WEBVTT
Kind: captions
Language: en
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So todays lecture well continue with the discussion
on maxima minima and well formalize a statement
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we made last time So
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so for a what we did last time if you recall
for a possible maximum or minimum at x0 where
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x0 is a stationary point a critical point
we have decided to call it critical point
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some call it stationary point Critical point
that is grad f at x0 is 0 We wanted to do
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the second derivative test and what we did
We said something like sign of f of x0 plus
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H for H small minus fx0 H small norm of H
small depends on sign of QH
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That was informally what we did and what was
QH QH we defined half of the second term in
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the Taylors series sation i equal to 1 to
n j equal to 1 to n del square f del xi del
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xj hj hi where h is the factor h1 h2ů hn
prime Now as far the sign is concerned this
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fellow is doing nothing Right This is simply
a constant so this is we have also written
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as H prime So everything evaluated at x0 I
keep on forgetting that
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So this is H prime Hessian f at x0 H This
is Hessian I mean one should not confuse with
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this H Hf at x is del f del square f del xi
del xj at x0 1 to n 1 to n
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Okay So we makes two assumption here okay
And well see what assumption is necessary
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that in our so were actually discuss about
it last time We have also assumed it Assume
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that second order partial derivatives are
all continuous Continuous at x0 And also assume
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that determinant of the Hessian at fx0 is
non zero I will come back to this assumption
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later why we need that May be this lecture
in the next lecture or next lecture So if
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determinant of H of fx0 is zero we call x0
a degenerate critical point And in this case
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the second derivative case test will fail
It will not give you anything
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As a proof well suggest youll see if you see
the proof you will see that if this determinant
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is zero the second derivative test doesnt
workIts simply like that if you do the function
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of one variable if f double prime at that
critical point 0521 function from interval
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to R f prime 0525 x0 is 0 and if double prime
x0 is also 0 then second derivative test for
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one variable also fails Similar situations
also occur here and in this case degenerate
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case we have to do for that test either by
observation or by going to higher derivatives
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Anyways so since the we have only conce concerned
about sign we forget about this half part
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and take our QH since you are conce it half
will not change any sign so we will take QH
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0605 to be this thing If you want to keep
half no problem but this time I have to write
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a half so let us forget about the half part
Okay
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The first observation you made here
that if you have any lambda in real then Q
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of lambda H for any H So this is lambda will
be multiplied with each of the component here
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So lambda each of the component with 0643
it will come out to be this So if I multiply
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with any lambda sign of QH doesnt change it
remains same Sign of Q lambda H is sign of
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QH for any lambda in RThis is needed for scaling
as you will see So let us suppose that first
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case one Q of H is greater than zero Suppose
for all H in Rn okay Then write this way f
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of x0 plus H minus f of x0 is equal to okay
I have 0746 taken out half part so may be
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I write half QH why Im writing Taylors series
expansion remember that it had to have grad
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f at x0 dot H but grad fx0 this is 0 so it
will actually start at QH
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correct
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Plus I will have error term so I will do it
only for
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second order term Second order Taylors series
expansion Okay Now look at this fellow Infimum
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of QH infimum sorry infimum of QH H in Rn
and norm H equal to 1 Lets call it m My claim
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is this m is bigger than 0 Why Well QH is
always bigger than 0 for all H in Rn Norm
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H equal to 1 this is a circle right Now in
circle in the very first second lecture or
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first lect second lecture of this course we
have shown that if I have a continuous function
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then it attains its maxima and minima over
a compact set The circle S1 is a closed and
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bounded ci circle
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This is a set of all H such that norm H equal
to 1 This is a compact set so on that compact
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set infimum will be attained Because Q is
a continuous function of H from definition
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And infimum is greater than 0 because QH is
greater than 0 for all H and on the compact
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if 1013 continuous function is greater than
0 on a compact set its infimum is also greater
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than 0 Thats very easy to prove Okay So what
does it say That if I take any H in Rn then
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if I look at QH by norm H H by norm H is on
the circle because H by norm H has norm 1
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So this will be greater than equal to m because
m in the infimum Which implies from this equality
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that for any H QH is greater than equal to
m into norm of H square for all H in Rn Very
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good So what about fx0 plus H minus of fx0
from there this is greater than equal to m
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into norm of H square plus norm of H square
EH Now EH is the error term So can choose
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a delta greater than 0 such that if norm of
H is less than delta then E of H m is a fixed
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number which is infimum this is less than
lets say I should have put a half here na
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if I have 1212 Doesnt matter So just say anything
1 4 by m So that should it show that f of
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x0 plus H minus of fx0 is bigger than equal
to so it is a half here 1 4 m H square which
