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Hello everyone , i am excited to learn more!
Helllo all
I write The MATLAB Blog and have covered various enhancements to MATLAB's ODE capabilities over the last couple of years. Here are a few such posts
- The new solution framework for Ordinary Differential Equations (ODEs) in MATLAB R2023b
- Faster Ordinary Differential Equations (ODEs) solvers and Sensitivity Analysis of Parameters: Introducing SUNDIALS support in MATLAB
- Solving Higher-Order ODEs in MATLAB
- Function handles are faster in MATLAB R2023a (Faster function handles led to faster ode45 and friends)
- Understanding Tolerances in Ordinary Differential Equation Solvers
Everyone in this community has deeply engaged with all of these posts and given me lots of ideas for future enhancements which I've dutifully added to our internal enhancment request database.
Because I've asked for so much in this area, I was recently asked if there's anything else we should consider in the area of ODEs. Since all my best ideas come from all of you, I'm asking here....
So. If you could ask for new and improved functionality for solving ODEs with MATLAB, what would it be and (ideally) why?
Cheers,
Mike
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Yesterday I had an urgent service call for MatLab tech support. The Mathworks technician on call, Ivy Ngyuen, helped fix the problem. She was very patient and I truly appreciate her efforts, which resolved the issue. Thank you.
Hi. I'm interested to learn more about MATLAB.
excited to learn more on Mathworks
Looking forward to the Expo!
I saw an interesting problem on a reddit math forum today. The question was to find a number (x) as close as possible to r=3.6, but the requirement is that both x and 1/x be representable in a finite number of decimal places.
The problem of course is that 3.6 = 18/5. And the problem with 18/5 has an inverse 5/18, which will not have a finite representation in decimal form.
In order for a number and its inverse to both be representable in a finite number of decimal places (using base 10) we must have it be of the form 2^p*5^q, where p and q are integer, but may be either positive or negative. If that is not clear to you intuitively, suppose we have a form
2^p*5^-q
where p and q are both positive. All you need do is multiply that number by 10^q. All this does is shift the decimal point since you are just myltiplying by powers of 10. But now the result is
2^(p+q)
and that is clearly an integer, so the original number could be represented using a finite number of digits as a decimal. The same general idea would apply if p was negative, or if both of them were negative exponents.
Now, to return to the problem at hand... We can obviously adjust the number r to be 20/5 = 4, or 16/5 = 3.2. In both cases, since the fraction is now of the desired form, we are happy. But neither of them is really close to 3.6. My goal will be to find a better approximation, but hopefully, I can avoid a horrendous amount of trial and error. It would seem the trick might be to take logs, to get us closer to a solution. That is, suppose I take logs, to the base 2?
log2(3.6)
I used log2 here because that makes the problem a little simpler, since log2(2^p)=p. Therefore we want to find a pair of integers (p,q) such that
log2(3.6) + delta = p + log2(5)*q
where delta is as close to zero as possible. Thus delta is the error in our approximation to 3.6. And since we are working in logs, delta can be viewed as a proportional error term. Again, p and q may be any integers, either positive or negative. The two cases we have seen already have (p,q) = (2,0), and (4,-1).
Do you see the general idea? The line we have is of the form
log2(3.6) = p + log2(5)*q
it represents a line in the (p,q) plane, and we want to find a point on the integer lattice (p,q) where the line passes as closely as possible.
[Xl,Yl] = meshgrid([-10:10]);
plot(Xl,Yl,'k.')
hold on
fimplicit(@(p,q) -log2(3.6) + p + log2(5)*q,[-10,10,-10,10],'g-')
plot([2 4],[0,-1],'ro')
hold off
Now, some might think in terms of orthogonal distance to the line, but really, we want the vertical distance to be minimized. Again, minimize abs(delta) in the equation:
log2(3.6) + delta = p + log2(5)*q
where p and q are integer.
