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PROFESSOR: Hi.
In today's lesson, hopefully we will begin to reap the
rewards of our digression into the subject of linear algebra.
Recall that in the last few lectures, what we have been
dealing with is the problem of inverting
systems of linear equations.
And what we would like to do today is to tackle the more
general problem of inverting systems of equations, even if
the equations are not linear.
And with this in mind, I simply entitle today's lesson
"Inverting More General Systems of Equations."
And by way of a very brief review, recall that given the
linear system y1 equals a1 1 x1 plus et cetera a1 n xn up
to yn equals an1 x1 plus et cetera an n xn.
We saw that that system was invertible, meaning what?
That we could solve this system for the x's as linear
combinations of the y's if-- and only if--
the inverse of the matrix of coefficients
of this system exists.
And in terms of determinants, recall that that means if and
only if the determinant of the matrix of
coefficients is not 0.
If the determinant of the matrix of coefficients was 0,
then we saw that y1 up to yn are not independent.
In fact, in that context, that's where we came to grips
with the concept of a constraint, that the
constraint actually turned out to be what?
The fact that the y's were not independent.
Meaning that we could express a linear combination of the
y's equal to 0.
So what situation were we at then?
Given a linear system, if the matrix of coefficients does
not have its determinant equal to 0, the system is
invertible.
We can solve for the x's in terms of the y's.
If the determinant of the matrix of coefficients is 0's,
the y's are not linearly independent.
In other words, the system is not invertible.
And now what we would like to do is to tackle the more
general problem of inverting any system of equations.
And by any, I mean what?
Now we have y1 is f sub 1 of x1 up to xn, et cetera. y sub
n is f sub n of x1 up to xn.
And now what we're saying is we do not know whether the f's
are linear or not.
In fact, if they are linear, we're back to-- as a special
case, what we've tackled before.
But now we're assuming that these need not be linear.
And what we would like to do is to invert this system,
assuming, of course, that such an inversion is possible.
Again, what do we mean by the inversion?
We mean somehow, we would like to know that given this system
of n equations and n unknowns where the y's are expressed
explicitly in terms of the x's, can we invert this, and
express the x's in terms of the y's--
either explicitly or implicitly?
That's the problem that we'd like to tackle.
And what we're going to use to tackle this is our old friend,
the differential, the linear approximation, again, that
motivated our whole study of linear systems
in the first place.
Remember, we already know that if y1 is a differentiable
function of x1 up to xn, that delta y1 sub tan is exactly
equal to the partial of f1 with respect to x1 times delta
x1 plus et cetera the partial of f1 with respect to
xn times delta xn.
And in terms of reviewing the notation, because we will use
it later in the lecture, notice that delta y1 tan was
what we abbreviated to be dy1.
And that delta x1 up to delta xn are abbreviated
respectively as dx1 up to dxn.
Generalization of what we did for the differential in the
case of one independent variable and we went through
this discussion when we talked about exact differentials in
block three.
And in this similar way, we could of course express delta
y, 2, tan, et cetera, and we can talk about the linear
approximations.
All right?
Not the true delta y's now, but the linear part of the
delta y's, the delta y1 sub tan, or if you prefer, delta
y1 sub lin, L-I-N. All right?
And the key point now is what?
If the y's happen to be continuously differentiable
functions of the x's in the neighborhood of the point x
bar equals a bar--
in other words, x1 up to xn is equal to a1 comma et cetera up
to an, x1 equals a1, x2 equals a2, et cetera--
then near that point, what we're saying is what?
That the error term goes to 0 very rapidly.
And as long as the functions are continuously
differentiable, it means what?
That the change in y is approximately equal to the
change in y sub tan.
