ans =

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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.)

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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

Is anyone else disappointed with uifigures? It seems apparent that these will eventually replace traditional figures, but there is still so much that doesn't quite work. I've tinkered with uifigures since their introduction in release 2016a, but even in 2020a I can't reliably export a uifigure to a graphics file. Sure it works sometimes, but not if a uipanel is present. The exportgraphics command is not as powerful as the print command, leaving some graphic formats (such as *.svg and *.eps) unsupported. How do you generate graphic files of a specific size? You can't even use subplots without overriding the default AutoResizeChildren setting!

Everything with uifigures seems to be slower and less stable than the figure variant. App Designer is much better than GUIDE, but that is not exactly high praise. I would rather generate "apps" programatically across several files instead of dealing with a single-file class definition containing 1000+ lines.

Where is this transition going? MATLAB graphics are moving away from Java in favor of JavaScript, and I'm not sure that we are at all ready for that.

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

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