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Let $$f(x)=\lim_{n\to\infty}\text{Im}\left( -\frac{2\lfloor n\rfloor}{\pi}H_{-x}^{\left( -\frac1{2\lfloor n \rfloor} \right)} \right)$$ where $H$ is harmonic number and $x\in\mathbb{R}^+$. I do not know much about harmonic numbers, but I am wondering if this function is same as floor function, i.e. is it true that $f(x)=\lfloor x \rfloor$ for all $x\in\mathbb{R}^+$? I plotted the floor function and $f(x)$ for $n=1000$ and functions overlap. Is it just a coincidence or there is a proof that $f(x)=\lfloor x \rfloor$?

Edit
Also, I noticed that $f(x)$ is continuous on whole interval $(0,\infty)$ only if $n$ doesn't approach $\infty$. At $n\to\infty$ this function isn't continuous on integer values of $x$. Can it be proved?

rubik
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  • Which definition of $H$ are you using? – Regret Feb 20 '15 at 11:48
  • @Regret. I think there is only one definition of harmonic numbers. See this. –  Feb 20 '15 at 12:32
  • But here the $H$ has negative subscript, not covered by the "one defintion" cited. – GEdgar Feb 20 '15 at 14:32
  • @GEdgar. It says that $H_n=\gamma+\psi_0(n+1)$ for all real $n$ and digamma function is definad for all real numbers. –  Feb 20 '15 at 14:45
  • When I plot your function for $n=1000$, it looks nothing like $\lfloor x \rfloor$. – GEdgar Feb 21 '15 at 18:23
  • @GEdgar. What program you have used to plot this function? I know only two programs which can plot harmonic number functions for all complex number. These programs are Mathematica and Maple. This is my plot in Mathematica. As you can see, only difference is that $f(x)$ is continuous, but it is only beacuse $n$ is finite. –  Feb 22 '15 at 20:03

1 Answers1

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OK, from the Mathematica code shown, I see that I had misinterpreted $H_{-x}^{(r)}$ as a power of $H_{-x}$. But in fact it is a two-argument generalized Harmonic number, denoted $H_{-x,r}$ on the Mathworld page.

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According to Maple, near $r=0$ we have $$ H_{k,r} = k - \big(\ln\Gamma(k+1)\big)r+O(r^2)\qquad\text{ as } r \to 0 $$ Therefore, for any $x > 0$ we have $$ \frac{-2n}{\pi}H_{-x,{-1/(2n)}} = \frac{-2n}{\pi}\left(-x+\ln\Gamma(-x+1)\frac{-1}{2n}+O\left(\frac{1}{n^2}\right)\right) $$ The $-x$ term is real, so we get $$ \lim_{n\to\infty}\mathrm{Im}\left(\frac{-2n}{\pi}H_{-x,{-1/(2n)}}\right) =\frac{-1}{\pi}\mathrm{Im}\big(\ln\Gamma(-x+1)\big) $$ So the question is whether this is $\lfloor x \rfloor$. It no longer has anything to do with harmonic numbers. It will follow from the functional equation for $\Gamma$.

GEdgar
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