so you want to calculate h minus 1, one thing you can do, is take the limit z

goes to alpha, of alpha minus zh is z. and there are, the, approximation and I

add alpha to the function is approximated by h minus 1 over alpha minus z.

so then we can just go ahead and expand it.

So 1 over alpha, so it's h minus 1 over 1 minus 0 over alpha, 1 minus 0 over alpha,

is sum zdn over alpha to the n. and that gives us the, uh, [COUGH] the

coefficients that, the asymptotic growth is 1 over alpha to the n, and then the

constant is, well in this case m equals 1.

so it's minus 1 times f of alpha over, well, we'll look at this proof of exactly

where this constant comes from in a minute.

that's all going to be done on the next slide.

So just a couple of notes about this. So so I'll complete this, proof on the

next slide. So as I mentioned, the error is

exponentially small. now, if the, if they're close to the same

distance, there might be periodicities for some problems that you might have to

look at. you know, often the next term does

involve some kind of periodicities. now and the other thing to notice is, if

you go back and look at our transfer theorem for rational functions it's the

same. well and it ought to be the same, because

if f of g and g of 0 polynomials, you get the same kind of result.

but still, it's quite different paths, to this same result.

the residue calculus in Koshi's theorem and so forth, gives us the asymptotic

result that we need for a meromorphic functions.

And again, we've been through quite a bit of complicated abstractions in this

lecture. but really what you need to remember is

these simple formulas. and in very many cases, as we'll see in

the next lecture, it's simply a matter of, of plugging into these formulas to

get the coefficient asymptotic's. The essence of analytic commonotorics, is

we start with a commenotorial construction, and that gives us a

generating function equation. And then an analytic transfer theorem

like this takes, the generating function equation and immediately gives us the

coefficient asymptotic's. now before I go on, I want to justify

this constant that's the coefficient. so I mentioned before that when we

compute the constant if the poles of order 1, just by taking the limit of

alpha minus zh of z. Well, if you take that limit, then you

can use L'Hopital's rule, take the derivative of both sides.

Uh,and so that's alpha minus z times the derivative of f, plus f times the deriv

alpha minus c, which introduces a minus sign, on the bottom is g prime over z.

But, the pole of order 1, g prime of z is not going to be 0, so we can just plug in

alpha. It kills the first term, and we get minus

f of alpha over g prime of alpha. So that's an easy way to compute the

coefficient. if it's of order 2, then we have to apply

l'Hôpital's rule twice, and so we can do the expansion as before.

We have two terms now but we just take the leading term, and now our leading

term we're expanding 1 minus z over alpha squared.

and then that, because of the square, brings in a factor of n, into the leading

term, simply by elementary generating function exponentially.

and then the rest of the calculation is using l'Hôpital's rule twice.

take the first derivative, take the second derivative.

and if it's of order 2, the second derivative is non 0, and then taking z

equals alpha, kills all of the terms except 1.

and we're left with 2 f of alpha over g double prime of alpha, and then you can

convince yourself from that, that's of order m.

Then our coefficient that we need is minus 1 to the m, and f of alpha nth

derivative of a g, evaluated at alpha over alpha n, that's the coefficient.

and then the growth is, n to the n minus 1 over n, where m's the order of the pole

1 over alpha to the n. so a simple formula for the coefficient,

in terms of the order of the pole. And again there's a lot of calculations

here but I really think the bottom line is you can see some light at the end of

the tunnel. we've done a lot of calculations but what

we have in the end is for many, many examples.

of, First of all, we have an analytic transfer to take a meromorphic generating

function, into its asymptotic growth form.

In most of the cases that we're going to consider, that we have considered in

terms of commenotorial classes, the poles are order 1, so the growth is a constant

times Beta to the m, where beta's 1 over the closest route to the origin.

So all we have to do, is compute the dominant pole.

Now that's the smallest root where the denominator's 0.

maybe we have to use a ah, [COUGH] math package to do that but many times it's

simple, it's elementary. and you should check that no other no

other poles have the same magnitude, that's usually very easy, you usually

just do it with a plot. and then what's the residue from the

theorem? It's just f of alpha over g prime of

alpha an easy calculation. And, then the constant is just now, it

might, you might worry by g prime being 0, well if your prime is 0, then its not

order of 1, so you'll just have to adjust to do the order's of mk's, but we have a

formula for that, too. again that, doesn't come up, in a great

many of the commenotorial classes that we consider.

So the constant then, is just h minus 1 over alpha.

And the exponential growth factor, beta, is just 1 over alpha.

so that's the bottom line, is we've got a very simple method to transfer from

generating a function equation into constant times exponential growth factor.

so there has been no doubt a great deal of difficult and complicated mathematical

material in this lecture. but really the purpose of it, is to

persuade you that there's rigor in our being able to do these kinds of analytic

transfers. most of the time in applications we can

just use it without going back to the detail in difficult theorems that we've

talked about. so, just to give, just a couple of

examples so if our meromorphic function is rational like z over 1 minus z

squared, well there's, alpha, that's the root.

So it's phi hat, which is the same as 1 over phi, so the exponential growth's

going to be phi to the n. and what's the constant?

it's phi hat over 1 plus, 2 ha, 2 phi hat, which is 2 phi hat squared 5 minus

1. So that's just square root of 5.

and then, to get the coefficient we divide by phi n, which takes that out.

So that immediately gives the asymptotic's for that generating

function, which comes up in a couple of combinatorial classes.

this is the combinate generating functions for derangements.

so the pole is at 1, so f of alpha is e to the minus 1, is 1 over e.

g prime of alpha, is just 1. So, it's just 1 over e divided by 1, 1 to

the n. coefficient of z to the n, and hz is 1

over e. or to generalize, to generalize

derangements. so in, in that case f of, alpha again is

1. That's where the pole is, there's only 1

of them. evaluate the numerator at 1.

you get e to the minus h3, or one over e to the h3.

exponential growth is 1 to the n, which goes away, divide by alpha as 1.

so immediately we get the coefficient with a very simple calculation.

so just as an example, here, here's what we're going to be doing for the entire

next lecture is looking at how analytic combinatorics gets us from a

combinatorial specification, down to asymptotic results.

So we talked about generalized derangements.

That's a combinatorial construction and the symbolic method immediately gives

that generating function. And now, we have an analytic transfer, to

take us immediately from that generating function, right down to the asymptotic's

of the coefficients. you don't need to know too much about

Koshi's theorem or residues, or any of these things in order to be able to apply

this analytic transfer theorem. so, in the next lecture we're going to

see many, many, many more examples of this general idea that we can have an

analytic transfer that works for a, a broad variety of generating function

equations, and takes us right to the coefficient asymptotic's.