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So, today we are going to
continue looking at critical
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00:00:28 --> 00:00:31
points,
and we'll learn how to actually
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00:00:31 --> 00:00:33
decide whether a typical point
is a minimum,
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maximum, or a saddle point.
So, that's the main topic for
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today.
So, remember yesterday,
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we looked at critical points of
functions of several variables.
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And, so a critical point
functions, we have two values,
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x and y.
That's a point where the
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partial derivatives are both
zero.
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00:01:11 --> 00:01:15
And, we've seen that there's
various kinds of critical
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points.
There's local minima.
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So, maybe I should show the
function on this contour
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plot,there is local maxima,
which are like that.
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And, there's saddle points
which are neither minima nor
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maxima.
And, of course,
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if you have a real function,
then it would be more
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complicated.
It will have several critical
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00:01:45 --> 00:01:48
points.
So, this example here,
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well, you see on the plot that
there is two maxima.
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And, there is in the middle,
between them,
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00:01:58 --> 00:02:00
a saddle point.
And, actually,
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00:02:00 --> 00:02:02
you can see them on the contour
plot.
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00:02:02 --> 00:02:07
On the contour plot,
you see the maxima because the
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00:02:07 --> 00:02:12
level curves become circles that
now down and shrink to the
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maximum.
And, you can see the saddle
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point because here you have this
level curve that makes a figure
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00:02:18 --> 00:02:20
eight.
It crosses itself.
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00:02:20 --> 00:02:25
And, if you move up or down
here, so along the y direction,
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00:02:25 --> 00:02:28
the values of the function will
decrease.
36
00:02:28 --> 00:02:32
Along the x direction,
the values will increase.
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00:02:32 --> 00:02:37
So, you can see usually quite
easily where are the critical
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00:02:37 --> 00:02:42
points just by looking either at
the graph or at the contour
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00:02:42 --> 00:02:44
plots.
So, the only thing with the
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00:02:44 --> 00:02:47
contour plots is you need to
read the values to tell a
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00:02:47 --> 00:02:51
minimum from a maximum because
the contour plots look the same.
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00:02:51 --> 00:02:53
Just, of course,
in one case,
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00:02:53 --> 00:02:56
the values increase,
and in another one they
44
00:02:56 --> 00:03:03
decrease.
So, the question -- -- is,
45
00:03:03 --> 00:03:17
how do we decide -- -- between
the various possibilities?
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00:03:17 --> 00:03:23
So, local minimum,
local maximum,
47
00:03:23 --> 00:03:26
or saddle point.
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00:03:26 --> 00:03:38
49
00:03:38 --> 00:03:44
So, and, in fact,
why do we care?
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00:03:44 --> 00:03:55
Well, the other question is how
do we find the global
51
00:03:55 --> 00:04:05
minimum/maximum of a function?
So, here what I should point
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00:04:05 --> 00:04:07
out, well,
first of all,
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00:04:07 --> 00:04:09
to decide where the function is
the largest,
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00:04:09 --> 00:04:12
in general you'll have actually
to compare the values.
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00:04:12 --> 00:04:14
For example,
here, if you want to know,
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00:04:14 --> 00:04:16
what is the maximum of this
function?
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00:04:16 --> 00:04:19
Well, we have two obvious
candidates.
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00:04:19 --> 00:04:22
We have this local maximum and
that local maximum.
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00:04:22 --> 00:04:24
And, the question is,
which one is the higher of the
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00:04:24 --> 00:04:26
two?
Well, in this case,
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actually, there is actually a
tie for maximum.
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00:04:30 --> 00:04:32
But, in general,
you would have to compute the
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00:04:32 --> 00:04:34
function at both points,
and compare the values if you
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know that it's three at one of
them and four at the other.
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Well, four wins.
The other thing that you see
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00:04:40 --> 00:04:43
here is if you are looking for
the minimum of this function,
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00:04:43 --> 00:04:47
well, the minimum is not going
to be at any of the critical
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points.
So, where's the minimum?
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00:04:49 --> 00:04:53
Well, it looks like the minimum
is actually out there on the
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boundary or at infinity.
So, that's another feature.
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The global minimum or maximum
doesn't have to be at a critical
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point.
It could also be,
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00:05:01 --> 00:05:05
somehow, on the side in some
limiting situation where one
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variable stops being in the
allowed rang of values or goes
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to infinity.
So, we have to actually check
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the boundary and the infinity
behavior of our function to know
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where, actually,
the minimum and maximum will
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be.
So, in general,
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00:05:27 --> 00:05:37
I should point out,
these should occur either at
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00:05:37 --> 00:05:48
the critical point or on the
boundary or at infinity.
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00:05:48 --> 00:05:52
So, by that,
I mean on the boundary of a
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domain of definition that we are
considering.
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And so, we have to try both.
OK, but so we'll get back to
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that.
For now, let's try to focus on
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the question of,
you know, what's the type of
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the critical point?
So, we'll use something that's
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00:06:16 --> 00:06:21
known as the second derivative
test.
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00:06:21 --> 00:06:25
And, in principle,
well, the idea is kind of
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similar to what you do with the
function of one variable,
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00:06:29 --> 00:06:32
namely, the function of one
variable.
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00:06:32 --> 00:06:34
If the derivative is zero,
then you know that you should
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00:06:34 --> 00:06:38
look at the second derivative.
And, that will tell you whether
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00:06:38 --> 00:06:41
it's curving up or down whether
you have a local max and the
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00:06:41 --> 00:06:44
local min.
And, the main problem here is,
95
00:06:44 --> 00:06:46
of course, we have more
possible situations,
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00:06:46 --> 00:06:48
and we have several
derivatives.
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00:06:48 --> 00:06:52
So, we have to think a bit
harder about how we'll decide.
98
00:06:52 --> 00:06:56
But, it will again involve the
second derivative.
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00:06:56 --> 00:07:01
OK, so let's start with just an
easy example that will be useful
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00:07:01 --> 00:07:06
to us because actually it will
provide the basis for the
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general method.
OK, so we are first going to
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00:07:10 --> 00:07:15
consider a case where we have a
function that's actually just
103
00:07:15 --> 00:07:20
quadratic.
So, let's say I have a
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00:07:20 --> 00:07:28
function, W of (x,y) that's of
the form ax^2 bxy cy^2.
