Plotting functions of more than one variable, f(x,y)

Question

Conor Pierce 2024-3-5

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评论： Conor Pierce 2024-3-6

Hi there i am trying to plot two functions on matlab which consist of two variables, x and y. I have written a Newton-Raphson m-file to find these points but i would like to have the two functions plotted so that i can make better initial estimates for the Newton-Raphson code to work with initially so that it will converge accurately.

How would I plot the following functions in order to have a visual of their points of intersection?

x^3 - y^2 = 1

0.5 + cos(x)tanh(y) = 0

Any help would be greatly appreciated!

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Answer 1

Manikanta Aditya 2024-3-5

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在 MATLAB Online 中打开

Hey Conor,

I checked and tried to plot using the 'fimplicit' function for the equations you mentioned.

Here is a sample code to plot your functions:

% Define the functions

f1 = @(x, y) x.^3 - y.^2 - 1;

f2 = @(x, y) 0.5 + cos(x).*tanh(y);

% Create a new figure

figure;

% Plot the first function

fimplicit(f1, [-2, 2, -2, 2], 'r');

hold on;

% Plot the second function

fimplicit(f2, [-2, 2, -2, 2], 'b');

% Add title and labels

title('Intersection of Two Functions');

xlabel('x');

ylabel('y');

% Add a legend

legend('x^3 - y^2 = 1', '0.5 + cos(x)tanh(y) = 0', 'Location', 'best');

% Hold off the figure

hold off;

This code will plot the two functions in the range [-2, 2] for both x and y. The points of intersection between the red and blue curves represent the solutions to the system of equations.

Please adjust the range [-2, 2, -2, 2] as per your requirements to get a better view of the intersection points. This should give you a good starting point for your Newton-Raphson method.

'fimplicit' function: Plot implicit function - MATLAB fimplicit (mathworks.com)

Hope it helps!

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Manikanta Aditya 2024-3-6

在 MATLAB Online 中打开

To find and mark the points of intersection on your plot, you can use MATLAB's numerical solvers like 'fsolve' to find the intersections based on your initial guesses, which you can fine-tune by visually inspecting the plot.

Here's how you can integrate this into your code:

% Define the functions

f1 = @(x, y) x.^3 - y.^2 - 1;

f2 = @(x, y) 0.5 + cos(x).*tanh(y);

% Plot the functions

figure;

fimplicit(f1, [-2, 2, -2, 2], 'r');

hold on;

fimplicit(f2, [-2, 2, -2, 2], 'b');

title('Intersection of Two Functions');

xlabel('x');

ylabel('y');

legend('x^3 - y^2 = 1', '0.5 + cos(x)tanh(y) = 0', 'Location', 'best');

% Define a function for the system of equations

systemOfEquations = @(v) [f1(v(1), v(2)); f2(v(1), v(2))];

% Initial guess for the intersection points

initialGuess = [0, 1]; % This is just an example. Adjust based on the plot.

% Use fsolve to find the intersection point

options = optimoptions('fsolve', 'Display', 'none'); % Suppress fsolve output

intersectionPoint = fsolve(systemOfEquations, initialGuess, options);

% Mark the intersection point on the plot

plot(intersectionPoint(1), intersectionPoint(2), 'ko', 'MarkerFaceColor', 'g');

% Hold off the figure

hold off;

% Display the intersection point in the Command Window

disp('Intersection Point:');

Intersection Point:

disp(intersectionPoint);

0.0000 -0.1614

'fsolve' finds the roots of a system of nonlinear equations, which in this case are your two functions.
You need to provide an initial guess. By inspecting the plot you've generated, you can make a reasonable guess where the functions intersect.
The options for 'fsolve' are set to suppress output messages for cleaner execution. You can remove or adjust these as needed.
This example finds one intersection point. If your plot suggests there are multiple intersections, you'll need to run 'fsolve' with different initial guesses to find the other points.

Hope this clarifies!

Dyuman Joshi 2024-3-6

编辑：Dyuman Joshi 2024-3-6

在 MATLAB Online 中打开

As can be observed from the plot, fsolve() does not find an intersection point of the curves in subject. Because the initial guess provided is not good.

As mentioned above, fsolve only provides a single output at a time as per the provided input. If you want to get all the solutions, you can solve the equations symbolically. Or utilize FEX ubsmissions on obtaining curve intersections.

Or compare values directly, as done in the other answer.

% Define the functions
f1 = @(x, y) x.^3 - y.^2 - 1;
f2 = @(x, y) 0.5 + cos(x).*tanh(y);
% Plot the functions
figure;
fimplicit(f1, [-20, 20], 'r');
hold on;
fimplicit(f2, [-20, 20], 'b');
title('Intersection of Two Functions');
xlabel('x');
ylabel('y');
legend('x^3 - y^2 = 1', '0.5 + cos(x)tanh(y) = 0', 'Location', 'best');
% Define a function for the system of equations
systemOfEquations = @(v) [f1(v(1), v(2)); f2(v(1), v(2))];
%Original initial guess
out = fsolve(systemOfEquations, [0 1])
No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.
out = 1×2
    0.0000   -0.1614

%Updated initial guess

initialGuess = [1, 1];

intersectionPoint = fsolve(systemOfEquations, initialGuess)

Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.

intersectionPoint = 1×2

2.0981 2.8699

% Mark the intersection point on the plot

plot(intersectionPoint(1), intersectionPoint(2), 'ko', 'MarkerFaceColor', 'g');

% Hold off the figure

hold off;

Conor Pierce 2024-3-6

Thanks for all your help guys!

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Answer 2

Aquatris 2024-3-5

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编辑：Aquatris 2024-3-6

在 MATLAB Online 中打开

A simple code:

y = -20:0.001:20; % define y

y(y==0) = []; % remove y == 0 for to prevent division by zero in x2 equation

x1 = (1+y.^2).^(1/3); % solve for x in equation x^3 - y^2 = 1

x2 = acos(-0.5./tanh(y));% solve for x in equation 0.5 + cos(x)tanh(y) = 0

all(abs(x1.^3-y.^2-1)<1e-9) % check if x1 y is solution to x^3 - y^2 = 1

ans = logical

1

all(abs(0.5+cos(x2).*tanh(y))<1e-9) % check if x2 y is solution to 0.5 + cos(x)tanh(y) = 0

ans = logical

1

x1(imag(x1)~=0) = nan; % remove solutions with imaginary numbers

x2(imag(x2)~=0) = nan; % remove solutions with imaginary numbers

plot(x1,y,'b.',x2,y,'r.')

xlabel('x'),ylabel('y')

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Walter Roberson 2024-3-5

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y = -20:0.001:20; % define y

y(y==0) = []; % remove y == 0 for to prevent division by zero in x2 equation

x1 = (1+y.^2).^(1/3); % solve for x in equation x^3 - y^2 = 1

x2 = acos(-0.5./tanh(y));% solve for x in equation 0.5 + cos(x)tanh(y) = 0

all(abs(x1.^3-y.^2-1)<1e-9) % check if x1 y is solution to x^3 - y^2 = 1

ans = logical

1

all(abs(0.5+cos(x2).*tanh(y))<1e-9) % check if x2 y is solution to 0.5 + cos(x)tanh(y) = 0

ans = logical

1

x1(imag(x1)~=0) = nan;

x2(imag(x2)~=0) = nan;

plot(x1,y,'b.',x2,y,'r.')

xlabel('x'),ylabel('y')

Aquatris 2024-3-6

Thank you for pointing it out @Walter Roberson. I also changed the answer to integrate your addition.

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