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I'm reading Data Structures and Algorithms by Goodrich. The explanation that he gives for Big Oh notation is given below:

Let $f(n)$ and $g(n)$ be functions mapping positive integers to positive real numbers. We say that $f(n)$ is $O(g(n))$ if there is a real constant $c > 0$ and integer constant $n_0\geq1$ such that $|f(n)| \leq cg(n)$, for $n\geq n_0$. This definition is referred to as the "big-oh" notation, and is sometimes pronounced as "$f(n)$ is big-oh of $g(n)$".

If I understood this definition right, it would mean that I can say; for a function $f(n) = n$ then $n$ is big-oh of $n^2$ because $n \leq 1\cdot n^2$ for any $n_0$. But, that is not accurate because big-oh notation for a function '$n$' is $O(n)$. What am I missing here? Would it be accurate for me to say that '$n$' is big-oh of $n^2$?

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    $\begingroup$ It causes less confusion to say that "$f$ is a $O(g)$" or what is the same $f\in O(g)$. Here $O(g)$ is the set of all functions (over the integers in this case) that satisfy your definition. You have that the identity function $f(n)=n$ satisfies $f\in O(n)$ and it also satisfy $f\in O(n^2)$. The two sets intersect and $f\in O(n)\cap O(n^2)$. $\endgroup$
    – plop
    Sep 4 '20 at 10:38
  • $\begingroup$ It is actually true that $n$ is big oh of $n^2$, in most formal texts. It is just that in informal usage, people don't say this because people often use big oh to mean big theta, not big oh. $\endgroup$
    – 6005
    Sep 4 '20 at 13:01
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The statement $n = O(n^2)$ is true. There is nothing wrong with it. Maybe you're thinking of the Theta notation.

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  • $\begingroup$ Would it be accurate to represent the big-oh notation of a linear algorithm as O(n^2)? Shouldn't big-oh notation place a tight upper bound on the function? $\endgroup$
    – Leks
    Sep 4 '20 at 10:22
  • $\begingroup$ It would be accurate. Saying that $f(n)=O(g(n))$ just tells you that, asymptotically, $g(n)$ is some upper bound to $f(n)$, up to multiplicative constants and lower-order additive terms. As you can see from the definition, there is no need for $g(n)$ to be tight. Perhaps you are interested in the big Omega and Theta notations from the reference I posted. In your case, the statement $n=\Theta(n)$ is correct but $n=\Theta(n^2)$ is not. $\endgroup$
    – Steven
    Sep 4 '20 at 11:02
  • $\begingroup$ It is encouraged, however, to make the big O bound as tight as reasonably possible. It does not make much sense to give an algorithm that clearly runs in $\Theta(n^3)$ but give it a running time of $O(n^5)$, instead $O(n^3)$ would be the preferred running time. $\endgroup$
    – STanja
    Sep 4 '20 at 14:15
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    $\begingroup$ @STanja, no "encouragement" at all. $O(\cdot)$ expresses one thing, use it to say that. If you want to express "tight", use $\Theta(\cdot)$ or (even better) $f(n \sim g(n)$. $\endgroup$
    – vonbrand
    Sep 4 '20 at 19:31
  • $\begingroup$ @STanja it is done all the time. sometimes it is hard to give the tightest big-O runtime, and so a coarse one is given instead. Yes, you're right that if one is "clearly" Theta n^3, then it wouldn't be described as O(n^5) but it's not always "clear". Bottom line is the big-O is an upper bound, so n is definitely O(n^2) $\endgroup$
    – JimN
    Sep 5 '20 at 1:38
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Firstly let me bring little more exactness. In sentence

If I understood this definition right, it would mean that I can say; for a function $f(n) = n$ then $n$ is big-oh of $n^2$ because $n \leq 1\cdot n^2$ for any $n_0$.

is used "any $n_0$", but in definition of $O$ we have $\exists n_0$ - existence is essential.

To clear your doubts, hope, will be helpful, if you look at $O(n)$ as set of functions. $O(n^2)$ is also set of functions, but more wide. So $O(n) \subset O(n^2)$. And of course as $f(n)=n \in O(n)$, then also $f(n)=n \in O(n^2)$.

By the way, we can say, than for any $k\geqslant 2, k \in \mathbb{N}$ is true $f(n)=n \in O(n^k)$, but this is not more accurate in sense, that we went to more wide set of functions $O(n) \subset O(n^k)$. Most accurate for $n$, in this sense, is $O(n)$.

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