The code will not run when I tried to write the code successfully, so try fixing the code, please.

Problem 1:

Find the solutions to the equation:

x log(x)=3

Part A.

Plot the appropriate function and find initial estimates, or intervals, for the location of the solution.

% Your code goes here:
% Define the function
f = @(x)x.*log(x) - 3;
% Plot the function
x = linspace(0.5, 10, 100);
y = f(x);
plot(x, y);
grid on;

Part B

Write a script that uses the Bisection method to find the solution to the equation.

%  Your code goes here:
% Bisection method function
a = 0.5;
b = 1;
for n=1:20
    c=(a+b)/2;  % Bisection
    if abs(f(c))<1.e-10
        a=c;b=c; break
    elseif f(a)*f(c)<0
        b=c;
    else
        a=c;
    end
end
c=(a+b)/2

c = 1.0000

c = 1.0000

c = 1

function x = Bisection(f, a, b, ~)
error = 1e-4;
% Check if the interval contains a root
if f(a)*f(b) > 0
    error('The interval does not contain a root.');
end
% Initialize the midpoint
x = (a + b)/2;
% Iterate until the error is within tolerance
while abs(f(x)) > error
    if f(x)*f(a) < 0
        b = x;
    else
        a = x;
    end
    x = (a + b)/2;
end
% Display the solution
fprintf('The solution to the equation is x = %f.\n', x);

Part C Use the appropiate MATLAB function to solve the same problem and verify you solutions in Part B.

%  Your code goes here:
% Use the fzero function to solve the equation
x = fzero(f, [0.5, 1]);
% Display the solution
fprintf('The solution to the equation is x = %f.\n', x);

Problem 2:

Find the solutions to the equation:

sin(x)+1x+1=5

Part A.

Plot the appropriate function and find initial estimates, or intervals, for the location of the solutions.

% Your code goes here:
% Define the function
f = @(x) sin(x) + 1/(x + 1) - 5;
fplot(f,[-2,2]);grid on
% Bisection
a=0;b=5;  %c=0.2608
a=0;b=5; %c=0.9349
a=1.2;b=2;
for n=1:30
    c=(a+b)/2;  % Bisection
    if abs(f(c))<1.e-10
        a=c;b=c; break
    elseif f(a)*f(c)<0
        b=c;
    else
        a=c;
    end
end
c=(a+b)/2
xsoln=fzero(f,[0,2])
xsoln=fzero(f,[0,5])
xsoln=fzero(f,[0,2])
hold on;plot(c,f(c),'*r');hold off

Part B

Write a script that uses the False Position method to find the solutions to the equation.

%  Your code goes here:
% Your code goes here
clear
f = @(x) sin(x) + 1/(x + 1) - 5;
fplot(f,[1.2,2]);grid on
% Bisection
a=0;b=0.5;  %c=0.2608
a=0.5;b=1.2; %c=0.9349
a=1.2;b=2;
for n=1:80
    %c=(a+b)/2;  % Bisection
    m=(f(b)-f(a))/(b-a); % False Position
    c=a-f(a)/m;
    if abs(f(a)*f(c))<1.e-10
        a=c;b=c; break
    elseif f(a).*f(c)<0
        b=c;
    else
        a=c;
    end
end
c=(a+b)/2
disp([a,b])
xsoln=fzero(f,[0,0.5])
xsoln=fzero(f,[0.5,1.2])
xsoln=fzero(f,[1.2,2])
hold on;plot(c,f(c),'*r');hold off

Part C

Use the appropiate MATLAB function to solve the same problem and verify you solutions in Part B.

%  Your code goes here:
% Use the fzero function to solve the equation
x = fzero(f, [-10, -1]);
% Display the solution
fprintf('The solution to the equation is x = %f.\n', x);

Problem 3:

Find the solutions to the equation:

x (log(x)+x)=5

Part A.

Plot the appropriate function and find initial estimates, or intervals, for the location of the solution.

% Your code goes here:
clc;clear all;close all;
%Part A
f=@(x)x.*(log(x)+x)-5;
df=@(x)log(x)+2+2*x;
x=1:0.01:3;
plot(x,f(x))

Part B

Write a script that uses the Newton's method to find the solution to the equation.