So that what we're saying is--
and remember this is what motivated our linear systems
in the first place--
that delta y1 is approximately the partial of y1 with respect
to x1, evaluated when x bar is a bar times delta x1 plus et
cetera the partial of y1 with respect to xn also evaluated
when x bar equals a bar times delta x sub n et cetera, down
to delta y sub n is approximately equal to the
partial of y sub n with respect to x1 times delta x1,
plus et cetera the partial of y sub n with respect to x sub
n times delta xn, where all of these partials are evaluated
specifically at the point x bar equals a bar.
The point being what?
That since we're evaluating all these partials at a
particular point, every one of these
coefficients is a constant.
You see in general, these partial derivatives are
functions, but as soon as we evaluate them at a given
value, they become specific numbers.
So this is now what?
On the right-hand side, we have a linear system, and the
point is that we are using the fundamental result that we can
use the linear approximation as being very nearly equal to
the true change in y.
And it's in that sense that we have derived our system of n
linear equations and n unknowns.
You see, this is a linear system.
The approximation--
again, let me emphasize that, because it's very important.
The approximation hinges on the fact that what this is
exactly equal to would be delta y1 tan et
cetera, delta y sub n.
But we're assuming that these are close enough in a small
enough neighborhood of x bar equals a bar so that we can
make this particular statement.
So this is a linear system.
And because it is a linear system, we're back to our
special case that this system is invertible if and only if
the determinant of the coefficients--
matrix coefficients--
is not 0.
And what is that matrix coefficient?
It consists of n rows and n columns and the row is
determined by the subscript on the y.
And the column is determined by the subscript on the x.
So we write that matrix as the partial of y sub i with
respect to x sub j.
In other words, the i-th row involves the y's and the j-th
column, the x.
All right?
And that's exactly, then, how we handle
this particular system.
OK?
Quite mechanically.
And let's just summarize that then very quickly.
If f sub 1 et cetera and f sub n are continuously
differentiable functions of x1 up to xn near x bar equals a
bar, then the system y1 equals f1 of x1 up to xn et cetera,
yn equals f sub n of x1 up to xn--
that system is invertible if and only if the determinant--
the n by n determinant whose entry in i-th row, j-th column
is the partial of y sub i with respect to x sub j if and only
if that determinant is not 0.
Now what does that mean, to say that it's invertible?
It means that we can solve for the x's in terms of the y's.
Now we may not be able to do that explicitly.
The best we may be able to do is to do that implicitly,
meaning this.
Let me just come back to something I said before to
make sure that there's no misunderstanding about this.
What we're saying is, that in this particular linear system
of equations, as long as this determinant of coefficients is
not 0, we can explicitly solve for delta x1 up to delta xn in
terms of delta y1 up to delta yn, even if we may not be able
to solve explicitly for the x's in terms of the y's.
In other words, the crucial point is, we can solve for the
changes in x in terms of the changes in y.
And that is implicitly enough to see what the x's look like
in terms of the y's.
Once we know what the change of x looks like in terms of
the change in y, then we really know what x itself
looks like in terms of the y's, even as I say it may be
implicitly rather than explicitly.
At any rate, this matrix is so important that it's given a
very special name, definition.
The matrix whose entry in the i-th row, j-th column is the
partial of y sub i with respect to x sub j is called
the Jacobian.
I put "matrix" in quotation marks here, because some
people refer to the Jacobian meaning a matrix.
Other people call the Jacobian the determinant of the
Jacobian matrix.
I'm not going to make any distinction this way.
It'll be clear from context.
Usually when I say the "Jacobian," I will mean the
Jacobian matrix.
I might sometimes mean the Jacobian determinant.
And so to avoid ambiguity, I will hopefully say "Jacobian
matrix" when I mean the matrix, and "Jacobian
determinant" when I mean the determinant.
But should I forget to do this or should you read a textbook
where the word Jacobian is used without the proper noun
after it, it should be clear from context which is meant.
But at any rate, that's what we mean by the Jacobian of y1
up to yn with respect to x1 up to xn.
And this Jacobian matrix is often abbreviated by--
you either write J for Jacobian of y1 up to yn over
x1 up to xn.
