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I have a very basic question: What is the difference between $\lfloor f \rfloor (x)$ and $\lfloor{f(x)}\rfloor$? Are $\lfloor f \rfloor (x)$ and $f(\lfloor x \rfloor)$ equivalent?

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What is the difference between $f\circ g(x)$ and $f(g(x))$? – leo Dec 22 '12 at 18:04

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You seem to be asking two different questions. $\lfloor f \rfloor (x)$ is defined as $\lfloor f(x) \rfloor$. Hence, it is indeed true that $\lfloor f \rfloor (x) = \lfloor f(x) \rfloor$.

However, $\lfloor f(x) \rfloor$ is different from $f(\lfloor x \rfloor)$. For instance, consider $f(x) = x^2$. Then $\lfloor f(x) \rfloor = \lfloor x^2 \rfloor$, while $f(\lfloor x \rfloor) = \lfloor x \rfloor ^2$. These two are not equal. For instance, choosing $x = 2.5$, we get that $$\lfloor f(2.5) \rfloor = \lfloor 2.5^2 \rfloor= \lfloor 6.25 \rfloor = 6$$ whereas $$f(\lfloor 2.5 \rfloor) = f(2) = 2^2 = 4$$ Hence, in general $$\lfloor f(x) \rfloor \neq f(\lfloor x \rfloor)$$

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Hmm yeah if ⌊f⌋(x) is defined as ⌊f(x)⌋, it is obvious that the second claim fails. The first assumption is also true for the ceiling function as well right?. – mathamania Dec 22 '12 at 17:53
May I ask you why some old authors like Richard A. Silverman who is a famous in analytic modern calculus used $[x]$ instead? Thanks Marvis. – Babak S. Dec 22 '12 at 17:55
@mathamania Yes. Typically any function on $f$ i.e. $H(f)(x)$ denotes $H(f(x))$ i.e. what I mean is, for instance, $(f^2 + 3f)(x) = f(x)^2 + 3f(x)$. In your case, $H$ is the floor function. It is true for the ceiling function as well. – user17762 Dec 22 '12 at 17:55
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@BabakSorouh I guess it is just a question of notation. I would guess that initially people used only the floor function and hence they had the notation $[x]$. But once the ceil function also began to be used, people wanted to somehow adapt the earlier notation $[x]$ to indicate floor or ceil. This is just my interpretation. – user17762 Dec 22 '12 at 17:59
Notation $[x]$ (the greatest integer in $x$) was used until quite recently. Certainly Silverman was not the first or the only one. It is what I learned in school back in my younger days. – GEdgar Dec 23 '12 at 13:31
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I know an example showing that the last claim fails. $f(x)=|x|$

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helpful counterexample! + – amWhy Mar 4 at 0:44

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