0:18

So recall will be proofed last class.

Given the power series of this form, ak(z- z0) to the k, with the ak's

are complex numbers, there exists a number R that's between zero and infinity.

And both zero and infinity are allowable values for R.

Such that this series converges absolutely in fact for

all zs that are within R of the center of z0 and it diverges for

all z's whose distance from z0 is bigger than R.

So we prove the existence of this disk in other words, centered at z0

0:56

and of radius uppercase R, such that whenever you

pick a z inside, for that z, the series converges.

And whenever you pick one on the outside, for that z, the series diverges.

1:11

On the boundary we had no conclusive answer.

Either convergence or divergence could happen.

But how do you find this R?

We know it exist but how do you find it?

Here's a first idea on how to find such an R.

If the sequence of quotients ak divided by ak + 1 has a limit.

Then this limit is the radius of convergence of the power series.

1:40

This theorem called the Ratio Test does not say that necessarily the sequence

of quotients of successive coefficients has a limit, it just says if that

sequence has limit then that limit is the radius of convergence of the power series.

So, let's look at some examples.

By the way, infinity is an allowable limit.

So if ak over ak+1 absolute value goes infinity as k goes to infinity,

then the radius of convergence r of the power series is infinity,

in other words it converges for all z in the complex plane.

2:33

to the k where k goes from zero to infinity.

So if we're just looking at z to the k, and this z0 here is equal to zero and

the ak's are all equal to one.

So if I look at ak divided by the next one, by ak + 1, they're all equal to one.

So it's one over one that's equal to one in absolute value and

one as k goes to infinity remains one, and so it converges to one.

Therefore, by the previous theorem, the radius of convergence is one.

3:35

Here's another example, z to the k divided by k factorial.

So ak now is one over k factorial.

So if I look at ak and divide it by the next one, ak + 1,

well ak is 1 over k factorial and I divide that by 1 over k + 1 factorial.

So I look at 1 over k factorial divided by 1

over k + 1 factorial, that's ak + 1.

And so I multiply with the reciprocal and

end up with k + 1 factorial over k factorial.

But k + 1 factorial = k + 1 times k times k- 1 and

so fourth all the way down to 1.

And k factorial, that's k times k- 1 and so forth all the way down to 1.

And so all these terms cancel out except for the k + 1 term that's left.

So k + 1 factorial divided by k factorial is simply k + 1 and

as k goes to infinity, that goes to infinity.

Therefore in this case the radius of convergence R is infinity.

So this series converges for all z's in the complex plane.

Let's look at another example.

4:57

Z to the k divided by k to the k.

So now, ak is equal to one over k to the k.

So again, ak divided by ak + 1, is going to be 1 over k to the k.

Divided by 1 over k + 1 to the k + 1.

And again I multiply with a reciprocal and I get k + 1 to the k + 1 over k to the k.

5:25

Now this is a little bit less clear on how to find the limit.

I can actually find the limit, and I'll quickly show you how but

the ratio test is a little harder to apply here because this limit is not as obvious.

So I'm going to show you how you could find this limit.

You could write this as k + 1 to the power of k.

6:19

And k + 1 divided by k, I could split that up into 1 + 1 over k and

then raise that to the k power.

You may or may not have seen this in calculus but

this sequence, 1 + 1 over k to the k actually has limit.

And the limit of that is e.

That's Euler's number e.

So, everything in pink goes to e as k goes to infinity.

But k + 1 the orange term goes to infinity.

6:50

Since the second term goes to e and no to zero, the infinity just wins and the whole

expression goes to infinity as k goes to infinity.

And therefore, ak over ak + 1 actually has a limit.

The limit is infinity and

therefore, the radius of convergence of this power series right here is infinity.

7:22

So this last example we just looked at, ak = 1 over k to the k,

is actually easier to treat using the root test.

And it goes as follows, instead of looking at ak divided by the next coefficient

ak + 1, you simply look at the kth root of the kth coefficient ak.

