ans = 
主要内容
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I was browsing the MathWorks website and decided to check the Cody leaderboard. To my surprise, William has now solved 5,000 problems. At the moment, there are 5,227 problems on Cody, so William has solved over 95%. The next competitor is over 500 problems behind. His score is also clearly the highest, approaching 60,000.
D.R. Kaprekar was a self taught recreational mathematician, perhaps known mostly for some numbers that bear his name.
Today, I'll focus on Kaprekar's constant (as opposed to Kaprekar numbers.)
The idea is a simple one, embodied in these 5 steps.
1. Take any 4 digit integer, reduce to its decimal digits.
2. Sort the digits in decreasing order.
3. Flip the sequence of those digits, then recompose the two sets of sorted digits into 4 digit numbers. If there were any 0 digits, they will become leading zeros on the smaller number. In this case, a leading zero is acceptable to consider a number as a 4 digit integer.
4. Subtract the two numbers, smaller from the larger. The result will always have no more than 4 decimal digits. If it is less than 1000, then presume there are leading zero digits.
5. If necessary, repeat the above operation, until the result converges to a stable result, or until you see a cycle.
Since this process is deterministic, and must always result in a new 4 digit integer, it must either terminate at either an absorbing state, or in a cycle.
For example, consider the number 6174.
7641 - 1467
We get 6174 directly back. That seems rather surprising to me. But even more interesting is you will find all 4 digit numbers (excluding the pure rep-digit nmbers) will always terminate at 6174, after at most a few steps. For example, if we start with 1234
4321 - 1234
8730 - 0378
8532 - 2358
and we see that after 3 iterations of this process, we end at 6174. Similarly, if we start with 9998, it too maps to 6174 after 5 iterations.
9998 ==> 999 ==> 8991 ==> 8082 ==> 8532 ==> 6174
Why should that happen? That is, why should 6174 always drop out in the end? Clearly, since this is a deterministic proces which always produces another 4 digit integer (Assuming we treat integers with a leading zero as 4 digit integers), we must either end in some cycle, or we must end at some absorbing state. But for all (non-pure rep-digit) starting points to end at the same place, it seems just a bit surprising.
I always like to start a problem by working on a simpler problem, and see if it gives me some intuition about the process. I'll do the same thing here, but with a pair of two digit numbers. There are 100 possible two digit numbers, since we must treat all one digit numbers as having a "tens" digit of 0.
N = (0:99)';
Next, form the Kaprekar mapping for 2 digit numbers. This is easier than you may think, since we can do it in a very few lines of code on all possible inputs.
Ndig = dec2base(N,10,2) - '0';
Nmap = sort(Ndig,2,'descend')*[10;1] - sort(Ndig,2,'ascend')*[10;1];
I'll turn it into a graph, so we can visualize what happens. It also gives me an excuse to employ a very pretty set of tools in MATLAB.
G2 = graph(N+1,Nmap+1,[],cellstr(dec2base(N,10,2)));
plot(G2)
Do you see what happens? All of the rep-digit numbers, like 11, 44, 55, etc., all map directly to 0, and they stay there, since 0 also maps into 0. We can see that in the star on the lower right.
G2cycles = cyclebasis(G2)
G2cycles{1}
All other numbers eventually end up in the cycle:
G2cycles{2}
That is
81 ==> 63 ==> 27 ==> 45 ==> 09 ==> and back to 81
looping forever.
Another way of trying to visualize what happens with 2 digit numbers is to use symbolics. Thus, if we assume any 2 digit number can be written as 10*T+U, where I'll assume T>=U, since we always sort the digits first
syms T U
(10*T + U) - (10*U+T)
So after one iteration for 2 digit numbers, the result maps ALWAYS to a new 2 digit number that is divisible by 9. And there are only 10 such 2 digit numbers that are divisible by 9. So the 2-digit case must resolve itself rather quickly.
What happens when we move to 3 digit numbers? Note that for any 3 digit number abc (without loss of generality, assume a >= b >= c) it almost looks like it reduces to the 2 digit probem, aince we have abc - cba. The middle digit will always cancel itself in the subtraction operation. Does that mean we should expect a cycle at the end, as happens with 2 digit numbers? A simple modification to our previous code will tell us the answer.
