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Okay good morning So till now we were looking
at concept of vector spaces or fundamental
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of vector spaces and we revised many concepts
all are very useful in subsequent development
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where we will be talking about solving problems
numerically So we started with concept of
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a general vector space We qualified sets which
can be denoted as
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So the nice things that we like in three dimensions
have now been made available in any other
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vector space You have orthogonality you have
orthogonal functions you have orthogonal sets
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you have orthogonal polynomials and now we
are poised to start using them to solve problems
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numerically Well what kind of problems In
my first lecture I talked about a classification
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of problems
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And also on the moodle web page I have uploaded
one module module 0 which talks about different
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problems in chemical engineering Now almost
all of you are post graduate students so I
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am not really talked about this problem in
the board We through assignments or even through
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when we do some developments
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So physical problems you already know where
you get partial differential equations where
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you get set of algebraic equations nonlinear
algebraic equations ordinary differential
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equations and so on So these things I am assuming
that you already know to some extent that
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is the motivation from the physics or chemical
engineering point of view where do these problems
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are rise
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My motivation or my next aim is to look at
problem transformations So first of all what
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I am going to do is using the language of
vector spaces that we developed till now I
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am going to represent different equations
that you are familiar with or that you are
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going to solve as a part of this course and
then I am going to show you that actually
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we cannot solve them in most of the cases
exactly analytically
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We have to construct numerical solutions but
these numerical solutions are not in the same
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space as that of the original problem So the
spaces associated with the original problem
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and the spaces associated with the solutions
are different So intrinsically we are finding
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an approximation to the truth and you should
be aware of this reality So by second module
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which we will almost we about 12 to 14 lectures
will be devoted to problem discretization
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Now approximation theory is a branch of applied
mathematics and we obviously cannot do a justice
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to this in few lectures We are just going
to see only some part of it not everything
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whatever is relevant to this course but between
the problem which you are defined from physics
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your real world problem and the solution stands
this approximation theory In most of the cases
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we cannot solve the problems analytically
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We have to resort to some kind of numerical
approximations and so what is involved what
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is conceptually happening when you go from
the original problem to the transform problem
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First of all can we represent all these problems
Now that we have defined this new concept
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of space can I generalize my idea of you know
how a problem is in general defined in applied
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mathematics
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Can I say that all these problems are and
some says the same problems can I classify
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these problems What kind of problems that
can arise in applied mathematics or in numerical
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analysis for solving engineering problem So
let us look at this classification and then
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so we will get an overview bird’s eye view
and then we will start looking at specific
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solutions So everything that we developed
till now you will start seeing its applications
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as we go along in the next 10 to 14 lectures
okay
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So first of all you should know that you have
an original problem here
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and then you use approximation theory
and then you get a transform problem and this
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transform problem is something that we solve
using different numerical techniques okay
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So we applied different numerical techniques
or numerical tools I would say so you use
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different numerical tools and then use all
an approximate numerical solution
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So this is very very important and if you
get a bird’s eye view of what is really
