WEBVTT
Kind: captions
Language: en
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Okay So let us continue from where we umm
left the proof yesterday So we wrote this
00:00:25.579 --> 00:00:33.410
theorem for the maxima minima saddle point
test in terms of the Hessian matrix Instead
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of calculating the i-n values we just look
at the principal minors of the umm Hessian
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matrix and then decide well be able to decide
the some some critical point is a maxima minima
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or a saddle point So what was the theorem
So I have the same setup f is from an open
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set to R P is a saddle point in U Instead
of writing Hfp repeatedly Ill just denote
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it for this theorem A equal to Hfp because
once P is fixed f is fixed I can just call
00:01:05.040 --> 00:01:12.280
it A and Ak be the Kth principal minor of
A So youve defined Kth principal minor that
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is a first K row and first K column you take
and leave rest of the part
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And of course you can see that An is equal
to A nth principal minor is the entire matrix
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A So what is the test Well if for some two
case of even number the principal minor A2k
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the determinant is 0 then P is a saddle point
Now if we have determinant of A not equal
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to 0 that is to say P is a non degenerate
critical point that is An is invertible matrix
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the Hessian is an invertible matrix at P then
P is a local minimum if and only if every
00:01:59.080 --> 00:02:10.070
principal minor of Ak has positive determinant
Similarly P is a local maximum if minus 1
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power k determinant Ak for every K is greater
than 0 And of course if determinant of An
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is equal to 0 then P is a non P is a degenerate
critical point and in that case we cannot
00:02:25.090 --> 00:02:28.560
conclude anything about second derivative
test Anything from second derivative test
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Second derivative test fails we have to do
something else there We have to go for higher
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derivatives or may be we have to look do something
by observation Umm I forgot to note something
00:02:41.900 --> 00:02:45.700
yesterday that actually were dealing with
a non constant functions because constant
00:02:45.700 --> 00:02:51.270
function grad f is 0 for every every P so
there is no test for that
00:02:51.270 --> 00:02:57.190
And so actually I can add here that P is a
strict local minima P is a strict local maxima
00:02:57.190 --> 00:03:04.459
that is theres no point nearby in the neighbourhood
where it is equal to fp Okay so what about
00:03:04.459 --> 00:03:11.570
the proof In the proof lets in a part So let
us have determinant of A2k less than 0 for
00:03:11.570 --> 00:03:19.989
some 2k between 2 and n Ive denoted P with
these co ordinates P1 to P2k 2k plus 1 Pn
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and Q is P1 to P2k so Ive chopped off this
2k plus 1 to Pn part Then well define this
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open set in R2k which is x1 x2 x2k 0330 such
that x1 x2 x2k 0334 with P2k plus 1 to Pn
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See that P is fixed so these co ordinates
are fixed that belongs to U And we considered
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this function g from V to R which is gx1 x2
x2k is equal to f of x1 x2 x2k this is a free
00:03:49.910 --> 00:03:58.879
variable and this 2k plus 1 to n are fixed
Then it is very easy to note that since these
00:03:58.879 --> 00:04:08.980
are fixed that Hessian matrix of g at Q is
del square f del xi del xj k cross k matrix
00:04:08.980 --> 00:04:12.459
which is precisely the 2k principal minor
00:04:12.459 --> 00:04:22.770
Now under assu assumption that okay we have
assumed that all I think yesterday we wrote
00:04:22.770 --> 00:04:28.630
down that all de all derivatives all second
order partial derivatives are continuous these
00:04:28.630 --> 00:04:36.070
are symmetric matrix again so A2k is a symme
real symmetric matrix It has 2k many i-n values
00:04:36.070 --> 00:04:43.190
may be with repetition beta 1 beta 2 beta
2k some of them may be equal But I can write
00:04:43.190 --> 00:04:47.860
counting multiplicity I can write 2k many
i-n values and all of you know the determinant
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of 2k is a product of the i-n values
