Floor function

  • #1
391
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does the floor function satisfy

[tex] floor(x)= x + O(x^{1/2}) [/tex]

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

Answers and Replies

  • #2
236
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Why would the order of [tex]x-\lfloor x\rfloor[/tex] depend on the order of x?
 
Last edited:
  • #3
CRGreathouse
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does the floor function satisfy

[tex] floor(x)= x + O(x^{1/2}) [/tex]

Yes. It also satisfies

[tex]\lfloor x\rfloor=x+O(2^{2^x})[/tex].

But both are needlessly weak.
 
  • #4
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Yes. It also satisfies

[tex]\lfloor x\rfloor=x+O(2^{2^x})[/tex].

But both are needlessly weak.

what do you mean by 'weak' , is there a proof for [tex] \lfloor x\rfloor=x+O(x^{1/2}) [/tex].
 
  • #5
CRGreathouse
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what do you mean by 'weak' , is there a proof for [tex] \lfloor x\rfloor=x+O(x^{1/2}) [/tex].

[tex]O(2^{2^x})[/tex] is weaker than [tex]O(\sqrt x)[/tex] 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 [tex]\lfloor x\rfloor=x+O(\sqrt x)[/tex].
 
  • #6
391
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[tex]
\lfloor x\rfloor-x
[/tex] can not be bigger than one by the definition of floor function and fractional part so

perhaps [tex]
\lfloor x\rfloor=x+O(x^{e})
[/tex] fore any e=0 or bigger than 0 is this what you meant ??
 
  • #7
Petek
Gold Member
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I think what the others are trying to say is that, since [itex]\lfloor x\rfloor-x[/itex] is bounded, then [itex]\lfloor x\rfloor=x+O(1)[/itex].

Petek
 
  • #8
CRGreathouse
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I think what the others are trying to say is that, since [itex]\lfloor x\rfloor-x[/itex] is bounded, then [itex]\lfloor x\rfloor=x+O(1)[/itex].

Indeed.
 

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