By analyzing the first equation, I was able to find all the solutions.
Firstly, we know $3$ things about the floor functions :
- $\forall x \in \mathbb R$, $x-1\le\lfloor x\rfloor\le x$
- $\forall x \in \mathbb R_+$, $x^2-x\le x\lfloor x\rfloor\le x^2$
- $\forall x \in \mathbb R$, $x^2-1\le\lfloor x^2\rfloor\le x^2$
Ok now, let's use those inequalities to solve the first equation !
Case $1$ : $x\ge 0$
$$x^2-x\le x\lfloor x\rfloor\le x^2$$
$$-2x^2\le -2x\lfloor x\rfloor\le -2x^2+2x$$
$$3x^2-2x^2\le 3x^2-2x\lfloor x\rfloor\le 3x^2-2x^2+2x$$
$$x^2+4x^2-4\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor\le x^2+2x+4x^2$$
$$5x^2-4+x\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x\le 5x^2+2x+x$$
$$5x^2-4+x-4x\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor\le 5x^2+2x+x-4x+4$$
$$5x^2-3x-4-\frac{7}{2}\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}\le 5x^2-x+4-\frac{7}{2}$$
$$5x^2-3x-\frac{15}{2}\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}\le 5x^2-x+\frac{1}{2}$$
Ok now, solve for $5x^2-3x-\frac{15}{2}=0$ and for $5x^2-x+\frac{1}{2}=0$
We get $0$ solution for the second one and $2$ for the first one. Caution, the solution must be greater or equal than $0$. And the solution which satisfies this rule is $\frac{6+2\sqrt{159}}{20}$.
So, if there is a solution greater or equal to $0$, $x$ needs to be between $0$ and $\frac{6+2\sqrt{159}}{20}$.
Now we have $2$ possibilities : $\lfloor x \rfloor = 0$ or $\lfloor x \rfloor = 1$
Case $1.1$ : $\lfloor x \rfloor = 0$
So the function $3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}$ is simplified into $3x^2+x-\frac{7}{2}=0$
And we get the solution (greater than $0$) : $$\frac{-1+\sqrt{43}}{6}$$
which is equal to $0$ when we use the floor function.
Case $1.2$ : $\lfloor x \rfloor = 1$
After some simplification, we get :
$3x^2-x+4\lfloor x^2 \rfloor-\frac{15}{2}$
$\lfloor x^2 \rfloor$ could be equal to $1$ or $2$ because here $1\le x\le\frac{6+2\sqrt{159}}{20}$ so $1\le x^2 \le\frac{6+2\sqrt{159}}{20}^2$
Case $1.2.1$ : $\lfloor x^2 \rfloor=1$
We get : $3x^2-x+4-\frac{15}{2}=3x^2-x-\frac{7}{2}$
There is one solution greater or equal than $0$, where the floor is equal to $1$ and where the floor of the square is equal to $1$. It is : $$\frac{1+\sqrt{43}}{6}$$
Case $1.2.2$ : $\lfloor x^2 \rfloor=2$
There isn't a solution which satifies every parameters (greater or equal than $0$, the floor of the number is equal to $1$ and the floor of the square is equal to $2$)
Case $2$ : $x\lt 0$
We get : $$5x^2-x-\frac{15}{2}\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}\le 5x^2-3x+\frac{1}{2}$$
Ok now, solve for $5x^2-x-\frac{15}{2}=0$ and for $5x^2-3x+\frac{1}{2}=0$
We get $0$ solution for the second one and $2$ for the first one. Caution, the solution must be less than $0$. And the solution which satisfies this rule is $\frac{2-2\sqrt{151}}{20}$.
So, if there is a solution less than $0$, $x$ needs to be between $\frac{2-2\sqrt{151}}{20}$ and $0$.
