WEBVTT
Kind: captions
Language: en
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So we start the second module of this course
This week we will be mainly concerned with
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the properties of derivatives So in the last
lecture we have seen given a linear map operators
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from RN to RM how it is to be realized as
a matrix
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So we continue with that that suppose I have
F again U in RN to RM and X not in U is differentiable
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according to our definition at X not So how
to compute matrix of this derivative Remember
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derivative is a linear map of RN to RM Okay
we follow the same pattern as we did in the
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case of any linear map from RN to RN We observe
the Matrix will be a M cross N matrix
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What will be the first column If I fix basis
fix Canonical
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bases E1N ENN this is for RN And ENM ENEEM
this is for RM correct With that from the
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previous calculation we know the first column
of DF of matrix of DF X not this will be DF
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X not acting at E1N If you recall the calculation
this is what we have But we have already proved
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something that E1N is a direction in RN right
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For instance if I having R3 this is the direction
E11 E2 E13 E33 E23 so this is the direction
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E13 And you know this fellow will be equal
to the directional derivative who is in the
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direction EN at the point X not Remember we
have proved in the last lecture So what is
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this now I apply a definition this is Limit
H going to 0 H F of X not plus HE1N minus
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F of X not We denote it by some special notation
so this is a special notation for this particular
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direction
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If I fix the Canonical bases we make the special
notation for this Limit and we call it dell
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F dell X1 at the point X not Now F is from
RN to RM so I know F can be written as F1
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F2 FN where each FI or FK is from U to R So
F is actually the vector so maybe I should
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write in the column vector So if I write dell
F dell X1 at X not this will be a column vector
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dell F1 dell X1 at X not dell F2 dell X1 at
X not and so on up to dell FM dell X1 at X
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not
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So I got the first column of the matrix of
DF X not Now there is nothing special about
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first column If I change 1 to any variable
XK then I will get the Jth column
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So what I mean here that Jth column let us
say of matrix of DF X not will be DF E J N
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at X not and according to this description
I will write at dell F dell X J which is dell
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F1 dell F1 dell X J at X not dell F2 dell
X J at X not dell FM dell X J at X not Now
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I got all the columns Now I put J equal to
1 to N I got all the columns So finally I
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get
matrix of DF X not is equal to first column
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sorry second column and finally the Nth column
This is I equal to 1 to M J equal to 1 to
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M
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So this is M cross N matrix which is the matrix
of DF X not and again if you see in books
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or notes they have again a special notation
for this 1 usually the book do not write in
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this form matrix DF X not they write it as
JF at X not and it has a name it is called
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Jacobian of F at X not So look at it very
carefully and then I want to make a comment
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So what we have done here we compute the matrix
of DF X not assuming if it is differentiable
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F differentiable then this will be I look
at at I wanted to first confine the first
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column or Kth column I evaluated at EKN and
I know by previous theorem that this is directional
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derivative in the direction of EKL and I calculate
that put the columns together I get it
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But there is a point here Thing is that suppose
I get any function if I give you any function
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F from RN to RM or some sub set of RN to RM
you know in the first example one of the examples
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the directional derivatives in all direction
may exists but still the function may not
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be differentiable in our sense Because it
may not fail to be continuous so given any
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function F maybe all this fellows exist dell
F I dell XJ all of them exist
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But if we still fail to be differentiable
so given any F actually after computing this
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I have to check that this is a derivative
that is it satisfies our definition I will
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explain it through an example
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Let us take a very simple example let us say
F X from entire R2 to R3 and F of XY two variables
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let us say let us take something simple okay
So according to first I will write as F equal
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to F1 F2 F3 where F1 is from R2 to R equal
to X F2 XY is Y F3 XY is equal to X square
