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Radius and center.

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Well thus far, we've been considering

power series that are centered around zero.

By which I mean, we've been looking at power series look like this.

The sum, n goes from zero, say to infinity.

Of C sub n times just X to the n.

But we can center our power series around

some other point around some other point a.

So instead of writing this.

Let's say I want to

write down a power series that's centered around

some other point say a. Well then I'd write my power series like

this. The sum n goes from 0 to infinity of C sub

n again, but now times, not just x to the n, but x

minus a to the nth power. In this case, the interval of convergence

is centered around a as well.

Well, let's find an interval on which this series converges.

And the goal is to show that interval is really centered around a so

it legitimately makes sense to think of this power series as centered around a.

And to find the interval on which this power

series converges I am going to use the ratio test.

So let's, let's do that now. So how does the ratio test work here?

Well, it asks me to look at the

limit as n approaches infinity, of the n plus first term, c sub n plus

1 times x minus a to the n plus first power, divided by the nth term.

Which is c sub n times x minus a to the n.

And, I've got absolute value bars here because I'm really

applying the ratio test to the sum of the absolute values.

I'm using it to analyze absolute convergence.

Okay, well, what's this limit?

And well, I can simplify this limit

a bit.

I could write this as the limit, n goes to infinity, let me

pull off the C sub n plus 1 and the C sub n.

That'll be the fraction C sub n plus 1, over C sub n.

And in here I've got n plus 1 copies of x minus a.

Here I've got n copies of x minus a. Most of those will cancel.

All that I'm left with is just one copy of x minus

a in the numerator, so I'll write times absolute value of x

minus a.

And the ratio test tells me that if this is less than 1, then the

original series converges absolutely, and if this is

bigger than 1, then the original series diverges.

2:28

Well, how can I really analyze this.

I don't really know anything about C sub n plus 1 and C sub n.

But let's just pretend, right?

Let's just pretend that the limit just of this C sub

n plus 1 over C sub n ends up being 1

over R.

And in this part here, the absolute value

of x minus a, that's just a constant anyway.

So I'll write times absolute value x minus a.

I'm wondering when is that less than 1.

Well, I mean I'm just playing make believe here.

R is some positive number, I'll multiply both sides of this inequality by R.

So it's positive, doesn't affect the inequality at all.

Now I've got the absolute value x minus a

is less than R after multiplying through by R.

Well, what does this mean?

This means the distance between x and a is less than R.

Well, that really means that x is between a minus R and a plus R.

And then, at least when x is between a minus R

and a plus R, I know that the series converges absolutely.

I don't know what happens

at the end points, but the big deal here is that look, this thing

legitimately looks like an interval centered at the point a with radius R.

In what's to come, I'll usually be talking about power series

that are centered around zero, but it's important to realize that

we can talk about power series That are centered around some

other point and there are certainly cases where that'll come in handy.

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