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For the computation of the log factorial, i.e., $\log(n!)$, is the Big(O) run time for this $n\log(n)$? How can this be assumed graphically?

***Update: How does the sum of $\log(n!)$ work then? I.e., the sum of log Factorial?

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    The sum comes from log rules. See my answer below. – 2017-02-20
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    I see the properties of the log rule, but is sum(log(n!)) then O(n *log(n)) as well?? – 2017-02-20
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    No. Note that $$\log(n!)=\sum_{k=1}^n\log(k)=\mathcal O(n\log(n))$$ – 2017-02-20
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    I added the graphical part. :D – 2017-02-20

2 Answers 2

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Notice that:

$$\log(ab)=\log(a)+\log(b)$$

Thus,

$$\log(n!)=\sum_{k=1}^n\log(k)<\sum_{k=1}^n\log(n)=n\log(n)$$

A lower bound may be done with Riemann sums:

$$\sum_{k=1}^n\log(k)>\int_1^n\log(x)\ dx=n\log(n)-n+1$$

Here's the graphical component:

enter image description here

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$$\log(n!) = \sum_{k=1}^n \log(k)$$ Can you figure it out from here using properties of Big-$O$ notation? You'll find that the answer to your conjecture is a resounding yes; graphically, this is because $n^n$ is a decent estimator of $n! $. Look up Stirling Approximation for more detail

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    Yes I can. Specifically, is the run time of the log(n!) nlog(n) in bigO? – 2017-02-20
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    @NinaMcGyver yes. Notice that $\lim_{n \to \infty} \frac{\log(n!)}{n\log(n)} = 1$ – 2017-02-20
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    @NinaMcGyver graphically this is because $n^n$ is a decent estimator of $n! $. Look up Stirling Approximation for more detail – 2017-02-20
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    made an update above. Thanks – 2017-02-20
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    @BrevanEllefsen Isn't Stirling approximation overkill for this? – 2017-02-20
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    @SimplyBeautifulArt yes... but just take the simplest form of it and you're fine. Knowing the exact constants isn't important. This is more or less what you do in your post. – 2017-02-20
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    @SimplyBeautifulArt even your answer is overkill if you can take limits and already have the OP's conjecture for the record, as I point out in my comment above – 2017-02-20
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    Hm, now that I think about it, I feel like there should be different names to the approximations of the factorial. Who's idea was it to name them all "Stirling's approximation"? – 2017-02-20
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    @BrevanEllefsen :-/ Those limits aren't trivial though. – 2017-02-20
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    @BrevanEllefsen $n^n$ is not a decent estimator of $n!$. Actually, we have $n! \in o(n^n)$. – 2017-02-20