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The topic for today is --
Today we're going to talk,
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I'm postponing the linear
equations to next time.
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Instead, I think it's a good
idea, since in real life,
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most of the
differential equations
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are solved by numerical methods
to introduce you to those right
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away.
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Even when you see
the compute where
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you saw the computer screen,
the solutions being drawn.
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Of course, what
really was happening
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was that the computer was
calculating the solutions
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numerically, and
plotting the points.
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So, this is the main
way, numerically,
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it is the main way differential
equations are actually solved,
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if they are of any
complexity at all.
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So, the problem is, that
initial value problem,
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let's write up the first
order problem the way
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we talked about it on Wednesday.
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And now, I'll specifically add
to that, the starting point
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that you used when you did
the computer experiments.
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And, I'll write the
starting point this way.
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So, y of x0 should be y0.
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So, this is the
initial condition,
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and this is the first order
differential equation.
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And, as you know,
the two of them
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together are called an IVP,
an initial value problem,
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which means two things,
the differential
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equation and the
initial value that you
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want to start the solution at.
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Okay, now, the method we
are going to talk about,
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the basic method of
which many others
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are merely refinements
in one way or another,
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is called Euler's method.
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Euler, who did, of course,
everything in analysis,
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as far as I know,
didn't actually
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use it to compute solutions
of differential equations.
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His interest was theoretical.
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He used it as a method of
proving the existence theorem,
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proving that solutions existed.
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But, nowadays, it's used
to calculate the solutions
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numerically.
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And, the method is very
simple to describe.
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It's so naive that
you probably think
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that if you that
living 300 years ago,
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you would have discovered
it and covered yourself
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with glory for all eternity.
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So, here is our starting
point, (x0, y0).
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Now, what information
do we have?
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At that point all we have is
the little line element, whose
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slope is given by f of (x, y).
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So, if I start the
solution, the only way
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the solution could
possibly go would
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be to start off in
that direction, since I
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have no other information.
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At least it has the correct
direction at (x0, y0).
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But, of course, it's not likely
to have the correct direction
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anywhere else.
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Now, what you do, then,
is choose a step size.
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I'll try just two
steps of the method.
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That's, I think, good enough.
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Choose a uniform step size,
which is usually called h.
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And, you continue that
solution until you
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get to the next point, which
will be x0 + h, as I've
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drawn it on the picture.
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So, we get to here.
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We stop at that
point, and now you
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recalculate what the
line element is here.
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Suppose, here, the line
element, now, through this point
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goes like that.
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Well, then, that's
the new direction
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that you should start
out with going from here.
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And so, the next step of the
process will carry you to here.
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That's two steps
of Euler's method.
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Notice it produces a broken line
approximation to the solution.
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But, in fact, you only
see that broken line
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if you are at a computer if
you are looking at the computer
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visual, for example, whose
purpose is to illustrate
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for you Euler's method.
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In actual practice,
what you see is,
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the computer is
simply calculating
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this point, that point, and
the succession of points.
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And, many programs
will just automatically
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connect those points by
a smooth looking curve
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if that's what
you prefer to see.
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Well, that's all there
is to the method.
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What we have to do now is derive
the equations for the method.
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Now, how are we
going to do that?
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Well, the essence of it is
how to get from the nth step
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to the n plus first step?
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So, I'm going to draw a picture
just to illustrate that.
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So, now we are not at x0.
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But let's say we've
already gotten to (xn, yn).
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How do I take the next step?
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Well, I take the line
element, and it goes up
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like that, let's say,
because the slope is this.
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I'm going to call
that slope A sub n.
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Of course, A sub n is the
value of the right hand side
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at the point (xn, yn), and we
will need that in the equation,
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-- -- but I think it will be
a little bit clearer if I just
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give it a capital
letter at this point.
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Now, this is the new point,
and all I want to know
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is what are its coordinates?
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Well, the x n plus one is there.
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The y n plus one is here.
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Clearly I should
draw this triangle,
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complete the triangle.
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This side of the triangle,
the hypotenuse has slope An.
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This side of the
triangle has length h.
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h is the step size.
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Perhaps I'd better
indicate that,
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actually put that up so
that you know the word step.
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It's the step size
on the x axis,
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how far you have to go to get
from each x to the next one.
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What's this?
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Well, if that slope has this,
the slope An, this is h.
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Then this must be h times
An, the length of that side,
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in order that the ratio of
the height to this width
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should be An.
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And, that gives us the method.
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How do I get from, clearly, to
get from xn to x n plus one,
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I simply add h.
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So, that's the
trivial part of it.
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The interesting thing is, how
do I get the new y n plus one?
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And so, the best way to write
it as, that y n plus one
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minus yn divided
by h, well, sorry,
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y n plus one minus yn is this
line, the same as the line
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h times An.
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So, that's the way to write it.
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Or, since the computer is
interested in calculating
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y n plus one itself, put
this on the other side.
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You take the old yn,
the previous one,
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and to it, you add h times An.
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And, what, pray tell, is An?
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Well, the computer has to be
told that An is the value of f.
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So, now, with that,
let's actually
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write the Euler program,
not the program,
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but the Euler-- the
Euler method equations,
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let's just call it
the Euler equations.
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What will they be?
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First of all, the new
x is the old x plus h.
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The new y is just what
I've written there,
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the old y plus h times a
certain number, An, and finally,
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An has the value-- It's the
slope of the line element here,
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and therefore by definition,
that's f of (xn,yn).
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So, if these three equations
which define Euler's method.
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I assume in 1.00 you must
be asked to, at some point,
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as an exercise in the term
at one point to program
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the computer in C or whatever
they're using, Java, now,
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I guess, to do Euler's method.
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And, these would be
the recursive equations
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that you would
put in to do that.
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Okay, let's try
an example, then.
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So, what would be a
good color for Euler?
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Well, a purple.
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I assume nobody can see purple.
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Is that correct?
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Can anyone in the back of the
room see that that's purple?
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Okay.
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Sit closer.
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So, let's calculate.
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The example, I'll
use a simple example,
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but it's not entirely trivial.
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My example is going
to be the equation,
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x squared minus y squared
on the right hand side.
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And, let's start with y of
zero equals one, let's say.
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And so, this is my
initial value problem,
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that pair of equations.
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And, I have to
specify a step size.
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So, let's take the
step size to 0.1.
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You choose the step size,
or the computer does.
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We'll have to talk about
that in a few minutes.
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Now, what do you do?
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Well, I say this is a nontrivial
equation because this equation,
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as far as I know, cannot be
solved in terms of elementary
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functions.
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So, this equation
would be, in fact,
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a very good candidate for a
numerical method like Euler's.
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And, you had to use it, or maybe
it was the other way around,
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I forget.
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On your problem set, you drew a
picture of the direction field
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and answered some questions
about the isoclines,
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how the solutions behave.
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Now, the main thing
I want you to get,
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this is not just
for Euler's, talking
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about Euler's equations.
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But in general, for
the calculations
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you have to do in this course,
it's extremely important
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to be systematic because
if you are not systematic,
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you know, if you just scribble,
scribble, scribble, scribble,
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scribble, you can do
the work, but it becomes
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impossible to find mistakes.
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00:11:25,600 --> 00:11:28,921
You must do the work
in a form in which it
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can be checked, which you
can look over it and find,
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and try to see where mistakes
are if, in fact, there are any.
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So, I strongly suggest,
this is not a suggestion,
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00:11:41,220 --> 00:11:46,744
this is a command, that you make
a little table to do Euler's
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00:11:46,744 --> 00:11:51,213
method by hand, I'd only
ask you for a step or two,
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00:11:51,213 --> 00:11:53,832
but since I'm just
trying to make
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sure you have some
idea of these equations
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and where they come from.
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So, first, the value of
n, then the value of xn,
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then the value of
the yn, and then,
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00:12:06,768 --> 00:12:09,840
a couple of more
columns which tell you
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how to do the calculation.
