Because in this case it's not approximation.

the thing that's raised to the, e is raised to that power is log of 1 over 1

minus z. So that's alpha equals 1.

Beta equals zero. There's nothing added on.

And row equals 1. So we just plug in alpha equals 1.

Gamma of 1 is 1. Beta equals 0 so that's a 1.

rho equals 1. So it's 1 to the N.

and all that's left is 1, from exp-log. So the coefficient of z to the N of P of

z is asymptotic to 1. And it's exponential, so number of

permutations as sub type n factorial. that doesn't seem like much of of much

interest. But remember the correlary says the

expected number of components. So that's the expected number of cycles.

Which does maybe require some [INAUDIBLE].

Some calculation it's alpha log in. Alpha's one so that's immediate through

singularity analysis prove that not just the number of permutations.

But also the average number of components comes immediately log in.

and again for The basic problem like this we have lots

of ways to proceed to improve the log in result.

Our point is that for variance of this structure or for other similar structures

we can use precisely the same process to get an answer where direct calculations

might be extremely complicated. Were prohibitive.

So like, examples. So we looked at derangements.

So we can look at derangements. Or we can look at a problem like, how

many cycles are there in a random derangement.

might be hard to think about how to even approach a problem like that.

but it's immediate from the X blog schema.

And precisely the same process works for more general settings.

Like maybe we want to know about cycles and derangements.

We talked about derangements, which are the set of all permutations having no

singleton cycle so we take out the singleton cycles.

So that's the construction. It's a set of cycles none of which, all

of which are lengths greater than one, greater than zero and so, all we do is

subtract off, one from log of one minus z to take out the singleton cycles.

D is, equals x log, [INAUDIBLE], or one minus z minus one.

that's x blog. The only difference is that, now that

beta is minus one. So it's the x blog form.

it's just that beta is minus one. So,

When we now, apply the theorem, we get the same answer.

except that, 'cuz of the e to the beta, we get e to the minus 1.

and that's a familiar result. the number of [INAUDIBLE] have to

multiply by n factorial. It's n factorial, over e.

and when you looked, we looked at this as an early example in part 1 of the course.

and that 1 over e started out as a finite sum and we had to show that the tail's

negligable, and so forth. this higher level way of looking at the

same problem really gives insight in where this form of the solution comes

from. it's all about the minus one.

And we subtracted off the minus one. That translated right through the

theorem, down to the exponent of e. and not only that, we have the

correlative theorem which tells us, average number of cycles in a random

derangement. But still log in, doesn't involve data,

data so the average number of cycles in a random derangement, is proportional to

the log in still. And this generalizes to include any

restrictions on the cycle links. So you could pick t different integers

and say I want to study the class of all permutations having no cycles of length

w1, w2, down through wt. the uh,construction is, just sets the

cycles with that restriction on the cycle length, immediately translates to a

generating function equation where we subtract off the term corresponding to

each one of those cycle lengths from log of one over one minus z.

but when applying the explog theorem all that does is when we approximate that

function near the singularity, which is rho equals 1.

Those Z to the Ws become 1. it all comes to the beta.

so applying the theorem then we're just going to get e to the minus theta.

and then it's just e to that constant and whatever those constraints are, you can

compute the constant, and you can know that the number of generalized

derangements of that form are n factorial over e to that constant, very easy to

compute. pick 2, 3, 5, 9, and 11, and you can

compute the number and know that average number of derangements that don't have

cycles of length 2, 3, 5, 11. You might be challenged to try to derive

a result like that using elementary techniques.