WEBVTT
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PROFESSOR: Welcome
back to recitation.
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In this video I
want us to practice
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using Newton's Method to find
the solution to an equation.
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So what we're going
to do in particular
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is we're going to
use Newton's Method
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to approximate a solution
to the following equation,
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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, x_0, be
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pi over 6.
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And I want you to find x_2.
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So why don't you pause the
video, take a little time
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to work on that, and
then I'll come back
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and I will show
you 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
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to approximate a solution
to this equation.
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And so what I want to
point out first is,
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I want to point
out why pi over 6
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is a reasonable first
value to choose,
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and I want to point out that
this, in fact, has only one
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solution.
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So what I'm going to do, to
give us a reason for that,
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is I'm going to draw a
rough sketch of two curves
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and show where they intersect.
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And so I want us to notice that
if I were to look at the two
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curves, y equals cosine
x and y equals 3/2 x
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and I draw them on
the same xy-plane,
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that where they intersect
will be where I have solutions
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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 3/2 x.
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So let me give you a rough
sketch of those two curves
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and 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 something like this.
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Maybe not the most
perfect, but again,
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it's kind of a 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 one and a half.
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So the 1 is about here, One
and a half 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 3/2 x
looks something like this.
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So it's fairly
straightforward to see
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that these two curves
intersect at one spot, whatever
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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
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is, why is pi over
six potentially
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a good guess to start with?
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Well, the value, this
is the x-value 1,
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and this is the x-value 0.
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We know for a fact
that we have to have
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this x-value line
between 0 and 1
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because of where my point is
at the time that x equal 1,
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I'm all the way up at
y 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
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try some other values like
pi over 3 and pi over 4
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and put those in and
see how they compared.
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But at least, we'll
just say, at least we
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know x is between 0
and 1, and pi over 6
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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 start to do some work.
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If we want to
solve the equation,
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2 cosine x equals 3x,
what we're really doing
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is we're looking for
zeros of this function.
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So we find the zeros of
this function, which we
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know there's only one of them.
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We find the zero
of this function,
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then we actually have
solved 2 cosine x equals 3x.
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So hopefully that
makes sense to you
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that we're actually going
to apply Newton's Method
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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
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of this is going to 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
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at the previous value divided
by the derivative evaluated
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at the 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
x_0 to x_1 and then x_1 to x_2.
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So in our case, we have x_1
equals, well, x_0 is pi over 6.
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And then we have
minus the function
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evaluated at pi over 6,
and then the derivative
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evaluated at pi over 6.
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So the function evaluated at
pi over 6-- cosine of pi over 6
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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-- we
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get negative 1 and
then a negative 3.
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And if you simplify
this, you get
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that this is approximately
0.564, or around that.
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OK?
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And now from here, you
would then, for x_2,
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you're going to take 0.564
minus these things evaluated
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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
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get somewhere around, depending
on how many decimal places
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you kept, you
should get something
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around one of these two values.
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So you actually get,
after x_0, by x_1,
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you have something that is at
least fixed to the first two
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decimal places.
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And then this third
decimal place, maybe it's
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going to be a 4
or a 3 in the end.
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But depending on what
value we choose here,
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we might get slightly
different values here based
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on the rounding.
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So just suffice it
to say, I got x_1.
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Your x_2 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
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to find a solution
to an equation
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that I had written up here,
this 2 cosine x equals 3x,
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and I pointed out
a couple things.
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I pointed out that finding
a solution to this equation
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is the same as finding a
solution to the equation
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cosine x equals 3/2 x.
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And so I did that as
a graph to sort of see
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if I could get an initial
idea of what kind of solution
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I was looking for.
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And then we just started
using Newton's Method
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on a particular function.
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And that function was
this side of the equation
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minus this side.
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Because if 2 cosine
x minus 3x equals 0,
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then 2 cosine x equals 3x.
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So we had this function over
here, 2 cosine x minus 3x,
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and I said I was looking
for zeros 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.