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00:12:11,888 --> 00:12:15,999
You are going to need
the value of the slope,
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00:12:15,999 --> 00:12:18,454
and it's probably
a good idea, also,
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00:12:18,454 --> 00:12:23,000
because otherwise you'll
forget it, to put in h An
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00:12:23,000 --> 00:12:26,000
because that occurs
in the formula.
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00:12:26,000 --> 00:12:30,000
All right, let's start doing it.
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00:12:30,000 --> 00:12:32,541
The first value of n is zero.
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That's the starting point.
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00:12:34,000 --> 00:12:38,000
At the starting point, (x0,
y0), x has the value zero,
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00:12:38,000 --> 00:12:42,000
and y has the value
one, so, zero and one.
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00:12:42,000 --> 00:12:45,000
In other words, starting,
I'm carrying out
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00:12:45,000 --> 00:12:47,800
exactly what I drew
pictorially only now
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00:12:47,800 --> 00:12:51,664
I'm doing it arithmetically
using a table
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00:12:51,664 --> 00:12:55,000
and substituting
into the formulas.
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00:12:55,000 --> 00:13:00,000
Okay, the next thing we
have to calculate is An.
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00:13:00,000 --> 00:13:03,000
Well, since An is the value
of the right hand side,
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00:13:03,000 --> 00:13:07,000
at the point zero one,
you have to plug that in.
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00:13:07,000 --> 00:13:11,000
The right hand side is x
squared minus y squared.
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00:13:11,000 --> 00:13:14,000
So, it's 0 squared
minus 1 squared.
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00:13:14,000 --> 00:13:20,000
The value of the slope, there,
is minus one, negative one.
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Now, I have to
multiply that by H.
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00:13:22,000 --> 00:13:22,999
h is 0.1.
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00:13:22,999 --> 00:13:25,452
So, it's negative,
I'll never learn that.
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00:13:25,452 --> 00:13:28,333
The way you learn to
talk in kindergarten
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00:13:28,333 --> 00:13:33,250
is the way you learn to
talk the rest of your life,
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00:13:33,250 --> 00:13:34,500
unfortunately.
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00:13:34,500 --> 00:13:38,800
In kindergarten, we said minus.
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Negative 0.1.
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n is one now.
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What's the value of xn?
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Well, to the old
value I add 1/10.
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What's the value of y?
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00:13:50,775 --> 00:13:56,000
Well, at this point, you
have to do the calculation.
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00:13:56,000 --> 00:13:59,000
It's the old value of y.
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To get this new value, it's
the old value plus this number.
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00:14:06,000 --> 00:14:12,816
Well, that's this plus
that number is nine tenths.
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00:14:12,816 --> 00:14:19,180
An, now I have to calculate
the new slope at this point.
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00:14:19,180 --> 00:14:25,500
Okay, that is one tenth squared
minus nine tenths squared.
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00:14:25,500 --> 00:14:38,000
That's 0.01 minus 0.81, which
makes minus 0.80, I hope.
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00:14:38,000 --> 00:14:40,270
Check it on your calculators.
243
00:14:40,270 --> 00:14:43,428
Whip them out and
press the buttons.
244
00:14:43,428 --> 00:14:46,832
I now multiply that
by h, which means
245
00:14:46,832 --> 00:14:51,428
it's going to be minus 0.08,
perhaps with a zero after.
246
00:14:51,428 --> 00:14:55,000
I didn't tell you how
many decimal places.
247
00:14:55,000 --> 00:14:59,000
Let's carry it out to
two decimal places.
248
00:14:59,000 --> 00:15:03,000
I think that will
be good enough.
249
00:15:03,000 --> 00:15:10,000
And finally, the last step, 2,
here, add another one tenth,
250
00:15:10,000 --> 00:15:15,000
so the value of x
is now two tenths.
251
00:15:15,000 --> 00:15:18,000
And finally, what's
the value of y?
252
00:15:18,000 --> 00:15:22,000
Well, I didn't tell
you where to stop.
253
00:15:22,000 --> 00:15:27,328
Let's stop at y of
0.2 because there's
254
00:15:27,328 --> 00:15:31,000
no more room on the blackboard.
255
00:15:31,000 --> 00:15:34,500
About approximately
how big is that?
256
00:15:34,500 --> 00:15:37,816
In other words,
this is, then, this
257
00:15:37,816 --> 00:15:43,800
is going to be the old y
plus this number, which
258
00:15:43,800 --> 00:15:48,000
seems to be 0.82 to me.
259
00:15:48,000 --> 00:15:51,000
So, the answer is,
the new value is 0.82.
260
00:15:51,000 --> 00:15:52,665
Okay, we got a number.
261
00:15:52,665 --> 00:15:55,200
We did what we are
supposed to do.
262
00:15:55,200 --> 00:15:56,000
We got a number.
263
00:15:56,000 --> 00:15:57,000
Next question?
264
00:15:57,000 --> 00:16:00,000
Now, let's ask a few questions.
265
00:16:00,000 --> 00:16:02,625
One of the first,
most basic things
266
00:16:02,625 --> 00:16:04,998
is, you know, how right is this?
267
00:16:04,998 --> 00:16:09,800
How can I answer such a question
if I have no explicit formula
268
00:16:09,800 --> 00:16:11,000
for that solution?
269
00:16:11,000 --> 00:16:15,000
That's the basic problem
with numerical calculation.
270
00:16:15,000 --> 00:16:17,400
In other words, I
have to wander around
271
00:16:17,400 --> 00:16:19,800
in the dark to some
extent, and yet
272
00:16:19,800 --> 00:16:24,000
have some idea when I've arrived
at the place that I want to go.
273
00:16:24,000 --> 00:16:28,000
Well, the first question
I'd like to answer,
274
00:16:28,000 --> 00:16:30,800
is this too high or too low?
275
00:16:30,800 --> 00:16:36,000
Is Euler, sorry, he'll forgive
me in heaven, I will use him.
276
00:16:36,000 --> 00:16:38,500
By this, I mean,
is the result, let
277
00:16:38,500 --> 00:16:43,375
me just say something first,
and that I'll criticize it.
278
00:16:43,375 --> 00:16:46,000
Is Euler too high or too low?
279
00:16:46,000 --> 00:16:50,333
In other words, is the result
of using Euler's method,
280
00:16:50,333 --> 00:16:53,428
i.e. is this number
too high or too low?
281
00:16:53,428 --> 00:16:58,500
Is it higher than the right
answer, what it should be?
282
00:16:58,500 --> 00:17:02,750
Or, is it lower than
the right answer?
283
00:17:02,750 --> 00:17:06,776
Or, God forbid, is
it exactly right?
284
00:17:06,776 --> 00:17:09,600
Well, it's almost
never exactly right.
285
00:17:09,600 --> 00:17:12,000
That's not an option.
286
00:17:12,000 --> 00:17:16,000
Now, how will we
answer that question?
287
00:17:16,000 --> 00:17:19,000
Well, let's answer
it geometrically.
288
00:17:19,000 --> 00:17:24,454
Basically, if the solution
were a straight line, then
289
00:17:24,454 --> 00:17:29,000
the Euler method would be
exactly right all the time.
290
00:17:29,000 --> 00:17:31,775
But, it's not a line.
291
00:17:31,775 --> 00:17:34,000
Then it's a curve.
292
00:17:34,000 --> 00:17:38,000
Well, the critical
question is, is it curved?
293
00:17:38,000 --> 00:17:39,284
Is the solution?
294
00:17:39,284 --> 00:17:41,000
So, here's a solution.
295
00:17:41,000 --> 00:17:44,600
Let's call it y1
of x, and let's say
296
00:17:44,600 --> 00:17:46,776
here was the starting point.
297
00:17:46,776 --> 00:17:49,000
Here, the solution is convex.
298
00:17:49,000 --> 00:17:52,000
And, here the solution
is concave, right?
299
00:17:52,000 --> 00:17:57,000
Concave up or concave down,
if you learn those words,
300
00:17:57,000 --> 00:18:01,000
but I think those have,
by now, I hope pretty well
301
00:18:01,000 --> 00:18:03,000
disappeared from the curriculum.
