- #1

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Code:

`f(x) = O(x[SUP]-1/4[/SUP])`

Code:

`n[SUP]-1/4[/SUP]`

Thanks in advance

- Thread starter kungal
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- #1

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Code:

`f(x) = O(x[SUP]-1/4[/SUP])`

Code:

`n[SUP]-1/4[/SUP]`

Thanks in advance

- #2

arildno

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An O-notation requires a limit to which the independent variable is supposed to go.

- #3

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Note that the result f(x) = O

- #4

arildno

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Yes it is._{p}'s (my mistake) and x is a set of random variables. I don't think this changes the nature of the question drastically.

Note that the result f(x) = O_{p}(x^{-1/4}) is quite common so it can't be a meaningless statement.

Is it meant that f(x) is O(x^(-1/4) ) as x goes to zero, or as x goes to, say, infinity?

That is two entirely different situations, and needs, therefore, to be specified.

Hence, the meaninglessness of your first line.

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HallsofIvy

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- #6

arildno

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Well, if that is the general default notation, I'll make a note of that.

In my own applied maths books, they dutifully make explicit what limiting operation we are speaking about

- #7

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if f = O

- #8

arildno

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Not at all.

if f = O_{p}(n^{-1/4}) then for large n f grows at the same rate as n^{-1/4}?

f might, for example, become more and more strongly oscillatory as x goes to infinty, even though f's

For example, let

[tex]f(x)=Ax^{-\frac{1}{4}}\cos(x^{2})[/tex]

This f is definitely O(x^(-1/4)), but its rate of growth will, be:

[tex]\frac{df}{dx}\to{-2A}x^{\frac{3}{4}}\sin(x^{2}), x\to\infty[/tex]

- #9

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Thanks

- #10

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Well, O's should actually be Op's (my mistake) and x is a set of random variables.

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