And again, you take the absolute value.

If that sequence has a limit as k goes to infinity,

then the radius of convergence is one over that limit.

7:52

We have to be a little careful.

So, what if this limit is equal to zero?

If the limit is equal to zero, then we say,

the radius of convergence is infinity, and if the kth root of the ak's goes to

infinity Then the rate is of convergence is 0.

So this is how this is supposed to be read.

Let's look at some examples again.

So here's that series that we had some trouble with using the ratio test,

z to the k divided k to the k.

So here the ak is where 1/k to the k, and if I look at the kth root

of ak, it's the kth Root of 1 / k to the k.

But that's just 1/k.

And as k goes to infinity, that goes to 0.

Since that goes to 0, now for the root test, we have to take 1 over the limit.

And if the limit is 0, then we said 1 over the limit is infinity.

So in this case, the radius of convergence is infinity.

We had seen that by actually finding the limit in the ratio test.

8:57

So the ak's are equal to k and

if I take the k through to ak, that's the kth root of k.

And it turns out the kth root of k actually goes to 1 as k goes to infinity.

And therefore, the radius of convergence is 1.

We saw this using the ratio test as well.

So, often it is the case that it's not just one test that works.

You can use multiple tests.

So both the ratio test and the root test worked in this case and

gave us that the radius of convergence is 1.

9:56

Let's write out a few terms of this series.

When k is equal to 0, I just get everything to the power of 0 so that is 1.

When k is equal to one I get -1 divided by 2,

so minus 1 half times z to the two k.

So z squared.

11:01

So what are my coefficients?

Well a2k can be even ones Are of the form (-1) to the k over 2 to the k.

So these are even coefficients, because I have a 2k here.

So this is my a2k, what I see right there,

because the name of the coefficient goes by the power of z.

Since the exponent of z is 2k.

This must be the 2k coefficient.

This is how the coefficients are labeled whereas the odd coefficients.

Well, they're not there, they're all equal to zero.

So, a2k+1 they're all equal to 0.

2k+1 is always an odd number.

So, these are all equal to 0 So now I want to take the kth root of ak.

But that seems to depend on whether I'm taking an even coefficient or

an odd coefficient.

So let's look at the 2kth root of a2k and the 2k+1st root a2k+1.

The 2kth root of a2k So I need to take the 2kth root

of the absolute value of (-1) to the k/2 to the k.

So I need to take the 2kth root of |(-1) to the k|/2 to the k.

So the numerator is just 1, so I need to take the 2kth root

of 1 over 2 to the k.

So that equals 1 over 2 to

the k divided by 2k so

that's 1 over 2 to the one-half.

So the 2kth root of a2k is 1/ 2 to the 1/2.

And the 2k + first root of a2k + 1,

well that That's equal to 0 because a2k + 1 = 0.

And so, if I look at the whole sequence,

the kth root of ak, that needs to include the even and the odd coefficients.

While for odd coefficients, I get all these 0's and for

even coefficients I get all these 1/root 2, 1/root 2, 1/root 2.

So this sequence bounces, between these two numbers, but then never

comes to a limit because it keeps bouncing back and forth between these two numbers.

So the sequence does not have a limit.

The sequence not having a limit means I cannot use the root test.

13:31

Note, ak over ak+1 has no limit either because again

If I have an even coefficient in

the numerator then we're going to have a zero in the denominator that's no good.

And if I have an odd coefficient in the numerator that's zero divided by something

non-zero that bounces backward to zero so that's not really good either.

So neither the root test nor the ratio test work in this case.

14:21

And so now the series s of series and w and power series of w looks like

a perfectly reasonable series because this whole even odd thing is now gone.

I can take the k root of the coefficients of this new series,

the least of the coefficients, the k root of the absolute value of that is the kith

root of one over two to the k so it's one half, sk goes to infinity

that remains one half Has a perfectly fine limit, and so the series in

the Ws converges for absolute value of W less than 1/1/2 which is 2.