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happening in approximation theory then given
original problem you will be able to think
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of a transformation There is no unique way
what you will realize when you do these transformations
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I will take the same problem and show you
how it can be transformed in multiple ways
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okay I will start with the boundary value
problem
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I will show you that this you know transformation
A transformation B finite difference method
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orthogonal collocation method There is a different
way of transforming the problem and you know
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finite element method and so on So same problem
can be approximated in different ways well
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each one of them has advantages and disadvantages
There is one way which is unique in such that
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you know text case of all the difficulties
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There is something to be given and something
to be taken So it is always a given take So
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you get a transform problem and then you use
the known numerical tools like algebraic equations
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solving nonlinear algebraic equation solving
and get approximate solutions Then well as
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a chemical engineer or the physicist you should
go back and see whether the physical solution
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that makes sense here
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There are so many issues when you do from
here to here For example the transform problem
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could be ill condition The solution that you
get may not be good solution because of you
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know ill conditioning problems and so on So
just because I have a very good computer and
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I have a good you know powerful program with
me to solve it does not mean that my solution
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which I get is finally correct
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There could be problems on the way there could
be problems here there could be problems here
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okay So again for some time you will see generic
things and then we will start getting into
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specifics My first task is to define a transformation
so what is the transformation Transformation
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is any operation or any rule that text and
element from a space x and transform it into
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an element in space y okay
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So I have here two vector spaces I have vector
space x
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x and y are vector spaces and if some subset
of x well I am still writing generic things
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but as we go along all these ideas will become
fall in place will become clear What it means
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in concrete terms and terms of So transformation
is a rule that associates with every x that
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belongs to m and element y which is
okay
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I have not written the words but the bare
definition If I am given two vector spaces
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x and y and I take a subset m belonging to
x okay and then if I pick any element x then
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a transformation is a rule that gives me element
y from space Y okay Before I move on to give
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you examples of set transformations well you
are aware of some transformations and I will
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show you that everything that we need to solve
as a part of this course actually can be put
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into the generic framework okay
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Before I move on I am going to define two
concepts one
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concept is a linear transformation okay So
linear transformation is a special class of
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transformations not all transformations are
linear okay Linear transformations if I give
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you x and y that belong to set m we are still
talking about the same set m subset of x and
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so on and we have defined this transformation
okay
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And if I pick up any two scalars alpha beta
say belong two set of real numbers are the
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field on which you are working with let us
for the time being that real numbers okay
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So if I can write transformation of alpha
x plus beta y equal to alpha times Tx plus
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beta times Ty If I can do this if a transformation
allows you to write transformation of this
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combine vector alpha x plus beta y as alpha
times transformation of x plus beta times
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transformation of y okay
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Then such a transformation would be a linear
transformation Give you a simplest example
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let x be Rn and y be Rm okay Y equal to Ax
where A is a
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m cross n matrix is this a linear transformation
Just applied the definition I take A times
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alpha x so let me take two vectors say v and
u that belong to x I am taking two vectors
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v and u belong to x okay What will happen
