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I have a function $f(x) = \cos(x)$.

I am given $\lim_{x \to 0} \cos(x) = 1$.

Is it correct to say $f(x)$ is of order $n^2$? What about $f(x)$ is $O(n^2)$

Or should I say $f(x)$ is of order $x^2$ or $f(x)$ is $O(x^2)$


I think $O(x^2)$ is correct because it depends on $x$, but this sounds awkward. I have not seen $O(x^2)$ elsewhere.

I think $O(n^2)$ is incorrect because $n$ is not defined anywhere (maybe implicitly?).


$f(x) = \cos(x)$ was just an example. If we let $g(x) = x^2$, would it be $O(x^2)$ or $O(n^2)$?

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None of this is correct! For a function, Big-O describes how a function grows as it approaches infinity or some value. In your case, as x goes to $0$ so:

$$ f(x)= O(1) \text{ as } x \rightarrow 0$$

Interestingly, $ \cos(x)= O(1) \text{ as } x \rightarrow \infty$ because of the definition of Big-O. Although the function continues to oscillate between -1 and 1, not approaching anything, the definition of Big-O is that $f(x)$ is Big-O of $g(x)$ if there is some constant, $C$ such that $\lvert f(x) \rvert < C \cdot g(x)$. Now we can make $C$ any number we want. So usually we want $g(x) = 1$ to indicate a function that does not grow with $x$ so just imagine $C$ is a number just larger than $1$. We can now state $$ \cos(x)= O(1) \text{ as } x \rightarrow \infty$$


As far as using $x$ or $n$ for Big-$O$ notation, for mathematical functions, Big-$O$ describes behavior as the function approaches some value $x$ even if $x$ is $\infty$ so $x$ would be used.

In describing time complexity of computer algorithms, Big-$O$ is used to describe how the time of computation grows as the data size grows. Traditionally, the variable $n$ is used as the size of the data set so $n$ would be used. Example, average case time complexity of QuickSort is $O(n \cdot \log (n))$.

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  • $\begingroup$ I understand your answer, but $f(x) = \cos(x)$ was just an example. I am asking whether to use $x$ or $n$ in Big-O notation. $O(x^2)$ or $O(n^2)$. $\endgroup$
    – qsdyebbv
    May 25 '18 at 4:27
  • $\begingroup$ Ah. For functions I believe $x$ is normal. $n$ is used in Big-O when describing computer algorithms. As in how much time increases as data size, n, grows. $\endgroup$
    – Dan Sp.
    May 25 '18 at 13:42
  • $\begingroup$ @qsdyebbv whichever one you use has to be the same on both sides of the $=$ sign. $\endgroup$ May 25 '18 at 20:14
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First lets look at the definition : $f(n) = O(g(n))$ means there are positive constants c and k, such that $0 ≤ f(n) ≤ cg(n)$ for all $n ≥ k$. Simply we have to find a asymptotic upper bound in a long run (not at a point less than k).

As far your $f(x)$ is concerned i.e $f(x)=cos(x)$ We know that $cos(x)<x$ after some $x>k$ (You can see the graph). But to strenthen things $cos(x)<1 \forall x$ So it is said that $cos(x)=O(1)$.Generally speaking $O(x^2)$(after some $n>k$) is correct but not a good bound than $O(1)$ .

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