105
00:07:28 --> 00:07:32
OK, so this guy has a critical
point at the origin because if
106
00:07:32 --> 00:07:36
you take the derivative with
respect to x,
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00:07:36 --> 00:07:38
well, and if you plug x equals
y equals zero,
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00:07:38 --> 00:07:42
you'll get zero,
and same with respect to y.
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00:07:42 --> 00:07:44
You can also see,
if you try to do a linear
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00:07:44 --> 00:07:47
approximation of this,
well, all these guys are much
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00:07:47 --> 00:07:50
smaller than x and y when x and
y are small.
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00:07:50 --> 00:07:55
So, the linear approximation,
the tangent plane to the graph
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00:07:55 --> 00:07:59
is really just w=0.
OK, so, how do we do it?
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00:07:59 --> 00:08:03
Well, yesterday we actually did
an example.
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00:08:03 --> 00:08:09
It was a bit more complicated
than that, but let me do it,
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00:08:09 --> 00:08:13
so remember,
we were looking at something
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00:08:13 --> 00:08:19
that started with x^2 2xy 3y^2.
And, there were other terms.
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00:08:19 --> 00:08:23
But, let's forget them now.
And, what we did is we said,
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00:08:23 --> 00:08:28
well, we can rewrite this as (x
y)^2 2y^2.
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00:08:28 --> 00:08:31
And now, this is a sum of two
squares.
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00:08:31 --> 00:08:35
So, each of these guys has to
be nonnegative.
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00:08:35 --> 00:08:40
And so, the origin will be a
minimum.
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00:08:40 --> 00:08:44
Well, it turns out we can do
something similar in general no
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00:08:44 --> 00:08:47
matter what the values of a,
b, and c are.
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00:08:47 --> 00:08:50
We'll just try to first
complete things to a square.
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00:08:50 --> 00:08:55
OK, so let's do that.
So, in general,
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00:08:55 --> 00:09:01
well, let me be slightly less
general, and let me assume that
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00:09:01 --> 00:09:08
a is not zero because otherwise
I can't do what I'm going to do.
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00:09:08 --> 00:09:20
So, I'm going to write this as
a times x^2 plus b over axy.
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00:09:20 --> 00:09:25
And then I have my cy^2.
And now this looks like the
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00:09:25 --> 00:09:28
beginning of the square of
something, OK,
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00:09:28 --> 00:09:31
just like what we did over
there.
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00:09:31 --> 00:09:39
So, what is it the square of?
Well, you'd start with x plus I
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00:09:39 --> 00:09:45
claim if I put b over 2a times y
and I square it,
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00:09:45 --> 00:09:52
then see the cross term two
times x times b over 2a y will
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00:09:52 --> 00:09:57
become b over axy.
Of course, now I also get some
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00:09:57 --> 00:10:01
y squares out of this.
How many y squares do I get?
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00:10:01 --> 00:10:05
Well, I get b^2 over 4a^2 times
a.
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00:10:05 --> 00:10:11
So, I get b2 over 4a y^2.
So, and I want,
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00:10:11 --> 00:10:17
in fact, c times y^2.
So, the number of y^2 that I
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00:10:17 --> 00:10:22
should add is c minus b^2 over
4a.
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00:10:22 --> 00:10:27
OK, let's see that again.
If I expand this thing,
143
00:10:27 --> 00:10:33
I will get ax^2 plus a times b
over 2a times 2xy.
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00:10:33 --> 00:10:39
That's going to be my bxy.
But, I also get b^2 over 4a^2
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00:10:39 --> 00:10:44
y^2 times a.
That's b^2 over 4ay^2.
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00:10:44 --> 00:10:47
And, that cancels out with this
guy here.
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00:10:47 --> 00:10:52
And then, I will be left with
cy^2.
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00:10:52 --> 00:10:58
OK, do you see it kind of?
OK, if not, well,
149
00:10:58 --> 00:11:04
try expanding this square
again.
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00:11:04 --> 00:11:06
OK, maybe I'll do it just to
convince you.
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00:11:06 --> 00:11:11
But, so if I expand this,
I will get A times,
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00:11:11 --> 00:11:16
let me put that in a different
color because you shouldn't
153
00:11:16 --> 00:11:19
write that down.
It's just to convince you again.
154
00:11:19 --> 00:11:25
So, if you don't see it yet,
let's expend this thing.
155
00:11:25 --> 00:11:35
We'll get a times x^2 plus a
times 2xb over 2ay.
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00:11:35 --> 00:11:42
Well, the two A's cancel out.
We get bxy plus a times the
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00:11:42 --> 00:11:53
square of that's going to be b^2
over 4a^2 y^2 plus cy^2 minus
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00:11:53 --> 00:11:59
b^2 over 4ay^2.
Here, the a and the a
159
00:11:59 --> 00:12:06
simplifies, and now these two
terms simplify and give me just
160
00:12:06 --> 00:12:09
cy^2 in the end.
OK, and that's kind of
161
00:12:09 --> 00:12:12
unreadable after I've canceled
everything,
162
00:12:12 --> 00:12:19
but if you follow it,
you see that basically I've
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00:12:19 --> 00:12:24
just rewritten my initial
function.
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00:12:24 --> 00:12:29
OK, is that kind of OK?
I mean, otherwise there's just
165
00:12:29 --> 00:12:32
no substitute.
You'll have to do it yourself,
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00:12:32 --> 00:12:38
I'm afraid.
OK, so, let me continue to play
167
00:12:38 --> 00:12:43
with this.
So, I'm just going to put this
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00:12:43 --> 00:12:48
in a slightly different form
just to clear the denominators.
169
00:12:48 --> 00:12:56
OK, so, I will instead write
this as one over 4a times the
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00:12:56 --> 00:13:03
big thing.
So, I'm going to just put 4a^2
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00:13:03 --> 00:13:10
times x plus b over 2ay squared.
OK, so far I have the same
172
00:13:10 --> 00:13:13
thing as here.
I just introduced the 4a that
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00:13:13 --> 00:13:19
cancels out, plus for the other
one, I'm just clearing the
174
00:13:19 --> 00:13:28
denominator.
I end up with (4ac-b^2)y^2.
175
00:13:28 --> 00:13:32
OK, so that's a lot of terms.
But, what does it look like?
176
00:13:32 --> 00:13:35
Well, it looks like,
so we have some constant
177
00:13:35 --> 00:13:38
factors, and here we have a
square, and here we have a
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00:13:38 --> 00:13:39
square.
