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In our last lecture we were discussing about the finding the multiple roots for an equation
fx is equal to zero.
What we are really looking for was the multiple root of fx is equal to zero. We know the definition
of multiple root. Let us say x is equal to xi which is a multiple root of multiplicity
m. So let us take this multiplicity also. It is a multiple root of multiplicity m. Then we know that f of xi would be zero, and
then its derivative is also zero; f prime of xi is equal to zero and so on up to m minus
1th derivative. It is equal to zero and nth derivative is not equal to zero. This is the
basic definition of the multiple root of multiplicity m.
Now to find this particular root we shall first of all attempt whether we can use the
methods that we have derived for simple roots by some modification or some other criteria;
such that we can use those methods without having to make much changes in the methods
that we have. Let us first of all attempt in this particular form. We know that if f
of x has got a root, x is equal to xi of multiplicity m and then its derivative f prime of x has
got the same root x is equal to xi of multiplicity m minus one. Therefore we would like to consider
a function which is a ratio of these two and define hx as fx upon f dash x. Then this particular ratio i.e. this new function
that we have defined as hx has got a simple root at x is equal to xi. Therefore for this
function x is equal to i is a simple root. What does this imply? This implies that since
x is equal to xi is a simple root, I can now apply the all the available methods that we
have discussed for finding the simple root; to find the simple root of hx is equal to
zero and hence the multiple root of fx is equal to zero which is the multiplicity m.
Let us try to get the Newton Raphson method and its application to find the multiple root.
Now what we are saying is, we are applying the Newton Raphson method of hx. Therefore
the iterative method would be xk plus one is equal to xk minus h at xk divided by h
prime at xk. Now I need these two quantities to put here. So h at xk will be simply equal
to f at xk divided by f prime at xk, so that is from the definition of hx. Now differentiate
h prime. So let's get out derivative of this that will be d upon dx of f upon f dash. So
let us differentiate it. So if I differentiate you will get f prime into f prime i.e. f prime
square minus f into f double prime divided by f prime whole square. So that is the simple
derivative of f upon f prime. Therefore I can now evaluate h prime at x is equal to
xk and then substitute it over here. So I can put this here and get it as xk plus one
is equal to xk minus f of k by f prime of k; that is our value of h at xk. In the denominator
we have got h prime. So this will go up as f prime k whole square divided by f prime
k whole square minus f of k fk double prime, which we can simplify in one more step i.e.
by cancelling one of f prime k. So I can write this as xk minus f evaluated at k, f prime
at k from here, and the denominator is the same denominator f prime k whole square minus
fk f second prime at k. Therefore I can obtain the simple root of hx is equal to zero using
this Newton Raphson method modification and therefore this gives me the multiple root
of multiplicity of the original eqation.
Therefore we can say that x is equal to xk approximation. This is xi. It is an approximation
to the multiple root. Therefore we obtain the root without any furher modifications in the
Newton Raphson method. Now if this is always so we do not need any other method, but if
you look at the cost of the computation; the cost of the computation needs to be evaluated,
f at k f prime at k f double prime at k. So we have got this fk f prime k and f double
prime k. So I need to evaluate three evaluations to get the Newton Raphson method and the order
of the method is same because we have evaluated the simple root of hx is equal to zero. The
order of convergence is two. So the order of convergence still remains as two and only
thing that we have done is one extra computation. If this extra computation is allowed and we
do not have any problem, we can definitely use this particular method; but however we
shall give an alternative method wherein we do not need evaluation of three, we need evaluation
of only two, provided you know the order or multiplicity in advance.
Now this method can also be applied for secant method. We can apply this secant method. Let
us see how you put the same thing into the secant method. Let us take the way in which
we had written the computational way of writing the formula. We have written it as xk plus
one is equal to xk minus one f at k minus xk fk at minus one divided by fk minus fk
minus one.This is the computational form of secant mehod that we have used earlier. So
therefore I need to only insert into this the values of fk fk minus one; this is equal
to x at k minus one.
Now I need to write this for the evaluation of the simple root for hx is equal to zero.
