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Hi.

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Welcome back to recitation.

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We've been talking
about Taylor series

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for a number of
functions and rules

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by which you can
compute Taylor series.

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I have here an example that I
don't think we did in lecture.

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So this is the function f
of x equals secant of x.

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Now, unlike some of the
other ones you've seen,

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there's not a really simple
formula for the whole Taylor

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series of secant x.

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So what I'd like you
to do is not to find,

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you know, a formula
for the general term,

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but rather, just to
use some of the tools

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that we've learned to
compute the first few terms

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of the Taylor series for
f of x equals secant x.

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Say, up through the x to the
fourth term, if you wanted,

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or even a little further if
you were feeling ambitious.

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So why don't you pause the
video, have a go at that,

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come back, and we
can do it together.

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So welcome back.

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I asked you to compute the first
few terms of a Taylor series.

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One thing you can always do
in this case, is you can go

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and you can apply the
general formula that we

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have for Taylor
series, and use it

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to compute the series that way.

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So in order to do
that, you just need

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to compute a few derivatives
of your function.

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So remember that the general
formula for a Taylor series

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is the Taylor series for f of
x is equal to the sum from n

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equals 0 to infinity
of the nth derivative

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of f taken at zero divided by
n factorial times x to the n.

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And since we're only interested
in the first few terms,

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this is f of 0
plus f prime of 0 x

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plus f double prime of 0 over
2 x squared, plus dot dot dot.

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I'm not going to write
out the next few terms.

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And if you want to apply
this formula to secant

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of x, what you
would have to do, is

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you would have to compute
these derivatives.

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And so we could try doing that.

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So in our case, f of x
is equal to secant of x.

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So f of 0 is secant
of 0, which is just 1.

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So then we need to know
f prime of x, so that's

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the derivative of secant of x.

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So this is one you
should remember.

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This is equal to
secant x times tan x.

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And when you plug in x equal
to 0, well, the tan x is 0.

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So f prime of 0 is equal to 0.

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And then you could keep going.

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So you could compute
f double prime of x.

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So we would have to compute the
derivative of secant x tan x.

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And you would do that
by the product rule.

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You know, you take
derivative of secant

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of x, and that gives you
secant x tan x times tan

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x plus-- and then you, OK.

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So you leave secant of
x, and you multiply it

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by the derivative of tangent,
which is secant squared x.

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And now when you put in 0 here,
you get f double prime of 0.

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Well, OK.

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So this has a tan x
in it, so that part's

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going to give you 0.

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And here we end up with secant
of 0 times secant squared of 0,

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so that's 1 times 1 times 1.

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So that's just 1.

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So f double prime of 0 is 1,
and you could keep doing this.

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Now one thing you'll notice
is that this is getting

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more and more complicated.

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I mean, we can simplify this
expression a little bit.

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We could write it as
secant x tan squared

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x plus secant cubed x,
and there are, you know,

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all sorts of trig
manipulations you could do,

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if you wanted to try and rewrite
that in some simpler form.

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But fundamentally,
it's more complicated

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than the first derivative
was, and that's

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more complicated
in the function.

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And it will keep getting
more complicated as you

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compute more derivatives.

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So we can do this.

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So so far this shows us, by the
way, that this is equal to 1

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plus 0x plus x squared
over 2 plus dot dot dot.

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And if you wanted
to compute, you

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know, up through the fourth
degree term or something

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like this, that's something
that's manageable.

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But I want to suggest that
there are maybe some nicer ways

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to do it.

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So one thing to notice
is that secant of x

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is closely related to
the function cosine of x.

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And you know the Taylor
series for cosine of x.

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So one thing you
could think to do,

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is to leverage the information
that you have about cosine

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of x in order to use it to get
some information about secant

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of x.

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So one simple way to do that, is
that you know that cosine of x

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is even.

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It's an even function.

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Cosine of minus x is
equal to cosine of x.

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So that means secant
of x is also even.

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Secant of x is an even function.

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And you've seen that even
functions, their Taylor

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series, all the odd
powers have coefficient 0.

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So what that means is
without ever computing

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the third derivative,
we can know already

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that the next term
in this Taylor series

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is going to be 0 x cubed over 6.

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OK?

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So that's nice.

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So the odd terms of the Taylor
series for secant of x are 0.

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OK.

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So that's one thing
you can get right away.

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So that gives you,
if you like, that

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gives you half of the
terms of the Taylor series.

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It's a little bit
of a joke, but.

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OK, so then you only need to
figure out the even terms.

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That's one way we can leverage
the relationship between secant

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and cosine.

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The other way is
that we can remember

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that Taylor series multiply
just like polynomials do.

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So if secant is 1
over cosine, that

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means secant times
cosine is equal to 1.

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OK?

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So idea, secant of x times
cosine of x is equal to 1.

