WEBVTT
Kind: captions
Language: en
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Okay I will come to the last lecture of this
module Here we will discuss about well in
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several variables calculus you don’t stop
at taking first derivative right You then
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take second derivative and third derivative
and finally arrive finally you talk about
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function which has all derivatives and then
you talk about Taylor’s theorem and Taylor’s
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formula We will do that next week for several
variables but for time being this lecture
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we concentrate on higher order derivatives
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So the same setup F is from U open connected
set maybe connectedness is not essential for
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this lecture RM and we know we can define
for some X in U we can define this derivative
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DFX which is a linear map from RN to RM let
me write it in this way DFX belongs to L RN
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to RM so this is the set of all linear maps
from RN to RM and all in the if I write in
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matrix form these are all if I fix bases from
RN and RM then this will give us all M cross
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matrixes real matrixes
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So if you talk about second derivative you
must consider this X going to DFX but now
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you will see this is the map from U to so
DF if I call this map DF this is the map from
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U to DFX belongs to L of RN to RM which is
itself a vector’s space If you now want
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to talk about the second derivative that is
this square F so X in U this square F at X
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that is a derivative of this map this will
be a linear map from U is in RN to L of RN
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to RM
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Well you see it is already matrix value this
fellow is already matrix Now you are defining
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maps from RN to matrixes so it will be again
you can say it is a matrix huge matrix where
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each entry is a matrix of order So total matrix
will be M cross N into N So this is very difficult
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to write down and people still do analysis
but let us not bother too much about this
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thing because I think in matrix form it doesn’t
help much
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If you want to do some analysis you have to
think of some other ways maybe we encounter
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or may not in this course but there is one
thing if we take real valued function this
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becomes very easier
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In the sense that suppose of F from U in RN
to R real value and you know actually to talk
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about differentiable function in F you can
write F as F1 F2 FM and you understand the
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properties or differentiability of each component
FI and then you can talk about derivative
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of F in terms of those So enough to from all
practical purpose if you want to talk about
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higher derivative you consider F from RN to
R and for RN to RM you start from here and
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then go up with the formula we have it
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But now this DFX for X in U DFX is a linear
map from RN to R so this is a row vector which
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we have actually written as actually it is
we have fixed notation Grad F at X which is
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you know del F dell X1 del F del X2 so on
del X del XN this is the reserved rotation
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for Grad So you see now X going to Grad F
at X this now a map from U which is in RN
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to RN again so D2 of FX maybe we will write
it as a special notation I don’t want to
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use this notation because this is reserved
for something else
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Okay let me write for this course I am not
going to talk about this This will be in a
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linear map from RN to RM and this I can write
it as M cross N matrix how well Grad FX is
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from U to RN Grad F a vector in RN now I will
consider this is a function so what I will
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do…
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I will write Grad F at X as it is a RN to
RN del F is map from RN to RN so I can write
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it as N component we have done it before Well
each FI is Grad F del XI Now I know what is
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the second derivative derivative of this map
this is we have already known so that will
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be del FI so this will be del F1 del X1 del
F2 del X1 so on del FN del X1 del F1 del X2
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del F2 del X2 del X2 so on del FN del X2 del
F1 del XN so one del FN del XN We know this
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from before because I have writtenActually
this is Jacobian of this map del F at X correct
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But now I know what is FI so this matrix will
be I can write it as I don’t have place
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to write here maybe I go back to that board
So let me keep the setup
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so Grad square F at X is now I write what
is definition of FI F1 is Grad F del X1 so
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del square F del X1 del X1 del X1 square del
F2 is del F del X2 so that is del X1 del X2
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so on del X del X1 del XN next del F del X1
del X2 del square del square F del X2 square
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so one del square F del X2 del XN del square
F del X1 del XN so one del square F del XN
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square
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Which is written as del FI sorry del F del
X JXI I equal to 1 to N J equal to 1 to N
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Some books use a special notation for this
special notation for this we will use it in
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our next calculation next time it is called
H F at X and written as Hession
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Okay so if you don’t remember which way
to follow del XJ or del Xi always go back
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to this way that you write del FX is equal
to F1 F2 FN FI equal to this thing and then
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apply the derivative formula so there is nothing
much to remember here In one minute you can
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just write it down if you know how to write
a derivative of a map from RN to RM by component
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wise
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Now this matrix has this property You see
1 2 del X2 del X1 and 21 del X1 del X2 so
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order is changed So what I can see from here
is that this matrix H is a symmetric matrix
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if del square F del XJ XI equal to del square
F del XI del XJ for all IJ So if the mixed
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partial derivative they are called mixed partial
derivative They are two different components
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involved if changing the order are equal then
the matrix becomes symmetric
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But this is not always the case For example
I give you very elementary example let us
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say F from entire R2 to R F of XY equal to
X into Y X square minus Y square divided by
