PROFESSOR: Welcome back to recitation.
In this video I want to finish up our work with the ratio
test, and in particular in this video, I'd like to
address something that Professor
Jerison talked about.
In, when he was talking about these Taylor series, he was
talking about the radius of convergence at some point.
And he didn't go into it very specifically, but he was
essentially saying for some values of x-- you'll have a
series that has x to the n in it-- for some values of x it
will converge, and for some values it will diverge.
So let me just remind you of something you already know.
So, if we consider the series x to the n.
Right?
This is the geometric series and we know that this is equal
to 1 over 1 minus x if the absolute value of x
is less than 1.
Right?
And so we know that this series is finite.
This sum converges.
To whatever value, If I plug in x here, it converges when
absolute value of x is less than 1.
And it diverges when absolute value of x is bigger than 1.
And we're not going to address when the absolute
value of x equals 1.
We won't address that case.
But, the point I want to make it is that for some values
this series converges, and for some
values this series diverges.
And those values can give us the radius of convergence.
So the radius of convergence of this series is actually 1,
because x goes from 0 up to 1, and then from 0 down to 1.
If you think about it, radius might be a confusing term, but
can think about it as a circle and one dimension less than
maybe you usually think about it as a circle.
If this is the number line, we're going from 0 up to 1 and
down to minus 1.
So you're going in one direction up to 1 and one
direction down to minus 1.
So this is radius 1.
r is equal to 1.
So what I want to do is figure out how I can use the ratio
test to tell me the radius of convergence of other series,
other power series, besides this one.
OK?
So we're really interested in for what values of x do these
power series converge.
OK?
And how we're going to do that, is we're going to
directly use the ratio test. And one thing that I mentioned
in the other case, in the other ratio test video, is I
said you want to have your terms be positive.
And so just to make this easy on ourselves, when we do the
ratio test we're just going to take the absolute
value of the ratio.
And that will be sufficient for our purposes to determine
the radius of convergence.
I don't want to go into anything more
complicated than that.
So we'll see as we do these examples, we're just gonna,
we're just gonna take the absolute value of the ratio.
So let's actually start off right away
with doing some examples.
And each time I do an example, I'm asking
the following question.
Actually, I was going to say for what values of x, but
that's not quite true.
I just want to know what is the radius of convergence for
each power series?
So I'm not going to write it up every time, but this is the
question we want to be thinking about.
So I'm going to write down some power series and we're
going to see what is the radius of
convergence for each of those.
I was going to write find the values of x for which the
power series converge, but that's not quite true because
sometimes it will actually converge on one
or both of the endpoints.
We're not going to deal with that one.
But sometimes it will converge on one or the
other of the endpoints.
So I'm going to be asking the question, find the radius of
convergence.
So let's do an example, and I'll show you how it applies
to the ratio test. Then we'll go from there and do some
other examples.
So let's consider the series x over 2 to the n.
This will be an easy one.
Now the ratio test told us that we
examine a certain limit.
We examine the limit as n goes to infinity.
It said a sub n plus 1 over a sub n.
Well if we think of this whole thing as a sub n, then a sub n
plus 1 is x over 2 to the n plus 1.
And then we have to divide by x over 2 to the n.
So let me just make sure we understand that.
The way we, the way we want to think about this is as if the
a sub n is actually a function of x that also depends on n.
OK, so a sub n, we think of it as x over 2 raised to the n.
So a sub n plus 1 is x over 2 raised to the n plus 1.
a sub n is x over 2 raised to the n.
OK, and so this looks exactly like the ratio test we had
before, I told you I'm going to put absolute values on it.
And so let me simplify this.
I put the division sign maybe kind of funny 'cause I wanted
to have it in a row there.
I apologize if that made it more confusing.
So this equals, well I have x to the n plus 1 over x to the
n times 2 to the n-- this come to the numerator because it's
in the denominator of the denominator--
over-- and this goes to the denominator--
2 to the n plus 1.
Now what is this?
This whole thing converges to what?
2 to the n over 2 to the n plus 1 simplifies to 1/2, and
x to the n plus 1 over x the n simplifies to x.
So this is equal to absolute value of x over 2.
Now we haven't shown how the ratio test is going to help us
find the radius of convergence.
And here's where we're going to see it working.
If we want to know that, we want to draw a conclusion
about the convergence of this series.
We know that if this limit is less than 1, that the series
definitely converges.
So if I take my limit when I'm all done, and I set it less
than 1, and I solve for absolute value of x, then that
will tell me exactly what the radius of convergence is.
Let me go through that one more time.
