Order notation: f(x)=O(x^n) as x-> 0

[coverband]

Hello,

I am having difficulty getting to grips with “the order notation”. I have the following definition…

“A function f(x) is said to be of order x^n as x -> 0 if there is a non-zero constant C such that |f(x)|<C|x^n| for all x in an interval around x=0. This is written as
f(x) = O(x^n) as x->0”

some examples:

x(1+x^2)^1/2 = x + x^3/2 +… = O(x)
x/(1+x) = x(1 – x + x^2 +…) = O(x)
(x+b)^a – x^a = x^a(1 + ab/x +…) –x^a = O[x^(a-1)]

 

[CRGreathouse]

As an example, sin(x) = x + O(x^3) as x -> 0. sin(0.01) is about 0.01 - 0.01^3 * 1/6, sin(0.001) is about 0.001 - 0.001^3/6, and so forth. In this case C is something like 1/6 for an appropriate interval.

I'm not sure exactly what you want. The intuition is that the function near 0 is closely approximated by the given polynomial.

 

[DrGreg]

The idea behind the "big O" notation is to make rigorous the idea that, say

$$e^x = 1 + x + \mbox{ terms involving } x^2 \mbox{ or smaller}$$​

which you would write as

$$e^x = 1 + x + O(x^2) \mbox{ as } x \rightarrow 0$$​

and which means that

$$|e^x - 1 - x| < C|x^2|$$​

for some C, for all small x.

You are, in effect, treating x² as negligible for the purpose of your calculation, but instead of just ignoring it, you keep track of what you have ignored. By the end of your calculation, you should find the ignored term goes to zero as $x \rightarrow 0$. If it doesn't, you've neglected too much, and you'll have to go back and add another term.