ans =
結果:
Imagine that the earth is a perfect sphere with a radius of 6371000 meters and there is a rope tightly wrapped around the equator. With one line of MATLAB code determine how much the rope will be lifted above the surface if you cut it and insert a 1 meter segment of rope into it (and then expand the whole rope back into a circle again, of course).
Hello,
I am using BEAR TOOLBOX to obtain impulse response function of outcome variable to the 25 basis point monetary policy shock. The problem is there is no option in App of BEAR toolbox. how can i do it . please suggest
Hello everybody. I'm using Newton's method to solve a liner equation whose solution should be in [0 1]. Unfortunately, the coe I'm using gives NaN as a result for a specific combination of parameters and I would like to understand if I can improve the code I wrote for my Newton's method. In the specific case I'm considering, I reach the maximum iterations even if the tolerance is very low.
function [x,n,ier] = newton(f,fd,x0,nmax,tol)
% Newton's method for non-linear equations
ier = 0;
for n = 1:nmax
x = x0-f(x0)/fd(x0);
if abs(x-x0) <= tol
ier = 1;
break
end
x0 = x;
end
% % % % Script for solving NaN
mNAN= 16.1;
lNAN= 10^-4;
f= @(x) mNAN*x+lNAN*exp(mNAN*x)-lNAN*exp(mNAN);
Fd= @(x) mNAN*(1+lNAN*exp(mNAN*x));
tolNaN=10^-1;
nmax=10^8;
AB0 = 0.5;
[amNAN,nNAN,ierNAN]=newton(f,Fd,AB0,nmax,tolNaN);
amNAN
Llimit=f(0)
Ulimit=f(1)
fplot(@(x) mNAN*x+lNAN*exp(mNAN*x)-lNAN*exp(mNAN),[0 1.1])
This project discusses predator-prey system, particularly the Lotka-Volterra equations,which model the interaction between two sprecies: prey and predators. Let's solve the Lotka-Volterra equations numerically and visualize the results.% Define parameters
% Define parameters
alpha = 1.0; % Prey birth rate
beta = 0.1; % Predator success rate
gamma = 1.5; % Predator death rate
delta = 0.075; % Predator reproduction rate
% Define the symbolic variables
syms R W
% Define the equations
eq1 = alpha * R - beta * R * W == 0;
eq2 = delta * R * W - gamma * W == 0;
% Solve the equations
equilibriumPoints = solve([eq1, eq2], [R, W]);
% Extract the equilibrium point values
Req = double(equilibriumPoints.R);
Weq = double(equilibriumPoints.W);
% Display the equilibrium points
equilibriumPointsValues = [Req, Weq]
% Solve the differential equations using ode45
lotkaVolterra = @(t,Y)[alpha*Y(1)-beta*Y(1)*Y(2);
delta*Y(1)*Y(2)-gamma*Y(2)];
% Initial conditions
R0 = 40;
W0 = 9;
Y0 = [R0, W0];
tspan = [0, 100];
% Solve the differential equations
[t, Y] = ode45(lotkaVolterra, tspan, Y0);
% Extract the populations
R = Y(:, 1);
W = Y(:, 2);
% Plot the results
figure;
subplot(2,1,1);
plot(t, R, 'r', 'LineWidth', 1.5);
hold on;
plot(t, W, 'b', 'LineWidth', 1.5);
xlabel('Time (months)');
ylabel('Population');
legend('R', 'W');
grid on;
subplot(2,1,2);
plot(R, W, 'k', 'LineWidth', 1.5);
xlabel('R');
ylabel('W');
grid on;
hold on;
plot(Req, Weq, 'ro', 'MarkerSize', 8, 'MarkerFaceColor', 'r');
legend('Phase Trajectory', 'Equilibrium Point');
Now, we need to handle a modified version of the Lotka-Volterra equations. These modified equations incorporate logistic growth fot the prey population.
