends to a limit, namely 0, ie if $\cos(n!)$ does not tend to a limit, then $e+1/(2\pi)$ is irrational. The latter is open, isn't it? –  SJR Nov 12 '10 at 6:31
    
@SJR, yes. –  Gerry Myerson Nov 12 '10 at 13:41
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