N = (0:999)';
Ndig = dec2base(N,10,3) - '0';
Nmap = sort(Ndig,2,'descend')*[100;10;1] - sort(Ndig,2,'ascend')*[100;10;1];
G3 = graph(N+1,Nmap+1,[],cellstr(dec2base(N,10,2)));
plot(G3)
This one is more difficult to visualize, since there are 1000 nodes in the graph. However, we can clearly see two disjoint groups.
We can use cyclebasis to tell us the complete story again.
G3cycles = cyclebasis(G3)
G3cycles{:}
And we see that all 3 digit numbers must either terminate at 000, or 495. For example, if we start with 181, we would see:
811 - 118
963 - 369
954 - 459
It will terminate there, forever trapped at 495. And cyclebasis tells us there are no other cycles besides the boring one at 000.
What is the maximum length of any such path to get to 495?
D3 = distances(G3,496) % Remember, MATLAB uses an index origin of 1
D3(isinf(D3)) = -inf; % some nodes can never reach 495, so they have an infinite distance
plot(D3)
The maximum number of steps to get to 495 is 6 steps.
find(D3 == 6) - 1
So the 3 digit number 100 required 6 iterations to eventually reach 495.
shortestpath(G3,101,496) - 1
I think I've rather exhausted the 3 digit case. It is time now to move to the 4 digit problem, but we've already done all the hard work. The same scheme will apply to compute a graph. And the graph theory tools do all the hard work for us.
N = (0:9999)';
Ndig = dec2base(N,10,4) - '0';
Nmap = sort(Ndig,2,'descend')*[1000;100;10;1] - sort(Ndig,2,'ascend')*[1000;100;10;1];
G4 = graph(N+1,Nmap+1,[],cellstr(dec2base(N,10,2)));
plot(G4)
cyclebasis(G4)
ans{:}
And here we see the behavior, with one stable final point, 6174 as the only non-zero ending state. There are no circular cycles as we had for the 2-digit case.
How many iterations were necessary at most before termination?
D4 = distances(G4,6175);
D4(isinf(D4)) = -inf;
plot(D4)
The plot tells the story here. The maximum number of iterations before termination is 7 for the 4 digit case.
find(D4 == 7,1,'last') - 1
shortestpath(G4,9986,6175) - 1
Can you go further? Are there 5 or 6 digit Kaprekar constants? Sadly, I have read that for more than 4 digits, things break down a bit, there is no 5 digit (or higher) Kaprekar constant.
We can verify that fact, at least for 5 digit numbers.
N = (0:99999)';
Ndig = dec2base(N,10,5) - '0';
Nmap = sort(Ndig,2,'descend')*[10000;1000;100;10;1] - sort(Ndig,2,'ascend')*[10000;1000;100;10;1];
G5 = graph(N+1,Nmap+1,[],cellstr(dec2base(N,10,2)));
plot(G5)
cyclebasis(G5)
ans{:}
The result here are 4 disjoint cycles. Of course the rep-digit cycle must always be on its own, but the other three cycles are also fully disjoint, and are of respective length 2, 4, and 4.
I've been working on some matrix problems recently(Problem 55225)
and this is my code
It turns out that "Undefined function 'corr' for input arguments of type 'double'." However, should't the input argument of "corr" be column vectors with single/double values? What's even going on there?
isequaln exists to return true when NaN==NaN.
unique treats NaN==NaN as false (as it should) requiring NaN to be replaced if NaN is not considered unique in a particular application. In my application, I am checking uniqueness of table rows using [table_unique,index_unique]=unique(table,"rows","sorted") and would prefer to keep NaN as NaN or missing in table_unique without the overhead of replacing it with a dummy value then replacing it again. Dummy values also have the risk of matching existing values in the table, requiring first finding a dummy value that is not in the table.
uniquen (similar to isequaln) would be more eloquent.
Please point out if I am missing something!
Following on from my previous post The Non-Chaotic Duffing Equation, now we will study the chaotic behaviour of the Duffing Equation
P.s:Any comments/advice on improving the code is welcome.
