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Good morning, till last class we have discussed
single steps methods to solve initial
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problems. And we have reviewed some of the
methods and then try to attempt some
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problems. So, in the single steps methods
what we have seen to compute value at a
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particular lid point. We need one value at
one past point. So this is single steps method.
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That is means to compute y of x n plus 1,
we need a y of x n. So, as the name suggest
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as
in multi step method maybe we could expect
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that instead of one past point we need a
several past points. So, the further these
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multi steps methods can be classified into
different varieties. So, let us look at into
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these multi step methods.
.
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So, multi step methods. So, what it does this
uses past values of y of x and or y dashed
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of
x
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to construct a polynomial that approximate
the derivatives, and extrapolates into the
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next intervals. So, what it does? It uses
past values of y x and or. That means could
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be
sometimes y dash or sometimes just y alone.
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To construct a polynomial that
approximates the derivatives and extrapolates
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into the next interval.
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.So, what is the general, the general methods?
y n plus 1 plus h times some processor phi
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x n plus x n k plus 1 then the derivative
terms. So, look at it the value of n plus
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1 stage
demands values at n, n minus 1, n minus k
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plus 1. Plus also demands the derivative
values these points. So, this is more general
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case right? So, this can also be put it in
a
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little different sums as follows.
.
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So, this is functional use at past points
plus h. So, explicitly I am writing the phi.
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So,
which can be written as… So, here i is 1
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to k, here we are including 0 as well. So,
for the
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derivative. So, this the more general method.
So, once we write such a general method
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few remarks. If b 0 is 0. Example if b 0 is
0, example y n plus 1, a 1 y n plus a 2 y
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n
minus 1 plus b 1 y n prime, if b 0 that is
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b 0. So, look at it what is b 0 is coefficient
of y n
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plus 1 prime. And we are trying to compute
y n plus 1. Now what is y n plus one prime?
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X n plus 1.
So, it demands y n plus 1 that means to compute
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y n plus 1 right hand side demands y n
plus 1. So, now if b 0 is 0, then the right
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hand side does not demand y n plus 1. So,
to
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compute y n plus 1, y n, n minus 1, n minus
k plus 1. And since b 0 is 0 we need from
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y
n prime up to y n prime n minus k plus 1.
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So, that means if you know the past values
n to
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n minus k plus 1, 1 can compute y n plus 1.
Hence this method is explicit method, this
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method is explicit method. Now if b 0 is non
0. Say example y n plus 1 is say some 4 y
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n minus 6 y n plus 1 plus 2 h y n plus prime
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.plus 7 h. So, look at that. We are computing
y n plus 1 and right hand side demands y n
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1
1, because to compute y n plus 1 prime we
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need y n plus 1. So, that means this is implicit
method why? Because right hand side… See
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you are computing y n plus 1, but right
hand side also demands y n plus 1. So, therefore,
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this is implicit method. So let us try to
discuss explicit methods. So let us start
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explicit method and try to understand.
.
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So, explicit methods. So, consider our initial
value problem with. Now
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integrating
between one interval that is x n to x n plus
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1 we get. So, what we are done? So, we have
integrated between just one interval. So,
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the method is. Suppose this is x n, so x n
minus
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1 n minus k plus 1. So, the method we are
talking about explicit methods, that means
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to
compute y here, so we would like to compute
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y there. So, we need we need the past
points y. x n, x n minus 1, x n minus k plus
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1 right? So, for that what we are doing we
are
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just integrating between only one step.
Now where are we trying to use the past points?
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So, the past points look at this. Now we
have to approximate f by a suitable polynomial.
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So, here we will use this past points. So,
what we do? To evaluate two one can
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approximate f of x y by a
polynomial that
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interpolates f of x y at k points. What are
they? x n, y n, x n minus 1, y n minus 1,
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x n
minus k.
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So look, we have to evaluate this integran
the integration so integration. So, one can
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approximate by a polynomial that interpolates
f at k point. And what are they? x n minus
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.1, x n minus k. So, these are all past points
so there is nothing implicit everything is
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explicit. Now how do we approximate? One you
can use forward or backward or
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difference formulas right? So let us try to
use one of them to approximate f.
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.
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So, we use Newton’s backward difference
formula of. Well if there are k past points
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what will be the degree of the polynomial
that could be approximated k minus one? Now
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in order to use this we have to assume the
that. If f has k constant derivatives or
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polynomial of degree k minus 1 will be f n
plus x minus x n by h delta f n, x minus x
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n, x
minus x n minus 1 factorial 2 h squared. Then
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the second differences plus. x minus x n
minus 1 n minus k plus 2. This k minus 1 turn
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h power k minus 1 and this. So this is
backward operator where, this is backward
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operator, so that is this on f n backward
operator. Now plus
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k factorial and this, this is the reminder
term. So, this where f k is the
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k th derivative of f evaluate at some zeta
in the interval.
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So look at it, we have used Newton’s backward
difference formula and this is given by,
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this is given by f n plus x minus x n by h.
This is a first differences second order
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differences that ((Refer Time: 19.43)) order
differences and this is the k minus 1 term.
