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Floor function

  1. Mar 10, 2010 #1
    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]
     
  2. jcsd
  3. Mar 10, 2010 #2
    Why would the order of [tex]x-\lfloor x\rfloor[/tex] depend on the order of x?
     
    Last edited: Mar 11, 2010
  4. Mar 10, 2010 #3

    CRGreathouse

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    Yes. It also satisfies

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

    But both are needlessly weak.
     
  5. Mar 11, 2010 #4
    what do you mean by 'weak' , is there a proof for [tex] \lfloor x\rfloor=x+O(x^{1/2}) [/tex].
     
  6. Mar 11, 2010 #5

    CRGreathouse

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    [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].
     
  7. Mar 11, 2010 #6
    [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 ??
     
  8. Mar 11, 2010 #7
    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
     
  9. Mar 11, 2010 #8

    CRGreathouse

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    Indeed.
     
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