ˆ Solution for Ex 1.

ˆ Solution for Ex 1.

The aim is the one of studying the speed of convergence of the methods. In practice we need to compare the number of iterations needed by each algorithm to reach the solution up to the desired tolerance. Note that g

2

(x) = g

₁

(g

₁

(x)) . Thus, the rst iterative method gives

x

_k

= g

₁

(x

_k−1

) = g

^(k)₁

(x

₀

), while the second one ˜x

k

= g

₂

(˜ x

_k−1

) = g

₂^(k)

(˜ x

₀

).

If we take ˜x

0

= x

₀

, we obtain

˜

x

_k

= g

^(k)₂

(˜ x

₀

) = g

^(2k)₁

(˜ x

₀

).

This means that the second algorithm will converge to the solu- tion in about half of the iteration needed by the rst scheme. Write the script Esercizio1 as follows.

1

^{close all}

2

^{clear all}

3

^clc

4

format long

5

^{c = 2;}

6

g1 = @(x) x

−

(x.^2

−

_c)./(2*x);

7

g2 = @(x) g1(g1(x));

8

[x1 iter1] = fixedpoint(g1,1,1.e

−

06,50)

9

[x2 iter2] = fixedpoint(g2,1,1.e

−

06,50)

10

^{x1 =}

11

1.414213562373095

12

^{iter1 =}

13

⁵

ˆ Solution for Ex 2. First verify (on paper) that the xed point it- eration and newton methods converge. Then, write on Esercizio2 the following code.

1

^{clear all}

2

^{close all}

3

^clc

4

f = @(x) x.^2

−

log(x.^2+2);

5

g = @(x)sqrt(log(x.^2+2));

6

fp = @(x) 2*x

−

2*x./(x.^2+2);

7

x0 = 2; tol = 1.e

−

10; nmax = 100;

8

sol = fzero(f,1);

9

[x,t] = bisection(f,0,2,tol,nmax);

10

^{for i=1:t}

11

x_bs(i)=bisection(f,0,2,tol,i);

12

x_fp(i)=fixedpoint(g,x0,tol,i);

13

x_nt(i)=newton(f,fp,x0,tol,i);

14

x_ai(i)=aitken(g,x0,tol,i);

15

err_bs(i)=abs(x_bs(i)

−

sol);

16

err_fp(i)=abs(x_fp(i)

−

sol);

17

err_nt(i)=abs(x_nt(i)

−

sol);

18

err_ai(i)=abs(x_ai(i)

−

sol);

19

^end

20

figure(1), semilogy(1:t,err_bs,'

−

_b*',1:t,err_fp,...

21

^'

−

go',1:t,err_nt,'

−

k.', 1:t,err_ai,'

−

ro'),...

22

legend('Bisection','Fixed point method','Newton',...

23

'Aitken''s method');

24

xlabel('Number of iterations'), ylabel('Error');

The results are displayed in the next page.

0 5 10 15 20 25 30 35 Number of iterations

10^-20 10^-15 10^-10 10^-5 10⁰

Error

Bisection

Fixed point method Newton

Aitken's method

ˆ Solution for Ex 3. On a script Esercizio3 write 1

^{clear all}

2

^{close all}

3

^clc

4

f =@(x) tan(x)

−

log(x.^2+2);

5

g=@(x) atan(log(x.^2+2));

6

x0=1; tol=1.e

−

10; nmax=90;

7

sol=fzero(f,1);

8

[x,t]=fixedpoint(g,x0,tol,nmax);

9

^{for i=1:t}

10

x_fp(i)=fixedpoint(g,x0,tol,i);

11

x_ai(i)=aitken(g,x0,tol,i);

12

err_fp(i)=abs(x_fp(i)

−

sol);

13

err_ai(i)=abs(x_ai(i)

−

sol);

14

^end

15

^for i =1:length(err_ai)

16

^if err_ai(i)==0

17

err_ai(i) = 1.e

−

16; % only for the plot

18

^end

19

^end

20

figure(1), semilogy(1:t,err_fp,...

21

^'

−

go', 1:t,err_ai,'

−

ro'),...

22

legend('Fixed point method',...

23

'Aitken''s method');

24

xlabel('Number of iterations'), ylabel('Error');

The results are displayed in the next page.

0 2 4 6 8 10 12 14 16 18 20 Number of iterations

10^-20 10^-15 10^-10 10^-5 10⁰

Error

Fixed point method Aitken's method