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Hello, so in the last class we have learnt
explicit multi step method, so when do we
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say it
is explicit. Suppose in order to compute value
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at n plus 1 stage if the process is asking
k
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passed points; that means n n minus 1 n minus
k plus 1. So, then we say it is explicit
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multi step method, so know in this lecture
we are going to discuss implicit. So, when
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do
we say implicit as I mentioned before to compute
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value at n plus 1 stage, if the processor
on the right hand side demanding the value
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at n plus 1 stage. So, then that is called
implicit, so let us see how are we can derive
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implicit expressions.
.
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So, implicit methods y dashed is f of x y,
so this is our initial value problem, now
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for the
explicit method we have integrator from only
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within one interval, for the implicit we are
stretching to the left. So, we integrate
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x n minus j to that means more than one interval,
so more than this is the interval of integration
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then we get right. Further, how are going
to approximate this f if you recall in the
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explicit method, we have interpolate this
using k
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passed points. Now, that we are calling implicit
that means in the interpolation we need
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to consider n plus point as well approximate
f of x y by a polynomial that interpolates
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at
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.k plus 1 points x n plus one x n. So, please
make a note x n plus 1 included, x n plus
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1 is
included, now similar to the explicit method
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we have to use Newton’s backward.
.
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So, let us use, using Newton’s backward
difference formula, with this variable we
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are
going to get of cause k-th degree polynomial,
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we have k plus 1 point. So, I am not
writing in the terms of the x variable, so
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directly I am writing in terms of u, so this
indeed. So, this is the reminded term again
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I am not giving the polynomial in terms of
variable x because we have done explicitly
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when we have discussed explicative methods,
so similarly one can do it, so this can be
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left to the exercise.
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..
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Now, this can be simplified as follows this
is the polynomial, so this is the reminder
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term, now as before
substituting 3 into we get
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y of x n minus j look the transformation.
So, here the x 1 minus j accordingly when
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we use the variable change of variable x
minus by h n is u we get a different limit
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that is minus j to 1 plus T k plus 1. So,
say T k
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plus 1 star until the interval sign, so further
of this can be simplified as no other
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approximation n minus j plus h submission
comes out the codes in this case.
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So, star within the integration and if you
remove the integration it is this that is
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a notation
I follow, now look at in the explicit method
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if you could recall we have defined gamma
m 0 and in this case delta n j. So, what is
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the difference, so it is only one interval
x n to x
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n plus 1, whereas here x n minus j to x n
plus 1, so this is the indication the length
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of the
interval. So, if the length of the interval
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is just 1 we are using 0 and if it is x n
minus j-th
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x n we are using j.
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..
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Now, the polynomial become
the approximation becomes where the codes
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this is the
error, so this is the error. Now, one can
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compute the codes 1 plus j, 1 plus j square
etcetera, so when we are using more than one
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interval to the left if we fix a particular
j,
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so that will determine the codes. So, these
are functions of j for example, x minus 1
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minus 2, minus 3, so lengths of the interval
codes vary.
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.
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Now, let us take a simple case j is 0 that
is again with a one interval, so j equals
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to 0 case
if this is a case where 1 etcetera accordingly
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the approximation becomes
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so accordingly
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.this is the approximation. Now, what is the
option in our hand k, so how many points we
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are using to interpolate the polynomial, suppose
we use 3 points that includes x n plus 1,
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so we get quadratic equation suppose these
are known.
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.
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Then we get the polynomial as follows this
is f n
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and the error T 3, so this method is
called
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Adams Moulton method. So, earlier one is Adams
backward that is explicit
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method and this is Adams Moulton method, implicit
method why it is implicit, the
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implicit nature come from
because in order to calculate f of. So, this
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is implicit in this
signs, so we have two different concepts that
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is if the interpolating is including the n
plus
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n-th point then we end of with a implicit
for the interpolation. If, we are not using
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a
correct point only past points then we end
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up with explicit in general these are the
popular methods.
