Floor function

does the floor function satisfy

$$floor(x)= x + O(x^{1/2})$$

the idea is the floor function would have an 'smooth' part given by x and a oscillating contribution with amplitude proportional to $$x^{1/2}$$

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 Why would the order of $$x-\lfloor x\rfloor$$ depend on the order of x?

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 Quote by zetafunction does the floor function satisfy $$floor(x)= x + O(x^{1/2})$$
Yes. It also satisfies

$$\lfloor x\rfloor=x+O(2^{2^x})$$.

But both are needlessly weak.

Floor function

 Quote by CRGreathouse Yes. It also satisfies $$\lfloor x\rfloor=x+O(2^{2^x})$$. But both are needlessly weak.
what do you mean by 'weak' , is there a proof for $$\lfloor x\rfloor=x+O(x^{1/2})$$.

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 Quote by zetafunction what do you mean by 'weak' , is there a proof for $$\lfloor x\rfloor=x+O(x^{1/2})$$.
$$O(2^{2^x})$$ is weaker than $$O(\sqrt x)$$ in the sense that there are functions which are in the former but not the latter, but none in the latter but in the former.

You should be able to give a one-line proof of a statement stronger than $$\lfloor x\rfloor=x+O(\sqrt x)$$.

 $$\lfloor x\rfloor-x$$ can not be bigger than one by the definition of floor function and fractional part so perhaps $$\lfloor x\rfloor=x+O(x^{e})$$ fore any e=0 or bigger than 0 is this what you meant ??
 Recognitions: Gold Member I think what the others are trying to say is that, since $\lfloor x\rfloor-x$ is bounded, then $\lfloor x\rfloor=x+O(1)$. Petek

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 Quote by Petek I think what the others are trying to say is that, since $\lfloor x\rfloor-x$ is bounded, then $\lfloor x\rfloor=x+O(1)$.