GILBERT STRANG: OK, what I want
to do today is show you
two different ways that
derivatives are used.
In one of them, the problem is
to find a close approximation
to the value f at a point x.
f of x.
The second application
is to solve an
equation, where often--
and I use a different letter,
capital F, just because it's a
different function and there
will be different
examples for this one.
So I just chose capital F
to keep them separate.
This is the problem of
solving an equation.
And Newton had an idea and
it's survived all these
centuries and it's still
the good way.
OK, what's this based
on, both of them?
That's the point.
They're both based
on the same idea.
Suppose at a point, which is
near the x we want, or near
the solution to this problem, at
some point, let me call it
a, suppose we know the slope,
the derivative, at that point.
So I'm using f prime
for the derivative.
At that point, well, we know
what the definition is.
But I'm also supposing that
we've got that number.
And now I want to use this
knowledge of the slope at that
nearby point a to come close to
the solution, the f of x,
or the x there.
OK, well you remember this is
delta f divided by delta x.
You recognize that before we
take the limit this is two
nearby points, delta x apart.
Their values, the values
of f at those points.
It's the change in f divided
by the change in x--
that's what the derivative
is--
as one point approaches
the point a
where we know the slope.
Here's the idea.
I'm just going to erase
that stuff.
Well, now I won't have an
equal sign anymore.
I'll have an approximately
equal.
So the slope--
this is delta f over delta x--
is not the same as df
dx, which is this.
But if x is close to a this will
normally be close to the
correct slope, the instant
slope at the point.
I'm just going to use this
approximation to find a good
approximation for f of x.
So in this application on the
left I know the x and I want
to find what is f
at that point?
So I look at this.
I multiply up by x minus a.
I move the f of a to
the other side.
And what do I have?
I still have an approximation
sign.
So when I move the f of a to the
other side, there it is.
And then, the other part is x
minus a times f prime at a.
That's my formula.
Let me just talk about that
formula for a minute and then
give examples.
What that says is that if I want
to find the value of f at
a nearby point, a good thing
to do is use this linear
approximation.
I use the word "linear"
because the graph
of that is a line.
I'm following a straight line
instead of following the
curved graph.
That's the message of
today's lecture.
Follow the line.
So it's a line.
It starts out at the correct
point at x equal a.
It has the corrects slope
f prime of a.
And if I don't go too far
that line won't be too
far from the curve.
Good.
You'll see it now in examples.
Let me get the corresponding
idea here.
Now I have to remember to use
capital F. So let me create
the formula that helps to solve,
approximately solves,
this equation.
What's the difference?
Now x is what I'm looking for.
x is what I'm looking for and
F of x is what I know.
I know it's to be 0.
So I'm going to use this with
capital F. I know F of x
should be 0 and x
is the unknown.
So can I just move that equation
around again to get
an equation?
I'll bring the x minus
a up as I did before.
And I have to use an
approximation sign as always.
The F of x is 0, so I have minus
F of a, and then I'm
dividing by F prime of a.
There you have it.
That is Newton's insight.
Newton and then somebody named
Raphson helped out, made it
work for general functions.
And once again, I forgot.
I should be using
capital F there.
Because in the right side of
the board I'm calling the
function capital F. This is the
little movement of x away
from a, which will bring us
closer to the answer.
You'll see is this formula
work in a picture.
So I'm ready for an example
starting with approximation.
So here's my problem.
Here's my example.
Find the square root of--
so my problem is going to be
find the square root or
approximate square root of 9.
I'm going to shift away
from 9 a little bit.
9.06.
OK, how does that fit
this example?
My function is the square root
function, x to the 1/2.
It's derivative, f prime.
I know the derivative of
that is 1/2 x to one
lower power minus 1/2.
So that's 1 over 2
square root of x.
Good.
I know my function.
I know its derivative.
Now I pick a point a, which
is close and easy.
So close, I'm looking for a
point a, which is near 9.06
and has a nice square root.
Well, 9.
So I'll choose a to be 9.
The correct value f at
the point a is the
square root of 9.
3.
That's the point.
