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Hi.

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Welcome back to recitation.

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In class, Professor Jerison
and Professor Miller have

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taught you a little bit about
Taylor series and some of the

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manipulations you can do with
them, and have computed a

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bunch of examples for you.

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So I have three more examples.

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Here of functions whose Taylor
series are nice to compute.

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So the first one is cosh x.

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That's the hyperbolic cosine.

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So just to remind you, this can
be written in terms of the

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exponential function as e
to the x plus e to the

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minus x over 2.

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The second one is the function 2
times sine of x times cosine

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of x, just your regular
sine and cosine here.

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And the third one is x times the
logarithm of the quantity

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1 minus x cubed.

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So why don't you pause the
video, take some time to work

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out the Taylor series for these
three functions, come

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back, and we can work
them out together.

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So here we have three functions
whose Taylor series

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we're trying to compute.

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Let's start with the first
one and go from there.

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So this first one is the
hyperbolic cosine that's given

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by the formula e to the x plus
e to the minus x over 2.

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So there are a couple different
ways you could go

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about this one.

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This is actually, the hyperbolic
cosine is very

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susceptible to the method of
just using the formula that

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you have. So if you remember,
the derivative of the

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hyperbolic cosine is the
hyperbolic sine.

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The derivative of the hyperbolic
sine is the

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hyperbolic cosine again.

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So this function has very
easy-to-understand

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derivatives, which you can see,
you know, just by looking

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at its formula.

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It's easy to understand,
because the exponential

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function has very simple
derivatives, and e to the the

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minus x also has very
simple derivatives.

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So you could do it like that.

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The other thing you could do,
is that you already know the

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Taylor series for e to the x.

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And I believe you've also seen
the Taylor series for e to the

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minus x, and even if you
haven't, you can figure it out

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just by substitution.

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So if you remember, so e to the
x is given by the sum from

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n equals 0 to infinity of x
to the n over n factorial.

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I'm going to pull the
1/2 out in front.

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And e to the minus x is given by
the same thing, if you put

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in minus x for x.

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So it's n equals 0 to infinity,
so that works out to

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minus 1 to the n x to the
n or n factorial.

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Now, when you add these two
series together, what you see

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is that when n is even, over
here, you have x to the n over

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n factorial, and over
here, you have x to

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the n over n factorial.

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So what you get is, well, you
get 2 x to the n over n

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factorial, and then you multiply
by a half, so you

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just get x to the n
over n factorial.

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When n is odd, here you have x
to the n over n factorial, and

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here you have minus 1 x to
the n over n factorial.

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So you add them and you get 0.

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So what happens is that this
series looks just like the

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series for e to the x, except
the odd terms have died off.

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So we're left with just 1 plus
x squared over 2 factorial

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plus x to the fourth over 4
factorial plus x to the sixth

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over 6 factorial, and so on.

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And if you wanted to write this
in summation notation,

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you could write it as the sum
from n equals 0 to infinity of

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x to the 2n over 2n factorial.

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So this is the Taylor
series for the

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hyperbolic cosine function.

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Also, if you wanted, say, the
hyperbolic sine function, you

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could do something very similar,
or you could remember

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that the hyperbolic sine is
the derivative of the

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hyperbolic cosine, and just take
a derivative right from

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this expression.

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One other thing that you should
notice is that this

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looks very similar to the
expression of the Taylor

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series for cosine of x.

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So more of our sort of funny
coincidences between regular

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trig functions and hyperbolic
trig functions.

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All right.

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That's the first one.

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How about the second one?

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So here we have just some
regular trig functions.

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We have 2 sine x cosine x.

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Let me see where I've
got some space.

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I can do it right here.

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Let me box off a little
space for myself.

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So 2 sine x cosine x-- there
are a couple different ways

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you could proceed with
this function.

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So one is, you know the Taylor
series for sine x

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and cosine x already.

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So if all you wanted was a few
terms of this Taylor series,

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one natural thing to do would be
to take the series for sine

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x, take the series for cosine x,
multiply them together like

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you would multiply polynomials,
and what you

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would get is the Taylor
series for this

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expression, for this function.

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That's one way to proceed.

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That works perfectly well.

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Another thing you could do,
is you could try taking

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derivatives.

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You could have a situation where
every time you take a

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derivative, you apply
product rule.

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It's going to get more
and more complicated.

