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I noticed that $\ln(x)-\ln(\ln(x))$ seems to by asymptotic with the Lambert W Function. Is this true?

Jonathan
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    Yes it is - https://en.wikipedia.org/wiki/Lambert_W_function#Asymptotic_expansions By applying $z\mapsto z e^z$ to such expression we get something pretty close to the identity function. – Jack D'Aurizio Mar 15 '17 at 00:55
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    From page 13 in https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function

    For $x>e$ the expansion : $$W(x)= \ln(x)-\ln\left( \ln(x)- \ln\left( \ln(x)-\ln(...) \right)\right)$$ gives successive approximates : $$W(x)\simeq \ln(x)$$ $$W(x)\simeq \ln(x)-\ln\left( \ln(x)\right)$$ $$W(x)\simeq \ln(x)-\ln\left( \ln(x)- \ln\left( \ln(x) \right)\right)$$ $$W(x)\simeq \ln(x)-\ln\left( \ln(x)- \ln\left( \ln(x)-\ln\left( \ln(x) \right) \right)\right)$$ etc. // which is consistent with Simply Beautiful Art's answer.

     – JJacquelin Mar 15 '17 at 07:08

1 Answers1

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By definition:

$$x=W(x)e^{W(x)}$$

It thus follows that

$$e^{W(x)}=\frac x{W(x)}$$

$$W(x)=\ln\left(\frac x{W(x)}\right)$$

By applying the fixed point method, we find that for any initial condition greater than $e$ and $x>e$, we have

$$W(x)=\ln\left(\frac x{W(x)}\right)=\ln\left(\frac x{\ln\left(\frac x{W(x)}\right)}\right)=\dots=\ln\left(\frac x{\ln\left(\frac x{\ln\left(\frac x{\vdots}\right)}\right)}\right)$$

And by expanding with log rules with some obvious bounds on $W(x)$, we find that

$$W(x)\sim\ln(x)-\ln(\ln(x)-\ln(\ln(x)-\ln(\dots)))\sim\ln(x)-\ln(\ln(x))+o(1)$$

Here is a visualization of this with the dotted line being the Lambert W function:

enter image description here

Simply Beautiful Art
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    Note that one does not need to carry this process on forever, We can already see that W(x)<ln(x)W(x)<ln⁡(x) from the original definition, thus$$W(x)=\ln\left(\frac x{W(x)}\right)>\ln\left(\frac x{\ln(x)}\right)$$But$$W(x)=\ln\left(\frac x{\ln\left(\frac x{W(x)}\right)}\right)<\ln\left(\frac x{\ln\left(\frac x{\ln(x)}\right)}\right)$$which is all you really need. – Simply Beautiful Art Mar 15 '17 at 23:14
  • Note that for $|W(z)|<1$ the same formula holds, but with $\exp$ instead of $\log$. ( https://en.wikipedia.org/wiki/Lambert_W_function#Asymptotic_expansions#Representations ) – Thomas Ahle Jul 16 '19 at 16:58