The Original Duffing Equation is the following:
Let 
. This implies that
. This implies thatThen we rewrite it as a System of First-Order Equations
Using the substitution 
for 
, the second-order equation can be transformed into the following system of first-order equations:
for , the second-order equation can be transformed into the following system of first-order equations:
Exploring the Effect of γ.
% Define parameters
gamma = 0.1;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;
% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) - alpha*y(1) - beta*y(1)^3 + gamma*cos(omega*t)];
% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0
% Time span
tspan = [0 200];
% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);
% Plot the results
figure;
plot(t, y(:, 1));
xlabel('Time');
ylabel('x(t)');
title('Solution of the nonlinear system');
grid on;
% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait');
grid on;
% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail
% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), 'r', 'LineWidth', 1.5);
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait - Tail of the Solution');
grid on;
% Define parameters
gamma = 0.318;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;
% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) - alpha*y(1) - beta*y(1)^3 + gamma*cos(omega*t)];
% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0
% Time span
tspan = [0 800];
% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);
% Plot the results
figure;
plot(t, y(:, 1));
xlabel('Time');
ylabel('x(t)');
title('Solution of the nonlinear system');
grid on;
% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait');
grid on;
% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail
% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), 'r', 'LineWidth', 1.5);
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait - Tail of the Solution');
grid on;
% Define parameters
gamma = 0.338;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;
% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) - alpha*y(1) - beta*y(1)^3 + gamma*cos(omega*t)];
% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0
% Time span with more points for better resolution
tspan = linspace(0, 200,2000); % Increase the number of points
% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);
% Plot the results
figure;
plot(t, y(:, 1));
xlabel('Time');
ylabel('x(t)');
title('Solution of the nonlinear system');
grid on;
% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait');
grid on;
% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail
% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), 'r', 'LineWidth', 1.5);
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait - Tail of the Solution');
grid on;
ax = gca;
chart = ax.Children(1);
datatip(chart,0.5581,-0.1126);
% Define parameters
gamma = 0.35;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;
% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) - alpha*y(1) - beta*y(1)^3 + gamma*cos(omega*t)];
% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0
% Time span with more points for better resolution
tspan = linspace(0, 400,3000); % Increase the number of points
% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);
% Plot the results
figure;
plot(t, y(:, 1));
xlabel('Time');
ylabel('x(t)');
title('Solution of the nonlinear system');
grid on;
% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait');
grid on;
% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail
% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), 'r', 'LineWidth', 1.5);
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait - Tail of the Solution');
grid on;
Studying the attached document Duffing Equation from the University of Colorado, I noticed that there is an analysis of The Non-Chaotic Duffing Equation and all the graphs were created with Matlab. And since the code is not given I took the initiative to try to create the same graphs with the following code.
- Plotting the Potential Energy and Identifying Extrema
% Define the range of x values
x = linspace(-2, 2, 1000);
% Define the potential function V(x)