So, basically,
179
00:13:39 --> 00:13:44
we've written this as a sum of
two squares, well,
180
00:13:44 --> 00:13:47
a sum or a difference of two
squares.
181
00:13:47 --> 00:13:51
And, maybe that's what we need
to figure out to know what kind
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00:13:51 --> 00:13:55
of point it is because,
see, if you take a sum of two
183
00:13:55 --> 00:13:57
squares,
that you will know that each
184
00:13:57 --> 00:14:01
square takes nonnegative values.
And you will have,
185
00:14:01 --> 00:14:04
the function will always take
nonnegative values.
186
00:14:04 --> 00:14:07
So, the origin will be a
minimum.
187
00:14:07 --> 00:14:10
Well, if you have a difference
of two squares that typically
188
00:14:10 --> 00:14:13
you'll have a saddle point
because depending on whether one
189
00:14:13 --> 00:14:18
or the other is larger,
you will have a positive or a
190
00:14:18 --> 00:14:24
negative quantity.
OK, so I claim there's various
191
00:14:24 --> 00:14:32
cases to look at.
So, let's see.
192
00:14:32 --> 00:14:34
So, in fact,
I claim there will be three
193
00:14:34 --> 00:14:37
cases.
And, that's good news for us
194
00:14:37 --> 00:14:40
because after all,
we want to distinguish between
195
00:14:40 --> 00:14:45
three possibilities.
So, let's first do away with
196
00:14:45 --> 00:14:52
the most complicated one.
What if 4ac minus b^2 is
197
00:14:52 --> 00:14:56
negative?
Well, if it's negative,
198
00:14:56 --> 00:15:00
then it means what I have
between the brackets is,
199
00:15:00 --> 00:15:06
so the first guy is obviously a
positive quantity,
200
00:15:06 --> 00:15:10
while the second one will be
something negative times y2.
201
00:15:10 --> 00:15:13
So, it will be a negative
quantity.
202
00:15:13 --> 00:15:23
OK, so one term is positive.
The other is negative.
203
00:15:23 --> 00:15:31
That tells us we actually have
a saddle point.
204
00:15:31 --> 00:15:35
We have, in fact,
written our function as a
205
00:15:35 --> 00:15:40
difference of two squares.
OK, is that convincing?
206
00:15:40 --> 00:15:42
So, if you want,
what I could do is actually I
207
00:15:42 --> 00:15:47
could change my coordinates,
have new coordinates called u
208
00:15:47 --> 00:15:50
equals x b over 2ay,
and v, actually,
209
00:15:50 --> 00:15:55
well, I could keep y,
and that it would look like the
210
00:15:55 --> 00:16:02
difference of squares directly.
OK, so that's the first case.
211
00:16:02 --> 00:16:12
The second case is where
4ac-b^2 = 0.
212
00:16:12 --> 00:16:18
Well, what happens if that's
zero?
213
00:16:18 --> 00:16:21
Then it means that this term
over there goes away.
214
00:16:21 --> 00:16:25
So, what we have is just one
square.
215
00:16:25 --> 00:16:29
OK, so what that means is
actually that our function
216
00:16:29 --> 00:16:32
depends only on one direction of
things.
217
00:16:32 --> 00:16:36
In the other direction,
it's going to actually be
218
00:16:36 --> 00:16:38
degenerate.
So, for example,
219
00:16:38 --> 00:16:40
forget all the clutter in
there.
220
00:16:40 --> 00:16:45
Say I give you just the
function of two variables,
221
00:16:45 --> 00:16:49
w equals just x^2.
So, that means it doesn't
222
00:16:49 --> 00:16:53
depend on y at all.
And, if I try to plot the
223
00:16:53 --> 00:16:58
graph, it will look like,
well, x is here.
224
00:16:58 --> 00:17:04
So, it will depend on x in that
way, but it doesn't depend on y
225
00:17:04 --> 00:17:10
at all.
So, what the graph looks like
226
00:17:10 --> 00:17:18
is something like that.
OK, basically it's a valley
227
00:17:18 --> 00:17:22
whose bottom is completely flat.
So, that means,
228
00:17:22 --> 00:17:24
actually, we have a degenerate
critical point.
229
00:17:24 --> 00:17:28
It's called degenerate because
there is a direction in which
230
00:17:28 --> 00:17:30
nothing happens.
And, in fact,
231
00:17:30 --> 00:17:38
you have critical points
everywhere along the y axis.
232
00:17:38 --> 00:17:42
Now, whether the square that we
have is x or something else,
233
00:17:42 --> 00:17:46
namely, x plus b over 2a y,
it doesn't matter.
234
00:17:46 --> 00:17:48
I mean, it will still get this
degenerate behavior.
235
00:17:48 --> 00:17:56
But there's a direction in
which nothing happens because we
236
00:17:56 --> 00:18:02
just have the square of one
quantity.
237
00:18:02 --> 00:18:06
I'm sure that 300 students
means 300 different ring tones,
238
00:18:06 --> 00:18:09
but I'm not eager to hear all
of them, thanks.
239
00:18:09 --> 00:18:18
[LAUGHTER]
OK, so, this is what's called a
240
00:18:18 --> 00:18:28
degenerate critical point,
and [LAUGHTER].
241
00:18:28 --> 00:18:33
OK, so basically we'll leave it
here.
242
00:18:33 --> 00:18:38
We won't actually try to figure
out further what happens,
243
00:18:38 --> 00:18:42
and the reason for that is that
when you have an actual
244
00:18:42 --> 00:18:44
function,
a general function,
245
00:18:44 --> 00:18:46
not just one that's quadratic
like this,
246
00:18:46 --> 00:18:50
then there will actually be
other terms maybe involving
247
00:18:50 --> 00:18:54
higher powers,
maybe x^3 or y^3 or things like
248
00:18:54 --> 00:18:56
that.
And then, they will mess up
249
00:18:56 --> 00:19:00
what happens in this valley.
And, it's a situation where we
250
00:19:00 --> 00:19:03
won't be able,
actually, to tell automatically
251
00:19:03 --> 00:19:06
just by looking at second
derivatives what happens.
252
00:19:06 --> 00:19:09
See, for example,
in a function of one variable,
253
00:19:09 --> 00:19:12
if you have just a function of
one variable,
254
00:19:12 --> 00:19:14
say, f of x equals x to the
five,
255
00:19:14 --> 00:19:18
well, if you try to decide what
type of point the origin is,
256
00:19:18 --> 00:19:20
you're going to take the second
derivative.