We can evaluate the function
easily at 9.
And f prime at a is 1 over
2 square root of a.
9.
Square root of 9.
So what do I have for that?
That's the easy number, 3.
That's 1/6.
In other words, I know what's
happening at x equal 9.
And that's a particular point
I'm calling a and I'm working
from there.
Then what does my approximation
say?
It says that f at this
nearby point.
So x is 9.06.
This is the x where I want
to know the square root.
So this is saying that the
square root of 9.06--
that's my f of x--
is approximately f at a.
That's the square root of 9.
That's the 3.
So far reasonable.
That's the approximation I
started with just using the 9.
But now I'm improving
it by the difference
between x minus 3.
Oh, what is x?
So I'm plugging in x is 9.06.
That's where I really want
the square root.
That's my x.
Minus a, which is the 9.
Times f prime, which we
figured out as 1/6.
That's the linear approximation
following this
line to this number,
which is what?
3 plus--
That's 0.06 divided by 6.
That's 3.
What do I have there?
This difference is 0.06.
Divided by 6 and its 0.01.
That's the approximation.
Closer than 3.
That comes from following
the line.
Let me draw a graph to show
you what I mean by
following the line.
Here is my square
root function.
The square root looks
something like that.
And here is the point x equals
9, where I know that the
height is 3.
What else do I know?
Here is 9.06, a little
further over, and I'm
looking for that point.
I'm looking for the square
root of 9.06.
So this was 9 here and
this was 9.06.
And how am I getting close
to that point?
Well, I'm not going to follow
the curve to do square root of
9.06 exactly, any
more than your
calculator or computer does.
I'm going to follow this line.
So that line is the
tangent line.
It's the line that goes through
the right thing at a
with the right slope.
So you can see that by
following the line
that gave me the 3.
And then here is the little--
you see what I'm--
I'm missing by a very small
amount, practically
too small to see.
I'm picking the point on the
line and that was this across,
this delta x was the 0.06.
The little tiny bit
here is the 0.06.
And this delta f is the
little piece that I
added on the 0.01.
How did that come from?
It came from the fact that
the correct slope is 1/6.
If I go over 0.06, I
should go up 0.01.
So you see what I'm doing?
I'm taking that point as my
close approximation to the
square root.
Closer than 3 to the square
root of 9.06.
OK.
That's a first example.
I'll give a second one.
But first, I'd like to give an
example that's like this one
of Newton's method.
May I change now to
Newton's method?
So I want to create
an equation.
And actually, I want it to
solve the same problem.
So I'm going to take my function
to be x squared minus
9.06 and I'll set that to 0.
I'm just keeping
my two examples
close because the answer--
in both problems, I'm looking
for the square root of 9.06.
Of course, that's the solution
to this equation, the square
root of 9.06.
OK again, I pick a point a
close to the solution.
And again, I'm going
to take a to be 3.
So 3 is close to the
correct solution.
The correct x is the square
root of 9.06.
But I'm starting close to it.
OK, at that point I
figure out F of a.
Newton wants to know
F of a, the value.
And Newton also wants
to know the slope.
So the value at a is--
3 squared is 9.
9 minus 9.06 is minus 0.06.
F prime of a is-- what's
the derivative of F?
Capital F. Well, the derivative
is certainly 2x.
And at the point a it's 6.
This is the 2x.
2a.
This is the 2a because I'm
evaluating the slope
at the point a.
Actually, if you want a
picture, it's quite
interesting to see
the picture.
Let me graph this f of x now.
What does my function
f of x look like?
x squared.
That's a parabola going up.
It starts below here.
It starts somewhere down here
and it curves up like so.
So this is the point I want.
There is the square
root of 9.06.
I've graphed a function
x squared minus 9.06.
That's the point I'd
like to find.
I'm not expecting to
find it exactly.
What I do is I have
a nearby point 3.
At 3 I do know the exact value
and I know the slope.
Ha.
Even my bad art is telling me
what's going to happen.
Let me write that 3 better.
a equals 3.
This is where I know
what's happening.
I know the value of F. It's
actually pretty small, but
it's negative.