V = -x.^2 / 2 + x.^4 / 4;
% Plot the potential function
figure;
plot(x, V, 'LineWidth', 2);
hold on;
% Mark the minima at x = ±1
plot([-1, 1], [-1/4, -1/4], 'ro', 'MarkerSize', 5, 'MarkerFaceColor', 'g');
% Add LaTeX title and labels
title('Duffing Potential Energy: $$V(x) = -\frac{x^2}{2} + \frac{x^4}{4}$$', 'Interpreter', 'latex');
xlabel('$$x$$', 'Interpreter', 'latex');
ylabel('$$V(x)$$','Interpreter', 'latex');
grid on;
hold off;
- Solving and Plotting the Duffing Equation
% Define the system of ODEs for the non-chaotic Duffing equation
duffing_ode = @(t, X) [X(2);
X(1) - X(1).^3];
% Time span for the simulation
tspan = [0 10];
% Initial conditions [x(0), v(0)]
initial_conditions = [1; 1];
% Solve the ODE using ode45
[t, X] = ode45(duffing_ode, tspan, initial_conditions);
% Extract displacement (x) and velocity (v)
x = X(:, 1);
v = X(:, 2);
% Plot both x(t) and v(t) in the same figure
figure;
plot(t, x, 'b-', 'LineWidth', 2); % Plot x(t) with blue line
hold on;
plot(t, v, 'r--', 'LineWidth', 2); % Plot v(t) with red dashed line
% Add title, labels, and legend
title(' Component curve solutions to $$\ddot{x}-x+x^3=0$$','Interpreter', 'latex');
xlabel('t','Interpreter', 'latex');
ylabel('$$x(t) $$ and $$v(t) $$','Interpreter', 'latex');
legend('$$x(t)$$', ' $$v(t)$$','Interpreter', 'latex');
grid on;
hold off;
% Phase portrait with nullclines, equilibria, and vector field
figure;
hold on;
% Plot phase portrait
plot(x, v,'r', 'LineWidth', 2);
% Plot equilibrium points
plot([0 1 -1], [0 0 0], 'ro', 'MarkerSize', 5, 'MarkerFaceColor', 'g');
% Create a grid of points for the vector field
[x_vals, v_vals] = meshgrid(linspace(-2, 2, 20), linspace(-1, 1, 20));
% Compute the vector field components
dxdt = v_vals;
dvdt = x_vals - x_vals.^3;
% Plot the vector field
quiver(x_vals, v_vals, dxdt, dvdt, 'b');
% Set axis limits to [-1, 1]
xlim([-1.7 1.7]);
ylim([-1 1]);
% Labels and title
title('Phase-Plane solutions to $$\ddot{x}-x+x^3=0$$','Interpreter', 'latex');
xlabel('$$ (x)$$','Interpreter', 'latex');
ylabel('$$v(v)$$','Interpreter', 'latex');
grid on;
hold off;
An attractor is called strange if it has a fractal structure, that is if it has non-integer Hausdorff dimension. This is often the case when the dynamics on it are chaotic, but strange nonchaotic attractors also exist. If a strange attractor is chaotic, exhibiting sensitive dependence on initial conditions, then any two arbitrarily close alternative initial points on the attractor, after any of various numbers of iterations, will lead to points that are arbitrarily far apart (subject to the confines of the attractor), and after any of various other numbers of iterations will lead to points that are arbitrarily close together. Thus a dynamic system with a chaotic attractor is locally unstable yet globally stable: once some sequences have entered the attractor, nearby points diverge from one another but never depart from the attractor.
The term strange attractor was coined by David Ruelle and Floris Takens to describe the attractor resulting from a series of bifurcations of a system describing fluid flow. Strange attractors are often differentiable in a few directions, but some are like a Cantor dust, and therefore not differentiable. Strange attractors may also be found in the presence of noise, where they may be shown to support invariant random probability measures of Sinai–Ruelle–Bowen type.
Lorenz
% Lorenz Attractor Parameters
sigma = 10;
beta = 8/3;
rho = 28;
% Lorenz system of differential equations
f = @(t, a) [-sigma*a(1) + sigma*a(2);
rho*a(1) - a(2) - a(1)*a(3);
-beta*a(3) + a(1)*a(2)];
% Time span
tspan = [0 100];
% Initial conditions
a0 = [1 1 1];
% Solve the system using ode45
[t, a] = ode45(f, tspan, a0);
% Plot using scatter3 with time-based color mapping
figure;
scatter3(a(:,1), a(:,2), a(:,3), 5, t, 'filled'); % 5 is the marker size
title('Lorenz Attractor');
xlabel('x(t)');
ylabel('y(t)');
zlabel('z(t)');
grid on;
colorbar; % Add a colorbar to indicate the time mapping
view(3); % Set the view to 3D
Sprott
% Define the parameters
a = 2.07;
b = 1.79;
% Define the system of differential equations
dynamics = @(t, X) [ ...
