A nice equation with the floor function

[Gp7417]

Hi all!
I should solve the following equation involving the ceil or floor function:
$$x \cdot \lceil x \cdot \lceil x \rceil \rceil = 82$$
I have found just one solution 41/9, but I have no idea about how finding the other solutions.
Thanks you in advance!

 

[jbriggs444]

I should solve the following equation involving the ceil or floor function:
$$x \cdot \lceil x \cdot \lceil x \rceil \rceil = 82$$
I have found just one solution 41/9, but I have no idea about how finding the other solutions.

What convinces you that there are other solutions?

 

[mfb]

Can the expression stay the same or decrease if you increase x?
Can the expression stay the same or increase if you decrease x?

 

[Gp7417]

What convinces you that there are other solutions?

The problem says: 'Find all possibile solutions'. I find this solution numerically.

Can the expression stay the same or decrease if you increase x?
Can the expression stay the same or increase if you decrease x?

The expression must be the same.

 

[mfb]

The expression must be the same.

Right, so can this be possible with a different x?

The problem says: 'Find all possibile solutions'.

That does not tell you how many solutions there are.

 

[Gp7417]

I propose this solution, check it out!
We must find solution on the positive real axis.
Let be $x= n + \delta$, where $n$ is the integer part and $\delta$ is the fractional part with $0 \leq \delta < 1$.
Given that $x \lfloor x \lfloor x \rfloor \rfloor=82$ and $x \lfloor x \lfloor x \rfloor \rfloor \leq x^3$ , we get $x \geq 82^{\frac{1}{3}} \approx 4.344$. Therefore $n \geq 4$.
If 5 is not solution the equation, then every number greater than 5 are not solutions. Therefore $n=4$.
With the previous statements, the equation becames:
$(4+\delta)\cdot \lfloor 16+4\delta \rfloor=82$
This equation is equivalent to the following equations:
$0 \leq \delta < 1/4$ and $(4+\delta)\cdot(16+0)=82$
$1/4 \leq \delta < 2/4$ and $(4+\delta)\cdot(16+1)=82$
$2/4 \leq \delta < 3/4$ and $(4+\delta)\cdot(16+2)=82$
$3/4 \leq \delta < 4/4$ and $(4+\delta)\cdot(16+3)=82$
Only the third equation admits a solution that is $\delta=5/9$.

 

[fresh_42]

What I don't understand here is why there is a solution at all?
Obviously x has to be between 4 and 5, excluding 4 and 5. So the inner ceiling is 5 and the outer at least 21. But four times twenty-one is already 84.

 

[mfb]

You write the ceil function and then derive things for the floor function? For the floor function that works.

So the inner ceiling is 5 and the outer at least 21.

See the floor/ceil mixup.

 

[fresh_42]

$$x \cdot \lceil x \cdot \lceil x \rceil \rceil = 82$$

See the floor/ceil mixup.

Which mix?
(https://en.wikipedia.org/wiki/Floor_and_ceiling_functions#Definition_and_properties)

 

[Gp7417]

With $\lceil x \rceil$ I want to denote the floor function of $x$.

 

[SammyS]

With $\lceil x \rceil$ I want to denote the floor function of $x$.

That's how to denote the ceiling function.

Use lfloor and rfloor for the floor function. $\ \lfloor x \rfloor$

 

[Ray Vickson]

The problem says: 'Find all possibile solutions'. I find this solution numerically.

The expression must be the same.

No. The function $f(x) = x \lfloor x \lfloor x \rfloor \rfloor$ is strictly increasing in $x$, sometimes linearly and sometimes through a jump discontinuity.
To see this, let $x = n + r$ where $n \geq 0$ is an integer and $0 \leq r < 1$. Then the inner floor function is $\lfloor n + r \rfloor = n$, so
$$f(n+r) = (n+r) \lfloor (n+r)n \rfloor$$
The function $g(n,r) = \lfloor (n+r)n \rfloor$ is positve and nondecreasing in $n$ and $r$, so $f(n+r) = (n+r) g(n,r)$ is strictly increasing.

That means that $f(x) = 82$ has at most one root.