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JOEL LEWIS: Hi.

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Welcome back to recitation.

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We've been talking about Taylor
series and different

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sorts of manipulations you can
do with them, and different

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examples of Taylor series.

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So I have an example here
that I don't think

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you've seen in lecture.

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So this is the Taylor series 1
plus 2x plus 3x squared plus

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4x cubed plus 5x to the fourth,
and, and so on.

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I'm going to tell you that this
is a Taylor series for a

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fairly nice function.

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And what I'd like you to do,
is try and figure out what

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that function is.

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Now, I'm kind of sending you
off onto this task without

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giving you much guidance, so let
me give you a little bit

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of a hint, which is that the
thing to do here, is you have

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a bunch of different tools
that you know how to

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manipulate Taylor series by.

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So you know how to, you know, do
calculus on Taylor series,

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you can take derivatives and
integrals, you know how to add

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and multiply and perform these
sorts of manipulations.

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So you might think of a
manipulation you could perform

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on this Taylor series that'll
make it a simpler expression,

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or something you're already
familiar with, or so on.

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So let me give you some
time to work on that.

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Think about it for a while.

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Try and come up with
the expression

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for this Taylor series.

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And come back, and we can
work on it together.

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So hopefully you had some luck
working on this problem, and

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figured out what the right
manipulation to use is.

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I didn't, I kind of tossed it at
you without a whole lot of

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guidance, and I don't think
you've done a lot of examples

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like this before.

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So it's a little tricky.

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So what I suggested was that
you think about things that

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you, you know how to
do that could be

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used to simplify this.

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So looking at it, it's just
one Taylor series.

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So it's not clear that sort of
Taylor series arithmetic is

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going to help you very much.

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It's not obviously a
substitution of some value

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into some other Taylor series
that you're familiar with.

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It's not obviously, say, a sum
of two Taylor series that you

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already know very well.

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So those things don't, aren't
immediately, it's not clear

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where to go with them.

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One thing that does pop
out-- well, OK.

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So let's talk about
some of the other

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tools that I mentioned.

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We have calculus on
Taylor series.

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So we have differentiation.

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And if you take-- so we see here
that the coefficients are

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sort of a linear polynomial.

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If you take a derivative, what
happens is, well, this becomes

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2 plus 6x plus 12x squared
plus 20x cubed.

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And those coefficients, 2, 6,
12, 20, those are given by a

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quadratic polynomial.

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So that makes our life kind of
more complicated, somehow.

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But if we look at this, we see
that integrating this power

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series is really easy to do.

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This power series has a really
nice antiderivative.

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So what I'm going to do, is I'm
going to call this power

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series by the name f of x.

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And then what I'd like you
to notice, is that the

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antiderivative of f of x dx--
well, we can integrate power

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series termwise.

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And so what we get is--
well, so all right.

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So I'm going to do, I'm going
to put the constant of

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integration first. So
the antiderivative

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of this is c plus--

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well, 1, you take its integral
and you get x.

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And 2x, you take its
integral, and you

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just get plus x squared.

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And 3x squared, you take its
integral and you get plus x

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cubed, and plus x to the fourth
from the next one, and

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plus x to the fifth,
and so on.

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So, OK.

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So I put the c here, right?

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So any power series of this form
is an antiderivative of

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the power series that
we started with.

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Any power series of this form
has a, has the derivative

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equal to thing that we're
interested in.

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And since we really care about
what f is, and not what its

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antiderivative is, we
can choose c to be

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any convenient value.

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So I'm gonna, in particular,
look at the power series where

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c is equal to 1.

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And why am I going to
make that choice?

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Well, because we've seen a power
series that looks very

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much like this before,
with the 1 there.

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So we know 1 plus x plus x
squared plus x cubed plus x to

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the fourth and so on.

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We know that this is equal
to 1 over 1 minus x.

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OK?

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So since we know that this is
the case, that means that our

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power series, f of x, is just
the derivative of this.

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That's what we just
showed here.

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So, f of x is equal to d over
dx of 1 over-- whoops, that

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should be--

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over 1 minus x.

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OK?

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And notice that, you know if I'd
chosen a different choice

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of constant here, it would
be killed off by this

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differentiation.

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So it really was irrelevant.

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So d over dx of 1
over 1 minus x.

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Yes, of d over dx of 1 minus x.

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And, OK.

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Well, this is an easy derivative
to compute.

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This is 1 minus x to the minus
1, so you do a little chain

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rule thing, and I think what you
get out is that this is 1

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over 1 minus x squared.

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So this gives me a nice
formula for this

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function f of x.

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If you wanted to check, one
thing you could do, is you

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could set about computing a few
terms, the power series

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for this function, for 1 over
1 minus x squared, 1 over 1

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minus x quantity squared
I should say.

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So you could do that either by
using your derivative formula,

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which is easy enough to do.

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Another thing you could do, is
you could realize that this is

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1 over 1 minus x times 1 over
1 minus x, so you could try

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multiplying that polynomial by
itself and that power series

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by itself, and see if it's easy
to get back this formula

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that we had.

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But any of those ways is a good
way to double check that

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this is really true, that this
function, that this power

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series here, f of x, has
this functional form.

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Now of course, I haven't said
anything about the radius of

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convergence, but the thing to
remember when you're doing

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calculus on power series,
is that when you take a

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derivative or an antiderivative
of a power

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series, you don't change the
radius of convergence.

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Sometimes you can fiddle with
what happens at the endpoints,

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but the radius of convergence
stays the same.

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So this is going to be true
whenever x is between

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negative 1 and 1.

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And in fact, this series
diverges at this endpoint.

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That's pretty easy to check.

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So there we go.

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So this is the functional form
for this power series.

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It's valid whenever x is between
negative 1 and 1.

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That's its range
of convergence.

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And so here we have a cute
little trick for figuring out

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some forms of power series.

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And of course if you were
interested, so like I said,

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now you had a, ypu could look
at the derivative of this,

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which I mentioned had some
coefficients that were given

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by some quadratic polynomial.

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Now that you have a functional
form, you could figure out,

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you know, "Oh what, you know,
what is the function whose

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power series has that quadratic
polynomial as

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coefficients?" And you can do
a whole bunch of other stuff

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by similarly taking derivatives
of other power

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series that you're familiar
with, or integrals.

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So I'll stop there.

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