WEBVTT
Kind: captions
Language: en
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Okay in the last lecture we have derived the
second derivative test in terms of this umm
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QH that is umm second term in the umm Taylors
series expansion So what we have done is this
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thing thats for f on an open set R and its
not in U I have a critical point
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that is grad f of x0 0 Then weve taken Q of
H with half or without half H prime Hessian
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at x0 H Okay And what are the test that QH
greater than 0 for all H implies x0 local
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minimum QH less than 0 for all H x0 local
maximum and QH1 greater than 0 for some H1
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and QH2 less than 0 for some H2 implies x0
is a saddle
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Thats what the test was second derivative
test Now we also had noticed that in the proof
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also that we have used mainly the scaling
property So because of the scaling property
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any H can be brought down to norm 1 That has
no nothing wrong in assuming instead of for
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all H for all H norm 1 for all H for all H
norm 1 and H1 norm 1 H2 norm 1 right So what
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we can do here is that that we can write here
again QH is greater than 0 for all H in Rn
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This is to same as saying minimum of lets
say let me write in this form now instead
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of H
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y in Rn norm y equal to 1 yHfx0 this is greater
than 0 This is if and only if Similarly QH
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is less than 0 for all H in Rn this is if
and only if maximum y in Rn norm y equal to
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1 y prime Hfx0 y is less than 0 And QH1 less
than 0 and QH2 the third statement there greater
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than 0 if and only if minimum of y in Rn norm
y equal to 1 of y prime Hy is less than Hfx0y
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is less than 0 less than maximum y in Rn norm
y equal to 1 y prime Hfx0y Right Because this
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is because of scaling property you can deduce
that So look at these two board cleaned At
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its end point you will see that what youve
written is is just because of the Qlambda
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H in lambda square QH But why I have written
like this
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Well you see every time I have to check I
have to compute this QH and see if I want
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to use this board for checking minima maxima
or saddle point I have to kind of check for
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every edge And if I write it in this way well
I have reduced the problem to norm of if you
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look at this form at norm of y only 1 you
thats the reduction but its not a great reduction
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But if you now I apply some results from linear
algebra then this will give us a great advantage
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in actually checking minima maxima or saddle
point And what is that Under the assumption
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we made that del square f del xi xj they are
continuous at x0 this Hfx0 under the assumption
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we have proved that mixed partial derivatives
are equal So Hfx0 is a real symmetric matrix
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And what do you know by real symmetric matrix
That they have real i-n values and there i-n
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vectors can be chosen to be real as well
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And this quantity has a special meaning here
this quantity has special meaning in terms
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of i-n values So if you use your linear algebra
the first course in linear algebra
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minimum y in Rn norm y equal to 1 y prime
Hfx0y equal to lambda min the minimum i-n
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value of Hfx0 and the minimum is attained
at an i-n vector y corresponding to lambda
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min that is Hfx0y equal to lambda min y Similarly
maximum y prime Hfx0y norm y equal to 1 this
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is maximum of the lambda max that is maximum
i-n value of H of Hfx0 and the maximum is
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attained at an i-n vector y corresponding
to
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lambda max that is a sorry Hfx0y equal to
lambda max y This is a simple result from
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linear algebra You prove it This can be verified
with calculus as well And well be able to
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do it after the next week when we do it Lagrange
Multiplier Technique may be Ill do it or may
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be Ill umm give it as an assignment But for
time being let us recall linear algebra and
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this result
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So what he says now here that if the minimum
i-n value is greater than 0 and all i-n values
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are greater than 0 then its a local minima
If maximum i-n value is less than 0 then all
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i-n values has to be less than 0 then its
a local maximum And if theres some i-n values
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less than 0 and some i-n values bigger than
0 then its a saddle So in terms of i-n values
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this is a test of secon secon second derivative
test reduces to this pa this thing that 0952
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looking for the i-n values But in general
what happens that i-n values if n is large
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n equal to 2 no problem n equal to 3 okay
Not take too much of problem But n equal to
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say 100 then calculating the i-n values may
be a difficult task in general So from this
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observation and this linear algebra result
well now write down a practical test for actually
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checking minima maxima and saddle points in
case of non degenerate critical point that
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is when Hessian has non zero determinant So
instead of checking QH for every H we just
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go through this check and that makes life
little easier
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Now if you can calculate i-n values nothing
better Okay so lets write it as a theorem
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So again the same setup f is from open set
to R x0 in U and grad fx0 equal to 0 that
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is x0 is a critical point Instead of avoiding
writing grad fx0 every time let me put a notation
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for it A equal to because I am looking at
one x0 so let me write it as A just say n
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cross in matrix A Okay So from that this same
