In short: support varying color in at least the plot, plot3, fplot, and fplot3 functions.
This has been a thing that's come up quite a few times, and includes questions/requests by users, workarounds by the community, and workarounds presented by MathWorks -- examples of each below. It's a feature that exists in Python's Matplotlib library and Sympy. Anyways, given that there are myriads of workarounds, it appears to be one of the most common requests for Matlab plots (Matlab's plotting is, IMO, one of the best features of the product), the request precedes the 21st century, and competitive tools provide the functionality, it would seem to me that this might be the next great feature for Matlab plotting.
I'm curious to get the rest of the community's thoughts... what's everyone else think about this?
Hello,
I already asked this question before but it is really important to have an answer since the last answer i got wasn't helpful.
i will post the whole code:
"hold on
load(Filename.mat')
...
I was in a meeting the other day and a coworker shared a smiley face they created using the AI Chat Playground. The image looked something like this:
And I suspect the prompt they used was something like this:
"Create a smiley face"
I imagine this output wasn't what my coworker had expected so he was left thinking that this was as good as it gets without manually editing the code, and that the AI Chat Playground couldn't do any better.
I thought I could get a better result using the Playground so I tried a more detailed prompt using a multi-step technique like this:
"Follow these instructions:
- Create code that plots a circle
- Create two smaller circles as eyes within the first circle
- Create an arc that looks like a smile in the lower part of the first circle"
The output of this prompt was better in my opinion.
These queries/prompts are examples of 'zero-shot' prompts, the expectation being a good result with just one query. As opposed to a back-and-forth chat session working towards a desired outcome.
I wonder how many attempts everyone tries before they decide they can't anything more from the AI/LLM. There are times I'll send dozens of chat queries if I feel like I'm getting close to my goal, while other times I'll try just one or two. One thing I always find useful is seeing how others interact with AI models, which is what inspired me to share this.
Does anyone have examples of techniques that work well? I find multi-step instructions often produces good results.
To solve a surface integral for example the over the sphere easily in MATLAB, you can leverage the symbolic toolbox for a direct and clear solution. Here is a tip to simplify the process:
Use Symbolic Variables and Functions: Define your variables symbolically, including the parameters of your spherical coordinates θ and ϕ and the radius r . This allows MATLAB to handle the expressions symbolically, making it easier to manipulate and integrate them.
Express in Spherical Coordinates Directly: Since you already know the sphere's equation and the relationship in spherical coordinates, define x, y, and z in terms of r , θ and ϕ directly.
Perform Symbolic Integration: Use MATLAB's `int` function to integrate symbolically. Since the sphere and the function are symmetric, you can exploit these symmetries to simplify the calculation.
Here’s how you can apply this tip in MATLAB code:
% Include the symbolic math toolbox
syms theta phi
% Define the limits for theta and phi
theta_limits = [0, pi];
phi_limits = [0, 2*pi];
% Define the integrand function symbolically
integrand = 16 * sin(theta)^3 * cos(phi)^2;
% Perform the symbolic integral for the surface integral
I am often talking to new MATLAB users. I have put together one script. If you know how this script works, why, and what each line means, you will be well on your way on your MATLAB learning journey.
% Clear existing variables and close figures
clear;
close all;
% Print to the Command Window
disp('Hello, welcome to MATLAB!');
% Create a simple vector and matrix
vector = [1, 2, 3, 4, 5];
matrix = [1, 2, 3; 4, 5, 6; 7, 8, 9];
% Display the created vector and matrix
disp('Created vector:');
disp(vector);
disp('Created matrix:');
disp(matrix);
% Perform element-wise multiplication
result = vector .* 2;
% Display the result of the operation
disp('Result of element-wise multiplication of the vector by 2:');
disp(result);
% Create plot
x = 0:0.1:2*pi; % Generate values from 0 to 2*pi
y = sin(x); % Calculate the sine of these values
% Plotting
figure; % Create a new figure window
plot(x, y); % Plot x vs. y
title('Simple Plot of sin(x)'); % Give the plot a title
xlabel('x'); % Label the x-axis
ylabel('sin(x)'); % Label the y-axis
grid on; % Turn on the grid
disp('This is the end of the script. Explore MATLAB further to learn more!');
Although, I think I will only get to see a partial eclipse (April 8th!) from where I am at in the U.S. I will always have MATLAB to make my own solar eclipse. Just as good as the real thing.
