WEBVTT
Kind: captions
Language: en
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What we observed that we could achieve our
goal to have every level curve locally as
00:00:27.860 --> 00:00:35.200
a graph that if I have this for every fixed
x in the neighborhood what we have to decide
00:00:35.200 --> 00:00:40.500
I can have a neighborhood of y such that y
going to f x y
00:00:40.500 --> 00:00:48.440
this function it is as zero here zero here
means it changes sign they are the two end
00:00:48.440 --> 00:00:53.930
points and the uniqueness will be guaranteed
if this is strictly increasing This is one
00:00:53.930 --> 00:00:54.930
of the sufficient conditions
00:00:54.930 --> 00:01:05.040
There could be many but we have to attack
whatever we have in position so if it is strictly
00:01:05.040 --> 00:01:08.210
decreasing or strictly increasing then I will
have unique zero
00:01:08.210 --> 00:01:12.070
And what does it mean by strictly decreasing
or strictly increasing That means I should
00:01:12.070 --> 00:01:21.020
have del f x so I should have d f x d y
00:01:21.020 --> 00:01:27.399
this is greater than zero or strictly less
than zero here it is strictly increasing or
00:01:27.399 --> 00:01:41.170
strictly decreasing for each fixed x in this
interval so that is to say
00:01:41.170 --> 00:01:53.070
in terms of f del f del y x y is strictly
greater than zero or strictly less than zero
00:01:53.070 --> 00:02:04.079
for x y in the interval U
00:02:04.079 --> 00:02:12.240
Ok
Well what I am trying to do given each x in
00:02:12.240 --> 00:02:16.870
this interval we can try to find the unique
y This is sufficient condition
00:02:16.870 --> 00:02:19.040
You understand very well Ok that we could
have
00:02:19.040 --> 00:02:26.200
instead of defining function of x you can
define function of y; that will also do So
00:02:26.200 --> 00:02:33.170
the role of x and y could be interchanged
So first I depending on the situation that
00:02:33.170 --> 00:02:43.010
suppose I do not achieve this but I have del
f del x x y is greater than zero in some interval
00:02:43.010 --> 00:02:56.110
and for fixed y f x y changes sign in the
interval of x naught minus epsilon to x plus
00:02:56.110 --> 00:03:01.540
epsilon then I could have achieved this So
this is
00:03:01.540 --> 00:03:08.409
for writing function of x Instead of that
we could try to write function of y as well
00:03:08.409 --> 00:03:17.430
So role of x and y could have interchanged
SO what we actually mean here is that conclusion
00:03:17.430 --> 00:03:36.500
is that I need one of this either del f del
x or del f del y and x y in this interval
00:03:36.500 --> 00:03:39.720
is non-zero
00:03:39.720 --> 00:03:56.890
Or x naught y naught point non-zero and del
f del x del f del y continuous at x naught
00:03:56.890 --> 00:04:00.090
y naught
Then what will happen it is non-zero so one
00:04:00.090 --> 00:04:01.849
of them is non-zero
00:04:01.849 --> 00:04:05.450
So let’s say positive or negative; If it
is positive or negative then continuous and
00:04:05.450 --> 00:04:12.879
I can decide the interval around where it
remains the same sign
00:04:12.879 --> 00:04:16.440
Ok
So I need continuous partial derivatives of
00:04:16.440 --> 00:04:23.030
f and del f del x del f del y at that point
non-zero
00:04:23.030 --> 00:04:30.020
So if you look at it geometrically that is
a very nice interpretation
00:04:30.020 --> 00:04:39.130
So what we so this is this fellow is what
this fellow according to our notation is grad
00:04:39.130 --> 00:04:42.130
f x naught y naught
00:04:42.130 --> 00:04:48.760
Now what does x naught y naught belongs This
belongs to this level curve f x y equal to
00:04:48.760 --> 00:05:04.510
some constant of x y equal to zero
If I have a level curve let us say this
00:05:04.510 --> 00:05:25.010
and I fix the point x naught y naught here
then you can check very easily that this fellow