So the root test works for the series in w.

And it converges for absolute value w less than 2.

15:04

But w is simply z squared.

So if the series converges when I plug in a w less than 2,

then it also converges when I plug in a z squared less than 2.

It's the same series.

So the original series therefore converges when the absolute value of z

squared is less than 2.

And that means the absolute value of z is less than root 2.

15:24

So the radius of convergence for this series is actually root 2.

But neither of my two formulas worked.

There's another formula that would find this radius and there is.

It's called the Cauchy Hadamard Formula.

The radius of convergence of the power series ak(z-z0)k equals

1 over the lim sup of the kth root of the absolute value of ak.

And this formula always works.

There's no if this exists.

The lim sup always exists.

And this always gives you the radius of convergence.

Just many people are not so comfortable with a lim sup, and therefore,

this is a less quoted result.

But this result always works.

So, if the root tests fail and the ratio test fails,

you can always use the Cauchy-Hadamard criterian, it always works.

Remember, the lim sup is just the greatest accumulation point of a sequence.

We had calculated the kth root of the a case.

The kth root of the a case was simply bouncing back and

forth between 0 and 1 / root 2, 0,

1 / root 2, 0, 0, 1 / root 2, and so forth.

16:44

And the lim sup is the greatest possible limit of a sub sequence basically.

If you pick this sub sequence that consisted of these 1 over square root of

2s, then the limit of that is 1 / square root of 2,

so R is 1 over that, and that equals root 2.

We will be using the lim sup again in this lecture series.

And so if you're not really comfortable with a lim sup, and

don't want to catch up on that right now, that's all right.

17:11

Now, what does all this have to do with analytic functions and complex analysis?

Remember, we had the theorem that sets the power series is analytic for

all those zs that are in the disk of convergence of the power series.

So that's nice, you can build analytic functions using power series.

18:07

And moreover, these a-ks can be calculated, so

I can calculate this power series.

The a-ks are exactly the kth derivative of f at z0 divided by k factorial.

This power series converges in some disk of convergence and

the radius of that disk is upper case r.

We can calculate it using the theorems from the previous pages.

An the r that you're going to get is at least the radius little r of the disk in

which f was analytic.

So in each disk, in which a function is analytic,

it has a power series representation.

Here's another disk that just barely fits into u from a different center.

And again f is going to have a power series representation there.

It's going to be a different power series than the one

that I have in the green disk.

Nonetheless, I have a representation as a power series.

So locally an analytic function is represented by a power series.

Let's look at some examples.

19:03

So here, again, as reminded you of the formula, f(z) can be written as a power

series in the disc where it's analytic and the a-ks can be calculated by

taking the kth derivative of f at the center, divided by k factorial.

19:17

Let's start with the function e to the z,

because its derivatives are really easy to calculate.

The kth derivative of f at z is also e to the z, and if we plug in z0 = 0,

then we know the kth derivative of f at z0 is e to the 0 and so that's 1, for all k.

So if I wanted to calculate these coefficients a-k,

the numerators are all going to be 1.

So a-k is going to be 1 / k factorial.

And therefore, I find that e to the z is the series of a-k,

which is 1 / k factorial, times z minus 0 to the k,

so z to the k, for all z in c.

Because the disk for the exponential function which is analytic

in the entire plane is the disk centered at the origin of radius infinity.

So I can make this disk as large as I want to, and the function is analytic in there,

so I might just as well choose the radius infinity.

Therefore this power series representation holds in the entire complex plane.

20:37

And therefore the a-ks, now all the numerators here are equal to e.

So a-k = e / k factorial, so that e to the z becomes a-k,

which is e / k factorial (z- z0), 1 minus z0 is 1, so (z- 1) to the k.

There would have been another way to see this.

Another way to see this would have been to write e to the z,

as e to the z- 1, that's not true, times e, but e to the z- 1,

I can use the power series representation that I already found.

