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In the last video, we
took the Maclaurin series

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of cosine of x.

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We approximated it
using this polynomial.

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And we saw this pretty
interesting pattern.

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Let's see if we can
find a similar pattern

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if we try to approximate sine
of x using a Maclaurin series.

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And once again, a
Maclaurin series

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is really the same thing
as a Taylor series,

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where we are centering
our approximation

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around x is equal to 0.

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So it's just a special
case of a Taylor series.

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So let's take f of
x in this situation

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to be equal to sine of x.

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And let's do the same thing
that we did with cosine of x.

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Let's just take the
different derivatives

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of sine of x really fast.

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So if you have the first
derivative of sine of x,

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is just cosine of x.

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The second derivative
of the sine of x

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is the derivative of cosine of
x, which is negative sine of x.

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The third derivative is going
to be the derivative of this.

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So I'll just write
a 3 in parentheses

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there, instead of doing
prime, prime, prime.

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So the third derivative
is the derivative

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of this, which is
negative cosine of x.

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The fourth derivative
is the derivative

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of this, which is
positive sine of x again.

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So you see, just like cosine
of x, it kind of cycles

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after you take the
derivative enough times.

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And we care-- in order to do
the Maclaurin, series-- we care

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about evaluating the function,
and each of these derivatives

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at x is equal to 0.

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So let's do that.

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So for this, let me do this
in a different color, not

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that same blue.

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I'll do it in this purple color.

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So f-- that's hard
to see, I think

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So let's do this
other blue color.

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So f of 0, in this
situation, is 0.

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And f, the first derivative
evaluated at 0, is 1.

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Cosine of 0 is 1.

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Negative sine of 0
is going to be 0.

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So f prime prime, the second
derivative evaluated at 0 is 0.

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The third derivative
evaluated at 0 is negative 1.

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Cosine of 0 is 1.

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You have a negative out there.

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It is negative 1.

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And then the fourth
derivative evaluated at 0

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is going to be 0 again.

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And we could keep going,
but once again, it

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seems like there's a pattern.

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0, 1, 0, negative
1, 0, then you're

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going to go back to positive 1.

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So on and so forth.

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So let's find its
polynomial representation

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using the Maclaurin series.

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And just a reminder,
this one up here,

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this was approximately
cosine of x.

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And you'll get closer and
closer to cosine of x.

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I'm not rigorously
showing you how

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close, in that it's definitely
the exact same thing as cosine

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of x, but you get closer
and closer and closer

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to cosine of x as you
keep adding terms here.

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And if you go to
infinity, you're

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going to be pretty
much at cosine of x.

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Now let's do the same
thing for sine of x.

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So I'll pick a new color.

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This green should be nice.

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So this is our new p of x.

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So this is approximately
going to be

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sine of x, as we add
more and more terms.

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And so the first term here, f
of 0, that's just going to be 0.

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So we're not even going
to need to include that.

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The next term is
going to be f prime

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of 0, which is 1, times x.

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So it's going to be x.

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Then the next term is f
prime, the second derivative

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at 0, which we see here is 0.

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Let me scroll down a little bit.

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It is 0.

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So we won't have
the second term.

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This third term right
here, the third derivative

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of sine of x evaluated
at 0, is negative 1.

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So we're now going
to have a negative 1.

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Let me scroll down
so you can see this.

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Negative 1-- this is
negative 1 in this case--

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times x to the third
over 3 factorial.

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And then the next
term is going to be 0,

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because that's the
fourth derivative.

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The fourth derivative evaluated
at 0 is the next coefficient.

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We see that that is going to be
0, so it's going to drop off.

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And what you're
going to see here--

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and actually maybe I haven't
done enough terms for you,

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for you to feel good about this.

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Let me do one more
term right over here.

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Just so it becomes clear.

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f of the fifth
derivative of x is

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going to be cosine of x again.

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The fifth derivative-- we'll
do it in that same color,

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just so it's consistent-- the
fifth derivative evaluated at 0

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is going to be 1.

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So the fourth derivative
evaluated at 0 is 0,

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then you go to the fifth
derivative evaluated at 0,

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it's going to be positive 1.

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And if I kept doing this,
it would be positive

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1-- I have to write the 1
as the coefficient-- times x

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to the fifth over 5 factorial.

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So there's something
interesting going on here.

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For cosine of x, I had 1,
essentially 1 times x to the 0.

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Then I don't have x
to the first power.

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I don't have x to the
odd powers, actually.

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And then I just essentially have
x to all of the even powers.

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And whatever power it is, I'm
dividing it by that factorial.

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And then the sines
keep switching.

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I shouldn't say this is an even
power, because 0 really isn't.

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Well, I guess you can
view it as an even number,

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because-- well I won't
go into all of that.

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But it's essentially 0, 2,
4, 6, so on and so forth.

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So this is
interesting, especially

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when you compare to this.

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This is all of the odd powers.

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This is x to the first
over 1 factorial.

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I didn't write it here.

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This is x to the
third over 3 factorial

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plus x to the fifth
over 5 factorial.

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Yeah, 0 would be an even number.

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Anyway, my brain is in a
different place right now.

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And you could keep going.

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If we kept this process
up, you would then

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keep switching sines.

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X to the seventh over
7 factorial plus x

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to the ninth over 9 factorial.

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So there's something
interesting here.

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You once again see this
kind of complimentary nature

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between sine and cosine here.

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You see almost
this-- they're kind

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of filling each
other's gaps over here.

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Cosine of x is all
of the even powers

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of x divided by that
power's factorial.

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Sine of x, when you take its
polynomial representation,

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is all of the odd powers of
x divided by its factorial,

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and you switch sines.

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In the next video,
I'll do e to the x.

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And what's really
fascinating is that e

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to the x starts to look like
a little bit of a combination

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here, but not quite.

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And you really do
get the combination

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when you involve
imaginary numbers.

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And that's when it starts to
get really, really mind blowing.