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According to the Wikipedia entry on elementary functions, the trigonometric functions and their inverses are elementary functions.

It doesn't seem to me that the floor and ceiling functions should be elementary, since they don't seem too natural. However, here is an expression of the floor function using only elementary functions

$$ \text{floor}(x) = (x - 0.5)-\frac{\arctan(\tan(\pi(x-0.5)))}{\pi} $$

As pointed out below, this function is undefined for the integers.

Here is a plot of this function using Desmos. I am aware that this is not proof of this identity, and that this identity depends on the choice of range for $\arctan$. However, given that $\arctan$ was listed as an elementary function on the Wikipedia entry, I assumed that this was independent of the choice of range.

Is there anything I am missing here? Or is floor, and therefore ceiling and modulo all elementary functions?

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No. All the "basic" elementary functions (trig functions, exponentials, etc.) are continuous at every point in their domain, and the operations of addition, multiplication and composition all preserve continuity; it follows that all elementary functions are continuous everywhere on their domain. The floor function is defined for all real numbers but is discontinuous on the integers, so cannot be elementary. Note that your function is elementary, and so it is continuous on its domain - but is undefined on the integers, precisely where the floor function is discontinuous.

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  • $\begingroup$ Ah that's rather inconvenient... but it does seem that you are correct in the statement that all of the elementary functions are continuous and I want to create a non-continuous one. Thanks. $\endgroup$ – Enrico Borba Aug 1 '18 at 5:21

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