257
00:19:20 --> 00:19:23
It will be zero,
and then you can conclude.
258
00:19:23 --> 00:19:26
Those things depend on higher
order derivatives.
259
00:19:26 --> 00:19:29
So, we just won't like that
case.
260
00:19:29 --> 00:19:34
We just won't try to figure out
what's going on here.
261
00:19:34 --> 00:19:40
Now, the last situation is if
4ac-b^2 is positive.
262
00:19:40 --> 00:19:45
So, then, that means that
actually we've written things.
263
00:19:45 --> 00:19:52
The big bracket up there is a
sum of two squares.
264
00:19:52 --> 00:20:00
So, that means that we've
written w as one over 4a times
265
00:20:00 --> 00:20:08
plus something squared plus
something else squared,
266
00:20:08 --> 00:20:12
OK?
So, these guys have the same
267
00:20:12 --> 00:20:18
sign, and that means that this
term here will always be greater
268
00:20:18 --> 00:20:22
than or equal to zero.
And that means that we should
269
00:20:22 --> 00:20:24
either have a maximum or
minimum.
270
00:20:24 --> 00:20:29
How we find out which one it is?
Well, we look at the sign of a,
271
00:20:29 --> 00:20:30
exactly.
OK?
272
00:20:30 --> 00:20:35
So, there's two sub-cases.
One is if a is positive,
273
00:20:35 --> 00:20:40
then, this quantity overall
will always be nonnegative.
274
00:20:40 --> 00:20:54
And that means we have a
minimum, OK?
275
00:20:54 --> 00:20:58
And, if a is negative on the
other hand,
276
00:20:58 --> 00:21:01
so that means that we multiply
this positive quantity by a
277
00:21:01 --> 00:21:04
negative number,
we get something that's always
278
00:21:04 --> 00:21:10
negative.
So, zero is actually the
279
00:21:10 --> 00:21:18
maximum.
OK, is that clear for everyone?
280
00:21:18 --> 00:21:29
Yes?
Sorry, yeah,
281
00:21:29 --> 00:21:34
so I said in the example w
equals x^2, it doesn't depend on
282
00:21:34 --> 00:21:37
y.
So, the more general situation
283
00:21:37 --> 00:21:44
is w equals some constant.
Well, I guess it's a times (x b
284
00:21:44 --> 00:21:48
over 2a times y)^2.
So, it does depend on x and y,
285
00:21:48 --> 00:21:51
but it only depends on this
combination.
286
00:21:51 --> 00:21:54
OK, so if I choose to move in
some other perpendicular
287
00:21:54 --> 00:21:58
direction,
in the direction where this
288
00:21:58 --> 00:22:02
remains constant,
so maybe if I set x equals
289
00:22:02 --> 00:22:06
minus b over 2a y,
then this remains zero all the
290
00:22:06 --> 00:22:08
time.
So, there's a degenerate
291
00:22:08 --> 00:22:11
direction in which I stay at the
minimum or maximum,
292
00:22:11 --> 00:22:15
or whatever it is that I have.
OK, so that's why it's called
293
00:22:15 --> 00:22:17
degenerate.
There is a direction in which
294
00:22:17 --> 00:22:29
nothing happens.
OK, yes?
295
00:22:29 --> 00:22:31
Yes, yeah, so that's a very
good question.
296
00:22:31 --> 00:22:33
So, there's going to be the
second derivative test.
297
00:22:33 --> 00:22:36
Why do not have derivatives yet?
Well, that's because I've been
298
00:22:36 --> 00:22:39
looking at this special example
where we have a function like
299
00:22:39 --> 00:22:41
this.
And, so I don't actually need
300
00:22:41 --> 00:22:43
to take derivatives yet.
But, secretly,
301
00:22:43 --> 00:22:46
that's because a,
b, and c will be the second
302
00:22:46 --> 00:22:49
derivatives of the function,
actually, 2a,
303
00:22:49 --> 00:22:52
b, and 2c.
So now, we are going to go to
304
00:22:52 --> 00:22:54
general function.
And there, instead of having
305
00:22:54 --> 00:22:57
these coefficients a,
b, and c given to us,
306
00:22:57 --> 00:23:00
we'll have to compute them as
second derivatives.
307
00:23:00 --> 00:23:03
OK, so here,
I'm basically setting the stage
308
00:23:03 --> 00:23:07
for what will be the actual
criterion we'll use using second
309
00:23:07 --> 00:23:13
derivatives.
Yes?
310
00:23:13 --> 00:23:16
So, yeah, so what you have a
degenerate critical point,
311
00:23:16 --> 00:23:20
it could be a degenerate
minimum, or a degenerate maximum
312
00:23:20 --> 00:23:23
depending on the sign of a.
But, in general,
313
00:23:23 --> 00:23:26
once you start having
functions, you don't really know
314
00:23:26 --> 00:23:30
what will happen anymore.
It could also be a degenerate
315
00:23:30 --> 00:23:36
saddle, and so on.
So, we won't really be able to
316
00:23:36 --> 00:23:40
tell.
Yes?
317
00:23:40 --> 00:23:43
It is possible to have a
degenerate saddle point.
318
00:23:43 --> 00:23:46
For example,
if I gave you x^3 y^3,
319
00:23:46 --> 00:23:49
you can convince yourself that
if you take x and y to be
320
00:23:49 --> 00:23:53
negative, it will be negative.
If x and y are positive,
321
00:23:53 --> 00:23:55
it's positive.
And, it has a very degenerate
322
00:23:55 --> 00:23:59
critical point at the origin.
So, that's a degenerate saddle
323
00:23:59 --> 00:24:01
point.
We don't see it here because
324
00:24:01 --> 00:24:04
that doesn't happen if you have
only quadratic terms like that.
325
00:24:04 --> 00:24:12
You need to have higher-order
terms to see it happen.
326
00:24:12 --> 00:24:23
OK.
OK, so let's continue.
327
00:24:23 --> 00:24:27
Before we continue,
but see, I wanted to point out
328
00:24:27 --> 00:24:30
one small thing.
So, here, we have the magic
329
00:24:30 --> 00:24:34
quantity, 4ac minus b^2.
You've probably seen that
330
00:24:34 --> 00:24:37
before in your life.
Yet, it looks like the
331
00:24:37 --> 00:24:40
quadratic formula,
except that one involves
332
00:24:40 --> 00:24:43
b^2-4ac.