board whatever we have I will write down the
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statement Umm okay Dont 1146 I just want to
make it P because I have to use this notation
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x in the proof Okay So dont worry In the x0
I will use P instead of y y point x0 I will
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use some point P Umm okay So let H is equal
to Hessian and K equal to 1 to n Ak be the
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Kth principal min Kth principal minor
of A What does it mean So A is this n cross
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n matrix You look at the one by one first
one by one matrix This is A1 The first row
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first uh entry of the first row first column
Then in the first row you get two elements
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and then second row two elements
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So I think you all know what is principal
minor So this is Ak is the A matrix where
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K plus 1 to n and K plus 1 to n entries of
rows and columns are deleted Okay So conclusion
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is like this If for some K determinant of
A2k is negative okay Not A Ak but so K has
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to be between 1 to n then P is a saddle b
if determinant of An An is A so in is not
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equal to 0 then so let determinant of A is
not equal to 0 the determinant the Hessian
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is non zero determinant If for all K determinant
of Ak is greater than 0 then P is a local
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minimum Okay
In fact I can write it in this form Determinant
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of Ak is greater than 0 for all K if and only
if P is a local minimum Okay
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Minus 1 power K determinant of Ak is greater
than 0 for all K umm greater than 0 or less
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than 0 Hmm greater than 0 for all K if and
only if P is a local maximum
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And finally if determinant of An is 0 P is
a we have already called it degenerate critical
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point and in this case the test fails Okay
Before I prove it let me give an example of
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how this test can be applied So lets take
umm there are many examples here lets take
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one Umm okay x square y y square z z square
z minus 2x fx example xyz x square y y square
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z z square x minus 2x So first search for
critical point Correct Grad fP equal to 0
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What is the solution Youll see the only solution
is if you solve this the only solution is
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P equal to 1 1 and 1 1 and minus 1 I have
written something wrong right Umm yeah Sorry
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Del f del y is x square plus 2yz and del f
del z is I have missed this part y square
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plus 2zx 2z 2x 2x No 2zx Thike Del f del z
is umm y square plus 2zx x square plus 2yz
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y square plus 2zx yeah That will give us this
thing
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So let us calculate the Hessian Hf at P which
is A in that which is equal to del square
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f del x square del square f del x del y del
square f del x del z so on You calculate how
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much is that First column del square f del
x x square that is gives you 2y del square
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f del x del y umm del x del y that gives you
2x and this gives you del f square del de
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del z that is 0 Now del square
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so I know what will be here this will be 2x
here because symmetric matrix and here it
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will be 0 again okaySo del square f del y
square that is 2z del square f del y square
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2z and del square f del z del square f del
z del y is umm just del del del square f del
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y del z so this quarter is del square f del
y del z which is equal to how much 2y so 2y
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here it will be and del square f del z square
is equal to simply 2 is that okay That at
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the point P so which is equal to the matrix
2 2 0 2 minus 1 2 0 2 2 Okay And you see determinant
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of A2 which is this part minus 2 minus 6 this
is minus 6 less than 0 so P is a saddle okay
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So I dont have to check for Q and all those
things Check the calculation carefully but
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okay Theorem2127 okay
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So for today I will start the proof because
time is running out and I will complete the
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proof in the next lecture Proof is very interesting
So first a part So let me try if I can finish
00:21:48.540 --> 00:22:04.420
the proof of a part today Okay What I have
assumption So determinant of A2k is less than
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0 for some 2k less than equal to n less than
equal to 2 Okay So let me have this point
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P critical point Let me write it as P1 P2
P2k P2k plus 1 Pn Okay Lets take this set
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V This is set of all x1 x2 x2k this set such
that x1 x2 x2k P2k plus 1 Pn belongs to U
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that is I take the first 2k section of U with
the last 2k plus 1 to n coordinate fixed Okay
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And I define so this is half set in R2k I
define g from this V this is an open set in
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R2k to R by g of x1 x2k equal to f of x1 x2
x2k Pk plus 1 Pn Okay Then what I want you
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to do so P is this let Q is this
you calculate Hg at Q You find out since Pk
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plus 1 Pns are fixed this is del square f
del xi del xj i equal to 1 to K j equal to
00:24:22.720 --> 00:24:32.571
1 to K which is A2k So now what do I do
00:24:32.571 --> 00:24:38.710
I know determinant of A2k is given to be negative
But what is determinant of A2k A2k is a real
00:24:38.710 --> 00:24:53.100
symmetric matrix So let beta 1 beta 2 beta
2k are i-n values of A2k And then all of you
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know this result determinant of a matrix symmetric
2456 have the i-n values this is product of
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the i-n values 2k is a even number Even number
of real number real terms their product is
00:25:12.150 --> 00:25:19.410
less than 0 What does it mean All of them
cannot be negative because even then itll
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be greater than 0 All of them cannot be positive
Even they will be positive So there exist
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beta i such that and beta j such that beta
i is positive and beta j is negative Let me
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complete it here and continue from this point
So from this point well continue in the next
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lecture