We knew that if we had this series in terms of a sub n,
and we looked at the ratio and we took the limit of these
ratios, and in the limit the value was will strictly less
than 1, we know the series converges.
So now we have this thing in terms of x, but ultimately
it's the same process.
And we get to a place where we know the limit.
It depends on x.
And so the limit will be less than 1, exactly where this
thing is less than 1.
Right?
So it's now in terms of x.
This less than 1 I'm putting in at the end.
This is where the ratio test is happening.
So what does this tell us?
Where does this converge?
It converges when absolute value of x over 2 is less than
1, which means absolute value of x is less than 2.
Right?
I just multiply by 2 'cause absolute value of 2 is 2.
I don't have to worry about if 2 is negative.
So the absolute value of x is less than 2.
That tells me the radius of convergence is actually 2.
Now this shouldn't surprise us really.
Because this is, in fact, a geometric series, and we know
that the usual geometric series converges when the
absolute value of x is less than 1.
So it's not surprising, and actually fits in with what we
already know, that if this thing on the inside, if its
absolute value is less than 1, that the thing will converge.
So that that means absolute value of x has to
be less than 2.
So again, hopefully this fits in with the knowledge we
already have. Hopefully it makes sense to you, what we're
trying to do.
And so let's do another couple of examples to make this, make
this solidify.
All right what were my other examples?
OK, let's do this one.
x to the n over n factorial.
All right, so now our a sub n, when we look at this
geometric-- geometric? this is not geometric-- when we look
at the series, our a sub n is going to be x to
the n over n factorial.
And so when we take our whole limit, the thing that's going
to happen is we're going to have something in terms of x
at the end.
Our goal is to find for what values of x is that thing we
have less than 1?
Again this is where we're using the ratio test.
So let's look at this thing.
We want to look at the limit as n goes to infinity.
Well, a sub n plus 1 is going to be x to the n plus 1 over
the quantity n plus 1 factorial.
I'm going to put the absolute value over there.
Now I'm going to multiply it by 1 over a sub n.
So this is actually a sub n, so 1 over that is n factorial
over x to the n.
Now we have to be a little careful here.
I want to make sure everybody understands something.
n factorial is n times n minus 1 times n minus 2 times n
minus 3 all the way down.
n plus 1 factorial is n plus 1 times n times n minus 1 times
n minus 2 all the way down.
So n plus 1 factorial--
I'll just write this here-- n plus 1 factorial equals n plus
1 times n factorial.
That's very important that you recognize that OK?
Because the division of factorials is a little more
complicated than the division of straight polynomials or
something like this.
Alright?
Because when I take this limit, what am i going to get?
I'm going to get x to the n plus 1 over x to the n.
That's going to give me 1x.
And then n factorial divided by n plus 1 factorial, well
based on the fact that this is equal to n plus 1 times n
factorial, the n factorials divide, and I'm left with x
over n plus 1.
That's actually what this limit equals.
Let me go through that one more time.
This part is easy, x to the n plus 1 over x to the m
gives me the x.
And then n factorial over n plus 1 factorial is just 1
over n plus 1.
And I have be careful because I wrote equals, but there's
still an n here.
So let me erase that.
I'm not gonna, I don't think, my studio audience didn't say
anything yet, but I don't think they were going to let
me get away with that.
This is a limited as n goes to infinity of x over n plus 1.
Now what is that?
Well as n goes to infinity, for any fixed x I pick--
we have to be careful here-- for any fixed x I pick as n
goes to infinity, this quantity is equal to 0.
OK?
If I were moving x around, if I were moving x with n, this
will be a problem, but x is fixed.
When I do this sum, I fix my x at the beginning.
So for any fixed x n plus 1 is getting arbitrarily large.
So x over n plus 1 is getting arbitrarily small.
So this limit is 0.
This is strictly less than 1.
What does this mean?
For any x I pick, this whole thing is less than 1.
Right?
Any fixed x, this ratio is always less than 1.
What does that mean?
That means the radius of convergence is infinite.
So the radius is infinity.
OK.
The radius of convergence is actually infinite.
What series is this?
This is actually the Taylor series for e to the x.
So we know that the Taylor series for e to the x that
this converges for any x.
That's a nice thing.
So now we've used the ratio test to tell us how the Taylor
series behaves for a function that we know.
OK, we know its radius of convergence is infinite.
So that's pretty nice.
All right let's do maybe one or two more examples,
depending on how much room I have.
OK, let's try this one.
x to the n over n times 2 to the n.
