I have tried to define the ceiling function of $x$ in terms of its floor function; I thought this might be easy, but it isn't. I can easily do this with a piecewise equation, but I need to do without it.

Attempts

I have tried three methods, one of which ($\mathbf{(3)}$) almost always works:

$$\left\lfloor x\right\rfloor+1\not= \left\lceil x\right\rceil,\tag 1$$ when $x=3$, because: $$\left\lfloor 3\right\rfloor+1=4,\text{and }4\not=\lceil3\rceil.$$ $$\left\lfloor x+1\right\rfloor\not= \left\lceil x\right\rceil,\tag2$$ when $x=3$. $$\left\lfloor x+\frac{x}{x+1}\right\rfloor\not= \left\lceil x\right\rceil,\tag3$$ when $\varepsilon$ is a number extremely close to $0$ and $x=(8+\varepsilon)$, i.e $x=8.099999999999987$. In that event: $$\left\lfloor 8.099999999999987+\frac{8.099999999999987}{9.099999999999987}\right\rfloor=\left\lfloor8.09999999999998+0.89010989010989\right\rfloor=\lfloor 8.99010989010987\rfloor=8,\text{and }8\not=\left\lceil8.099999999999987\right\rceil$$

Question

Is it possible to define $\left\lceil\ldots\right\rceil$ in terms of $\left\lfloor \ldots\right\rfloor$?

You can calculate ceiling as:

$$\lceil{x}\rceil=-\lfloor{-x}\rfloor$$

You can calculate floor itself as:

$$\lfloor{x}\rfloor=x-\frac12\left(\frac{\ln\left(e^{2\pi i\left(x-\frac12\right)}\right)}{\pi i}+1\right)$$

This is because $e^{2\pi i\left(x-\frac12\right)}$ has a period of $1$.

Well-known lemma: $$ \left\lceil x \right\rceil = - \left\lfloor -x \right\rfloor $$*Proof.* Let $ x = a + k $, where $ a = \left\lfloor x \right\rfloor $. If $k=0$ the proof is trivial ,so let $ 0 < k < 1 $. Then, the LHS is $a+1$ and $$ \begin {align*} \text {RHS} &= - \left\lfloor - \left( a + k \right) \right\rfloor \\&= - \left\lfloor -a - k \right\rfloor \\&= - \left( -a - 1\right) \\&= a + 1. \end {align*} $$So we are done. $\Box$

$$ \lceil x \rceil = -\lfloor-x\rfloor $$

If $a$ is infinitesimal positive number, and you assume that $\lceil{a}\rceil=1$, then...

$$\lfloor{x+1-a}\rfloor=\lceil{x}\rceil$$