Question

Write one a MATLAB function that implements the Bisection method, Newton’s method and Secant Method (all in one function). Your function must have the following signature

function output = solve(f,options)

% your code here

end

where

the input is

• • f: the function in f(x) =0.
  • options: is a struct type with the following fields o method: bisection, newton or secant
  • tol: the tolerance for stopping the iterations.
  • maximum_iterations: the maximum number of iterations allowed.
  • initial_guess: that is P_0; if the method needs it
  • df: the derivative of f if the method needs it
  • initial_interval: if the method needs it

the output is also a struct type with the following fields

• • message: either ‘success’ or an error message.
• • root: the solution in case of success.
• • iterations: an array that saves all iterations of the algorithm. Each row represents an iteration of the algorithm. Each row must contain P_n, f(P_n) and |P_n-P_n-1|.
• Write a script file that tests your function using the following equations

                                  600x^4 – 550x^3 +200x^2 – 20x -1 = 0

Answer 1

function dummy=bisc_newton_secant()
clc;
clear all;

f=@(x)600*x^4 - 550*x^3 +200*x^2-20*x -1; %function
fp=@(x)600*4*x^3 - 550*3*x^2 +200*2*x-20; % derivative of unction

tol=1e-8;

a=0;
b=2; % interval
x0=0.5;
disp('Root by Bisection method')
y=bisecion(f,a,b,tol) % function calling

% a1=0;
% b1=1; % interval
% disp('Root by fixed point method')
% C1=fixedp(g,x0,tol) % function calling

disp('Root by Newton method')
y2=newt(f,x0,tol)% function calling

a2=0.5;
b2=2; % interval

disp('Root by Secant method')
y1=secn(f,a2,b2,tol)% function calling

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%bisection method%%%%%%%%%%%
function c=bisecion(f,a,b,tol)
if(f(a)*f(b)>0)
fprintf('Root is not applicable in [%f, %f]\n',a,b)
else

disp('_____________________________________________________')
disp('n p_n ea')
disp('_____________________________________________________')
err=0.1;
k=1;
while ( err>tol )

c = (a + b)/2;
if ( f(c) == 0 )
break;
elseif ( f(a)*f(c) < 0 )
b = c;
else
a = c;
end
  
err=abs( (b-a)/2);

k=k+1;
fprintf('%f\t%10f \t %10f\n',k,c,err)
end

end
end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Secant method %%%%%%%%%%%%%%%%%%%%%%

function x1=secn(f,a,b,tol)
x(1)=a;
x(2)=b;
err=0.1;
%secont method
i=2;

disp('_____________________________________________________')
disp('x f(x) error')
disp('_____________________________________________________')
while (abs(err) > tol & i<=100)
x(i+1)=x(i)-f(x(i))*(x(i)-x(i-1))/(f(x(i))-f(x(i-1)));
err (i-1)= abs(x(i+1)-x(i));
  
i = i+1;

  
end
for j=1:i-2
fprintf('%f\t%10f \t %10f \n',x(j),f(x(j)),err(j))
end
x1=x(end);
err1=err(end);
  
  

end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Newton method %%%%%%%%%%%%%%%%%%%%%%

function x=newt(f,x0,tol)
k=1;
err=0.1;
x=x0;
disp('_____________________________________________________')
disp('x f(x) error')
disp('_____________________________________________________')
while(err>tol)

x1=x-(f(x)/fp(x)); %newton method
  
err=abs(x-x1);

x=x1;

k=k+1;
fprintf('%f\t%10f \t %10f \n',x,f(x),err)
end

end
end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Result%%%%%%%%%%%%

Root by Bisection method
_____________________________________________________
n p_n ea
_____________________________________________________
2.000000   1.000000    0.500000
3.000000   0.500000    0.250000
4.000000   0.250000    0.125000
5.000000   0.125000    0.062500
6.000000   0.187500    0.031250
7.000000   0.218750    0.015625
8.000000   0.234375    0.007813
9.000000   0.226563    0.003906
10.000000   0.230469    0.001953
11.000000   0.232422    0.000977
12.000000   0.231445    0.000488
13.000000   0.231934    0.000244
14.000000   0.232178    0.000122
15.000000   0.232300    0.000061
16.000000   0.232361    0.000031
17.000000   0.232330    0.000015
18.000000   0.232346    0.000008
19.000000   0.232353    0.000004
20.000000   0.232349    0.000002
21.000000   0.232351    0.000001
22.000000   0.232352    0.000000
23.000000   0.232353    0.000000
24.000000   0.232353    0.000000
25.000000   0.232353    0.000000
26.000000   0.232353    0.000000
27.000000   0.232353    0.000000
28.000000   0.232353    0.000000

y =

0.2324

Root by Newton method
_____________________________________________________
x f(x) error
_____________________________________________________
0.385185   2.745637    0.114815
0.281283   0.714026    0.103903
0.234849   0.034931    0.046434
0.232358   0.000068    0.002491
0.232353   0.000000    0.000005
0.232353   0.000000    0.000000

y2 =

0.2324

Root by Secant method
_____________________________________________________
x f(x) error
_____________________________________________________
0.500000   7.750000    1.501953
2.000000   5959.000000    0.001923
0.498047   7.619192    0.113670
0.496124   7.492449    0.063088
0.382454   2.674183    0.063160
0.319366   1.337855    0.021495
0.256206   0.339699    0.002312
0.234710   0.032990    0.000045
0.232398   0.000636    0.000000
0.232353   0.000001    0.000000

y1 =

0.2324

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