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Big O notation (for calculus, not computer science)

  1. Jul 21, 2010 #1
    I understand the formal definition for big O notation but is there an intuitive interpretation?. For example, if

    Code (Text):
    f(x) = O(x[SUP]-1/4[/SUP])
    is it reasonable to say that for large n f(x) grows at the same rate as
    Code (Text):
    n[SUP]-1/4[/SUP]
    ?

    Thanks in advance
     
  2. jcsd
  3. Jul 21, 2010 #2

    arildno

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    Your first line is meaningless.
    An O-notation requires a limit to which the independent variable is supposed to go.
     
  4. Jul 21, 2010 #3
    The O's should actually be Op'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) = Op(x-1/4) is quite common so it can't be a meaningless statement.
     
  5. Jul 21, 2010 #4

    arildno

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    Yes it is.

    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.
     
  6. Jul 21, 2010 #5

    HallsofIvy

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    I hate to disagree with Arildno but it is common practice (though perhaps "abuse of notation") to use just O(f(x)) to mean "as x goes to infinity".
     
  7. Jul 21, 2010 #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
     
  8. Jul 21, 2010 #7
    I'm glad we've cleared that up but is it reasonable to say that
    if f = Op(n-1/4) then for large n f grows at the same rate as n-1/4?
     
  9. Jul 21, 2010 #8

    arildno

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    Not at all.

    f might, for example, become more and more strongly oscillatory as x goes to infinty, even though f's magnitude will be bounded by some constant multiplied by x^(-1/4).

    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]
     
  10. Jul 21, 2010 #9
    Thanks
     
  11. Jul 22, 2010 #10
    Well, O's should actually be Op's (my mistake) and x is a set of random variables.
     
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