Can we do that using MATLAB? The skill about about mathematics often lies in formulating a word problem, and then turning the word problem into a problem of mathematics that we know how to solve. We are almost there now. I next want to formulate this into a problem that intlinprog can solve. The problem at first is intlinprog cannot handle absolute value constraints. And the trick there is to employ slack variables, a terribly useful tool to emply on this class of problem.
Rewrite delta as:
delta = Dpos - Dneg
where Dpos and Dneg are real variables, but both are constrained to be positive.
prob = optimproblem;
p = optimvar('p',lower = -50,upper = 50,type = 'integer');
q = optimvar('q',lower = -50,upper = 50,type = 'integer');
Dpos = optimvar('Dpos',lower = 0);
Dneg = optimvar('Dneg',lower = 0);
Our goal for the ILP solver will be to minimize Dpos + Dneg now. But since they must both be positive, it solves the min absolute value objective. One of them will always be zero.
r = 3.6;
prob.Constraints = log2(r) + Dpos - Dneg == p + log2(5)*q;
prob.Objective = Dpos + Dneg;
The solve is now a simple one. I'll tell it to use intlinprog, even though it would probably figure that out by itself. (Note: if I do not tell solve which solver to use, it does use intlinprog. But it also finds the correct solution when I told it to use GA offline.)
solve(prob,solver = 'intlinprog')
The solution it finds within the bounds of +/- 50 for both p and q seems pretty good. Note that Dpos and Dneg are pretty close to zero.
2^39*5^-16
and while 3.6028979... seems like nothing special, in fact, it is of the form we want.
R = sym(2)^39*sym(5)^-16
vpa(R,100)
vpa(1/R,100)
both of those numbers are exact. If I wanted to find a better approximation to 3.6, all I need do is extend the bounds on p and q. And we can use the same solution approch for any floating point number.
Check out how these charts were made with polar axes in the Graphics and App Building blog's latest article "Polar plots with patches and surface".
Nine new Image Processing courses plus one new learning path are now available as part of the Online Training Suite. These courses replace the content covered in the self-paced course Image Processing with MATLAB, which sunsets in 2026.
New courses include:
- Work with Image Data Types
- Image Registration
- Edge, Circle, and Line Detection
- Manage and Process Multiple Images
The new learning path Image Segmentation and Analysis in MATLAB earns users the digital credential Image Segmentation in MATLAB and contains the following courses:
Do you have a swag signed by Brian Douglas? He does!
Apparently, the back end here is running 2025b, hovering over the Run button and the Executing In popup both show R2024a.
ver matlab
Registration is now open for MathWorks annual virtual event MATLAB EXPO 2025 on November 12 – 13, 2025!
Register now and start building your customized agenda today!
Explore. Experience. Engage.
Join MATLAB EXPO to connect with MathWorks and industry experts to learn about the latest trends and advancements in engineering and science. You will discover new features and capabilities for MATLAB and Simulink that you can immediately apply to your work.
all(logical.empty)
Discuss!
I just noticed that MATLAB R2025b is available. I am a bit surprised, as I never got notification of the beta test for it.
This topic is for highlights and experiences with R2025b.
I came across this fun video from @Christoper Lum, and I have to admit—his MathWorks swag collection is pretty impressive! He’s got pieces I even don’t have.
So now I’m curious… what MathWorks swag do you have hiding in your office or closet?
- Which one is your favorite?
- Which ones do you want to add to your collection?
Show off your swag and share it with the community! 🚀
“Hello, I am Subha & I’m part of the organizing/mentoring team for NASA Space Apps Challenge Virudhunagar 2025 🚀. We’re looking for collaborators/mentors with ML and MATLAB expertise to help our student teams bring their space solutions to life. Would you be open to guiding us, even briefly? Your support could impact students tackling real NASA challenges. 🌍✨”
Since R2024b, a Levenberg–Marquardt solver (TrainingOptionsLM) was introduced. The built‑in function trainnet now accepts training options via the trainingOptions function (https://www.mathworks.com/help/deeplearning/ref/trainingoptions.html#bu59f0q-2) and supports the LM algorithm. I have been curious how to use it in deep learning, and the official documentation has not provided a concrete usage example so far. Below I give a simple example to illustrate how to use this LM algorithm to optimize a small number of learnable parameters.