So I would now use the method as xk plus one is equal to xk minus one h evaluated at xk
minus xk, h evaluated at k minus one divided by hk minus hk minus one. Now we have written
everything in terms of the function hx. So let us substitute the value of what is hk
and what is hk minus one. So that will simply give us xk minus one into f of k minus f prime
of k i.e. the numerator minus xk, f k minus one is fk minus one upon f prime at k minus
one and then we shall divide the whole thing by hk minus hk minus one, so fk minus f prime
at k minus fk minus one f prime at k minus one. Now it is just a matter of simplication
over here. We can see that from the denominator first of all we can write this as fk into
f prime k minus one minus fk minus one f prime of k. So I first simplified the denomianator;
when it goes to the numerator we can see f prime k f prime k minus one cancels with the
f prime k from this. Therefore I would not repeat that particular factor. So the numerator
would be simply xk minus one from here, fk from here and f prime at k minus one minus
xk fk minus one into f prime of k. So that is the simplication of the numerator. So this
will be the secant formula as applied to hx is equal to zero. Therefore the order of the
method is the same as 1.618. Therefore rate of convergence would not change. In the secant
method also we are now using f prime whereas in the original secant method for finding
a root of fx is equal to zero we have been using only fk, fk minus one. So we have got
only one evaluation at any particular stage but here we need fk. We need f prime at k
plus one and f prime k. Therefore at any stage I need to evaluate two evaluations i.e. your
fk is required and f prime of k is also required.
So we have now again increased one function evaluation as the cost. Therefore it will
be more expensive than the normal secant method. As I said earlier if the evaluation of f prime
k is not very costly, we can definitely go about using the secant and the Newtan Raphson
methods. However if the muliplicity is known; let me asume that the multiplicity m is known.
Now I would like to use the definition of multiplicity as we have written earlier. Let
us repeat it once more; the f of xi is equal to zero, f prime of xi is equal to zero upto
m minus 1th derivative; all of them are zero and nth derivative at xi is not equal to zero.
So this is the definition of the multiplicity at the root x is equal to xi. I would like
to have a look at what happens in the error analysis as we have done in the Newtan Raphson
method. Let us start with the what is the definition of fk. Definition of xk is, f of
fx and xk is the exact solution plus the error. Now I would expand this in Taylor series.
We are just following the steps that we have done for showing the error analysis of the
Newtan Raphson method. When I expand it, I will use this information. Therefore when
I write f of xhi the first term is zero and second term is epsilon k f prime of xi i.e.
zero and so on. All the terms, the first m terms are zero because of this information.
The first non vanishing term in the expansion is containing f of m nth derivative of xi.
Now if I write that one, the first non vashing term is epsilon k to the power of m divided
by factorial m, nth derivative at xi plus (the next term is) epsilon k m plus one by
m one plus derivative f of m plus one of xi. Before I add the denominator let us first
simplify this expression. Take out whatever is possible for us. This the common factor
here. So I will take out the factor epsilon k to the power of m by factorial m; fm xi
also I will take it out and write this as one plus
(I have taken out epsilon k to the power of m) I have epsilon k left out here; denominator
is m plus one and this ratio let us just call it as f. Let us call the ratio F as the ratio
of f m plus one of xi diveded by nth derivative with respect to xi. So I have just taken the
common factor here and then I have written this remaining part as one plus epsilon k
by m plus one; f is this ratio and whatever we are writing here are second order terms
and higer order terms.
Now the same thing we we would like to do for the denominator also. Now if I take the
denominator, the denominator will contain f prime of xk. So this is f prime of xi plus
epsilon k. Now again expand this by Taylor series. Again I would be using the information
that all these derivatives are zero upto m minus 1th derivative and this will be the
non vashing term. Therefore the first non vanishing term will be containing f nth derivative
but this will be epsilon k to the power of m minus one by m minus one factorial and f
of m xi plus the next term is epsilon k to the power of m by factorial m fm plus one
xi plus so on. Now as we have done in the previous case let us now take out the common
factor here also; epsilon k to the power of m minus one by m minus one factorial, fm of
xi is again a common factor. Now whatever is left out is, one epsilon k here, m here;
this ratio is exactly the same f; fm plus one divided by fm xi is same as f and order
of epsilon k square. Now I will take the ratio of f of xk upon f prime of xk. Now let me
just put this slide back once more; we have here fk, the factor outside is epsilon k to
the power m by factorial m fm of xi and the factor here is epsilon k to the power of m
minus one m minus one factorial mth derivative of xi.
