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Hi.

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Welcome back to recitation.

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In class, Professor Jerison
and Professor Miller

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have taught you a little
bit about Taylor series

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and some of the manipulations
you can do with them,

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and have computed a bunch
of examples for you.

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So I have three
more examples here

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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

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written in terms of the
exponential function as e

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

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The second one is
the function 2 times

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sine of x times cosine of
x, just your regular sine

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and cosine here.

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And the third one is x times
the logarithm of the quantity 1

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minus x cubed.

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

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to work out the Taylor series
for these three functions,

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come 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

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that's given by the formula e
to the x plus e to the minus

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x over 2.

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So there are a
couple different ways

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you could go about this one.

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This is actually,
the hyperbolic cosine

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is very susceptible to
the method of just using

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the formula that you have.

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So if you remember,
the derivative

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of the hyperbolic cosine
is the hyperbolic sine.

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The derivative of
the hyperbolic sine

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is the hyperbolic cosine again.

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So this function has very
easy-to-understand derivatives,

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which you can see, you know,
just by looking at its formula.

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It's easy to understand, because
the exponential function has

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very simple derivatives,
and e to the the minus

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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

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know the Taylor
series for e to the x.

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

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to the 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 n

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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,

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if you put in minus x for x.

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So it's n equals 0 to infinity,
so that works out to minus 1

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to the n x to the
n or n factorial.

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Now, when you add these
two series together,

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what you see is that when
n is even, over here,

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you have x to the n over n
factorial, and over here,

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you have x to the
n over n factorial.

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So what you get is,
well, you get 2 x

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to the n over n factorial, and
then you multiply by a half,

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so you 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,

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and 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

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like the series for e to the
x, except the odd terms have

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died off.

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

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to the fourth over 4 factorial
plus x to the sixth over 6

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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

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equals 0 to infinity of x
to the 2n over 2n factorial.

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So this is the Taylor series for
the hyperbolic cosine function.

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

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you could do something
very similar,

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or you could remember that
the hyperbolic sine is

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the derivative of the
hyperbolic cosine,

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and just take a derivative
right from this expression.

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One other thing that
you should notice

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is that this 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

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between regular 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 and cosine x

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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 x,

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take the series for cosine
x, multiply them together

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like you would
multiply polynomials,

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and what you would get
is the Taylor series

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for this 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

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could try taking derivatives.

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You could have a situation
where every time you

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take a 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,

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if you wanted more
than just a few terms.

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

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identities.

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So if you look at
this expression,

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

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So once you realize
that this is sine of 2x,

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there's a much,
much shorter path

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available to you, which is that
you already know the Taylor

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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.

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So sine of x is x--
so in this case,

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that's going to be 2x--
then minus, so in sine of x,

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we have x cubed
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

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minus-- so on.

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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.

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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 this logarithm

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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,

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and 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

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of anything we've done before?

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Well, we have a Taylor series
for a logarithm function,

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right?

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We've already seen in
lecture, I believe,

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we've seen that
ln of 1 plus x is

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equal to x minus x squared
over 2 plus x cubed over 3,

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minus x to the fourth over four,
and so on, 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 is what
log of 1 plus x--

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this is the Taylor series
for 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 x,

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with the appropriate
substitution.

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So in particular,
we just have to put

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minus x 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,

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we 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 4,

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and so on.

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And so finally, x ln
of 1 minus x cubed,

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we just get by multiplying this
whole expression through by x.

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

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

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x to 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

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to figure out how to write
this in summation notation,

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if you wanted.

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So just quickly to summarize,
we had these three power series,

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these three functions
that we started out with,

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and we used a bunch of
different techniques

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that we've learned in order
to compute their power series.

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

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

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that 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 definition,

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if we had wanted to.

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

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we recognized that
as something that

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is susceptible to a
substitution, although also,

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with a little more,
work, we could

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have done 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

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of 1 minus x cubed, we first
saw the substitution here

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

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by a polynomial, which is
a relatively easy thing

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to do for power series.

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So that's what we did
in this recitation,

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and I'll leave it at that.

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