00:05:25.010 --> 00:05:34.650
this is precisely the normal at x naught y
naught
00:05:34.650 --> 00:05:47.190
For a level curve grad at some point is the
geometrically is the normal to that point
00:05:47.190 --> 00:05:51.550
And this condition means normal is non-zero
normal non-zero means it is not parallel to
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x axis
Ok so this condition grad f x naught y naught
00:06:00.660 --> 00:06:20.350
or let’s say n not equal to zero that is
n is not parallel to x axis This is what geometrically
00:06:20.350 --> 00:06:22.730
the condition we are imposing
00:06:22.730 --> 00:06:29.000
And what it means and also normal is changing
continuously
00:06:29.000 --> 00:06:33.670
Now if normal is not parallel to x-axis why
we achieve it You forget everything else We
00:06:33.670 --> 00:06:42.630
have this curve and normal is not parallel
to x-axis then the tangent
00:06:42.630 --> 00:06:55.830
this is not parallel to y axis Because normal
and tangent are perpendicular to each other
00:06:55.830 --> 00:07:00.630
If at a point tangent is not parallel to y
axis what does it mean
00:07:00.630 --> 00:07:09.150
If I move little bit from here this side or
this side and fix x here say x naught plus
00:07:09.150 --> 00:07:14.330
something some little bit I move from x naught
so in that neighborhood I have to decide
00:07:14.330 --> 00:07:26.470
Then I draw this line; this is the line x
equal to x naught plus delta x
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Then this line will intersect the curve at
unique point Similarly here x naught minus
00:07:34.850 --> 00:07:40.690
delta x
So this line x naught minus delta x y equal
00:07:40.690 --> 00:07:47.530
to x naught minus delta x this line this will
intersect this curve at unique point
00:07:47.530 --> 00:07:49.860
And that is my unique y; that is what we are
looking at
00:07:49.860 --> 00:07:52.410
So geometrically that is it what is going
on
00:07:52.410 --> 00:07:55.840
And it is very easy that is why I said that
is very easy to understand in function of
00:07:55.840 --> 00:08:04.300
2 variables taking values in R
What we all demand is normal exist non-parallel
00:08:04.300 --> 00:08:13.940
to x axis and it is moving continuously If
I have that we have it
00:08:13.940 --> 00:08:20.440
And of course you can see instead of this
why we cannot achieve it everywhere why I
00:08:20.440 --> 00:08:21.530
need this
00:08:21.530 --> 00:08:31.080
Suppose I have picture like this and I am
looking at this point x naught This normal
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is outward normal this way
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So tangent is this line itself
00:08:38.140 --> 00:08:42.459
Now if move from x naught moving from this
side does not make sense because there is
00:08:42.459 --> 00:08:49.550
no portion of the curve if I move this side
any point you move you will always have two
00:08:49.550 --> 00:08:52.990
points on the curve
00:08:52.990 --> 00:08:56.180
So you cannot decide y uniquely
00:08:56.180 --> 00:08:57.980
So that is the geometry of implicit function
theorem
00:08:57.980 --> 00:09:02.730
Now I hope we have understood what is needed
00:09:02.730 --> 00:09:08.339
Once you have understood we just put them
whatever we need as assumption and prove the
00:09:08.339 --> 00:09:10.329
theorem
00:09:10.329 --> 00:09:19.699
So let me state implicit function theorem
for function of two variables
00:09:19.699 --> 00:09:37.389
For f from R2 to R
So you state your theorem you prove your theorem
00:09:37.389 --> 00:09:41.459
you have to do it carefully
So far what we have been doing we have been
00:09:41.459 --> 00:09:45.720
doing informally whatever we need
So we put it formally now
00:09:45.720 --> 00:09:47.519
So let us put the set up
00:09:47.519 --> 00:10:04.410
Let U in R2 be an open set
