How would I prove that the statement


is true/false?

  • 3
    $\begingroup$ There is no "equation" here. Either the function $n \mapsto 10n^3 + 3n$ is in $\Theta(n^3)$ or not. Review the appropriate definitions. $\endgroup$
    – dkaeae
    Commented Mar 13, 2019 at 14:38
  • $\begingroup$ You should start by looking at the relevant definitions. $\endgroup$
    – Juho
    Commented Mar 13, 2019 at 16:45
  • $\begingroup$ Have this in mind: $10n^3+3n=Θ(n^3)$ is an abuse of notation. This is what they intend to say: $10n^3+3n$ belongs to $Θ(n^3)$. $\endgroup$
    – rranjik
    Commented Mar 13, 2019 at 17:13
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    – Raphael
    Commented Mar 13, 2019 at 19:59

2 Answers 2


Remember that, $10n^3+3n=Θ(n^3)$ is an abuse of notation. Think of $Θ(n^3)$ as a set/class of functions. You are prove to that $10n^3+3n$ belongs to that set.

Simply put, every element in the set has a property, and everything that has the property belongs to the set ("if and only if" is the mathy way they say this). A function $f(n)$ belongs to $Θ(g(n))$ if that property holds.

So, what is that property exactly ?

$C_1g(n) ≤ f(n) ≤ C_2 g(n)$ whenever $n > k$

Well, what does that mean ? For a decently large $n$, every element (called $f(n)$ here) in the set (which is $Θ(g(n))$ here), is smaller than a constant ($C_2$) times $g(n)$, and also it is always larger than a different constant ($C_1$) times $g(n)$.

Note that that the inequality implies, $\frac{f(n)}{g(n)} ≤ C_2$ and $C_1 ≤ \frac{f(n)}{g(n)}$. All we have left to do is to find such a $C_1$ and $C_2$.



this is always less than or equal to $\frac{10n^3+3n^3}{n^3}$ - because we assume $n$ is a positive integer and so $n^3$ is always at least $n$ - which is less than or equal to $13$. This proves the $\frac{f(n)}{g(n)} ≤ C_2$ part.

Moreover $\frac{10n^3+3n}{n^3}$ is greater than or equal to $\frac{10n^3}{n^3}$. Again as n is positive dropping $3n$ could only make it smaller. This implies the constant $C_1$ is $10$, the second part.

The first part proves that $f(n)$ belongs to $O(g(n))$ and the second proves $\Omega(g(n))$ which implies $Θ(g(n))$ usually.

So this gives a sloppy bound that this function could only be 13 times as big, and 10 times as small. $10n^3 ≤ 10n^3 + 3n ≤ 13n^3$. To be precise we assumed $n>1$ and $k=1$ here. Use the limit definition for an exact proof. But this is pragmatic and should be enough in most cases.

  • $\begingroup$ Surely that's $O$, not $\Theta$? $\endgroup$
    – Draconis
    Commented Mar 13, 2019 at 18:14
  • $\begingroup$ Oopsies! Sorry for the wrong answer. I thought it was big O. Let me delete it now. $\endgroup$
    – rranjik
    Commented Mar 13, 2019 at 18:14
  • $\begingroup$ @Draconis thank you very much for pointing out. I did not want to delete the answer. I have edited this to prove $\Theta$. I totally thought it was $O$ when I wrote this the first time. Please check this if you get a chance. $\endgroup$
    – rranjik
    Commented Mar 13, 2019 at 18:57
  • 1
    $\begingroup$ All seems correct; I'd use some different notation but that's just a matter of convention. $\endgroup$
    – Draconis
    Commented Mar 13, 2019 at 19:23

This isn't really an equation. The equals sign in the middle is an abuse of notation: $\Theta(n^3)$ is a set of functions, not a single function, so it should more properly be written $(10n^3+n) \in \Theta(n^3)$.

The formal definition of $\Theta(n^3)$ is something like this:

$\Theta(n^3)$ is the set of all functions $f(n)$ such that, for some specific choice of constants $N, C, D$, then $n > N \implies Cn^3 < f(n) < Dn^3$.

Basically, when $n$ gets really big, $\Theta(n^3)$ means that the function starts to "look like" $n^3$, up to a constant factor. There's a useful lemma here which helps more:

For a function $f(n)$, if $\lim_{n \rightarrow \infty} \frac{f(n)}{n^3}$ exists and is a positive constant, then $f(n) \in \Theta(n^3)$.

This is much easier to use here, in my opinion. With a bit of algebra, we can say that $\lim_{n \rightarrow \infty} \frac{10n^3+n}{n^3} = \lim_{n \rightarrow \infty} \frac{10n^3}{n^3} + \lim_{n \rightarrow \infty} \frac{n}{n^3} = 10+0 = 10$. Since ten is a positive constant, we've shown that $(10n^3+n) \in \Theta(n^3)$.


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