Question: I need a set-builder notation for $\{1 , 3 , 5 , -7 , -9 , -11 , 13 , 15 , 17 , \ldots\}$. It is an infinite set of odd positive integers, which are multiplied by $-1$ three by three. One of the terms of this question is not to use floor function.

Answer: 
If you are allowed to use square roots, then something like: $$\left\{(2n+1) \cdot(-1)^{\tfrac{1}{\sqrt{5}} \left(\left(\tfrac{1+\sqrt{5}}{2}\right)^{n-1} - \left(\tfrac{1-\sqrt{5}}{2}\right)^{n-1}\right) - (n-1)} \big| n \in \mathbb{N}\right\}$$
will work. (Here, we use the convention that $0 \in \mathbb{N}$.)

This answer has the advantage that it doesn't use the floor/ceiling functions, modular arithmetic, or trigonometric functions (as OP requested), but the downside is that it's complicated.
Define $f:\mathbb{N} \to \mathbb{Z}$ as follows. $f(1) = 1$, $f(2) = 3$ $f(3) = 5$ and $$f(n) = -\text{sgn}(f(n-3))\cdot(2n-1)$$

for $n \geq 4$

Then this set is $\{c \in \mathbb{Z}: \exists n \in \mathbb{N} \ \text{such that} \ f(n) = c\}$. That is, the set is the image of the function.