to alpha u plus beta v times A
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I am constructing a new vector alpha u plus
beta v alpha and beta are two scalars okay
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Now this equal to alpha times Au plus beta
times Av right Is this a linear transformation
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Just applied these definitions See this is
the transformation why It is obtained when
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A operates on so what is the operator here
T transformation is A A operates on x Give
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any element x A will operate on that and give
you an element
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It will take an element from n dimensional
space will give you an m dimensional vector
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okay So this is the linear transformation
right Now tell me whether this is the linear
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transformations So next one well my x is still
Rn my y is Rm okay I am defining a transformation
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y equal to Ax plus b where b is a constant
vector Just do it is this a linear transformation
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Professor student conversation starts Who
said yes sir Why plus b Professor student
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conversation ends So if you actually take
a vector if I take alpha v plus beta u I would
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write when I am operating on this vector I
would write like this is not it y equal to
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okay Now this part cannot be split as sum
of two transformations I cannot write this
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I cannot write this as what will be Tu
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Tu y equal to transformation of u will be
Au plus b okay What will be transformation
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of b Av plus b okay So for this particular
transformation I cannot split I cannot write
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T of alpha x plus beta y equal to well one
mistake which I have made here in notation
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let me correct it here well let us not take
this x and y let us take this u and v Otherwise
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you will get confuse between this y and y
here okay
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So u and v here let us write u and v here
So u and v are two vectors for m not x and
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y because we are using y to denote the element
in the range space What is range space Which
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is the domain space X is the domain space
and m is the domain and y is the range space
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So now let us u and v are the elements from
m and y we are using to denote the range space
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So this definition stands corrected
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So what actually would appear to be a linear
transformation is not really a linear transformation
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You cannot satisfy the basic condition of
linearity for this particular transformation
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okay Now
everything that is not a linear transformation
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any transformation that is not a linear transformation
is a nonlinear transformation the simple definition
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okay
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Whichever transformation does not obey this
simple law when you apply operator T on alpha
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u plus beta v you should be able to write
alpha times T operating on u plus beta times
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T operating on v it is not a linear transformation
okay So let me give you some more examples
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of the transformations and I just want you
to think that understand that these new examples
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that we are going to talk about are not different
from
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So which is the equation which is probably
most familiar to you The most familiar equation
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that you have from your undergraduate or right
from your 12th standard or 11th standard is
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this or may be now it is introduced in 9th
and 10th standard right Solving linear algebraic
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equations so well after sometime you realize
that it is nothing but a matrix equation So
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you are thought about a matrix
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So what is the matrix Matrix is an operator
operating on x giving you an element y in
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the range space What I want to show you is
that after we have generalize the concepts
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of vector spaces almost every problem that
we encounter in engineering can be viewed
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in the same manner it is not different okay
It is an operator operating on a vector giving
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you another vector in the range space okay
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So now let me moved to the third example so
one example I give you was Ax plus b my third
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example here is
so my x here is C1 set of continuously differentiable
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functions on interval a b set of once differential
functions on interval a and b So independent
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variable t belongs to a b So yt equal to dbydt
of xt
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okay I am taking a vector from set of continuously
once differentiable functions operating this
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operator dbydt on it
