What's the partial of N with respect to x?

The partial of N, with respect to x, is 2.

The partial of M with respect to y is equal to the partial of N with respect to x.

The partial of this with respect to y is equal to the partial of this with respect to N.

The partial of psi, with respect to y, is equal to N.

It was different than N with respect to x.

So the partial of this, with respect to y, is different than the partial of N, with respect to x.

So psi, with respect to y, is equal to N.

And the partial derivative of psi, with respect to y, is equal to N.

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

The partial of N, with respect to x is equal to well, there's no x here, right?

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?

Take the partial with respect to y, and set that equal to our N expression.

Holy. worthy of respect.

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.

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.

And out of respect, I'm...

That's a sign of respect.

Have a bit of respect.

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

n times n minus one times n minus two.

Respect....

Respect.

Respect.

Respect.

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.

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