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is always bigger than 0 as long as H is non
zero
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And H equal to 0 this is fx0 That shows for
norm of H less than delta f of x0 plus H is
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always bigger than f of x0 Correct So thats
for for all y in a neighbourhood of x0 this
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sign is bigger than so x0 is a local minimum
So if QH is greater than 0 for all H in Rn
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then x0 is a local minimum Now maximum part
is easy Next if I have QH is less than 0 for
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all H in Rn Hmm Then apply one this part for
minus f Okay How does it help Because you
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will see that H minus f at x0 is equal to
minus Hf at x0 So and so a local minima for
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minus f is a local maxima for f So if this
is so I will get 1426 same consideration x0
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is a local maxima Apply everything this analysis
for minus f So that solves the question for
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minu f as x0 is a local minima or maxima when
QH is either greater 0 or less than 0 So third
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possibility there exist H1 in Rn such that
Q of H1 is positive and H2 in Rn such that
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Q of H2 is negative that is H attains both
positive and negative sign In that case what
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happens In that case my claim is that x0 will
be a saddle
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How Well So again the same situation Grad
of x0 is 0 so for a fixed H and for any lambda
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in R you look at this fellow f of x0 plus
lambda H minus f of x0 See now two there is
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U okay here is my U here is some x0 H1 is
somewhere here and H2 may be somewhere here
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I dont care Actually without loss of generality
by translating I can say that x0 is 0 So this
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is translation so this is 0 Without loss of
generality you just look at U minus x0 it
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will have same kind of property Okay what
is that Follow geometry This is equal to Q
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half okay lambda H plus lambda square H square
Elambda H Which is equal to half of lambda
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square QH plus lambda square So lets take
the lambda square part out Now gain the same
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idea If I choose lambda lets say mod lambda
less than some delta here is a lambda square
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there so Elambda H is the error term so I
can have Elambda H this is less than one fourth
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of QH right So in particular if I apply to
H1 so I will have H1 here H1 here everything
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H1 here so sign of this fellow in this case
fx0 plus lambda H1 minus f of x0 Q of H1 is
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positive so this will be positive
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Similarly choosing lambda some less than some
delta prime I can make Elambda H2 less than
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one fourth of QH2 and in that case I will
have f of x0 plus lambda H2 minus f of x0
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less than 0 So what is happening Now lambda
is any arbitrary number less than delta H
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is somewhere Assume x0 is 0 Now multiplying
by lambda I can always bring it to a neighbourhood
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around 0 H Hone whatever we H1 multiplied
by suitable lambda I can always bri bring
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Elambda H1 sorry lambda H1 near a neighbourhood
around 0 So that means I can make so if I
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take delta0 to be minimum of delta and delta1
I can bring x0 plus lambda Hi in Bx0 delta0
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So I have a point x0 plus lambda H1 minus
fx0 is greater than 0 so if this is greater
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than 0 1957 x0 plus lambda H1 is greater than
fx0 and fx0 plus lambda H2 is less than x0
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And that is precisely the definition for x0
is a saddle point So this is the so called
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second derivative test for function of n variable
Okay So go through the proof again You have
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to atleast go you have to you have have to
go through it atleast twice then you understand
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what is going on here Very easy but still
go through it once again And Ill do an example
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for you to end todays class Let me write down
the example I have already written down for
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it f from say R3 to R f of xyz equal to xyz
x plus y plus z minus y So I have to look
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for critical point first There may be many
and in this case there are uncountably many
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actually As you see this will be equal to
2xyz minus 2xyz minus yz right Here Del f
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del y this will be 2xyz minus zx and del f
del z will be 2xyz minus xy Okay So if I put
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grad f at x equal to 0 then if I assume x0
equal to 0 y0 equal to 0 z0 equal to 0 then
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the solution is only solution is you can find
x equal to half y equal to half z equal to
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half x equal to 0 y not equal to 0 z not equal
to 0 x equal to 0 y not equal to 0 z not equal
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to 0 but this is equal to 0 so this is not
possible Similarly x not equal to 0 y equal
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to 0 z not equal to 0 x not equal to 0 y not
equal to 0 and z equal to 0
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they are all not possible
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Okay So one solution but if you see any of
two from x y z is 0 are 0 and other anything
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other arbitrary possible That is are all critical
points for any z For all x for all y all z
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If you calculate the Hessian f at x that comes
out to be 2yz 2yz minus z 2yz minus y 2yz
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minus z 2 just a minute Hfx fx x del f del
y 2xz sorry 2xz and 2yz minus y 2xy minus
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x 2xy minus x 2xy I think I made a mistake
here del fx del fx del x x 2yz minus z 2yz
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del f del y del x minus z del f del z x 2yz
minus y Thike Yeah So Hf at half half half
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is equal to half 0 0 0 half 0 0 0 half So
QH is always equal to how much You can check
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QH is always QH half of H prime Hf at half
half half Horlicks this is always greater
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than 0 Okay
So this is local minima whereas Hf at lets
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say x00 if you check determinant of this its
same as determinant of Hf at 0y0 same as determinant
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of Hf at 00z equal to 0 So these critical
points are all degenerate and we have not
00:26:58.140 --> 00:27:12.990
so far deduced any test for degenerate critical
points So well talk about this test in some
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other language something from linear algebra
is happening in the nex next lecture