In this video, I'm
going to present
the main idea of the entire
course, so listen up carefully.
So let's start with a problem.
Let's suppose that
you have $1 and you
invested in some stock
that doubles every year,
darn good investment.
And you wonder how much money
you'll have after n years.
Now, we write this in
terms of equations.
We say x of n is how much
money you have after n years,
and our initial condition
is you start off with $1.
And our equation is that the
amount that you have each year
is twice what you had
the previous year.
Some people write
it as the amount
that you'll have next
year is twice the amount
that you have now.
It's totally equivalent.
This n is 1 less than this n.
And now we just go ahead
and figure out the amount
that you have after one
year is twice what you had
to start with, so that's $2.
The amount you have
after two years
is twice that, so
that's 2 times $2.
The amount that you have after
three years is twice that,
so that's 2 times 2
times $2, and in general,
the amount that you have after
n years is going to be 2 times
2 times 2 times 2.
And how many factors
of 2 do we have?
Well, starting at what
you started off with,
you multiply it by 2 n times.
So of course, this just
gives you 2 to the n.
So the amount of money
you have after n years
is 2 to the n dollars.
Great, and it's pretty
easy to generalize this.
If instead of starting with x
of n is twice x of n minus 1,
if we said x of n is some other
constant times x of n minus 1,
then at every year,
we would multiply by a
and we would pick a
factor of a to the n.
Also, if we didn't start off
with $1, if we started off
with $7, we would have winded
up with 2 to the n times 7,
or in general, wind up
with a to the n times
whatever you started with.
Every year, you multiply by
a, you do that for n years,
you pick a factor of a to the n.
Great, easy problem.
Now, we're going
to make it harder.
Now we're going to look
at coupled equations.
Instead of having
one quantity x1,
we have two
quantities, x1 and x2.
And I'll let you think of
your own hokey word problem
that they fit.
But the rules are
that x1 each year
is [INAUDIBLE] previous
year, plus whatever
x2 was the previous year.
And x2 each year is x1
of the previous year
plus twice x2 of
the previous year.
And now you say,
Okay, how in blazes
am I going to solve that?
Because it seems that to
solve the equations for x1,
you need to know x2.
To solve the equations for
x2, you need to know x1.
Everything is coupled.
Coupled is hard.
Now you can try writing things
in terms of linear algebra.
You can package x1
and x2 into a vector,
and then you write down
a matrix a 2, 1, 1, 2.
And let's suppose
that we started off
with x1 being 4 and x2 equal 2.
And now, our equations can
be written in matrix form
as the vector time
n is just 2, 1, 1,
2 times a vector
of time n minus 1.
We can follow our
previous reasoning.
x at time 1 is a
times x times 0.
x at time 2 is a
times x at time 1.
So that's a times a
times x at time 0.
And in general, x at time n is
a to the n times x at time 0.
There's only one
trouble with that.
How in blazes are you
going to figure out
the nth power of a matrix?
It's not 2 to the n, 1 to the
n, 1, to the n, and 2 to the n.
You don't just take the
nth power of the entries,
you have to somehow figure out
the nth power of the matrix.
And at this stage of the game,
we don't know how to do that.
So we got a solution that was
correct but absolutely useless.
Later on, we'll get some insight
on taking powers of matrices,
but until we do that, this
line raising doesn't help us.
We have to find another way to
solve our coupled equations.
And the answer is to
switch to new coordinates.
So instead of working
with the x-coordinates,
I'm going to find some new
coordinates, y1 and y2,
such that x1 is the sum of the
y's, and x2 is the difference
of the y's.
Or equivalently, y1 is the
average of the two x's, and y2
is 1/2 the difference
of the two x's.
Pretty easy to get these
equations from these equations.
And we're going to rewrite our
equations in terms of the y's.
So here are our equations.
And if we take the sum
of those two equations,
we get that x1 plus x2 at
time n is 3x 1 plus 3x 2
at time n minus 1.
And if you divide by 2, this
becomes y1 atom and this
becomes 3y 1 atom n minus 1.
So lo and behold, y satisfies
a decoupled equation.
You don't have to know anything
about y2 to solve for y1.
Likewise, if you take the
difference of the equations,
and divide by 2, you
get that y2 at time
and is y2 at time n minus 1.
So the evolution of y1 is easy,
the evolution of y2 is easy.
This is all
summarized in a chart.
This picture is the
course in a nutshell.
If you want to solve for x at
time n in terms of x at time 0,
that's a coupled.
Coupled problems are hard.
But if you switch to
the y-coordinates,
you have to learn how
to change coordinates.
You have to learn to go
from one set of coordinates
to another set of coordinates.
And this is going to involve
change of basis matrices.
In the case of this
particular problem,
it was a simple
sum and difference.
In general, it's a little
bit more complicated.
But you convert to
other coordinates
and then in the
other coordinates,
your problem is
totally decoupled.
You solve it one
variable at a time.
And that's easy.
You figure out what
y is at time n,
and then you do
another change of basis
back to the original
coordinates,
and you figure out
what x is at time n.
So you don't go straight
across the hard way,
you ride the elevator, cross
the bridge, and come back down.
So let's do that for
this particular problem.
Riding the first
elevator is easy.
We said that y1 was
x1 plus x2 over 2.
So 4 plus 2 over 2 is 3, 4
minus 2 over 2 is 1, check.
So this is 3 1.
Then we said that y1 of n
was 3 times y1 of n minus 1.
So y1 triples every time, so
y1 of n is 3 to the n times
y1 of 0.
So 3 to the n times 3
is 3 to the n plus 1.
So y2 of n is 1 times
y2 of n minus 1,
you multiply by 1 each time.
So you multiply by 1 to the
n, and of course, that's 1.
And so up here, this is
3 to the n plus 1, 1.
And finally, we convert
back y1 of n plus y2 of n
is 3 to the n plus 1.
y1 of n minus y2 of n is
3 to the n plus 1 minus 1.
That's our answer.
So what's the
moral of the story?
Well, the kind of
coupled problem
I gave you isn't the only
kind of coupled system
that you'll ever see.
You sometimes see systems
of first order differential
equations.
Sometimes, you see systems
of second order differential
equations.
And the point is if the
matrix is 2, 1, 1, 2,
you would always use
these particular changes
of coordinates.
The change of coordinates
doesn't depend on
whether you're talking
about a difference
equation or first order
differential equation,
or second order differential
equation, whatever.
It's really a property
of the matrix only.
Now once you've done the
change of coordinates,
you still have to solve
the equations one variable
at a time, but that's
relatively easy.
The right coordinates
depend only on the matrix,
and we're going to see how
to figure out the right
coordinates in terms
of the eigenvalues
the eigenvectors of a matrix.
This matrix happens
have eigenvectors
1, 1, and 1 minus 1.
That tells you what the
change of coordinates
is, the eigenvalues
of 3 and 1, and that
tells you what are
the numbers that
appear in equations after you
do the change of coordinates.
There's a machine for
doing this no matter
what matrix you started
with, and that's
what we're going to develop
in the course of the course.