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This is my operator t it gives me another
vector y which is from the set of continuous
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functions the resultant need not be differentiable
okay I am getting set of continuous functions
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okay So this is my space x this is my space
y this is the operator t which operates on
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a vector this equation is fundamentally not
different from y equal to Ax A is the matrix
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which operates on a vector x gives me a vector
y No difference
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Here I am getting a continuous function when
operated by a differentiation operator on
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you know a vector what is this vector This
is a differentiable function or defined on
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some interval interval could be whatever It
could be 0 to 1 it could be 0 to infinity
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it could be minus infinity to infinity depending
upon their application of your choice the
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interval could be different
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But conceptually y equal to Ax and this equation
are not different they are one and the same
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and so is the case if my operator is little
bit different if I write this is third another
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example Yt will be 3 This is another operation
this is another operator of t 3 times dbydt
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plus 5 so actually you will get 3dxbydt plus
15x This is an operator operating on this
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vector giving me another vector okay No difference
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It is an operator exactly in the same sense
as okay Now here we are going from n dimension
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to m dimension m could be smaller n could
be larger depending upon what kind of matrix
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we are talking about when you deal with matrix
equations Let me give you another example
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of transformation In fact when you start understanding
this language of vector spaces very very powerful
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language developed in the beginning of 20th
century or end of 19th century
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Everything you know falls logically into the
place You will start seeing you know one unified
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structure in all the applied mathematics that
you have been studied okay Now let me look
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at this next example You know my x
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so a b is some interval and I am taking set
of all integrable functions on the interval
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and then I am defining a transformation alpha
equal to a to b what is this You are familiar
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with this
definite integral
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Area under a curve so what is the range space
R right So this is an operator so I would
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say that this operator a to b something dt
this is my operator T it takes a vector from
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the space and maps it onto R okay So it is
a function from let us called this whatever
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this x which is defined here okay x to R This
operator T operates on this vector okay gives
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me a one scalar number okay
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So this is from see just like in this case
you can have an operator A which is from higher
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dimensions to lower dimensions okay You can
have operator A which is only where y is a
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scalar you can have that right You can have
the operator which will give you as a scalar
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value in R So likewise this is an operator
which takes a function from this infinite
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dimensional vector space and gives you scalar
value in R
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This is a transformation is it a linear transformation
So what will happen if I do aft plus bgt no
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no not a b
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We are taking a b here know we will taking
delta and gamma are the two scalars and this
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goes from a to b so I can write this as delta
times ab ftdt plus gamma times integral over
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ab gtdt okay So this is delta tft plus gamma
tgt everyone let me on this so this is the
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linear transformation
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I can write this as a linear combination of
two vectors in the product or in the range
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space okay This solution plus this solution
a linearly add up so linear transformation
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In the same sense this is the linear transformation
try to understand this okay No difference
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between what you have done here and what you
have done here All these are linear transformations
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So if linear transformations have some nice
properties they not only hold for finite dimensional
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vector spaces you can translate them into
any other vector space That is the power of
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using the vector space approach okay So for
so good let us move on to the problems that
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we encounter more frequently one is boundary
value problem can I represent it in the new