Now if I take the ratio of these two you can see that fm xi cancels, epsilon k to the power
of m minus one cancels and m minus one factorial cancels. So whatever is left out is simply
epsilon k in the numerator here and m in the denominator, so remaining parts get cancelled
out of this particular ratio and what we have here is one plus (I am writing this particular
term) epsilon k upon m plus one into f plus higher order terms; in the denominator is
this particular factor which i will take it up and write this as epsilon k by m to the
F to the power of minus one. So I have taken the denominator up.
Now let us open it up or simplify it further. I will retain the first term as it is, epsilon
k by m plus one of F. Then I expand it out by binomial expansion; one minus epsilon k
by m into F plus order of epsilon k square times. Now let us multiply it out and simplify
it, so what I will have here is epsilon k by m into one and epsilon k term is one upon
m plus one from here, minus one upon m from here, epsilon k into f plus order of epsilon
k square. This is epsilon k square; this is epsilon k square and this is epsilon k square.
So whatever is left out are all the remaining terms of order of epsilon k square.
Now let us simplify it further. So I would therefore have this as epsilon k by m into
(now here I am simplifying this, therefore I will have here m minus m minus one). So
I will have one minus epsilon k by m into m plus one into f plus order of epsilon k
square. So this is what I have from the simplification. May be if you just want to have both the terms
together; I will bring this slide down once and we can just have a look at what we have.
This is one minus epsilon k by m into m plus one into fk square.
Now let us put this into our Newton Raphson method and see why the difficulty was arising.
So the method was xk plus one is equal to xk minus fk by f prime k.
Therefore we have the error as xi plus epsilon k plus one is xi plus epsilon k minus (now
we have just now obtained the ratio - this and this; this is ratio) therefore this is
epsilon k by m into one minus epsilon k by m into m plus one into f plus order of epsilon
k square. Now let us simplify; xi cancels off from here, so the left hand side is epsilon
k plus one and on the right hand side I will connect epsilon k from here, so one minus
one by m into epsilon k. This is plus, we can write down this quantity; one upon m,
m plus one f epsilon k square plus order of epsilon k cubed. So we just collected the
terms and wrote this as one minus one upon m into epsilon k; this is plus one upon m
into m plus one f into epsilon k square plus one.
Now the error for the Newton Raphson method; the right side is epsilon k. Therefore order
is one. Therefore Newton Raphson method as applied to finding a multiple root would drop
down the order by one from quadratic convergence to linear convergence only. As I said this
can easily be experienced when you are actually doing the computation. When you know that
the method should converge fast, after few iterations what would happen in your computation
when look at the results is, you find that after even large number of iterations the
error is reducing it very slowly. If you know the exact solution of the problem and then
see the error, the error is reducing by only a factor of one decimal place or so. Therefore
this slow convergence will imply that you are possibly going at a multiple root. Now
we have shown that for a multiple root Newton Raphson method has only order one; then what
do we do? We can do a very simple modification to the Newton Raphson method.
That simple modification is; suppose I start the Newton Raphson method with xk plus one
is equal to xk minus alpha. Alpha is a constant. So I would try to start a method like this.
Now let us just go back to this slide once. Why I have introduced an? I have introduced
an alpha for this, because this ratio is fk by prime k. So let me go to slides behind;
we have taken the ratio of f by prime k at this quantity which you have finally brought
it to the form of this as fk by f prime k. Now if I put an alpha over here, what I am
doing here is; I am bringing alpha over here. If I am bringing an alpha over here this is
going to be an alpha here, that means this error expression would therefore look like,
(I would just write the same thing straight away because it is trivial quantity) one minus
alpha by m epsilon k plus alpha m into m plus one F epsilon k square plus order of epsilon,
epsilon k cubed. Therefore if I introduce this alpha over here, I am introducing an
alpha here in this bracket, one minus alpha by m and I am multiplying a by alpha in this
one.
