Approximation of a quotient that involves the Lambert function.

Question

I would like to find an asymptotic upper bound for $$\frac{-\ln n}{W(- \ln^{-c}n)}$$ where $c$ is positive and $W$ is the Lambert function. More precisely, I want something which dominates this quotient asymptotically. The bound should be quite tight.

Can we say that $\frac{-\ln n}{W(- \ln^{-c}n)}=O\left(\frac{\ln n}{\ln \ln n}\right)$ ?

Thank you very much.

Answer 1

A more accurate asymptotic series can be obtained with more terms.

but, before, one can observe that $y=\frac{-\ln n}{W(- \ln^{-c}n)}$ is not real if $n<e^{e^{1/c}}$

This means that, for small values of $c$, $n$ has to be very large :

In case of $c=0.2$ for example, $y$ is not real if $n<10^{64}$ (roughly)

In case of $c=0.5$ for example, $y$ is not real if $n<1618$

So, I doubt that small values of $c$ ( i.e.: $c<<1$) be encountered in real world. If they are, what does mean so big values of $n$ and correlatively of $y$ ?

To be accurate, the number of terms depends on $c$. The power of last term $\ln(n)$ must be negative. For example with the above series, the power of the fourth term is negative only if $c>0.5$

Other example : if $c=0.2$ the series requires more than 7 terms.

The next table shows some numerical results :

SECOND BRANCH OF THE W-function (noted $W_{-1}$ ) :

An approximate can be found in page 13 of the paper : https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function

$$W_{-1}(X)=-\theta-\ln\left(\theta+\ln\left(\theta+\ln\left(\theta+\ln\left( \theta+...\right) \right) \right) \right) \qquad\text{where}\qquad\theta=-\ln(-X)$$ Here we have : $$X=-\ln^{-c}(n)\quad\to\quad\theta=c\:\ln(\ln(n))$$

The next table of numerical examples shows that a large number of terms are necessary to a good accuracy, especially in case of small $c$.

NOTE :

The therm $\frac{\ln(n)}{\ln(\ln(n))}$ appears in the simplest form of approximate (with only one term): $$\frac{-\ln n}{W_{-1}(- \ln^{-c}n)}\simeq \frac{\ln(n)}{c\:\ln(\ln(n))}$$ Of course, the accuracy is much better in using the above formula with as many terms as possible.

Answer 2

For $c>0$ and $n$ sufficient large the dominator becomes $W(0)$ :

$W(-(\ln n)^{-c})\to W(0)=0$ for $n\to \infty$.

It’s $\frac{1}{W(x)}= \frac{1}{x}e^{W(x)}$ by definition.

Substitute $x$ by $-(\ln n)^{-c}$.

Then your term becomes $(\ln n)^{c+1}$; that’s very different to $\frac{\ln n}{\ln\ln n}$.

EDIT:

I don’t know exactly what you mean, but I will try.

$x n^{\frac{1}{x}}=y_n$ with $y_n:=(\ln n)^{1+c}$

Lambert is not necessary, it’s enough to know:

$n^{\frac{1}{y_n}}=(n^{\frac{1}{x}})^{(n^{-\frac{1}{x}})}\le e^{\frac{1}{e}}$

Let be $n^{\frac{1}{y_n}}>1$.

Then for $n^{\frac{1}{x}}$ you’ve two solutions, if $1<n^{\frac{1}{y_n}}<e^{\frac{1}{e}}$, one solution if $n^{\frac{1}{y_n}}=e^{\frac{1}{e}}$ and no solution, if $n^{\frac{1}{y_n}}>e^{\frac{1}{e}}$.

Two solutions: (1) $1<n^{\frac{1}{x}}<e$ => $x>\ln n$ , (2) $e<n^{\frac{1}{x}}$ => $x<\ln n$

One solution: $n^{\frac{1}{x}}=e$ => $x=\ln n$

And: $1<n^{\frac{1}{y_n}}\le e^{\frac{1}{e}}$ => $y_n>e\ln n$ => $n>e^{e^{\frac{1}{c}}}$

If you choose solution (1), you get $x>e^{\frac{1}{c}}$. For $c\to 0$ only solution (2) remains.

Is this the answer to your question (more or less) ?!