Is it possible to represent this function as a polynomial, by removing the ceiling function?

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I've been working through a derivation and have arrived at the following expression:

$$E = 1 - \frac{x}y \left( \bigg\lceil \dfrac{x}{y} \bigg\rceil \right)^{-1}$$

where $x,y \in \mathbb{R^+}$.

I would like to know whether this can be reduced further, by converting the floor function to a polynomial to simplify expression?

furthermore, lets say there was a constant k, where $k \in \mathbb{N^+}$ hence,

$$E = 1 - \frac{x}{k.y} \left( \bigg\lceil \dfrac{x}{k.y} \bigg\rceil \right)^{-1}$$

could this be reduced to,

$$E = 1 - \frac{x}y \left( \bigg\lceil \dfrac{x}{y} \bigg\rceil \right)^{-1}$$

Kind regards!

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If I understand your question correctly then the answer is No. The function $$ f(x)=\frac{x}{\lceil x \rceil} $$ has infinitely many (jump) discontinuities.

If we can somehow convert $f(x)$ into a polynomial or a rational function then it would be continuous on the whole $\Bbb R$ except for only finitely many points. This is impossible since it'd contradict the fact that $f(x)$ has infinitely many discontinuity.