%  Your code goes here:
%Part B
f=@(x)x.*(log(x)+x)-5;
df=@(x)log(x)+2+2*x;
es=0.001;
x=2;
x=Newton(f,df,x,es);
function x=Newton(f,df,x0,es)
while (1)
    x=x0-f(x0)/df(x0);
    if abs((x-x0)/x)*100<es
        break
    end
    x0=x;
    end
end

Part C

Use the appropiate MATLAB function to solve the same problem and verify you solutions in Part B.

%  Your code goes here:
%part C
X=fzero(f,2);

Problem 3:

Find the solutions to the system of equations:

(x+1)2+(2y−1)2=3

(x−1)2+(y−1)2−xy=2

Part A.

Plot the appropriate function and find initial estimates, or intervals, for the location of the solutions.

% Your code goes here:
f1 = @(x, y) (x + 1)^2 + (2*y - 1)^2 - 3;
f2 = @(x, y) (x - 1)^2 + (y - 1)^2 - x*y - 2;
% Create a grid of x and y values for plotting
[x, y] = meshgrid(-2:0.1:2, -2:0.1:2);
% Calculate the values of the functions over the grid
z1 = f1(x, y);
z2 = f2(x, y);
% Create contour plots to visualize the functions
figure;
contour(x, y, z1, [0 0], 'r', 'LineWidth', 2);
hold on;
contour(x, y, z2, [0 0], 'b', 'LineWidth', 2);
xlabel('x');
ylabel('y');
title('Contour Plots of the Equations');
legend('f1(x, y) = 0', 'f2(x, y) = 0');
grid on;

Part B

Write a script that uses the Newton's method to find the solutions to the equation.

%  Your code goes here:
% Define the system of equations as functions
f1 = @(x, y) (x + 1)^2 + (2*y - 1)^2 - 3;
f2 = @(x, y) (x - 1)^2 + (y - 1)^2 - x*y - 2;
% Define the Jacobian matrix of the system
Jacobian = @(x, y) [2*(x+1), 4*(2*y-1); 2*(x-1) - y, 2*(y-1) - x];
% Define tolerance and maximum number of iterations
tolerance = 1e-6;
maxIterations = 100;
% Initial guess
x0 = [0; 0];
% Initialize variables
x_k = x0;
for k = 1:maxIterations
    % Calculate the values of the functions and Jacobian at x_k
    f_k = [f1(x_k(1), x_k(2)); f2(x_k(1), x_k(2))];
    J_k = Jacobian(x_k(1), x_k(2));
    
    % Newton's method update
    delta_x = -J_k\f_k;
    x_k = x_k + delta_x;
    
    % Check for convergence
    if norm(delta_x) < tolerance
        break;
    end
end
% Display the solution
fprintf('Approximate Solution: x = %.6f, y = %.6f\n', x_k(1), x_k(2));

Part C

Use the appropiate MATLAB function to solve the same problem and verify you solutions in Part B.

%  Your code goes here:
% Define the system of equations as a function handle
fun = @(x) [f1(x(1), x(2)); f2(x(1), x(2))];
% Initial guess
x0 = [0; 0];
% Solve the system of equations using fsolve
options = optimoptions('fsolve', 'Display', 'off');
[x_fsolve, ~, exitflag] = fsolve(fun, x0, options);
if exitflag > 0
    fprintf('Solution found using fsolve: x = %.6f, y = %.6f\n', x_fsolve(1), x_fsolve(2));
else
    fprintf('fsolve did not converge to a solution.\n');
end
% Compare with the solution from Part B
fprintf('Solution from Part B: x = %.6f, y = %.6f\n', x_k(1), x_k(2));

Problem 4:

Find the solutions to the system of equations:

(x+1)2+y=5

(x−1)2+(y−1)2−sin(x)=2Part A.

Plot the appropriate function and find initial estimates, or intervals, for the location of the solutions.

% Your code goes here:
% Define the equations
f1 = @(x, y) (x + 1)^2 + y - 5;
f2 = @(x, y) (x - 1)^2 + (y - 1)^2 - sin(x) - 2;
% Create a grid of points for visualization
[x, y] = meshgrid(-5:0.1:5, -5:0.1:5);
% Calculate the values of the functions over the grid
z1 = f1(x, y);
z2 = f2(x, y);
% Create contour plots to visualize the functions
figure;
contour(x, y, z1, [0 0], 'r', 'LineWidth', 2);
hold on;
contour(x, y, z2, [0 0], 'b', 'LineWidth', 2);
xlabel('x');
ylabel('y');
title('Contour Plots of the Equations');
legend('f1(x, y) = 0', 'f2(x, y) = 0');
grid on;

Part B

Write a script that uses the Newton's method to find the solutions to the equation.