302
00:18:03,000 --> 00:18:06,888
Call it, if you
haven't up until now,
303
00:18:06,888 --> 00:18:10,000
what mathematicians
call it, convex is that,
304
00:18:10,000 --> 00:18:11,632
and the other one is concave.
305
00:18:11,632 --> 00:18:13,400
Well, how do Euler's
solutions look?
306
00:18:13,400 --> 00:18:15,000
Well, I'll just sketch.
307
00:18:15,000 --> 00:18:18,684
I think from this you can see
already, when you start out
308
00:18:18,684 --> 00:18:22,000
on the Euler's solution,
it's going to go like that.
309
00:18:22,000 --> 00:18:23,500
Now you are too low.
310
00:18:23,500 --> 00:18:25,900
Well, let's suppose after
that, the line element
311
00:18:25,900 --> 00:18:29,998
here is approximately the
same as what it is there,
312
00:18:29,998 --> 00:18:32,000
or roughly parallel.
313
00:18:32,000 --> 00:18:34,000
After all, they are
not too far apart.
314
00:18:34,000 --> 00:18:37,000
And, the direction
field is continuous.
315
00:18:37,000 --> 00:18:39,331
That is, the directions
don't change drastically
316
00:18:39,331 --> 00:18:40,900
from one point to another.
317
00:18:40,900 --> 00:18:43,333
But now, you see
it's still too low.
318
00:18:43,333 --> 00:18:47,284
It's even lower as it
pathetically tries to follow.
319
00:18:47,284 --> 00:18:50,125
It's losing territory,
and that's basically
320
00:18:50,125 --> 00:18:52,000
because the curve is convex.
321
00:18:52,000 --> 00:18:53,800
Exactly the opposite
what would happen
322
00:18:53,800 --> 00:18:59,284
if the curve were concave,
if the solution curve were
323
00:18:59,284 --> 00:19:00,000
concave.
324
00:19:00,000 --> 00:19:02,904
Now it's too high,
and it's not going
325
00:19:02,904 --> 00:19:07,632
to be able to correct that
as long as the solution
326
00:19:07,632 --> 00:19:09,000
curve stays concave.
327
00:19:09,000 --> 00:19:13,000
Well, that's probably
too optimistic.
328
00:19:13,000 --> 00:19:15,220
It's probably more like this.
329
00:19:15,220 --> 00:19:21,000
So, in other words, in this
case, if the curve is convex,
330
00:19:21,000 --> 00:19:25,000
Euler is going to be too
high, sorry, too low.
331
00:19:25,000 --> 00:19:28,125
Let's put E for Euler.
332
00:19:28,125 --> 00:19:30,000
How about that?
333
00:19:30,000 --> 00:19:31,140
Euler is too low.
334
00:19:31,140 --> 00:19:34,500
If it's concave, then
Euler is too high.
335
00:19:34,500 --> 00:19:36,000
Okay, that's great.
336
00:19:36,000 --> 00:19:40,000
There's just one little
problem left, namely,
337
00:19:40,000 --> 00:19:44,000
if we don't have a
formula for the solution,
338
00:19:44,000 --> 00:19:47,178
and we don't have
a computer that's
339
00:19:47,178 --> 00:19:51,270
busy drawing the picture
for us, in which case
340
00:19:51,270 --> 00:19:54,583
we wouldn't need
any of this anyway,
341
00:19:54,583 --> 00:20:01,000
how are we supposed to tell
if it's convex or concave?
342
00:20:01,000 --> 00:20:02,284
Back to calculus.
343
00:20:02,284 --> 00:20:04,000
Calculus to the rescue!
344
00:20:04,000 --> 00:20:05,665
When is a curve convex?
345
00:20:05,665 --> 00:20:09,089
A curve is convex if its
second derivative is positive
346
00:20:09,089 --> 00:20:13,713
because the first to be convex
means the first derivative is
347
00:20:13,713 --> 00:20:16,000
increasing all the time.
348
00:20:16,000 --> 00:20:18,000
And therefore, the
second derivative,
349
00:20:18,000 --> 00:20:22,000
which is the derivative
of the first derivative,
350
00:20:22,000 --> 00:20:23,284
should be positive.
351
00:20:23,284 --> 00:20:26,000
Just the opposite
here; the curve,
352
00:20:26,000 --> 00:20:29,000
the slope is, the
first derivative,
353
00:20:29,000 --> 00:20:33,000
is decreasing all the time and
therefore the second derivative
354
00:20:33,000 --> 00:20:34,000
is negative.
355
00:20:34,000 --> 00:20:39,500
So, all we have to do is decide
what the second derivative
356
00:20:39,500 --> 00:20:41,000
of the solution is.
357
00:20:41,000 --> 00:20:43,000
We should probably
call it a solution.
358
00:20:43,000 --> 00:20:46,000
y of x is a little too vague.
359
00:20:46,000 --> 00:20:49,000
y1 means the solution
started at this point.
360
00:20:49,000 --> 00:20:51,625
So, in fact, probably
it would have
361
00:20:51,625 --> 00:20:54,856
been better from the beginning
to call that y1, except there's
362
00:20:54,856 --> 00:20:57,000
no room, y1, let's say.
363
00:20:57,000 --> 00:20:59,800
That means the solution
which started out
364
00:20:59,800 --> 00:21:01,750
at the point, (0, 1).
365
00:21:01,750 --> 00:21:05,332
So, I'm still talking about
at a solution like that.
366
00:21:05,332 --> 00:21:08,855
All right, so I want to
know if this is positive,
367
00:21:08,855 --> 00:21:12,220
the second derivative is
positive at the starting point,
368
00:21:12,220 --> 00:21:14,000
zero, or it's negative.
369
00:21:14,000 --> 00:21:17,000
Now, again, how you can
regulate the second derivative,
370
00:21:17,000 --> 00:21:20,000
if you don't know what the
solution is explicitly,
371
00:21:20,000 --> 00:21:24,070
then the answer is you can do it
from the differential equation
372
00:21:24,070 --> 00:21:24,570
itself.
373
00:21:24,570 --> 00:21:26,000
How do I do that?
374
00:21:26,000 --> 00:21:30,250
Well: easy. y prime equals
x squared minus y squared.
375
00:21:30,250 --> 00:21:34,416
Okay, that tells
me how to calculate
376
00:21:34,416 --> 00:21:39,000
y prime if I know
the value of x and y,
377
00:21:39,000 --> 00:21:41,000
in other words, the 0.01.
378
00:21:41,000 --> 00:21:45,000
What would be the value
of y double prime?
379
00:21:45,000 --> 00:21:48,000
Well, differentiate
the equation.
380
00:21:48,000 --> 00:21:52,000
It's two x minus two y y prime.
381
00:21:52,000 --> 00:21:55,000
Don't forget to
use the chain rule.
382
00:21:55,000 --> 00:22:00,000
So, if I want to calculate
at (0, 1), in other words,
383
00:22:00,000 --> 00:22:04,000
if my starting point is that
curve convex or concave,
384
00:22:04,000 --> 00:22:10,000
well, let's calculate.
y of zero equals one.
385
00:22:10,000 --> 00:22:12,000
Okay, what's y prime of zero?
386
00:22:12,000 --> 00:22:16,000
Well, I don't have to
repeat that calculation.
387
00:22:16,000 --> 00:22:18,333
Using this, I've
already calculated
388
00:22:18,333 --> 00:22:20,000
that it was negative one.
389
00:22:20,000 --> 00:22:24,000
And now, the new thing,
what's y double prime of zero?
390
00:22:24,000 --> 00:22:26,000
Well, it is this.
391
00:22:26,000 --> 00:22:27,200
I'll write it out.
392
00:22:27,200 --> 00:22:31,000
It's two times zero minus two
times negative y, which is one,
393
00:22:31,000 --> 00:22:36,000
two times one times y prime,
which is negative one.
394
00:22:36,000 --> 00:22:43,400
You want to see we are pulling
ourselves up by our own boot
395
00:22:43,400 --> 00:22:45,000
straps, which is impossible.
396
00:22:45,000 --> 00:22:49,000
But, it is not impossible
because we are doing it.
397
00:22:49,000 --> 00:22:50,712
So, what's the answer?
