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For any positive integer n, what is the value of t* that maximises the following expression?

$$\displaystyle \sum_{j=1}^{n-t^*}\left(\frac{t^*-j+2}{t^*+j}\right)$$

where $t^*$ is some integer in the set $\{0,1,2...,n-1\}$.

Clearly $t^*$ = $f(n)$ but I am unable to find what this function is? E.g. $t^*=\frac{3}{5}(n-1)$ rounded to the nearest whole number is a decent estimate but not exact. Perhaps there is no closed-form solution. Thanks for your help!

awwlaz
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2 Answers2

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Reindex with $j\mapsto j-t^*$ and you have $$\sum_{j=1+t^*}^n\frac{2t^*-j+2}{j}$$ or just $$\sum_{j=1+t^*}^n\left(\frac{2t^*+2}{j}-1\right)$$ which is $$(2t^*+2)\sum_{j=1+t^*}^n\frac{1}{j}-(n-t^*-1)$$ The difference between this expression and the same expression with $t^*$ replaced by $t^*-1$ is $$(2t^*+2)\sum_{j=1+t^*}^n\frac{1}{j}-(n-t^*-1)-2t^*\sum_{j=t^*}^n\frac{1}{j}+(n-t^*)$$ which simplifies to $$2\sum_{j=1+t^*}^n\frac{1}{j}-1$$ The critical value for $t^*$ will happen when this difference is $0$. Either the floor or the ceiling of that number is the answer you are seeking. Representing $H(m)$ as the $m$th partial Harmonic summ, we are trying to solve $$2H(n)-2H(t^*)=1$$ or rather $$H(t^*)=H(n)-\frac{1}{2}$$ If you can find an inverse function for $H$, you have your answer. For large $n$, $H(n)\approx \log(n)+\gamma$, where $\gamma$ is the Euler-Mascheroni constant, so if we take that and run with it, $$t^*\approx\exp\left(\log(n)-\frac{1}{2}\right)=\frac{1}{\exp(1/2)}n\approx0.6065n$$ agreeing with your empirical estimate of $3/5$. This is more accurate the larger the value of $n$.

2'5 9'2
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  • thank you for your help! very helpful indeed! – awwlaz Oct 05 '13 at 03:00
  • sure; check that you refresh. I went through several edits where the end part wasn't quite worked out well. – 2'5 9'2 Oct 05 '13 at 03:03
  • You can improve this a bit by using $H(m)\approx\log(m)+\gamma+\frac{1}{m^2}$. Since we already have that $t^$ is in the ballpark of $\frac{1}{\exp(1/2)}n$, this more detailed replacement of the partial Harmonic sum yields $t^\approx \frac{1}{\exp(1/2+(e-1)/n^2)}n$. – 2'5 9'2 Oct 05 '13 at 03:09
  • Is it safe to say there is no closed-form solution, even if we involve floor/ceiling functions? – awwlaz Oct 05 '13 at 03:11
  • I think that would be safe to say. Well, you could have a "closed form solution" involving $H^{-1}$. Also, not that I'm chasing rep points, but since you are new to this site: if you find someone's posts even mildly useful, whether it's for your own questions or someone else's, it's good practice to upvote. It's how the system organically floats useful responses for future readers. – 2'5 9'2 Oct 05 '13 at 03:14
  • Of course I did try to vote up :) - but only users with 15 reputation points and above can vote up! Soon enough hopefully! – awwlaz Oct 05 '13 at 03:17
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In the edit of my answer to this question, I proposed for the inverse of the harmonic number

$$n\sim -\frac{1}{2}+\sqrt{\frac{1}{6 W(k)}-\frac{1}{4}}\quad \text{where} \qquad k=\frac{1}{2}e^{2(\gamma-H_n)}$$

So, after @Alex Jordan's answer,

$$H(t^*)=H(n)-\frac{1}{2} \implies t^*=-\frac{1}{2}+\frac{1}{6} \sqrt{\frac{6-9 W\left(\frac{1}{6} e^{1-2 \psi (n+1)}\right)}{W\left(\frac{1}{6} e^{1-2 \psi (n+1)}\right)}}$$ Expanding as a series for large $n$, this gives $$t^*=\frac{n}{\sqrt{e}}\Bigg[1+\frac{1-\sqrt{e}}{2 n}+\frac{1-e}{24 n^2}+\frac{e-1}{48 n^3}+O\left(\frac{1}{n^4}\right) \Bigg]$$

For $n=10$, this would give $$t^*=\frac{50419-2400 \sqrt{e}-19 e}{4800 \sqrt{e}}=5.864447$$ while the exact solution is $5.864453$

Claude Leibovici
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