PROFESSOR: Welcome
back to recitation.
In this video I
want us to practice
using Newton's Method to find
the solution to an equation.
So what we're going
to do in particular
is we're going to
use Newton's Method
to approximate a solution
to the following equation,
2 cosine x equals 3x.
And I'm going to tell
you where to start.
We're going to the have
our initial value, x_0, be
pi over 6.
And I want you to find x_2.
So why don't you pause the
video, take a little time
to work on that, and
then I'll come back
and I will show
you how I did it.
OK.
Welcome back.
Again, what we're going to
do is use Newton's Method
to approximate a solution
to this equation.
And so what I want to
point out first is,
I want to point
out why pi over 6
is a reasonable first
value to choose,
and I want to point out that
this, in fact, has only one
solution.
So what I'm going to do, to
give us a reason for that,
is I'm going to draw a
rough sketch of two curves
and show where they intersect.
And so I want us to notice that
if I were to look at the two
curves, y equals cosine
x and y equals 3/2 x
and I draw them on
the same xy-plane,
that where they intersect
will be where I have solutions
to this equation.
And that's because I just
divide both sides by 2.
Whatever solves this equation
solves the equation, cosine x
equals 3/2 x.
So let me give you a rough
sketch of those two curves
and we'll see what the
intersections look like.
So I'm going to do
that right down here.
OK.
So let me, let me first
draw-- make this y equals 1.
Make this y equals minus 1.
And I'm going to draw cosine x
first, y equals cosine x first,
because I'm most likely to
have a hard time with that,
and I'll do my x
scale once I'm done.
And so cosine x, y equals cosine
x, looks something like this.
Maybe not the most
perfect, but again,
it's kind of a rough sketch.
That's pretty good.
Something like this.
So this is y equals cosine x.
And now I want to graph
y equals 3x over 2.
And that goes through
the point (0, 0).
It also goes through the
point one comma three halves.
Well, this is pi
over 2 right here.
So 1 is about here, we'll say.
Because pi over 2 is a little
bigger than one and a half.
So the 1 is about here, One
and a half is about here.
Or 3/2, if you need
to remind yourself.
So the line y equals 3/2 x
looks something like this.
So it's fairly
straightforward to see
that these two curves
intersect at one spot, whatever
this spot is.
OK?
And notice, to the left
they don't intersect.
So we are just looking
for a single solution.
And then the other thing
I want to point out
is, why is pi over
six potentially
a good guess to start with?
Well, the value, this
is the x-value 1,
and this is the x-value 0.
We know for a fact
that we have to have
this x-value line
between 0 and 1
because of where my point is
at the time that x equal 1,
I'm all the way up at
y equals 3/2 up here.
So at least we know
we're between 0 and 1.
And then from there
you could even
try some other values like
pi over 3 and pi over 4
and put those in and
see how they compared.
But at least, we'll
just say, at least we
know x is between 0
and 1, and pi over 6
is certainly in that region.
So that's a good
first starting point.
Now I'm going to come over
here and start to do some work.
If we want to
solve the equation,
2 cosine x equals 3x,
what we're really doing
is we're looking for
zeros of this function.
So we find the zeros of
this function, which we
know there's only one of them.
We find the zero
of this function,
then we actually have
solved 2 cosine x equals 3x.
So hopefully that
makes sense to you
that we're actually going
to apply Newton's Method
to this function.
And so when we apply Newton's
Method, we need the function.
We also need to the derivative.
So let me remind
you, to derivative
of this is going to be
negative 2 sine x minus 3.
Right?
The derivative of cosine
x is negative sine x.
And so this is exactly
the derivative.
And then let me remind you
what Newton's Method says.
It says the next x-value is
equal to the previous x-value
minus the fraction of
the function evaluated
at the previous value divided
by the derivative evaluated
at the previous value.
Right?
So this is the formula you
have for Newton's Method.
So let's see if we can get from
x_0 to x_1 and then x_1 to x_2.
So in our case, we have x_1
equals, well, x_0 is pi over 6.
And then we have
minus the function
evaluated at pi over 6,
and then the derivative
evaluated at pi over 6.
So the function evaluated at
pi over 6-- cosine of pi over 6
is root 3 over 2.
So root 3 over 2 times
2-- we get a root 3.
Separate that out.
And then here, pi over
6 times 3 is pi over 2.
So we get a minus pi over 2.
Sine pi over 6 is 1/2.
So we get negative
2 times 1/2-- we
get negative 1 and
then a negative 3.
And if you simplify
this, you get
that this is approximately
0.564, or around that.
OK?
And now from here, you
would then, for x_2,
you're going to take 0.564
minus these things evaluated
at 0.564.
This ratio, f of 0.564
divided by f prime at 0.564.
But I'm not going to do
that because you should
get somewhere around, depending
on how many decimal places
you kept, you
should get something
around one of these two values.
So you actually get,
after x_0, by x_1,
you have something that is at
least fixed to the first two
decimal places.
And then this third
decimal place, maybe it's
going to be a 4
or a 3 in the end.
But depending on what
value we choose here,
we might get slightly
different values here based
on the rounding.
So just suffice it
to say, I got x_1.
Your x_2 should
be about the same.
It should be one of these two.
OK.
So let me just remind you
what we were doing here.
We were trying to
use Newton's Method
to find a solution
to an equation
that I had written up here,
this 2 cosine x equals 3x,
and I pointed out
a couple things.
I pointed out that finding
a solution to this equation
is the same as finding a
solution to the equation
cosine x equals 3/2 x.
And so I did that as
a graph to sort of see
if I could get an initial
idea of what kind of solution
I was looking for.
And then we just started
using Newton's Method
on a particular function.
And that function was
this side of the equation
minus this side.
Because if 2 cosine
x minus 3x equals 0,
then 2 cosine x equals 3x.
So we had this function over
here, 2 cosine x minus 3x,
and I said I was looking
for zeros of that function.
And that's where
Newton's Method comes in.
So I think that is
where I will stop.