These equations are:
% Define parameters
alpha = 1.0;
K = 100; % Carrying Capacity of the prey population
beta = 0.1;
gamma = 1.5;
delta = 0.075;
% Define the symbolic variables
syms R W
% Define the equations
eq1 = alpha*R*(1 - R/K) - beta*R*W == 0;
eq2 = delta*R*W - gamma*W == 0;
% Solve the equations
equilibriumPoints = solve([eq1, eq2], [R, W]);
% Extract the equilibrium point values
Req = double(equilibriumPoints.R);
Weq = double(equilibriumPoints.W);
% Display the equilibrium points
equilibriumPointsValues = [Req, Weq]
% Solve the differential equations using ode45
modified_lv = @(t,Y)[alpha*Y(1)*(1-Y(1)/K)-beta*Y(1)*Y(2);
delta*Y(1)*Y(2)-gamma*Y(2)];
% Initial conditions
R0 = 40;
W0 = 9;
Y0 = [R0, W0];
tspan = [0, 100];
% Solve the differential equations
[t, Y] = ode45(modified_lv, tspan, Y0);
% Extract the populations
R = Y(:, 1);
W = Y(:, 2);
% Plot the results
figure;
subplot(2,1,1);
plot(t, R, 'r', 'LineWidth', 1.5);
hold on;
plot(t, W, 'b', 'LineWidth', 1.5);
xlabel('Time (months)');
ylabel('Population');
legend('R', 'W');
grid on;
subplot(2,1,2);
plot(R, W, 'k', 'LineWidth', 1.5);
xlabel('R');
ylabel('W');
grid on;
hold on;
plot(Req, Weq, 'ro', 'MarkerSize', 8, 'MarkerFaceColor', 'r');
legend('Phase Trajectory', 'Equilibrium Point');
Does your company or organization require that all your Word Documents and Excel workbooks be labeled with a Microsoft Azure Information Protection label or else they can't be saved? These are the labels that are right below the tool ribbon that apply a category label such as "Public", "Business Use", or "Highly Restricted". If so, you can either
- Create and save a "template file" with the desired label and then call copyfile to make a copy of that file and then write your results to the new copy, or
- If using Windows you can create and/or open the file using ActiveX and then apply the desired label from your MATLAB program's code.
For #1 you can do
copyfile(templateFileName, newDataFileName);
writematrix(myData, newDataFileName);
If the template has the AIP label applied to it, then the copy will also inherit the same label.
For #2, here is a demo for how to apply the code using ActiveX.
% Test to set the Microsoft Azure Information Protection label on an Excel workbook.
% Reference support article:
% https://www.mathworks.com/matlabcentral/answers/1901140-why-does-azure-information-protection-popup-pause-the-matlab-script-when-i-use-actxserver?s_tid=ta_ans_results
clc; % Clear the command window.
close all; % Close all figures (except those of imtool.)
clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format compact;
% Define your workbook file name.
excelFullFileName = fullfile(pwd, '\testAIP.xlsx');
% Make sure it exists. Open Excel as an ActiveX server if it does.
if isfile(excelFullFileName)
% If the workbook exists, launch Excel as an ActiveX server.
Excel = actxserver('Excel.Application');
Excel.visible = true; % Make the server visible.
fprintf('Excel opened successfully.\n');
fprintf('Your workbook file exists:\n"%s".\nAbout to try to open it.\n', excelFullFileName);
% Open up the existing workbook named in the variable fullFileName.
Excel.Workbooks.Open(excelFullFileName);
fprintf('Excel opened file successfully.\n');
else
% File does not exist. Alert the user.
warningMessage = sprintf('File does not exist:\n\n"%s"\n', excelFullFileName);
fprintf('%s\n', warningMessage);
errordlg(warningMessage);
return;
end
% If we get here, the workbook file exists and has been opened by Excel.
% Ask Excel for the Microsoft Azure Information Protection (AIP) label of the workbook we just opened.
label = Excel.ActiveWorkbook.SensitivityLabel.GetLabel
% See if there is a label already. If not, these will be null:
existingLabelID = label.LabelId
existingLabelName = label.LabelName
% Create a label.
label = Excel.ActiveWorkbook.SensitivityLabel.CreateLabelInfo
label.LabelId = "a518e53f-798e-43aa-978d-c3fda1f3a682";
label.LabelName = "Business Use";
% Assign the label to the workbook.
fprintf('Setting Microsoft Azure Information Protection to "Business Use", GUID of a518e53f-798e-43aa-978d-c3fda1f3a682\n');
Excel.ActiveWorkbook.SensitivityLabel.SetLabel(label, label);
% Save this workbook with the new AIP setting we just created.
Excel.ActiveWorkbook.Save;
% Shut down Excel.
