Is it n, n 1 over n, n over n 1, n divided by n 1, n 1 divided by n, or is it n squared 1 over n 1?

Θ(n²), O(n²), and O(n³).

Okay. So basically (log n)⁷ beats 9n(log n)² and n² ³ beats (log n)⁷ and 9n(log n)² beats n² ³.

Now, if K is n 2 that's n times n 2 where n².

So I subscript n by n is the n by n identity matrix.

So altogether we're talking about n n² or n³.

n times n minus one times n minus two.

If we divide through by n, we get 2 n n and divided by n is actually n.

Russian(n, n) for lots of different values of n.

n

N

N

N

N

N

N

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Log n is piddly as n gets big compared to n, and n is actually pretty piddly as n gets big compared to n².

Keep shoot 'n. Keep shoot 'n.

Here we have n n 1.

Could there be n 1 ways, n 1 squared, n 1 factorial, or 2 n 1?

So, it's really n times n plus m which is (n² nm).

We're much better with n log n.

With the very long list that were wanting to tap values of, you might as well just sort the whole thing. n, well what happens with n? n where look we're comparing n log n to n( n) which is n³ ². n³ ² is asymptotically larger than n log n so we're still better off just sorting the whole list.

Is the depth that we get this way log n, n, or 2 n, and are the number of leaves at the bottom log n, n, or 2 n?

Adding up these three parts we get 2 n n(n 1) 2.

Ctrl N

N Go

PAL N

N no.

Stop! n

And... n?

Uu n...

N o!

F(n) is in O(g(n)) there's our friend O that's really like saying f(n) g(n).

n here has to match this n here.

It's n times the factorial of n 1.

So it's Θ(n²). It's order of n², as well as order of n³.

Or hear f(n) is some basic function, like for example n log n.

But imagine if it was, you know, an n by n by n cube.

We could write it as n factorial it s a lower case n n factorial over n minus k factorial.

If f(n) grows roughly the same as g(n), then g(n) should grow roughly the same as f(n).

For small N, it's 0.25, but for larger N like N 10, it's now 0.025.

Line up everybody. Line up. Line up.

When n is 4, n minus 1 is 3.