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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Are there useful lower and upper (preferably sharp or asymptotic) bounds for the following fractional part-power summation? Given arbitrary reals $a, b > 1$ and integer $n \geqslant 1$, let \begin{align} f_{n}(x) = \sum_{i = 0}^{\lfloor \log_a x \rfloor} \left( \text{frac}( \log_{a} (b^{-i} x) ) \right)^{n}, \end{align} where $\text{frac}$ denotes the fractional part function. Naive bounds include $0 \leqslant f_n(x) \leqslant \cdots \leqslant f_{1}(x) \leqslant f_{0}(x) = \lfloor \log_a x \rfloor + 1$ for $x \geqslant 1$, so $f_{n}(x) = O(\log x)$, but I suspect that this can be improved substantially for certain special values of $a$ and $b$.

As the order of divergence should depend on the algebraic nature of the factor $\log_{a} b$ by Elementary Equidistribution Theory, I expect a more rapid divergence in the limit $x \to \infty$ if $\log_{a} b$ is rational, perhaps $O(\log x)$. My guess is $O(\log \log x)$ or something similar if $\log_a b$ is irrational.

Any help is certainly appreciated!

share|cite|improve this question
Clearly we can write $\mathcal{O}(\log x)$ instead of just $\mathcal{O}(x)$. – anon Mar 2 '12 at 3:58
Can you throw in a few parentheses so I can tell whether you have $(\log_ab)^{-i}x$ or $\log_a(b^{-i}x)$? frac$(u^n)$ or $({\rm frac}(u))^n$? – Gerry Myerson Mar 2 '12 at 4:59

Your Answer


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

Browse other questions tagged or ask your own question.