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Then k th term is the reminder term that involves
derivatives of f k thought derivative of
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f. Now our task is that to simplify further
and substitute this in the integrant. So before
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we substitute let us try to simplify this
further. So note that you have... See p minus
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k 1 is
a function of x right. So you have this variables
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x minus x n your variable is now here x
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.that is x minus x n, x minus x n, minus 1.
So, 1 so far so we try to introduce change
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of
variable as follows.
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.
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So, what is the change of variable? Changing
the variable in three by u equals to x minus
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x n by h. Look at that. So you have backward
Newton difference formula, Newton’s
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backward different formula. So this is x minus
x n by h we are transforming it to new
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variable u. Hence we try to convert everything
in terms of u. So if, note that if this is
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u
what will be x minus x n minus 1. What will
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be that? Let us see this is… that means
u h
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plus x n is x. So, we need a next term is
x minus x n h. So we need to compute this.
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So what will be this to this notation? This
will be u plus 1. So, you can verify. So,
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hence
with this our polynomial become p k minus
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1 x n plus u h because x is this. Equals f
n
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approximate u times first different. So, what
could be the next term? Again we go back.
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See this is x minus x n by h is u, so x minus
x n by h is 1 u. Now we have x minus x n
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minus 1 by h. So that must be u plus 1.
So u, u plus 1 by 2 factorial second order
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differences. Now the next term
u plus k minus
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1 k minus k minus 1 factorial, k minus 1 factorial,
this will be k minus 2 because we
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need a one previous factorial term. So u,
u plus 1, plus 2, plus k minus 2, k minus
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1
factorial, k minus 1, f n plus the next term
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k factorial, two h power k. Look what did
we
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do?
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..
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We have transform this using variable
u so correspondingly one h goes to x minus
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x n by
h another goes to this. But if you observe
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there is no h here in this term in the
denominator. So, therefore, we have to supply
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1 h 2 and n minus k plus 1 further this
term. So that means total how many k? So,
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accordingly we get a numerator h power k
and the denominator h power k. So, that denominator
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got adjusted into u, u plus 1 u plus
2 this under the numerator determined.
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.
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.So, now further this can be simplified as
follows, P k minus 1 x n plus. So what is
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our
new variable now? Its u, u is our new variable.
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So, this equals m 0 2, k minus 1 minus m
power m, minus u m plus
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u c k h power k. So we have to define where
minus u c m is?
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So what we have done we have approximated
and we have put it in a simpler form and
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this is a reminder term. Now what is our concern?
Our concern was to substitute the polynomial
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that has been approximated into this
integrant, which was two. So we have approximated
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f by using k past points and that was
four. So, now what we have suppose to do now
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using four into with this notation b x will
be h d u. This is because of the change of
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the variable. Now when we substitute this
in
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the integrant the expression two reduces as
follows.
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.
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Plus h 0 to 1. So, I will explain why this
0 to 1. This is the reminder term and the
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variable is t u. So, why the…
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..
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So, our limits x n 2 x n plus 1. That what
was our transformation? Our transformation
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was
x equals to x n plus u h. So, therefore, when
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x is an x u is varying from 0 and one x
is, x n plus 1 so this will be. See when x
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is x n this implies u is 0. And when x is
x n plus
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1, x n plus 1 minus x n is h.
.
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So, this is so the change of variable brings
these limits. So, now this is equals plus
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h I
take the integrant inside summation outside,
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which is permitted. So, 0 k minus 1 so some
new notation. So I introduce a new notation
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look this same thing I am writing, where this
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.has got in. So integrant, integral has gone
inside and the differences have retained as
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it is
and this is definitely t k. So that means
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when the summation has come out what is left
integral 0 to 1 minus 1 power n and minus
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u c m. So that must be therefore, let us
defined where.
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So, these are some kind of codes equals this.
So, where this and this h because h power
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k
there is one h, h power k plus 1. So this
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is the
reminder right? So, try to follow carefully
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so y x n plus 1 is we have approximated the
integrant by polynomial using k past points.
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And we got a polynomial of degree k minus
1. And that has been substituted in the, in
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the integrant. So that has been substituted
that has a simplified form and these are codes.
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.
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So now let us simplify further so the approximation
is given by. So, after removing the
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error. I have removed the error time and this
approximation. So, this our formula
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however these codes are to be computed right.
So, then the next issue is calculating this
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codes. So, let us say m 0. So, m 0 because
what was our formula our formula was n
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minus u c m also defined, so using that this
will be 1. So, you can compute and to realize
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that it is this. Now these codes will be computed
to get the formula. So, I am doing that
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we get the formula.
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..
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So, therefore, because what was the formula?
1 0 and 0 difference that is f n 1 0 is just
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1
0 gamma 1 0 just 1 therefore, first term should
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be f n. This is then sorry gamma 0 0. So,
then gamma 1 0 is half therefore, the next
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term should be half dell f n right? So, half
of
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and 5 by 12 the second order differences 3
by 8. So, this is the general formula
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now why
did we say this is explicit? Look to compute
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y n plus we need past points because you
see these backward differences it will ask
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for n minus 1 so on.
