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Good morning. So, the second lecture is single
step methods for initial value problems.
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So, we have learnt formally, what is an initial
value problem? Now, we would like to
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learn single step methods which are numerical
methods to solve initial value problem.
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.
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So, let us consider an IVP as follows y dash
equals to f of x y, y of x 0 is y 0. Now,
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we
would like to define single step method, single
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step method. So, what is this? Basically
the numerical methods are the algorithms in
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some sense, they are the algorithms to get
values at discrete points. So, your single
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step method looks like this, we have input
and
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we get output. So, what is input? We are giving
y n, y n prime, and the step size. Then
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we get y n pulse 1.
Now, what is reason why are we are calling
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this is single step method. So, let us define
the process. So, then we will understand y
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of x n plus 1 equals f of x n, y n, y n prime
h
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and y of x n equals y n, n equals to 0, 1
say some N h minus 1. So, why we are calling
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this is a single step method. If you see carefully,
there are two indices n plus 1 and n
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.involved. So, the right hand side expects
values at n, and left hand side we are obtaining
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at n plus 1 that means the process is demanding
only one past value. Only one past value
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that means at x n, then it is generating y
of x n plus 1. So, this is the motivation
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to say
that this process is a single step method.
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Why it is single step, now you can correlate
y of
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n plus 1 is in terms of some y of n. So, only
one past value is required. So therefore,
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it is
single step method.
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.
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So, example
say y n plus 1 is y n plus
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x n h. Suppose there is a method like this.
So, this
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is single step method because it is asking
only x n and y n, it is generating y n plus
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1.
Suppose, say 2 y n minus 3 y n prime. So,
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this is also single step method because it
is
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asking only information at 1 pass 2 valves
that is at x n. So, y n of x n y n prime is
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y dash
of x n. So, we get y n plus 1. So, this are
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the single step method.
Now, we would like to learn a particular method
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to solve I V Ps and this particular
method is a single step method. So, what is
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that it is very popularly known method
Taylor series method. So, we consider the
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I V P d y by d x equal to f of x y, y of x
0
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equals to y 0 x belongs to x naught b. Now,
when we say Taylor series, definitely we are
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going to use Taylor series expansion of the
solution about some point, but before we do
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that it will be better to know the assumptions
and at which Taylor series expansion
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possible.
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..
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So we would like to write down assumption.
So, what is assumption, the differential
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equation which equation say differential equation
this is 1. So, the differential equation 1
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has a unique solution y of x on x 0 b and
y of x has continuous partial derivatives
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of
order say p plus 1 on. So, this is assumption
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what is that the differential equation 1 has
a
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unique solution y of x on x 0 b. Essentially
this is existence the uniqueness theorem and
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y of x has continuous partial derivatives
of order say p plus 1 on the central. So,
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then the
solution y of x of 1, can be expanded in a
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Taylor series about any point say x equal
to x
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not as follows.
.
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.We will write down Taylor series about x
equal to x naught. So, it is given by
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y prime of
x naught plus factorial 2 y double of x not
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plus, so on plus, p
so I switched over from this
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notation to this notation. Please try to follow
y 1 y 2 and then p is. So, it is a customary
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to put it in a parenthesis, this plus where
this is since we are expanding around x 0
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within
this. So, let us call 2. So, in case we are
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expanding around x n then zeta varies x n
to x.
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Now, this is standard Taylor series expansion
about x equal to x naught. Now, definitely
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when we use Taylor series expansion to compute
a solutions at the particular point we
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have lot of restriction, what are the restrictions.
The first restriction is how long we
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consider, how long we take terms into account
like 5 terms 10 terms 20 terms. So, that
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means we have to consider the number of terms
in that account.
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Naturally one would guess easily that the
accuracy of your method as direct consequence
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in terms of the number of accounts. So, let
us see the last terms we have written in a
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very
specific form because we have not written
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in a terms of a x 0 rather we have put it
as zeta
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n, where zeta n lies within some interval.
Now, this last term has a particular name
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and
then it has a vital role.
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.
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So, let us see what is that? So, this is reminded
term R n. So, this the reminded term. So,
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if we generalize y of x n plus 1 y of x n
and x h y dash of x n p plus R n. So, what
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is
notation we are adopting now. Earlier we have
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written if you recall, we have written x
minus x 0 x minus x 0 so on.