I know the value of the slope.
So this is like I've blown
up this picture here.
This F is the negative 0.06.
Is the value of F.
The slope is 6.
And what's the x?
What is my improved guess
at the solution?
My improved guess, I don't
follow that curve because that
would be perfect.
But curves are hard to follow.
I'll follow the straight line
and that's my next x.
My better x.
You see, that's a lot better.
Let me do the numbers
here and you'll see
that it's a lot better.
What is the x that Newton's
method gives us?
I'm using Newton's
formula here.
So Newton's formula says that
x minus the a is 3.
And Newton's formula is going
to take equal to get an
approximate x.
Minus F of a.
So what's minus F of a?
That will be plus 0.06.
Divided by F prime of a,
the slope, which was 6.
What do I get?
I did 0.01.
Actually, the two examples,
of course, are parallel.
In a way, the graph of that
square root function kind of
got just flipped to this graph
of the x squared function.
The square root function and
the x squared function are
just sort of inverses
to each other.
Their graphs flip and so now the
slope is 6 instead of 1/6.
And do you see that that
distance, this is the x minus
a distance that has
to be 0.01.
That's what I concluded.
I go across 1 if I'm going
up/down by 6 because
the slope is 6.
OK, good.
Now, how close am I?
Well, I can't resist asking.
Suppose I multiply
3.01 by 3.01.
Am I close to the 9.06 that
was my whole goal?
And how close?
Of course, I'm not going
to be exact.
If I do that multiplication
I get 301, 903.
Combined I get 9 and the decimal
put it in there, 0601.
So by that method or by that
method I ended up with 3.01 as
much closer to the square
root of 9.06.
You see when I square 3.01,
it slightly overshot.
It slightly overshot the 9.06.
This little error there, this
is the overshoot error.
And when I square it, it's way
out in the ten thousandths
place, the fourth
decimal place.
OK.
So those are the two
parallel examples.
May I go back to the linear
approximation?
Because I think another
example's appropriate there.
And then I'll come back and give
another Newton example.
So that's my plan here.
Two examples of each.
So those examples were parallel,
the next two
examples will be a
little different.
So let me show you
example two.
So linear approximation
example two.
So I want--
OK, I'm going to look
for something that
I don't know exactly.
Let me take e to the 0.01.
What's the value of e to
the power of 0.01?
So the function is e to the x.
And I'm looking at x is 0.01.
That's what I want.
I am not going to get an
exact number here.
I'm going to follow
the tangent line.
I'm going to get
a close number.
So where shall I start?
I'll start with a number that's
close to that and where
I do you know the correct e.
And a number close
to that is take--
I'll choose a to be 0.
That's close to 0.01.
Then f of a is e
to the 0 power.
So that's 1.
Now for the straight line
approximation, I also need the
correct slope.
Correct slope at a?
Well, I do know the
slope of this.
f prime at 0.
That's my a.
f prime at 0.
0 is my a.
Well, I know the derivative
of e to the x.
That's one thing I like.
It's e to the x.
So at x equals 0, again I get a
1 for the derivative, which
is the same, I get e to the 0.
I get also a 1.
So now I know what's happening
at a equals 0.
I want to know approximately
what's happening at the nearby
point, 0.01.
e to the 0.01, e to the x.
This is the e to the x.
That's my function.
And I'm only going to get it
approximately, is the value at
this known point.
The value at the known point 0,
the exact exponential is 1.
Plus x minus a times the slope,
the corrects slope, at
this not quite perfect
point, 0.
And the correct slope is 1.
And of course, a was 0.
So you see I'm using all the
facts at a to get an
approximate fact at x.
And what have I got here?
I've just got 1 plus x.
You know, in a way,
that's perfect.
Because it shows what the linear
approximation is doing.
You remember the series
for e to the x?
My correct function
is e to the x.
My approximate function
is 1 plus x.
What's the connection?
You remember that e to the x,
the series for e to the x
started out 1 plus x plus 1/2
x squared, 1/6 x cubed.
Those are the guys, those are
the higher order corrections
that following the
line misses.
