And to compute them, let's first compute the various components here.

In some cases it's even been possible to distinguish the murderer's face at the very moment he committed the crime.

And for the case of when we have a large training set, not just a training set of one example, here's what we do.

So that's 1, 2, 3, 4, 5, 6, 7, 8, if we include the global frame.

We're given a recursive formula for the Stirling numbers, which is S(n, k) kS(n 1, k) S(n 1, k 1).

We're doing remove.mn, which is a (logn) operation, so we get n (logn) for this loop plus (L) for the build heap.

Mu, sigma squared.

There's our value v, and now, we're going to run the partitioning algorithm on this, and there is a couple of different cases that can happen.

We really don't want this to be n here because our inductive step is going to assume this true for everything smaller than n and not equal to n but the reality is this is actually a tiny little bit less than n.

That is they need a constant times in the squared number of operations to sort an input array of length N. Merge sort by contrast needs at most a constant times N times log N, not N squared but N times

So what is the logorithm? Well, for our purposes, we can just think of a logorithm as follows, okay?

So, that's what n 2 looks like.

Because now all of a sudden, that equity could be worth a lot.

This case also applies to other cases.

So graphically, what the logarithm is going to look like is it's going to look like.

So my question to you is how many of these

Then we connect the corresponding nodes, but there's just one of them, so it's just that.

I can't find my vanity case.

I'll discuss two very promising approaches.

How many total environment frames will we be juggling at the same time? Fill in the blank.

As we look at these trends, do we become despondent, or will we become energized?

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

It grows less slowly asymptotically.

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.

We're now ready to start evaluating some cosmological models. We'll consider several special cases.

Or, we could even come up with a formula for the nth figure.

So this space is at position 3.

(Laughter)

Alright, so next we're going to look at some recursively generated graphs.

Hit him again, hit him again!

In this case, we're done we locked out.

And it turns out we need to figure out what is the partial derivative of two cases for J equals 0 and for J equals 1 want to figure out what is this partial derivative for both the theta(0) case and the theta(1) case.

But let's just make show if c₁ is equal to 1, then we need n to be less than or equal to this expression.

Give me the case!

I think by 6 .

The running time for that approach is going to be big Θ of n, the number of nodes times the time that it takes to do a shortest path search.

If we lower the price, the quantity demanded will go up.

If we make this equal to our u, what's our u prime? u prime is just going to be n times t to the n minus 1.

That's P of H.

Let's move up to n 8.

So we need to actually add a second base case.

An equality is reflexive, so Big Theta ought to be reflexive.

And I think you'll see in a second, or maybe you already realize why I'm doing this.

So this altogether is going to get run once for each node so this part of the algorithm actually takes time and logn.

I want to know what's in the case. Is it a bomb? Go to Macy's.