398
00:22:50,712 --> 00:22:54,000
Zero here, two, I've
calculated without having
399
00:22:54,000 --> 00:23:00,270
the foggiest idea of what the
solution is or how it looks.
400
00:23:00,270 --> 00:23:03,571
I've calculated that
its second derivative
401
00:23:03,571 --> 00:23:07,000
at the starting point is two.
402
00:23:07,000 --> 00:23:13,000
Therefore, my solution is
convex at the starting point.
403
00:23:13,000 --> 00:23:17,000
And therefore, this
Euler approximation,
404
00:23:17,000 --> 00:23:22,000
if I don't carry it out
too far, will be too low.
405
00:23:22,000 --> 00:23:25,000
So, it's convex Euler, too low.
406
00:23:25,000 --> 00:23:32,000
Now, you could argue, yeah,
well, what about this?
407
00:23:32,000 --> 00:23:36,000
[LAUGHTER] So, you
could go like this,
408
00:23:36,000 --> 00:23:39,000
and then you can
see it catches up.
409
00:23:39,000 --> 00:23:45,875
Well, of course, if the curve
changes from convex to concave,
410
00:23:45,875 --> 00:23:52,125
then it's really impossible
to make any prediction at all.
411
00:23:52,125 --> 00:23:54,000
That's a difficulty.
412
00:23:54,000 --> 00:24:02,000
So, all this analysis is
only if you stay very nearby.
413
00:24:02,000 --> 00:24:04,763
However, I wanted to show
you, the main purpose
414
00:24:04,763 --> 00:24:08,496
of it in my mind
was to show you how
415
00:24:08,496 --> 00:24:11,000
do you use, it's
these equations,
416
00:24:11,000 --> 00:24:14,000
how to use the differential
equation itself to get
417
00:24:14,000 --> 00:24:17,885
information about the solutions,
without actually being
418
00:24:17,885 --> 00:24:20,500
able to calculate the solutions?
419
00:24:20,500 --> 00:24:23,888
Now, so that's the
method, and that's
420
00:24:23,888 --> 00:24:27,000
how to find out
something about it.
421
00:24:27,000 --> 00:24:32,000
And now, what I'd like
to talk about is errors.
422
00:24:32,000 --> 00:24:34,000
How do I handle, right?
423
00:24:34,000 --> 00:24:38,500
So, in a sense, I've
started the error analysis.
424
00:24:38,500 --> 00:24:43,600
In other words, the error,
by definition, the error
425
00:24:43,600 --> 00:24:46,000
is this difference, e.
426
00:24:46,000 --> 00:24:50,000
So, in other words,
what I'm asking here,
427
00:24:50,000 --> 00:24:52,000
is the error positive?
428
00:24:52,000 --> 00:24:55,816
It depends which we measure it.
429
00:24:55,816 --> 00:24:59,571
Usually, you take
this minus that.
430
00:24:59,571 --> 00:25:04,000
So, here, the error would
be considered positive,
431
00:25:04,000 --> 00:25:07,600
and here it would be
considered negative,
432
00:25:07,600 --> 00:25:13,908
although I'm sure there's a book
somewhere in the world, which
433
00:25:13,908 --> 00:25:15,270
does the opposite.
434
00:25:15,270 --> 00:25:20,080
Most hedge by just using the
absolute value of the error
435
00:25:20,080 --> 00:25:23,888
plus a statement that the
method is producing answers
436
00:25:23,888 --> 00:25:27,000
which are too low or too high.
437
00:25:27,000 --> 00:25:30,714
The question, then,
is, naturally, this
438
00:25:30,714 --> 00:25:35,000
is not the world's best method.
439
00:25:35,000 --> 00:25:36,750
It's not as bad as it seems.
440
00:25:36,750 --> 00:25:39,600
It's not the world's best
method because that convexity
441
00:25:39,600 --> 00:25:42,600
and concavity means that
you are automatically
442
00:25:42,600 --> 00:25:45,000
introducing a systematic error.
443
00:25:45,000 --> 00:25:47,565
If you can predict
which way the error is
444
00:25:47,565 --> 00:25:50,250
going to be by just knowing
whether the curve is
445
00:25:50,250 --> 00:25:52,665
convex or concave,
it's not what you want.
446
00:25:52,665 --> 00:25:55,250
I mean, you want
to at least have
447
00:25:55,250 --> 00:25:57,000
a chance of getting
the right answer,
448
00:25:57,000 --> 00:25:59,800
whereas this is telling
you you're definitely
449
00:25:59,800 --> 00:26:01,750
going to get the wrong answer.
450
00:26:01,750 --> 00:26:04,332
All it tells you is,
and it's telling you
451
00:26:04,332 --> 00:26:09,000
whether your answer is going
to be too high or too low.
452
00:26:09,000 --> 00:26:14,220
We've like a better chance
of getting the right answer.
453
00:26:14,220 --> 00:26:20,227
Now, so the question is, how
do you get a better method?
454
00:26:20,227 --> 00:26:23,600
A search is for a better method.
455
00:26:23,600 --> 00:26:26,888
Now, the first
method, which will
456
00:26:26,888 --> 00:26:31,665
occur, I'm sure, to anyone
who looks at that picture,
457
00:26:31,665 --> 00:26:36,000
is, look, if you
want this yellow line
458
00:26:36,000 --> 00:26:40,000
to follow the white
one, the white solution,
459
00:26:40,000 --> 00:26:43,000
more accurately,
for heaven's sake,
460
00:26:43,000 --> 00:26:47,375
don't take such big steps.
461
00:26:47,375 --> 00:26:54,908
Take small steps, and then
it will follow better.
462
00:26:54,908 --> 00:27:00,000
All right, let's draw a picture.
463
00:27:00,000 --> 00:27:02,000
Excuse me.
464
00:27:02,000 --> 00:27:08,000
My little box of
treasures, here.
465
00:27:08,000 --> 00:27:13,000
[LAUGHTER] So, use
a smaller step size.
466
00:27:13,000 --> 00:27:18,332
And the picture, roughly,
which is going to justify that,
467
00:27:18,332 --> 00:27:21,000
will look like this.
468
00:27:21,000 --> 00:27:24,328
If the solution curve
looks like this, then
469
00:27:24,328 --> 00:27:29,500
with a big step size,
I'm liable to have
470
00:27:29,500 --> 00:27:33,000
something that looks like that.
471
00:27:33,000 --> 00:27:36,072
But, if I take a
smaller step size,
472
00:27:36,072 --> 00:27:38,333
suppose I halve the step size.
473
00:27:38,333 --> 00:27:40,428
How's it going to look, then?
474
00:27:40,428 --> 00:27:44,000
Well, I better switch
to a different color.
475
00:27:44,000 --> 00:27:50,000
If I halve the step size, I'll
get a littler, goes like that.
476
00:27:50,000 --> 00:27:52,500
And now it's following closer.
477
00:27:52,500 --> 00:27:55,714
Of course, I'm
stacking the deck,
478
00:27:55,714 --> 00:28:00,000
but see how close it follows?
479
00:28:00,000 --> 00:28:02,000
I'm definitely not to
be trusted on this.
480
00:28:02,000 --> 00:28:05,000
Okay, let's do the opposite,
make really big steps.
481
00:28:05,000 --> 00:28:06,500
Suppose instead
of the yellow ones
482
00:28:06,500 --> 00:28:09,125
I used the green one
of double step size.
483
00:28:09,125 --> 00:28:11,400
Well, what would
have happened then?
484
00:28:11,400 --> 00:28:15,000
Well, I've started out, but now
I've gone all the way to there.
485
00:28:15,000 --> 00:28:17,000
And now, on my
way up, of course,
486
00:28:17,000 --> 00:28:19,125
it has a little further to go.
487
00:28:19,125 --> 00:28:22,000
But, if for some reason, I
stop there, you could see,
488
00:28:22,000 --> 00:28:23,250
I would be still lower.
489
00:28:23,250 --> 00:28:29,000
In other words, the bigger the
steps size, the more the error.
490
00:28:29,000 --> 00:28:33,000
And, where are the errors
that we are talking about?