X(2) + a * X(1) * X(2) + X(1) * X(3); % dx/dt
1 - b * X(1)^2 + X(2) * X(3); % dy/dt
X(1) - X(1)^2 - X(2)^2 % dz/dt
];
% Initial conditions
X0 = [0.63; 0.47; -0.54];
% Time span
tspan = [0 100];
% Solve the system using ode45
[t, X] = ode45(dynamics, tspan, X0);
% Plot the results with color gradient
figure;
colormap(jet); % Set the colormap
c = linspace(1, 10, length(t)); % Color data based on time
% Create a 3D line plot with color based on time
for i = 1:length(t)-1
plot3(X(i:i+1,1), X(i:i+1,2), X(i:i+1,3), 'Color', [0 0.5 0.9]*c(i)/10, 'LineWidth', 1.5);
hold on;
end
% Set plot properties
title('Sprott Attractor');
xlabel('x(t)');
ylabel('y(t)');
zlabel('z(t)');
grid on;
colorbar; % Add a colorbar to indicate the time mapping
view(3); % Set the view to 3D
hold off;
Rössler
% Define the parameters
a = 0.2;
b = 0.2;
c = 5.7;
% Define the system of differential equations
dynamics = @(t, X) [ ...
-(X(2) + X(3)); % dx/dt
X(1) + a * X(2); % dy/dt
b + X(3) * (X(1) - c) % dz/dt
];
% Initial conditions
X0 = [10.0; 0.00; 10.0];
% Time span
tspan = [0 100];
% Solve the system using ode45
[t, X] = ode45(dynamics, tspan, X0);
% Plot the results
figure;
scatter3(X(:,1), X(:,2), X(:,3), 5, t, 'filled');
title('Rössler Attractor');
xlabel('x(t)');
ylabel('y(t)');
zlabel('z(t)');
grid on;
colorbar; % Add a colorbar to indicate the time mapping
view(3); % Set the view to 3D
Rabinovich-Fabrikant
%% Parameters for Rabinovich-Fabrikant Attractor
alpha = 0.14;
gamma = 0.10;
dt = 0.01;
num_steps = 5000;
% Initial conditions
x0 = -1;
y0 = 0;
z0 = 0.5;
% Preallocate arrays for performance
x = zeros(1, num_steps);
y = zeros(1, num_steps);
z = zeros(1, num_steps);
% Set initial values
x(1) = x0;
y(1) = y0;
z(1) = z0;
% Generate the attractor
for i = 1:num_steps-1
x(i+1) = x(i) + dt * (y(i)*(z(i) - 1 + x(i)^2) + gamma*x(i));
y(i+1) = y(i) + dt * (x(i)*(3*z(i) + 1 - x(i)^2) + gamma*y(i));
z(i+1) = z(i) + dt * (-2*z(i)*(alpha + x(i)*y(i)));
end
% Create a time vector for color mapping
t = linspace(0, 100, num_steps);
% Plot using scatter3
figure;
scatter3(x, y, z, 5, t, 'filled'); % 5 is the marker size
title('Rabinovich-Fabrikant Attractor');
xlabel('x(t)');
ylabel('y(t)');
zlabel('z(t)');
grid on;
colorbar; % Add a colorbar to indicate the time mapping
view(3); % Set the view to 3D
References
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).
A library of runnable PDEs. See the equations! Modify the parameters! Visualize the resulting system in your browser! Convenient, fast, and instructive.
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');
Swimming, diving
16%
Other water-based sport
4%
Gymnastics
20%
Other indoor arena sport
15%
track, field
24%
Other outdoor sport
21%
346 个投票
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.)
Hello, MATLAB enthusiasts! 🌟
Over the past few weeks, our community has been buzzing with insightful questions, vibrant discussions, and innovative ideas. Whether you're a seasoned expert or a curious beginner, there's something here for everyone to learn and enjoy. Let's take a moment to highlight some of the standout contributions that have sparked interest and inspired many. Dive in and see how you can join the conversation or find solutions to your own challenges!
Interesting Questions
How can i edit my code which works on r2014b version at work but not on my personal r2024a version? by Oluwadamilola Oke
Oluwadamilola Oke is seeking assistance with a MATLAB code that works on version r2014b but encounters errors on version r2024a. The issue seems to be related to file location or the use of specific commands like movefile. If you have experience with these versions of MATLAB, your expertise could be invaluable.