Excel.ActiveWorkbook.Close;
Excel.Quit;
% Excel is now closed down. Delete the variable from the MATLAB workspace.
clear Excel;
% Now check to see if the AIP label has been set
% by opening up the file in Excel and looking at the AIP banner.
winopen(excelFullFileName)
Note that there is a line in there that gets an AIP label from the existing workbook, if there is one at all. If there is not one, you can set one. But to determine what the proper LabelId (that crazy long hexadecimal number) should be, you will probably need to open an existing document that already has the label that you want set (applied to it) and then read that label with this line:
label = Excel.ActiveWorkbook.SensitivityLabel.GetLabel
This stems purely from some play on my part. Suppose I asked you to work with the sequence formed as 2*n*F_n + 1, where F_n is the n'th Fibonacci number? Part of me would not be surprised to find there is nothing simple we could do. But, then it costs nothing to try, to see where MATLAB can take me in an explorative sense.
n = sym(0:100).';
Fn = fibonacci(n);
Sn = 2*n.*Fn + 1;
Sn(1:10) % A few elements
For kicks, I tried asking ChatGPT. Giving it nothing more than the first 20 members of thse sequence as integers, it decided this is a Perrin sequence, and gave me a recurrence relation, but one that is in fact incorrect. Good effort from the Ai, but a fail in the end.
Is there anything I can do? Try null! (Look carefully at the array generated by Toeplitz. It is at least a pretty way to generate the matrix I needed.)
X = toeplitz(Sn,[1,zeros(1,4)]);
rank(X(5:end,:))
Hmm. So there is no linear combination of those columns that yields all zeros, since the resulting matrix was full rank.
X = toeplitz(Sn,[1,zeros(1,5)]);
rank(X(6:end,:))
But if I take it one step further, we see the above matrix is now rank deficient. What does that tell me? It says there is some simple linear combination of the columns of X(6:end,:) that always yields zero. The previous test tells me there is no shorter constant coefficient recurrence releation, using fewer terms.
null(X(6:end,:))
Let me explain what those coefficients tell me. In fact, they yield a very nice recurrence relation for the sequence S_n, not unlike the original Fibonacci sequence it was based upon.
S(n+1) = 3*S(n) - S_(n-1) - 3*S(n-2) + S(n-3) + S(n-4)
where the first 5 members of that sequence are given as [1 3 5 13 25]. So a 6 term linear constant coefficient recurrence relation. If it reminds you of the generating relation for the Fibonacci sequence, that is good, because it should. (Remember I started the sequence at n==0, IF you decide to test it out.) We can test it out, like this:
SfunM = memoize(@(N) Sfun(N));
SfunM(25)
2*25*fibonacci(sym(25)) + 1
And indeed, it works as expected.
function Sn = Sfun(n)
switch n
case 0
Sn = 1;
case 1
Sn = 3;
case 2
Sn = 5;
case 3
Sn = 13;
case 4
Sn = 25;
otherwise
Sn = Sfun(n-5) + Sfun(n-4) - 3*Sfun(n-3) - Sfun(n-2) +3*Sfun(n-1);
end
end
A beauty of this, is I started from nothing but a sequence of integers, derived from an expression where I had no rational expectation of finding a formula, and out drops something pretty. I might call this explorational mathematics.
The next step of course is to go in the other direction. That is, given the derived recurrence relation, if I substitute the formula for S_n in terms of the Fibonacci numbers, can I prove it is valid in general? (Yes.) After all, without some proof, it may fail for n larger than 100. (I'm not sure how much I can cram into a single discussion, so I'll stop at this point for now. If I see interest in the ideas here, I can proceed further. For example, what was I doing with that sequence in the first place? And of course, can I prove the relation is valid? Can I do so using MATLAB?)
(I'll be honest, starting from scratch, I'm not sure it would have been obvious to find that relation, so null was hugely useful here.)
Hi,
I have a plot as attached herewith in which the value of the point as shown by arrow mark is to be determined and compared to a reference value. It is plotted at a time step of 999 (t ranges from 1 to 1000).
global data;
cp=0;
for i=999:max(length(data.variable.t))
for j=60:max(length(data.variable.x))-1
if data.variable.curvepressure(i,j) <= -10.2661
disp(data.variable.curvepressure(i,j))
cp=1;
break
end
end
end
The above code is not working and need your advice please.
hello i found the following tools helpful to write matlab programs. copilot.microsoft.com chatgpt.com/gpts gemini.google.com and ai.meta.com. thanks a lot and best wishes.