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..
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Now, if you need at a particular point x 1
y of x 1 is y of x 0 x 1 minus x 0 y dash
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x 0
plus x 1 minus x 0 square by factorial 2 y
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double of x 0 and x 1 minus x 0 for p, p
factorial y p of x 0 plus the reminder. So,
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strictly speaking this should be R 1 because
we
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are expanding at. So, what is x 1 minus x
0, this is h the step size therefore, x n
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plus 1
minus x n this is our h. So, what I just have
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written is at x n plus 1 the Taylor series
expansion at x n plus 1.
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So, this is a Taylor series expansion at x
n plus 1 y of x n plus h y dash perfection
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plus x
square by 2 factorial this and reminder terms
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right. Now, who is going to supply us as the
higher order derivative, you see suppose somebody
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who like to use Taylor series
expansion up to to 5 terms. So, then 5 terms
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means 1 2 3 4 will be h cube 5 will be h 4.
That means derivative of y 5 are required.
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So, who is going to give us these the initial
value problem which is going to give us.
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..
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So, the remark here is the higher order derivatives.
So, the higher order derivatives
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should be there the higher order derivatives
are to be computed from the I V P, but how
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do we compute. If you look it your I V P is
y dash equals to f of x y. So, we require
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high
order derivatives means under assumptions
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what is assumption see, higher orders for
example, y 2 will be first derivatives of
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f right. So, y 3 will be second derivatives
of f.
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So, we required higher order derivatives of
y which means we required higher order
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derivatives of f.
.
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.Now, to have that of what is assumption,
the assumption is f of x y is differentiable
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as
many times as we require. So, these are the
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two assumptions. The first assumptions is
the
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higher order derivatives have to become computed
from the initial problem and may this
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not assumption this is a the process, but
for that what is assumptions required f of
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x y is
differentiable as many times as we require.
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Now, let us try to compute. So, y dash of
x n
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is f of x n y right now we need. So, in shorthand
notation this is now we required y
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double. So, this is d f by d x.
So, this now we compute using chain rule,
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tau f by tau x plus tau f by tau y into d
y by d
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x, but what is d y by d x, yes you are true
d y by d x is f. So, it is f right. So, we
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need y
double of x n this will be tau f by tau x
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plus f tau f by tau y at the point x n y n.
Similarly,
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all other higher order terms can be computed.
Now, which these let us see what happens
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to our Taylor series.
.
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It is y of x n plus 1 x n plus h plus square
the factorial 2 tau f by tau x plus f tau
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f by tau
y at x n y n plus the higher order comes h
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p, p factorial f p minus 1 plus. So, the reminder
I would like to introduce term like this.
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So, this expression I am calling T S. That
is
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Taylor series where epsilon p plus 1 equals
h p plus 1. So, this is residual, residual
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means
the error. So, quick remarks observe this
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term carefully the derivative is p minus 1
x h
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power p, p minus 1 y is that because actual
it should have been h for p y p, but y p is
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nothing but f p minus 1.
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.So, this is residual. So, what do you mean
by residual, that is error of course, I have
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not
defined formulary what is error. So, we will
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definitely defined. So, in what way we are
going to define this error, yes as I mention
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somebody is doing up to 3 term. That means
considering Taylor series up to 3 terms and
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somebody else considering up to 10 terms
somebody 50 terms. So, we expect there will
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be a difference with the solution. So, y is
that definitely see you consider only 3 terms
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other person consider 50 terms another
person consider 100 terms.
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So, the added terms should definitely refine
your solutions. So, that it more accurate
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compare to the solution which is obtained
with only 3 term right. So, when you are
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considering let us say 25 terms of your Taylor
series, then what are we doing, we are
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force to neglect the terms from 26 onwards
right. So, you expect that your solutions
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is
accurate up to that level what is that level
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up to 25 terms 26 and onwards we are
neglected.
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So, if at all there is a deviation in your
solution with that true solution you expect
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that
the deviations contributions is from 26 terms
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onwards. Suppose, somebody considers 50
terms than the deviation from the exact solution
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will be from 50 first onwards right. So,
we are going to define this residual and discuss
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about it in a more regress way. So, that
means if somebody says I do not mind my solutions
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is if it is a accurate only up to 1
decimal I do not bother.
