When are "normal" functions normal?

Question

I expected that the fractional part of f(n), n being an integer, would be distributed uniformly over [0,1] (for positive functions - otherwise take [-1,1]) for any run-of-the-mill function, except there is a good reason otherwise. I experimented a bit with MATHEMATICA and found to be dead wrong. Methodics: I computed f(n) for $0<n<=100000$ and the n-th central moments of that set. I multiplied them with $(2n+1)2^{2n}$ to normalize since a uniformly distributed U[0,1] after this operation has 1 for even and 0 for odd moments. Call f random if the moments of the set ${f(1),...,f(10⁶)}$ are about as close to 1 and 0 as those of the random number set. Some random :-) results:
$f(n)=n^r$ (trivial for r integer or r<0): Sqrt[n] is random (funnily, more random than a genuine random series :-) and I guess so is any f with non-integer r>0.
$f(n)=log(n)$: Very non-random, possibly because it's so flat.
$f(n)=n*log(n)$: Random.
$f(n)=sin(n)$: Non-random but the moments are $(2n+1)!/(n!)^2$. (This should be easy to prove since sin is periodic and you can approximate by the respective moment integrals.)
$f(n)=exp(n)$: Checked only to n<=100 for obvious reasons, slightly non-random.

Do you have a reference for me? At least the sqrt part was already analyzed 40 years ago. (And could someone verify my results, at least on sin(n)?)

Answer 1

You are asking for which functions $f$ the sequence $f(n)$ is equidistributed modulo 1. This is a whole area of mathematics, which began with the work of Weyl in 1916, who discovered the connection between equidistribution and estimates for exponential sums. One of the standard references is the book by Kuipers and Niederreiter, "Uniform distribution of sequences", which can be found here:

http://web.maths.unsw.edu.au/~josefdick/preprints/KuipersNied_book.pdf

If you are more interested in special cases and quantitative aspects, you might have to study exponential sums in more detail. In this case I would recommend "Van der Corput's Method of Exponential Sums" by Graham and Kolesnik (which apparently is not free online).

Answer 2

In the case of $\log(n)$, if $\epsilon > 0$ is small we have $\text{frac}(\log(N) - \log(n)) < \epsilon$ for $Ne^{-j} > n > N e^{-\epsilon-j} $, $j = 0, 1, 2, \ldots$, so the fraction of $n \in \{1,2, \ldots, N\}$ with $\text{frac}(\log(N) - \log(n)) < \epsilon$ is approximately $(1-e^{-\epsilon})/(1-e^{-1})$, and this is significantly different from $\epsilon$.