Question

Problem 2: Compare performance of Newton's method and Muller's method on the problem of finding roots of a polynomial with real co- efficients by the method of deflation ·Write a code implementing deflation method for finding all roots of a polynomial using (a) Newton's method, (b) Muller's method . On the example of P(x)+2+4r+3, show that Newton's method can not produce complex roots when starts from real On the example of P(x) = x3+4x2 +4x+3, show that Muller's ·Show that Newton's method gives correct roots of P(x) = 2.3 + initial approximation. method is not sensitive to a real initial approximation. 4r2+4x +3, if the initial approximation has a nonzero complex component. . Write a report. Include results of the calculations done in previ- ous items, add minor comments explaining the results. Attach

Answer 1

Problem-1

Newton method

a) f(x) =0 in [-0.9,1], f(x)=x²sin(x)

We assume initial solution x0=1 and calculate improved solution successively as follows:

$\small x_{imp}=x_0-\frac{f(x_0)}{f'(x_0)}$

We assign x₀ = x_imp after above step and compare f(x₀) with 0

It the error is less than 1E-6 we stop and see the no of iterations for convergence.

Code and output are given below:

f=@(x) x^2*sin(x);
fd=@(x) 2*x*sin(x)+x^2*cos(x);
x0=1;
iter=0;
while f(x0)>1e-6;
ximp=x0-f(x0)/fd(x0);
x0=ximp;
iter=iter+1;
if iter>100;
break;
end;
end;
x0
iter

Output

x0 =

0.0069

iter =

12

Hence root is 0.0069 wit 12 iterations.

b ) g(x) =0 in [-0.9,1], g(x)=x²sin(x)-x

Code:

f=@(x) x^2*sin(x)-x;
fd=@(x) 2*x*sin(x)+x^2*cos(x)-1;
x0=0.1;
iter=0;
while abs(f(x0))>1e-6;
ximp=x0-f(x0)/fd(x0);
x0=ximp;
iter=iter+1;
if iter>100;
break;
end;
end;
x0
iter

Output

x0 =

1.7352e-08

iter =

2

c) h(x)=0 on [-0.9,1],

$\small h(x)=\sqrt[3]{x}$

The code and output are given below:

f=@(x) x^(1/3);
fd=@(x) 1/3*x^(-2/3);
x0=.1;
iter=0;
while abs(f(x0))>1e-6;
ximp=x0-f(x0)/fd(x0);
x0=ximp;
iter=iter+1;
if iter>100;
break;
end;
end;
x0
iter

Output

x0 =

-2.5353e+29 + 5.9881e+14i

iter =

101

It is seen that the solution does not converge as h^'(x) tends to be 0 near the actual solution x=0 hence 1/h'(x) tends to be infinity.

Bisection method

a) Code is given below:

fun=@(x) x^2*sin(x);
iterr=0;
xa=-.9;xb=1;
xm=(xa+xb)/2;
fm=fun(xm);
while abs(fm)>1e-6;
iterr=iterr+1;
xm=(xa+xb)/2;
fm=lab4(xm);
fa=lab4(xa);
fb=lab4(xb);
if xa*fm>=0;
xa=xm;
else
xb=xm;
end
if iterr>999;
break;
end
end;
iterr
xm

Output

terr =

5

xm =

-0.0094

b) Output:

iterr =

5

xm =

-0.0094

c) Output

iterr =

5

xm =

-0.0094

Here the solution converges

Fixed point iteration

a) This function can not be put in the form x=g(x) hence solution by this method is not possible.

b) We modify the function as

$\small x=x^2sin(x)$

Starting with initial value of 0.1 we successively calculate the improved values till we get difference between two successive values as less than 1E-6

The code is given below:

f=@(x) x^2*sin(x);
x0=0.6
iter=0;
for ii=1:100
ximp(ii)=f(x0);
if abs(x0-ximp(ii))>1e-6;
x0=ximp(ii);
iter=iter+1;
end;
end
x0
iter

Output

x0 =

5.8036e-07

iter =

3

c) This equation can not be expressed in the form

x=g(x)

Hence solution by this method is not possible

Comparison

1. Newton method does not converge when the deriavative tends to be 0 within the interval of solution.

2. Newton method may require more iterations for the type of function a)

3. Bisection method appears to be less sensitive to type of functions.

4. For the type of functions a) and c) it is not possible to express the equation in the form

x=g(x).

Hence fixed point iteration method can not be used for these functions.

5. The no of iterations required for fixed point iteration is comparable to bisection and Newton method.