Yohay has been working on a simulation to measure particle speed and fit it to the Maxwell-Boltzmann distribution. However, the fit isn't aligning perfectly with the data. Yohay has shared the code and histogram data for community members to review and provide suggestions.
Alessandro Livi is toggling between C++ for Arduino Pico and MATLAB App Designer. They suggest an enhancement where typing // for comments in MATLAB automatically converts to %. This small feature could improve the workflow for many users who switch between programming languages.
Popular Discussions
Athanasios Paraskevopoulos has started an engaging discussion on Gabriel's Horn, a shape with infinite surface area but finite volume. The conversation delves into the mathematical intricacies and integral calculations required to understand this paradoxical shape.
Honzik has brought up an interesting topic about custom fonts for MATLAB. While popular coding fonts handle characters like 0 and O well, they often fail to distinguish between different types of brackets. Honzik suggests that MathWorks could develop a custom font optimized for MATLAB syntax to reduce coding errors.
From the Blogs
Guy Rouleau addresses a common error in Simulink models: "Derivative of state '1' in block 'X/Y/Integrator' at time 0.55 is not finite." The blog post explores various tools and methods to diagnose and resolve this issue, making it a valuable read for anyone facing similar challenges.
Guest writer Gianluca Carnielli, featured by Adam Danz, shares insights on creating time-sensitive animations using MATLAB. The article covers controlling the motion of multiple animated objects, organizing data with timetables, and simplifying animations with the retime function. This is a must-read for anyone interested in scientific animations.
Feel free to check out these fascinating contributions and join the discussions! Your input and expertise can make a significant difference in our community.
Hi everyone,
I've recently joined a forest protection team in Greece, where we use drones for various tasks. This has sparked my interest in drone programming, and I'd like to learn more about it. Can anyone recommend any beginner-friendly courses or programs that teach drone programming?
I'm particularly interested in courses that focus on practical applications and might align with the work we do in forest protection. Any suggestions or guidance would be greatly appreciated!
Thank you!
"What are your favorite features or functionalities in MATLAB, and how have they positively impacted your projects or research? Any tips or tricks to share?
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?
Something that had bothered me ever since I became an FEA analyst (2012) was the apparent inability of the "camera" in Matlab's 3D plot to function like the "cameras" in CAD/CAE packages.
For instance, load the ForearmLink.stl model that ships with the PDE Toolbox in Matlab and ParaView and try rotating the model.
clear
close all
gm = importGeometry( "ForearmLink.stl" );
pdegplot(gm)
Things to observe:
- Note that I cant seem to rotate continuously around the x-axis. It appears to only support rotations from [0, 360] as opposed to [-inf, inf]. So, for example, if I'm looking in the Y+ direction and rotate around X so that I'm looking at the Z- direction, and then want to look in the Y- direction, I can't simply keep rotating around the X axis... instead have to rotate 180 degrees around the Z axis and then around the X axis. I'm not aware of any data visualization applications (e.g., ParaView, VisIt, EnSight) or CAD/CAE tools with such an interaction.
- Note that at the 50 second mark, I set a view in ParaView: looking in the [X-, Y-, Z-] direction with Y+ up. Try as I might in Matlab, I'm unable to achieve that same view perspective.
Today I discovered that if one turns on the Camera Toolbar from the View menubar, then clicks the Orbit Camera icon, then the No Principal Axis icon:
That then it acts in the manner I've long desired. Oh, and also, for the interested, it is programmatically available: https://www.mathworks.com/help/matlab/ref/cameratoolbar.html
I might humbly propose this mode either be made more discoverable, similar to the little interaction widgets that pop up in figures:
Or maybe use the middle-mouse button to temporarily use this mode (a mouse setting in, e.g., Abaqus/CAE).
I've noticed is that the highly rated fonts for coding (e.g. Fira Code, Inconsolata, etc.) seem to overlook one issue that is key for coding in Matlab. While these fonts make 0 and O, as well as the 1 and l easily distinguishable, the brackets are not. Quite often the curly bracket looks similar to the curved bracket, which can lead to mistakes when coding or reviewing code.
So I was thinking: Could Mathworks put together a team to review good programming fonts, and come up with their own custom font designed specifically and optimized for Matlab syntax?
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