But that's really the same
333
00:24:43 --> 00:24:47
thing.
OK, so let's see,
334
00:24:47 --> 00:24:57
where does the quadratic
formula come in here?
335
00:24:57 --> 00:25:00
Well, let me write things
differently.
336
00:25:00 --> 00:25:03
OK, so we've manipulated
things, and got into a
337
00:25:03 --> 00:25:08
conclusion.
But, let me just do a different
338
00:25:08 --> 00:25:14
manipulation,
and write this now instead as
339
00:25:14 --> 00:25:23
y^2 times a times x over y
squared plus b(x over y) plus c.
340
00:25:23 --> 00:25:28
OK, see, that's the same thing
that I had before.
341
00:25:28 --> 00:25:35
Well, so now this quantity here
is always nonnegative.
342
00:25:35 --> 00:25:39
What about this one?
Well, of course,
343
00:25:39 --> 00:25:43
this one depends on x over y.
It means it depends on which
344
00:25:43 --> 00:25:45
direction you're going to move
away from the origin,
345
00:25:45 --> 00:25:48
which ratio between x and y you
will consider.
346
00:25:48 --> 00:25:51
But, I claim there's two
situations.
347
00:25:51 --> 00:25:57
One is, so, let's try to
reformulate things.
348
00:25:57 --> 00:26:04
So, if a discriminate here is
positive, then it means that
349
00:26:04 --> 00:26:10
these have roots and these have
solutions.
350
00:26:10 --> 00:26:19
And, that means that this
quantity can be both positive
351
00:26:19 --> 00:26:24
and negative.
This quantity takes positive
352
00:26:24 --> 00:26:31
and negative values.
One way to convince yourself is
353
00:26:31 --> 00:26:37
just to, you know,
plot at^2 bt c.
354
00:26:37 --> 00:26:43
You know that there's two roots.
So, it might look like this,
355
00:26:43 --> 00:26:48
or might look like that
depending on the sign of a.
356
00:26:48 --> 00:26:52
But, in either case,
it will take values of both
357
00:26:52 --> 00:26:54
signs.
So, that means that your
358
00:26:54 --> 00:26:56
function will take values of
both signs.
359
00:26:56 --> 00:27:04
360
00:27:04 --> 00:27:13
The value takes both positive
and negative values.
361
00:27:13 --> 00:27:21
And, so that means we have a
saddle point,
362
00:27:21 --> 00:27:28
while the other situation,
when b^2-4ac is negative -- --
363
00:27:28 --> 00:27:36
means that this equation is
quadratic never takes the value,
364
00:27:36 --> 00:27:39
zero.
So, it's always positive or
365
00:27:39 --> 00:27:42
it's always negative,
depending on the sign of a.
366
00:27:42 --> 00:27:48
So, the other case is if
b^2-4ac is negative,
367
00:27:48 --> 00:27:53
then the quadratic doesn't have
a solution.
368
00:27:53 --> 00:27:58
And it could look like this or
like that depending on whether a
369
00:27:58 --> 00:28:03
is positive or a is negative.
So, in particular,
370
00:28:03 --> 00:28:12
that means that ax over y2 plus
bx over y plus c is always
371
00:28:12 --> 00:28:21
positive or always negative
depending on the sign of a.
372
00:28:21 --> 00:28:23
And then, that tells us that
our function,
373
00:28:23 --> 00:28:25
w, will be always positive or
always negative.
374
00:28:25 --> 00:28:28
And then we'll get a minimum or
maximum.
375
00:28:28 --> 00:28:40
376
00:28:40 --> 00:28:44
OK, we'll have a min or a max
depending on which situation we
377
00:28:44 --> 00:28:47
are in.
OK, so that's another way to
378
00:28:47 --> 00:28:51
derive the same answer.
And now, you see here why the
379
00:28:51 --> 00:28:55
discriminate plays a role.
That's because it exactly tells
380
00:28:55 --> 00:28:59
you whether this quadratic
quantity has always the same
381
00:28:59 --> 00:29:04
sign,
or whether it can actually
382
00:29:04 --> 00:29:12
cross the value,
zero, when you have the root of
383
00:29:12 --> 00:29:16
a quadratic.
OK, so hopefully at this stage
384
00:29:16 --> 00:29:20
you are happy with one of the
two explanations,
385
00:29:20 --> 00:29:23
at least.
And now, you are willing to
386
00:29:23 --> 00:29:26
believe, I hope,
that we have basically a way of
387
00:29:26 --> 00:29:30
deciding what type of critical
point we have in the special
388
00:29:30 --> 00:29:32
case of a quadratic function.
389
00:29:32 --> 00:29:58
390
00:29:58 --> 00:30:05
OK, so, now what do we do with
the general function?
391
00:30:05 --> 00:30:19
Well, so in general,
we want to look at second
392
00:30:19 --> 00:30:24
derivatives.
OK, so now we are getting to
393
00:30:24 --> 00:30:27
the real stuff.
So, how many second derivatives
394
00:30:27 --> 00:30:29
do we have?
That's maybe the first thing we
395
00:30:29 --> 00:30:32
should figure out.
Well, we can take the
396
00:30:32 --> 00:30:39
derivative first with respect to
x, and then again with respect
397
00:30:39 --> 00:30:44
to x.
OK, that gives us something we
398
00:30:44 --> 00:30:54
denote by partial square f over
partial x squared or fxx.
399
00:30:54 --> 00:31:00
Then, there's another one which
is fxy, which means you take the
400
00:31:00 --> 00:31:05
derivative with respect to x,
and then with respect to y.
401
00:31:05 --> 00:31:09
Another thing you can do,
is do first derivative respect
402
00:31:09 --> 00:31:12
to y, and then with respect to
x.
403
00:31:12 --> 00:31:17
That would be fyx.
Well, good news.
404
00:31:17 --> 00:31:22
These are actually always equal
to each other.
405
00:31:22 --> 00:31:26
OK, so it's the fact that we
will admit, it's actually not
406
00:31:26 --> 00:31:30
very hard to check.
So these are always the same.
407
00:31:30 --> 00:31:33
We don't need to worry about
which one we do.
408
00:31:33 --> 00:31:36
That's one computation that we
won't need to do.
409
00:31:36 --> 00:31:43
We can save a bit of effort.