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It still works.

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It's a little complicated to do
it that way, if you wanted

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more than just a few terms.

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The other thing you could do,
is you could remember your

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trig identities.

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So if you look at this
expression, this should be

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familiar to you, because
it's just sine of 2x.

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So once you realize that this is
sine of 2x, there's a much,

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much shorter path available to
you, which is that you already

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know the Taylor series for sine
of x, so what you can do,

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is you can just plug in 2x
into that Taylor series.

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So sine of x is, well so OK,
x, so in this case, that's

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going to be 2x, then minus, so
in sine of x, we have x cubed

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over 3 factorial.

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So here we're going to have 2x
cubed over 3 factorial plus--

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OK.

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So then, you know, and so on.

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So here we'll have 2x to the
fifth over 5 factorial minus--

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so on.

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If you wanted to write this in
summation notation, you could

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write it as the sum from
n equals 0 to infinity.

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Well, the denominator has got
to be 2n plus 1 factorial,

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because we want it to
go through the odds.

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And then we've got minus 1 to
the n times 2 to the 2n plus 1

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times x to the 2n plus 1.

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So this is 2x.

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What we've got here, if you
didn't have the 2's there,

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that would just be the series
for the regular sine.

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OK.

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So this is the series for this
function, 2 sine x cosine x.

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And I'll go over here
to do the third one.

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So what is the third one?

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It's x ln 1 minus x cubed.

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Well, what can we do
with this series?

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The x out front is
just multiplying

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this logarithm part.

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That's something we can
save until the end.

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If we can figure out what the
Taylor series for the ln of 1

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minus x cubed part is, then we
just multiply x into it, and

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that'll give us the Taylor
series for this whole thing.

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So the x out front
is pretty simple.

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So now what about this ln of
1 minus x cubed stuff?

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Well, a thing to remember is,
does it remind you of anything

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we've done before?

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Well, we have a Taylor
series for a

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logarithm function, right?

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We've already seen in lecture,
I believe, we've seen that ln

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of 1 plus x is equal to x minus
x squared over 2 plus x

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cubed over 3, minus x to the
fourth over four, and so on,

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alternating signs.

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Notice that the denominators,
when you have a logarithm,

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these are not factorials.

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These are just the integer 2,
the integer 3, the integer 4,

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unlike for exponentials
and trig functions.

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So this what log of 1 plus x,
this is the Taylor series for

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log of 1 plus x.

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Well, how does that help us?

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Well, log of 1 minus x cubed we
can get from log of 1 plus

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x, with the appropriate
substitution.

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So in particular, we just
have to put minus x

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cubed in for x here.

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So what does that give us?

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It gives us the ln of 1 minus
x cubed is equal to-- well,

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minus x cubed minus, so we put
minus x cubed in here, we

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square it, and we just
get x to the sixth.

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x to the sixth over 2.

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Then, all right.

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So minus x cubed quantity cubed
is minus x to the ninth.

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So minus x to the ninth over 3,
minus x to the twelfth over

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4, and so on.

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And so finally, x ln of 1 minus
x cubed, we just get by

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multiplying this whole
expression through by x.

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So this is equal to minus x to
the fourth minus x to the

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seventh over 2 minus x-- whoops,
not ten,-- minus x to

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the tenth over 3, minus x to
the 13 over 4, and so on.

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And I'll leave it as an exercise
for you to figure

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out, how to write this
in summation

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notation, if you wanted.

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So just quickly to summarize,
we had these three power

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series, these three functions
that we started out with, and

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we used a bunch of different
techniques that we've learned

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in order to compute their
power series.

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So over here, we took the
function that we'd seen, and

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we knew a formula for it in
terms of other functions that

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we already knew, and so we
plugged in those power series,

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and used our addition rule
for power series.

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We could have also done this
one directly from the

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definition, if we
had wanted to.

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For the second one, for the 2
sine x cosine x, we recognized

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that as something that is
susceptible to a substitution,

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although also, with a little
more, work, we could have done

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it by a couple of different
methods.

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For example, by multiplying
two power series together.

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And finally, for this third one,
for the x ln of 1 minus x

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cubed, we first saw the
substitution here that we

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could make, and then we just
did a multiplication by a

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polynomial, which is a
relatively easy thing to do

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for power series.

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So that's what we did in
this recitation, and

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I'll leave it at that.

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