JOEL LEWIS: Hi.
Welcome back to recitation.
We've been talking about Taylor series and different
sorts of manipulations you can do with them, and different
examples of Taylor series.
So I have an example here that I don't think
you've seen in lecture.
So this is the Taylor series 1 plus 2x plus 3x squared plus
4x cubed plus 5x to the fourth, and, and so on.
I'm going to tell you that this is a Taylor series for a
fairly nice function.
And what I'd like you to do, is try and figure out what
that function is.
Now, I'm kind of sending you off onto this task without
giving you much guidance, so let me give you a little bit
of a hint, which is that the thing to do here, is you have
a bunch of different tools that you know how to
manipulate Taylor series by.
So you know how to, you know, do calculus on Taylor series,
you can take derivatives and integrals, you know how to add
and multiply and perform these sorts of manipulations.
So you might think of a manipulation you could perform
on this Taylor series that'll make it a simpler expression,
or something you're already familiar with, or so on.
So let me give you some time to work on that.
Think about it for a while.
Try and come up with the expression
for this Taylor series.
And come back, and we can work on it together.
So hopefully you had some luck working on this problem, and
figured out what the right manipulation to use is.
I didn't, I kind of tossed it at you without a whole lot of
guidance, and I don't think you've done a lot of examples
like this before.
So it's a little tricky.
So what I suggested was that you think about things that
you, you know how to do that could be
used to simplify this.
So looking at it, it's just one Taylor series.
So it's not clear that sort of Taylor series arithmetic is
going to help you very much.
It's not obviously a substitution of some value
into some other Taylor series that you're familiar with.
It's not obviously, say, a sum of two Taylor series that you
already know very well.
So those things don't, aren't immediately, it's not clear
where to go with them.
One thing that does pop out-- well, OK.
So let's talk about some of the other
tools that I mentioned.
We have calculus on Taylor series.
So we have differentiation.
And if you take-- so we see here that the coefficients are
sort of a linear polynomial.
If you take a derivative, what happens is, well, this becomes
2 plus 6x plus 12x squared plus 20x cubed.
And those coefficients, 2, 6, 12, 20, those are given by a
quadratic polynomial.
So that makes our life kind of more complicated, somehow.
But if we look at this, we see that integrating this power
series is really easy to do.
This power series has a really nice antiderivative.
So what I'm going to do, is I'm going to call this power
series by the name f of x.
And then what I'd like you to notice, is that the
antiderivative of f of x dx-- well, we can integrate power
series termwise.
And so what we get is-- well, so all right.
So I'm going to do, I'm going to put the constant of
integration first. So the antiderivative
of this is c plus--
well, 1, you take its integral and you get x.
And 2x, you take its integral, and you
just get plus x squared.
And 3x squared, you take its integral and you get plus x
cubed, and plus x to the fourth from the next one, and
plus x to the fifth, and so on.
So, OK.
So I put the c here, right?
So any power series of this form is an antiderivative of
the power series that we started with.
Any power series of this form has a, has the derivative
equal to thing that we're interested in.
And since we really care about what f is, and not what its
antiderivative is, we can choose c to be
any convenient value.
So I'm gonna, in particular, look at the power series where
c is equal to 1.
And why am I going to make that choice?
Well, because we've seen a power series that looks very
much like this before, with the 1 there.
So we know 1 plus x plus x squared plus x cubed plus x to
the fourth and so on.
We know that this is equal to 1 over 1 minus x.
OK?
So since we know that this is the case, that means that our
power series, f of x, is just the derivative of this.
That's what we just showed here.
So, f of x is equal to d over dx of 1 over-- whoops, that
should be--
over 1 minus x.
OK?
And notice that, you know if I'd chosen a different choice
of constant here, it would be killed off by this
differentiation.
So it really was irrelevant.
So d over dx of 1 over 1 minus x.
Yes, of d over dx of 1 minus x.
And, OK.
Well, this is an easy derivative to compute.
This is 1 minus x to the minus 1, so you do a little chain
rule thing, and I think what you get out is that this is 1
over 1 minus x squared.
So this gives me a nice formula for this
function f of x.
If you wanted to check, one thing you could do, is you
could set about computing a few terms, the power series
for this function, for 1 over 1 minus x squared, 1 over 1
minus x quantity squared I should say.
So you could do that either by using your derivative formula,
which is easy enough to do.
Another thing you could do, is you could realize that this is
1 over 1 minus x times 1 over 1 minus x, so you could try
multiplying that polynomial by itself and that power series
by itself, and see if it's easy to get back this formula
that we had.
But any of those ways is a good way to double check that
this is really true, that this function, that this power
series here, f of x, has this functional form.
Now of course, I haven't said anything about the radius of
convergence, but the thing to remember when you're doing
calculus on power series, is that when you take a
derivative or an antiderivative of a power
series, you don't change the radius of convergence.
Sometimes you can fiddle with what happens at the endpoints,
but the radius of convergence stays the same.
So this is going to be true whenever x is between
negative 1 and 1.
And in fact, this series diverges at this endpoint.
That's pretty easy to check.
So there we go.
So this is the functional form for this power series.
It's valid whenever x is between negative 1 and 1.
That's its range of convergence.
And so here we have a cute little trick for figuring out
some forms of power series.
And of course if you were interested, so like I said,
now you had a, ypu could look at the derivative of this,
which I mentioned had some coefficients that were given
by some quadratic polynomial.
Now that you have a functional form, you could figure out,
you know, "Oh what, you know, what is the function whose
power series has that quadratic polynomial as
coefficients?" And you can do a whole bunch of other stuff
by similarly taking derivatives of other power
series that you're familiar with, or integrals.
So I'll stop there.