%  Your code goes here:
f1 = @(x, y) (x + 1)^2 + y - 5;
f2 = @(x, y) (x - 1)^2 + (y - 1)^2 - sin(x) - 2;
% Define the Jacobian matrix of the system
Jacobian = @(x, y) [2*(x+1), 4*(2*y-1); 2*(x-1) - y, 2*(y-1) - x];
% Define tolerance and maximum number of iterations
tolerance = 1e-6;
maxIterations = 100;
% Initial guess
x0 = [0; 0];
% Initialize variables
x_k = x0;
for k = 1:maxIterations
    % Calculate the values of the functions and Jacobian at x_k
    f_k = [f1(x_k(1), x_k(2)); f2(x_k(1), x_k(2))];
    J_k = Jacobian(x_k(1), x_k(2));
    
    % Newton's method update
    delta_x = -J_k\f_k;
    x_k = x_k + delta_x;
    
    % Check for convergence
    if norm(delta_x) < tolerance
        break;
    end
end
% Display the solution
fprintf('Approximate Solution: x = %.6f, y = %.6f\n', x_k(1), x_k(2));

Part C

Use the appropiate MATLAB function to solve the same problem and verify you solutions in Part B.

%  Your code goes here:
% Define the system of equations as a function handle
fun = @(x) [f1(x(1), x(2)); f2(x(1), x(2))];
% Initial guess
x0 = [0; 0];
% Solve the system of equations using fsolve
options = optimoptions('fsolve', 'Display', 'off');
[x_fsolve, ~, exitflag] = fsolve(fun, x0, options);
if exitflag > 0
    fprintf('Solution found using fsolve: x = %.6f, y = %.6f\n', x_fsolve(1), x_fsolve(2));
else
    fprintf('fsolve did not converge to a solution.\n');
end
% Compare with the solution from Part B
fprintf('Solution from Part B: x = %.6f, y = %.6f\n', x_k(1), x_k(2));
end

[Torsten]

That's all ? No more problems for us to solve ?

[James]

That is all. Please fix the code so it can run again in MATLAB, please. I keep getting warnings for this code.

[dpb]

That's all???? We just do your homework for you? Don't think so, no...not without at least your showing what you've attempted to do to fix whatever the issues are first.

And, you haven't even formatted the code where it would run in the forum interpreter...

[James]

Help me with this homework, please. I tried to write the code for the problems, but the answers did not show up.

[Walter Roberson]

"The code will not run" is not sufficiently precise. You should be more specific about what you observe.

Note by the way that you never call your bisection function -- and that you are not asked to write any functions.

[James]

I mean that it did not show the graph to these problems. It did not show the answers to the code.

Answer 1

[Walter Roberson]

You need to split the code into different files.

Problem 1:

Note that you never call the function Bisection.

Part A

% Your code goes here:
% Define the function
f = @(x)x.*log(x) - 3;
% Plot the function
x = linspace(0.5, 10, 100);
y = f(x);
plot(x, y);
grid on;

Part B

%  Your code goes here:
% Bisection method function
a = 0.5;
b = 1;
for n=1:20
    c=(a+b)/2;  % Bisection
    if abs(f(c))<1.e-10
        a=c;b=c; break
    elseif f(a)*f(c)<0
        b=c;
    else
        a=c;
    end
end
c=(a+b)/2

c = 1.0000

c = 1.0000

c = 1

function x = Bisection(f, a, b, ~)
error = 1e-4;
% Check if the interval contains a root
if f(a)*f(b) > 0
    error('The interval does not contain a root.');
end
% Initialize the midpoint
x = (a + b)/2;
% Iterate until the error is within tolerance
while abs(f(x)) > error
    if f(x)*f(a) < 0
        b = x;
    else
        a = x;
    end
    x = (a + b)/2;
end
% Display the solution
fprintf('The solution to the equation is x = %f.\n', x);
end

[Walter Roberson]

Problem 2:

Find the solutions to the equation:

sin(x)+1x+1=5

Part A.

Plot the appropriate function and find initial estimates, or intervals, for the location of the solutions.

% Your code goes here:
% Define the function
f = @(x) sin(x) + 1/(x + 1) - 5;
fplot(f,[-2,2]);grid on

Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.