491
00:28:33,000 --> 00:28:36,744
Well, the way to think
of the errors, this
492
00:28:36,744 --> 00:28:39,776
is the error, that
number the error.
493
00:28:39,776 --> 00:28:42,555
You can make it
positive, negative,
494
00:28:42,555 --> 00:28:48,713
or just put it automatically an
absolute value sign around it.
495
00:28:48,713 --> 00:28:51,000
That's not so important.
496
00:28:51,000 --> 00:28:53,568
So, in other words,
the conclusion
497
00:28:53,568 --> 00:28:57,330
is, that the error
e, the difference
498
00:28:57,330 --> 00:29:01,496
between the true value
that I should have gotten,
499
00:29:01,496 --> 00:29:05,332
and the Euler value that
the calculation produced,
500
00:29:05,332 --> 00:29:10,500
that the error e,
depends on the step size.
501
00:29:10,500 --> 00:29:14,856
Now, how does it depend
on the step size?
502
00:29:14,856 --> 00:29:18,284
Well, it's impossible to
give an exact formula,
503
00:29:18,284 --> 00:29:20,999
but there's an approximate
answer, which is,
504
00:29:20,999 --> 00:29:22,400
by and large, true.
505
00:29:22,400 --> 00:29:27,000
So, the answer is, e is
going to be a function of h.
506
00:29:27,000 --> 00:29:28,500
What function?
507
00:29:28,500 --> 00:29:31,776
Well, asymptotically,
which is another way
508
00:29:31,776 --> 00:29:35,200
of putting quotation marks
around, what did I say?
509
00:29:35,200 --> 00:29:40,000
It's going to be a constant,
some constant, times H.
510
00:29:40,000 --> 00:30:14,220
[LAUGHTER] It looks like
this, and for this reason
511
00:30:14,220 --> 00:30:21,000
it's called a first order, the
Euler is a first order method.
512
00:30:21,000 --> 00:30:26,918
And now, first order does
not refer to the first order
513
00:30:26,918 --> 00:30:29,714
of the differential equation.
514
00:30:29,714 --> 00:30:35,776
It's not the first order
because it's y prime
515
00:30:35,776 --> 00:30:38,000
equals f of (x, y).
516
00:30:38,000 --> 00:30:41,856
The first order
means the fact that h
517
00:30:41,856 --> 00:30:44,000
occurs to the first power.
518
00:30:44,000 --> 00:30:46,912
The way people
usually say this is
519
00:30:46,912 --> 00:30:50,332
since the normal way of
decreasing the step size,
520
00:30:50,332 --> 00:30:54,500
as you'll see as is you
try to use a computer
521
00:30:54,500 --> 00:30:58,000
visual that deals
with the Euler method,
522
00:30:58,000 --> 00:31:01,108
which I highly
recommend, by the way,
523
00:31:01,108 --> 00:31:04,499
so highly recommended
that you have to do it,
524
00:31:04,499 --> 00:31:12,000
is that the way to say it, each
new step halves the step size.
525
00:31:12,000 --> 00:31:16,305
That's the usual way to do it.
526
00:31:16,305 --> 00:31:23,180
If you halve the step size,
since this is a constant,
527
00:31:23,180 --> 00:31:33,000
if I halve the step size, I
halve the error, approximately.
528
00:31:33,000 --> 00:31:36,000
Halve the step size,
halve the error.
529
00:31:36,000 --> 00:31:40,905
That tells you how the
error varies with step
530
00:31:40,905 --> 00:31:44,000
size for Euler's method.
531
00:31:44,000 --> 00:31:47,875
Please understand,
that's what people
532
00:31:47,875 --> 00:31:52,428
say, and please understand
the grammatical construction.
533
00:31:52,428 --> 00:31:56,636
Since everyone in
the math department
534
00:31:56,636 --> 00:32:03,000
has a cold these days
except me for the moment,
535
00:32:03,000 --> 00:32:09,000
everyone goes around
chanting this mantra.
536
00:32:09,000 --> 00:32:11,284
This is totally irrelevant.
537
00:32:11,284 --> 00:32:15,000
This whole mantra, feed
a cold, starve a fever.
538
00:32:15,000 --> 00:32:18,200
And if you asked
them what it means,
539
00:32:18,200 --> 00:32:21,999
they say eat a lot
if you have a cold.
540
00:32:21,999 --> 00:32:25,815
And if you have a fever,
don't eat very much, which
541
00:32:25,815 --> 00:32:29,000
is not what it means at all.
542
00:32:29,000 --> 00:32:33,000
Grammatically, it's exactly
the same construction as this.
543
00:32:33,000 --> 00:32:37,000
What this means is if
you halve the step size,
544
00:32:37,000 --> 00:32:38,500
you will halve the error.
545
00:32:38,500 --> 00:32:41,600
That's what feed a cold,
starve a fever means.
546
00:32:41,600 --> 00:32:44,750
And, remember this for
the rest of your life.
547
00:32:44,750 --> 00:32:48,850
If you feed a cold, if you eat
too much when you have a cold,
548
00:32:48,850 --> 00:32:53,600
you will get a fever and end up
still having to starve yourself
549
00:32:53,600 --> 00:32:57,000
because, of course, nobody,
when you have a fever,
550
00:32:57,000 --> 00:33:02,000
nobody feels like eating,
so they don't eat anything.
551
00:33:02,000 --> 00:33:05,330
All right, you got that?
552
00:33:05,330 --> 00:33:06,000
Good.
553
00:33:06,000 --> 00:33:14,000
I want all of you to go home
and tell that to your mothers.
554
00:33:14,000 --> 00:33:21,000
You know, that's the way
we always used to speak.
555
00:33:21,000 --> 00:33:25,666
Grimmer ones: spare
the rod, spoil
556
00:33:25,666 --> 00:33:33,750
the child does not mean that
you should not hit your kid.
557
00:33:33,750 --> 00:33:40,665
It means that if you
fail to hit your kid,
558
00:33:40,665 --> 00:33:45,800
he or she will be spoiled,
whatever that means.
559
00:33:45,800 --> 00:33:50,000
So, you don't want to do that.
560
00:33:50,000 --> 00:33:56,000
I guess the mantra today
would be, I don't know.
561
00:33:56,000 --> 00:34:00,081
Okay, so the first
line of defense
562
00:34:00,081 --> 00:34:05,500
is simply to keep having
the step size in Euler.
563
00:34:05,500 --> 00:34:10,100
And, what people do is, if
they don't want to use anything
564
00:34:10,100 --> 00:34:13,904
better than Euler's method, is
you keep having the step size
565
00:34:13,904 --> 00:34:16,665
until the curve doesn't
seem to change anymore.
566
00:34:16,665 --> 00:34:20,000
And then you say, well,
that must be the solution.
567
00:34:20,000 --> 00:34:22,921
And, I asked you on the
problems set, how much would
568
00:34:22,921 --> 00:34:26,571
you continue to have to
halve the step size in order
569
00:34:26,571 --> 00:34:30,000
for that good thing to happen?
570
00:34:30,000 --> 00:34:33,330
However, there are
more efficient methods
571
00:34:33,330 --> 00:34:36,000
which get the results faster.
572
00:34:36,000 --> 00:34:39,571
So if that's our
good method, let's
573
00:34:39,571 --> 00:34:43,000
call this our still
better method.
574
00:34:43,000 --> 00:34:47,000
The better methods
aim at being better.
575
00:34:47,000 --> 00:34:51,000
They keep the same
idea as Euler's method,
576
00:34:51,000 --> 00:34:56,600
but they say, look, let's try
to improve that slope, An.
577
00:34:56,600 --> 00:35:00,776
In other words, since the
slope, An, that we start with
578
00:35:00,776 --> 00:35:05,000
is guaranteed to be wrong if
the curve is convex or concave,
579
00:35:05,000 --> 00:35:06,875
can we somehow correct it?
580
00:35:06,875 --> 00:35:10,000
So, for example, instead of
immediately aiming there,
581
00:35:10,000 --> 00:35:13,632
can't we somehow aim it
so that by luck, we just,
582
00:35:13,632 --> 00:35:17,000
at the next step just lands
us back on the curve again?