00:04:50.199 --> 00:04:57.139
And this is given to be less than 0 Now here
comes the catch that I have even number of
00:04:57.139 --> 00:05:06.340
real numbers here i-n values whose product
is less than 0 So that says that there must
00:05:06.340 --> 00:05:14.270
be one which is positive and there must be
one because if all of them are negative since
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there are even number of terms the product
will be positive And if all are positive product
00:05:19.040 --> 00:05:28.620
is anyway positive So I can conclude from
here that there exist i and j such that beta
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i is negative and beta j is positive So one
i-n value negative one i-n value positive
00:05:39.000 --> 00:05:44.430
okay Given i-n value I can of course talk
of I have an corresponding i-n
00:05:44.430 --> 00:05:45.490
vector
00:05:45.490 --> 00:06:13.030
So let u be a i-n vector for beta i and v
be i-n vector for beta j That is HgQu equal
00:06:13.030 --> 00:06:34.210
to beta iu and H here HgQv is beta jv What
is u and v u is a vector in R2k v is a vector
00:06:34.210 --> 00:06:52.850
in R2k right Okay Now just consider this vector
Y1 u1 u2 u2k and 0 upto nth co ordinate 2k
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plus 1 to n And Y2 equal to v1 v2 v2k and
rest of the part 0 So these two vectors are
00:07:10.020 --> 00:07:22.060
in Rn So 2k plus 1 to n co ordinates who are
missing there so I put them 0 Okay Now look
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at what happens Y1 prime AY1 Y1 has this co
ordinate 0 So this will be how much You just
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calculate this will be uH uA2k which is HQ
u which is equal to so this is A I should
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have written A here right A2k here Which is
equal to beta i u prime u Now it doesnt matter
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if you take u norm 1 or not u prime u is always
positive or all the time you can take norm
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of i-n vector is always norm 1 Usually we
take i-n vectors norm 1 In that case u one
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u is 1 which is beta i this is less than 0
And what about Y2A Y2 prime Y2 for the same
00:08:51.959 --> 00:09:03.560
reason this is v prime A2kv which is beta
j v prime v which is beta j bigger than 0
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So what we conclude
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Okay now something is less than 0 something
is bigger than 0 I can make u was norm 1 so
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Y Y1 and Y2 both are same norm as u because
I have just added 0 That implies minimum of
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Y norm 1 YA Y prime AY is less than 0 less
than maximum v equal to 1 sorry Y equal to
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1 Y prime AY Because there is something positive
one vector for which is positive so maximum
00:09:43.429 --> 00:09:48.910
is bigger than 0 One vector which is negative
so minimum is less than 0 And you remember
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this is precisely the condition for P is a
saddle point Okay Thats the proof for the
00:10:08.009 --> 00:10:15.069
a part b and b part one and two is very easy
Thats direct from linear algebra You can check
00:10:15.069 --> 00:10:25.660
it in your linear algebra notes that this
is simple linear algebra Determinant of Ak
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greater than 0 for all k implies all i-n values
of A are positive That is to say if and only
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if if and only if minimum that the minimum
of the i-n values is positive This is if and
00:10:56.470 --> 00:11:06.449
only if P is a local minima Okay
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And once we have one part from local maximum
consider minus f and we have already observed
00:11:17.290 --> 00:11:37.769
Hf H of minus f at P equal to minus Hf of
P So A2k for minus A2k corresponding to the
00:11:37.769 --> 00:11:43.259
principal minor the principal minors of sorry
principal minors of order K corresponding
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to f is negative of principal minor corresponding
to minus f and you know determinant of this
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is simply minus 1 power K determinant of Ak
So apply one You will get P is local maxima
00:12:13.800 --> 00:12:25.079
because a local maximum for f is a local minimum
for minus f Okay Thats the proof Very good
00:12:25.079 --> 00:12:39.040
So this gives you a ve uh one way to check
the nature of this critical point saddle minimum
