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What is the big-O of the function $2\log(\log(n)) + 3n\log(n) + 5\log(n)$?
Is it just $O(n\log(n))$ for the whole function? I'm not sure how to represent $2\log(\log(n))$.

ALute
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    "Big O" is time complexity that describes the worst case scenario.. so, you want to look for the term that will produce the highest values when considering values of n while approaching infinity. As for the other two terms, they will "fall to the side", or really, become so small in contrast to the overall resulting value that the terms are trivial to consider. So with that being said, which of the three terms will always produce the largest values "down the road"? –  Jul 29 '17 at 03:26
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    @Charles BigO is not a time complexity but a mathematical notation that describes how two real functions are related in respect to increasing arguments. It often will be used to describe the worst case running time of an algorithm but it also will be used to describe the average running time of an algorithm or anything else where I wanted to know how this "anything else" is bounded by a simple function if the argument increases. – miracle173 Jul 29 '17 at 03:54
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    @miracle173 No, Big O is worst case, Theta is average, and Omega is best case. They are classifications for time complexities that are described most simply by a single function that bounds the algorithm's worst run time scenario. I am familiar with this topic, but thank you. I was more trying to get the OP to answer their own question by offering leading questions and instructions on how to do so, than I was trying to give a wholesome definition of time complexity classifications. –  Jul 29 '17 at 03:57
  • https://en.wikipedia.org/wiki/Big_O_notation – miracle173 Jul 29 '17 at 04:37
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    @Charles "No, Big O is worst case, Theta is average, and Omega is best case." - No, no no no, NO! You couldn't be more wrong. – Raphael Jul 29 '17 at 06:27
  • Duplicate? – Raphael Jul 29 '17 at 06:30
  • Let us continue this discussion in chat. – Raphael Jul 30 '17 at 20:37
  • There is no mention of “time complexity” either in the question or in any of the correct commentaries. – gnasher729 Apr 18 '22 at 21:17

5 Answers5

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Starting with an analytical assessment of the function's terms, we'll first format all terms as similarly as possible to each other, in order to more easily compare with one another. To do so, all constant factors will be removed, leaving only the variable elements of each term. This is acceptable to do since we're ultimately seeking Big-$O$ classification, which typically sets all constants within its expression to a value of $1$.

  • $2\log(\log(n)) \to \log(\log(n))$
  • $3n\log(n) \to n\log(n)$
  • $5\log(n) \to \log(n)$

Rewriting,

$\quad f(n) \to \log(\log(n)) + n\log(n) + \log(n)$,

where $n \in \mathbb{N}.$

Now, since each term varies only by their logarithmic arguments, we can establish an inequality by direct comparison of those arguments, resulting in

$\quad \log(\log(n)) \lt \log(n) \lt n \log(n),\ \forall n \gt 1$

Therefore, it follows that

$\quad f(n) \in O(n\log(n))$

John L.
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Please note that "the big-O of the function" isn't a correct formulation. We assume $\log$ is the binary logarithm $\log_2$. But actually the proof can be extended to any base.

We have

$$\log(x)\lt x, \forall x>0$$ and, if we plug in $\log(n)$ for $x$ $$\log(\log(n))\lt \log(n).\tag 1$$

Recall the definition of $f(n)=O(g(n))$: $$|f(n)|\le M |g(n)|,\forall n\ge n_0$$ for appropriate $M$ and $n_0$.