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Now, that means that
the Taylor series

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for secant of x times the
Taylor series for cosine of x

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has to simplify just to 1.

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So we can write
down that product

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as a product of two
infinite polynomials,

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and we can start
multiplying term by term.

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And that'll allow us to
solve for a bunch of terms,

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just by solving some
simple linear equations.

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So let me show you what I mean.

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So we know-- let me write
it as cosine times secant.

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So we know that cosine of x is
1 minus x squared over 2 plus x

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to the fourth over 24--
that's 4 factorial-- minus x

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to the sixth over 720, which is
6 factorial, plus dot dot dot.

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And so we know that
if we multiply this

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by the series for
secant of x-- well,

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what is the series
for secant of x?

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Well, we've already
computed a few terms.

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We know that it's 1
plus x squared over 2,

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we computed that already.

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And we know that the
third degree term is 0.

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So there's some fourth
degree coefficient, a_4, x

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to the fourth, or 4 factorial.

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And there's some-- well,
we know, we said it's even,

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so we know the fifth
degree coefficient is 0.

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So then the sixth degree
coefficient we don't know yet.

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So this is plus a sub 6 times x
to the sixth over 6 factorial,

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plus dot dot dot.

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So we know that when we multiply
these two things together,

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it has to give us just 1.

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All the higher-order
terms have to cancel,

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because over here we have a 1.

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So what you can do, is
you can actually try

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multiplying these things out.

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So it's easy to
see, for example,

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that the constant term of this
product is just 1 times 1,

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which is 1.

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Which is good.

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So that's a check on
what we've done so far.

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And there is no x-term,
because there are no x's.

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The x squared term
here is 1 times

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x squared over 2 minus x
squared over 2 times 1.

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Well, that gives us
0, so that's good.

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So this product is
equal to-- well, it's 1,

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plus we saw 0x,
plus 0 x squared.

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How about the x cubed term?

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Well, there are no odd
terms in this product,

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so there's no x cubed.

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How about the x to
the fourth term?

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Well, OK.

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So how do we get
an x to the fourth?

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We could have an x to the
fourth here times a constant.

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So that's x to the
fourth over 24.

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Or, we could have an x
squared times an x squared.

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So in this case, that gives us
minus x squared over 2 times

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x squared over 2, which is
minus x to the fourth over 4.

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Or we could have a constant
times an x to the fourth.

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So this is plus a_4 x
to the fourth over 24.

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And then we'll have a
sixth-degree term, and so on.

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Notice that there's
never any involvement

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from the higher-order terms
in the fourth power here.

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Right?

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If you ever took an
x to the sixth here,

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well then everything you
multiply it by has at least an

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x to the sixth power.

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So we don't have to worry
about that showing up

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in the x to the fourth term.

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Well, OK.

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We know this is
actually equal to 1,

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so we know that this
thing has to be 0.

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This is x to the fourth term.

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It has to be 0.

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So that means that 1 over 24
minus 1 over 4 plus a_4 over 24

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equals 0.

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So OK.

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So now you can multiply
everything through by 24

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and rearrange to figure
out that a_4 is equal to 5.

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So I've done that correctly.

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And then if you wanted, it would
be fairly easy to go back up

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and then you look at
the x to the sixth term.

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And from there, you
could figure out

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that a_6 was equal to, say,
61 or something like that.

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I think 61.

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And you could keep doing this.

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So this is one way
to compute more terms

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of the series for secant of x.

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Another thing you could
do-- which I'll just

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mention very briefly,
I'm not going to show you

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how to do it-- is that
you can do long division

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on power series.

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So it actually works out very--
it works out just like this.

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It's mathematically equivalent.

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The way you actually
do it looks different.

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It looks like long division.

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When you do long division
with polynomials,

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you start with the
highest-order term.

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Of course, power series don't
have highest-order terms.

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What you actually do
with a power series, is

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you start with the
lowest-order term.

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So to divide this
into 1, you'd say, oh,

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you need a factor of
1, and then you'll

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have a-- you know, you
subtract off 1 times this,

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and that gives you
an x squared plus 2.

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And so, OK, so you say, I need
a plus an x squared plus 2,

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and so on.

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So that was too brief for you
to understand it properly,

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but you can go and look
up somewhere the method

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of long division
on power series.

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So just to recap, we
talked about three methods

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for computing the coefficients
of a power series.

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There's the formula that
you were given, which works,

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and which you could use.

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In this case, it's a
little complicated.

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Then there's the method
of using a relationship

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between your power series
and other known power series.

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In this case, we can
use the relationship

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that we know our power series
satisfies a certain product.

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We know that our power series
times cosine of x equals to 1,

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and we can use that to solve
for some of the coefficients.

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Or you can also, similarly,
when you have that situation,

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you can also use long division
to compute the coefficients.

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I'll end there.