583
00:35:17,000 --> 00:35:20,666
In other words, with sort of
looking for the short path,
584
00:35:20,666 --> 00:35:22,997
a shortcut path,
which by good luck
585
00:35:22,997 --> 00:35:25,800
will end us up back
on the curve again.
586
00:35:25,800 --> 00:35:29,000
And, all the simple improvements
on the Euler method,
587
00:35:29,000 --> 00:35:32,284
and they are the
most stable in ways
588
00:35:32,284 --> 00:35:34,714
to solve differential
equations numerically,
589
00:35:34,714 --> 00:35:39,000
aim at finding a better slope.
590
00:35:39,000 --> 00:35:43,000
So, they find a better
value for a better slope,
591
00:35:43,000 --> 00:35:45,496
find a better value than An.
592
00:35:45,496 --> 00:35:49,000
Try to improve that
slope that you found.
593
00:35:49,000 --> 00:35:54,625
Now, once you have the idea that
you should look for a better
594
00:35:54,625 --> 00:35:59,000
slope, it's not very difficult
to see what, in fact,
595
00:35:59,000 --> 00:36:00,332
you should try.
596
00:36:00,332 --> 00:36:04,332
Again, I think most
of you would say, hey,
597
00:36:04,332 --> 00:36:07,000
I would have thought of that.
598
00:36:07,000 --> 00:36:09,541
And, you would be
closer in time,
599
00:36:09,541 --> 00:36:11,614
since these methods
were only found
600
00:36:11,614 --> 00:36:15,625
about around the turn of the
last century is when I place
601
00:36:15,625 --> 00:36:19,375
them, mostly by some
German mathematicians
602
00:36:19,375 --> 00:36:22,000
interested in solving
equations numerically.
603
00:36:22,000 --> 00:36:25,000
All right, so what
is the better method?
604
00:36:25,000 --> 00:36:29,666
Our better slope, what should
we look for in our better slope?
605
00:36:29,666 --> 00:36:34,500
Well, the simplest
procedure is, once again, we
606
00:36:34,500 --> 00:36:38,080
are starting from there,
and the Euler slope would
607
00:36:38,080 --> 00:36:41,000
be the same as a line element.
608
00:36:41,000 --> 00:36:44,000
So, the line element
looks like this.
609
00:36:44,000 --> 00:36:47,200
And, our yellow slope,
A, and I'll still
610
00:36:47,200 --> 00:36:51,000
continue to call it An, goes
like that, gets to here.
611
00:36:51,000 --> 00:36:55,000
Okay, now if it were convex,
if the curve were convex,
612
00:36:55,000 --> 00:36:57,140
this would be too low.
613
00:36:57,140 --> 00:36:59,815
And therefore, the
next step would be,
614
00:36:59,815 --> 00:37:03,875
I'm going to draw this
next step in pink.
615
00:37:03,875 --> 00:37:08,800
Well, let's continue in here,
would be going up like that.
616
00:37:08,800 --> 00:37:11,332
I'll call this Bn,
just because it's
617
00:37:11,332 --> 00:37:14,000
the next step of Euler's method.
618
00:37:14,000 --> 00:37:19,000
It could be called An prime
or something like that.
619
00:37:19,000 --> 00:37:20,200
But this will do.
620
00:37:20,200 --> 00:37:24,800
And now what you do is,
let me put an arrow on it
621
00:37:24,800 --> 00:37:28,500
to indicate parallelness,
go back to the beginning,
622
00:37:28,500 --> 00:37:31,000
draw this parallel to Bn.
623
00:37:31,000 --> 00:37:32,452
So, here is Bn.
624
00:37:32,452 --> 00:37:35,375
Again, just a line
of that same slope.
625
00:37:35,375 --> 00:37:39,998
And now, what you should use
as the simplest improvement
626
00:37:39,998 --> 00:37:45,815
on Euler's method, is take
the average of these two
627
00:37:45,815 --> 00:37:48,800
because that's
more likely to hit
628
00:37:48,800 --> 00:37:52,800
the curve than An will,
which is sure to be
629
00:37:52,800 --> 00:37:55,712
too low if the curve is convex.
630
00:37:55,712 --> 00:37:58,200
In other words,
use this instead.
631
00:37:58,200 --> 00:37:59,000
Use that.
632
00:37:59,000 --> 00:38:02,498
So, this is our better slope.
633
00:38:02,498 --> 00:38:06,285
Okay, what will we
call that slope?
634
00:38:06,285 --> 00:38:08,428
We don't have to
call it anything.
635
00:38:08,428 --> 00:38:11,856
What were the equations
for the method be?
636
00:38:11,856 --> 00:38:17,000
Well, x n plus one is gotten
by adding the step size.
637
00:38:17,000 --> 00:38:20,000
So, here's my step size
just as it was before.
638
00:38:20,000 --> 00:38:23,072
Just as it was
before, the new thing
639
00:38:23,072 --> 00:38:26,332
is how to get the
new value of y.
640
00:38:26,332 --> 00:38:31,267
So, y n plus one should be
the old yn, plus h times not
641
00:38:31,267 --> 00:38:37,000
this crummy slope, An, but
the better, the pink slope.
642
00:38:37,000 --> 00:38:39,000
What's the formula
for the pink slope?
643
00:38:39,000 --> 00:38:41,499
Well, let's do it in two steps.
644
00:38:41,499 --> 00:38:44,000
It's the average of An and Bn.
645
00:38:44,000 --> 00:38:48,285
Hey, but you didn't tell me, or
I didn't tell you what Bn was.
646
00:38:48,285 --> 00:38:52,664
So, you now must tell the
computer, oh yes, by the way,
647
00:38:52,664 --> 00:38:56,400
you remember that An
was what it always was.
648
00:38:56,400 --> 00:38:59,284
The interesting
thing is, what is Bn?
649
00:38:59,284 --> 00:39:03,331
Well, to get Bn, Bn is
the slope of the line
650
00:39:03,331 --> 00:39:05,000
element at this new point.
651
00:39:05,000 --> 00:39:08,000
Now, what am I going
to call that new point?
652
00:39:08,000 --> 00:39:12,000
I don't want to call this
y value, y n plus one,
653
00:39:12,000 --> 00:39:15,500
because that's, it's
this up here that's
654
00:39:15,500 --> 00:39:17,400
going to be the y n plus one.
655
00:39:17,400 --> 00:39:19,776
All this is, is
a temporary value
656
00:39:19,776 --> 00:39:23,665
used to make another
calculation, which will then
657
00:39:23,665 --> 00:39:27,000
be combined with the
previous calculations
658
00:39:27,000 --> 00:39:29,000
to get the right value.
659
00:39:29,000 --> 00:39:31,500
Therefore, give it
a temporary name.
660
00:39:31,500 --> 00:39:34,614
That point, we'll
call it, it's not
661
00:39:34,614 --> 00:39:38,000
going to be the final,
the real y n plus one.
662
00:39:38,000 --> 00:39:41,499
We'll call it y n plus
one twiddle, y n plus one
663
00:39:41,499 --> 00:39:41,998
temporary.
664
00:39:41,998 --> 00:39:44,000
And, what's the formula for it?
665
00:39:44,000 --> 00:39:47,630
Well, it's just going to
be what the original Euler
666
00:39:47,630 --> 00:39:51,192
formula; it's going to be y n
plus what you would have gotten
667
00:39:51,192 --> 00:39:53,600
if you had calculated,
in other words,
668
00:39:53,600 --> 00:39:56,400
it's the point that the
Euler method produced,
669
00:39:56,400 --> 00:40:01,000
but it's not, finally,
the point that we want.
670
00:40:01,000 --> 00:40:04,000
Now, do I have to
say anything else?
671
00:40:04,000 --> 00:40:08,000
Yeah, I didn't tell the
computer what Bn was.
672
00:40:08,000 --> 00:40:11,688
Okay, Bn is the slope
of the direction
673
00:40:11,688 --> 00:40:15,428
field at the point n plus one.