And then, we have the last one,
410
00:31:43 --> 00:31:51
namely, the second partial with
respect to y and y fyy.
411
00:31:51 --> 00:32:00
OK, so we have three of them.
So, what does the second
412
00:32:00 --> 00:32:02
derivative test say?
413
00:32:02 --> 00:32:16
414
00:32:16 --> 00:32:22
It says, say that you have a
critical point (x0,
415
00:32:22 --> 00:32:27
y0) of a function of two
variables, f,
416
00:32:27 --> 00:32:34
and then let's compute the
partial derivatives.
417
00:32:34 --> 00:32:41
So, let's call capital A the
second derivative with respect
418
00:32:41 --> 00:32:45
to x.
Let's call capital B the second
419
00:32:45 --> 00:32:49
derivative with respect to x and
y.
420
00:32:49 --> 00:32:55
And C equals fyy at this point,
OK?
421
00:32:55 --> 00:32:59
So, these are just numbers
because we first compute the
422
00:32:59 --> 00:33:02
second derivative,
and then we plug in the values
423
00:33:02 --> 00:33:04
of x and y at the critical
point.
424
00:33:04 --> 00:33:14
So, these will just be numbers.
And now, what we do is we look
425
00:33:14 --> 00:33:21
at the quantity AC-B^2.
I am not forgetting the four.
426
00:33:21 --> 00:33:26
You will see why there isn't
one.
427
00:33:26 --> 00:33:31
So, if AC-B^2 is positive,
then there's two sub-cases.
428
00:33:31 --> 00:33:39
If A is positive,
then it's local minimum.
429
00:33:39 --> 00:33:50
430
00:33:50 --> 00:33:56
The second case,
so, still, if AC-B^2 is
431
00:33:56 --> 00:34:04
positive, but A is negative,
then it's going to be a local
432
00:34:04 --> 00:34:11
maximum.
And, if AC-B^2 is negative,
433
00:34:11 --> 00:34:17
then it's a saddle point,
and finally,
434
00:34:17 --> 00:34:24
if AC-B^2 is zero,
then we actually cannot
435
00:34:24 --> 00:34:28
compute.
We don't know whether it's
436
00:34:28 --> 00:34:33
going to be a minimum,
a maximum, or a saddle.
437
00:34:33 --> 00:34:37
We know it's degenerate in some
way, but we don't know what type
438
00:34:37 --> 00:34:40
of point it is.
OK, so that's actually what you
439
00:34:40 --> 00:34:43
need to remember.
If you are formula oriented,
440
00:34:43 --> 00:34:45
that's all you need to remember
about today.
441
00:34:45 --> 00:34:53
But, let's try to understand
why, how this comes out of what
442
00:34:53 --> 00:34:59
we had there.
OK, so, I think maybe I
443
00:34:59 --> 00:35:05
actually want to keep,
so maybe I want to keep this
444
00:35:05 --> 00:35:06
middle board because it actually
has,
445
00:35:06 --> 00:35:09
you know, the recipe that we
found before the quadratic
446
00:35:09 --> 00:35:12
function.
So, let me move directly over
447
00:35:12 --> 00:35:16
there and try to relate our old
recipe with the new.
448
00:35:16 --> 00:35:43
449
00:35:43 --> 00:35:50
OK, you are easily amused.
OK, so first,
450
00:35:50 --> 00:35:57
let's check that these two
things say the same thing in the
451
00:35:57 --> 00:36:01
special case that we are looking
at.
452
00:36:01 --> 00:36:12
OK, so let's verify in the
special case where the function
453
00:36:12 --> 00:36:22
was ax^2 bxy cy^2.
So -- Well, what is the second
454
00:36:22 --> 00:36:28
derivative with respect to x and
x?
455
00:36:28 --> 00:36:31
If I take the second derivative
with respect to x and x,
456
00:36:31 --> 00:36:34
so first I want to take maybe
the derivative with respect to
457
00:36:34 --> 00:36:37
x.
But first, let's take the first
458
00:36:37 --> 00:36:46
partial, Wx.
That will be 2ax by, right?
459
00:36:46 --> 00:36:50
So, Wxx will be,
well, let's take a partial with
460
00:36:50 --> 00:36:54
respect to x again.
That's 2a.
461
00:36:54 --> 00:37:02
Wxy, I take the partial respect
to y, and we'll get b.
462
00:37:02 --> 00:37:06
OK, now we need,
also, the partial with respect
463
00:37:06 --> 00:37:13
to y.
So, Wy is bx 2cy.
464
00:37:13 --> 00:37:17
In case you don't believe what
I told you about the mixed
465
00:37:17 --> 00:37:21
partials, Wyx,
well, you can check.
466
00:37:21 --> 00:37:24
And it's, again, b.
So, they are,
467
00:37:24 --> 00:37:30
indeed, the same thing.
And, Wyy will be 2c.
468
00:37:30 --> 00:37:39
So, if we now look at these
quantities, that tells us,
469
00:37:39 --> 00:37:46
well, big A is two little a,
big B is little b,
470
00:37:46 --> 00:37:55
big C is two little c.
So, AC-B^2 is what we used to
471
00:37:55 --> 00:38:04
call four little ac minus b2.
OK, ooh.
472
00:38:04 --> 00:38:07
[LAUGHTER]
So, now you can compare the
473
00:38:07 --> 00:38:10
cases.
They are not listed in the same
474
00:38:10 --> 00:38:14
order just to make it harder.
So, we said first,
475
00:38:14 --> 00:38:20
so the saddle case is when
AC-B^2 in big letters is
476
00:38:20 --> 00:38:26
negative, that's the same as
4ac-b2 in lower case is
477
00:38:26 --> 00:38:30
negative.
The case where capital AC-B2 is
478
00:38:30 --> 00:38:35
positive, local min and local
max corresponds to this one.
479
00:38:35 --> 00:38:40
And, the case where we can't
conclude was what used to be the
480
00:38:40 --> 00:38:44
degenerate one.
OK, so at least we don't seem
481
00:38:44 --> 00:38:48
to have messed up when copying
the formula.
482
00:38:48 --> 00:38:56
Now, why does that work more
generally than that?
483
00:38:56 --> 00:39:03
Well, the answer that is,
again, Taylor approximation.
484
00:39:03 --> 00:39:16
Aww.
OK, so let me just do here
485
00:39:16 --> 00:39:22
quadratic approximation.
So, quadratic approximation
486
00:39:22 --> 00:39:25
tells me the following thing.