Now you can see that we started with alpha as a constant. If I now chose alpha is equal
to m, then epsilon k vanishes. Now we shall say, choose alpha is equal to m. If I choose
alpha is equal to m, then epsilon k square is order of epsilon k square or we can call
it as C epsilon k square, where now C is this quantity alpha by m into m plus one; if you
open it out (what we had written it earlier) fm plus one xi by fm of xi. So the order has
immediately gone up to two and now we have got the rate of convergence of this method
as two. That means again we have got back our quadratic convergence by simply using
the order of multiplicity that is known to us in advance in the Newton Raphson method
here. If we are able to know it therefore we are now avoiding evaluation of one more
function. As we have seen earlier we wanted evaluation of f at xk, f prime at xk and f
double prime at xk for the Newton Raphson method to have the required second order convergence.
So here I don't need any extra evaluation. The cost is the same; identically same as
the cost as the earlier method for simple root except that we know in advance the order
of multiplicity. So that factor can be brought in here and then use that as the required
Newton Raphson method, the order of convergence will be remain same. Of course if you do not
know the multiplicity of the root, we have to follow the other way of getting the root.
Now let us take a simple example of this. Now I take this equation x cubed plus seven
x square plus fifteen x plus nine is equal to zero. It is a double root. It has double
root at x is equal to minus three. I start the initial approximation x0 is equal to minus
2.5 and the find the root corrected to three decimal places and I would be using the Newton
Raphson method.
Now the first thing is, the multiplicity given to us; it is a double root. Therefore m is
equal to two. Now since m is equal to two is given to us I would now write down my method
as xk plus one is equal to xk minus two times fk by f prime k. So I would now be using that
xk plus one is equal to xk minus two times fk upon f prime k. Now let us substitute the
values of f of xk, f prime xk i.e. xk plus one is xk, two f of k is from here, that is
xk cube plus seven times xk square plus fifteen times xk plus nine. We are differentiating
with respect to x. So I will have three x square plus fourteen xk plus fifteen. I have
differentiated this; this is three x square plus fourteen xk plus fifteen in the denominator.
Now we start the iteration. It is given to us the initial approximation as minus 2.5.
So I can substitute it here; k is equal to zero, I have the right hand side as x0, two
times x0 whole cube, seven time x0 square, fifteen times x at zero plus nine divided
by three x0 square plus fourteen x0 plus fifteen. So we just substitute the value of minus 2.5
in the expression and then evaluate it from here. I substitute it over; this is minus
2.5 and I evaluate the numerator, two times minus 0.375, the denominator is minus 1.5
and that is equal to the value minus 3.1. So minus 2.5; the iteration has gone to minus
3.1; I substitute x2 is equal to x1 minus two times everything evaluated at x1 numerator
and denominator is also evaluated at x1. So I have here minus 3.1 minus two times of what
I evaluated; I evaluate this at x1 i.e. at the value minus 3.1, I get minus 0.021 and
the denominator is 0.43. I simplify the whole thing. I get minus 3.0023. I repeat the same
thing by taking this as x2 two and numerator evaluated at x2; denominator evaluated at
x2. So I have got here minus 3.0023 minus two times the numerator by denominator and
this is minus 3.5000001.
Now we can see that in three iterations we have got an accuracy of five decimal places.
Indeed it would be very interesting for you to just take it as an exercise and solve the
same problem by Newton Raphson method without inserting the factor two. You will be able
to know just after three iterations with your computations that we are converging very slowly
in this one. So hence if I do not know the multiplicity that is given to us, I would
go to the alternative way of finding the second derivative also and then using the Newton
Raphson method.
