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PROFESSOR: Welcome back
to recitation.

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In this video I want us to
practice using Newton's Method

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to find the solution
to an equation.

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So what we're going to do in
particular is we're going to

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use Newton's Method to
approximate a solution to the

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following equation, 2
cosine x equals 3x.

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And I'm going to tell
you where to start.

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We're going to the have
our initial value,

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x0, be pi over 6.

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And I want you to find x2.

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So why don't you pause the
video, take a little time to

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work on that, and then I'll come
back and I will show you

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how I did it.

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OK.

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Welcome back.

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Again, what we're going to do
is use Newton's Method to

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approximate a solution
to this equation.

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And so what I want to point out
first is, I want to point

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out why pi over 6 is a
reasonable first value to

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choose, and I want to point out
that this, in fact, has

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only one solution.

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So what I'm going to do to give
us a reason for that is

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I'm going to draw a rough sketch
of two curves and show

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where they intersect.

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And so I want us to notice that
if I were to look at the

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two curves, y equals cosine x
and y equals 3/2x and I draw

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them on the same x, y plane,
that where they intersect will

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be where I have solutions
to this equation.

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And that's because I just
divide both sides by 2.

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Whatever solves this equation
solves the equation, cosine x

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equals 3/2x.

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So let me give you a rough
sketch of those two curves and

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we'll see what the intersections
look like.

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So I'm going to do that
right down here.

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OK.

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So let me, let me first draw,
make this y equals 1.

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Make this y equals minus 1.

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And I'm going to draw cosine x
first, y equals cosine x first

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because I'm most likely to have
a hard time with that,

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and I'll do my x scale
once I'm done.

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And so cosine x, y equals
cosine x looks

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something like this.

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Maybe not the most perfect,
but again, it's kind of a

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rough sketch.

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That's pretty good.

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Something like this.

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So this is y equals cosine x.

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And now I want to graph
y equals 3x over 2.

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And that goes through
the point, 0, 0.

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It also goes through the point
one comma three halves.

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Well, this is pi over
2 right here.

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So 1 is about here, we'll say.

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Because pi over 2 is a little
bigger than 1 and 1/2.

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So the 1 is about here, 1
and 1/2 is about here.

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Or 3/2, if you need to
remind yourself.

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So the line, y equals 3/2x looks
something like this.

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So it's fairly straightforward
to see that these two curves

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intersect at one spot, whenever
this spot is.

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OK?

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And notice, to the left
they don't intersect.

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So we are just looking for
a single solution.

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And then the other thing I want
to point out is, why is

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pi over six potentially a good
guess to start with?

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Well, the value, this is
the x value 1, and this

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is the x value 0.

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We know for a fact that we have
to have this x value line

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between 0 and 1 because of where
my point is at the time

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that x equal 1, I'm all
the way up at y

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equals 3/2 up here.

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So at least we know we're
between 0 and 1.

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And then from there you could
even try some other values

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like pi over 3 and pi over 4 and
put those in and see how

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they compared.

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But at least, we'll just say, at
least we know x is between

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0 and 1, and pi over 6 is
certainly in that region.

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So that's a good first
starting point.

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Now I'm going to come
over here and

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start to do some work.

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If we want to solve the
equation, 2 cosine x equals

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3x, what we're really doing
is we're looking

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for 0's of this function.

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So we find the 0's of this
function, which we know

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there's only one of them.

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We find the 0 of this function,
then we actually

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have solved 2 cosine
x equals 3x.

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So hopefully that makes sense
to you that we're actually

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going to apply Newton's Method
to this function.

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And so when we apply Newton's
Method, we need the function.

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We also need to the
derivative.

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So let me remind you, to
derivative of this is going to

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be negative 2 sine x minus 3.

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Right?

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The derivative of cosine
x is negative sine x.

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And so this is exactly
the derivative.

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And then let me remind you what
Newton's Method says.

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It says the next x value is
equal to the previous x value

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minus the fraction of the
function evaluated at the

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previous value divided by the
derivative evaluated at the

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previous value.

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Right?

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So this is the formula you
have for Newton's Method.

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So let's see if we can
get from x0 to x1

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and then x1 to x2.

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So in our case, we have x1
equals, well, x0 is pi over 6.

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And then we have minus the
function evaluated at pi over

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6, and then the derivative
evaluated at pi over 6.

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So the function evaluated at pi
over 6-- cosine of pi over

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6 is root 3 over 2.

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So root 3 over 2 times
2-- we get a root 3.

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Separate that out.

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And then here, pi over 6
times 3 is pi over 2.

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So we get a minus pi over 2.

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Sine pi over 6 is 1/2.

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So we get negative
2 times 1/2--

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we get negative 1 and
then a negative 3.

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And if you simplify this, you
get that this is approximately

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0.564, or around that.

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OK?

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And now from here, you would
then, for x2, you're going to

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take 0.564 minus these things
evaluated at 0.564.

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This ratio, f of 0.564 divided
by f prime at 0.564.

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But I'm not going to do that
because you should get

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somewhere around, depending on
how many decimal places you

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kept, you should get something
around one

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of these two values.

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So you actually get, after x0,
by x1, you have something that

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is at least fixed to the first
two decimal places.

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And then this third decimal
place, maybe it's going to be

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a 4 or a 3 in the end.

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But depending on what about you
we choose here, we might

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get slightly different values
here based on the rounding.

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So just suffice it
to say, I got x1.

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Your x2 should be
about the same.

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It should be one of these two.

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OK.

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So let me just remind you
what we were doing here.

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We were trying to use Newton's
Method to find a solution to

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an equation that I had written
up here, this 2 cosine x

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equals 3x, and I pointed
out a couple things.

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I pointed out that finding a
solution to this equation is

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the same as finding a solution
to the equation, cosine x

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equals 3/2 x.

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And so I did that as a graph to
sort of see if I could get

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an initial idea of what kind of
solution I was looking for.

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And then we just started using
Newton's Method on a

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particular function.

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And that function was
this side of the

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equation minus this side.

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Because if 2 cosine x minus
3x equals 0, then 2

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cosine x equals 3x.

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So we had this function over
here, 2 cosine x minus 3x, and

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I said I was looking for
0's of that function.

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And that's where Newton's
Method comes in.

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So I think that is where
I will stop.

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