Check out the LLMs with MATLAB project on File Exchange to access Large Language Models from MATLAB.
Along with the latest support for GPT-4o mini, you can use LLMs with MATLAB to generate images, categorize data, and provide semantic analyis.
function ans = your_fcn_name(n)
n;
j=sum(1:n);
a=zeros(1,j);
for i=1:n
a(1,((sum(1:(i-1))+1)):(sum(1:(i-1))+i))=i.*ones(1,i);
end
disp
I am trying to earn my Intro to MATLAB badge in Cody, but I cannot click the Roll the Dice! problem. It simply is not letting me click it, therefore I cannot earn my badge. Does anyone know who I should contact or what to do?
I define the class in matlab as:
classdef Myclass
properties
Content
end
methods
function obj = Myclass(content)
obj.Content = content;
end
function disp(obj)
A = symmatrix('A(1/3,[0,0,1])');
disp(A);
end
end
end
When we run this class in live editor return 'A(1/3,[0,0,1])' rather than latex form.
Myclass(1)
% return 'A(1/3,[0,0,1])'
A = symmatrix('A(1/3,[0,0,1])');
% return latrx form A(1/3,[0,0,1])
Hi,
I'm trying to write a code which can determine the gradient change in a given profile. However the code is unable to determine the same correctly and giving incorrect results. For the pic1 below it should give me the width of the region where it is a steep gradient determined.
However for pic2 below it shouldn't as there is no steep gradient compared to pic1
The code is as below:
clear width;
global width;
global data;
for i=1:max(length(data.variable.t))
for j=1:max(length(data.variable.x))-1
change_p = (abs(data.variable.gradpressure(i,j+1))-abs(data.variable.gradpressure(i,j)))/abs(data.variable.gradpressure(i,j));
%disp(change_p)
if change_p > 0.1
disp("steep gradient found")
width(j)=1-data.variable.x(j);
disp(width)
else
disp("no steep gradient found")
end
end
end
the data.variable.gradpressure is a 1000x100 matrix with t along the vertical and x along the horizontal.
with regards,
rc
Sub aspenaAdorption()
' Declare variables for the ACM application, document, and simulation
Dim ACMApp As Object
Dim ACMDocument As Object
Dim ACMSimulation As Object
' Create an instance of the ACM application
Set ACMApp = CreateObject("ACM Application")
' Use "ACM Application" for Aspen Custom Modeler
' Use "ADS Application" for Aspen Adsorption
' Make the ACM application visible
ACMApp.Visible = True
' Open the specified simulation document
Set ACMDocument = ACMApp.OpenDocument("C:\Users\user\Desktop\H2_Purification.acmf")
' Set the simulation object to the current simulation in the application
Set ACMSimulation = ACMApp.Simulation
' Set the simulation to run in dynamic mode
ACMSimulation.RunMode = "Dynamic"
' Run the simulation
ACMSimulation.Run (True)
' Check if the simulation was successful and display a message box
If ACMSimulation.Successful Then
MsgBox "Simulation Complete"
Else
MsgBox "Simulation Failed"
End If
' Quit the ACM application
ACMApp.Quit
End Sub
I have an old application that gives me an error when I run it. The error message states: "Could not find version 7.13 of the MCR. Attempting to load mclmcrrt713.dll. Please install the correct version of the MCR." I tried to install this version, but it is no longer available. Any help would be highly appreciated. Thanks!
In the program given below I fail to obtaine real pole as title in intger format if anyone know please guide me
num1=[1 -1];
den1=conv([1 1],conv([1 2+2j],[1 2-2j]));
G=tf(num1,den1);
P=pole(G);
Z = zero(G);
formatSpec='%s,%i,%f+%fi,%i';
a="Root Locus of ";
b='step response of';
figure(17)
rlocus(G)
p=sprintf(formatSpec,a,Z,P/1i,P(3,1));
title(p);
Hi,
I have a condition where at the left boundary and the right boundary I need the flux to be non-zero.
Thus, the equations
pL + qL*fL = 0 and pr + qr*fr = 0 are such that fL and fr being non zero is making my equation hard to determine the values for pl, ql as well as pr, qr.
