Find the domain of this function involving fractional part and greatest integer function

Question

Consider a function $$y=\log_{10}\{\log_{10}\lfloor\log_{10}(\log_{10}x)\rfloor\}$$ where $\{x\}$ denotes the fractional part of $x$ and $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.

Find its domain.

I have tried traditional ways of solving this. But I'm not able to handle the fractional part function and the greatest integer function.

Also, in the domain, we have to exclude values that can be written as continuous powers of $10$.

I have tried Wolfram Alpha and I got one solution. It is $x=23^{{272}^{56}}$. For this, $y$ equals $-1$.

Any help is greatly appreciated.

Note: By "continuous powers of $10$", I mean expressions like $10^{{10}^{{10}^{10}}} $.

Answer 1

Let's work the expression from the inside out.

For $\log_{10}(x)$ to be defined we need $x>0$.

For $\log_{10}(\log_{10}(x))$ we need $\log_{10}(x)>0$. That is, $x>1$.

The floor function is defined for all reals, so no new restriction for $\lfloor\log_{10}(\log_{10}(x))\rfloor$.

Then for $\log_{10}(\lfloor\log_{10}(\log_{10}(x))\rfloor)$ we need $\lfloor\log_{10}(\log_{10}(x))\rfloor > 0$, which means $\log_{10}(\log_{10}(x))\ge 1$. That is $\log_{10}(x) \ge 10$, $x\ge 10^{10}$.

The fractional part function is defined for all reals, so no new restriction for $\{\log_{10}(\lfloor\log_{10}(\log_{10}(x))\rfloor)\}$.

Finally, for $\log_{10}(\{\log_{10}(\lfloor\log_{10}(\log_{10}(x))\rfloor)\})$ we need $\{\log_{10}(\lfloor\log_{10}(\log_{10}(x))\rfloor)\} >0$. Equivalently, $\log_{10}(\lfloor\log_{10}(\log_{10}(x))\rfloor) \not\in\Bbb Z$. Let see which values of $x$ are excluded:

\begin{align} &\log_{10}(\lfloor\log_{10}(\log_{10}(x))\rfloor)=n\\ &\iff \\ &\lfloor\log_{10}(\log_{10}(x))\rfloor = 10^n\\ &\iff \\ &10^n\le\log_{10}(\log_{10}(x))< 10^n+1\\ &\iff \\ &10^{10^n}\le\log_{10}(x)< 10^{10^n+1}\\ &\iff \\ &10^{10^{10^n}}\le x< 10^{10^{10^n+1}}\\ \end{align}

So you need to exclude those intervals.

The domain is $\displaystyle [10^{10},\infty) \setminus \bigcup_{n\in \Bbb Z} [10^{10^{10^n}}, 10^{10^{10^n+1}})$, which can be seen to be equal to $\displaystyle \bigcup_{n=0}^\infty [10^{10^{10^n+1}}, 10^{10^{10^{n+1}}})$

Answer 2

As compound functions, I restrict the domain of definition from outside to inside In other words if $f=\circ^n_{i=1} f_i $. And if $D_{f_i} $ is the denoted as the domain of definition of $f_i$.

1

$\log_{10}(\cdot x )$ accepts $x \in ]0,\infty[$

2

$\log_{10}\{ \cdot x \}$ accepts $x \in \bigcup_{k=0}^\infty ]k,k+1[$

3

$\log_{10}(\{\log_{10}(\cdot x)\})($ accepts $\log_{10}(\cdot x) \in \bigcup_{k=0}^\infty ]k,k+1[$ i.e. $(\cdot x) \in \bigcup_{k=0}^\infty ]10^k,10^{k+1}[$

4

$\log_{10}(\{\log_{10}(\lfloor \cdot x \rfloor)\})$ accepts $\log_{10}(\lfloor \cdot x \rfloor) \in \bigcup_{k=0}^\infty ]k,k+1[$ i.e. $(\lfloor \cdot x \rfloor) \in \bigcup_{k=0}^\infty ]10^k,10^{k+1}[$ i.e. $(\cdot x) \in \cup_{k=0}^\infty ]1+10^k,10^{k+1}[ $.

5

$\log_{10}(\{\log_{10}(\lfloor \log_{10}(\log_{10})(\cdot x) \rfloor)\})$ accepts $\log_{10}(\log_{10})(\cdot x) \in \cup_{k=0}^\infty ]1+10^k,10^{k+1}[ $ i.e. $(\log_{10})(\cdot x) \in\cup_{k=0}^\infty ]10^{1+10^k},10^{10^{k+1}}[$ i.e. $(\cdot x) \in \cup_{k=0}^\infty ]10^{10^{1+10^k}},10^{10^{10^{k+1}}}[$

Hence $$D_f=\cup_{k=0}^\infty ]10^{10^{1+10^k}},10^{10^{10^{k+1}}}[$$

If everything is ok.