So if we choose $f(n)=\log(\log(n))$, $g(n)=\log(n)$, $M=1$ ,$n_0=2$

we see that $(1)$ is $$\log(\log(n))=O(\log(n))$$ and of course $$\log(\log(n))=O(n\log(n)).$$ So all three function in your expressions are $O(n\log(n))$ and therefore every linear combination of them $$a\log(\log(n)) + b \, n\log(n) + c\log(n), \quad a,b,c \in \mathrm R$$ is $O(n\log(n))$.

miracle173
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  • @zaph The notation of this answer is correct. Depending on who is doing the writing, a can be used, or an $ –  Jul 30 '17 at 16:27
  • @zaph The notation of this answer is correct. When writing these expressions, an $=$ can be used, as well as $\in$. Mathematicians tend to use $=$ more, whereas I myself prefer $\in$ set notation because I feel it's more descriptive of the situation. Both are perfectly acceptable though! –  Jul 30 '17 at 16:30
  • If < is used in log(log(n)) < log(n) why not use < in log(log(n)) = O(log(n))? Wouldn't log(log(n)) < O(log(n)) be more consistent? – zaph Jul 30 '17 at 16:53
  • @zaph Please notice that the r.h.s. of the expressions you've questioned have $O()$. I believe your confusion stems from neglecting to recognize this. :) –  Jul 30 '17 at 17:26
  • I see that and it is my ignorance of math that is my problem. Not all computer software developers understand such the math notation details but we may still find big-o information useful. And in this case the overall concept of the question. – zaph Jul 30 '17 at 19:24
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    @zaph No worries. To pull off Wiki's page for Big O: "Note that "=" is not meant to express "is equal to" in its normal mathematical sense, but rather a more colloquial "is". So, you can read it as a property of the function. Ex: "He is a boy" :: "$f$ is (has) a Big O of $n : log(n)$". Or at least, that's how I perceive it. Critiques are welcome of course. –  Jul 30 '17 at 23:04
  • Obviously I'm not the audience this site addresses not having a degree in CS and only some 5 decades of software development experience. I did not realize this site was oriented to the mathematically advanced which I am not. I now find that Computer Science is not something I am part of. Note: The answer by Charles manages to be clear without confusion. – zaph Jul 31 '17 at 00:22
  • @Charles, rather than just "is", more formally $O(f)$ is a set of all functions that satisfy the definition of Big O for a function $f$ and you would probably say "is an element of" or "is in" e.g. $g \in O(f)$ for some functions $g$ and $f$. This is why the use of the comparison or equality operators ($<, \leq, =, \geq, >$) can sometimes be misleading because it implies a symmetry which it does not have. – ryan Jul 31 '17 at 06:27
  • @ryan Yeah, I read that off the wiki page too. And yes, I agree with that, which is why I said that I myself use $\in$ instead of $=$. –  Jul 31 '17 at 09:10
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It is obvious that $\log n\le an$ for some constant $a$ because the number of digits of $n$ cannot exceed $n$. Then substituting $n$ for $\log n$, $\log\log n\le a\log n$. (For the base-2 logarithm, $a=1$ works.)

Combining, we have

$$2\log\log n+3n\log n+5\log n\le (2a+3n+5)\log n\le cn\log n$$ because for $n\ge1$ and $c=2a+5+3$,

$$\frac{2a+5}n+3\le c.$$

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Yes, the expression is $O(n \log n)$. For large $n$, $n \log n$ dominates $\log n$ and $\log \log n$.

Ashwin Ganesan
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The best way to find big-o of a function like this:

$$f(n) = \sum_{i=1}^k f_i(n)$$

is to find an i where:

$$\forall j \in [1,k], j\neq i \rightarrow \lim_{n->\infty} \frac{f_j(n)}{f_i(n)} =0$$

therefor big-o is $$n\log(n)$$

Mostafa Barmshory
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    $\lim_\limits{n->\infty} \frac{f_j(n)}{f_i(n)} =0$ is too restictive, $\lim_\limits{n->\infty} \frac{f_j(n)}{f_i(n)} <\infty$ will be sufficient, too. $\lim_\limits{n->\infty}\sup \frac{f_j(n)}{f_i(n)} <\infty$ is exact. And if $f_j(n)<f_i(n)$ this is already sufficient, you dont have to calculate the limit. So in this case I can't see why it should be the best method. – miracle173 Jul 30 '17 at 20:17
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    Nevertheless your argument is correct. – miracle173 Jul 31 '17 at 04:10