It tells me,
487
00:39:25 --> 00:39:30
if I have a function,
f of xy, and I want to
488
00:39:30 --> 00:39:37
understand the change in f when
I change x and y a little bit.
489
00:39:37 --> 00:39:40
Well, there's the first-order
terms.
490
00:39:40 --> 00:39:43
There is the linear terms that
by now you should know and be
491
00:39:43 --> 00:39:51
comfortable with.
That's fx times the change in x.
492
00:39:51 --> 00:39:56
And then, there's fy times the
change in y.
493
00:39:56 --> 00:40:00
OK, that's the starting point.
But now, of course,
494
00:40:00 --> 00:40:03
if x and y, sorry,
if we are at the critical
495
00:40:03 --> 00:40:09
point, then that's going to be
zero at the critical point.
496
00:40:09 --> 00:40:16
So, that term actually goes
away, and that's also zero at
497
00:40:16 --> 00:40:22
the critical point.
So, that term also goes away.
498
00:40:22 --> 00:40:24
OK, so linear approximation is
really no good.
499
00:40:24 --> 00:40:27
We need more terms.
So, what are the next terms?
500
00:40:27 --> 00:40:35
Well, the next terms are
quadratic terms,
501
00:40:35 --> 00:40:38
and so I mean,
if you remember the Taylor
502
00:40:38 --> 00:40:42
formula for a function of a
single variable,
503
00:40:42 --> 00:40:46
there was the derivative times
x minus x0 plus one half of a
504
00:40:46 --> 00:40:51
second derivative times x-x0^2.
And see, this side here is
505
00:40:51 --> 00:40:55
really Taylor approximation in
one variable looking only at x.
506
00:40:55 --> 00:40:57
But of course,
we also have terms involving y,
507
00:40:57 --> 00:41:00
and terms involving
simultaneously x and y.
508
00:41:00 --> 00:41:10
And, these terms are fxy times
change in x times change in y
509
00:41:10 --> 00:41:17
plus one half of fyy(y-y0)^2.
There's no one half in the
510
00:41:17 --> 00:41:20
middle because,
in fact, you would have two
511
00:41:20 --> 00:41:24
terms, one for xy,
one for yx, but they are the
512
00:41:24 --> 00:41:26
same.
And then, if you want to
513
00:41:26 --> 00:41:29
continue, there is actually
cubic terms involving the third
514
00:41:29 --> 00:41:32
derivatives, and so on,
but we are not actually looking
515
00:41:32 --> 00:41:34
at them.
And so, now,
516
00:41:34 --> 00:41:39
when we do this approximation,
well, the type of critical
517
00:41:39 --> 00:41:45
point remains the same when we
replace the function by this
518
00:41:45 --> 00:41:48
approximation.
And so, we can apply the
519
00:41:48 --> 00:41:53
argument that we used to deduce
things in the quadratic case.
520
00:41:53 --> 00:41:55
In fact, it still works in the
general case using this
521
00:41:55 --> 00:41:57
approximation formula.
522
00:41:57 --> 00:42:12
523
00:42:12 --> 00:42:26
So -- The general case reduces
to the quadratic case.
524
00:42:26 --> 00:42:31
And now, you see actually why,
well, here you see,
525
00:42:31 --> 00:42:36
again, how this coefficient
which we used to call little a
526
00:42:36 --> 00:42:41
is also one half of capital A.
And same here:
527
00:42:41 --> 00:42:47
this coefficient is what we
call capital B or little b,
528
00:42:47 --> 00:42:52
and this coefficient here is
what we called little c or one
529
00:42:52 --> 00:42:57
half of capital C.
And then, when you replace
530
00:42:57 --> 00:43:02
these into the various cases
that we had here,
531
00:43:02 --> 00:43:06
you end up with the second
derivative test.
532
00:43:06 --> 00:43:08
So, what about the degenerate
case?
533
00:43:08 --> 00:43:11
Why can't we just say,
well, it's going to be a
534
00:43:11 --> 00:43:16
degenerate critical point?
So, the reason is that this
535
00:43:16 --> 00:43:20
approximation formula is
reasonable only if the higher
536
00:43:20 --> 00:43:24
order terms are negligible.
OK, so in fact,
537
00:43:24 --> 00:43:27
secretly, there's more terms.
This is only an approximation.
538
00:43:27 --> 00:43:30
There would be terms involving
third derivatives,
539
00:43:30 --> 00:43:34
and maybe even beyond that.
And, so it is not to generate
540
00:43:34 --> 00:43:37
case,
they don't actually matter
541
00:43:37 --> 00:43:39
because the shape of the
function,
542
00:43:39 --> 00:43:42
the shape of the graph,
is actually determined by the
543
00:43:42 --> 00:43:45
quadratic terms.
But, in the degenerate case,
544
00:43:45 --> 00:43:49
see, if I start with this and I
add something even very,
545
00:43:49 --> 00:43:53
very small along the y axis,
then that can be enough to bend
546
00:43:53 --> 00:43:56
this very slightly up or
slightly down,
547
00:43:56 --> 00:44:00
and turn my degenerate point in
to either a minimum or a saddle
548
00:44:00 --> 00:44:03
point.
And, I won't be able to tell
549
00:44:03 --> 00:44:06
until I go further in the list
of derivatives.
550
00:44:06 --> 00:44:14
So, in the degenerate case,
what actually happens depends
551
00:44:14 --> 00:44:20
on the higher order derivatives.
552
00:44:20 --> 00:44:38
553
00:44:38 --> 00:44:42
So, we will need to analyze
things more carefully.
554
00:44:42 --> 00:44:45
Well, we're not going to bother
with that in this class.
555
00:44:45 --> 00:44:52
So, we'll just say,
well, we cannot compute,
556
00:44:52 --> 00:44:54
OK?
I mean, you have to realize
557
00:44:54 --> 00:44:57
that in real life,
you have to be extremely
558
00:44:57 --> 00:45:02
unlucky for this quantity to end
up being exactly 0.
559
00:45:02 --> 00:45:03
[LAUGHTER]
Well, if that happens,
560
00:45:03 --> 00:45:05
then what you should do is
maybe try by inspection.
561
00:45:05 --> 00:45:08
See if there's a good reason
why the function should always
562
00:45:08 --> 00:45:11
be positive or always be
negative, or something.
563
00:45:11 --> 00:45:16
Or, you know,
plot it on a computer and see
564
00:45:16 --> 00:45:23
what happens.
