I would like to solve the equation $x \log\log x = n$. I've seen a lot of post about the equation $x \log x$ but here I have a composition of $\log$. How can I solve it ?

Thank you very much.

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    $\begingroup$ Numerically? There is no reason to think the inverse of $x \log(\log x)$ can be writtenn in any simpler form. Even the inverse of $x\log(x)$, which you mention, has inverse in terms of a non-elementary function, the Lambert W function. $\endgroup$ – GEdgar Apr 8 '16 at 14:19
  • $\begingroup$ There is no analytical solution, even using the Lambert $W$ function. Just numerical resolution. $\endgroup$ – Yves Daoust Apr 8 '16 at 14:21
  • $\begingroup$ Thanks @GEdgar. In fact I was wondering if we can use Lambert function. I prefer an asymptotic approximation, rather than a numerical resolution. $\endgroup$ – Dingo13 Apr 8 '16 at 14:21
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    $\begingroup$ The solution is the fixed point of $x\mapsto\frac{n}{\log\log x}$ hence it is quite close to $\frac{n}{\log\log n}$ or to $\frac{n}{\log\log\frac{n}{\log\log n}}$. $\endgroup$ – Jack D'Aurizio Apr 8 '16 at 14:26
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    $\begingroup$ @Dingo13: sure. Since $x\log\log x=n$, $$ \frac{x\log\log n}{n} = \frac{\log\log n}{\log \log x},$$ but the function $\log\log z$ grows so slowly (its derivative is $\frac{1}{z\log z}$) that the last ratio is clearly very close to $1$. $\endgroup$ – Jack D'Aurizio Apr 8 '16 at 15:04

With the estimate $x=\dfrac n{\log(\log(n))}$, you have

$$\dfrac n{\log(\log(n))}\log\left(\log\left(\dfrac n{\log(\log(n))}\right)\right)=\dfrac n{\log(\log(n))}\log\left(\log(n)-\log\left(\log(\log(n))\right)\right)$$

which is asymptotically $n$.

For example, for $n=100$,




  • $\begingroup$ Thank you @YvesDaoust. Can we prove this estimation? Using $O$ or $\theta$? Thank you again. $\endgroup$ – Dingo13 Apr 8 '16 at 14:33
  • $\begingroup$ That obvious as $\log(\log(t))=o(t)$. $\endgroup$ – Yves Daoust Apr 8 '16 at 15:12

Let $x_n$ denote the solution of $x \log \log x = n$. Here is an asymptotic expansion for $x_n$.

Since $x \log\log x$ is monotone increasing and goes to $\infty$, it is clear that $x_n$ is motonone increasing and $\lim_n x_n = \infty$.

Since $0=\lim_n\frac{1}{\log \log x_n}=\lim_n \frac{x_n}{n}$, we have the estimate $x_n = o(n)$

Sharpening this is easy: $\ln x_n + \ln\ln\ln x_n = \ln n$ hence $\ln x_n + o(\ln x_n) =\ln n$ and $\ln x_n = \ln n +o(\ln n)$

Plugging the last estimate in the equation $\ln x_n + \ln\ln\ln x_n = \ln n$, $$\ln x_n + \ln\ln(\ln n +o(\ln n))=\ln n\\\ln x_n + \ln\ln [\ln n(1+o(1))]= \ln n\\ \ln x_n + \ln[\ln\ln n +\ln(1+o(1))]=\ln n\\ \ln x_n + \ln[\ln\ln n +o(1)]=\ln n\\ \ln x_n + \ln\ln\ln n + o\left(\frac1{\ln\ln n} \right)=\ln n\\$$

Hence $\displaystyle x_n = \exp(\ln n)\exp(-\ln\ln\ln n))\exp\left(o\left(\frac1{\ln\ln n} \right)\right)=\frac{n}{\ln \ln n}+o\left( \frac{n}{\ln \ln n}\right)$

Plugging the estimate $\ln x_n + \ln\ln\ln n + o\left(\frac1{\ln\ln n} \right)=\ln n$ back into $\ln x_n + \ln\ln\ln x_n = \ln n$ yields an even more precise estimate, ad libitum: $$ x_n = \frac{n}{\ln \ln n} + \frac{ \ln\ln\ln n}{\ln n (\ln\ln n)^2}+o\left( \frac{ \ln\ln\ln n}{\ln n (\ln\ln n)^2}\right)$$


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