674
00:40:15,428 --> 00:40:19,800
And the computer
knows what that is.
675
00:40:19,800 --> 00:40:24,000
And, this point, y n
plus one temporary.
676
00:40:24,000 --> 00:40:31,000
So, you make a temporary choice
of this, calculate that number,
677
00:40:31,000 --> 00:40:34,000
and then go back,
and as it were,
678
00:40:34,000 --> 00:40:39,625
correct that value to this value
by using this better slope.
679
00:40:39,625 --> 00:40:44,500
Now, that's all there
is to the method,
680
00:40:44,500 --> 00:40:48,000
except I didn't
give you its name.
681
00:40:48,000 --> 00:40:52,000
Well, it has three names,
four names in fact.
682
00:40:52,000 --> 00:40:54,664
Which one shall I give you?
683
00:40:54,664 --> 00:40:56,000
I don't care.
684
00:40:56,000 --> 00:41:00,000
Okay, the shortest
name is Heun's method.
685
00:41:00,000 --> 00:41:05,000
But nobody pronounces
that correctly.
686
00:41:05,000 --> 00:41:07,664
So, it's Heun's method.
687
00:41:07,664 --> 00:41:13,000
It's called, also, the
Improved Euler method.
688
00:41:13,000 --> 00:41:19,000
It's called Modified Euler,
very expressive word,
689
00:41:19,000 --> 00:41:21,250
Modified Euler's method.
690
00:41:21,250 --> 00:41:25,000
And, it's also called RK2.
691
00:41:25,000 --> 00:41:31,000
I'm sure you'll
like that name best.
692
00:41:31,000 --> 00:41:33,000
It has a Star Wars
sort of sound.
693
00:41:33,000 --> 00:41:37,576
RK stands for Runge Kutta,
and the reason for the two
694
00:41:37,576 --> 00:41:42,307
is not that it uses, well, it
is that it uses two slopes,
695
00:41:42,307 --> 00:41:47,000
but the real reason for the two
is that it is a second order
696
00:41:47,000 --> 00:41:47,500
method.
697
00:41:47,500 --> 00:41:52,724
So, that's the most important
thing to put down about it.
698
00:41:52,724 --> 00:41:57,500
It's a second order method,
whereas Euler's was only
699
00:41:57,500 --> 00:42:00,000
a first order method.
700
00:42:00,000 --> 00:42:03,000
So, Heun's method, or
RK2, let's write it,
701
00:42:03,000 --> 00:42:08,000
is the shortest thing to write,
is a second order method,
702
00:42:08,000 --> 00:42:11,632
meaning that the error
varies with the step
703
00:42:11,632 --> 00:42:15,000
size like some
constant, it will not
704
00:42:15,000 --> 00:42:20,500
be the same as the constant
for Euler's method, times
705
00:42:20,500 --> 00:42:22,000
h squared.
706
00:42:22,000 --> 00:42:25,227
That's a big saving
because it now
707
00:42:25,227 --> 00:42:28,908
means that if you
halve the step size,
708
00:42:28,908 --> 00:42:34,250
you're going to decrease
the error by a factor of one
709
00:42:34,250 --> 00:42:34,875
quarter.
710
00:42:34,875 --> 00:42:38,000
You will quarter the error.
711
00:42:38,000 --> 00:42:41,142
Now, you say, hey, why should
anyone use anything else?
712
00:42:41,142 --> 00:42:44,000
Well, think a little second.
713
00:42:44,000 --> 00:42:48,000
The real thing which determines
how slowly one of these methods
714
00:42:48,000 --> 00:42:51,927
run is you look at the
hardest step of the method
715
00:42:51,927 --> 00:42:54,452
and ask how long
does the computer
716
00:42:54,452 --> 00:42:57,666
take, how many of those
hardest steps are there?
717
00:42:57,666 --> 00:43:01,500
Now, the answer is, the
hardest step is always
718
00:43:01,500 --> 00:43:05,452
the evaluation of the function
because the functions that
719
00:43:05,452 --> 00:43:09,800
are common use are not x
squared minus y squared.
720
00:43:09,800 --> 00:43:14,856
They take half a page and
have, as coefficients,
721
00:43:14,856 --> 00:43:18,250
you know, ten decimal
place numbers, whatever
722
00:43:18,250 --> 00:43:22,750
the engineers doing it,
whatever their accuracy was.
723
00:43:22,750 --> 00:43:26,600
So, the thing that controls
how long a method runs
724
00:43:26,600 --> 00:43:33,000
is how many times the slope,
the function, must be evaluated.
725
00:43:33,000 --> 00:43:37,000
For Euler, I only have
to evaluate it once.
726
00:43:37,000 --> 00:43:40,000
Here, I have to
evaluate it twice.
727
00:43:40,000 --> 00:43:44,571
Now, roughly speaking, the
number of function evaluations
728
00:43:44,571 --> 00:43:48,000
will each give you the exponent.
729
00:43:48,000 --> 00:43:52,440
The method that's called
Runge Kutta fourth order
730
00:43:52,440 --> 00:43:56,000
will require four
evaluations of slope,
731
00:43:56,000 --> 00:44:05,000
but the accuracy will be like
h to the fourth: very accurate.
732
00:44:05,000 --> 00:44:09,863
You halve the step size, and
it goes down by a factor of
733
00:44:09,863 --> 00:44:10,363
16.
734
00:44:10,363 --> 00:44:10,863
Great.
735
00:44:10,863 --> 00:44:14,000
But you had to evaluate
the slope four times.
736
00:44:14,000 --> 00:44:19,000
Suppose, instead, you halve
four times this thing.
737
00:44:19,000 --> 00:44:21,496
And, what would you have done?
738
00:44:21,496 --> 00:44:26,500
You would have decreased it
to 1/16th to what it was.
739
00:44:26,500 --> 00:44:31,500
You still would increase the
number of function evaluations
740
00:44:31,500 --> 00:44:38,200
you needed by four, and you
would have decreased the error
741
00:44:38,200 --> 00:44:40,000
by a 16th.
742
00:44:40,000 --> 00:44:42,625
So, in some sense,
it really doesn't
743
00:44:42,625 --> 00:44:46,600
matter whether you use a very
fancy method, which requires
744
00:44:46,600 --> 00:44:48,500
more function evaluations.
745
00:44:48,500 --> 00:44:50,000
That's true.
746
00:44:50,000 --> 00:44:52,664
The error goes down
faster, but you are
747
00:44:52,664 --> 00:44:55,125
having to more work to get it.
748
00:44:55,125 --> 00:44:57,000
So, anyway, nothing is free.
749
00:44:57,000 --> 00:44:59,000
Now, there is an RK4.
750
00:44:59,000 --> 00:45:02,362
I think I'll skip
that, since I wouldn't
751
00:45:02,362 --> 00:45:06,000
dare to ask you any
questions about it.
752
00:45:06,000 --> 00:45:08,904
But, let me just
mention it, at least,
753
00:45:08,904 --> 00:45:10,600
because it's the standard.
754
00:45:10,600 --> 00:45:13,000
It uses four evaluations.
755
00:45:13,000 --> 00:45:15,800
It's the standard
method, if you don't
756
00:45:15,800 --> 00:45:18,200
want to do anything fancier.
757
00:45:18,200 --> 00:45:21,000
It's rather
inefficient, but it's
758
00:45:21,000 --> 00:45:23,444
very accurate, standard
method, accurate,
759
00:45:23,444 --> 00:45:27,000
and you'll see when
you use the programs,
760
00:45:27,000 --> 00:45:31,000
it's, in fact, a program
which is drawing those curves,
761
00:45:31,000 --> 00:45:34,816
the numerical method which
draws all those curves
762
00:45:34,816 --> 00:45:38,625
that you believe in
on the computer screen
763
00:45:38,625 --> 00:45:41,125
is the RK4 method.
764
00:45:41,125 --> 00:45:46,000
The Runge Kutta, I should
give them their names.
765
00:45:46,000 --> 00:45:49,400
Runge Kutta, fourth
order method.