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So, then how many terms we can consider and
keep quite right we will be interested in
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because somebody is asking you to have accuracy
of only 1 decimal. So, why do you
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bother to compute 100 terms 200 terms it is
not required right. So, similarly, somebody
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would say with this step size, if you consider
this many terms, then what will be the
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accuracy. So, these kind of a trade off various
trade off, what does the tradeoffs between
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if you know the error, means if you know up
to this accuracy I need the solution. So,
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that
is 1 quantity the number of terms another
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quantity, then the step size is another quantity.
So, there is a kind of a trade of among this
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quantity.
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..
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So, lets us have careful look. So, remarks
1 if epsilon is a pre-assigned number, then
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nod
epsilon p plus 1 is less than, see this is
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the condition on the arrow means epsilon is
preassigned we expect that you must continue
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up to this condition. That is your residual
must be less than epsilon. Then if h is given
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step size p can be calculated, if p is given
h
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can be calculated theoretically right, provided
we replace by
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is this if you look at your
epsilon p plus 1 it is h for h for p plus
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1 by p plus 1 factorial and y p plus 1 of
zeta n. So,
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this is hp plus 1 by plus 1 factorial f p
of zeta n y n. So, this our residual, now
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this
residual is less than pre assigned number.
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.
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.So, mod of less than epsilon pre assigned.
So, if h is known f is our hand. So, we have
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to
take the maximum of this and epsilon is known.
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So, we should compute p. Suppose p
known and epsilon is off course known you
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should be able to compute it theoretically.
So, this is trade of essentially that means,
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if the error is in your hand means if you
know
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up to this accuracy I need the solution and
what are the other left out parameters is
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step
size h and number of term. So, you have to
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play with this two. The more number of
terms you expect more accuracy, same time
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if you jump. So, you have taken at 1 this
point then with large step size your jumping.
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Then one would expect less accurate solution.
So, this trade of is very much important.
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So, when we come across examples specific
examples we will discuss how to do this.
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Now, what is the process involved in Taylor
series. So, let us define h times f of x n
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y n
then y n 1 y n p minus 1 h as y dash y double
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y p. So, this nothing but the right hand side
of Taylor series, but for one term. So, if
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we define like this, then what was a number
we
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have given T s, then T s become y n plus 1
equals y n plus h f. So, where we need y we
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need all the quantities. So, it is n is 0
1 say something.
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So, this is the corresponding single step
processor. So, we believe that now Taylor
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series
method is a single step method as you can
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see it will ask only 1 past value however
higher order derivatives. So, we are not denying
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that. So, that as that as do with number
of terms, where as all they values whether
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it is first derivatives second derivatives
50 th
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derivatives. All the terms are expected only
at 1 past point, so that is motivation for
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single step method. Therefore, will believe
that Taylor series method is a single step
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method, before we discuss more about the error.
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..
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So, let us look at once specific example and
see how do we compute the solution using
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Taylor series. The example
say solve y dash d s 3 x y square
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and y of 0 is minus 1. To
compute y of 0.2 using Taylor series with
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h equals 0.1, then consider only first 3 terms
of
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the Taylor series. So, suppose this is an
example. So, what it says for this o d this
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is the
initial condition, we compute y of 0.2 using
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Taylor series with step size 0.1 now what
is
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remaining the number of terms. So, consider
only first 3 terms of the Taylor series. So,
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let us briefly look at the solutions. So,
since it is suggesting to use only first 3
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terms. So,
let us look at the Taylor series. The Taylor
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series is given by y n plus 1 is y n plus
h
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factorial 2 this is our Taylor series first
3 terms 1 2 3.
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Now, let us identify what is known data from
their from the problems the known data is
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f. This is our f, f of x y equals 3 x plus
y square and what else is known x 0, x 0 is
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0 then
what is known step size 0.1. Now, we need
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to compute y at 0.2. So, what is required
see
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with this h is 0.1 with x 0 is 0 what is x
1 x 0 plus h that is. So, this suggest that
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our x 2 is
0.2 that means we need to compute the solution
204
00:36:13,890 --> 00:36:29,109
at y o f 0.1 and y of 0.2. So, that means
we have to solve for y of 0.1 y of 0.2. So,
205
00:36:29,109 --> 00:36:38,490
we solve let us in order to solve what we
required, we require to compute this term
206
00:36:38,490 --> 00:36:43,109
and this term rest in hand. So, y dash and
y
207
00:36:43,109 --> 00:36:49,110
double dash you have to compute. So, let us
do that.