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X square plus Y square when X and Y both is
not equal to 0 is it no origin and 0 at the
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origin What you can calculate here so do this
calculation yourself that del F del X this
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is equal to you just calculate it del F del
X equal to well let us calculate it doesn’t
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matter
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Let us calculate it at 0 Y this will be equal
to minus Y So del square F del y del X at
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0 0 so you verify this calculation is equal
to minus 1 whereas del F del Y at X 0 this
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will be X so del square F del X del Y this
will be equal to 1 clearly this two are not
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equal so del square F del Y del X and del
square F del X del Y at 0 0 they are not equal
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one is minus 1 and as in Y You do this calculation
very easy to do the derivative
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But this is not a very likely nice situation
for us to go ahead with calculation to go
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ead with calculus we want the mix partial
derivative be equal so that I get this matrix
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symmetric It becomes very handy when you next
time we talk about Taylor’s theorem and
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also when you talk about maximum and minimum
real valued function So come up with certain
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criteria of check maximum and minimum this
symmetric is essential otherwise it gets total
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mess
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So we actually want condition for equality
of mix partial derivative So we want condition
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such there del F del XI del F del XJ and del
square F del XJ del XI for given I J Let me
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state it as a theorem I state it for F from
U I state it more general RN into RM U is
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open connected state but open is enough here
Condition says so let’s fix a point I want
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to always check at a point suppose given I
and J given I from I in 1 to M and J in 1
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to N del F del XI del F del XJ both exists
on a ball around delta That is for all points
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in the ball around delta for some delta
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See when X not is in U and U is open there
will be always a ball but I want a ball such
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that this del F XI and del XJ both exists
on the entire ball not only at X not on the
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entire ball for some delta greater than 0
and both differentiable at X not that is del
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F and del XJ differentiable at X not Then
del square F del XJ XI equal to del square
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F del XI XJ mix partial derivative at equal
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So what is the condition they must be of course
this must two must exist so they must be differentiable
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at X not but the condition is there both exists
I missed most important and continuous Why
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I did not write because I am saying that there
is a differentiable but differentiable X not
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means it is already continuous at X not but
I want it continuous on the entire B X not
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delta but if anyways you can have this or
may not have this continuity of X not is enough
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But it has to exists on the entire ball that
is important so you add N continuous or not
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it does not make any different that is what
I mean Okay the proof is little long and the
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proof is not going to we are not going to
use the entire idea of the proof any way But
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still you need a proof once you write a statement
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So what I write here is proof but I will write
a sketchy proof I will give you the idea of
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the proof you complete the lines in between
Okay Do the sketching and you go like this
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First may your reduction that I am only concerned
about 2 XI given XI and XJ so let us only
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consider because nothing to be changed for
R1 and RM F from You in R2 because two variables
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are involved at R For RM what you do you apply
component wise two variables are involved
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do R2 is enough If we have RN you do component
wise
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Okay and without loss of your X not is 0 0
this is just to make the calculation easy
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because U is any set in R2 here is the origin
but you can always shift origin such a way
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that 0 0 belongs here There is a transformation
of origin and that does not change any property
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differentiability continuity anything
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So we choose we want to go into hypothesis
so some H greater than 0 such that H 0 cross
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0 H this set is in U so here is my U from
You around origin I choose H such that this
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square of length H and height H this is completely
named With that what you do for a fixed H
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define G of X equal to F of X H minus F of
XJ
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Then we apply Mean Value theorem to get this
is equal to H into G prime at some Z1 right
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Z1 is 0 to H but immediately you can write
G prime Z1 equal to del F del X1 at some Z1H
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minus del F del X1 F at Z1 0 and apply the
definition of what is del X1 That was my second
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step See verify all those steps I am not writing
the entire proof
00:22:39.970 --> 00:22:58.090
Third step is you again write del F del X
as applying MVT del square del X square at
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0 0 at Z not applying applying the condition
of derivative sorry applying this is differentiable
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del F del x at X 0 as del square F del X square
0 0 at Z plus H into del square F del Y del
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X 0 0 plus mod Z H into U and H U and H will
go to 0 as H goes to 0 because as H goes to
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0 Z also goes to 0 Z is between 0 and H
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Okay from that what we do we calculate GH
minus G 0 it will come out to be del square
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F del okay let me skip this step because this
need some technicality what I actually want
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to show is that this divided by H square this
goes to del square F by del Y del X at 0 0
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And what you do in the next step you repeat
the entire thing by defining G1 Y equal to
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F of HY minus F of ZY so here I take XH XO
with X and I did G1 and G2
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And you again show by the same step G1H minus
G10 divided by H square it will be just ulta
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just other way round sorry this goes to 0
this thing as H goes to 0 this as H goes to
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0 and final observation is GH minus G0 equal
to G1H minus G10 so this two Limits will be
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equal because numerator are equal denominators
are H square so there is the idea of the proof
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You fill up the details
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Thank you!