Does the floor function satisfy floor(x) = x + O(x^(1/2))?

[zetafunction]

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}

 

[Tinyboss]

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

 

[CRGreathouse]

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.

 

[zetafunction]

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}).

 

[CRGreathouse]

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).

 

[zetafunction]

<br /> \lfloor x\rfloor-x<br /> can not be bigger than one by the definition of floor function and fractional part so

perhaps <br /> \lfloor x\rfloor=x+O(x^{e})<br /> fore any e=0 or bigger than 0 is this what you meant ??

 

[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).

Petek

 

[CRGreathouse]

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).

Indeed.