Can this be solved to remove the floor function and simplify the answer?

Question

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

$$E = 1 - \frac{x}y \left( \bigg\lfloor \dfrac{2x+yx-2}{2y} \bigg\rfloor \right)^{-1}$$

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

I would like to know whether this can be reduced further, by removing the floor function and reduce the number of terms to simplify expression?

Kind regards!

Answer 1

An observation rather than a general answer

Because you haven't said what you mean by "simplify" (fewer operations? Fewer characters? Fewer divisions? ...) this isn't really answerable in its current form. But it's still possible to make at least one observation:

For a moment, fix $y = 2$. Then your formula becomes $$ E(x) = 1 - \frac{x}{2\lfloor x-\frac{1}{2} \rfloor} $$ At every point $x$ of the form $0.5, 1.5, 2.5, \ldots$, $E$ has a discontinuity.

If you're hoping for a simpler expression, it needs to be one that can be discontinuous at a countable number of locations. You could probably generate this using something like the "signum" function and a "mod", but that's really no less complex than "floor". But if you're hoping to generate it with polynomials, trig functions, etc., I don't see an obvious way to make it happen.

Answer 2

In general, a floor function cannot be simplified away: it creates an often-infinite series of discontinuities that make it impossible to characterize the function without a way to get all the discontinuities in. For this, the floor function is ideal, and so we use it.

In this particular case, we have a large division-by-zero area when $2y>2x+yx-2$, equal to when $y > 2\frac{x-1}{2-x}, x < 2$, and a series of curves of the form $y=2\frac{x-1}{2k-x}, k\in\mathbb{Z}$ where the function is discontinuous. Without the floor function, or something similar (and frankly, the floor function is way nicer than the alternatives), we cannot describe these discontinuities and thus the behavior of the function in a finite fashion.

In this image, I have plotted the area where the function is undefined, and many of the curves where the function is discontinuous. As $k$ increases, the discontinuity curves get closer and closer to the $x$ axis; I've only plotted to $k=100$.