Or else you use a modification of the
partial derivative notation.
And you sort of read this as if it said the partial of y1
up to yn divided by the partial of x1 up to xn.
And again, there is the same analogy between why this
notation was invented and why the notation dy divided by dx
was invented.
But in terms of giving you a general overview of what we're
interested in, I think I would like to leave the discussion
of why we write this in a
fractional form to the homework.
In other words, as either a supplement to the learning
exercises or else as part of the supplementary notes in one
form or another, we will take care of all of the
computational aspects of how one handles the Jacobian.
But what I wanted to do now was to emphasize how one uses
the Jacobian matrix and differentials to invert
systems of n-equations and n-unknowns.
And I will use the technique that's used right in the
textbook and which is part of the
assignment for today's unit.
The example that I have in mind--
I simply picked the usual case, n equals 2, so that
things don't get that messy.
Using again, the standard notation when one deals with
two independent variables, let u equal x
squared minus y squared.
Let v equal to 2xy.
Let's suppose now that I would like to find the partial of x
with respect for u, treating v as the
other independent variable.
You see, again, I want to review this thing.
When I say find the partial of x with respect to u holding v
constant, it is not the same as finding the partial of u
with respect to x from here and then just inverting it.
Namely the partial of u with respect to x here assumes that
y is being held constant.
And if you then invert that, recall that what you're
finding is the partial of x with respect to u treating y
as the other variable.
We want the partial of x with respect to u treating u and v
as the pair of independent variables.
Why?
Because that's exactly what you mean by
inverting this system.
This system as given expresses u and v in terms of the pair
of independent variables x and y.
And now what you'd like to do is express the pair of
variables x and y in terms of the independent variables u
and v, assuming of course, that u and v are indeed
independent variables.
The mechanical solution is simply this.
Using the language of differentials, we
write down du and dv.
Namely, du is what?
The partial of u with respect to x times dx, plus the
partial of u with respect to y times dy.
And from the relationship that u equals x squared minus y
squared, we see that du is 2x dx minus 2y dy.
Similarly, since the partial of v with respect to x is 2y,
and the partial of v with respect to y is 2x, we see
that dv is 2y dx plus 2x dy.
If we now assume that this is evaluated at some point x0,
y0, what do we have over here?
Once we've picked out a point x0 y0 to evaluate this at--
and I left out that because it simply would make the notation
too long, but I'll talk about that more later.
Assuming that we've evaluated this at a particular fixed
value of x and y, we have what?
du is some constant times dx plus some constant times dy.
dv is some constant times dx plus some constant times dy.
In other words, du and dv are expressed as linear
combinations of dx and dy.
We know how to invert this type of a system, assuming
that it's invertible.
Sparing you the details, what I'm saying is what?
I could multiply, say, the top equation by x, the bottom
equation by y.
And when I add them, the terms involving dy will drop out.
And I will get x du plus y dv is 2x dx plus 2y dx.
In other words twice--
it's 2x squared.
I multiply the top equation by x, the bottom equation by y.
So the right-hand side here becomes 2x squared dx plus 2y
squared dx, which is twice x squared plus y squared dx.
I now divide both sides of the equation through by twice x
squared plus y squared.
I wind up with the fact that dx is x over twice the
quantity x squared plus y squared du, plus the quantity
y over twice x squared plus y squared dv.
Recall that I also know by definition that dx is the
partial of x with respect to u times du plus the partial of x
with respect to v times dv, recalling from our lecture on
exact differentials that the only way two differentials in
terms of the du and dv can be equal is if they're equal
coefficient by coefficient.
I can therefore equate the two coefficients of du to conclude
that the partial of x with respect to u is x over 2 times
the quantity x squared plus y squared.
In fact, I can get the extra piece of information even
though I wasn't asked for that in this problem, that the
partial of x with respect to v is y over twice the quantity x
squared plus y squared.
By the way, observe purely algebraically that the only
time I would be in any difficulty with this procedure
is if x squared plus y squared happened to equal 0.
