ceiling-and-floor-functionstrigonometry
I've wanted to know the solution to this for years since I couldn't figure out how to get the answer, not even from wolfram alpha and this has also made me question whether if it was even possible to make t the subject of this equation. In other words, I've been trying to change it from:
$$p = s(\left \lfloor t\right \rfloor+\frac{\sin(\frac{\pi\left \lfloor t\right \rfloor}{t})}{\sin(\frac{\pi}{t})})$$
To:
$t =\text{equation in terms of }p\text{ and }s$
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 $\epsilon$).
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
As shown by this graph:
$$p = s\left(\left \lfloor t\right \rfloor+\frac{\sin(\frac{\pi\left \lfloor t\right \rfloor}{t})}{\sin(\frac{\pi}{t})}\right)\iff\frac ps-k=\frac{\sin\left(\frac{\pi k}{x}\right)}{\sin\left(\frac\pi x\right)},0<x-k<1$$
Rewriting with a Chebyshev U series:
$$\frac ps-k=U_{k-1}\left(\cos\left(\frac\pi x\right)\right)=\sum_{n=0}^{\left\lfloor\frac{k-1}2\right\rfloor}\frac{(-1)^n \Gamma(k-n)}{n!\Gamma(k-2n)} \left(2\cos\left(\frac\pi x\right)\right)^{k-2n-1} $$
Now it is just about inverting a $\cos\left(\frac\pi x\right)$ polynomial of degree $\left\lfloor\frac{k-1}2\right\rfloor$. There are elementary inverses for the $k=1,\dots 5,7,9$ cases. However, cases, like when $k=7$ and $k=9$ do not show major simplifications in radical form.