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or maximum For the rest of the todays lecture
and thats the end of this third module well
00:12:47.929 --> 00:12:58.879
specialize this theorem with examples of course
to n equal to 2 I will directly apply this
00:12:58.879 --> 00:13:13.689
theorem and see what happens So specialized
to R2 that is now I have a function f from
00:13:13.689 --> 00:13:25.260
an open set U in R2 1315 are only two variable
to R In R2 well see most of the books use
00:13:25.260 --> 00:13:42.170
some special notation So if I have some x0y0
in U this grad f at x0y0 which is actually
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del f del x at x0y0 del f del y at x0y0 people
use some special notation for this
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This one is written as this way So this is
notation only fx x0y0 is a partial derivative
00:14:04.470 --> 00:14:10.480
del f del x instead of writing this one writes
fx for R2 So this is so Im following some
00:14:10.480 --> 00:14:20.019
book notations standard book notation here
Okay Similarly del square f del x square this
00:14:20.019 --> 00:14:29.959
is denoted by notation again fxx and del square
f del y square equal to fyy and del square
00:14:29.959 --> 00:14:39.730
f del xy del x del y equal to f first you
doing y so yx But we are actually assuming
00:14:39.730 --> 00:14:48.959
that they are continuous in our so this is
fxy first x then y So these are the notations
00:14:48.959 --> 00:14:58.519
special notations for usually used in books
theyre same thing right And this Hessian matrix
00:14:58.519 --> 00:15:10.639
at x0y0 is usually again denoted by this delta
x0y0 which will be equal to now in this notation
00:15:10.639 --> 00:15:26.589
assuming second order partial derivatives
are continuous so that the matrix is symmetric
00:15:26.589 --> 00:15:50.339
Okay So what other test says Apply the theorem
Okay What will it say if I apply the theorem
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That you see A2k even but there is only one
princ there are only two principal minor here
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delta 1 which is fxx and delta 2 which is
delta itself
00:16:08.139 --> 00:16:17.449
And I should have some delta 2k less than
0 for saddle point so test says A if delta
00:16:17.449 --> 00:16:34.749
2k but only one possibility that is a delta
itself is less than 0 then x0y0 is a saddle
00:16:34.749 --> 00:16:51.360
Okay Next delta not equal to 0 All the principal
minors are positive that is delta greater
00:16:51.360 --> 00:17:05.100
than 0 and fxx greater than 0 that is if and
only if x0y0 So this is the first part So
00:17:05.100 --> 00:17:20.709
this was the assumption b so one if and only
if x0y0 is local minimum Similarly delta what
00:17:20.709 --> 00:17:29.639
will happen All the principal minors are and
what was the condition there Minus one power
00:17:29.639 --> 00:17:42.190
K remember so maximum one power K delta Ak
So now here I have delta 1 K equal to 1 so
00:17:42.190 --> 00:17:51.799
I should have fxx less than 0 and 2 so minus
1 power
00:17:51.799 --> 00:17:58.769
2 so this was the condition is 1 so delta
must be gre and that is second principal minus
00:17:58.769 --> 00:18:07.490
delta itself it is greater than 0 if and only
if x0y0 local maximum So you just go back
00:18:07.490 --> 00:18:16.970
to the theorem apply it you get this result
And of course third condition is there c delta
00:18:16.970 --> 00:18:28.779
equal to 0 I should have written deter determinant
everywhere Determinant determinant determinant
00:18:28.779 --> 00:18:46.720
Delta equal to 0 then Okay So this is so easy
to remember So max delta determinant greater
00:18:46.720 --> 00:18:58.509
than less than 0 saddle Both the cases determinant
is determinant is positive And so it is determined
00:18:58.509 --> 00:19:07.029
by the sign of fxx fxx greater than 0 local
minima fxx less than 0 local maxima Now I
00:19:07.029 --> 00:19:16.191
will do two quick example Instead of doing
just writing one function I do it with some
00:19:16.191 --> 00:19:22.820
problem which where we can apply these things
So this is a problem I found it in some one
00:19:22.820 --> 00:19:44.350
of the books Find minimum distance
from the point 0b b greater than 0 that is
00:19:44.350 --> 00:20:01.240
on the y axis to the parabola x square minus
4y equal to 0 So what is the problem Here
00:20:01.240 --> 00:20:13.299
is the point 0b this is b okay And here is
a parabola
00:20:13.299 --> 00:20:20.360
y umm y equal to x square by 4 okay
00:20:20.360 --> 00:20:24.120
I want to find so so you can draw different
lines from this to the parabola different
00:20:24.120 --> 00:20:30.570
points You have to find the distance that
is minimum distance that is the length which
00:20:30.570 --> 00:20:38.580
poi which le length has the minimum distance
So for that I have to find out a point on
00:20:38.580 --> 00:20:46.389