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language of spaces
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Can I view it as a transformation from vector
space to another vector space how will I represent
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this Let we will do a little bit more work
we are to go back to the definition of product
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spaces but still you can do it You can show
it is a transformation from particular type
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of one domain space to another range space
okay So this particular Ax equal to b y equal
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to Ax I am going to wipe it out I am going
to just leave it
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So this is the reference we know this very
well and I just want to map everything to
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00:30:06,679 --> 00:30:17,371
this particular What about ODE IVP ordinary
differential equation initial value problem
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00:30:17,371 --> 00:30:31,289
Can I represent it as a transformation okay
Well my ODE IVP is given by say dxbydt equal
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00:30:31,289 --> 00:30:45,770
to f of xtt where f is some this kind of equation
we encounter say solving the batch reactor
212
00:30:45,770 --> 00:30:48,710
problem or batch distillation problem
213
00:30:48,710 --> 00:30:56,860
So you will have equations which are of this
type dxbydt equal to f of xt where x states
214
00:30:56,860 --> 00:31:05,130
and then t would come because you are giving
some input policy you like cooling policy
215
00:31:05,130 --> 00:31:09,990
or hitting policy and so on So this function
of t will appear so this kind of equations
216
00:31:09,990 --> 00:31:18,680
we very very frequently have to use Let us
look right now one dimensional equation so
217
00:31:18,680 --> 00:31:20,539
one dimensional means with the only one straight
218
00:31:20,539 --> 00:31:27,070
Let us not complicate the life by getting
into n differential equations and n unknowns
219
00:31:27,070 --> 00:31:32,630
which will be anyway hitting into later but
right now for representation purpose let us
220
00:31:32,630 --> 00:31:40,100
look at so what should x belongs to the solution
x where should it belongs to C1 Professor
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00:31:40,100 --> 00:31:52,190
student conversation starts She says C1 do
you agree What is C1 Set of once differentiable
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00:31:52,190 --> 00:31:53,190
functions Professor student conversation ends
223
00:31:53,190 --> 00:31:59,620
So the solution x should belong to the set
of once differentiable functions Is it clear
224
00:31:59,620 --> 00:32:05,559
it has to otherwise you know the equation
is not defined okay So my solution x so my
225
00:32:05,559 --> 00:32:24,490
y should be
226
00:32:24,490 --> 00:32:37,440
now this independent variable t it belongs
to some set a b okay so y should be but this
227
00:32:37,440 --> 00:32:45,580
problem it comes up we are normally asked
to solve this problem Solve from initial condition
228
00:32:45,580 --> 00:32:57,260
at xa right okay
229
00:32:57,260 --> 00:33:04,820
We are normally asked to solve this problem
for some specified initial condition xa at
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00:33:04,820 --> 00:33:09,200
point a let us say that this is the initial
point of the interval if it is time 0 to infinity
231
00:33:09,200 --> 00:33:12,800
at time 0 you specify what is the initial
condition and then you want to solve this
232
00:33:12,800 --> 00:33:20,020
problem from 0 to t infinity okay So you are
typically given so that is why it becomes
233
00:33:20,020 --> 00:33:24,750
initial value problems ordinary differential
equations initial value problem okay
234
00:33:24,750 --> 00:33:32,000
So the solution actually belongs to the product
space R you just think about it where R appears
235
00:33:32,000 --> 00:33:37,770
because of the specified initial condition
The solution should satisfy two things it
236
00:33:37,770 --> 00:33:45,710
should be once differentiable function initial
value should be equal to R or initial value
237
00:33:45,710 --> 00:33:51,559
should be equal to xa xa is some number from
real life You are talking of real differentiable
238
00:33:51,559 --> 00:33:57,250
equations real valued functions we are getting
into complex okay real valued functions
239
00:33:57,250 --> 00:34:05,280
So the first value initial value is a real
number okay So the range space is nothing
240
00:34:05,280 --> 00:34:14,260
but set of once differentiable functions The
product space form by this and R R comes from
241
00:34:14,260 --> 00:34:41,509
here okay and what about x
so domain is first I must check out here whether
242
00:34:41,509 --> 00:34:51,200
y also has to be once differentiable Yeah
I think this description is correct so my
243
00:34:51,200 --> 00:34:53,649
x is a set of once differentiable functions
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00:34:53,649 --> 00:35:04,509
My y is a set which is once differentiable
and cross R so this is the map from so this
245
00:35:04,509 --> 00:35:11,579
transformation so what is this transformation
how will I represent this so I would represent
246
00:35:11,579 --> 00:35:35,589
this as dbydt minus f
operating on xt see this xt operates dxbydt
247
00:35:35,589 --> 00:35:41,539
f operates on x you will get f of xt okay
f is the operator which operates on vector
248
00:35:41,539 --> 00:35:45,969
x okay and what should be the right hand side
249
00:35:45,969 --> 00:36:02,770
Right hand side should be 0 vector and
the initial condition if you ask me to represent
250
00:36:02,770 --> 00:36:10,279
this as a transformation in the same sense
as y equal to ax I would write it like this
251
00:36:10,279 --> 00:36:23,239
okay This operator what is this operator Dxbydt
minus f of xt okay Given any vector this operator
252
00:36:23,239 --> 00:36:29,279
operates on x okay When I take this on the
left hand side what is on the right hand side
253
00:36:29,279 --> 00:36:34,049
but it is not 0 it is 0 vector okay
254
00:36:34,049 --> 00:36:43,150
It is 0 vector in this space set of continuously
differentiable functions and what should be
255
00:36:43,150 --> 00:36:48,289
the solution what is the requirement that
the solution should have for the solution
256
00:36:48,289 --> 00:36:59,930
that initial condition should be xa okay So
this is my vector in the cross space here
