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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 sorts

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of manipulations you
can do with them,

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and different examples
of Taylor series.

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So I have an example
here that I don't

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think you've seen in lecture.

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

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

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

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nice function.

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

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what that function is.

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Now, I'm kind of sending
you off onto this task

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without giving
you much guidance,

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so let me give you a
little bit of a hint, which

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is that the thing
to do here, is you

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have a bunch of different
tools that you know how

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to 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,

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you know how to add and
multiply and perform

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these sorts of manipulations.

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So you might think
of a manipulation

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you could perform on this
Taylor series that'll

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make it a simpler expression,
or something you're already

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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 for this Taylor

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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,

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and 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

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of 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

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that you know how to do that
could be 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

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is going to help you very much.

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It's not obviously
a substitution

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of some value into some
other Taylor series

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that you're familiar with.

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

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that you already know very well.

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So those things don't,
aren't immediately,

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it's not clear 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 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,

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well, this becomes
2 plus 6x plus 12x

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squared plus 20x cubed.

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And those coefficients,
2, 6, 12, 20, those

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are given by a
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

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to call this power series
by the name f of x.

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And then what I'd
like you to notice,

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is that the
antiderivative of f of x

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dx-- well, we can integrate
power 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

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the constant of
integration first.

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So the antiderivative of
this is c plus-- well, 1,

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you take its integral
and you get x.

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And 2x, you take its
integral, and you just

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get plus x squared.

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And 3 x squared, you
take its integral

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and you get plus x
cubed, and plus x

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

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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

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of the power series
that we started with.

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Any power series of this
form has the derivative equal

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to thing that we're
interested in.

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And since we really
care about what f is,

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and not what its
antiderivative is,

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we can choose c to be
any convenient value.

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So I'm gonna, in
particular, look

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at the power series
where 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

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that looks very much like
this before, with the 1 there.

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

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plus x to 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,

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that means that our
power series, f of x,

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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

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over-- whoops that should
be 1 over 1 minus x.

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OK?

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And notice that,
you know, if I'd

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chosen a different
choice of constant here,

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it would be killed off
by this 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,

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so you do a little
chain rule thing,

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and I think what you get
out is that this is 1 over 1

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minus x squared.

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So this gives me a nice formula
for this function f of x.

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

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is you could set about
computing a few terms, the power

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

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1 over 1 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

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could realize that this is 1
over 1 minus x times 1 over 1

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minus x, so you
could try multiplying

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that polynomial by itself and
that power series by itself,

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and see if it's easy to get
back this formula that we had.

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But any of those
ways is a good way

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to double check
that this is really

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true, that this function, that
this power series here, f of x,

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has this functional form.

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Now of course, I
haven't said anything

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about the radius of
convergence, but the thing

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to remember when you're doing
calculus on power series,

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is that when you take a
derivative or an antiderivative

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of a power series,
you don't change

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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

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x is between 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-- you 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

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is the function
whose power series

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has that quadratic
polynomial as coefficients?"

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And you can do a whole
bunch of other stuff

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by similarly taking derivatives
of other power series

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that you're familiar
with, or integrals.

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So I'll stop there.