Why does the floor function $x \mapsto \lfloor x \rfloor$ have expansion $x + O(1)$?

Question

Shouldn't it just be the largest previous integer?

Why is there a remainder term $O(1)$?

Thanks,

Edit: I am working on a problem that uses the Abel summation formula, and the integration part of the formula integrates a discrete object (against a $C^1$ function), which in examples I have seen get converted into a floor function.

Answer 1

A function of type $O(1)$ is by definition simply a function $f$ such that there is some $M > 0$ such that $|f(x)| \leq M$ for all suitable $x$. If $x \in \Bbb{R}$, and if $d_{x} := x - \lfloor x \rfloor$, then $0 \leq d_{x} < 1$; hence the map $x \mapsto d_{x}$ is $O(1)$, and we can formally write $x = \lfloor x \rfloor + O(1)$ (and the equation $\lfloor x \rfloor = x + O(1)$ is, of course, also valid).