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Big O notation provides an asymptotic upper bound on a function, and omega notation provides an asymptotic lower bound. Three asymptotic notations are used to represent the time complexity of algorithms. They are: Big-O notation, Omega notation and Theta notation.

The given equation is:

$\log \left(\mathrm{n}!\right)=\mathrm{胃}\left(\mathrm{nlogn}\right)$

Let the right-hand side derivation be log(n!).

$$\begin{gathered} \log \left(\mathrm{n}!\right)=\log \left(1脳2脳3............\left(\mathrm{n}-1\right)脳\mathrm{n}\right) \\ \log \left(\mathrm{n}!\right)=\log 1+\log 2+\log 3..........+\mathrm{logn} \end{gathered}$$

From here, it is clear that $\log 1鈮�\mathrm{logn}$

$$\begin{gathered} \log \left(\mathrm{n}!\right)鈮�\mathrm{logn}+\mathrm{logn}+\mathrm{logn}.....+\mathrm{logn} \\ \log \left(\mathrm{n}!\right)鈮�\mathrm{nlogn} \end{gathered}$$

Now, prove for omega notation.

$\log \left(\mathrm{n}!\right)=\mathrm{惟}\left(\mathrm{nlogn}\right)$

Let the left-hand side derivation be${\left(\mathrm{n}!\right)}^2$.

$$\begin{gathered} {\left(n!\right)}^2=n!脳n! \\ {\left(\mathrm{n}!\right)}^2=\left(\mathrm{n}脳1\right)脳\left(\left(\mathrm{n}-1\right)脳2\right)脳\left(\left(\mathrm{n}-2\right)脳3\right)脳..........脳\left(1脳\mathrm{n}\right) \\ {\left(\mathrm{n}!\right)}^2=\underset{\mathrm{k}=1}{\overset{\mathrm{n}}{鈭�}}\left(\mathrm{n}-\mathrm{k}+1\right)脳\mathrm{n} \\ {\left(\mathrm{n}!\right)}^2=\underset{\mathrm{k}=1}{\overset{\mathrm{n}}{鈭�}}\left(-{\mathrm{k}}^2+\mathrm{nk}+\mathrm{k}\right) \end{gathered}$$

Here a quadratic equation is obtained, which is equal to data-custom-editor="chemistry" $\mathrm{f}\left(\mathrm{k}\right)$.

$\mathrm{f}\left(\mathrm{k}\right)=-\mathrm{k}+\mathrm{nk}+\mathrm{k}$where data-custom-editor="chemistry" $\left(1鈮�\mathrm{k}鈮�\mathrm{n}\right)$鈥︹赌�.鈥�(1)

Here, put the value of k in the equation (1).

$$\begin{gathered} \mathrm{f}{\left(\mathrm{k}\right)}_{\min }=-1^2+\mathrm{n}脳1+1 \\ \mathrm{f}{\left(\mathrm{k}\right)}_{\min }=\mathrm{n} \end{gathered}$$

Equate both the equations:

$$\begin{gathered} {\left(n!\right)}^2=\underset{k=1}{\overset{n}{鈭�}}f\left(k\right) \\ {\left(\mathrm{n}!\right)}^2鈮�\underset{\mathrm{k}=1}{\overset{\mathrm{n}}{鈭�}}\mathrm{f}{\left(\mathrm{k}\right)}_{\min } \\ {\left(\mathrm{n}!\right)}^2鈮�\underset{\mathrm{k}=1}{\overset{\mathrm{n}}{鈭�}}\mathrm{n} \\ {\left(\mathrm{n}!\right)}^2鈮�{\mathrm{n}}^{\mathrm{n}} \end{gathered}$$

Taking $\log$on both sides:

$$\begin{gathered} \log {\left(\mathrm{n}!\right)}^2鈮�\left(\log \right){\mathrm{n}}^{\mathrm{n}} \\ 2\log \left(\mathrm{n}!\right)鈮�\mathrm{n}\left(\log \right)\mathrm{n} \\ \log \left(\mathrm{n}!\right)鈮�\frac{\mathrm{n}}{2}\left(\log \right)\mathrm{n} \\ \log \left(\mathrm{n}!\right)鈮�\mathrm{n}\left(\log \right)\mathrm{n} \end{gathered}$$

So, the final answer obtained is $\log \left(\mathrm{n}!\right)=\mathrm{胃}\left(\mathrm{nlogn}\right)$

Hence data-custom-editor="chemistry" $\log \left(\mathrm{n}!\right)=\mathrm{胃}\left(\mathrm{nlogn}\right)$is proved.