# Floor function

1. Mar 10, 2010

### 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 $$x^{1/2}$$

2. Mar 10, 2010

### Tinyboss

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

Last edited: Mar 11, 2010
3. Mar 10, 2010

### CRGreathouse

Yes. It also satisfies

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

But both are needlessly weak.

4. Mar 11, 2010

### zetafunction

what do you mean by 'weak' , is there a proof for $$\lfloor x\rfloor=x+O(x^{1/2})$$.

5. Mar 11, 2010

### CRGreathouse

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

6. Mar 11, 2010

### zetafunction

$$\lfloor x\rfloor-x$$ 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 ??

7. Mar 11, 2010

### 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

8. Mar 11, 2010

Indeed.