So this is equal to this one e times, and

now I use my power series representation that I already found in the first example.

I plug in z- 1 for z right there, so I get the sum, k from 0 to infinity.

And instead of z, because this is true for all z, I can plug in (z- 1).

So (z- 1) to the k / by k factorial.

If I now bring the e inside, I get this representation,

e / k factorial times (z- 1) to the power k.

22:12

So f(z) = sine z.

If I evaluate that at 0 to get the 0th coefficient, f(0) = 0.

The first derivative is cosine z.

The first derivative at 0 is 1.

The second derivative is- sine z, at 0, that's 0.

Third derivative is- cosine z at 0 is -1.

Fourth derivative, we're back to sine z.

Fourth derivative at 0 is 0.

And now we're going to keep repeating because these derivatives keep cycling

through, sine, cosine,- sine,- cosine, and so the numerator of the a-k cycle through

0, 1, 0, -1, 0, 1, 0, -1, and the denominators are these k factorials.

So, I find a0 = 0 divided by 0 factorial.

a1 = 1 over 1 factorial.

a2 = 0.

a3 = -1 / 3 factorial.

a4 = 0.

23:11

a5 = 1 / 5 factorial, a6 is 0 and so forth.

All together, I find sine z is a0, this is 0, this a1,

that's my a2, this is a0, this is a3, a4, a5, so

all the even coefficients are actually equal to 0.

So I might just as well not write those down and get only odd powers of z and

they have alternating signs and the denominators keep getting bigger and

bigger there, the factorials of the odd numbers.

So I find z- z cubed over 3 factorial,

that's this term right here, + z to the 5th / 5 factorial,

then- z to the 7th / 7 factorial, and so forth.

If you wanted to write that as one formula,

it's the sum of only odd powers of z.

So, that's why we write the exponent as 2k + 1.

The factorial that I need is the same number, so 2k + 1 factorial, and

the numerator gives me an alternating sign, alternating between +1 and -1.

24:23

Now let's look at cosine z.

That's also analytic in the entire complex plane.

And if I choose the origin as the center of my power series, and

again I get a radius of convergence of infinity.

We could do the exact same analysis, just find all the derivatives at 0.

And then write down this series or

you could remember that cosine z is the derivative of sine z.

And a power series within its disk of convergence can be differentiated

term by term, I can pull the derivative to the inside of the series.

And therefore I find that it's the sum -1 to the k over 2k + 1

factorial of the derivative of z to the 2k + 1.

But the derivative of z to the 2k + 1 is (2k + 1) times z to the 2k.

25:49

So, I'm left with -1 to the k divided 2k factorial times Z to the 2k.

So this time, I only have even powers of Z, and they alternate and

sign again, and again I have to divide by a factorial.

So, FI one minus Z squared over 2 factor Plus either the fourth of

a 4 factorial minus Z to the six over six factorial and so forth.

That is the power series representation of Co Sine Z centered at the origin.

26:20

So this theorem actually implies that an analytic function is entirely

determined in a disk by all of its derivatives at the center of the disk.

Because the ak's can be calculated by taking the case derivative of f at

the center of the disk divided by k factorial.

And the ak is determined by power series.

It consists of ak times z- z0 to the k.

The ak is entirely determined the power series and

are entirely determined by the kth derivatives of f at the center.

And therefore, knowing all the derivatives of an analytic function at the center of

a disk determines the function.

27:02

A corollary of that observation is the following.

Suppose you have two functions, f and g.

That they are both analytic in this disk of radius r centered at some point z0.

So suppose you also know that all their derivatives agree at this one point,

z0 then the two functions must actually agree.

27:25

By having the same derivatives of 1 point of all orders,

not just the same first derivatives but the same derivatives of all orders.

If all derivatives agree at z0,

then no matter what z you plug in, the functions must agree there.

So f and g are actually exact same function.

So that's quite a powerful theorem.