Now we have $2$ possibilities : $\lfloor x \rfloor = -1$ or $\lfloor x \rfloor = -2$
Case $2.1$ : $\lfloor x \rfloor = -1$
We get : $3x^2+3x+4\lfloor x^2\rfloor + \frac{1}{2}=0$
Case $2.1.1$ : $\lfloor x^2\rfloor=1$
If that's the case, then $x=-1$. However, when $x =-1$, it's not equal to $0$. So it's wrong.
Case $2.1.2$ : $\lfloor x^2\rfloor=0$
Then we have : $3x^2+3x + \frac{1}{2}=0$. And here there is 2 solutions which satisfies everything (less than $0$, floor equal to $-1$ and floor of the square to $0$.
And it's : $$\frac{-3-\sqrt{3}}{6}, \frac{-3+\sqrt{3}}{6}$$
Case $2.2$ : $\lfloor x \rfloor = -2$
We have : $3x^2+5x+4\lfloor x^2\rfloor + \frac{9}{2}=0$
$\lfloor x^2 \rfloor$ is equal to $1$ because here $\frac{2-2\sqrt{151}}{20}\le x\le -1$ so $1\le x^2 \le\frac{2-2\sqrt{151}}{20}^2$
We get now : $3x^2+5x + \frac{17}{2}=0$. However, there is no solutions.
So finally, there is $4$ solutions : $$x=\left\{\frac{-3\pm\sqrt{3}}{6},\frac{\pm 1+\sqrt{43}}{6}\right\}$$
Ok so after a lot of thinking, I found a way to solve for the second equation.
First, we have inside the floor function : $\lfloor x\rfloor$ and $\lfloor x^2\rfloor$.
This allows us to deduce when we have discontinuities in the function.
For $x\ge 0$, we have at $1$,$\sqrt{2}$,$\sqrt{3}$,... discontinuities.
Now, let us recall something I said earlier.
For $x\ge 0$ :
$$5x^2-3x-\frac{15}{2}\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}$$
$$5x^2-3x-\frac{15}{2}\ge 1\text{ for }x\ge \frac{6+2\sqrt{179}}{20}$$
So we know for sure that for $x\ge \frac{6+2\sqrt{179}}{20}$, $3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}\ge 1$.
The solutions needs to be in the interval $[0,\frac{6+2\sqrt{179}}{20}[$.
- For $x\in[0,1[$, we have this : $3x^2+x-\frac{7}{2}$ (it's easy to show that it is increasing in this interval)
$$3x^2+x-\frac{7}{2}\ge 1\text{ for }x \ge \frac{-2+2\sqrt{91}}{12}$$
But $\frac{-2+2\sqrt{91}}{12}\gt 1$, plus because this function is increasing and we know it is equal to $0$ at $x=\frac{-1+\sqrt{43}}{6}$
The interval $[\frac{-1+\sqrt{43}}{6},1[$ is a solution to this equation.
- For $x\in[1,\sqrt{2}[$, we have this : $3x^2-x-\frac{7}{2}$ (it is increasing in this interval)
$$3x^2-x-\frac{7}{2}\ge 1\text{ for }x \ge \frac{1+\sqrt{55}}{6}$$
But $\frac{1+\sqrt{55}}{6}\lt \sqrt{2}$, plus because this function is increasing and we know it is equal to $0$ at $x=\frac{1+\sqrt{43}}{6}$
The interval $[\frac{1+\sqrt{43}}{6},\frac{1+\sqrt{55}}{6}[$ is a solution to this equation.
- For $x\in[\sqrt{2},\frac{6+2\sqrt{179}}{20}[$ because $\frac{6+2\sqrt{179}}{20}\lt\sqrt{3}$, we have this : $3x^2-x+\frac{1}{2}$ (it is increasing in this interval)
If $x=\sqrt{2}$ then we would have $-\sqrt{2}+\frac{13}{2}$ which is bigger than $1$. And because it's increasing, it'll always be bigger than $1$. So there is no solutions in this interval.