Now very important extension or application of the solution of a single nonlinear equation
is a system of nonlinear equations. Why it is very important here is, in most of the
practical applications the mathematical model of the physical system is either ordinary
differential equation or a partial differential equation and in most cases the process are
nonlinear processes. Therefore the differential equations that are produced are nonlinear
ordinary differential equations or nonlinear partial differential equations. Now we use
some numerical methods; finite difference method or finitely methods to solve this equation
and we produce a system of nonlinear algebraic equations which are the solutions of nodal
points; the points of intersection of the mesh. Therefore what we have is thousands
of nonlinear algebraic equations and how do we solve it? It is possible for us to extend
directly the methods that we have discussed now for solving the system of nonlinear equations.
So let us take the system of nonlinear equations as this. So I have got n equations as unknowns.
Now as I said it is not unusual to get few lakhs of equations, particularly in finite
difference methods. If you have got a three dimensional problem, in the three dimensional
problem it is not difficult at all to get the thousands or even few lakhs of nonlinear
equations. Now one comment before we give the method is that, we have already seen that
if Newton Raphson method is to converge it is necessary to have some good initial approximation;
we cannot be very far away from the exact solution of the problem. Now the case becomes
much more important or discussion becomes more important in the case of system of nonlinear
equations. Wherein let us say you have got ten variables and if one variable or two variables
are given very accurately but other variables given are very far away from the exact solution.
By the time those have been modified the solutions which are close enough could be spoiled. Therefore
it is necessary that all these values, through some other procedure or from physical considerations,
one must have an idea of what could be the order of magnitude or what the type of solution
could be, so that it can be used as your starting solution.
We shall follow the same procedure as used in deriving the Newton Raphson method earlier
in one of our lectures. So let us take the exact solution of this particular problem.
Let us say the exact solution is xi1, xi2 and xin. So this is exact solution of this
problem. We have used the Taylor series expansion to get the Newton Raphson method. One of the
ways of deriving is using the Taylor series method. We shall use that method to derive
the method for the nonlinear equations. Now let us take any initial approximation; x1k,
x2 k and xnk. So this is any approximation. I would now add to these suitable quantities,
so that they become the exact solution, which means I will add xi1 is equal to x1 k plus
some h1, xi2 is equal to x2k plus h2 and so on. I will put xink plus hn is equal to xin.
So I am adding the suitable quantities so that I get the exact solutions. Therefore
if I insert this here, xi1, xi2 and xin are exact solutions. Therefore when I substitute,
it is satisfied identically. So if I take any equation, any ith equation, this will
read as x1k plus h1 x2k plus h2 plus xnk plus hn is equal to zero. I have substituted these
values in this. So i is going from one two three n.
What I now do here is, I expand this in Taylor series; i.e. the function of n variables and
retain only the first order terms, that means I would now write this as fi evaluated as
x1k, x2k, xnk plus; from earlier expansion this will be the h1 into partial derivative
with respect to x1 evaluated at the point, plus h2 partial derivative with respected
to x2 evaluated at the point and so on. So this will be having h1 into delta fi upon
by delta x1. All of them are evaluated at k, so you can put it as 'at k" if you like.
Then we will have this as h2 partial derivative of fi with respected to x2 evaluated at xk
plus so on and hn partial derivative of fi by xn evaluated at k is equal to zero.
Now I have a collection of these equations one, two, three, n all these equations. I
take this. This is a quantity that is known to us; x1k, x2k, xnk is a previous iteration.
So the computed values are all known to us. So I can take this to the right hand side
and then I will write this in the matrix format for each one of them. So let us take i is
equal to one, two, three, so on. So I will have delta f upon delta x1, delta f of upon
delta x2, delta f1 by delta xn, f2 by delta x1, delta f2 by delta x2, delta xn. So we
are putting in the matrix format. So this will be delta fn by delta x1 delta, fn by
delta x2. This is delta fn by delta xn. Multiply this by h1, h2, hn. Now I write this particular
thing; this is taken to the right hand side, therefore I will have this as minus f1, f2,
fn, all of them evaluated at the iterate k i.e. kth iterate. This matrix if you denote
this by J; this by vector h, this is a matrix h; this is the right hand side, let us call
it has f; f is evaluated at k. So I can put it as f evaluated at k. This J is called the
Jacobian matrix of the system. So it has a name called the Jacobian matrix of the system.