the parabola for which this distance between
these two points are minimum Thats what we
00:20:46.389 --> 00:21:07.620
have to do So whats how do I start So let
x0y0 be such a point Be the point on the parabola
00:21:07.620 --> 00:21:24.470
where minimum distance is attained Okay Now
what is the distance I have to minimize this
00:21:24.470 --> 00:21:36.529
distance Distance from 0b to x0y0 which is
equal to x0 x square any point on the parabola
00:21:36.529 --> 00:21:50.259
x square plus y minus b square where y square
equal to 1 over 4x square because it has to
00:21:50.259 --> 00:21:57.429
be a point on the parabola Now if you see
theres a distance is a positive function so
00:21:57.429 --> 00:22:03.490
instead of minimizing this distance I could
as well minimize distance square that will
00:22:03.490 --> 00:22:09.190
minimize distance as well so I can get rid
of this root over
00:22:09.190 --> 00:22:17.399
So that I have some so derivative calculation
is easy So this information will give us this
00:22:17.399 --> 00:22:30.340
is x square plus x square by 4 minus b square
which is x square plus x power 4 by 16 minus
00:22:30.340 --> 00:22:38.889
xx square b by 2 plus b square So this is
a function of x only
00:22:38.889 --> 00:22:45.269
So you now see weve reduced the problem of
by minimizing a function of two variables
00:22:45.269 --> 00:22:53.559
x y to a function minimizing a function of
x only I want you to complete this yourself
00:22:53.559 --> 00:23:00.880
because now we have to check you have to find
a point x0 such that f prime x0 equal to 0
00:23:00.880 --> 00:23:10.899
and f double prime x0 is minimum so greater
than 0 Once you do it you will find theres
00:23:10.899 --> 00:23:19.409
some fun in this problem In a sense that this
position of x0 will depend on the position
00:23:19.409 --> 00:23:39.029
of b So complete it And I request all of you
to complete it That will help you in future
00:23:39.029 --> 00:23:42.460
in whatever way you understand it
00:23:42.460 --> 00:23:50.279
Okay I have reduced the problem of finding
minimizing by two variable by two finding
00:23:50.279 --> 00:23:57.299
the minimum of one variable function Second
example is also interesting in the sense that
00:23:57.299 --> 00:24:11.230
lets see what happens Lets say fxy defined
on entire R2 Two variable function Lets check
00:24:11.230 --> 00:24:23.970
for minimum maximum
on the entire R2 U equal to R2 Straightforward
00:24:23.970 --> 00:24:31.600
you can say theres no maximum Why Because
if you keep on increasing x and y this function
00:24:31.600 --> 00:24:40.980
goes to infinity You take x and y as big as
as you fxy you can make very close to infinity
00:24:40.980 --> 00:24:48.409
I mean it 2440 as big as you want So there
is no maximum So I have to check for saddle
00:24:48.409 --> 00:25:01.820
or minimum Okay lets see a critical point
What is grad fxy This is how much 4x cube
00:25:01.820 --> 00:25:19.809
plus 2xy and
grad f grad y is x square plus 2y So for critical
00:25:19.809 --> 00:25:33.710
point 4x cube plus 2xy equal to 0 and x square
plus 2y equal to 0 you see that only critical
00:25:33.710 --> 00:25:50.460
point is 00 That satisfy these two equation
Okay This is my fx this is my fy Lets see
00:25:50.460 --> 00:26:02.950
what is fxx I have to go go for the test This
is equal to 12x square plus 2y so fxx at 00
00:26:02.950 --> 00:26:09.299
is 0 fxy is how much
00:26:09.299 --> 00:26:20.360
First okay fyx Ill have to calculate first
x then y So fx is this fellow and then y that
00:26:20.360 --> 00:26:37.920
is 2x So fyx 00 is 0 and I dont need fyy still
I do it is 2 So delta equal to delta at 00
00:26:37.920 --> 00:26:49.770
is 0 0 0 2 Determinant of delta is 0 So 00
is a degenerate critical point So here we
00:26:49.770 --> 00:26:56.700
cannot apply the test but we can make a little
observation at as I was saying that if the
00:26:56.700 --> 00:27:03.590
second derivative test fails one tries to
make observation and see if we can still include
00:27:03.590 --> 00:27:13.000
some conclude something You look at this function
My claim is 00 is local minima in fact it
00:27:13.000 --> 00:27:28.070
is a global minimum because f of 00 is 0 and
you see x square y modulus of this this is
00:27:28.070 --> 00:27:35.039
of course the positive square root of x4 y
square correct So this is the geometric mean
00:27:35.039 --> 00:27:45.820
of x4 and y square So this is less than equal
to less than equal to x square plus y square
00:27:45.820 --> 00:27:53.190
by x power 4 by y square by 2 arithmetic mean
which is strictly less than both are positive
00:27:53.190 --> 00:28:05.090
x4 plus y square That says x4 plus x square
y plus y square is always greater than 0 for
00:28:05.090 --> 00:28:11.429
whatever your x and y are But f00 is 0 its
a local minima So in case of degenerate it
00:28:11.429 --> 00:28:19.229
is not still that hopeless You can do something
Thank you That ends up third module