Big O notation and polynomials

Question

I often see that the following polynomial can be written as such:

$f(x) = 6x^4+3x^3+O(x^2)$

where the big O collects all the lower order terms.

Yet, I also see this sometimes:

$f(x) = 6x^4+7x^9+O(x^{11})$

where the big O collects higher order terms.

How does this even make sense? For $f(x)$ to be in $O(g(x))$ doesn't it need to grow at least as fast as $g(x)$ at super large values of $x$?

Answer 1

You should write something like:

$f(x) \in 6x^4+3x^3+O(x^2)$ as $x \to \infty$

$f(x) \in 6x^4+3x^3+O(x^{11})$ as $x \to 0$

There are two points here. Firstly the Big-O notations denote classes of functions, not a single function, and it is a misleading abuse of notation to use "$=$" between a single function and a class. Don't forget that $O(x) - O(x) = O(x)$, not $O(x) - O(x) = 0$. Secondly, we cannot talk about limiting behaviour without specifying the region we are concerned with, just like it is meaningless to write "$\frac{2^n}{n^2} \to \infty$" just like that without specifying anything about $n$.

Answer 2

The Big-O terms in the function definitions are place-holders. They are just saying that there is a function that is asymptotically bounded above by $x^{11}$. The $O(x^{11})$ isn't collecting higher order terms. A function $g(x)$ containing $x^{12}$ for example, wouldn't be a part of $f(x)$.

The Big-O notation is commonly used as short-hand for this purpose.

Answer 3

To expand on what Bernard said:

In general, big-Oh and little-Oh terms are smaller than the explicit terms that they are together with.

When you have $f(x) = 6x^4+3x^3+O(x^2)$, this intends $x^2$ to be smaller than $x^3$ and $x^4$. This holds when $x$ is large, so when $x \to \infty$.

When you have $f(x) = 6x^4+7x^9+O(x^{11}) $, this intends $x^{11}$ to be smaller than $x^9$ and $x^4$. This holds when $x$ is small, so when $x \to 0$.