Now I think we better put a suffix here also, because delta fi by delta x1 all of them are
evaluated at xk. So these are all evaluated at xk. Therefore these are all numbers, these
are all constants. Therefore given at any iteration xk, we have a constant coefficient
matrix multiplied by the vector and the right hand side is known; therefore what this has
produced us is a system of linear of algebraic equations. So this is a system of linear algebraic
equations. Now therefore we need methods for solving the system of linear algebraic equations,
which is a topic which we shall be taking in the later lectures as to how to solve a
linear system of algebraic equations.
Let us now consider a particular case of a two into two system and show how we can obtain
the solution; we shall then apply it to get the solution in an example. Now let us take
simply a two by two system and see how we can solve a simple problem. Let us take a
two by two system, take the system as f(x, y) is 0; g(x, y) is 0. For this what we will
have here is delta f by delta x, delta f by delta y and the second row is delta g by delta
x, delta g by delta y; the vector is h1 and h2; right hand side is fk, gk; f is evaluated
at xk, yk and g is evaluated at xk. yk. What will be the next approximation? We should
have mentioned there. The next approximation will be, as you can see x1k plus one is equal
to x1k plus h1; x2k plus one is x2k plus h2; xnk plus one is xnk plus hn. So the remaining
part is the same as the Newton's method. So this is really an extension of Newton's method
to a system of nonlinear equations. So once we determine this h1 and h2 from here we can
go back, evaluate these partial derivatives again with respect to k; I can as well put
this as a suffix with k. I will evaluate everything at the current iterate again, solve the system,
get the increment and then again repeat the same thing. So that means for each iteration,
we shall be solving one linear algebraic system of equations. The order of the system depends
on the number of the equation that is given to us. Here we are solving two by two system;
in the general case we are solving the n by n linear algebraic system where we have one
particular system for each iteration.
Now let us take a simple example on this. We are given x square plus y square is 1.12.
I have taken a simple example in which you will be able to eliminate it, but we will
not eliminate it. Just to illustrate it we have a simple example; x square plus y square
is 1.12, xy is 0.23. The solution of this problem is close to 1 and 0.2, so that means
we are given initial approximation to the problem. Find the solution after one iteration
in the Newton's method. So what I need to build up is the Jacobian's matrix of this
first. So I build up the Jacobian. Now we will write this as f(x, y) as the first equation,
so we take everything to left hand side. Even when we are writing this, we will have to
write this as f1 of (x, y) is equal to zero or g of (x, y) is equal to zero. So these
are the two equations; therefore I will bring this right hand side to the left hand side
and define it as f. For the second equation I bring 0.23 to the left hand side, define
it as g (x, y), therefore f (x, y) will be x square plus y square minus 1.2 and g (x,
y) is equal to (x, y) minus 0.23. Then the Jacobian matrix will differentiate this partially
with respect to x; differentiate this partially with respect to y; then differentiate g also
partially with respect to x and y. So I have here two x, I have here two y and the derivative
of this will be y and x. Therefore the Jacobian matrix is this.
Now since it is two by two system what I have done here is, I have inverted it; because
once I invert it, I do not have to repeat the solution of the system of linear equations.
I just have to evaluate what it is. So this is a two by two system. So I will find the
determinant of this i.e. two x square minus two y square that goes into the denominator.
I will write down the co-factors for this. Co-factor for this is x; co-factor for this
two x; co factor for this and its transpose also, I take it as minus two y and minus y
here. So I get this as the inverse of this particular matrix J. Then once I write this
as the inverse of the matrix, I can now write down my system of equations.