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Now, for a given general multi step method
how do we compute the error, so well this
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is
the method and once we derive we know this
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is the error because we have derived it. But
on the other hand suppose there is a method
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given we do not know that what is the error,
so we have to compute that means the coefficients
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everything somebody gives you, this
is a multi step method. So, can you verify
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what could be the error for this approximation,
this is an important task one has to learn,
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so let us try to do that.
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..
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So, that is local truncation error we define
T n plus 1 as minus the general multi step
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method we have given a standard notation.
So, this is function values passed function
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values
and these are the derivate values, so this
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is exact minus. So, this is the
approximation right see what is this is exactly
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this is the method, so if you take minus
common this term plus this equals to y n plus
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1. Now, expanding in Taylor series let T n
plus 1 expressed as c 0 y of x n c 1 h okay
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that means you expand that in Taylor series
than expand this in Taylor series, so for
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a particular case we will do that.
But, in a general sense say for an example
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I can take an example and do it, see for an
example you need to expand n minus 2. So,
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this will be minus 2 h plus 2 double square
of double h n minus 2 h cube, so in that sense
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for a general term i, so for example i is
1
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then x n, so a n of x n. So, here y of x n
and here when i is 1 we get x n, so what will
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be
the coefficient from here 1 and from here
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a 1, so it is like that right.
Suppose i is 2, then a 2 will be multiplying
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of x n because we have seen here y of x n
is
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coefficient is 1 and a 2 will be multiplying.
Suppose y of x n minus 5, so what will be
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multiplying y of x n a 5, so that is how you
get therefore, if you take the difference
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and
define it as T n plus 1 and expand in Taylor
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series, put it in this form then we should
try
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to identify this coefficients in terms of
what in terms of a i and b i.
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..
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So, this can be done as follows c 0, what
is c 0 coefficient of y of x n, in this term
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y of x
n is 1, in this term coefficient of term x
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n is a i and in this nothing. Therefore, c
0 is 1
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minus
a i, so if we do it in a general sense we
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get slightly complex minus, so this is c q
and we get quite complex expression for T
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p plus 1 which is the error, so T p plus 1.
.
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We can write it, so it quite comparison, it
is not so much if we understand the logic
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that
would keep us comfortable just that this is
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an error term and it is of this form. Otherwise
it is very complex, so this must be of this
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form.
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..
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So, when do we say a method is of so much,
so the linear multi step method as defined
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earlier is said to be of order p if c 0, c
1 c p equals to 0 and c p plus 1 is non zero,
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how
come. Look at this, the way we have defined
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local truncation error exact minus the
approximation and that has been expanded in
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this form.
This is of order p if c 0 that means it agrees
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and this agrees up to p terms right c 0 rather
p plus 1, c 0, c 1, c 2, c p and from, here
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c p plus 1 is non zero. Hence, the error is
coming from this term, so that means in order
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to determine we have to force each of this
equals 0 and calls the system. So, let us
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look at a specific k s for how do we derive
for a
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given problem how do we determine the coefficients.
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..
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So, example derive a fourth order method
of the form y n plus 1, a y n minus 2 plus
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h b y
n prime plus e plus prime plus d minus 2 plus
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prime plus e y n minus 3 for prime and
find the number y n, y n minus 1, minus 2,
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n minus 3. So, we have to expand each of
them right, so let us start with see we need
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say for example y n minus 2, so this is y
of x
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n minus 2, h y dash of x n plus 1 of factorial
2 h square y w dash of x n minus. Next, y
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n
dash keep as it is then y n dash n minus 1,
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prime
y dashed of x n minus h y w of x n plus
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h square by 2 factorial, y 3 of x n minus
h cube by 3 factorial, y 4 of x n etcetera
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and 2,
so this is y dashed of x n minus 2 h then
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3, now our method is this.
.