208
00:36:49,110 --> 00:36:50,110
..
209
00:36:50,110 --> 00:37:01,650
So, y dash is also in hand f of x y which
is given by 3 x plus y square y double is
210
00:37:01,650 --> 00:37:08,859
3. So,
we can do using the formula because general
211
00:37:08,859 --> 00:37:14,240
expression I have computed, but since we
have particular case in hand. I would like
212
00:37:14,240 --> 00:37:18,430
to do the chain rule directly here. So, y
double
213
00:37:18,430 --> 00:37:36,240
derivatives of this respect x the 3 plus 2,
y 2 y dash. So, this is p plus 2 y, y dash
214
00:37:36,240 --> 00:37:53,001
is 3 x
plus. Now, consider the case n equals to 0.
215
00:37:53,001 --> 00:38:00,550
So, then n is equals to 0. In this Taylor
series
216
00:38:00,550 --> 00:38:07,780
y 1 is y 0 plus h y 0 prime plus h square
by factorial 2 y 0 double prime right.
217
00:38:07,780 --> 00:38:25,220
So, let us write down y 1 is y 0 h y 0 prime
h square by factorial 2 y 0 double prime.
218
00:38:25,220 --> 00:38:30,849
So,
we have y, y dash and y double, now we required
219
00:38:30,849 --> 00:38:48,540
y. So, y 0 is minus 1 y 0 prime is y
dash of x 0 which is y dash of 0 which is
220
00:38:48,540 --> 00:39:01,920
f of x 0 y 0 which is f of 0 y 0 is Minus
1. So,
221
00:39:01,920 --> 00:39:21,580
this is 3 time 0 minus 1 square equals 1 right
then y 0 double. So, you can do it. So, this
222
00:39:21,580 --> 00:39:27,609
is expression in hand. So, directly we can
substitute. So, I do not want to do this algebra
223
00:39:27,609 --> 00:39:39,050
here this is just for clarity.
Now, directly 3 plus 2 into y 0 is minus 1
224
00:39:39,050 --> 00:39:49,849
3 into x 0 is 0 plus minus 1 square. So, this
is
225
00:39:49,849 --> 00:40:00,640
this term is 0 this is 1. So, 3 minus 2 this
is 1. So, maybe I have taken example where
226
00:40:00,640 --> 00:40:06,640
the
calculations are simplified. So, we have the
227
00:40:06,640 --> 00:40:19,180
known data required in hand is this together
with this together with this and this.
228
00:40:19,180 --> 00:40:20,180
..
229
00:40:20,180 --> 00:40:37,150
Now, we would like to substitute in star.
So, we do it. So, y 1 is y of 0.1 that is
230
00:40:37,150 --> 00:40:50,180
y 0 plus
h. So, this is y 0 is minus 1 plus h is 0.1
231
00:40:50,180 --> 00:41:01,190
the y 0 prime is 1 plus h factor to y 0 1.
So, in
232
00:41:01,190 --> 00:41:26,310
this minus 1 plus 0.5. So, this is point so
this our y 1. So, this corresponding to n
233
00:41:26,310 --> 00:41:35,920
equal to
0. Now, let us take n equals to 1. So, if
234
00:41:35,920 --> 00:41:48,589
n equals to 1 then what do we need y 2 y 1
plus h
235
00:41:48,589 --> 00:42:04,510
y 1 prime h square by 2 factorial y 1.
So, we have to compute, now y 1 prime y 1
236
00:42:04,510 --> 00:42:14,839
double prime at x 1. So, y 1 prime is f of
y 1.
237
00:42:14,839 --> 00:42:40,230
So, this is 3 x 1. So, x 0 x 0 h is 0.1. So,
x 1 is 0 plus 0.1 therefore, and y 1 just
238
00:42:40,230 --> 00:42:53,960
now we
have computed therefore. So, this one can
239
00:42:53,960 --> 00:43:10,260
simplify we get this right now, next quantity
you have to compute this by 1 double.