00:10:04.410 --> 00:10:12.569
If from U to R
00:10:12.569 --> 00:10:18.500
Ok and x naught y naught belongs to U
00:10:18.500 --> 00:10:42.720
So suppose f has continuous partial derivative
00:10:42.720 --> 00:10:46.749
del f del y
00:10:46.749 --> 00:11:03.480
at x naught y naught and of course x naught
y naught is on the level curve equal to zero
00:11:03.480 --> 00:11:30.259
Then there exists a epsilon greater than zero
and a neighborhood let’s say theta of x
00:11:30.259 --> 00:11:49.759
naught y naught and a function
g of x naught minus epsilon x naught plus
00:11:49.759 --> 00:11:54.740
epsilon to R such that
what happens
00:11:54.740 --> 00:12:04.949
1 f x g x
00:12:04.949 --> 00:12:17.610
equal to zero for all x in x naught minus
epsilon x naught plus epsilon g of x belongs
00:12:17.610 --> 00:12:22.410
to theta and of course g of x naught equal
to y naught
00:12:22.410 --> 00:12:32.180
This is what we wanted to achieve
But we can actually achieve something more
00:12:32.180 --> 00:12:39.689
What that of course what we achieve what we
are going to write is much desirable
00:12:39.689 --> 00:12:43.279
We have taken f is very nice it is continuously
differentiable
00:12:43.279 --> 00:12:47.860
This is C1 function it is continuous for second
derivative
00:12:47.860 --> 00:12:53.130
So f is a smooth level curve
Whatever we have now defined what is smooth
00:12:53.130 --> 00:12:58.990
curve what you understand in layman's terms
that f is nice it is nice function of 2 variables
00:12:58.990 --> 00:13:18.509
And so we will also get g is differential
in the entire U
00:13:18.509 --> 00:13:22.660
So this is the statement of implicit function
from R2 to R
00:13:22.660 --> 00:13:28.500
Let us understand it and remember it from
R2 to R from here the generality statement
00:13:28.500 --> 00:13:30.579
is just mere formalities
00:13:30.579 --> 00:13:41.769
So this part is not really needed in the statement
but it is actually part of the proof; part
00:13:41.769 --> 00:13:45.279
of the proof I have just included it it is
not needed in the statement
00:13:45.279 --> 00:13:52.759
Whatever be the question is the question is
answered here without this statement this
00:13:52.759 --> 00:13:53.759
neighborhood
00:13:53.759 --> 00:13:59.329
But the way I have presented it I mean we
have motivated ourselves actually this will
00:13:59.329 --> 00:14:05.899
be true this theta will come and what will
happen is
00:14:05.899 --> 00:14:14.420
what these true statements means that actually
there exists a differentiable g such that
00:14:14.420 --> 00:14:25.579
f inverse of 0 intersection theta this is
the local part equal to x g x
00:14:25.579 --> 00:14:31.980
x in x naught minus epsilon x naught plus
epsilon
00:14:31.980 --> 00:14:35.139
So this part of the curve is the graph of
this function
00:14:35.139 --> 00:14:44.290
Ok now we actually know how to prove it
We have actually made the statement other
00:14:44.290 --> 00:14:54.800
way round that we have actually put the assumption
what we actually needed but still we need
00:14:54.800 --> 00:15:02.259
to write down the proof formally
So we know all the assumption
00:15:02.259 --> 00:15:07.829
This continuity is needed at x naught y naught
to maintain the same sign of del f del x or
00:15:07.829 --> 00:15:09.749
del f del y
00:15:09.749 --> 00:15:15.050
so that in the neighborhood for a fixed x
if f y is strictly increasing or strictly
00:15:15.050 --> 00:15:26.519
decreasing function of y and non-zero
Oh I have not put this assumption this here
00:15:26.519 --> 00:15:29.619
somehow I missed one assumption right
00:15:29.619 --> 00:15:35.860
Partial partial derivative and most important
assumption let us assume del f del y at x
00:15:35.860 --> 00:15:40.040
naught y naught sorry sorry sorry this is
greater than 0 or less than 0 does not matter
00:15:40.040 --> 00:15:45.790
But this was part of our assumption we need