For $x\lt 0$, we have at $-1$,$-\sqrt{2}$,$-\sqrt{3}$,... discontinuities.
$$5x^2-x-\frac{15}{2}\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}$$
$$5x^2-x-\frac{15}{2}\ge 1\text{ for }x\le \frac{1-3\sqrt{19}}{10}$$
So we know for sure that for $x\le \frac{1-3\sqrt{19}}{10}$, $3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}\ge 1$
The solutions needs to be in the interval $]\frac{1-3\sqrt{19}}{10},0[$.
- For $x\in ]-1,0[$, we have $3x^2+3x+\frac{1}{2}$ (it is decreasing from $-1$ to $-\frac{1}{2}$ and it is increasing from $-\frac{1}{2}$ to $0$).
This equation is equal to $0$ when $x=\frac{-3\pm\sqrt{3}}{6}$.
However, we know it's decreasing then increasing. So the intervals $]-1,\frac{-3-\sqrt{3}}{6}$ and $]\frac{-3+\sqrt{3}}{6},0[$ are others solutions.
For $x=-1$, we have $\frac{9}{2}\gt 1$. It's not a solution.
For $]\frac{1-3\sqrt{19}}{10},-1[$ because $\frac{1-3\sqrt{19}}{10}\gt -\sqrt{2}$, we have $3x^2+5x+\frac{17}{2}$ (it is decreasing in this interval)
It can be shown really easy that $\forall x\in\mathbb R$, $3x^2+5x+\frac{17}{2}\gt 1$.
So finally, we get :
$$x\in\left\{\left]-1,\frac{-3-\sqrt{3}}{6}\right]\cup\left[\frac{-3+\sqrt{3}}{6},0\right[\cup\left[\frac{-1+\sqrt{43}}{6},1\right[\cup\left[\frac{1+\sqrt{43}}{6},\frac{1+\sqrt{55}}{6}\right[\right\}$$
Hope this is the end...
Best Answer
It's just decimal representation. If you subtracta a $c$ from a number do you have to carry/borrow a one or not.
to wit....
So $x = 10^na_n + 10^{n_1}a_{n-1} + ..... + 10a_1 + a_0$
$\frac x{10} = 10^{n-1}a_n + 10^{n-2}a_{n-1} + ..... + a_1 + \frac {a_0}{10}$
$f(x) = \lfloor \frac x{10} \rfloor = 10^{n-1}a_n + 10^{n-2}a_{n-1} + ..... + a_1$
And $x - c = 10^na_n + 10^{n_1}a_{n-1} + ..... + 10a_1 + (a_0-c)$
And $\frac {x-c}{10} = 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 + \frac {(a_0-c)}{10}$
To find the least integer depends upon the value of $\frac {(a_0-c)}{10}$.
$-9 \le a_0 -c \le 9$ and $-.9\le \frac {(a_0-c)}{10} \le .9$
If $\frac {(a_0-c)}{10} \ge 0$ then
$\rfloor \frac {x-c}{10}\lfloor = 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 + \rfloor\frac {(a_0-c)}{10}\lfloor= 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 = f(x)$
If $\frac {(a_0-c)}{10} \le 0$ then
$\rfloor \frac {x-c}{10}\lfloor = 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 + \rfloor\frac {(a_0-c)}{10}\lfloor= 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 - 1= f(x)-1$
So $x$~$x-c$ if $a_0 -c \ge 0$. But $x$ !~ $x-c$ if $a_0 - c < 0$. So if $a_0 - c < 0 < 5$ this is not true.
....
You can do 2) the same way.
$f(x+c) = 10^{n-1}a_n + 10^{n-2}a_{n-1} + ..... + a_1 + \lfloor \frac {a_0+c}{10}\rfloor = f(x) + \lfloor \frac {a_0+c}{10}\rfloor$.
And $\lfloor \frac {a_0+c}{10}\rfloor = 0$ if $a_0+c< 10$ but $\lfloor \frac {a_0+c}{10}\rfloor = 1$ if $a_0 + c \ge 10$.
So 2) is true.