# Big O notation (for calculus, not computer science)

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

Code:
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:
n[SUP]-1/4[/SUP]
?

## Answers and Replies

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.

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.

arildno
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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.
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.

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".

arildno
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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".
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

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?

arildno
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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?
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
$$f(x)=Ax^{-\frac{1}{4}}\cos(x^{2})$$

This f is definitely O(x^(-1/4)), but its rate of growth will, be:
$$\frac{df}{dx}\to{-2A}x^{\frac{3}{4}}\sin(x^{2}), x\to\infty$$

Thanks

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