And now I look at this function, f(z) is given by the sum ak(z-

z0) to the k where z varies.

Now let z vary inside the disk of convergence.

And the claim is that the outcome is an analytic function

in that disk of convergence.

And more is true.

An analytic function has a derivative and

I can find that derivative by differentiating the series term by term,

so I can pull the derivative inside of this infinite sum.

We know that we can pull derivatives inside of finite sums.

But, for an infinite sum, that's a whole different matter.

So, that's something that needs to be proved.

But, it can be proved that, for this infinite sum,

which converges inside the disc of radius R.

I can differentiate inside the infinite sum and

therefore I find that f'(z) is ak(z- z0)k

which is k(z- z0) to the k- 1st power.

Notice I started the series at 1 here because when k = 0,

the derivative of that term goes 0, it goes away.

The second derivative can be found similarly.

I just take another derivative inside so

I find ak times k(z- z0) k- 1 to the k- 2.

And now to find to start the sum of 2.

In particular, if I look at the nth derivative of f, that's z.

So I keep taking derivative.

The nth derivative is going to be

something like the sum k from n to infinity and

then here I have ak times k(k- 1) and

so forth, through k- n + 1,

times z- z0 to the k- n.

When I plug in z0 for

z right there.

That makes this z as z0.

So as soon as this power k- n is at least 1.

The term vanishes because I have 0 to some positive power which is 0.

Only when this exponent is equal to 0, do I get a 1.

So when the exponent is equal to 0, that's the case when k is equal to n.

So the first term of my power's here.

The nth derivative of f at z0 is the first term of the power series and

that's going to be an times n, times n- 1 and so

forth all the way through n- n which is 0.

And so I just, all left with a 1.

That's the factorial of n, so that's an times m factorial.

I wrote this out back here again, the kth derivative, so

I can put it in terms of k instead of in terms of n.

The kth derivative of f at z0, is ak times k factorial and I can solve that for ak.

ak is there for the kth derivative of f at z0 divided by k factorial for all ks.

So that is amazing.

What this says that not only do I get an analytic function