Plz tell me how should I approach for the same.
with regards,
rc
What do you think about the NVIDIA's achivement of becoming the top giant of manufacturing chips, especially for AI world?
We are modeling the introduction of a novel pathogen into a completely susceptible population. In the cells below, I have provided you with the Matlab code for a simple stochastic SIR model, implemented using the "GillespieSSA" function
Simulating the stochastic model 100 times for 
Since γ is 0.4 per day, 
per day
per day% Define the parameters
beta = 0.36;
gamma = 0.4;
n_sims = 100;
tf = 100; % Time frame changed to 100
% Calculate R0
R0 = beta / gamma
% Initial state values
initial_state_values = [1000000; 1; 0; 0]; % S, I, R, cum_inc
% Define the propensities and state change matrix
a = @(state) [beta * state(1) * state(2) / 1000000, gamma * state(2)];
nu = [-1, 0; 1, -1; 0, 1; 0, 0];
% Define the Gillespie algorithm function
function [t_values, state_values] = gillespie_ssa(initial_state, a, nu, tf)
t = 0;
state = initial_state(:); % Ensure state is a column vector
t_values = t;
state_values = state';
while t < tf
rates = a(state);
rate_sum = sum(rates);
if rate_sum == 0
break;
end
tau = -log(rand) / rate_sum;
t = t + tau;
r = rand * rate_sum;
cum_sum_rates = cumsum(rates);
reaction_index = find(cum_sum_rates >= r, 1);
state = state + nu(:, reaction_index);
% Update cumulative incidence if infection occurred
if reaction_index == 1
state(4) = state(4) + 1; % Increment cumulative incidence
end
t_values = [t_values; t];
state_values = [state_values; state'];
end
end
% Function to simulate the stochastic model multiple times and plot results
function simulate_stoch_model(beta, gamma, n_sims, tf, initial_state_values, R0, plot_type)
% Define the propensities and state change matrix
a = @(state) [beta * state(1) * state(2) / 1000000, gamma * state(2)];
nu = [-1, 0; 1, -1; 0, 1; 0, 0];
% Set random seed for reproducibility
rng(11);
% Initialize plot
figure;
hold on;
for i = 1:n_sims
[t, output] = gillespie_ssa(initial_state_values, a, nu, tf);
% Check if the simulation had only one step and re-run if necessary
while length(t) == 1
[t, output] = gillespie_ssa(initial_state_values, a, nu, tf);
end
if strcmp(plot_type, 'cumulative_incidence')
plot(t, output(:, 4), 'LineWidth', 2, 'Color', rand(1, 3));
elseif strcmp(plot_type, 'prevalence')
plot(t, output(:, 2), 'LineWidth', 2, 'Color', rand(1, 3));
end
end
xlabel('Time (days)');
if strcmp(plot_type, 'cumulative_incidence')
ylabel('Cumulative Incidence');
ylim([0 inf]);
elseif strcmp(plot_type, 'prevalence')
ylabel('Prevalence of Infection');
ylim([0 50]);
end
title(['Stochastic model output for R0 = ', num2str(R0)]);
subtitle([num2str(n_sims), ' simulations']);
xlim([0 tf]);
grid on;
hold off;
end
% Simulate the model 100 times and plot cumulative incidence
simulate_stoch_model(beta, gamma, n_sims, tf, initial_state_values, R0, 'cumulative_incidence');
% Simulate the model 100 times and plot prevalence
simulate_stoch_model(beta, gamma, n_sims, tf, initial_state_values, R0, 'prevalence');
Twitch built an entire business around letting you watch over someone's shoulder while they play video games. I feel like we should be able to make at least a few videos where we get to watch over someone's shoulder while they solve Cody problems. I would pay good money for a front-row seat to watch some of my favorite solvers at work. Like, I want to know, did Alfonso Nieto-Castonon just sit down and bang out some of those answers, or did he have to think about it for a while? What was he thinking about while he solved it? What resources was he drawing on? There's nothing like watching a master craftsman at work.
I can imagine a whole category of Cody videos called "How I Solved It". I tried making one of these myself a while back, but as far as I could tell, nobody else made one.
Here's the direct link to the video: https://www.youtube.com/watch?v=hoSmO1XklAQ
I hereby challenge you to make a "How I Solved It" video and post it here. If you make one, I'll make another one.