% Bisection
a=0;b=5;  %c=0.2608
a=0;b=5; %c=0.9349
a=1.2;b=2;
for n=1:30
    c=(a+b)/2;  % Bisection
    if abs(f(c))<1.e-10
        a=c;b=c; break
    elseif f(a)*f(c)<0
        b=c;
    else
        a=c;
    end
end
c=(a+b)/2

c = 2.0000

try
    xsoln=fzero(f,[0,2])
catch ME
    disp(ME)
end

  MException with properties:

    identifier: 'MATLAB:fzero:ValuesAtEndPtsSameSign'
       message: 'Function values at the interval endpoints must differ in sign.'
         cause: {}
         stack: [4×1 struct]
    Correction: []

try
    xsoln=fzero(f,[0,5])
catch ME
    disp(ME)
end

  MException with properties:

    identifier: 'MATLAB:fzero:ValuesAtEndPtsSameSign'
       message: 'Function values at the interval endpoints must differ in sign.'
         cause: {}
         stack: [4×1 struct]
    Correction: []

try
    xsoln=fzero(f,[0,2])
catch ME
    disp(ME)
end

  MException with properties:

    identifier: 'MATLAB:fzero:ValuesAtEndPtsSameSign'
       message: 'Function values at the interval endpoints must differ in sign.'
         cause: {}
         stack: [4×1 struct]
    Correction: []

hold on;plot(c,f(c),'*r');hold off

Part B

Write a script that uses the False Position method to find the solutions to the equation.

%  Your code goes here:
% Your code goes here
clear
f = @(x) sin(x) + 1/(x + 1) - 5;
fplot(f,[1.2,2]);grid on

Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.

% Bisection
a=0;b=0.5;  %c=0.2608
a=0.5;b=1.2; %c=0.9349
a=1.2;b=2;
for n=1:80
    %c=(a+b)/2;  % Bisection
    m=(f(b)-f(a))/(b-a); % False Position
    c=a-f(a)/m;
    if abs(f(a)*f(c))<1.e-10
        a=c;b=c; break
    elseif f(a).*f(c)<0
        b=c;
    else
        a=c;
    end
end
c=(a+b)/2

c = 2.8512e+47

disp([a,b])

   1.0e+47 *

    5.7024    0.0000

try
    xsoln=fzero(f,[0,0.5])
catch ME
    disp(ME)
end

  MException with properties:

    identifier: 'MATLAB:fzero:ValuesAtEndPtsSameSign'
       message: 'Function values at the interval endpoints must differ in sign.'
         cause: {}
         stack: [4×1 struct]
    Correction: []

try
    xsoln=fzero(f,[0.5,1.2])
catch ME
    disp(ME)
end

  MException with properties:

    identifier: 'MATLAB:fzero:ValuesAtEndPtsSameSign'
       message: 'Function values at the interval endpoints must differ in sign.'
         cause: {}
         stack: [4×1 struct]
    Correction: []

try
    xsoln=fzero(f,[1.2,2])
catch ME
    disp(ME)
end

  MException with properties:

    identifier: 'MATLAB:fzero:ValuesAtEndPtsSameSign'
       message: 'Function values at the interval endpoints must differ in sign.'
         cause: {}
         stack: [4×1 struct]
    Correction: []

hold on;plot(c,f(c),'*r');hold off

[Walter Roberson]

Problem 3:

(the first time)

Find the solutions to the equation:

x (log(x)+x)=5

Part A.

Plot the appropriate function and find initial estimates, or intervals, for the location of the solution.

% Your code goes here:
clc;clear all;close all;
%Part A
f=@(x)x.*(log(x)+x)-5;
df=@(x)log(x)+2+2*x;
x=1:0.01:3;
plot(x,f(x))

Part B

Write a script that uses the Newton's method to find the solution to the equation.

%  Your code goes here:
%Part B
f=@(x)x.*(log(x)+x)-5;
df=@(x)log(x)+2+2*x;
es=0.001;
x=2;
x=Newton(f,df,x,es)

x = 1.9311

Part C

Use the appropiate MATLAB function to solve the same problem and verify you solutions in Part B.

%  Your code goes here:
%part C
try
    X=fzero(f,2)
catch ME
    disp(ME)
end

X = 1.9311

%function definitions must appear at the end
function x=Newton(f,df,x0,es)
while (1)
    x=x0-f(x0)/df(x0);
    if abs((x-x0)/x)*100<es
        break
    end
    x0=x;
    end
end