Now let us come to specific, so this is more
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general. So, let us try to fix up the number
of
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points. k is 3 and past points x n, y n, x
n minus 1, y n minus 1, x n minus 2. So, suppose
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these are the past points that means we are
approximating the polynomial using this past
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points. So, we have three points therefore,
we expect a quadratic and that quadratic has
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been integrated to get the formula. So, in
this case y n plus on1e is y n plus h.
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So we have only three points, so we get differences
up to second order plus there must be
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an error right? And what is a error t k, is
if k is three h power 4 and this must be
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evaluated as reminder. So I can write h power
4. Now we have to expand
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so first all
difference it has been backward operator has
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been expanded. So this is plus. So, we got
the approximation and which is explicit and
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which is multi step. Why multi step? To
compute y n plus we need n, n minus 1, n minus
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2. So let us write down this explicitly.
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..
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So, this error.... So, this an explicit method,
which is third, order why? The error is h
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power 4. So, order of the method is less so
that is third order and this has specific
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name
literature Adams-Bashforth method. So this
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is called Adams-Bashforth method. Now
naturally the question comes, can we reduce
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the error? So, what is the, see here we have
derived multi step method and it is third
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order method. What did we do? We have used
past points and how many? Three past points.
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So, we are approximated by a quadratic.
Now the natural question arises can we improve
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upon this method? So, what is the idea
instead of three past points if you use more
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past points then we get better polynomial
and
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then that that will be extrapolated right?
So, let us try with a four points. Now if
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you try
for four points. So, k equals 4 and the points
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are. Now I am not explaining the derivation
because all that we have to do is simply take
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our general expression, take our general
expression. And since we have one, two, three,
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four points then we can compute the
differences up to thirds order. So by doing
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so we get, already we have done the codes
after third order right? So, we get the following
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formula.
So this sum simplification 55 f n. 59 f n
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minus 1 coefficients are very big. So,
competition will be little traduces and the
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error. So, this is another method. So, this
Adams-Bashforth method. If you use more points
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we get the correspondence. Look at
that the coefficient are completely different.
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So, you keep on using more past points and
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.then you will get a different methods. So,
let us quickly look at some problem so that
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we
get some idea.
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.
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So, this is the u p compute y of point 6 using
Bashforth. So, let us the formula is given.
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What are we asking to compute y of point six
right? What was our x 0, 0 x 1, x 2 x 3. So,
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essentially we are asking to compute y 3,
right? Now to compute y 3 your AdamsBashforth
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is asking some past point right? y 1, y 2,
y 0 of course, see let us look at your
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method. Adams-Bashforth
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so y 3 y 2.
f 3, f 2 sorry f 2 because n is 2, f 1 5 f
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0. So, to compute f 1 and f 2 we need y 1
and y 2
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right? So, remark need y 0 of course, y 1,
y 2, to compute f 0, f 1, f 2. So, we have
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y 0 so
there is no issue with this, but who will
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give us f 1 and f 2. So, in the question itself
it
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should have been mentioned what are the past
values require compute using something.
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Using which kind of method? Of course, using
n explicit method then only we can
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compute. So, let us say it is mentioned compute
the past points say using Euler’s method.
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So let us compute using Euler’s method.
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..
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so, y dashed was… so x 0 is 0 y 0 is minus
1. Therefore, y 1 is so y 0 is minus 1 h is
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point 2 minus 2 x 0 is 0. So, this is then
y 2 h times minus 2 x 1 then minus. So, this
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is
minus point seven 2. So, we got y 1 y 2 using
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Euler method that means we have all the
data, we have y 0, y 1, y 2. Now what was
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our? Of course we need immediately f 0 so
f 0
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is, f 0 is
minus 1, f 0 is minus of minus 1. So, this
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is 1 then f 1 is minus 2, x 1 minus y 1.
So, f 1 is point 4 and f 2, f 2 is minus point
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not 8. So we have to compute, then y 3,
which is a f point 6. This is given by y 2
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00:52:17,430 --> 00:52:28,460
plus h by 12 23 f 2 minus 16 f 1 plus 5 f
0. So,
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00:52:28,460 --> 00:52:57,980
this is h is f 2 minus 16 f 1 plus 5. So,
we get approximate value, which is minus 1
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point
2 2 4. So, what we have done? We need the
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past values.
So, to compute using Adams-Bashforth we need
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three past values n, n minus 1, n minus
2. And in the question it should be given
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using, which method one should compute the
past values. In the present problem we have
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used Euler’s method the simple. But if you
really need better accuracy one should use
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00:53:41,190 --> 00:53:46,221
may be higher order tailor city method or
R K
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method. So, that will give more accurate results.
So you have seen what is the logic we
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have integrated between one interval and then
the polynomial has been fitted with a
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using k past point.
Now the larger the value of k the better approximation
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of the polynomial and hence the
arrive will be less. So, we have tried k three
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and then k four and k three there is a specific
name Adams-Bashforth, which is explicit method.
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And k four also is an explicit method.
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.Suppose instead of that if the right hand
side demands y and plus 1 as well then we
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end
up with something called implicit method.
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We, try to discuss in the next class these
methods.
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Thank you have a nice day.
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.