In other words, if x squared plus y squared happened to
equal 0, then to divide through by twice x squared
plus y squared is equivalent to dividing through by 0.
And division by 0 is not permissible.
In other words, somehow or other, I must take into
consideration that I am in trouble if x squared plus y
squared is 0.
Notice, by the way, that the only time that x squared plus
y squared can be 0 is if both x and y are 0.
And that means, again, that somehow or other, at the point
0 comma 0-- in a neighborhood of the point 0 comma 0 in the
neighborhood of the origin, I can expect to have a little
bit of trouble.
Now again, the main aim of the lecture is
to give you an overview.
The trouble that comes in at the origin will again be left
for the exercises.
In a learning exercise, we will discuss just what goes
wrong if you take a neighborhood of the origin to
discuss the change of variables u equals x squared
minus y squared v equals 2xy.
Suffice it to say for the time being that the system of
equations u equals x squared minus y squared, v equals 2xy,
is invertible in any neighborhood of a point x 0
comma y 0 except in that one possible case when you have
chosen as the point x 0 y 0, the origin.
At any rate, to take this problem away from the specific
concrete example that we've been talking about and to put
this in terms of a more general perspective, let's go
back more abstractly to the more general system.
Let's suppose now that u and v are any two continuously
differentiable functions of x and y.
Let u be f of xy.
Let v equal to g of xy.
And what we're saying is, if you pick a particular point x0
comma y0, then by mechanically using the total differential,
we have that du is the partial of f with respect to x
evaluated at x0 y0 times dx, plus the partial of f with
respect to y, evaluated at x0 y0 times dy.
We have that dv is the partial of g with respect to x
evaluated at x0 y0 times dx plus the partial of g with
respect to y, evaluated at x0 y0 times dy.
What is this now?
This is a linear system of two equations in two unknowns.
du and dv are linear combinations of the dx and dy.
The key point being again--
that's why I put this in here specifically with the x sub 0
and the y sub 0--
the key point is that as soon as you evaluate a partial
derivative at a fixed point, the value is a
constant, not a variable.
So this is what?
A linear system.
We have du as a constant times the x plus a
constant times dy.
dv is a constant times dx plus a constant times dy.
Again, to make a long story short, I can solve for dx in
terms of du and dv.
I can solve for dy in terms of du and dv, provided what?
That my matrix of coefficients does not have its determinant
equal to 0.
And to review this more explicitly so you see the
mechanics, all I'm saying is, to solve for dx, I can
multiply the top equation by the partial of g with respect
to y evaluated at x0 y0.
I can multiply the bottom equation by minus the partial
of f with respect to y evaluated at x0 y0.
And then when I add these two equations, the dy
term will drop out.
Again, leaving the details for you, it turns
out that dx is what?
The partial of g with respect to y--
and I've abbreviated this again-- this means what?
Evaluated at x0 y0 times du minus the partial of f with
respect to y at x0 y0 times dv over the partial of f with
respect to x times the partial with g with respect to y minus
the partial of f with respect to y times the partial of g
with respect to x.
And notice, of course, that this denominator is precisely
our matrix of coefficients.
f sub x, f sub y, g sub x, g sub y.
And the only place I've taken a liberty here is to use the
abbreviation of leaving out the x0 y0.
And the key point is what?
The only place I am going to get in trouble is if this
denominator happens to be 0.
In the two by two case-- in other words, in the case of
two equations and two unknowns, notice that we can
see explicitly what goes wrong when the determinant of
coefficients is 0.
The determinant of coefficients is just this
denominator.
And when that denominator is 0, we're in trouble.
In other words, the only time we cannot invert this system,
the only time we cannot find delta x and delta y in terms
of du and dv is when this determinant is 0.
Now you see, I think this is pretty straightforward stuff.
The textbook has a section on this, as you
will be reading shortly.
It is not hard to work this mechanically.