257
00:36:59,930 --> 00:37:10,089
okay So this operator operating on x should
give me 0 vector plus the initial value of
258
00:37:10,089 --> 00:37:17,859
x should be this is not x do not confuse this
0 to be x This is not x
259
00:37:17,859 --> 00:37:31,999
See when a transform okay let us look at it
okay See this is a dbydt minus f this whole
260
00:37:31,999 --> 00:37:46,739
thing when it operates on xt it gives me dxbydt
minus f of xtt this is what I get okay but
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00:37:46,739 --> 00:37:53,280
when I have taken this f on the left hand
side okay what is on the right hand side This
262
00:37:53,280 --> 00:38:11,900
equal to 0 but 0 means 0 where This is 0 for
the entire interval This is not one point
263
00:38:11,900 --> 00:38:20,650
right see this is the vector xt is the vector
defined from a to b okay
264
00:38:20,650 --> 00:38:26,259
So if the solution satisfied this equation
I should get 0 everywhere on the interval
265
00:38:26,259 --> 00:38:37,589
a to b So this is like solving you know operator
operating on the vector gives me 0 We have
266
00:38:37,589 --> 00:38:45,029
solved these kind of problems right We have
solved Ax equal to 0 We solved these kind
267
00:38:45,029 --> 00:38:54,450
of problems Ax equal to 0 right When columns
of A are linearly dependent you can get a
268
00:38:54,450 --> 00:39:02,940
solution we saw that we took an example we
got nonzero solutions okay
269
00:39:02,940 --> 00:39:08,890
So those of you who have been by now introduced
about null space range space you would no
270
00:39:08,890 --> 00:39:15,570
that the solution belongs to the null space
and so on So this is the problem which is
271
00:39:15,570 --> 00:39:21,710
similar to that Operator operating on x gives
me 0 vector so that is why here I am writing
272
00:39:21,710 --> 00:39:32,420
0 vector and x is addition should have initial
value to be equal to xa So this is the representation
273
00:39:32,420 --> 00:39:38,640
of so same thing is happening like this problem
okay
274
00:39:38,640 --> 00:39:47,270
See we solve for Ax equal to b where b is
the specific vector right This is the specific
275
00:39:47,270 --> 00:39:57,489
vector here 0 We solve for Ax equal to b okay
In fact I am trying to solve for Ax equal
276
00:39:57,489 --> 00:40:08,339
to 0 okay The parallel of that is this operator
operating on x gives me 0 vector okay Yeah
277
00:40:08,339 --> 00:40:13,710
Professor student conversation starts because
my solution should have initial condition
278
00:40:13,710 --> 00:40:16,109
equal to this I want that vector as a solution
279
00:40:16,109 --> 00:40:23,910
This is the way of representation I want that
vector as a solution for which initial value
280
00:40:23,910 --> 00:40:31,670
should be this and the right hand side should
be all 0 vector in the equation I take a vector
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00:40:31,670 --> 00:40:45,420
see I want the solution xt such that okay
let us take I think this would be easier Xa
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00:40:45,420 --> 00:41:00,180
equal to some alpha given that xa equal to
alpha okay So I will write here alpha I want
283
00:41:00,180 --> 00:41:11,809
a vector xt such that which is solution of
this problem such that xa equal to alpha condition
284
00:41:11,809 --> 00:41:13,109
number one okay
285
00:41:13,109 --> 00:41:16,930
Initial condition should be satisfied by the
solution right This is the condition number
286
00:41:16,930 --> 00:41:24,599
one what is the second condition if I operate
by this operator on this x I should get 0
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00:41:24,599 --> 00:41:32,470
vector I should get 0 everywhere on interval
a and b why because xt is defined on interval
288
00:41:32,470 --> 00:41:44,670
a to b okay Professor student conversation
ends This is the representation of the problem
289
00:41:44,670 --> 00:41:51,380
that Let us move onto boundary value problem
290
00:41:51,380 --> 00:41:58,729
Just think about it may not sink because you
know we are suddenly you are being asked to
291
00:41:58,729 --> 00:42:03,549
change the gear from three dimensions to some
infinite dimensional spaces and operator from
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00:42:03,549 --> 00:42:09,650
any space to any space it may take some time
to sink think about this okay Think about
293
00:42:09,650 --> 00:42:35,549
what I am saying Now consider boundary value
problem So a typical boundary value problem
294
00:42:35,549 --> 00:42:42,459
consists of a differential equation that holds
in the domain say 0 to 1 okay
295
00:42:42,459 --> 00:42:47,989
Z going from the independent variable going
from 0 to 1 1 is dimension less length let
296
00:42:47,989 --> 00:42:53,619
us say and there are two boundary condition
one at the z equal to 0 other one at z equal
297
00:42:53,619 --> 00:43:11,780
to 1 okay So some boundary condition f2 dubydz
at z equal to 1 u1 equal to 0 okay and you
298
00:43:11,780 --> 00:43:22,339
make an attempt to write what are the spaces
here
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00:43:22,339 --> 00:43:29,430
So this operator operating on u should give
you 0 not at one value where it should give
300
00:43:29,430 --> 00:43:45,469
you 0 everywhere between 0 to 1 excluding
the two boundary points
301
00:43:45,469 --> 00:43:46,469
okay
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00:43:46,469 --> 00:43:54,470
Now two boundary points you know you can actually
specify okay this instead of giving 0 we will
303
00:43:54,470 --> 00:44:01,329
put this as alpha 0 and alpha 1 so I could
specify some conditions here boundary conditions
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00:44:01,329 --> 00:44:09,059
but derivative plus some operator operating
on this First derivative and u0 gives me alpha
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00:44:09,059 --> 00:44:19,039
0 and this gives me alpha 1 So these are two
conditions okay So what is the underlying
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00:44:19,039 --> 00:44:20,039
domain here
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00:44:20,039 --> 00:44:28,099
Professor student conversation starts Twice
differentiable function on 0 to 1 Interval
308
00:44:28,099 --> 00:44:38,140
is 0 to 1 okay Professor student conversation
ends So my domain space here is C2 0 to 1
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00:44:38,140 --> 00:44:59,559
and my range space here is why is this R and
R coming Two boundary conditions the solution
310
00:44:59,559 --> 00:45:05,479
should satisfy this boundary condition at
initial point this boundary condition at z
311
00:45:05,479 --> 00:45:08,979
equal to 1 okay and this is the map
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00:45:08,979 --> 00:45:20,369
So what is the map So what is my transformation
My t corresponds to d2bydz square a times
313
00:45:20,369 --> 00:45:33,940
plus b one minute here we are not taking any
a b so b times dbydz plus c times whatever
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00:45:33,940 --> 00:45:46,809
it operates on so this operates on uz so this
is my t operating on uz okay Here it gives
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00:45:46,809 --> 00:45:53,150
me 0 vector here I am asking to find out solution
for 0 vector on the right hand side In general
316
00:45:53,150 --> 00:45:58,749
the right hand side here need not be a zero
vector
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00:45:58,749 --> 00:46:09,400
I can give you this equal to sine z for example
right So then it will be you know this operator
318
00:46:09,400 --> 00:46:17,019
t operating on uz giving me sine z with the
solution exist or not is the different story
319
00:46:17,019 --> 00:46:25,420
but I can ask you to do this okay and now
this is same as y equal to or Ax equal to
320
00:46:25,420 --> 00:46:33,770
b Conceptually no difference between Ax equal
to b and you know this operator operating
321
00:46:33,770 --> 00:46:41,190
on this vector giving me another vector which
is sine z okay
322
00:46:41,190 --> 00:46:50,869
I am trying to solve Ax equal to b okay No
difference the same thing So once you have
323
00:46:50,869 --> 00:46:55,410
this unified view of vector spaces you can
start looking at all these problems actually
324
00:46:55,410 --> 00:47:04,709
the same problem okay So now can I say what
is can I state a generic problem in a applied
325
00:47:04,709 --> 00:47:14,160
mathematics that are in or other saying applied
engineering mathematics Well I will do a generalization
326
00:47:14,160 --> 00:47:15,170
here
327
00:47:15,170 --> 00:47:22,359
And say that any problem so likewise here
in my notes I have also talked about how do
328
00:47:22,359 --> 00:47:28,630
look at a partial differential equation as
a transformation from particular space to
329
00:47:28,630 --> 00:47:37,400
another space domain space to range space
You can go through the example here okay So
330
00:47:37,400 --> 00:47:43,880
what is the problem that we have to solve
as a part of applied engineering mathematics
331
00:47:43,880 --> 00:48:04,900
We are given some domain some subset m which
is subset of x and y is range space okay
332
00:48:04,900 --> 00:48:21,199
And then what am I suppose to do I am supposed
to solve this equation
333
00:48:21,199 --> 00:48:32,940
Y equal to transformation of x okay such that
y belongs to y and x belongs to m which is
334
00:48:32,940 --> 00:48:46,789
subset of x
okay So y belongs to the range space and x
335
00:48:46,789 --> 00:48:50,859
belongs to so this is the problem this is
what we have to do and there are three problems
336
00:48:50,859 --> 00:48:59,910
that arise from this okay which I mean all
the problems that we encounter an engineering
337
00:48:59,910 --> 00:49:02,529
mathematics can be further classified into
three problems
338
00:49:02,529 --> 00:49:17,229
Direct problem in direct problem operator
T is given x is given you want to find out
339
00:49:17,229 --> 00:49:48,829
y Given T and x find y okay example definite
integrals
340
00:49:48,829 --> 00:50:02,249
okay Inverse problem
what would be inverse problem Just think alone
341
00:50:02,249 --> 00:50:10,150
y is given T is given you do not know x usual
problem Ax equal to b b is given A is given
342
00:50:10,150 --> 00:50:15,739
x is not given to you okay A differential
operator is given right hand side is given
343
00:50:15,739 --> 00:50:22,109
what is the vector that gives you the right
hand side okay
344
00:50:22,109 --> 00:50:29,709
And a third problem can you think of the third
problem third classification Find a transformation
345
00:50:29,709 --> 00:50:40,680
Given x yeah example fitting correlations
finding out rate equation for in reaction
346
00:50:40,680 --> 00:50:46,279
engineering You are given some measurements
you know you want to find out the module parameters
347
00:50:46,279 --> 00:50:58,229
okay So the third problem is identification
problem well a deal little bit more about
348
00:50:58,229 --> 00:50:59,849
this in our next lecture
349
00:50:59,849 --> 00:51:05,900
So the three essential problems that our three
essential classes of problems that we encounter
350
00:51:05,900 --> 00:51:10,589
in engineering mathematics out of which I
am not going to be so much worried about the
351
00:51:10,589 --> 00:51:15,589
direct problems they are relatively easy to
solve What are difficulty to solve to are
352
00:51:15,589 --> 00:51:21,200
inverse problems okay solutions to the so
called solutions of algebraic equations nonlinear
353
00:51:21,200 --> 00:51:25,589
algebraic equations boundary value problems
partial differential equations all these will
354
00:51:25,589 --> 00:51:26,970
fall into the inverse problems
355
00:51:26,970 --> 00:51:33,509
We are given a vector in the range space we
are given the operator we want to find out
356
00:51:33,509 --> 00:51:40,199
a vector in the domain space which satisfies
this particular equation okay and a third
357
00:51:40,199 --> 00:51:46,269
problem that we are going to look out little
more elaborately that is identification problem
358
00:51:46,269 --> 00:51:55,569
that is given y and x you want to find out
the operator t okay Given the observed data
359
00:51:55,569 --> 00:52:01,269
input given to a plant output given to a plant
I want to find out a transfer function
360
00:52:01,269 --> 00:52:04,440
In poses control if you are remember you are
finding out a transfer function finding out
361
00:52:04,440 --> 00:52:08,029
a transfer function is nothing but finding
out a differential equation that governs the
362
00:52:08,029 --> 00:52:15,359
input and output behavior okay This is the
problem of identification okay So these two
363
00:52:15,359 --> 00:52:21,619
classes I will once again briefly go over
this classification in my next class and then
364
00:52:21,619 --> 00:52:30,019
now we will start you know dealing with the
last two classes that is inverse problem and
365
00:52:30,019 --> 00:52:34,670
identification problem So how is approximation
theory going to be used here that is what
366
00:52:34,670 --> 00:52:35,719
we start from the next class