All right, so this is another power series we have. And we
want to know if I want to plug in some value of x and I want
to take the sum, will that converge for that particular
value of x?
We want to know what values of x can I plug in.
All right so again, we're going to look at the radius
of, find the radius of convergence based on the ratio
test. So the n plus first term is x to the n plus 1 over n
plus 1 2 to the n plus 1.
And then I have to multiply by the a sub m's, the a sub n
term, or multiply by 1 over that, sorry.
I'm dividing by a sub n, so I get n times 2 to the n
over x to the n.
All right, this gives me an x.
n plus 1 over n, let me just write out actually what we get
here, 2 to the n over 2 to the n plus 1 gives me an over 2.
and then n over n plus 1.
This is positive, so I don't have to worry
about anything there.
The limit as n goes to infinity absolute value of x
over 2 times n over n plus 1, what's that equal to?
Well this, as n goes to infinity is equal to 1, so
it's absolute value of x over 2.
Let me again remind you what we are doing.
We're saying I want to know the radius of convergence for
this series.
So I want to know what radius, so what values of x can I put
in to make this series converge?
I might miss the endpoints, but beyond that what values of
x make this series converge?
So I'm looking at the ratio test. I took the ratio test, I
got it all the way to a place where I have something in
terms of x.
As long as that thing is less than 1, I'm golden.
As long as that thing is less than 1, the series converges.
So again, I actually get another thing where the
absolute value of x is less than 2.
Alright?
I probably should have picked a different number there.
What do you think would happen if this was a 7?
Well everything would've been the same except this would
have been a 7, and the radius of convergence
would have been 7.
So I should have picked a different number there, so we
had a different radius of convergence, but you can see
how that works.
And the, I'm gonna do just, I have a room, I'm going to do
one more example.
'Cause this one's a good one to do also.
OK, I'm actually going to do another series that we know, x
to the 2n over 2n factorial.
I'm doing this one for a particular reason to help us
deal with when some things, when the exponents and the
factorials get a little more complicated.
I want to also point out what series is this.
You should know what series this is.
And I know, without thinking, that it's
either sine or cosine.
I guess I should say it's starting at
n equals 0 to infinity.
And then I might get nervous and say well, which one is it?
Well what's the first term?
x to the 0 is 1 over 0 factorial is 1.
So it looks like the first term is 1, so that makes me
know it's cosine.
Right?
If the first term were x I'd know it was sine.
So if I get nervous, so this, where it converges, is equal
to cosine x.
Let's look at this one.
All right, we got, now this you have to be a little more
careful because it's 2 times the quantity n plus 1 over 2n
plus 2 factorial.
So that's n plus 1 times 2 is 2n plus 2 times 2n factorial
over x to the 2n.
All right this is where it might get a little tricky.
This is 2n plus 2 divided by 2n.
So this is going to be the limit as n goes to infinity of
x squared, the absolute value squared which is just x
squared again, times this thing.
So let's figure out what this thing is.
OK?
What is 2n factorial?
I'm going to write out the first couple terms. OK, that's
going to be 2n times 2n minus 1 times, that's all
the way down to 1.
And then what's this?
This is 2n plus 2 times 2n plus 1 times 2n times, all the
way down to 1.
So we see, this is what I was talking about earlier actually
also, was that I've added 2 to this.
So it's not surprising I get 2 more terms, and then I'm down
to 2n factorial again.
Right?
So this 2n factorial.
This is 2n factorial.
So the 2n factorial here divides with the 2n factorial
here, and I'm left with these here.
I get 1 over 2n plus 2 times 2n plus 1.
Now what happens as n goes to infinity?
Obviously this ratio goes to 0.
So I'm actually in another case similar to the one I saw
with e to the x.
Is that this goes to 0.
So the limit is actually equal to 0, which is
always less than 1.
Any value of x I pick is less than 1.
And so this series actually converges for
any x that I pick.
So this is another case where I have the radius of
convergence is actually infinite.
OK, so we'll probably give you some problems where you have
some other things happening.
'Cause you can actually get the radius convergence is 0.
You can actually get that the only place it converges is at
x equals 0.
So you might actually get the limit as n goes
to infinity is infinity.
I didn't give you an example of that, but that's a case
where whatever x value you put in, your limit is
still bigger than 1.
Then you would always get a diverges, except when x is 0.
OK, so that's another thing that you can run into.
But, the point I want to make is that this is going to let
us determine, you know, at least the radius over which
these x values converge.
And it also helps us if we know certain things about a
function to tell us where this function is actually equal to
the series that we're dealing with.
So I think that's where I'll stop, and I hope this was
informative.