ceiling-and-floor-functions
I need some help to solve the next equation:
$$ \left \lfloor x^2 – x – 2 \right \rfloor = \left \lfloor x \right \rfloor $$
Where $ \left \lfloor \cdot \right \rfloor $ is the floor function.
What I've tried:
$$ x^2 – x – 2 – x < 1 $$
$$ x^2 – x – 2 \leq x < x^2 – x – 1 $$
When I try to solve this system, I don't get the right solution. Is it right what I'm doing? How can I do to solve this?
The equation is equivalent to
$$0\le\{\pi\}-\frac{\{a\pi\}}a<1.$$
The right inequality is always verified. The left one is certainly verified when
$$0\le\{\pi\}-\frac1a,$$ or $$a\ge\frac1{\{\pi\}},$$ which is a little more than $7$.
Assuming that $a$ is restricted to be a natural, it remains to try $a=1,2,\cdots7$. And as $\{\pi\}<\dfrac17$, all these values will work.
Your original statement is not true (please read the whole answer). But the following statement is correct:
$$\left \lfloor {\log_2 n} \right \rfloor = \left \lfloor {\log_2 \left \lceil \frac {n-1}{2}\right \rceil} \right \rfloor + 1\tag{1}$$
Proof:
Every number $n$ can be placed between two cosecutive powers of 2. In other words, there exists $k$ such that:
$$2^k\le n\le2^{k+1}-1\tag{1}$$
Obviously:
$$k\le\log_2n<k+1$$
$$k=\lfloor\log_2n\rfloor\tag{2}$$
On the other side from (1):
$$\frac{2^k-1}2 \le \frac{n-1}{2} \le \frac{2^{k+1}-2}2$$
$$2^{k-1}-\frac12 \le \frac{n-1}{2} \le 2^k-1$$
$$\lceil 2^{k-1}-\frac12\rceil \le \lceil\frac{n-1}{2}\rceil \le 2^k-1$$
$$2^{k-1} \le \lceil\frac{n-1}{2}\rceil \le 2^k-1$$
$${k-1} \le \log_2\lceil\frac{n-1}{2}\rceil \le \log_2 (2^k-1)<k$$
$${k-1} =\lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor$$
$$k=\lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor+1\tag{3}$$
By comparing (2) and (3) you get:
$$\lfloor\log_2n\rfloor = \lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor+1$$
...which completes the proof.
You can easily prove that the original statement is not true. You are basically saying that the function:
$$f(n)=\left \lfloor {\log n} \right \rfloor - \left \lfloor {\log \left \lceil \frac {n-1}{2}\right \rceil} \right \rfloor - 1$$
...is equal to zero for all values of $n$.
This is not true if "$\log$" stands for logartihm with base 10:
This is also not true if "$\log$" stans for natural logarithm "$\ln$":
If you don't trust these plots, calculate the value for $n=45$ and in both cases the result is -1, not 0.
Best Answer
Your system of equations isn't quite right (for example, $x$ may be the smaller of the two).
Here's one algebraic way to go about this:
To solve $\lfloor f(x) \rfloor = \lfloor x \rfloor$ write $x = n + \epsilon$ where $0\leq \epsilon <1$ and $n$ is an integer so that $\lfloor x\rfloor = n$.
Then the condition $\lfloor f(x)\rfloor = n$ is $n \leq f(n+\epsilon) < n+1$ (and the above conditions on $n$ and $\epsilon$).