a normal
00:15:45.790 --> 00:15:49.779
So this is the actual statement
00:15:49.779 --> 00:15:55.100
Ok so we need since it is greater than 0 it
will maintain continuity will tell you it
00:15:55.100 --> 00:16:01.410
will maintain the same sign and we know how
to actually go about it
00:16:01.410 --> 00:16:11.470
So let us write down the proof formally
what is the step 1
00:16:11.470 --> 00:16:17.670
Well what we do
00:16:17.670 --> 00:16:34.720
I have del f del x del y x naught y naught
is bigger than 0
00:16:34.720 --> 00:16:43.249
So I can choose a close interval
I can choose an open interval first and then
00:16:43.249 --> 00:16:45.069
inside that I can choose a closed interval
00:16:45.069 --> 00:16:59.320
You will see why I need a closed interval
00:16:59.320 --> 00:17:18.459
such that del f del y
at x y is let us say bigger than some number
00:17:18.459 --> 00:17:25.760
b Ok for all closed interval give it a name
00:17:25.760 --> 00:17:45.200
choose this interval what do you call it how
do you call it D small b here for all x y
00:17:45.200 --> 00:17:55.390
in D this is by continuity
00:17:55.390 --> 00:18:12.920
And del f del x x y is less than some number
a for all x y right in D
00:18:12.920 --> 00:18:18.130
It is a continuous function del f del x in
a closed interval so it will have a maxima
00:18:18.130 --> 00:18:26.940
and a minima so I check the maxima and there
it is the minima
00:18:26.940 --> 00:18:31.250
Ok
00:18:31.250 --> 00:18:42.320
Ok so let us look at f of x naught y naught
minus delta minus f of x naught y naught and
00:18:42.320 --> 00:18:48.260
apply MBT apply MBT for y naught x naught
is fixed
00:18:48.260 --> 00:18:50.390
So MBT on y naught variable
00:18:50.390 --> 00:19:24.300
What is it this is equal to how much del f
del x at some point x naught y 1 into uh Ok
00:19:24.300 --> 00:19:38.030
I can apply MBT to any x Ok
00:19:38.030 --> 00:19:39.270
Ok not x naught
00:19:39.270 --> 00:19:55.380
So x minus x naught plus dell f del y at x
1 y 1 into y naught minus delta minus y naught
00:19:55.380 --> 00:20:06.110
that is minus delta Ok
for x in this interval x naught minus delta
00:20:06.110 --> 00:20:11.960
x naught plus delta
Now what is f x naught y naught
00:20:11.960 --> 00:20:13.250
This is zero
00:20:13.250 --> 00:20:19.610
So left hand side is just this fellow f of
x naught y naught plus delta Ok
00:20:19.610 --> 00:20:32.820
Now what I will do I will make this delta
I will choose a smaller interval
00:20:32.820 --> 00:20:40.050
Epsilon is less than which epsilon epsilon
is less than delta such that I still have
00:20:40.050 --> 00:20:43.290
del f del x is less than a
00:20:43.290 --> 00:20:55.560
And I will also make epsilon minimum of b
delta by a
00:20:55.560 --> 00:21:04.820
Then if you look at this this fellow is less
than a epsilon less than equal to a epsilon
00:21:04.820 --> 00:21:17.630
or strictly less than a epsilon minus b delta
and this will be strictly less than 0; it
00:21:17.630 --> 00:21:23.040
will be less than or equal to that less than
yeah this is strictly less than
00:21:23.040 --> 00:21:30.370
So what is this this is f of x y naught minus
delta is less than 0
00:21:30.370 --> 00:21:36.770
So I could have I can apply MBT for each x
in x naught minus delta to x naught plus delta
00:21:36.770 --> 00:21:42.950
but I make smaller interval such that I get
f x for x in this smaller interval x naught
00:21:42.950 --> 00:21:48.320
plus x minus epsilon x naught plus epsilon
epsilon is less than delta and less than this
00:21:48.320 --> 00:21:50.170
fellow such that this is less than 0
00:21:50.170 --> 00:22:00.650
Now a similar calculation so this is x 1 y
1 you know how I get x 1 y 1 is in the line
00:22:00.650 --> 00:22:18.020
joining x y naught minus delta and x naught
y naught
00:22:18.020 --> 00:22:22.260
So I have to play a little strict I have to