One of the privileges of working at MathWorks is that I get to hang out with some really amazing people. Steve Eddins, of ‘Steve on Image Processing’ fame is one of those people. He recently announced his retirement and before his final day, I got the chance to interview him. See what he had to say over at The MATLAB Blog The Steve Eddins Interview: 30 years of MathWorking
Before we begin, you will need to make sure you have 'sir_age_model.m' installed. Once you've downloaded this folder into yourworking directory, which can be located at yourcurrent folder. If you can see this file in your current folder, then it's safe to use it. If you choose to use MATLAB online or MATLAB Mobile, you may upload this to yourMATLAB Drive.
This is the code for the SIR model stratified into 2 age groups (children and adults). For a detailed explanation of how to derive the force of infection by age group.
fprintf('Proportion of adults that became infected: %f\n', proportion_infected_adults);
Throughout this epidemic, 95% of all children and 92% of all adults were infected. Children were therefore slightly more affected in proportion to their population size, even though the majority of infections occurred in adults.
I would like to propose the creation of MATLAB EduHub, a dedicated channel within the MathWorks community where educators, students, and professionals can share and access a wealth of educational material that utilizes MATLAB. This platform would act as a central repository for articles, teaching notes, and interactive learning modules that integrate MATLAB into the teaching and learning of various scientific fields.
Key Features:
1. Resource Sharing: Users will be able to upload and share their own educational materials, such as articles, tutorials, code snippets, and datasets.
2. Categorization and Search: Materials can be categorized for easy searching by subject area, difficulty level, and MATLAB version..
3. Community Engagement: Features for comments, ratings, and discussions to encourage community interaction.
4. Support for Educators: Special sections for educators to share teaching materials and track engagement.
Benefits:
- Enhanced Educational Experience: The platform will enrich the learning experience through access to quality materials.
- Collaboration and Networking: It will promote collaboration and networking within the MATLAB community.
- Accessibility of Resources: It will make educational materials available to a wider audience.
By establishing MATLAB EduHub, I propose a space where knowledge and experience can be freely shared, enhancing the educational process and the MATLAB community as a whole.
In one line of MATLAB code, compute how far you can see at the seashore. In otherwords, how far away is the horizon from your eyes? You can assume you know your height and the diameter or radius of the earth.
The latest release is pretty much upon us. Official annoucements will be coming soon and the eagle-eyed among you will have started to notice some things shifting around on the MathWorks website as we ready for this.
The pre-release has been available for a while. Maybe you've played with it? I have...I've even been quietly using it to write some of my latest blog posts...and I have several queued up for publication after MathWorks officially drops the release.
The stationary solutions of the Klein-Gordon equation refer to solutions that are time-independent, meaning they remain constant over time. For the non-linear Klein-Gordon equation you are discussing:
Stationary solutions arise when the time derivative term, , is zero, meaning the motion of the system does not change over time. This leads to a static differential equation:
This equation describes how particles in the lattice interact with each other and how non-linearity affects the steady state of the system.
The solutions to this equation correspond to the various possible stable equilibrium states of the system, where each represents different static distribution patterns of displacements . The specific form of these stationary solutions depends on the system parameters, such as κ , ω, and β , as well as the initial and boundary conditions of the problem.
To find these solutions in a more specific form, one might need to solve the equation using analytical or numerical methods, considering the different cases that could arise in such a non-linear system.
By interpreting the equation in this way, we can relate the dynamics described by the discrete Klein - Gordon equation to the behavior of DNA molecules within a biological system . This analogy allows us to understand the behavior of DNA in terms of concepts from physics and mathematical modeling .
% Parameters
numBases = 100; % Number of spatial points
omegaD = 0.2; % Common parameter for the equation
% Preallocate the array for the function handles
equations = cell(numBases, 1);
% Initial guess for the solution
initialGuess = 0.01 * ones(numBases, 1);
% Parameter sets for kappa and beta
paramSets = [0.1, 0.05; 0.5, 0.05; 0.1, 0.2];
% Prepare figure for subplot
figure;
set(gcf, 'Position', [100, 100, 1200, 400]); % Set figure size
% Newton-Raphson method parameters
maxIterations = 1000;
tolerance = 1e-10;
% Set options for fsolve to use the 'levenberg-marquardt' algorithm
sgtitle('Stationary States for Different \kappa and \beta Values', 'FontSize', 16); % Super title for the figure
In the second plot, the elasticity constant κis increased to 0.5, representing a system with greater stiffness . This parameter influences how resistant the system is to deformation, implying that a higher κ makes the system more resilient to changes . By increasing κ, we are essentially tightening the interactions between adjacent units in the model, which could represent, for instance, stronger bonding forces in a physical or biological system .
