algebra-precalculusceiling-and-floor-functionselementary-number-theory
$$2001=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\\
(x>0 ,x\in\mathbb R)$$
find $x$ that satisfies the expression above.
My attempt
Since $$(x+1)^4\gt\text{(right side of the expression given)}\geq x^4\\
7\gt x\geq6$$
And, $\frac{2001}x$ needs to be a integer.
If $x$ is not a rational number, $\frac{2001}x$ is not a integer.
So, $x$ can be thought as $\frac ab(a,b\in\mathbb Z,a|2001)$
But, I failed to go further.
I find it useful to view your expression not within the realm of calculus, but rather in terms of constructing a digital signal. In particular, this repeating 3-valued signal is the goal:
i.e. $\cos(n\pi/2)$ for integer $n$. It is clear that we can achieve it as the sum of these two binary signals, each having half the frequency:
Note these two summands are, in form, hardly different from each other; the phase difference is worth noting, and that one is "down", the other "up".
The first summand is $\left\lfloor\frac{n}4\right\rfloor-\left\lfloor\frac{n+2}4\right\rfloor$ while the second is $1 + \left\lfloor\frac{n+1}4\right\rfloor-\left\lfloor\frac{n+3}4\right\rfloor$. We see that each binary oscillation is expressible as the difference of two (non-periodic) staircase signals, one lagging behind the other.
Solve for $x$ where both: $$\begin{align*} x\lfloor x\lfloor x\cdot 6\rfloor\rfloor &= 2001 \tag1\\ \lfloor x \rfloor &= 6 \tag2 \end{align*}$$
From the property that $\lfloor a \rfloor \le a$, the LHS of $(1)$ satisfies
$$\begin{align*} x\lfloor x\lfloor x\cdot 6\rfloor\rfloor &\le x\cdot x\lfloor x\cdot 6\rfloor \le x\cdot x\cdot x\cdot 6\\ x^3\cdot 6 &\ge 2001\\ x\cdot 6 &\ge \sqrt[3]{6^2\cdot2001} \approx 41.6 \end{align*}$$
From $(2)$,
$$\begin{align*} 6 \le x &< 7\\ x\cdot 6 &< 42 \end{align*}$$
So $\lfloor x\lfloor x\rfloor\rfloor = \lfloor x\cdot 6\rfloor = 41$.
Similarly, from $x\lfloor x\cdot41\rfloor = x\lfloor x\lfloor x\cdot 6\rfloor\rfloor = 2001$,
$$\begin{align*} x^2 \cdot 41 &\ge x\lfloor x\cdot41\rfloor = 2001\\ x \cdot 41 &\ge \sqrt{41\cdot 2001} \approx 286.4 \end{align*}$$
And from the previous step,
$$\begin{align*} 41.6\ldots \le x \cdot6 &< 42\\ 6.9\ldots \le x &< 7\\ 284.3\ldots \le x\cdot 41 &< 287 \end{align*}$$
So $\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = \lfloor x\cdot 41\rfloor = 286$.
One may say this step works, with $\lfloor x\cdot 6\rfloor$ narrowed down to one possibility, only because the lower and upper bounds of $x\cdot 6$ is near. This, I would agree.
This was also observed by some answers and comments in a similar question: find positive real number x that satisfies $2001=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor$.
For another example by @Sil, to solve for the positive $x$ where $$x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 31$$
Here $\lfloor x\rfloor = \left\lfloor\sqrt[4]{31}\right\rfloor = 2$, then on one side $2x\ge \sqrt[3]{2^2\cdot 31} \approx 4.99$ while $2x < 2\cdot 3 = 6$. The possibilities for the next step is $\lfloor x\lfloor x\rfloor\rfloor = \lfloor 2x\rfloor \in\{4,5\}$:
$$\begin{array}{r|c|c} \lfloor x\lfloor x\rfloor\rfloor = & 4 & 5 \\\hline \text{Bounds from $x\lfloor x\rfloor$:}\\ x\cdot 2 = x\lfloor x\rfloor \in & [4.99\ldots, 5) & [5,6)\\ \overset{/2}{\implies} x \in & \left[2.49\ldots, \frac52\right) & \left[\frac52, \frac62\right)\\ \overset{\cdot \lfloor x\lfloor x\rfloor\rfloor}{\implies} x\lfloor x\lfloor x\rfloor\rfloor \in & [9.97\ldots, 10) & \left[12.5, 15\right)\\\hline \text{Lower bound from the given:}\\ 31 = x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor \le x^2\lfloor x\lfloor x\rfloor\rfloor = & 4x^2 & 5x^2\\ x\lfloor x\lfloor x\rfloor\rfloor \ge & \sqrt{4\cdot31}\approx 11.1 & \sqrt{5\cdot31}\approx 12.4\\\hline \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor \in & \emptyset & \{12,13,14\} \end{array}$$
The next step even has 3 possibilities:
$$\begin{array}{r|c|c|c} \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = & 12 & 13 & 14\\\hline \text{Bounds from $x\lfloor x\lfloor x\rfloor\rfloor$:}\\ x\cdot 5 = x\lfloor x\lfloor x\rfloor\rfloor \in & [12.5, 13) & [13, 14) & [14, 15)\\ \overset{/5}{\implies} x \in & [2.5, 2.6) & [2.6, 2.8) & [2.8, 3)\\ \overset{\cdot \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor}{\implies} x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor \in & [30, 31.2) & [33.8, 36.4) & [39.2, 42)\\\hline \text{From the given:}\\ x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = & 31 & 31 & 31\\ x = \frac{31}{\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor} = & \frac{31}{12} \approx 2.5\\\hline \text{Verifications:}\\ \lfloor x\rfloor = & \lfloor2.5\ldots\rfloor\\ \lfloor x\lfloor x\rfloor\rfloor = & \lfloor5.1\ldots\rfloor\\ \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = & \lfloor12.91\ldots\rfloor \end{array}$$
So the only positive solution $x$ is $\frac{31}{12}$, which has $\lfloor x\lfloor x\rfloor\rfloor = 5 \ne \left\lfloor\sqrt[3]{2^2\cdot 31}\right\rfloor = 4$.
Best Answer
By letting $f(x)=x\lfloor x\lfloor x \lfloor x\rfloor\rfloor\rfloor $ we have $$ \lim_{x\to 7^-}f(x) = 7(7(7(7 - 1) - 1) - 1) = 2002 $$ and in a left neighbourhood of $x=7$ our function is a linear function with derivative $(7(7(7 - 1) - 1) - 1)=286.$ Since $f$ is increasing, the problem boils down to solving $$ 286(x-7)+2002 = 2001 $$ which leads to $x=\frac{2001}{286}$.