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.So, what is our T n plus 1 y
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minus h, so this is was our T n plus 1 and
this is written as c
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0, this is written as c 0 y of x n
plus c 1 h, so we have expanded this. So,
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the expansion of
this
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then minus a y minus 2 we have expanded, so
that is
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then minus h b y dash of x n
minus h c, this one we have expanded
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then minus h d minus h e. Now, what we have
to
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do we have to collect for c 0, c 1, c 2 as
powers of h power 0, h power 1, h power 2
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etcetera, so if we look at this collect the
coefficients of y of x n this 1 there and
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minus a
there.
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.
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So, if we do that, that means collect the
coefficients of equal powers of h, so to start
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with
h power 0, h power 0 is 1 minus a and there
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is no term, so 1 minus a this is our c 0 in
fact
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and we have to force it to 0 because we are
trying to obtain a fourth order. So, remark
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is
in order to obtain a fourth order method,
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we need to force c 0, c 1, c 2, c 3, c 4,
c 0, so h
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power 0, h power 1 look, so h power 1 this
is 1 here then h power 1 in this case minus
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and minus plus 2 a and h power 1.
So, there is minus b and here there is minus
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c and there is minus d and minus c, so I
repeat again h power this is 1 plus 2 a minus
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b minus b minus c minus d, so we get 1 plus
2 a minus b plus c plus d plus e. So, this
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is our c 0, this is our c 1 then h 2, the
coefficient
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of h 2 h square look at, here 1 by 2, now
here we have 4 h square, so 2 get cancelled,
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so 2
a, so if you take 1 by 2 common coefficient
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is 1 by 2. So, here 1 by 2 if you take
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00:41:27,119 --> 00:41:41,900
.common we have 4 minus 4 a minus 4 a, then
in this case plus c because minus h square
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minus c is minus plus.
So, just plus c then in this case h square
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that is 2, d 2 d and in this case coefficient
of x
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square is 3 e. So, we get the following equation
2 by 2 if I take 2 1 minus 4 a from where
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this is coming, this is 1 by 2, common 1 by
2 common. So, 1 from there minus 4 a then
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minus c plus 3 d plus 3 e, so how will we
get coefficient of h square, see h square,
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00:42:45,539 --> 00:42:53,819
so
minus plus and see this is h square and here
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plus 2 d and here plus 3 every time then
coefficient of h cube.
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00:43:03,339 --> 00:43:25,440
So, look at h cube 1 over 6 if, you take 1
over 6 common, 1 there and here h cube, so
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00:43:25,440 --> 00:43:36,269
the
next term is 1 over 3 factorial 2 h cube y
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00:43:36,269 --> 00:43:42,069
3. So, we have 8 h cube if you take 1 over
6
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common minus 8 a, so this is with minus sign,
this will be plus 8 a, so 1 over 6, 1 plus
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00:43:56,859 --> 00:44:01,420
8
a because it is see it is alternating its
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minus 2 plus and minus. So, this must be minus
1
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00:44:05,489 --> 00:44:17,789
over 3 factorial, that is 1 over 6 if you
take common and here and there 2 cube is 8
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00:44:17,789 --> 00:44:19,640
plus
8.
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00:44:19,640 --> 00:44:31,999
Now, look at the next, here h square and h
cube minus half c, because 1 over to c e and
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here we have, so if you take a half common,
here we have c 3 and here we have 4 d
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00:44:44,440 --> 00:45:10,469
minus 4 d and here we have 9 e minus 9 e.
So, we get minus c plus 4 d plus 9 e, now
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00:45:10,469 --> 00:45:30,940
h 4,
h power 4 look at it here, so 1 by 4 factorial
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00:45:30,940 --> 00:45:38,319
that will be taking common, so coefficient
is
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00:45:38,319 --> 00:45:52,160
1 there then we need to extend this one over
4 factorial 2 h power 4 y 4 x n.
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00:45:52,160 --> 00:45:53,160
.