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plus Y square correct Now I know so I am checking
at a point X not equal to X not Y not some
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arbitrary point
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So candidate I write candidate because I still
do not know let me write it let me say it
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I still do not know given arbitrary point
at X not Y not F is differentiable or not
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because I have not checked the definition
But if it is differentiable then candidate
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for D F X not is is from this formal is dell
F1 dell X then dell X2 dell Y and then dell
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sorry dell F2 dell X dell F3 dell X then dell
F1 dell Y dell F2 dell Y dell F3 dell Y
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Correct this will be a I equal to 1 to 3 J
equal to 1 to 2 because R2 to R3 so let us
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calculate the entries Dell F1 dell X which
is how much is that this is simply 1 and dell
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F 1 dell Y this is 0 dell F2 dell X this is
F2 is XY dell X is 0 Dell F2 dell Y this is
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1 dell F3 dell X this is equal to 2X and dell
F3 dell Y is equal to 2Y
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So again candidate for DF X not is how much
1001 2 X not 2 Y not I have to evaluate at
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X not Y not But this is only candidate now
I have to verify definition
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Verify through the definition that is I have
to put H EX not H equal to F of X not plus
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H minus F of X not minus so this is according
to all notation JF X not Y not J F X not acting
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at H and to show E X not H goes to 0 as H
goes to 0 Well let us check let H H is in
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R2 H1 H2 so norm of H is H2 square Let us
calculate this fellow F of X not plus H1 Y
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not plus H2 minus F of X not Y not minus J
F X not at H at H is H1 and H2 which is equal
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to if I write it in column form what is it
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X not plus H Y not plus H Y not plus H X not
plus H whole square plus Y not plus H whole
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square minus X not Y not X not square Y not
square minus JF X not this matrix acting at
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H1 H2 H1 H2 This matrix acting at H1 H2 you
can easily compute that will be how much H1
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then H2 and then 2 X not H1 plus 2 Y not H2
okay! Continue the calculation here
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This implies equal to how much this will get
cancel second column will get cancel third
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columns remains 0 0 H 1 square plus H2 square
so E X not H is simply 0 0 H 1 square divided
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by H 2 square divided by norm of H but norm
of H is
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this which is 00 root over of course this
goes to 0 as H goes to 0 because H goes to
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0 implies norm of H goes to 0 this is precisely
this quantity
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So after this calculation we can now surely
declare for any X not Y not in R2 F is differentiable
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at X not Y not and derivative at A Jacobian
matrix is 1001 2 X not 21 So you see there
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is some bit of work done it has to be done
Even if you calculate this matrix J F X not
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Jacobian matrix you have to actually check
that this thing this candidate is actually
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the derivative it may not be We have examples
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Let us look at another example I will let
you to do it very easy example F is from let
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us say RN to R so real value given by F of
X let us say X equal to X1 X2 XN equal to
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just norm of X square which is by delimitation
is X1 square X2 square XN square Now there
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is M equal to 1 so there is only one F so
Jacobian matrix will be according to our description
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this will be a matrix 1 cross N matrix that
is row vector
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And what is that row vector will be this is
dell F dell X1 dell F dell X2 dell F dell
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X N evaluated at the point X Again we will
see in the book people do not write JF X not
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for function from RN to R one usually write
it as Grad F X at X not So this is again candidate
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for DF at X not and we have to check that
H E X not H equal to F X not plus H minus
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F of X not minus Grad F X not at H and I have
to check E X not H goes to 0 as H goes to
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0
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I leave the verification on you so write down
everything and check it actually works And
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with one work I will finish the lecture today
that is you can easily see this Okay first
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of all what is this fellow A 1 cross N matrix
acting on a 1 N cross 1 vector that is actually
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inner product
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How does this vector X on H if H is H1 H2
HN then this is actually dell F dell X1 all
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the way to dell F dell XN dot H1 H2 HN which
is summation HI dell F dell XI I equal to
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1 to N so you use that and of course we can
calculate the dell F at X not is simply dell
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F X1 is 2 X1 2X2 2XN evaluated at X equal
to X1 X2 XN Okay So do the exercise and we
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will continue from here