240
00:43:10,260 --> 00:43:11,260
..
241
00:43:11,260 --> 00:43:37,890
So, y 1 double is 3 plus 2 y 1 3 x 1 plus
y 1 square equals 3 plus and 2 into minus
242
00:43:37,890 --> 00:43:56,580
0.8953
into 0.1 0.895 square. So, this is 3 3. So,
243
00:43:56,580 --> 00:44:14,240
let us give this minus here to this is 1.401.
So,
244
00:44:14,240 --> 00:44:23,619
let us you can also try you can also try.
So, this is this is of course some value we
245
00:44:23,619 --> 00:44:33,060
get it.
So, this is exactly y 1 prime right. So, we
246
00:44:33,060 --> 00:44:44,060
have it this is exactly 1.1010. So, this is
1.401
247
00:44:44,060 --> 00:45:11,609
and it is final we get something like this
1.4574 y 1 double right now. So, once we get
248
00:45:11,609 --> 00:45:23,339
this we have the solution in hand. So, it
is a straight substitution y 2 equals y of
249
00:45:23,339 --> 00:45:34,060
0.2
equals y 1 y 1 is minus 0.895 plus h into
250
00:45:34,060 --> 00:45:48,940
y 1 prime. Just now we got 1.1010 plus x square
by 21.4574.
251
00:45:48,940 --> 00:46:06,520
So, this one can compute. So, we get some
value. So, y 1 is a y of 0.1 and y 2 is y
252
00:46:06,520 --> 00:46:15,260
of 0.2.
So, you can compute at higher grid points.
253
00:46:15,260 --> 00:46:27,180
So, you can, but remember we have used only
the first 3 terms only right. So, if somebody
254
00:46:27,180 --> 00:46:32,310
would like to try further. So, there will
be
255
00:46:32,310 --> 00:46:39,930
slide modification in the values what we have
obtain. Now, quick remark here is, if you
256
00:46:39,930 --> 00:46:45,070
see while computing y 1 we would have rounded
off.
257
00:46:45,070 --> 00:46:51,940
For example, there is a 0.1 square is there.
So, we would have a rounded it some
258
00:46:51,940 --> 00:46:59,080
accuracy. Then we have put down that then
further y 1 has been used to compute y 2.
259
00:46:59,080 --> 00:47:07,010
So,
we again cut down to some terms and then put
260
00:47:07,010 --> 00:47:16,210
down. So, every time you cut down. So,
one kind of firm error is the number of terms
261
00:47:16,210 --> 00:47:24,750
the number of a terms. We are throwing
after a second level. So, the other is at
262
00:47:24,750 --> 00:47:32,070
each computation you are throwing some digits
and using that as an information to compute
263
00:47:32,070 --> 00:47:33,550
further values.
264
00:47:33,550 --> 00:47:41,079
.So, these are named with a specific terminology,
but of course they are the errors the first
265
00:47:41,079 --> 00:47:47,640
one that is neglecting after certain number
of terms is also termed of kind of errors.
266
00:47:47,640 --> 00:47:50,869
Then
neglecting few digits and then using that
267
00:47:50,869 --> 00:47:54,250
information further is also termed as a kind
of
268
00:47:54,250 --> 00:48:00,839
error. So, this more technical term exact
technical terms I would define little later.
269
00:48:00,839 --> 00:48:11,040
Now, let us look at the thrown part. In this
example we have calculated using the first
270
00:48:11,040 --> 00:48:15,960
3
terms that means we thrown from fourth term
271
00:48:15,960 --> 00:48:22,880
onwards. So, let us look at this fourth term.
So, that we understand what is happening.
272
00:48:22,880 --> 00:48:26,230
So, let us look the fourth term. So, what
is
273
00:48:26,230 --> 00:48:27,230
that?
.
274
00:48:27,230 --> 00:48:43,560
So, our p was 2 because h square, this is
p plus 1. So, this is epsilon 2 plus 1 this
275
00:48:43,560 --> 00:48:54,760
is
epsilon 3 this is. So, let us say we are considering
276
00:48:54,760 --> 00:49:15,480
this at first. So, that means while
computing 0.1 in this. So, what is this and
277
00:49:15,480 --> 00:49:32,200
y 3 is f 2, y 3 is f 2, y 3 is f 2 right f
2 of this.