But, otherwise we can't compute.
565
00:45:23 --> 00:45:33
OK, so let's do an example.
So, probably I should leave
566
00:45:33 --> 00:45:39
this on so that we still have
the test with us.
567
00:45:39 --> 00:45:42
And, instead,
OK, so I'll do my example here.
568
00:45:42 --> 00:46:20
569
00:46:20 --> 00:46:30
OK, so just an example.
Let's look at f of (x,
570
00:46:30 --> 00:46:37
y) = x y 1/xy,
where x and y are positive.
571
00:46:37 --> 00:46:39
So, I'm looking only at the
first quadrant.
572
00:46:39 --> 00:46:42
OK, I mean, I'm doing this
because I don't want the
573
00:46:42 --> 00:46:46
denominator to become zero.
So, I'm just looking at the
574
00:46:46 --> 00:46:50
situation.
So, let's look first for,
575
00:46:50 --> 00:46:55
so, the question will be,
what are the minimum and
576
00:46:55 --> 00:47:03
maximum of this function?
So, the first thing we should
577
00:47:03 --> 00:47:12
do to answer this question is
look for critical points,
578
00:47:12 --> 00:47:15
OK?
So, for that,
579
00:47:15 --> 00:47:19
we have to compute the first
derivatives.
580
00:47:19 --> 00:47:34
OK, so fx is one minus one over
x^2y, OK?
581
00:47:34 --> 00:47:39
Take the derivative of one over
x, that's negative one over x^2.
582
00:47:39 --> 00:47:44
And, we'll want to set that
equal to zero.
583
00:47:44 --> 00:47:50
And fy is one minus one over
xy^2.
584
00:47:50 --> 00:47:54
And, we want to set that equal
to zero.
585
00:47:54 --> 00:47:59
So, what are the equations we
have to solve?
586
00:47:59 --> 00:48:05
Well, I guess x^2y equals one,
I mean, if I move this guy over
587
00:48:05 --> 00:48:09
here I get one over x^2y equals
one.
588
00:48:09 --> 00:48:14
That's x^2y equals one,
and xy^2 equals one.
589
00:48:14 --> 00:48:18
What do you get by comparing
these two?
590
00:48:18 --> 00:48:21
Well, x and y should both be,
OK, so yeah,
591
00:48:21 --> 00:48:24
I agree with you that one and
one is a solution.
592
00:48:24 --> 00:48:27
Why is it the only one?
So, first, if I divide this one
593
00:48:27 --> 00:48:29
by that one, I get x over y
equals one.
594
00:48:29 --> 00:48:34
So, it tells me x equals y.
And then, if x equals y,
595
00:48:34 --> 00:48:40
then if I put that into here,
it will give me y^3 equals one,
596
00:48:40 --> 00:48:44
which tells me y equals one,
and therefore,
597
00:48:44 --> 00:48:50
x equals one as well.
OK, so, there's only one
598
00:48:50 --> 00:48:54
solution.
There's only one critical
599
00:48:54 --> 00:48:58
point, which is going to be
(1,1).
600
00:48:58 --> 00:49:09
OK, so, now here's where you do
a bit of work.
601
00:49:09 --> 00:49:18
What do you think of that
critical point?
602
00:49:18 --> 00:49:25
OK, I see some valid votes.
I see some, OK,
603
00:49:25 --> 00:49:28
I see a lot of people answering
four.
604
00:49:28 --> 00:49:30
[LAUGHTER]
that seems to suggest that
605
00:49:30 --> 00:49:34
maybe you haven't completed the
second derivative yet.
606
00:49:34 --> 00:49:37
Yes, I see someone giving the
correct answer.
607
00:49:37 --> 00:49:41
I see some people not giving
quite the correct answer.
608
00:49:41 --> 00:49:43
I see more and more correct
answers.
609
00:49:43 --> 00:49:49
OK, so let's see.
To figure out what type of
610
00:49:49 --> 00:49:52
point is, we should compute the
second partial derivatives.
611
00:49:52 --> 00:50:02
So, fxx is, what do we get what
we take the derivative of this
612
00:50:02 --> 00:50:11
with respect to x?
Two over x^3y, OK?
613
00:50:11 --> 00:50:25
So, at our point, a will be 2.
Fxy will be one over x^2y^2.
614
00:50:25 --> 00:50:37
So, B will be one.
And, Fyy is going to be two
615
00:50:37 --> 00:50:42
over xy^3.
So, C will be two.
616
00:50:42 --> 00:50:51
And so that tells us,
well, AC-B^2 is four minus one.
617
00:50:51 --> 00:51:02
Sorry, I should probably use a
different blackboard for that.
618
00:51:02 --> 00:51:06
AC-B2 is two times two minus
1^2 is three.
619
00:51:06 --> 00:51:10
It's positive.
That tells us we are either a
620
00:51:10 --> 00:51:17
local minimum or local maximum.
And, A is positive.
621
00:51:17 --> 00:51:21
So, it's a local minimum.
And, in fact,
622
00:51:21 --> 00:51:23
you can check it's the global
minimum.
623
00:51:23 --> 00:51:29
What about the maximum?
Well, if a maximum is not
624
00:51:29 --> 00:51:32
actually at a critical point,
it's on the boundary,
625
00:51:32 --> 00:51:35
or at infinity.
See, so we have actually to
626
00:51:35 --> 00:51:39
check what happens when x and y
go to zero or to infinity.
627
00:51:39 --> 00:51:42
Well, if that happens,
if x or y goes to infinity,
628
00:51:42 --> 00:51:44
then the function goes to
infinity.
629
00:51:44 --> 00:51:48
Also, if x or y goes to zero,
then one over xy goes to
630
00:51:48 --> 00:51:51
infinity.
So, the maximum,
631
00:51:51 --> 00:51:59
well, the function goes to
infinity when x goes to infinity
632
00:51:59 --> 00:52:05
or y goes to infinity,
or x and y go to zero.
633
00:52:05 --> 00:52:07
So, it's not at a critical
point.
634
00:52:07 --> 00:52:10
OK, so, in general,
we have to check both the
635
00:52:10 --> 00:52:13
critical points and the
boundaries to decide what
636
00:52:13 --> 00:52:15
happens.
OK, the end.
637
00:52:15 --> 00:52:18
Have a nice weekend.
638
00:52:18 --> 00:52:23