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.
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 in 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 comes 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
strictly less than 1,
we know the series converges.
So now we have this thing.
It's 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.
All right?
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 one x.
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 n 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 the limit 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 n'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.
All right?
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 then, I'm gonna do just--
I have 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 of x 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.