And then the question comes up is, how come when you pick up
a book on advanced calculus, there's usually a huge chapter
on Jacobians and inversion?
Why isn't it this simple in the advanced textbooks?
Why can we have it in our book this simply, but yet, in the
advanced book, why is there so much more to
this beneath the surface?
The answer behind all of this is quite subtle.
In fact, the major subtlety is this.
And that is that the notation du--
and for that matter, dv or dx or dy or dx1, dx2, whatever
you're using here-- is ambiguous.
And it's ambiguous for the following reason.
Recall how we defined the meaning of the symbol du.
If we're assuming that u is expressed as a function of the
independent variables x and y, then by du, we
mean delta u tan.
On the other hand, if we inverted this, du now--
in other words, what do I mean by inverted this?
What I mean first of all is if we assume now that x and y are
expressed in terms of u and v-- for example, suppose x is
some function h of u and v, what does du mean now?
Notice that now u is playing the role of
an independent variable.
For the independent variable, du just means delta u.
In other words, by way of a very quick review, notice that
if we're viewing u as being a dependent variable, then du
means delta u tan.
But if we're viewing u as being an independent variable,
then du means delta u.
And consequently, the results that we're using hinge very
strongly then-- in other words, the inversion that
we're using hinges very strongly on the requirement.
In other words, the inversion requires the validity of
interchanging delta u and delta u tan, et cetera.
Now let me show you what that means more explicitly.
Let's come back to something that we were talking about
just a few moments ago.
From a completely mechanical point of view, given that u
equals f of xy and g of xy, we very mechanically wrote down
that du was f sub x dx plus f sub y dy, dv was g sub x dx
plus g sub y dy, and then we just mechanically solved for
dx in terms of du and dv.
My claim is, is that if we translate this thing, if we
translate this thing into the language of delta u's, delta
x's, delta u tan's, delta x tans, et cetera, what we
really said was what?
That delta u tan was the partial of f with respect to x
times delta x, plus the partial of f with respect to y
times delta y.
And delta v tan was the partial of g with respect to x
times delta x, plus the partial of g with respect to y
times delta y.
And then when we eliminated delta y by multiplying the top
equation by g sub y and the bottom equation by minus f sub
y and adding, what we found was how to express delta x--
and catch this, this is the key point--
what we did was we expressed delta x as a linear
combination--
not of delta u and delta v, but of delta u
tan and delta v tan.
You see, notice that the result that we needed to have
to be able to use differentials was
not this, but this.
See, we found this, not delta x tan equals gy delta u minus
f sub y delta v over f sub x, g sub y minus f sub y g sub x.
To be able to say--
to invert this required that this was the expression that
we had, yet the expression that we were really evaluating
was this one.
In fact, let me come back for one moment, and make sure that
we see this.
You see, notice again that the subtlety of going from here to
here and inverting never shows us that we've interchanged the
roles of u and v from being the dependent variables to the
independent variables.
So the reason that there is so much work done in advanced
textbooks under the heading of inverting systems of equations
is to justify that being able to switch from delta x to
delta x tan or from delta u tan to delta u as we see fit,
whenever it serves our purposes.
The validity of being able to do that hinges on this more
subtle type of proof, that as far as I'm concerned, goes
beyond the scope of our text, other than for the fact that
in the learning exercises, I will find excuses to bring up
all of the situations that bring out where
the theory is important.
In other words, there will not be proofs of these more
difficult things.
Not because the proofs aren't important, but from the point
of view of what we're trying to do in our course, these
proofs tend to obscure the main stream of things.
So what I will do in the learning exercises is bring up
places that will show you why the theory is important, at
which point, I will emphasize what the result of the theory
is without belaboring and beleaguering you with the
proofs of these things.
At any rate, what I'd like to do now next time is to give
you an example where all of the material or the blocks of
material that we've done now on partial derivatives are
sort of pulled together very nicely.
But at any rate, we'll talk about that more next time.
And until next time, goodbye.
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