766
00:45:49,400 --> 00:45:53,500
Two mathematicians, I believe
both German mathematicians
767
00:45:53,500 --> 00:45:58,330
around the turn of the
last century, Runge Kutta
768
00:45:58,330 --> 00:46:01,875
fourth order method
requires four slopes,
769
00:46:01,875 --> 00:46:05,571
requires you to
calculate four slopes.
770
00:46:05,571 --> 00:46:09,500
I won't bother telling
you what you do,
771
00:46:09,500 --> 00:46:11,000
but it's a procedure like that.
772
00:46:11,000 --> 00:46:14,000
It's just a little
bit more elaborate.
773
00:46:14,000 --> 00:46:17,000
And you take two of these,
you make up a weighted average
774
00:46:17,000 --> 00:46:18,500
for the super slope.
775
00:46:18,500 --> 00:46:20,000
You use weighted average.
776
00:46:20,000 --> 00:46:23,000
What should I divide that
by to get the right...?
777
00:46:23,000 --> 00:46:24,000
Six.
778
00:46:24,000 --> 00:46:24,570
Why six?
779
00:46:24,570 --> 00:46:27,000
Well, because if all these
numbers were the same,
780
00:46:27,000 --> 00:46:32,330
I'd want it to come out to be
whatever that common value was.
781
00:46:32,330 --> 00:46:35,454
Therefore, in a
weighted average,
782
00:46:35,454 --> 00:46:40,000
you must always divide by
the sum of the coefficients.
783
00:46:40,000 --> 00:46:42,178
So, this is the super slope.
784
00:46:42,178 --> 00:46:45,332
And, if you plug that
super slope into here,
785
00:46:45,332 --> 00:46:48,500
you will be using the
Runge Kutta method,
786
00:46:48,500 --> 00:46:51,375
and get the best
possible results.
787
00:46:51,375 --> 00:46:54,571
Now, I wanted to
spend the last three
788
00:46:54,571 --> 00:46:58,600
minutes talking about pitfalls
of numerical computation
789
00:46:58,600 --> 00:46:59,800
in general.
790
00:46:59,800 --> 00:47:05,332
One pitfall I am leaving
you on the homework
791
00:47:05,332 --> 00:47:08,000
to discover for yourself.
792
00:47:08,000 --> 00:47:12,000
Don't worry, it won't
cause you any grief.
793
00:47:12,000 --> 00:47:16,000
It'll just destroy your
faith in these things
794
00:47:16,000 --> 00:47:22,000
for the rest of your life,
which is probably a good thing.
795
00:47:22,000 --> 00:47:25,800
So, pitfalls, number one,
you find, you'll find.
796
00:47:25,800 --> 00:47:29,725
Let me talk, instead,
briefly about number two,
797
00:47:29,725 --> 00:47:34,500
which I am not giving
you an exercise in.
798
00:47:34,500 --> 00:47:38,284
Number two is illustrated
by the following equation.
799
00:47:38,284 --> 00:47:40,000
What could be simpler?
800
00:47:40,000 --> 00:47:44,000
This is a very bad equation
to try to solve numerically.
801
00:47:44,000 --> 00:47:44,856
Now, why?
802
00:47:44,856 --> 00:47:47,250
Well, because if I
separate variables,
803
00:47:47,250 --> 00:47:49,000
why don't I save a little time?
804
00:47:49,000 --> 00:47:53,000
I'll just tell you what
the solution is, okay?
805
00:47:53,000 --> 00:47:55,000
You obviously
separate variables.
806
00:47:55,000 --> 00:47:57,000
Maybe you can do
it in your head.
807
00:47:57,000 --> 00:47:58,998
The solution will
be, the solutions
808
00:47:58,998 --> 00:48:01,332
will have an arbitrary
constant in them,
809
00:48:01,332 --> 00:48:03,555
and they won't be
very complicated.
810
00:48:03,555 --> 00:48:08,000
They will be one over c minus x.
811
00:48:08,000 --> 00:48:12,250
C is an arbitrary constant, and
as you give different values,
812
00:48:12,250 --> 00:48:15,000
you get, now, what do
those guys look like?
813
00:48:15,000 --> 00:48:16,360
Okay, so here I am.
814
00:48:16,360 --> 00:48:18,200
I start out at the point, one.
815
00:48:18,200 --> 00:48:20,452
And, I start out, I
tell the computer,
816
00:48:20,452 --> 00:48:23,855
compute for me the value
of the solution at one
817
00:48:23,855 --> 00:48:25,000
starting out at one.
818
00:48:25,000 --> 00:48:28,000
And, it computes and
computes a little while.
819
00:48:28,000 --> 00:48:31,600
But the solution, how does
this curve actually look?
820
00:48:31,600 --> 00:48:38,000
So, in other words, suppose I
say that y of zero equals one.
821
00:48:38,000 --> 00:48:39,875
Find me y of two.
822
00:48:39,875 --> 00:48:42,800
In other words, take a
nice small step size.
823
00:48:42,800 --> 00:48:45,713
Use the Runge Kutta
fourth order method.
824
00:48:45,713 --> 00:48:48,816
Calculate a little
bit, and tell me,
825
00:48:48,816 --> 00:48:51,500
I just want to know
what y of two is.
826
00:48:51,500 --> 00:48:53,000
Well, what is y of two?
827
00:48:53,000 --> 00:48:56,000
Well, unfortunately, how
does that curve look?
828
00:48:56,000 --> 00:48:59,125
The curve looks like this.
829
00:48:59,125 --> 00:49:02,332
At that point, it
drops to infinity
830
00:49:02,332 --> 00:49:05,800
in a manner of speaking,
and then sort of comes back
831
00:49:05,800 --> 00:49:07,000
up again like that.
832
00:49:07,000 --> 00:49:09,400
What is the value of y?
833
00:49:09,400 --> 00:49:11,222
This is the point, one.
834
00:49:11,222 --> 00:49:13,000
What is the value of y of two?
835
00:49:13,000 --> 00:49:15,000
Is it here?
836
00:49:15,000 --> 00:49:15,855
Is it this?
837
00:49:15,855 --> 00:49:19,448
Well, I don't know, but I do
know that the computer will not
838
00:49:19,448 --> 00:49:20,000
find it.
839
00:49:20,000 --> 00:49:22,724
The computer will
follow this along,
840
00:49:22,724 --> 00:49:25,750
and get lost in
eternity, in infinity,
841
00:49:25,750 --> 00:49:30,400
and see no reason whatever
why it should start again
842
00:49:30,400 --> 00:49:34,000
on this branch of the curve.
843
00:49:34,000 --> 00:49:36,800
Okay, well, can't we
predict that that's
844
00:49:36,800 --> 00:49:40,000
going to happen somehow,
avoid what I should have.
845
00:49:40,000 --> 00:49:44,500
The whole difficulty is, this
is called a singular point.
846
00:49:44,500 --> 00:49:49,400
The solution has a singularity,
in other words, a single place
847
00:49:49,400 --> 00:49:53,726
where it goes to infinity or
becomes discontinuous, maybe as
848
00:49:53,726 --> 00:49:54,815
a jump discontinuity.
849
00:49:54,815 --> 00:49:58,200
It has a singularity
at x equals c.
850
00:49:58,200 --> 00:50:02,220
This, in particular,
at x equals one here,
851
00:50:02,220 --> 00:50:06,725
but from the differential
equation, where is that c?
852
00:50:06,725 --> 00:50:10,333
There is no way
of predicting it.
853
00:50:10,333 --> 00:50:15,000
Each solution, in other words,
to this differential equation,
854
00:50:15,000 --> 00:50:17,000
has its own,
private singularity,
855
00:50:17,000 --> 00:50:21,400
which only it knows about, and
where it's going to blow up,
856
00:50:21,400 --> 00:50:25,500
and there's no way of telling
from the differential equation
857
00:50:25,500 --> 00:50:28,000
where that's going to be.
858
00:50:28,000 --> 00:50:32,000
That's one of the things that
makes numerical calculation
859
00:50:32,000 --> 00:50:37,000
difficult, when you cannot
predict where things are going
860
00:50:37,000 --> 00:50:39,450
to go bad in advance.