For example, consider the nonlinear function:
y_hat = @(a,t) a(1)*(t/100) + a(2)*(t/100).^2 + a(3)*(t/100).^3 + a(4)*(t/100).^4;
It represents a curve. Given 100 matching points (t → y_hat), we want to use least squares to estimate the four parameters a1–a4.
t = (1:100)';
y_hat = @(a,t)a(1)*(t/100) + a(2)*(t/100).^2 + a(3)*(t/100).^3 + a(4)*(t/100).^4;
x_true = [ 20 ; 10 ; 1 ; 50 ];
y_true = y_hat(x_true,t);
plot(t,y_true,'o-')
- Using the traditional lsqcurvefit-wrapped "Levenberg–Marquardt" algorithm:
x_guess = [ 5 ; 2 ; 0.2 ; -10 ];
options = optimoptions("lsqcurvefit",Algorithm="levenberg-marquardt",MaxFunctionEvaluations=800);
[x,resnorm,residual,exitflag] = lsqcurvefit(y_hat,x_guess,t,y_true,-50*ones(4,1),60*ones(4,1),options);
x,resnorm,exitflag
- Using the deep-learning-wrapped "Levenberg–Marquardt" algorithm:
options = trainingOptions("lm", ...
InitialDampingFactor=0.002, ...
MaxDampingFactor=1e9, ...
DampingIncreaseFactor=12, ...
DampingDecreaseFactor=0.2,...
GradientTolerance=1e-6, ...
StepTolerance=1e-6,...
Plots="training-progress");
numFeatures = 1;
layers = [featureInputLayer(numFeatures,'Name','input')
fitCurveLayer(Name='fitCurve')];
net = dlnetwork(layers);
XData = dlarray(t);
YData = dlarray(y_true);
netTrained = trainnet(XData,YData,net,"mse",options);
netTrained.Layers(2)
classdef fitCurveLayer < nnet.layer.Layer ...
& nnet.layer.Acceleratable
% Example custom SReLU layer.
properties (Learnable)
% Layer learnable parameters
a1
a2
a3
a4
end
methods
function layer = fitCurveLayer(args)
arguments
args.Name = "lm_fit";
end
% Set layer name.
layer.Name = args.Name;
% Set layer description.
layer.Description = "fit curve layer";
end
function layer = initialize(layer,~)
% layer = initialize(layer,layout) initializes the layer
% learnable parameters using the specified input layout.
if isempty(layer.a1)
layer.a1 = rand();
end
if isempty(layer.a2)
layer.a2 = rand();
end
if isempty(layer.a3)
layer.a3 = rand();
end
if isempty(layer.a4)
layer.a4 = rand();
end
end
function Y = predict(layer, X)
% Y = predict(layer, X) forwards the input data X through the
% layer and outputs the result Y.
% Y = layer.a1.*exp(-X./layer.a2) + layer.a3.*X.*exp(-X./layer.a4);
Y = layer.a1*(X/100) + layer.a2*(X/100).^2 + layer.a3*(X/100).^3 + layer.a4*(X/100).^4;
end
end
end
The network is very simple — only the fitCurveLayer defines the learnable parameters a1–a4. I observed that the output values are very close to those from lsqcurvefit.
I saw this YouTube short on my feed: What is MATLab?
I was mostly mesmerized by the minecraft gameplay going on in the background.
Found it funny, thought i'd share.
For the www, uk, and in domains,a generative search answer is available for Help Center searches. Please let us know if you get good or bad results for your searches. Some have pointed out that it is not available in non-english domains. You can switch your country setting to try it out. You can also ask questions in different languages and ask for the response in a different language. I get better results when I ask more specific queries. How is it working for you?