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00:45:53,160 --> 00:46:16,940
.So, here we have 8 there, so this is a plus
sign minus 8 a, so if we take common that
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00:46:16,940 --> 00:46:26,269
will
be
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00:46:26,269 --> 00:46:40,670
this 4 factorial if we take common, we get
mm, so look at carefully. So, one
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00:46:40,670 --> 00:47:03,729
coefficient then we 16, so we have plus 16
right this is a minus 16 a, so we get 1 by
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00:47:03,729 --> 00:47:11,969
let us
do it, 1 by 4 factorial is 24, 1 there 1 here,
186
00:47:11,969 --> 00:47:18,210
so 4 factorial taken common. So, this will
be
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00:47:18,210 --> 00:47:38,069
16 minus 16 a, so then here h 4 this will
be minus h cube by 3 factorial y 4, so h 4
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00:47:38,069 --> 00:47:45,539
with
the plus sign
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00:47:45,539 --> 00:48:04,849
this will be 6, 1 over 6 c, so this is plus
1 over 6 c then. Here, this is 1 over
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00:48:04,849 --> 00:48:16,059
6 common minus minus plus 8 d.
.
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00:48:16,059 --> 00:48:35,940
Then, here we have to go for one more
y 4 x n, so here one of 3 factorial common
192
00:48:35,940 --> 00:48:48,869
minus
minus plus 27 e, so this is equals to 0, this
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00:48:48,869 --> 00:49:02,380
is our c 4. So, the remark is since we are
asked
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00:49:02,380 --> 00:49:15,930
to determine fourth order method, we need
two 4 c two c 3 c 4 0, now we have a, b, c,
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00:49:15,930 --> 00:49:21,809
d,
e, so 1, 2, 3, 4, 5 equations let us call
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00:49:21,809 --> 00:49:26,160
this entire system.
197
00:49:26,160 --> 00:49:27,160
..
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00:49:27,160 --> 00:49:51,309
Now, on solving
we get a, on solving sys we get a equals to
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1 you may check please b is
21 by 8, c is minus 9 by 8, d is 15 by 8,
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00:50:02,180 --> 00:50:11,039
and e is minus 3 by 8 and hence the method
we
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00:50:11,039 --> 00:50:37,380
obtain plus h by 8, 21 y n prime
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00:50:37,380 --> 00:50:51,299
and if we take the next terms that means non
0 c 5 81 by
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00:50:51,299 --> 00:51:12,130
240 h 5. So, this is the fourth order method
that we have asked to derive, so that means
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00:51:12,130 --> 00:51:27,460
what did we do, we were given multi step method
were the coefficients are not known it
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00:51:27,460 --> 00:51:34,739
was asked derive a fourth order method.
So, then what did we do we have expanded each
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of the terms and then we have
considered the residual that is different
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between the exact approximation and substituted
the expansions. Then we have collected the
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equal powers of h and then we have
determined the coefficients that will desire
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the method. So, accordingly if suppose
somebody asks derive say third order method
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then we expand and collect the coefficients
of equal powers of h and force them to be
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0 of c, 0 c 1, c 2, c 3 and then c 4 onwards
will
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contribute the error right.
So, this is very important this will give
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an idea of how generally the local truncation
error is computed. If you do not know just
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a multistep method is given then what will
be
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00:52:41,160 --> 00:52:46,160
the error, so the same method and suppose
you do not know the coefficients and your
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asked to compute a method of order then also
similar method we do. But, then from the
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00:52:53,730 --> 00:53:01,039
system of equations we try to make them force
them to be 0 up to the desire order and
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00:53:01,039 --> 00:53:06,719
solve the system to determine the coefficients
and further determine the error.
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00:53:06,719 --> 00:53:15,720
.So, this gives a fairly a good idea of how
in general the implicit or explicit multi
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00:53:15,720 --> 00:53:20,700
step
methods are derived. Now, we have to discuss
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coming lectures may be some problems,
apart from problems we have to discuss about
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some theoretical considerations like
stability, convergence some things like that
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until then good day.
Thank you, bye.
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.