- #1

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[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]

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- Thread starter zetafunction
- Start date

- #1

- 391

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[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

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Why would the order of [tex]x-\lfloor x\rfloor[/tex] depend on the order of x?

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- #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

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\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

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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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