278
00:49:32,200 --> 00:49:45,760
So, for this example let us look at this quantity
we have only f 1 in hand. So, what is that
279
00:49:45,760 --> 00:50:11,839
f 1 is 3 plus 2 y 3 x plus y square. Now,
f 2 3 plus 2 y prime that is 2 y 3 x plus
280
00:50:11,839 --> 00:50:15,859
y square
and keeping y prime is f.
281
00:50:15,859 --> 00:50:38,950
So, that is 3 x plus y square plus 2 y into
3 plus 2 y by prime right. So, let us compute
282
00:50:38,950 --> 00:50:51,830
this f 2 at zeta 0 by 0. So, y 0 is for this
problem y 0 we have consider is minus 1. So,
283
00:50:51,830 --> 00:50:57,550
if
we 3 plus 2 into x 0 is 0 y 0 is minus 1.
284
00:50:57,550 --> 00:51:05,320
So, we get 1. So, this we get 1 plus y 0 is
minus
285
00:51:05,320 --> 00:51:23,660
1. So, minus 2 and here 3 plus this is minus
2. So, this not correct because x 0 3 plus
286
00:51:23,660 --> 00:51:43,250
2 x
zeta 0 plus 1 into 3 zeta 0 plus 1 minus 1
287
00:51:43,250 --> 00:51:52,660
3 plus minus 2 y this is wide prime.
288
00:51:52,660 --> 00:51:59,940
.So, this for the time being I am putting
next line. I can simplify 3 plus 2 into 3
289
00:51:59,940 --> 00:52:11,280
zeta plus 1
square minus 2 into 3 minus 2 y prime is 3
290
00:52:11,280 --> 00:52:19,940
zeta 0 plus 2. So, essentially this will be
some
291
00:52:19,940 --> 00:52:32,640
function of zeta 0 this will be some function
of zeta 0. Now, this zeta 0 is varying in
292
00:52:32,640 --> 00:52:34,720
this
interval. So, let us say you are computing
293
00:52:34,720 --> 00:52:41,590
zeta 0. So, 0 less than zeta 0 let us 0.1
here
294
00:52:41,590 --> 00:52:50,310
computing next solution it is within this
interval. Now, we have to analyze this. So,
295
00:52:50,310 --> 00:52:54,329
how
do we analyze naturally we have to look for
296
00:52:54,329 --> 00:52:56,510
the maximum.
.
297
00:52:56,510 --> 00:53:12,040
So, mod epsilon equals is less than are equals
to maximum of zeta 0 in the interval 0 to.1
298
00:53:12,040 --> 00:53:27,450
of g of this. So, this value if you compute
then 1 can say y of 0.1 is accurate up to.
299
00:53:27,450 --> 00:53:33,500
So,
this we get some say epsilon say we are getting
300
00:53:33,500 --> 00:53:44,380
this as some 3.54 into 10 minus 4. Say
not for this problem because say bit lengthy
301
00:53:44,380 --> 00:53:47,910
calculations. So, I am just a showing you
for
302
00:53:47,910 --> 00:54:02,600
example, that means y of 0.1 is accurate up
to 3.54 into 10 minus 4.
303
00:54:02,600 --> 00:54:09,480
So, this is the residual. So, I have not yet
mentioned what is the error different technical
304
00:54:09,480 --> 00:54:16,250
names of various errors, what kind of errors
I told you the number of terms there is 1
305
00:54:16,250 --> 00:54:25,520
name and then we cut the digits there is another
name. So, we formally learn coming
306
00:54:25,520 --> 00:54:31,340
lectures. So, this is how the Taylor series
method would give you approximate solution.
307
00:54:31,340 --> 00:54:38,630
So, the assumptions are first of all you would
able to compute higher order derivatives as
308
00:54:38,630 --> 00:54:47,010
time as possible then one can get approximate
solution up to some accuracy which
309
00:54:47,010 --> 00:54:53,020
depends on the number of terms one would consider.
Thank you.
310
00:54:53,020 --> 00:54:53,020
.