shrink the interval there
00:22:22.260 --> 00:22:33.920
Now what I do so similar calculation will
show it will follow that f of x y naught plus
00:22:33.920 --> 00:22:44.720
delta this will be bigger than 0 for x in
x naught minus epsilon x naught plus epsilon
00:22:44.720 --> 00:22:50.380
So plus delta bigger than 0 x naught minus
delta bigger than 0 so therefore
00:22:50.380 --> 00:23:11.220
there exists unique why unique because for
fixed x x naught minus epsilon x naught plus
00:23:11.220 --> 00:23:24.400
epsilon if f x of x y is strictly increasing
00:23:24.400 --> 00:23:32.710
this is implied by del f del x del f del y
bigger than 0
00:23:32.710 --> 00:23:43.080
There exists a unique y in the open interval
y naught plus minus delta y naught plus delta
00:23:43.080 --> 00:23:56.420
such that f of x y equal to 0
00:23:56.420 --> 00:24:05.700
So I define g of x equal to this y x in this
interval
00:24:05.700 --> 00:24:08.360
So I got this interval
00:24:08.360 --> 00:24:14.580
So I got definition of g
Only thing remains to prove is differentiability
00:24:14.580 --> 00:24:17.820
of g
But for differentiability we will show continuity
00:24:17.820 --> 00:24:22.310
and then differentiability I mean both will
come Ok
00:24:22.310 --> 00:24:25.970
So I got the first part right
00:24:25.970 --> 00:24:33.090
Second part differentiability
of g
00:24:33.090 --> 00:24:47.580
Well let’s check x 1 and x 2 in this interval
and let us put y 1 equal to g of x 1 and y
00:24:47.580 --> 00:24:53.000
2 equal to g of x 2 Ok
00:24:53.000 --> 00:25:06.580
Now look at this fellow
f of x 1 y 1 minus f of x 2 y 2; what is that
00:25:06.580 --> 00:25:11.540
Both of them are on the level surface so this
is 0 this is 0
00:25:11.540 --> 00:25:14.610
So this is 0
00:25:14.610 --> 00:25:26.480
Equal to I apply MBT again
del f del x at some x 3 y 3 at x 1 minus x
00:25:26.480 --> 00:25:30.950
2 Ok
00:25:30.950 --> 00:25:43.600
plus del f del y at some x 3 y 3 at y 1 minus
y 2
00:25:43.600 --> 00:25:53.110
Again same thing I have to write
x 3 y 3 in the line joining
00:25:53.110 --> 00:26:02.020
x 1 y 1 and x 2 y 2
What does this say this side is 0
00:26:02.020 --> 00:26:11.240
This fellow is less than a so this will say
g of x 1 minus g of x 2 which is equal to
00:26:11.240 --> 00:26:14.930
y 1 minus y 2
Bring everything to this side
00:26:14.930 --> 00:26:35.650
is equal to minus del f del x x 3 y 3 divided
by del f del y x 3 y 3 x 1 x 2
00:26:35.650 --> 00:26:39.240
Take mod take mod take mod here
Ok
00:26:39.240 --> 00:26:58.020
Keep it
This fellow is less than a
00:26:58.020 --> 00:26:59.400
This is greater than b
00:26:59.400 --> 00:27:07.440
So 1 upon this is less than b
00:27:07.440 --> 00:27:16.390
This of course implies g is continuous right
00:27:16.390 --> 00:27:31.710
And just for differentiability it also implies
g of x 1 minus g of x 2 divided by x 1 minus
00:27:31.710 --> 00:27:46.150
x 2 equal to minus del f del x at x 3 y 3
del f del y x 3 y 3
00:27:46.150 --> 00:27:50.330
Ok
So suppose x 1 goes to x 2 now
00:27:50.330 --> 00:27:54.800
I take limit x 1 goes to x 2 or x 2 goes to
x 1 let’s say
00:27:54.800 --> 00:28:05.260
What will happen
Limit x 2 goes to x 1 left-hand side
00:28:05.260 --> 00:28:12.190
which is actually derivative of g of x 1 this
is equal to
00:28:12.190 --> 00:28:18.500
x 1 goes to x 2 x 3 y 3 is in the line joining
x 1 x 2 so x 3 y 3 will also collapse to x
00:28:18.500 --> 00:28:19.500
1 x 2
00:28:19.500 --> 00:28:31.610
So this is minus del f del x x 1 y 1 by del
f del y x 1 y 1
00:28:31.610 --> 00:28:32.940
This is non-zero
00:28:32.940 --> 00:28:38.480
so this exists and g is differentiable
Not only differentiable we actually get what
00:28:38.480 --> 00:28:53.720
it is it is minus del f del x x 1 g x 1 divided
by del f del x x 1 That’s it That is the
00:28:53.720 --> 00:28:54.990
proof of implicit function theorem