Big O notation (for calculus, not computer science)

[kungal]

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

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

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

?

Thanks in advance

 

[arildno]

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

 

[kungal]

The O's should actually be O_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.

 

[arildno]

The O's should actually be O_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.

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]

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]

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

 

[kungal]

I'm glad we've cleared that up but is it reasonable to say that
if f = O_p(n^-1/4) then for large n f grows at the same rate as n^-1/4?

 

[arildno]

I'm glad we've cleared that up but is it reasonable to say that
if f = O_p(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

 

[kungal]

Thanks

 

[aprillove20]

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