WEBVTT
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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.