In the third plot the nonlinearity coefficient β is increased to 0.2 . This adjustment enhances the nonlinear interactions within the system, which can lead to more complex dynamic behaviors, especially in systems exhibiting bifurcations or chaos under certain conditions .
gives the solution for the Helmholtz problem. On the circular disc with center 0 and radius a. For the plot in 3-dimensional graphics of the solutions on Matlab for and then calculate some eigenfunctions with the following expression.
It could be better to separate functions with and as follows
diska = 1; % Radius of the disk
mmax = 2; % Maximum value of m
nmax = 2; % Maximum value of n
% Function to find the k-th zero of the n-th Bessel function
% This function uses a more accurate method for initial guess
If the answers are indeed AI generated, then the user didn't do "clearly indicating when AI generated content is incorporated".
That leads to my question that how do we enforce the guideline.
I am not against using AI for answers but in this case, I felt the answering text is mentioning all the relevant words but missing the point. For novice users who are seeking answers, this would be misleading and waste of time.
I have Line-to-line sinusoidal voltage reading as L1-L2, L2-L3 and L3-L1. I want to get RMS-like DC-signal that is equal to 400V when input voltage is 400V when there is no unbalance. I want to im...
Firstly, in order to obtain the first n decimal places of pi, we need to write the following code (to prevent inaccuracies, we need to take a few more tails and perform another operation of taking the first n decimal places when needed):
function Pi=getPi(n)
if nargin<1,n=3;end
Pi=char(vpa(sym(pi),n+10));
Pi=abs(Pi)-48;
Pi=Pi(3:n+2);
end
With this function to obtain the decimal places of pi, our visualization journey has begun~Step by step, from simple to complex~(Please try to use newer versions of MATLAB to run, at least R17b)
1 Pie chart
Just calculate the proportion of each digit to the first 1500 decimal places:
Imagine each decimal as a small ball with a mass of
For example, if, the weight of ball 0 is 1, ball 9 is 1.2589, the initial velocity of the ball is 0, and it is attracted by other balls. Gravity follows the inverse square law, and if the balls are close enough, they will collide and their value will become
After adding, take the mod, add the velocity direction proportionally, and recalculate the weight.
The digits of π are shown as a forest. Each tree in the forest represents the digits of π up to the next 9. The first 10 trees are "grown" from the digit sets 314159, 2653589, 79, 3238462643383279, 50288419, 7169, 39, 9, 3751058209, and 749.
BRANCHES
The first digit of a tree controls how many branches grow from the trunk of the tree. For example, the first tree's first digit is 3, so you see 3 branches growing from the trunk.
The next digit's branches grow from the end of a branch of the previous digit in left-to-right order. This process continues until all the tree's digits have been used up.
Each tree grows from a set of consecutive digits sampled from the digits of π up to the next 9. The first tree, shown here, grows from 314159. Each of the digits determine how many branches grow at each fork in the tree — the branches here are colored by their corresponding digit to illustrate this. Leaves encode the digits in a left-to-right order. The digit 9 spawns a flower on one of the branches of the previous digit. The branching exception is 0, which terminates the current branch — 0 branches grow!
LEAVES AND FLOWERS
The tree's digits themselves are drawn as circular leaves, color-coded by the digit.
The leaf exception is 9, which causes one of the branches of the previous digit to sprout a flower! The petals of the flower are colored by the digit before the 9 and the center is colored by the digit after the 9, which is on the next tree. This is how the forest propagates.
The colors of a flower are determined by the first digit of the next tree and the penultimate digit of the current tree. If the current tree only has one digit, then that digit is used. Leaves are placed at the tips of branches in a left-to-right order — you can "easily" read them off. Additionally, the leaves are distributed within the tree (without disturbing their left-to-right order) to spread them out as much as possible and avoid overlap. This order is deterministic.
The leaf placement exception are the branch set that sprouted the flower. These are not used to grow leaves — the flower needs space!
Let's still put the numbers in the form of circles, but the difference is that six numbers are grouped together, and the pure purple circle at the end is the six 9s that we are familiar with decimal places 762-767
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