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Question: Find a real value $y$ such that the following identity holds

$$\log(\log(\log(x)))=\int^x_y\frac{dx}{x\log(x)\log(\log(x))}\quad (x>y)$$

Then, find an algorithm to calculate

$$f_n(x)\quad (x>e^{n-1})$$

where $f_0(x)=x$ and $f_n(x)=\log(f_{n-1}(x))$.

Thanks in advanced! Help appreciated!!

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  • $\begingroup$ Hint: The derivative of $\log\log\log x$ is... $\endgroup$ – Lord Soth Apr 11 '13 at 4:56
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If you find the suitable lower bounds $y_n$ for $x$, then : $$f_n(x)=\log(\log(\log\dots(\log x)\dots))=\int_{y_n}^x\dfrac{dt}{\prod\limits_{k=1}^{n-1}f_k(t)}$$ for all $x>y_n$, and for all $n\geq1$

However, as the above said, if you want to define the iterated logarithm, then taking $x>e^{n-1}$ is not enough, but you have to consider :$$x>e^{e^{\dots ^e}}$$ For example take $\log(\log(\log(\log(\log x))))$. In order define it, you need :

$$\log(\log(\log(\log x)))>0\Rightarrow \log(\log(\log x)) >1$$ $$\Rightarrow \log(\log x)>e \Rightarrow \log x >e^e$$ $$\Rightarrow x>e^{e^e}\thickapprox 3.8\times10^6$$ while $e^{n-1}=e^{5-1}=e^4\thickapprox 54$

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Make the substitution $u=\log(\log x)$, We find that one antiderivative is $\log(|\log(\log x)|)$.

Substitute the two endpoints. So we want $\log(\log(\log y))=0$. That means $\log(\log y)=1$, so $\log y=e$, and therefore $y=e^e$.

Note that if we want the iterated logarithm to be defined, in general $x\gt e^{n-1}$ will not be enough. We need an iterated exponential, as in the calculation of $y$.

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