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.

Let $a$ and $b$ be positive reals both greater than $1$. I'd like to compute the limit of the summation \begin{eqnarray} \lim_{x \to \infty} \frac{1}{\log_a x} \sum_{k = 0}^{\lfloor \log_{a} x \rfloor} \left \lbrace (\log_{b} x) - k \log_b a \right \rbrace, \end{eqnarray} where $\{ \cdot \}$ denotes the fractional part function and $\lfloor \cdot \rfloor$ denotes the floor function. I believe that the limit is $\frac{1}{2}$ for almost all $a$ and $b$, and I believe the limit doesn't exist if $b = a^r$ where $r$ is a positive rational. I also have a feeling that Weyl's theorem may be playing a role here but I'm not sure if $\lbrace (\log_{b} x) - k \log_b a \rbrace$ is equidistributed on any interval.

Any help is certainly appreciated!

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Your notation is unclear, but I'll assume you meant $$ \log_b(a^{-k}x) = \log_b(x) - k\log_b a. $$ Weyl's theorem shows that if $\log_b a^{-1}$ is irrational then $\{k\log_b a^{-1}\}$ is asymptotically equidistributed. Shifting it by $\{\log_b x\}$ shouldn't make a difference, and so the limit approaches $1/2$.

share|improve this answer
    
Thanks. I added some parentheses to clarify the function. –  user02138 Feb 28 '11 at 22:40

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.