So h1, h2 are there. Now what we are writing here is the next step. We are writing this
vector h is equal to J inverts of (F)k. So this means h is equal to minus J inverts of
(F)k. Now we have h is equal to h1 h2. We have just evaluated the J inverts. So I substitute
it here; one upon two x minus x square minus y square and the Jacobian is this, we have
just written it. Now I will substitute the initial values x0 is one, y0 is equal to two,
then it is a simple matter of computation. So this is one divided by two times. I have
computed what is x square minus y square; set the value of x as one; two x is equal
to two; minus 2a minus 0.4; minus y is minus 0.2. Now this is the evaluation of f at xk
f at xk, yk; that means I have evaluated f (x, y) and g(x, y) from this and written these
values as minus 0.08 and minus 0.03. Now just multiply it out and divide it out by this
quantity and produce this as 0.0354, 0.0229 and we have x1 is equal to x0 plus h1, y1
is equal to 0.2 which is your initial approximation given to us plus 0.0229 that is 0.2229. This
is the solution of this particular problem. The initial approximation was given to us
as 1 and 0.2 and now we have got the new solution as 1.20354 and this.
Now it is possible for us to check the error and satisfy the given equation, that means
we can just put these values in the equation and see whether f is sufficiently close to
zero or g is sufficiently close to zero. The values are f at x1y1 is equal to x1 square
y1 square minus 0.00174 and g (x1y1) is this. So when we started it initially the values
of f and k were here; this is your f at xk, yk; g at xk, yk. The value of minus 0.08 is
now reduced to 0.00174 and 0.03 is reduced to 0.00079. So that means we are talking of
how the equation is satisfied. So we are now testing whether the equation has been satisfied
sufficiently. Now we are able to get the accuracy faster; probably you need just another three
or four iteration to get an accuracy of four or five decimals places. Therefore it is possible
for us to solve a system of nonlinear algebraic equation also by simple extension of the Newton's
method.
Now earlier we made a comment that of all the methods that are available the Newton
Raphson method is the most commonly used even today for any scientific and engineering applications.
The reason we have given was, the reasonable order of the method and the computational
cost. Now if you look at these nonlinear systems, you will find that the extension of secant
and other methods will be much more complicated and much more difficult and therefore the
Newton Raphson method is very valuable for the system of nonlinear equations also.
We all along talked of the real roots; a simple root and a multiple root. Now how do you get
the complex root; say you want to solve x square plus one is zero, so there is no real
roots; they are all complex roots, how do you get the complex roots? All the methods
that we have done, all of them work; except that you have to do complex arithmetic. You
have to start with the complex initial approximation and do the complex arithmetic. Every method
that has been done so for can be used. Therefore that is what we have. For example, I can write
down the Newton Raphson method as zk plus one is equal to zk minus f of zk by f prime
at zk; k is equal to 0, 1, 2 and so on. Now the only thing is that, a real approximation
cannot lead to the complex. So I need to have the complex initial approximation and I do
the complex arithmetic. So you can define in advance the computation as complex and
then go start with initial approximation and perform it; and the order of convergence retains
as the same, whether it is a real root or complex root there is no difference. Therefore
we do not need any special method for obtaining the complex root. However we can derive an
alternative method for the solution. We shall illustrate this method through an example.
Let fz is equal to z square plus z plus one is equal to zero be the given equation. Set
z is equal to x plus iy. Then we get f of z is equal to x plus iy whole square plus
x plus iy plus one is equal to x square minus y square plus x plus one plus i into two xy
plus y is equal to zero. Now setting the real part zero and imaginary part to zero, we get
the two equations; x square minus y square plus x plus one is equal to zero two xy plus
y is equal to zero. This gives a system of two nonlinear equations in two variables x
and y. We solve this system by the Newton's method that we have derived earlier. However
in this case the computation may take more time. We shall choose this procedure if we
cannot perform the complex arithmetic and using complex initial approximation. Therefore
the general technique in this particular case would be that substitute z is equal to x plus
iy in fz is equal to zero; separate the real and imaginary parts, put them equal to zero,
solve them as a system of two simultaneous nonlinear equations in two variables of x
and y. That gives the alternate method of the Newton's method that we have discussed
earlier. We shall stop it for today.