Those are the parts where the
function, the curve e to the
x, has left the line.
But if I don't go to far--So
this is 1.01.
x is 0.01.
That's my approximation.
1 plus x.
That's the thing to notice.
That linear approximations and
you could come back to that--
the formula for any
f and any a.
Linear approximations are just
like those power series.
Just like the e to the x equal
1 plus x plus 1/2 x squared
plus so on.
Except we cut them off after
just the constant term and the
linear term.
That's what this linear
approximation is about.
And you might say, what's
the next term?
And of course we know that the
next term is 1/2 x squared and
you could ask, what's the
next term in this?
In the general case, actually
let me tell you the next term.
Next would be--
but we're not using it.
Would be the 1/2.
It would be the x minus a
squared and it would be the
second derivative at a.
If we kept it that's
what we would keep.
OK, good.
You saw the main point of linear
approximation stop at
the linear term.
Finally, I go back to an example
of Newton's method.
A second example of
Newton's method.
So I've been thinking,
what should I do?
Let me use Newton's method
the way it's really used.
The way you use Newton's method
is you do this to come
close to the solution.
And then, you do it again.
You do it again starting
at 3.01.
So I planned to do just the same
thing for the next step
of Newton's method,
except the a.
I'm still aiming to solve this
same equation, but I'm going
to get closer than 3.01.
I got closer than 3 to 3.01.
Now I'm going to
restart there.
a is now going to be 3.01.
I need to compute F of a
for Newton's method.
So I have to do 3.01 squared and
take away that to see how
wrong I am.
HA.
We did 3.01 squared.
Actually, right there.
So if I take away the 9.06, the
F of a is-- well, that was
the whole point.
That it was pretty darn close.
But nothing compared to the
closeness we're going to get
at the second term.
And what's F prime of a?
Take the derivative 2x.
At the point a it's 2a.
And a is now 3.01.
So the slope is 6.02.
You see what I'm doing.
I'm just moving over to this
point, which that has become
now the a in the second try.
a in the second cycle
of Newton's method.
This is how Newton's method
really is used.
And now let's find the new
x, the highly improved
x, better than 3.01.
So Newton's method says
x, the new x, minus a.
I'm just using Newton's
formula.
x minus the a is supposed
to be minus the F of a.
So that's 0.0001 divided
by F prime of a, 6.02.
This is the delta
x you could say.
This is the little correction.
It's negative.
It means that we need to
pull back a little.
And you see that.
We slightly, slightly overshot
by following the tangent line.
The curve went up a little
across 0 a little before the
tangent line.
This is extremely close.
So now this gives me the new x
right here, 3.01 minus this
tiny little bit.
And so that's what the
calculator will do.
I hope you'll do it
on a calculator.
Just make the calculator take
that quantity, then
make it square it.
Then just go through this.
Find the new x.
Square it.
Subtract 9.06.
Let's see how close it is.
I believe that the error--
I don't know what it is.
That's more than I can do in
my head, squaring that 3.01
minus this little tiny bit.
But I am confident that the
error, the x new squared.
This is the formula for
x new, the second
cycle of Newton's method.
I think that minus the
9.06, I don't know--
can I just put a
bunch of zeros?
Somebody will want me to
put in a 1 here, so
I'll put in a 1.
I bet it's way out there.
So Newton's method is really
a terrific success.
Follow the line, then follow
the next tangent line.
Then follow the next one and you
home in very, very quickly
on the exact answer.
You get more and more decimal
places correct.
OK, that's two uses
of calculus coming
from the same idea.
The same idea delta
f over delta x.
In one case, it was f
that we didn't know.
In this case, it was x
that we didn't know.
In both cases that formula
gives a terrific and a
terrifically simple and
close approximation
to the exact answer.
Good.
Thank you very much.
ANNOUNCER: This has been
a production of MIT
OpenCourseWare and
Gilbert Strang.
Funding for this video was
provided by the Lord
Foundation.
To help OCW continue to provide
free and open access
to MIT courses, please
make a donation at
ocw.mit.edu/donate.
