Polynomial equal to the ceiling of x For a few days, I've been looking for a polynomial who's value is equal to the ceiling function of the only variable it contains. I thought about it for while and I haven't got a clue.

You have the ceiling function
$$c:\mathbb{R} \rightarrow \mathbb{R}$$
$$c(x)=\lceil x \rceil$$
and you're looking for a polynomial
$$c:\mathbb{R} \rightarrow \mathbb{R}$$
$$p(x)= \sum_{i=0}^n a_i x^i \text{ with } n\in \mathbb{N} \text{ and } a_i \in \mathbb{R} \text{ for all } i \in 0,...,n$$ 
One thing you should know about polynomials is that they are infinitely differentiable:
$$(a_i \cdot x^i)' = i \cdot a_i \cdot x^{i-1}$$
But $c$ isn't differentiable at $0, 1, 2, ...$ at all:
$$\lim_{\substack{x \rightarrow 0\\x > 0}} \frac{c(x)-c(0)}{x} = \lim_{\substack{x \rightarrow 0\\x > 0}} \frac{1}{x} = \infty \neq 0 = \lim_{h \rightarrow 0} \frac{0}{h} = \lim_{h \rightarrow 0} \frac{f(x_0+h)-f(x_0)}{h}$$
So $c$ can't be a polynomial.
Approximation
Taylor polynomials
However, you can approximate $c$ with a polynomial.
The following series will approximate $c$:
$$s_n(x)\frac{1}{2}+x+ \frac{1}{\pi} \cdot \sum_{k=1}^{n} \frac{\sin(2 k \pi x)}{k}$$
So:
$$\lim_{n \rightarrow \infty} s_n = c$$
As you can see, it is not a polynomial because the series is infinite (and includes sin).
When you take a finite part of it, e.g. the first 10 terms, you get $s_{10}$:

Source: Wolfram|Alpha
