Watch yourself, Lieu.

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

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

It's n times the factorial of n 1.

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?

Of course if you've got anything to drink, we'll take that in lieu of cash.

Is Fibonacci of N always lt or N 1?

Θ(n²), O(n²), and O(n³).

N by N, kind of like we have up here.

You may have noticed that I wrote 1 times N times factorial of N minus 1 instead of just N times factorial of N minus 1.

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?

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.

luminous, and or large objects, we use in lieu of standard candles, or, standard rulers.

Carl gets promoted in lieu of Demko, or along with Demko whichever, or I walk.

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.

And 8 n we can think of a just Θ( n).

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

So instead of just adding N plus 1 N times, we could say that this is just N times N plus 1.

If f(n) is in little o(g(n)), that's kind of like saying f(n) is strictly less than g(n).

So that reduces compute time from n squared, where n is number of particles, to n log n, which is much easier.

So the log of n, log base 2 of n is, you type the number N into your calculator, okay?

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.

n

N

N

N

N

N

N

? N? ?

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

It is Θ of a combination of these things n² and log n together?

For proofs the n 5 and n 6 reactors of Fukushima Daiichi

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?

We're going to take n 4 of the set of nodes and fully connect them with another set of n 4 nodes. n 4 from one side, n 4 from the other.

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

Here's the definition that there is some constant c₁ and c₂ bigger than 0 and a a threshold n₀ such that f(n) lies between c₁ of g(n) and c₂ of g(n) for all n bigger than the threshold n₀.

log base 2 of n, specifically.

N² kind of beats everything here.

I'm going to draw out a chart showing N and chart of N.