5
$\begingroup$

Question (corrected)

I managed to prove:

$$ f(z) \sim \left\{ \begin{array}{ll} - \ln |z| \int_0^{\frac{-N}{\ln|z|}} e^{-\frac{1}{|y|}} dy & |z|<< 1 \\ ? & |z| \approx 1 \\ ?? & |z|\gg 1 \\ \end{array} \right. $$

Where, $$ f(z) = z+ z^\frac{1}{2}+ z^\frac{1}{3}+ z^\frac{1}{4} +\dots + z^\frac{1}{N}$$

However, I do not have a asymptotic expression as $|z| \to 1$ or $|z| \to \infty$. Can someone provide that expressions as well?

Background

Consider the complex function:

$$ f(z) = z+ z^\frac{1}{2}+ z^\frac{1}{3}+ z^\frac{1}{4} +\dots + z^\frac{1}{N}$$

Let, us write $z= r e^{i \theta}$ where $1>> r > 0$.

$$ f(r e^{i \theta}) = r e^{i \theta}+ r^{1/2}e^{i \theta/2} + \dots +r^{1/n}e^{i \theta/n}$$

Consider, the function with $f(x) = e^\frac{-1}{|x|}$ and the integral:

$$ \int_{0}^{N \epsilon} f(y) dy = \lim_{\epsilon \to 0,N \to \infty} \Big(f(\epsilon) +f(2 \epsilon) + \dots + f(N\epsilon) \Big)\epsilon$$

with $N \epsilon = b$. We modify our considerations and use$^\ast$:

$$ \lim_{\epsilon \to 0,N \to \infty} \sum_{r=1}^{N} a_r f(r\epsilon) \epsilon = \lim_{s\to 1} \frac{1}{\zeta(s)} \times \sum_{r=1}^\infty \frac{a_r}{r^s} \int_0^{N\epsilon} f(y) dy$$

Choosing $a_r= e^{i\theta /r}$ and replacing $\epsilon = \frac{-1}{\ln \delta}$

$$ \lim_{\delta \to 0,N \to \infty} \Big(f(\frac{-1}{\ln \delta})e^{i \theta} + f(\frac{-2}{\ln \delta})e^{i\theta /2} + \dots + f(\frac{-N}{\ln\delta})e^{i \theta/N} \Big) \frac{-1}{{\ln\delta}} = \underbrace{\lim_{s\to 1} \frac{1}{\zeta(s)} \times\sum_{r=1}^\infty \frac{e^{i\theta /r}}{r^s}}_{=1} \int_{0}^{- \frac{N}{\ln \delta}} e^\frac{-1}{|y|} dy$$

Substituting with $f$, using asymptotics and solving the limit using this:

$$ \delta e^{i \theta} + \delta^{1/2} e^{i \theta/2} + \dots + \delta^{1/N} e^{i \theta/N} \sim - \ln \delta \int_0^{\frac{-N}{\ln \delta}} e^{-\frac{1}{|y|}} dy $$


$^\ast$We split into real and imaginary parts to apply the formula.

$\endgroup$
3
  • $\begingroup$ There must be some error in the $z >> 1$ case, as that integral doesn't ever converge. $\endgroup$ Aug 26, 2020 at 15:57
  • 2
    $\begingroup$ Clearly, $f(z)\sim z$ for large $z$ and $f(z) \sim z^{1/N}$ for $|z|\ll 1$. $\endgroup$
    – Gary
    Aug 26, 2020 at 16:00
  • $\begingroup$ @Gary but for some order of $N$ the expression will diverge. And to see an example with real $|z|<<1$ math.stackexchange.com/questions/3798963/… $\endgroup$ Aug 26, 2020 at 16:05

1 Answer 1

2
$\begingroup$

As $z$ approaches $1$, the value of $f$ clearly approaches $N$. To see the difference term, we observe that as $z\rightarrow 1$, we have $z^p = (1 + (z-1))^p = 1 + p(z-1) + O(|z-1|^2)$ by the binomial series. Hence $$ f(z) - N = \sum_{n=1}^N (z^{1/N} - 1) = \sum_{n=1}^N \frac1n(z-1) + O(|z-1|^2) = (H_N)(z-1) + O(|z-1|^2) $$ where $H_N$ is the $N$th harmonic number. One can use the binomial series similarly to get higher order terms if you want. Since the function is differentiable, for any point you will have more generally as $z\rightarrow c$, $f(z)-f(c) = f'(c) (z-c) + O(|z-c|^2)$.

For historical reasons this is the asymptotics in terms of $N$: We have $$ z^{1/N} = \sum_{k=0}^\infty \frac{(\log z)^k}{k!} \frac1{N^k} = 1 + \frac{\log z}{N} +O(1/N^2) $$ hence we have $$ \sum_{n=1}^N z^{1/n} = N + (\log z)\sum_{n=1}^N \frac1n +O(1) = N + (\log z)(\log N) + O(1) $$

$\endgroup$
2
  • $\begingroup$ By $|z| \to 1$. I meant $z= 1 + e^{i \theta} \epsilon$ where $\epsilon \to 0$ $\endgroup$ Aug 26, 2020 at 16:07
  • $\begingroup$ So you mean $z\rightarrow 1$ not $|z|\rightarrow 1$. I shall update accordingly. $\endgroup$ Aug 26, 2020 at 16:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .