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So welcome to the class. The course is introduction
to CFD.
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Introduction to Computational Fluid Dynamics.
My name is Ramakrishna. So the course is introductory
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in nature. So, basically what I am going to
do is, I am going to talk about
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fundamental concepts in computational fluid
dynamics. I am not going to talk about advance
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topics. I am not going to cover a lot of schemes
and so on. I am going to look more at a few
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of the simple schemes, why they work, how
they work.
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So but before we start off let us look at
the title and figure out little more about
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what we are going to talk about in this class.
I have told you what I meant by introduction.
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The course also title has computational and
fluid dynamics in it. It implies that, we
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are going to study differential equations
that come from that governs fluid dynamics
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problem using computers that of course is
a very general title.
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In reality what we will do is we will look
at differential equations and representing
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the differential equations somehow on computers
and solving the resulting equations that we
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get. If you say using computers in fluid dynamics
you could also interpret it as being you could
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take gather data using experimental techniques
for example, and you could post process that
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data that would really fall into the category
of computational fluid dynamics.
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So having said this, what does this involve?
What is the nature of solving things on the
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computer? So in order to understand the question,
solve on computer we have to look at what
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we mean by solve say an equation. So you are
familiar with this. If I give you an equation
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of the form f of x equal to 0 in algebraic
equation for example it could be ax square
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plus bx plus c equal to 0 you know that you
could find out if some x or some psi is a
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solution by substituting that psi into the
equations. This is what we do and to see whether
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the psi actually satisfies the equation.
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So if you substitute into the equation if
you substitute psi you get a psi square plus
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b psi plus c. If it happens to be 0 you have
a solution. If it happens to be not 0 then
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it leaves something called a residue. So if
there is a residue left over you do not have
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a solution. We expect the right hand side
to the 0, you are left with a residue you
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do not have a solution. So, the mechanism
that we have to find out whether something
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is a solution or not is to take the equation
that we are trying to solve and substitute.
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We have this, we have solved algebraic equations
in this case you know how to find as a solution
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of this quadratic equation and particular
if I say x square equal to or x square minus
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2 equal to 0 you know that you could come
up with a solution which is x equal to square
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root of there are 2 possible solutions. Am
I clear? Now what happens if the solution
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is not a number it is not always a number.
Take a situation where you have f of u of
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x, x equal to 0 and you have encountered situations
like this. One possibility is that you have
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x square plus u square equal to r square and
you could be asked to solve
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for u in terms of x.
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And you know one such solution is u of x is
R square minus x square square root is of
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course a negative solution to it also. So
what we have here is a solution to an equation
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which happens to be a function it is not a
number as we had in this case and this is
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the kind of equations that we are going to
looking at in this course. That is you may
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have seen situations where you either use
bisection method or Newton method to find
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roots to this polynomial.
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Or you try to solve equations of this form.
In this case, you have an equation in which
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the solution actually is a function. It is
not a number. So we have this issue that if
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I am going to solve fluid dynamics problems
on computers that I necessarily need to represent
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functions somehow on computers. We have to
address that issue. So the other situations
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where we have something like this.
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You could for example look at dp by rho plus
1 by 2 d of v square equal to 0 this is you
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are familiar with. This is called the Bernoulli's
equation in differential form. It is possible
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that in 3 dimensions along the stream line
that someone has given you at some location
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x that someone has give you the pressure and
you have the velocity field everywhere so
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you have pressure at x at this point x0 and
you would like to integrate this equation,
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density is a constant.
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You would like to integrate this equation
along this trajectory and what is the result
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which you get. The result that you get is
p as a function of the position vector x again
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a function. So in fluid mechanics this is
just an application to fluid mechanics. This
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is of course also an equation whose solution
is a function. This is just an application
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to fluid mechanics. What I am trying to say
is that we have whole series of problems in
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which the solution to the equations that we
seek are functions and numbers.
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So the first thing that we have to look at
is how do we go about representing these functions
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on the computer or in particular, how do we
represent anything at all on the computer.
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So we are going to look at computer representation
to start with, but before we get there you
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can ask the question why should I use computers
at all? Right one simple answer is in this
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(())(07:55) age what else do you expect.
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Everything is computerized we might well use
a computer, but there is a more serious reason
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for why we are going to use computers here.
If you have if you think about solutions to
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differential equations that you have learnt
before or solutions to other equations like
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this where they are like functions you talk
in terms of close form solutions or analytic
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solutions.
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You would have heard this expression called
closed form solutions
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or analytic solutions not to be confused with
anything that you learnt in complex variables.
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Closed form solutions. So basically meaning
that it is not an infinite series or something
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of that sort that you have something that
is the finite combination of what that is
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the question finite combination of something.
So what are all the functions? I mean if you
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think about it what are the functions that
we know?
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If you sit down and enumerate all the possible
functions that we know in the sense that we
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can write we know monomials that is something
of the form ax power n. We can use combinations
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of monomials using our standard algebraic
operations to come up with polynomials or
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rational polynomials possibly rational I will
put that in brackets. What I mean by that
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is you could have ax square as I have indicated
before bx plus c combination taken by using
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summation.
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Or you can have cx square plus dx plus e by
ax square plus x plus 1 whatever I am just
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making this up fine. So you can make up functions,
you can cook up functions using combinations
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of these, but I am not going to take an infinite
sector, infinite sum that is what I mean by
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closed form solutions. I am not going to take
an infinite sum I am just going to take combinations
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of these (())(10:07) so you can create function
like this. What are the other kinds that we
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know?
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We know transcendentals so that could be sin
of x, cosine of x, exponent of x, log x and
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so on, a power x I mean a raise tot eh poser
x. we really quickly run out of and of course
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the combinations again you could get tangent
of x which is sin x by cos x or you can see
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e power x sin x. you can take combinations.
You can take ln of mod cos x. see there are
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combinations that you can take to construct
more complex function, but this is all we
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have.
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So if you are going to turn around and you
are going to say that I have a solution to
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some function and I want a closed form solution
this is only choice that I have, I have nothing
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else that is left, I have no other combination
here that is going to work so if it is not
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possible, the question is if it is not possible
to represent my solution in terms of closed
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form solutions what do I do? So how does that
happen? How do we interpret that situation
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that I mean why does that happen at all? So
there are situations that u are encountered.
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So for instants if I have dy by dx equal to
fx. So this is a simple differential equation.
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I mean this can be just integrated and you
know get y an indefinite integral you can
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write it in terms of an indefinite integral
fx dx plus constant that is one way to do
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this. You can write it in terms of an indefinite
integral. The problem is that if f(x) equal
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to e power minus x square then they are in
difficulty then you have a problem, because
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unfortunately e power minus x square does
not have a closed form solution.
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So as a consequence what that means is e power
minus x square cannot be expressed in terms
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of any of these basic functions that we have.
So to repeat there are functions we have this
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set of (())(12:42) functions that we have
accessed to we can create an enormous number
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of enormous set of functions with these functions
taking linear combinations of these functions.
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However, there are still functions which cannot
be represented in terms of these basic functions.
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So they do not have a closed form. So this
is what I mean is y equal to integral e power
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minus x square dx unfortunately this integral
does not have a closed form representation.
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e power minus x square is okay, do not get
me around, but integral of e power minus x
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square dx does not have a closed form representation
and therein lies a problem. So we have no
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choice we have to use computers we have to
go through this process of somehow representing
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functions if not in terms of closed form solutions
on the computer in some fashion.
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The second reason why I want to use a computer
is if you look at what I said earlier right
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in the beginning so if you have 3 x square
plus 2 x plus 1 and you are looking for a
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solution one thing that I can do is I can
substitute I can guess values and I can substitute
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those values here so that to check whether
the residue is 0 or not. So you can do a blind
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search essentially. You can keep substituting
values and trying to figure out what the solution
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is.
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Of course you it has been proven that beyond
the quartic we cannot get even a closed off
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solution for this in the sense that you cannot
get an expression that will get a solution
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to a quartic equation this is I mean beyond
a quintic equation and so on. This is a quadratic
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equation. So you necessarily have to guess
or come up with an algorithm, a mechanism
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to generate guesses automatically which will
get you a solution to the equation that you
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see.
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And the process of automation which computers
of course a very good at this kind of (())(14:43)
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but what do you need a some kind of organization
or something of that sort so that you structure
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the search can be structured and an algorithm
can be developed. So we have now reduced our
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requirement to representing things on the
computer this thing is a generic term.
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So I will say representing we have now come
to what I said earlier in the class representing
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mathematical entities on the computers and
we will follow this in a sequence. I will
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make up a history. We will follow this in
a sequence which is very similar to what we
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do in calculus. We will start first by constructing
the real line. So towards that end we asked
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the question how do you represent anything
on the computer?
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What is the basic nature of how things are
represented on the computer? I am not going
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to dwell on computer architecture, I am not
going to dwell on other high performance related
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issues that are there that one would have
to that you know if you need by we need be
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we can visit at a later time, but the only
thing that I want you to know is that on a
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computer I will just recollect most of you
have already encountered this.
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Numbers are represented in a binary form which
means that a binary digit
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or a bit coming from a binary digit can either
be 1 or a 0 can be a 1 can take 2 case 1 or
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a 0 normally they will give you examples of
off and on and so on. It can be a 1 or a 0.
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The important thing is that it has 2 states
1 or a 0. So if I use 2 binary digits 1 next
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to the other just like we do our regular decimal
numbers then the 2 binary digits can represent
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4 combinations.
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So 2 bits so to speak end up with 4 combinations.
So these 4 combinations can be used to represent
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4 symbols of some kind and you can decide
what symbols that you want to use. The number
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of bits that we use then seems to relate to
the number of combinations that we have so
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3 of course 3 bits will give us 0 0 0 I will
just do this 1 more time 0 0 1, 0 1 0, 0 1
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1, and 1 0 0, 1 0 1, 1 1 1. Have I got everything?
1, 2, 3, 4, 5, 6, 7 and I missed something.
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0 1, 1 0 , 1 1, 1 1 0.
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Thank you very much, that is very important
so I will write that a little further up okay
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1 1 0 very important and 1 1 1 that is 8 of
them. So it is 3 bits. You are going to get
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in fact it seems 2 power 3 where this is 2
power 2. So we have something here. So if
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I use 4 bits I get 16 and I am not going to
work out the details of 4 bits obviously I
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am not able to write 3 bits (())(18:30) so
but you can actually write them you, you just
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have to be systematic and little organized
but you can write it out.
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So if there are 4 bits, so 4 bits corresponds
to 2 power 4 symbols which are 16 symbols.
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So there is a reason why I put 4 bits. 4 bits
incidentally in computer parlance is called
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a nibble simply this is just for fun it is
called a nibble because 8 bits are called
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a byte. So it is called a nibble there are
16 symbols. So 16 symbols if you can imagine
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0, 1 to 9 those are 10 symbols then you can
use so 10 of the symbols can be used to represent
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0 through 9.
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Which of course leaves 6 more symbols with
which it is possible for you to represent
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plus that by represent I mean you interpret
in that fashion minus revision and into that
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still leaves you 2 more may be an equal sign
and something else. So in actuality now you
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can just imagine that if I use 4 bits that
is 16 symbols I can interpret what I mean
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by interpret just coming back here what I
mean by interpret is I can say for instants
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0 0.
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This represents the number 0, 0 1 represents
the number 1 and so on. So each of these symbols
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when I have 8 combinations I can interpret
it as something just like when I write 1 the
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(())(20:10) 1, just like when I write the
(())(20:13) 1 you interpret it as 1. You give
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it the meaning 1. So in that sense these 16
symbols for each of these 16 symbols you can
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actually attribute some number associated
with those symbols and other operations.
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And you on the whole were able to build a
4 function calculator using a 4 bit computer
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if possible it is just 4 bits like the 4 bit
representation. It is actually possible for
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us to do this. So now what do we do in real
life so of course just for completion 8 bits
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gives me 2 power 8 256 symbols is called a
byte
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and most of the computers that you use these
days either have are 32 bit computer are increasingly
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64 bit computers.
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So you have 2 power 32 or 2 power 64 which
are enormous number of combinations, enormous
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number of symbols so 2 power 32 will give
you of the order of 4 billion symbols basically.
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So 3 useful things to learn the powers of
10 just as an (())(21:28). So 2 power 10 is
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1024 which is approximately like 1000 so 2
power 32 is 4 into 1 billion right it is in
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the order of 4 billion and it is very easy
to quickly estimate what these values are?
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So we have 2 power 32 symbols so what we can
do now is we can use these symbols first to
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count just like we went from 0 to 9 you can
actually go from 1 to 4 billion plus combinations
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or you can use them to represent integers
which basically means that you take 2 power
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31 going from minus 2 power 31 up to of course
there is 0 which is neither negative nor positive.
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So 0 1 and going up to 2 power 31 minus 1.
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So it is possible for us to interpret them
assign the symbol if one of them being minus
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2 power 31 and the other one to be 2 power
31 and you can represent basically integers
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using the 32 bits, but integers 32 bits it
is a very large number 4 billion, but it is
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not in the full set of countable numbers,
it is not the full set of integers so it is
199
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very large however large it is it is not even
the set of integers.
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So here I am I started of missing I want to
represent mathematical entities on the computer.
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I want to start with building the real line
is what I promised, but I am not even able
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to represent integers, the whole set of integers.
So there lies the problem. So from here how
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do I build the real line. So the real line
is greater difficulty. I mean real lines consists
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of rational and irrationals. Rationales are
of course countable set so you can map them
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to integers.
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Irrationals are an uncountable infinity. So
there is no hope of representing irrationals
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at all. So you could start the argument by
saying that thanks to (())(23:30) that I can
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represent any irrational as closely as I want
by a rational, but that is still does not
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have so we throw away all the rationals. They
are left with rationals. They are left with
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rational numbers and what we have is that
we have to look at basically may be representing
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rationals in terms of scaling a problem between
0 and 1.
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We scale a problem all our problems between
the interval 0 1. Then it is possible that
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we represent all our computations basically
work as a fraction on the interval 0 1 what
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I mean by that is that we only have the mantissa
which is somehow managed to scale the whole
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problem the problem that you are working on
so that the solution always lies between 0
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00:24:23,590 --> 00:24:26,740
and 1. So it is possible for us to do that.
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00:24:26,740 --> 00:24:30,409
You can use fix point arithmetic of course
if you can scale it you can always add a few
218
00:24:30,409 --> 00:24:34,639
decimals here. So it is always possible for
you to use 6 point arithmetic so as you say
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00:24:34,639 --> 00:24:39,139
0.125 or something of that sort. So it is
always your possibility to use fixed point
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arithmetic. So what you basically do is you
have the decimal point and you fix number
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of you use the 32 bits to represent the mantissa.
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00:24:49,890 --> 00:24:56,019
So as you can see what I have really done
is a cheat what I have really done is I have
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taken this integers that I have got I am just
interpreting the integers as a mantissa. I
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am just basically saying that it goes from
0 to 1 so instead of mapping it to minus 2
225
00:25:04,480 --> 00:25:09,820
power 31 to 2 power 31 minus 1 I am mapping
it to 0 up to 1 and representing the mantissa
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00:25:09,820 --> 00:25:17,519
as basically as that integer and all the arithmetic
that takes place is the integer arithmetic.
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So this looks like a good thing. This is called
fixed point arithmetic. So what is wrong with
228
00:25:30,290 --> 00:25:36,140
fixed point arithmetic? Why that should we
do fixed point arithmetic. Well, fixed point
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00:25:36,140 --> 00:25:42,799
arithmetic there are 2 issues that could crop
up. One is that if you do add numbers temporarily
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we may have numbers that exceed 1 you may
have to develop algorithms to take care of
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00:25:48,340 --> 00:25:52,519
that so you have 2 numbers that are added
up they exceed 1 then you turn around.
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And say that look I will take temporarily
take because I know my answer is finally going
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00:25:57,519 --> 00:26:01,169
to be between 0 and 1. I will store that excess
still it goes away that is one way to look
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00:26:01,169 --> 00:26:05,950
at it, but you have to do special things clearly
you have to do special things and there are
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00:26:05,950 --> 00:26:13,169
times there are known instances where in our
space program the French space program there
236
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are known instances where the over flows are
actually caused failures.
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00:26:17,659 --> 00:26:23,239
So using fixed point arithmetic there it is
very fast because it is effectively equivalent
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to integer arithmetic, but overflows that
can be a serious problem. There is another
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00:26:28,590 --> 00:26:36,779
problem which is worse. So you have just say
you are carrying only 3 decimal places. So
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00:26:36,779 --> 00:26:40,890
what is the problem? What is the issue? What
is the big deal? So whatever is here beyond
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00:26:40,890 --> 00:26:46,230
is whatever actual numbers whatever the actual
decimal representation that you have.
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00:26:46,230 --> 00:26:51,059
We will get to the issue of binary representation
and so on. Whatever you have beyond these
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00:26:51,059 --> 00:26:56,480
3 digits that you have put up here are actually
been thrown away. You are just essentially
244
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truncated in because you have an ability only
to represent 3 places. So this position here
245
00:27:03,809 --> 00:27:08,190
is suspect. You may have rounded off, you
may have rounded down, you may have truncated
246
00:27:08,190 --> 00:27:11,559
so this last number 5 is suspect.
247
00:27:11,559 --> 00:27:17,549
So if I start performing arithmetics so for
example if it turns out that from 0.125 I
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am actually going to subtract 0.12 as a course
as a part of my computation that I am performing
249
00:27:27,019 --> 00:27:34,730
this leaves me that 0.005 and it leaves me
with a number that I started off by saying
250
00:27:34,730 --> 00:27:41,970
a suspect that is I have 0.120 0 is also suspect,
but I am left with something that is not as
251
00:27:41,970 --> 00:27:46,200
good as what I started off with I lost my
most significant digits.
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00:27:46,200 --> 00:27:50,269
So anytime that you do subtraction this is
going to happen. This loss is going to happen
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00:27:50,269 --> 00:27:56,029
anytime that you do to subtraction. So there
are 2 extremes 1 extreme is when you are performing
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00:27:56,029 --> 00:28:00,149
additions then it is possible that you somehow
overflow that fixed point container that you
255
00:28:00,149 --> 00:28:05,570
have. The other extreme is that you have loss
of significance in the sense that you subtract
256
00:28:05,570 --> 00:28:06,570
out.
257
00:28:06,570 --> 00:28:09,159
Somehow due to this operation you knock of
the significant digits that you have the digits
258
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that you have whole meaning and you are left
with digits that are suspect. So if you say
259
00:28:14,700 --> 00:28:20,809
that I have only 3 places in which I can represent
this I have a problem and you can say well
260
00:28:20,809 --> 00:28:28,830
why should have to truncate, why should I
have to reduce or I always have to truncate
261
00:28:28,830 --> 00:28:31,249
is it possible not to truncate.
262
00:28:31,249 --> 00:28:41,470
You look at the issue, look at these numbers.
I have 1 by 2, 1 by 3, 1 by 4, 1 by 5, 1 by
263
00:28:41,470 --> 00:28:52,609
6, 1 by 7, 1 by 8, 1 by 9 and of course 1
by 10. I want to take it up until 10 in decimal.
264
00:28:52,609 --> 00:29:01,249
Look at this in decimal and ask yourself the
question which one of these fractions that
265
00:29:01,249 --> 00:29:11,460
I am representing here has a finite or a closed
I would say representation and which one of
266
00:29:11,460 --> 00:29:17,450
them has what we call recurring representations.
So this would be of course 0.5 whereas this
267
00:29:17,450 --> 00:29:21,240
would be 0.333 recurring.
268
00:29:21,240 --> 00:29:26,960
This would be 0.25, this would be 0.2, no
problem 0.1666 is recurring. So this would
269
00:29:26,960 --> 00:29:32,559
be recurring, this would be recurring, that
would be recurring, that would be recurring.
270
00:29:32,559 --> 00:29:38,059
So what does it that differentiates the recurring
ones in the nonrecurring ones. What differentials
271
00:29:38,059 --> 00:29:48,210
it is that these are all primes like and the
base 10 of our number system is a composition
272
00:29:48,210 --> 00:29:55,279
of 2 primes 2 and 5 and as long as you have
fractions that are made up of 2 and 5 they
273
00:29:55,279 --> 00:29:58,159
have a close representation.
274
00:29:58,159 --> 00:30:04,609
If they are primes that are not 2 or 5 then
you have problem that is the key. So this
275
00:30:04,609 --> 00:30:09,320
is 3 times 3 which is a problem, 1 by 7 is
a problem, 1 by 6 is a problem which is 2
276
00:30:09,320 --> 00:30:15,919
times 3 so though the 2 works the 3 does not
work and therein lies the problem. So you
277
00:30:15,919 --> 00:30:24,880
cannot get finite number of positions in decimal
for 1 by 3. So now by going to binary on computers
278
00:30:24,880 --> 00:30:35,090
we have done something worse. On binary you
cannot even represent 1/5 this is also recurring.
279
00:30:35,090 --> 00:30:41,039
This has unfortunate consequences because
being I would always say decimal I hold my
280
00:30:41,039 --> 00:30:47,789
hands up being of because of 10 digits we
have gone for a base 10 representation and
281
00:30:47,789 --> 00:30:53,409
as a consequence we are always going to take
when we say give me a set of points or give
282
00:30:53,409 --> 00:30:57,989
me a set of intervals we variable it in 110
and the tragedy is that one tenth cannot be
283
00:30:57,989 --> 00:31:00,730
represented exactly on a computer it is given.
284
00:31:00,730 --> 00:31:06,909
We never think to take one eighth or one fourth
or one sixteenth or whatever we tend to take
285
00:31:06,909 --> 00:31:15,080
one tenth. So back here so what it means coming
back here what it means is you have no choice.
286
00:31:15,080 --> 00:31:20,419
So if you are representing one third it is
going to be 0.333 and you know there are an
287
00:31:20,419 --> 00:31:24,960
infinite number of 3s here and you had to
cut it, you had to truncate, you had to truncate
288
00:31:24,960 --> 00:31:28,070
the series you have no choice.
289
00:31:28,070 --> 00:31:33,149
So you have an error associated with this
and as a consequence depending on what is
290
00:31:33,149 --> 00:31:38,820
happened here this digit as I said would be
suspect. It is not as large as it should be
291
00:31:38,820 --> 00:31:44,990
or it is not as small as it should be. So
fixed point arithmetic therefore has an issue.
292
00:31:44,990 --> 00:31:50,429
Fixed point arithmetic very clearly has an
issue. How do we fix this well we go to something
293
00:31:50,429 --> 00:32:01,139
called floating point arithmetic as many varieties
of floating point arithmetic that have existed,
294
00:32:01,139 --> 00:32:05,179
as computer manufacturers, once upon a time.
295
00:32:05,179 --> 00:32:23,840
But ever since floating point representation
I said representation. So now a days we would
296
00:32:23,840 --> 00:32:34,100
find we would use I triple E 654 754 you can
go look this up on popular versus you may
297
00:32:34,100 --> 00:32:39,570
go look for this I triple E 754.h of course
the header file most probably will be of the
298
00:32:39,570 --> 00:32:50,789
form I triple E 754.h. so the associated standards
were 654, 754, and the latest one 854. You
299
00:32:50,789 --> 00:32:52,909
can go find out the details about these.
300
00:32:52,909 --> 00:33:01,480
But the whole idea is that we ran into problems
earlier simply because the decimal point was
301
00:33:01,480 --> 00:33:09,559
fixed. So if I allow the decimal point to
move it is possible that something like this
302
00:33:09,559 --> 00:33:15,429
I may have ended up with more digits I do
not know. So that is what we are looking at.
303
00:33:15,429 --> 00:33:24,629
So what we will do is the 32 bits we will
assign these 32 bits in a fashion that we
304
00:33:24,629 --> 00:33:32,170
allow us to how should I put this that will
allow us a larger dynamic range we are not
305
00:33:32,170 --> 00:33:35,779
going to restrict ourselves to 0 to 1.
306
00:33:35,779 --> 00:33:39,950
So what we will take is we will take the real
line and I will show you what I mean. So I
307
00:33:39,950 --> 00:33:46,279
will assign we will pick the origin that is
very clear. We will pick one we know what
308
00:33:46,279 --> 00:33:54,730
we want to do there. We know that we cannot
pick all the points on the real line that
309
00:33:54,730 --> 00:34:02,489
is not possible so I have a largest number
that I can represent which is L I may choose
310
00:34:02,489 --> 00:34:09,310
to satisfy a symbol for infinity and we can
repeat the same thing on the other side.
311
00:34:09,310 --> 00:34:17,190
It is possible for us to have a minus L there
and the minus infinity at the other end. So
312
00:34:17,190 --> 00:34:22,149
we have 32 bits and we need to pick these
numbers. So, one other things of course as
313
00:34:22,149 --> 00:34:27,619
I said, we typically in and around between
1 and 0, we want to cluster of lot of numbers.
314
00:34:27,619 --> 00:34:35,889
So we want a certain amount of density here.
So we will divide up these 32 bits in the
315
00:34:35,889 --> 00:34:45,000
following fashion. We will assign 1 bit. I
will write this is bit 0. We will call this
316
00:34:45,000 --> 00:34:47,339
the sign bit.
317
00:34:47,339 --> 00:34:54,730
So this tells me whether the number is positive
or negative. So this tells me the number is
318
00:34:54,730 --> 00:35:01,840
positive or negative. The second thing that
we do is we assign the next 8 bits, these
319
00:35:01,840 --> 00:35:15,780
are 8 bits this goes from 1, 2, up to 8 bits
the next 8 bits for an exponent. We assign
320
00:35:15,780 --> 00:35:24,619
the next 8 bits to an exponent. So what is
the size of the exponent, it can go to 2 power
321
00:35:24,619 --> 00:35:29,650
8 so that is 256. So clearly just like we
did with integers you could have negative
322
00:35:29,650 --> 00:35:31,180
exponents and positive exponents.
323
00:35:31,180 --> 00:35:35,730
You can bias the exponents so that the negative
numbers and positive numbers. If you look
324
00:35:35,730 --> 00:35:42,290
up this file you will see that for the 32
bit representations they will actually have
325
00:35:42,290 --> 00:35:48,570
a they would define a bias in terms of the
hexadecimal I do not why I am writing this
326
00:35:48,570 --> 00:35:54,720
right now but in terms of hexadecimal they
will define the bias so that is half way through,
327
00:35:54,720 --> 00:36:01,030
half of it is negative, half of it is positive
of course there is 0 in between and then the
328
00:36:01,030 --> 00:36:02,030
remaining.
329
00:36:02,030 --> 00:36:11,520
How many are left 23 bits are left going from
bit 9, bit 9 remember we started to count
330
00:36:11,520 --> 00:36:21,270
at 0 up to 31 the remaining 23 bits are used
to represent the mantissa. The remaining 23
331
00:36:21,270 --> 00:36:33,359
bits are used to represent the mantissa is
that fine. So this is a standard that is used
332
00:36:33,359 --> 00:36:39,020
to represent floating point numbers. What
I want to point out here this is a particular
333
00:36:39,020 --> 00:36:46,950
set, particular type of floating point representation.
It corresponds to this standard.
334
00:36:46,950 --> 00:36:51,789
There are little more to it. If you got look
it up and you can check this out in any resource
335
00:36:51,789 --> 00:36:58,980
that you want given on the net. You go look
up the meaning of Big Endian and Little Endian
336
00:36:58,980 --> 00:37:07,820
because there is a humorous story is associated
with why this nomenclature Big Endian and
337
00:37:07,820 --> 00:37:15,710
Little Endian came but you can go look up
Endian as END Endian so there is a reason
338
00:37:15,710 --> 00:37:18,039
why these numbers are chosen this way.
339
00:37:18,039 --> 00:37:21,790
So the standard setting up of the standards
of course there is a little thought that is
340
00:37:21,790 --> 00:37:26,930
put into all of these. So if you perform a
comparison on a computer and you want to know
341
00:37:26,930 --> 00:37:30,950
if one number is larger than the other number
the first thing that you can do is just check
342
00:37:30,950 --> 00:37:35,539
the sign, but you just need to check the first
bit, if the first bit 1 is positive, the other
343
00:37:35,539 --> 00:37:39,099
negative then you know the positive number
is greater than the negative number just checking
344
00:37:39,099 --> 00:37:41,079
the first bit you will immediately know.
345
00:37:41,079 --> 00:37:44,480
The second thing that you can do is you can
check the exponent if the signs are the same,
346
00:37:44,480 --> 00:37:48,920
if 1 exponent is larger than the second exponent
then, you know that is larger. The exponents
347
00:37:48,920 --> 00:37:53,859
are the same and this is the same then you
go and check the mantissa to see whether that
348
00:37:53,859 --> 00:37:59,150
we will perform a comparison. So in that sense
the order of representation is very important.
349
00:37:59,150 --> 00:38:07,650
The sequence is very important because it
makes comparisons and so on very easy. So
350
00:38:07,650 --> 00:38:14,430
you would like to do now is I will just give
you a little detail on this tiny detail on
351
00:38:14,430 --> 00:38:21,740
just to show how clever we are. I will give
you tiny detail on this. So what we do in
352
00:38:21,740 --> 00:38:30,369
floating point arithmetic is that make sure
that we eliminate all leading zeros.
353
00:38:30,369 --> 00:38:38,450
So because I have an exponent now so the decimal
equivalent would be if I had 0.0125 then the
354
00:38:38,450 --> 00:38:48,099
decimal equivalent of that would be I would
convert it to 0.125 into 10 power minus 1.
355
00:38:48,099 --> 00:38:53,530
decimal equivalent of that would be I would
shift so my exponent what I would store in
356
00:38:53,530 --> 00:38:58,170
my exponent this is 1 and what I would store
in my mantissa would be 125 and I do not have
357
00:38:58,170 --> 00:39:01,990
to store this 0 and point because I know that
0 point is there.
358
00:39:01,990 --> 00:39:07,349
But choosing binary has an advantage having
chosen binary there is an advantage and what
359
00:39:07,349 --> 00:39:14,750
do I mean by that I mean the symbols can be
either 1 or 0 so if I have if I say that I
360
00:39:14,750 --> 00:39:23,150
have a binary number 0.0101 something of this
order and I want to do a shift I could have
361
00:39:23,150 --> 00:39:30,119
actually chosen this because this is actually
a power but it does not matter. I can shift
362
00:39:30,119 --> 00:39:40,650
so I would get binary 0101 so I have shifted
1 place and therefore my exponent has appropriately
363
00:39:40,650 --> 00:39:45,309
changed and my leading term is 1.
364
00:39:45,309 --> 00:39:49,270
The idea is that my leading term will always
be 1 because it is binary there is no other
365
00:39:49,270 --> 00:39:55,800
option. In decimal, the leading term can be
1 or a 2 or 3 or a 4 or whatever it is, but
366
00:39:55,800 --> 00:40:00,349
in binary the leading term will always be
1. So there is no reason why I should store
367
00:40:00,349 --> 00:40:05,310
the 1 what I am trying to say is in this case
there is no reason why I should have this
368
00:40:05,310 --> 00:40:09,880
1 here. Why should I have the 1 I know it
is going to be a 1. So I can actually get
369
00:40:09,880 --> 00:40:12,849
rid of this 1 and just store 0 1.
370
00:40:12,849 --> 00:40:20,390
I do not put the decimal point. What I mean
is that 1 is implied if it was the difference
371
00:40:20,390 --> 00:40:27,350
between this and this would be that this would
go all the way the 1 will shift all the way
372
00:40:27,350 --> 00:40:31,029
and of course I am not going to store it and
therefore I will store 0s. So what I am trying
373
00:40:31,029 --> 00:40:36,650
to say is this last the leading 1 that I have
does not need to be stored because the leading
374
00:40:36,650 --> 00:40:45,660
term they are always going to 1. So effectively
we have it looks like 24 bits.
375
00:40:45,660 --> 00:40:49,770
You can see where the 24 bit you can think
about where the 24 bit came from. 24 bits
376
00:40:49,770 --> 00:40:55,660
effectively. It is not that we got it free
it may look like we got it free. They are
377
00:40:55,660 --> 00:41:01,630
being clever but we just managed to get everything
that we can get. So the smallest number that
378
00:41:01,630 --> 00:41:07,450
we can represent smallest combination or the
number of combinations that we have is 2 power
379
00:41:07,450 --> 00:41:18,079
24, but you can imagine that if I get rid
of if I shift everything 2 power minus 24
380
00:41:18,079 --> 00:41:26,700
is what I can discriminate. What do I mean
by that? Well we can ask the question.
381
00:41:26,700 --> 00:41:36,410
We can ourselves a question. I have these
numbers here. This is the number line that
382
00:41:36,410 --> 00:41:43,280
I have. The number line that I have I am representing
certain numbers here so there are 2 power
383
00:41:43,280 --> 00:41:48,640
24 combinations in this from 0 to 1 that is
what we have basically shown 2 power 24 is
384
00:41:48,640 --> 00:41:52,280
large but it is only 16 million of the order
of 16 slightly over 16 million.
385
00:41:52,280 --> 00:41:56,260
So it is not really (())(41:53). It is not
definitely the continuum there. It is not
386
00:41:56,260 --> 00:42:00,970
all the rationals there either. So what I
have actually done when I say floating point
387
00:42:00,970 --> 00:42:05,210
representation and I am looking at this is
my real line what I have actually done what
388
00:42:05,210 --> 00:42:11,240
I have managed to do? What is that I am doing?
To answer that question I am going to ask
389
00:42:11,240 --> 00:42:13,550
myself.
390
00:42:13,550 --> 00:42:23,900
Is it possible for me to find a positive number,
and we get a positive number you can always
391
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generalize it later positive number so find
what is the nature of this positive number.
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So it should be this smallest positive number
such that since I have taken one 1 plus this
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00:42:43,840 --> 00:42:55,700
number epsilon is ! equal to 1. I want to
find this smallest positive number such that
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00:42:55,700 --> 00:43:01,549
1 plus epsilon ! equal to 1 and sort of get
normally the way I would like to do it is.
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I would like to find the largest or the flip
side of this is fine may be this will make
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00:43:06,859 --> 00:43:12,000
a little more sense and you can think about
it largest positive number I will say positive
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00:43:12,000 --> 00:43:21,700
number such that I will use a different epsilon
do not get confused 1 plus I will use that
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00:43:21,700 --> 00:43:28,059
epsilon equal to 1. So first question is,
is it possible for us to find an epsilon so
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00:43:28,059 --> 00:43:34,349
that 1 plus epsilon equal to 1 this is actually
possible. Well on our number line it turns
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00:43:34,349 --> 00:43:35,650
out it is possible.
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00:43:35,650 --> 00:43:42,950
I would suggest that you write a program to
find out what this value is and such exist.
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00:43:42,950 --> 00:43:46,619
You may be under the impression that a mathematics
of course if you write this you will assert
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00:43:46,619 --> 00:43:54,839
that epsilon equal to 0, but in our case this
is not really true I mean 1 plus epsilon equal
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00:43:54,839 --> 00:44:01,220
to 1 it is possible that you may be able to
find an epsilon positive numbers such that
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00:44:01,220 --> 00:44:07,690
1 plus epsilon equal to 1. A simple code that
you can write for this is for example, say
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while so you can set let us set epsilon.
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00:44:12,780 --> 00:44:18,789
Simple I am just going to write pseudo code
epsilon equal to 1.0 and I will ask the question
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00:44:18,789 --> 00:44:32,369
so the code while 1 plus epsilon. You can
say greater than 1 or ! equal to 1 greater
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00:44:32,369 --> 00:44:51,579
than 1 epsilon equal to epsilon into 0.5 you
have it. We make it one half so we are hunting.
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00:44:51,579 --> 00:44:56,280
So it is going to go through this and as long
as 1 plus epsilon is greater than 1 right
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00:44:56,280 --> 00:44:59,520
this process is going to keep going till you
get to a point that suddenly 1 plus epsilon
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00:44:59,520 --> 00:45:04,920
is not greater than 1. Epsilon is positive
so you can ask the question how is that possible?
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00:45:04,920 --> 00:45:10,380
Well it is possible print out epsilon and
see, what is that you get. So then you have
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00:45:10,380 --> 00:45:13,349
to ask the question you look at how the while
loop works and what is that you are actually
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00:45:13,349 --> 00:45:17,400
measuring and try to figure out are you getting
this, are you getting this or you are getting
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00:45:17,400 --> 00:45:23,839
something else which are the epsilons you
are getting. So the issue is that is it possible
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00:45:23,839 --> 00:45:29,200
that this actually works and what is the number
that you get. I suggest that you try this
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00:45:29,200 --> 00:45:33,980
out or 3 data types for those of you who are
in (())(45:33).
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00:45:33,980 --> 00:45:40,430
Please do this for float run this program
for double and if possible on you complier
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00:45:40,430 --> 00:45:48,010
long double. For those of you who are using
FORTRAN what I mean by that is real, real
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00:45:48,010 --> 00:46:04,369
into 8 and real into 16 is possible. These
do not correspond. We will start 16 if possible
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00:46:04,369 --> 00:46:08,910
and please let me know what is that you get.
So that is basically if something that you
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00:46:08,910 --> 00:46:14,780
can try out just run this on your favourite
computer and with whatever language that you
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00:46:14,780 --> 00:46:15,780
want.
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00:46:15,780 --> 00:46:18,750
If you have a scripting language also it is
fine just try it out and see what is it that
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00:46:18,750 --> 00:46:24,650
you get within this? So what we manage to
do in today's classes I have basically shown
427
00:46:24,650 --> 00:46:32,180
you that we need to in order to solve differential
equations that govern the fluid flow problems
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00:46:32,180 --> 00:46:37,190
we need to find the solutions or functions
we need to able to represent functions on
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00:46:37,190 --> 00:46:41,460
the computer. We also need to represent the
differential equations that we are going to
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00:46:41,460 --> 00:46:43,190
solve on the computer.
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00:46:43,190 --> 00:46:50,340
So that is basically what we have set out
to do and towards that direction we have started
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00:46:50,340 --> 00:46:56,099
off with like we do in calculus and mathematics
trying to represent the real line and the
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00:46:56,099 --> 00:47:02,270
real line we started with integers, we could
not represent all the integers of course there
434
00:47:02,270 --> 00:47:06,319
is no hope of really representing the whole
of the real line we have thrown away the irrationals
435
00:47:06,319 --> 00:47:08,119
we have represented some of the rationals.
436
00:47:08,119 --> 00:47:12,770
We have used the I triple E standard to do
that representation. We are now trying to
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00:47:12,770 --> 00:47:18,751
find out what exactly it is that we are doing
in representing numbers in this fashion. Once
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00:47:18,751 --> 00:47:24,280
we understand this, we will then in the next
class go on with representing other mathematical
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00:47:24,280 --> 00:47:32,040
entities like possibly vectors and matrices
and then see if we cannot represent functions.