Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Is it possible to solve the following equation for $\alpha$?

$$ M = \lfloor \alpha \rfloor + \alpha \sum_{k=\lfloor \alpha \rfloor +1}^N \frac{1}{k}$$

where $\alpha \geq 1$.

Intuitively, $M$ is a harmonic number (scaled by $\alpha$) and with the first $\lfloor \alpha \rfloor$ summands replaced by $1$. Or you can say that the summands are "capped at 1" after scaling by $\alpha$. For $\alpha=1$, $M$ is simply equal to the $N$th harmonic number. The summation may also be written as the difference of the harmonic numbers $H_N - H_{\lfloor \alpha \rfloor}$. However, I'm not yet quite sure how to approach this problem.

share|cite|improve this question

This feels like a problem that will need a numeric solution, so we can make some approximations. As you say, the summation can be written as $H_N - H_{\lfloor \alpha \rfloor} \approx \log N - \log {\lfloor \alpha \rfloor}$. Making that substitution, we have $M={\lfloor \alpha \rfloor}+\alpha(\log N- \log {\lfloor \alpha \rfloor})$ or $\alpha=\frac M{\log N−\log\lfloor \alpha \rfloor-1}$, which is a very nice form for iteration. $\log\lfloor \alpha \rfloor$ changes very slowly, so start by setting it to zero, calculating a guess at $\alpha$ and iterate to convergence. Then, if you want, tune it up by using the actual sum.

share|cite|improve this answer
Thanks a lot for the idea. Can you quantify how certain that feeling is (that there is no analytic solution at all)? I basically share your feeling, but I'm still not really certain that it's not possible. And a technical question: Is it no problem to replace $\frac{\lfloor \alpha \rfloor}{\alpha}$ by $1$ (the $-1$ term in the iteration equation)? – bluenote10 Nov 27 '13 at 12:32
It can't make a change in that term greater than $\frac 1\alpha. After a few iterations, floor)\alpha)$ won't be changing any more and you can put it in as a constant. At the end, if $\alpha$ is very close to a whole number, you need to check the other side – Ross Millikan Nov 27 '13 at 15:12

Your Answer


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.