# Solving an equation involving floor/ceiling as a summation bound

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.

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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.
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