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.
$$\left\{(-1)^{\lfloor n/3\rfloor}(2n+1)\mid n\in\mathbb N\right\}$$

Or, if you are using a finicky notion of set-builder notation:

$$\left\{x\in\mathbb Z\mid \exists n\in\mathbb N\left[x=(-1)^{\lfloor n/3\rfloor}(2n+1)\right]\right\}$$